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 3 
 
TREATISE 
 
 PRACTICAL MENSURATION, 
 
 IN TEN PARTS; 
 
 CONTAINING 
 
 The most approved Methods 
 of drawing Geometrical Figures; 
 Mensuration of Superficies ; 
 Land- Surveying ; IMensuration of 
 Solids; the Use of the Carpenter's 
 Rule; Timber Measure, in which 
 is shown the Method of measur- 
 ing and valuing standing Timber; 
 Artificers' Works, illustrated by 
 the Dimensions and Contents of 
 a House ; Mensuration of Hay- 
 stacks, Drains, Canals, Marlpits, 
 
 Ponds, Mill-dams, Embank- 
 ments, Quarries, Coal-heapSj and 
 Clay-heaps ; Conic Sections and 
 their Solids ; the most useful 
 Problems in Ganging, according 
 to the New Imperial Measures; 
 Plane Trigonometry with its 
 application to the Mensuration 
 of Heights and Distances ; 
 Trigonometrical Surveys ; and a 
 Dictionary of the Terms used in 
 Architecture. 
 
 BY A. WESBIT, 
 
 MASTER. OF THE CLASSICAL, COMMERCIAL, AND MATHEMATICAL 
 ACADEMY, KENNINGTON LANE, LAMBETH, LONDON; 
 
 AND AHTHOR OF 
 
 *' A Treatise on Practical Arithmetic ; " "A complete Treatise on Practical 
 
 Land- Surveying ; " " A Treatise on Practical Gauging ; " " Keys to the Arithmetic, 
 
 Mensuration, and Gauging;" "An Introduction to Parsing, adapted to 
 
 Murray's Grammar ;" " An Essay on Education ; " &c. &c. 
 
 NEW EDITION, ENLARGED, AND GREATLY IMPROVED. 
 TO WHICH IS ADDED, 
 
 A TREATISE OIT I.EVEI.X.ZN'G. 
 
 The whole illustrated by nearly seven hundred Practical Examples, and 
 nearly three hundred Woodcuts. 
 
 LONDON 
 
 LONGMAN, BROWN, GREEN, LONGMANS, & ROBERTS. 
 1859 
 
 Price Six Shilling:^, hound. 
 
sDU%-' 
 
 toyDOX : 
 
 PEir.IED BX SPOTTISWOODE AND CO. 
 NSW-STCliEX SQUAEE. 
 
TO 
 
 ELLIS CUJ^LIFFE LISTEE, ESQ., M.P. 
 
 OF 
 
 MANNINGHAM-HOUSE, 
 
 Near Bradford, 
 
 AS AN ACKNOWLEDGMENT OF GRATITUDE FOR NUMEROUS 
 
 FAVOURS RECEIVED, 
 
 AND AS 
 
 A TOKEN OF ESTEEM FOR HIS GENERAL KNOWLEDGH 
 
 OF SCIENTIFIC SUBJECTS, 
 
 THIS TREATISE 
 
 ON 
 
 PRACTICAL MENSURATION, 
 
 In all its Departments, 
 
 IS 
 MOST HUMBLY AND RESPECTFULLY DEDICATED, 
 BY HIS MUCH OBLIGED, 
 
 AND MOST OBEDIENT SERVANT, 
 
 A. NESBIT. 
 
 A. £t 
 
ADVEETISEMENT 
 
 TO THE TWELFTH EDITION. 
 
 In revising the present Edition of this work, the 
 Proprietors have spared neither labour nor expense 
 in order to render it as valuable as possible ; and for 
 this purpose they have engaged competent persons, 
 who have made several important and necessary im- 
 provements. They have commenced with the Geo- 
 metrical Definitions, Problems, and Theorems, by 
 arranging the subjects in i more concise, elegant, 
 and systematic order, the Geometrical Theorems 
 being much improved and rendered more complete 
 by prefixing a description to the respective theorems. 
 In the Mensuration of Superficies the same order of 
 arrangement has been strictly regarded, and several 
 additional and important notes have been added to 
 the respective problems. On the subject of Land 
 Surveying, great improvements have been made:, hy 
 introducing a simple method of surveying v/ithout 
 the use of the cross-staff. Similar arrangements and 
 miprovements have been observed throughout the 
 Mensuration of Solids, Conic Sections, and their 
 Solids, and the other parts of the book. Thus a large 
 
 A3 
 
ADVERTISEMENT. 
 
 saving of space has been effected, enabling the Pro- 
 prietors to add a complete Treatise on Trigonometry 
 without increasing the size of the book. They have, 
 therefore, great pleasure in introducing the present 
 Edition, believing it to be the cheapest and the best 
 school-book on the subject that has ever been pre- 
 sented to the Public. 
 
 In this Edition two new Parts have been intro- 
 duced. Part the Ninth treats of Plane and Oblique 
 Trigonometry, with a variety of illustrations and 
 practical examples. It will be seen, this subject is 
 treated, upon an entirely new, and improved method, 
 by substituting Arithmetical Complements in the 
 Logarithmic Numbers. 
 
 Part the Tenth treats of the Application of Plane 
 Trigonometry to the Mensuration of Heights and 
 Distances in a variety of useful questions ; the me- 
 thod of finding the Heights of Mountains by means 
 of the Barometer and Thermometer ; also numerous 
 examples for finding the magnitude and dimensions 
 of the earth ; the -vibration of pendulums in diiferent 
 latitudes, on the length of arcs of different meridians, 
 and a great variety of other important matter. 
 
 London, 1844. 
 
PREFACE, 
 
 Various have been the conjectures concerning the origin 
 of Geometry or Mensuration ; but as it is a Science of 
 general utility, there can, I think, be little doubt that its 
 existence is nearly coeval with the creation of Man. 
 Indeed I can see no reasonable objection why we may not 
 attribute its invention to our first parent Adam ; espe- 
 cially as we are informed in Holy Writ, that his son Cain 
 built a city ; to do which, it is evident, would require 
 some knowledge of a measuring unit, which is the first 
 principle of Mensuration. By the same infallible testi- 
 mony, we find that the Arts and Sciences were cultivated 
 to a considerable extent long before the Flood. Jubal was 
 the father of all such as handled the harp and organ ; and 
 Tubal-cain, an instructor of every artificer in brass and 
 iron. It is also more than probable that Noah was well 
 acquainted with the Art of Mensuration, as practised in 
 his day ; for it does not appear that he found any dif- 
 ficulty in building the Ark, which consisted of three 
 stories, according to certain dimensions given him by the 
 Lord of the Universe. 
 
 Diodorus, Herodotus, and Strabo, are of opinion that 
 the Science of Mensuration had its rise among the Egyp- 
 tians ; whom they represent as constrained, on account of 
 the removal or defacing of the land-marks by the annual 
 inundation of the Nile, to devise some method of ascer- 
 taining the ancient boundaries, after the waters had re- 
 tired. By Josephus, however, the invention is ascribed 
 to the Hebrews. This writer asserts that the Arts and 
 Sciences, for which Egypt was so long famous, were 
 carried into that country, by the Patriarch Abraham, from 
 Ur of the Chaldees ; but as Egypt was peopled by the 
 descendants of Ham, is it not more probable that they 
 had the rudiments of their Sciences originally from their 
 father ? 
 
 Be this as it may, it is well known that Egypt was, for 
 A 4 
 
VIU PREFACE. 
 
 many ages, the mother and nm-se of the Ai'ts and Sciences. 
 From this country they were conveyed into Greece by 
 Thales, about 600 years before the Christian JEra. This 
 Philosopher, aftev travelling into Egypt, and studying, 
 under its sages, Astronomy, Geometry, and other branches 
 of the Mathematics, returned to his own country, and em- 
 ployed himself in communicating the knowledge which he 
 had acquired. 
 
 The great utility of Mensuration, without which it is im- 
 possible rightly to conduct the affairs of civilized life, in- 
 duced many of the most celebrated Philosophers and 
 Mathematicians of antiquity to study its principles ; and 
 to Thales, Anaxagoras, Pythagoras, Socrates, Plato, 
 Apollonius, Philo, Ptolemy, Aristotle, EucKd, Ai'clii- 
 medes, &c. we are indebted for many substantial im- 
 provements. The moderns, likewise, have not been less 
 solicitous to enrich this Science, than the ancients ; ac- 
 cordingly Huygens, TVallis, Gregory, Halley, Euler, 
 Leibnitz, the Bernouilles, Yieta, Metius, Van Ceulen, 
 Barrow, Newton, Sharp, Machin, Moss, Leadbetter, 
 Simpson, Emerson, Plolliday, Fletcher, Robertson, Hutton, 
 Bonnycastle, Keith, Beckett, &c. have greatly improved 
 it by their labours. 
 
 After so many eminent men have written more or less 
 upon this Science, it may perhaps be thought presumption 
 in me to attempt to add any thing to its stores ; but as I 
 can say, without arrogance, that I have had considerable 
 experience in the Practical Part of Mensuration, in all its 
 Departments, I am persuaded that this work will be found 
 to contain many things both new and useful. 
 
 With regard to the Eules, indeed, nothing new can be 
 expected ; as they are to be found, with very little varia- 
 tion, in every modern Treatise on Mensuration ; but the 
 Questions, which amount to five hundred and eighty, are 
 almost wholly new ones ; and a great number of them 
 have been made from actual admeasurements. I have 
 likewise given copious directions for taking dimensions, 
 which art certainly forms a very essential part of Mensur- 
 ation ; for if the dimensions be improperly taken, the re- 
 sults must, of course, be incorrect. 
 
 The work is divided into Ten Parts ; and some of these 
 Parts are again subdivided into Sections. 
 
 Part the First contains Practical Geometry, and a few 
 of the most useful Geometrical Theorems ; most of which 
 
niEFACE. IX 
 
 are employed in solving Questions in this Work. The 
 Theorems are not demonstrated ; but references are given 
 to the elements of Euclid, Simpson, and Emerson, where 
 their demonstrations may be found. 
 
 Part the Second contains Mensuration of Superficies. 
 In the last Problem of tliis Part, I have given the inva- 
 luable Rule for finding the areas of curvilineal figures, by 
 means of equi-distant ordinates. This Rule was first de- 
 monstrated by the illustrious. Newton; but it is to Mr. 
 Thomas Moss that we owe its present simplicity. 
 
 Part the Third is divided into two Sections. In the 
 first are given the Methods of surveying and planning 
 single Fields, Woods, Eoads, and Rivers ; and also Rules 
 and Directions for Parting ofi', and Dividing Land. It 
 likewise exhibits foiu- of the most approved Methods of 
 siu-veying large Estates, illustrated by tliree distinct Plans, 
 and Field-Books. 
 
 Sometimes one of these Methods claims the preference, 
 and sometimes another, according to the different forms of 
 Estates ; but they are all approved of by oiu' best Land 
 Surveyors ; and are more accm-ate and practical than any 
 others, that have come to my knowledge. 
 
 This Section will be fomid to contain every thing that 
 is necessary for persons, in general, to know of Surveying. 
 In order to become a complete Surveyor, it is requisite 
 not only to study Works written professedly on the sub- 
 ject, but also to have a considerable portion oi Field-prac- 
 tice, under the direction of an able Tutor. 
 
 The second Section contains a collection of useful 
 Questions concerning Superficial Mensuration, which will 
 serve to exercise the ingenuity of the Learner ; and to 
 prove his knowledge of the Theorems and Rules given in 
 the first three Parts. 
 
 Part the Fourth is divided into four Sections. The first 
 contains the Mensm-ation of Solids ; the second, the De- 
 scription and Use of the Carpenter's Rule ; and the third. 
 Timber Measure. 
 
 The last Problem of this Section contains Rules and 
 Directions for measuring and valuing standing Timber ; 
 m.any of which were never before published. 
 
 In this part of the Work, I have been assisted by Mr. 
 Joseph Webster, of Farnley, near Leeds ; who has been 
 many years very extensively employed, as a valuer of tim- 
 ber, by the Earl of Cardigan, Lord Mexborough, &c. &c. 
 A 5 
 
PREFACE. 
 
 In order to render this Problem as useful as possible, I 
 have given a description of Timber Trees, and pointed 
 out the purposes for which their wood is best adapted ; 
 for it is impossible to become a valuer of timber without 
 being made acquainted with the properties of trees. 
 
 The fourth Section contains Miscellaneous Questions 
 concerning the Mensuration of SoUds. 
 
 Part the Fifth treats of the Method of measuring the 
 Works of Artificers ; viz. Bricklayers, Masons, Carpen- 
 ters and Joiners, Slaters and Tilers, Plasterers, Painters, 
 Glaziers, Plumbers, and Pavers. 
 
 The directions in this Part, for taking the dimensions, 
 making the deductions, &c. will be found to be very co- 
 pious. For some of these I am indebted to ^Ir. Benjamin 
 Jackson, se7iior, an able and experienced Architect in 
 Leeds ; and to Mr. Joseph Brooke, Teacher of the Mathe- 
 matics, at Wortley, near Leeds, who has had much ex- 
 perience in measuring the Works of Artificers. 
 
 This Part is concluded with a general Illustration, con- 
 taining the dimensions of a House ; and exhibiting the 
 methods of ruling the Book, entering the dimensions with 
 the contents ; and forming the bills for workmanship and 
 materials. 
 
 Part the Sixth treats of the Mensuration of Haystacks, 
 Drains, Canals, Marlpits, Embankments, Ponds, Mill- 
 dams, Quarries, Coal-heaps, Clay -heaps, and other irre- 
 gular figures, by means of equi-distant, parallel sections, 
 founded upon the method of equi-distant ordinates. 
 
 This method of finding the contents of irregular figures 
 is pointed out by Dr. Hutton, in his valuable Treatise on 
 Mensuration, octavo*, page 375 ; but Mr. Joseph Beckett 
 appears to have been the first who has applied it with any 
 degree of success. 
 
 This Part also contains the method of measuring the 
 circular Ponds made upon the Wolds in Yorkshire. This 
 was communicated to me by the Rev. W. Putsey, Master 
 of the Classical, Commercial, and Mathematical Academy, 
 Pickering. 
 
 In order to give the young Reader an idea of the great 
 improvements made in Agriculture and Commerce, by 
 means of Drains and Canals ; and also to make him ac- 
 quainted with some of the stupendous Works which have 
 been accomplished by the ingenuity, perseverance, and 
 
 * Price 18 5. in boards. 
 
PREFACE. XI 
 
 combined efforts of men ; I have concluded this Part with 
 a description -of a few of the principal Canals in England, 
 Scotland, France, and China ; and with an account of 
 some of the chief Drainages which have been executed in 
 the counties of York and Lincoln. 
 
 Part the Seventh treats of Conic Sections and their 
 Solids. It also contains a few of the leading properties 
 of the Ellipse, the Parabola, and the Hyperbola. Those 
 who desire more information on this subject, may consult 
 Simpson's, Emerson's, and Hutton's Conic Sections. 
 
 Part the Eighth displays the method of gauging all 
 kinds of open vessels used by Maltsters, Brewers, Dis- 
 tillers, Wine Merchants, Victuallers, &c. &c. 
 
 In this Part I have applied the method of equi-distant 
 ordinates or sections, to the gauging of vessels whose sides 
 are curved ; such as coppers, stills, &c. &c. This I have 
 not seen in any other Work. 
 
 I have also given the process of gauging and inching a 
 guile-tun, according to the method practised in the Excise. 
 Malt-gauging, and Cask -gauging are likewise treated of, 
 in this Part ; and it is concluded with a few Miscellaneous 
 Examples. 
 
 Part ninth treats of Plain and Oblique Trigonometry. 
 
 Part the tenth treats on the Application of Trigono- 
 metry to Heights and Distances, with a great variety 
 of useful Problems, and subjects on the Nature, Figure, 
 and Magnitude of the Earth, and of Levelling. 
 
 In order to render the Treatise as useful as possible, I 
 have given a Dictionary containing an explanation of the 
 most general terms made use of in Architecture. This 
 will be found a valuable addition by Bricklayers, Masons, 
 Joiners, and every other person concerned in measuring 
 or building. 
 
 The Work is brought to a conclusion by a number of 
 questions to be answered verbally by the Pupil ; for 
 nothing will tend more to make him an adept in Men- 
 suration, than committing the Definitions and Rules to 
 Memory. 
 
 Nearly the whole of the Rules upon which Mensuration 
 depends, require Algebra or Fluxions to demonstrate 
 them ; but as these Sciences are too abstruse and sublime 
 to be comprehended by the generality of those who are 
 concerned in Mensuration, it has been thought advisable 
 to give the Rules without the Demonstrations. Tliose 
 A 6 . 
 
Xii PREFACE. 
 
 who desire to penetrate the deepest recesses of scientific 
 knowledge, are referred to Simpson's, Emerson's and Hol- 
 liday's Fluxions, and to Doctor Hutton's Mensuration ; 
 in one or other of which "Works, all the Rules given in 
 this Treatise are demonstrated. 
 
 The Demonstrations would also have swelled the Work 
 very considerably, both in size and price ; but by with- 
 holding them, space has been obtained for a greater body 
 of useful matter, on the subject of Mensuration, than is to 
 be found in any other Work, of equal size, with wliicli I 
 am acquainted. 
 
 I should be wanting in respect, if I concluded this Pre- 
 face without paying a tribute to the memory of that pro- 
 found and indefatigable Mathematician, the late JMr. John 
 Ryley of Leeds ; — a man who has enriched almost every 
 periodical, mathematical publication, for nearly half a 
 century, with problems and demonstrations in science, 
 which have struck many of his contemporaries with un- 
 mixed admiration of his genius and attainments. To him 
 I submitted the Plan of this Work, which, I am happy to 
 say, received his entire approbation ; and I flatter myself 
 that it will be generally approved by the Public. 
 
 It now only remains for me to solicit the indulgence of 
 the candid Reader ; for in solving so large a number of 
 new questions as this Work contains, it is almost impos- 
 sible to avoid errors. These, however, it is hoped will be 
 found to be few, and of little moment ; for much care has 
 been employed in working the questions, and correcting 
 the press ; and I may add, that neither labour nor ex- 
 pense has been withheld in order to produce a Work of 
 general Utility. 
 
 ANTHONY NESBIT. 
 
 London-, Nov. 1841. 
 
 P. S. A KEY to the Mensuration, containing Solutions 
 to all the Questions which are not solved in that Work, 
 has been published for the use of Teachers and private 
 Students. This will be found of great advantage, as it 
 will render it quite unnecessary for the Tutor to inspect 
 every operation of his Pupil ; and will enable those who 
 may not have had an opportunity of acquiring a knovvdedge 
 of Mensuration in their Youth, and who cannot now have 
 the advantage of a living Instructor, to pursue their pri- 
 vate studies with ultimate success, if they apply them- 
 selves with diligence and perseverance. 
 
ADVERTISEMENT 
 TO THE FOURTH EDITION. 
 
 The Author lias carefully revised and corrected this 
 Edition, and made, in different parts of the Work, several 
 important and useful Additions and Improvements. 
 
 Part the Eighth, on Gauging, has been entirely re- 
 written, and much improved by the addition of New 
 Matter; and all the Eules and Examples have been 
 adapted to the New Imperial Standard Gallon and 
 Bushel. 
 
 The Author has calculated Three New Tables, v.rhich 
 exhibit the number of Cubic Inches in the different de- 
 nominations of Wine and Spirit Measure, Ale and Beer 
 Measure, and Corn or Dry Measure, according to the 
 New Imperial Standard Gallon and Bushel ; and also 
 given Rules and Directions for reducing the Old English, 
 Irish, and Scotch Measures to the New Imperial Measures ; 
 and vice versa. 
 
 He has also shewn the method of finding the New Im- 
 perial Multipliers, Divisors, and Gauge-Points ; and given 
 a Table in which they are all arranged in a convenient 
 order. 
 
 He has likewise described the New Imperial Sliding 
 Rule, and Diagonal Rod ; given particular and general 
 Rules for Cask Gauging; and also calculated a New 
 Table of Multipliers, by which the Content of any Cask 
 may be easily found, in Imperial Gallons. 
 
 These Alterations and Additions will be found very 
 great improvements to the Work ; and will tend to secure 
 and increase that popularity and patronage which it has 
 already obtained among Teachers and the Public ; as it 
 now contains such a body .of Practical Information, on 
 Mensuration in all its Departments, as cannot be found in 
 any other Work on the same Subject. 
 
 A. NESBIT. 
 
CONTENTS. 
 
 PART I. 
 
 Page 
 Geometrical Definitions - - - - 1 
 
 Geometrical Problems - - - - 7 
 
 To bisect a given line - - - - 7 
 
 To bisect a given angle - - - - 7 
 
 To draw a line parallel to a given line - 7 
 
 To erect a perpendicular from a given point, in a 
 given line - - - - - 8 
 
 From a given point, to let fall a perpendicular upon 
 a given line - - - - - 8 
 
 To find the centre of a given circle, or one already 
 described - - - - - 9 
 
 To make a triangle with three given lines - - 9 
 
 Having given the base, the perpendicular, and the 
 place of the perpendicular upon the base, to con- 
 struct a triangle - - - - - 10 
 
 To describe a square, whose side shall be equal to a 
 given line - - - - - 10 
 
 To describe a rectangle, whose length and breadth 
 
 shall be equal to two given lines - - - 11 
 
 Upon a given right line to construct a regular rhom- 
 bus - - - - - - 11 
 
 To construct an irregular rhombus, having given its 
 
 side and perpendicular height - - - 1 1 
 
 Having any two right lines given, to construct a 
 regular rhomboid - - - - 12 
 
 Having given the base, the perpendicular, and the 
 place of the perpendicular upon the base, to con- 
 struct an irregular rhomboid - - - 12 
 Having the transverse and conjugate diameters given, 
 to construct an ellipse - - - - 12 
 
 Upon a given line, to make a regular pentagon - 13 
 
 Upon a given line, to make a regular hexagon - 13 
 
XVI CONTENTS. 
 
 Page 
 
 Upon a given line, to construct a regular octagon - 14 
 In a given triangle, to inscribe a circle - - 14 
 
 About a given triangle, to circumscribe a circle ; or 
 to describe the circumference of a circle through 
 three given points - - - - 14 
 
 To make a triangle equal to a given trapezium - 15 
 
 To make a right angle by the line of chords on the 
 plane-scale - - - - - 15 
 
 To make an acute angle - - - - 15 
 
 To make an obtuse angle - - - - 15 
 
 To iind the number of degrees contained in any given 
 angle - - - - - - 16 
 
 Upon a given line, to make a regular polygon of any 
 
 proposed number of sides - - - 16 
 
 In any given circle, to inscribe a regular polygon of 
 any proposed number of sides; or to divide the 
 circumference into any number of equal parts - 16 
 To find a mean proportional between two given lines 17 
 To raise a perpendicular from any point, in a given 
 
 line, by a scale of equal parts - - - 17 
 
 Given the span or chord line, and height or versed 
 sine of the arch of a bridge or cellar, to find the 
 radius of the circle that will strike the arch - 17 
 
 Geometrical Theorems - - - - 18 
 
 An explanation of the principal Mathematical Cha- 
 racters - - - - - -23 
 
 PART II. 
 
 jMensuration of Superficies - - - 25 
 
 A Table of lineal measure - - - - 25 
 A Table of Square measm-e - - -26 
 
 To find the area of a square - - - 26 
 
 of a rectangle - - - 27 
 
 of a rhombus or rhomboides - 29 
 
 of a triangle, when the base and 
 
 perpendicular are given - - - - 30 
 of a triangle, the three sides only of 
 
 which are given - - - - - 31 
 Any two sides of a right-angled triangle being given, 
 
 to find the third side - - - - 33 
 To find the area of a trapezium - - - 35 
 of a trapezoid - - - 36 
 
CONTENTS. XVll 
 
 Page 
 
 To find the area of an irregular polygon of any 
 
 number of sides - - - - - 37 
 Given the side of a regular polygon, to find the 
 
 radius of its inscribed or circumscribing circle 39 
 
 To find the area of a regular polygon - - 40 
 
 of a regular polygon, when the side 
 
 only is given - - - - - 41 
 The diameter of a circle being given, to find the cir- 
 cumference ; or, the circumference being given, to 
 
 find the diameter - - - - 42 
 
 To find the length of any arc of a circle - - 44 
 
 To find the area of a circle - - - - 47 
 
 of a sector of a circle - - 48 
 
 of a segment of a circle - - 50 
 
 of a circular zone - - - 52 
 
 of a circular ring - - - 53 
 
 of a lune - - - - 54 
 
 of an ellipse - - - 55 
 
 of an elliptical segment - -56 
 
 of any curvilineal figure, by means 
 
 of equi-distant ordinates - - - - 57 
 
 A Table of circular segments - - - 58 
 
 PAET III. 
 
 Section I. 
 
 Land- Surveying - - - - - 63 
 
 Miscellaneous Instructions - - - 64 
 
 To measure a field of three sides - - - 65 
 
 a field of four sides - - - 70 
 
 a field of more than four sides - 76 
 
 an irregular, narrow piece of land - 80 
 
 a mere or wood - - - 82 
 
 To measure and plan roads, rivers, canals, 8cc. - 83 
 To find the breadth of a river - - ••86 
 To measure a line upon which there is an impedi- 
 ment - - - - - - 87 
 
 To part from a rectangular field, any proposed quan- 
 tity of land, by a line parallel to one of its sides - 87 
 To part from a trapezium or any irregular polygon 
 
 whatever, any proposed quantity of land - - 88 
 To reduce statute-measure to customary, and vice 
 
 tersa - - - - - - 92 
 
Xviii CONTENTS. 
 
 Page 
 
 To survey and plan Estates or Lordships - - 94 
 
 Method I. 94 
 
 II. 99 
 
 III. 100 
 
 IV. - - - - - 100 
 
 Miscellaneous Instructions - - - - 101 
 A Book of Dimensions, Castings, &c. belonging to 
 
 Plate III. - - - - - 104 
 Method of surveying single fields without the cross- 
 staff 10,9 
 
 Section II. 
 Miscellaneous Questions concerning Superficial Men- 
 suration - - - ° - - 117 
 
 PART IV. 
 
 Section I. 
 
 Mensuration op Solids - - - - 121 
 
 Definitions - - - - - -121 
 
 A Table of solid measure - - - - 124 
 
 To find the solidity and surface of a cube - - 124 
 
 the solidity and surface of a parallelopipedon 126 
 
 the solidity and surface of a prism - - 127 
 
 the solidity and surface of a cylinder - 1 29 
 
 the solidity and surface of a pyramid - 130 
 
 the solidity and surface of the frustum of a 
 
 pyramid ------ 131 
 
 the solidity and surface of a cone - - 132 
 
 the solidity and surface of the frustum of a 
 
 cone --.--- 133 
 
 the solidity and surface of a cuneus or wedge 135 
 
 . the solidity and surface of a prismoid - 136 
 
 the solidity and convex surface of a sphere 
 
 or globe ------ 138 
 
 the solidity and surface of the segment of a 
 
 sphere ------ 138 
 
 the solidity and convex surface of the frus- 
 
 tum of a sphere ----- 139 
 
 the solidity and surface of a circular spindle 140 
 
 the solidity of the middle frustum or zone of 
 
 a circular spindle - - - - 141 
 
 the solidity and surface of a cylindrical ring 142 
 
 the solidity or superficies of any regular body 143 
 
CONTENTS. XIX 
 
 Page 
 
 To find the solidity of an irregular solid - - 145 
 the magnitude or solidity of a body from its 
 
 weight ------ 147 
 
 A Table of the specific gravities of bodies - - 147 
 To find the weight of a body from its magnitude or 
 
 solidity - - - - - - 148 
 
 Section II. 
 
 Description of the Carpenter's Rule - - 149 
 
 The Use of the Sliding Rule - - - 150 
 
 Multiplication by the sliding rule - - - 150 
 
 Division by the sliding rule - - - 151 
 
 To square any number - - - - 151 
 
 To extract the square root of any number - - 151 
 
 To find a mean proportional between two numbers - 152 
 
 a fourth proportional to three numbers ; or 
 
 to perform the Rule of Three - - - 152 
 
 the areas of plane figures by the sliding rule 153 
 
 the contents of solids - - - 153 
 
 TiJiBER IVIeasure - - - - -154 
 
 To find the superficial content of a board or plank - 154 
 — the solidity of squared or four-sided timber 155 
 
 Section III. 
 
 To find the solidity of round or unsquared timber - 156 
 A Table for measuring timber - - - 158 
 
 To measure and value standing timber - - 160 
 
 IMiscellaneous Instructions for finding the dimensions 
 of standing timber, and for setting-out woods, and 
 valuing them previously to a fall - - 160 
 
 A Description of Timber Trees - - - 166 
 
 The Oak - - - - - - 166 
 
 The Ash 167 
 
 The Beech and Ehn - - - - 167 
 
 The Sycamore and Black Poplar - - - 167 
 
 The Alder, Weymouth Pine, and Fir - - 168 
 
 The Birch - - - - - - 169 
 
 The Yew - - - - - - 169 
 
 The Holly and Box - - - - 169 
 
 Section IV. 
 Miscellaneous Questions concerning Solids - - 171 
 
XX CONTENTS. 
 
 PART Y. 
 
 Page 
 
 Artificers' Work - - - 175 
 
 Work — Bricklayers - - - - 176 
 
 . Masons ----- 178 
 
 A Description of Portland, Yorkshire,, and Purbeck 
 
 Stone, and of various kinds of Marble - - 181 
 
 Work — Carpenters' and Joiners' - - - 182 
 
 Slaters' and Tilers' - - - 188 
 
 Plasterers' - - - - 189 
 
 Painters' - - - - - 190 
 
 Glaziers' - - - - - 192 
 
 Plumbers' - - - - 193 
 
 Pavers' - - - - - 194 
 
 Vaulted and Arched Rooes - - - 195 
 Prob. I. To find the content of the vacuity of a cir- 
 cular, an elliptical, or a gothic vaulted roof - 198 
 To find the concave sm-face of a circular, an elKptic, 
 or gothic vaulted roof - - - - 197 
 
 the solid content of a dome ; its height, and 
 
 the dimensions of its base being given - - 198 
 
 the superficial content of a spherical dome - 198 
 
 the solid content of a saloon - - 199 
 
 = the superficial content of a saloon - - 200 
 
 the solid content of the vacuity formed by a 
 
 groin arch, either circular or elliptical - - 200 
 the concave surface of a circular or an ellip- 
 tical groin - - - - - 201 
 General Iixustration of the W®rks of Arti- 
 ficers, FROM the Dl^ilENSIONS OF A HOUSE - 202 
 
 The Bricklayer's Work - - - 202 
 
 Abstract of the Brick Work - - - 204 
 
 The Bricklayer's Bill - - - - 205 
 
 The Mason's Work - - - - 206 
 
 Abstract of the Masonry - - - - 208 
 
 The Mason's Bill 210 
 
 The Carpenter's and Joiner's Work - - 213 
 
 Abstract of the Carpenter's and Joiner's Work - 213 
 
 The Carpenter's and Joiner's BiU. - - - 215 
 
 The Slater's Work and Bill - - - 215 
 
 The Plasterer's Work - - - 216 
 
 Abstract of the Plastei'ing - - - 217 
 
 The Plasterer's Bill - - - - 218 
 
 Useful Works, on Architecture, recommended - 218 
 
CONTENTS. XXl 
 
 PAET VI. 
 
 Page 
 The Mensuration of Hay- Stacks, Drains, Canals, 
 Marl-pits, Ponds, Mill-dams, Embankments, Quar- 
 ries, Coal-lieaps, and Clay-heaps - - 220 
 To measure a hay-stack, having a circular base - 220 
 
 a hay-stack, having a rectangular base - 224 
 
 To find the number of cubic yards, which have been 
 dug out of a drain or canal > - - 22S 
 
 • how many cubic yards have been dug out of 
 
 a marl-pit . _ - - - 233 
 
 Directions for measuring ponds and mill-dams - 237 
 
 for measuring embankments - - 238 
 
 for measuring quarries - - - 239 
 
 for measuring coal-heaps - - 240 
 
 for measuring clay-heaps - - 241 
 
 A description of some of the principal canals in 
 
 England, Scotland, France, and China - - 243 
 
 An account of some of the principal drains in the 
 county of York, and the agricultural improve- 
 ments made by them - - - - 247 
 An account of agricultural improvements made in the 
 
 county of Lincoln, by means of drains - - 249 
 
 lili-. Joseph Elkington's method of draining land - 250 
 
 PART YII. 
 
 Conic Sections and their Solids - - - 252 
 
 Definitions ...... 252 
 
 In an ellipse, to fmd the transverse, or conjugate, or 
 
 ordinate, or absciss, having the other three given - 25^ 
 
 Properties of the ellipse _ - - - 256 
 
 To describe a parabola - - - - 257 
 
 To find any parabohc absciss or ordinate - - 257 
 
 the length of the arc of a parabola, cut off by 
 
 a double ordinate _ . - . 258 
 
 the area of a parabola . - - 258 
 
 the area of a parabolic frustum - - 259 
 
 Properties of the parabola ' - - - - 259 
 
 To construct a hyperbola - - - - 259 
 
 In a hyperbola, to find the transverse, or conjugate, 
 
 or ordinate, or absciss _ - - - 260 
 
 To find the length of a hyperbolic curve - - 262 
 
XXll CONTENTS. 
 
 Page 
 
 To find the area of a hyperbola - - _ 263 
 
 Properties of the hyperbola _ _ _ 264 
 
 To find the solidity of a spheroid - - - 264 
 
 the solidity of the segment of a spheroid - 265 
 
 the content of the middle frustum of a 
 
 spheroid ------ 266 
 
 the solidity of an elliptic spindle - - 268 
 
 the solidity of a parabolic conoid - - 269 
 
 the solidity of the frustum of a paraboloid - 269 
 
 the solidity of a parabolic spindle - - 270 
 
 the solidity of the middle frustum of a para- 
 bolic spindle - - - - - 271 
 
 the solidity of a hyperboloid - - 271 
 
 the solidity of the frustum of a hyperboloid 272 
 
 the solidity of a hyperbolic spindle - - 273 
 
 the solidity of the middle frustum of an 
 
 elliptic or a hyperbolic spindle - - - 273 
 
 PART VIII. 
 
 GAUGING. 
 
 Gauging - - - - ^ - 275 
 
 New Imperial Tables - - - - 275 
 
 Tables of Wine and Spirit Measure - - 276 
 
 Tables of Ale and Beer Measure - - - 276 
 
 Tables of Corn or Dry Measure - - - 277 
 
 Kules for reducing the Old Measures to the New 
 
 Standard ; and vice versa . - _ 278 
 
 Multipliers for performing the same - - 279 
 
 for reducing imperial gallons and bushels 
 
 to old gallons and bushels . _ - 279 
 
 for reducino; Irish bushels and Scotch 
 
 bolls, to imperial bushels ; and vice versa - 280 
 
 A description of the sliding rule - - - 281 
 
 The method of finding the multipliers, divisors, and 
 
 gauge-points ----- 284 
 A table of multipliers, divisors, and gauge-points - 285 
 To find the content of a vessel in the form of a cube 286 
 the content of a vessel in the form of a 
 
 parallelopipedon ----- 287 
 the content of a vessel in the form of a 
 
 cylinder - - - > - 288 
 
CONTENTS. XXlll 
 
 Page 
 
 To find the content of a vessel in the form of a pris- 
 moid, or the frustum of a square pyramid, or a 
 cjlindroid - - - - - 289 
 
 — the content of a vessel, in the form of the 
 
 frustum of a cone . - _ - 290 
 
 '— the content of a circular vessel, with curved 
 
 sides - - - - - 291 
 
 To gauge and inch a round guile-tun, according to 
 
 the method practised in the Excise - - 294 
 
 a back or cooler - - - - 299 
 
 a cistern, couch, or floor of malt - - 301 
 
 Cask Gauging _ - - - - 304 
 
 Of the four varieties of casks _ - - 304 
 
 To take the dimensions of a standing cask - - 304 
 
 To take the dimensions of a lying cask - - 305 
 
 To find the content of a cask of any of the four 
 
 varieties ------ 306 
 
 General rules for finding the content of any cask, 
 
 without regard to its variety - - - 309 
 
 To find the content of a cask by the diagonal rod - 312 
 To ullage a standing cask - - - - 313 
 
 a lying cask - - - - 315 
 
 Miscellaneous Questions concerning Gauging - 316 
 
 The method of computing distances by the velocity 
 
 of sound ------ 318 
 
 Methods of ascertaining the tonnage of ships - 320 
 
 PART IX. 
 
 TRIGONOMETRY. 
 
 Explanations of the most useful properties - - 323 
 
 The doctrine of right-angled plane triangles, demon- 
 strations of different problems, and a variety of 
 practical examples - - - _ 328 
 
 The doctrine of oblique-angled plane triangles, de- 
 monstrations, and practical questions - - 335 
 
 PART X. 
 
 Trigonometry applied in ascertaining the heights and 
 distances of inaccessible objects in a great variety 
 of new and practical questions - - - 344 
 
XXIV CONTENTS. 
 
 Pago 
 
 The method of finding the heights of mountains by 
 
 means of the barometer and thermometer - 358 
 
 On the figure of the earth - - - - 363 
 
 Properties of the pendulum _ . - 359 
 
 On the figure and magnitude of the earth - - 369 
 
 On the true and elliptic latitude - - - 372 
 
 The arc of the meridian and the elliptic latitude - 372 
 
 The arc of the meridian and the true latitude - 373 
 
 The arc of the meridian between two latitudes - 374 
 
 The measure of a degree of the meridian - - 374 
 
 The length of a meridional degree - - - 374 
 
 The length of a mean degree _ - _ 374 
 
 To find distances on the earth's surface - - 375 
 
 The length of a degree of longitude - - - 375 
 
 The measurement of an arc of the meridian - - 376 
 
 The admeasurement of a base line - - - 378 
 
 Errors in taking angles - - - - 380 
 
 Of terrestrial refraction _ - - - 382 
 
 Of the reduction of angles - - - - 385 
 Of the reduction of angles, from one plane to another 387 
 Trigonometrical questions respecting the earth's 
 
 curvature - - - - -388 
 
 The dip or depression of the horizon at sea - - 389 
 
 Leyelling - - - - - 391 
 
 The Builder's Dictionaut - . - 409 
 
 Questions for the Examination of the Puph. - 434 
 
A TREATISE 
 
 ON 
 
 PRACTICAL MENSURATION. 
 
 PAKT I. 
 
 DEFINITIONS, TROBLEMS, AND THEOREMS, IN 
 GEOMETRY. 
 
 Geometry originally signified the Art of measuring 
 the Earth, or any distance or dimensions upon or within 
 it ; but it is now used for the Science of Quantity, Ex- 
 tension, or Magnitude, abstractedly considered. 
 
 1. 
 
 GEOMETEICAL DEFINITIONS. 
 
 A point is considered as having neither length, 
 
 breadth, nor thickness. 
 
 2. A line has length, but is considered as having neither 
 breadth nor tliickness. 
 
 3. Lines are either right, curved, or parallel. 
 
 4. A right or straight line lies 
 
 wholly in the same direction between 
 
 its extremities, and is the shortest 
 distance between two points. 
 
 o. A curved line continually 
 changes its direction between its 
 extremities. 
 
 6. Parallel lines always remain at 
 the same distance from each other, 
 though infinitely produced. 
 
GEOMETRICAL 
 
 PAET I. 
 
 . 7..Ji. super.Scies has length and 
 breacLtliVbut k consifeed -as having 
 no thickness. 
 
 8. A superficies may be contained 
 within one curved line, but cannot 
 be contained within fewer than tln-ee 
 straight lines. 
 
 9. A solid is a figure of three t 
 dimensions ; namely, length, breadth, 
 and thickness. 
 
 10. An angle is the inclination or opening of two lines, 
 having different directions, and meet- 
 ing in a point, which is called the 
 angular point, as at A; and when 
 tlu-ee letters are used, the middle one 
 denotes that point. 
 
 11. Angles are of three kinds; viz. right, acute, and 
 obtuse. 
 
 12. A right angle is made by two 
 right lines which are perpendicular 
 to each other. 
 
 13. An acute angle is less than a 
 right angle, as CAB. 
 
 14. The complement of an angle is 
 what it wants to complete a right 
 angle, as the angle DAB is the com- 
 plement of the angle CAB. 
 
 15. An obtuse angle is greater than 
 a ri^ht amrle, as BCD. 
 
 16. The supplement of an angle is what it wants of two 
 right angles, as the angle ACD is the supplement of the 
 angle BCD. 
 
 17. A triangle is a figure or superficies bounded by 
 three riglit hues, and admits of three varieties ; viz. equi- 
 lateral, isosceles, and scalene. 
 
TART I. 
 
 DEFINITIONS. 
 
 lb. An equilateral triangle lias all 
 its sides equal. 
 
 19. An isosceles triangle lias only 
 two of its sides equal. 
 
 A scalene triangle 
 
 20. 
 sides unequal. 
 
 lias all its 
 
 21. Triangles are also right-angled, acute-angled, and 
 obtuse-angled. 
 
 22. A right-angled triangle has one 
 right angle, the side opposite to which 
 is called the hypothenuse, the other 
 two being termed legs, or one the 
 perpendicular and the other the base ; 
 thus AC is the hypothenuse, BC the 
 perpendicular, and AB the base. 
 
 23. An acute-angled triangle has 
 all its angles acute. 
 
 24. An obtuse-angled triangle has 
 one of its an2;les obtuse, as ACB. 
 
 25. The base of any figure is that side upon which it is 
 supposed to stand, or upon which a perpendicular is let 
 fall from the vertex or opposite angle ; and the altitude of 
 a figure is its perpendicular height. In the last figure, 
 AB is the base, and CD the perpendicular. 
 
 26. 
 
 A fi^rure of four sides and angles is denominated a 
 
 quadrangle or quadrilateral figure. 
 
 27. A parallelogram is a quadrilateral figure, having 
 B 2 
 
 its 
 
GEOMETRICAL 
 
 PAJIT I. 
 
 opposite sides parallel and equal, and admits of four 
 varieties ; viz. the square, the rectangle, the rhombus, and 
 the rhomboid. 
 
 28. A square is an equilateral paral- 
 lelogram, having all its angles right 
 angles. 
 
 29. A rectangle is a parallelogram, 
 having its opposite sides equal, and 
 all its angles right angles. 
 
 30. A rhombus is an equilateral 
 parallelogram, having its opposite 
 angles equal, two of which are 
 acute, and two obtuse; which, in a 
 rcGTular rhombus, are 60 and 120 
 degrees. 
 
 31. A rhomboid or rhomboides 
 having its opposite sides and angles 
 equal, two of its angles being acute, 
 and two obtuse ; and when the figure 
 is regular, the angles are 60 and 120 
 degrees. 
 
 32. A trapezium is a quadi'ilateral 
 figure, whose opposite sides are not 
 parallel to each other. 
 
 33. A diagonal is a right line joininf 
 angles of a quadrilateral figure, as AB. 
 
 34. A trapezoid is a quadrilateral 
 figure, having two of its opposite sides 
 parallel. 
 
 parallelogram. 
 
 the opposite 
 
 35. Plane figures having more than four sides are 
 generally called polygons; and receive their particular 
 denominations from the number of their sides or angles. 
 
I'AliT 1. DEFINITIONS. O 
 
 36. A pentagon is a polygon of five sides ; a liexagou 
 of six ; a heptagon of seven ; an octagon of eight ; a 
 nonagon of nine ; a decagon of ten ; an undecagon of 
 eleven ; and a duodecagon of twelve sides. 
 
 37. A regular polygon has all its sides and angles 
 equal ; when they are unequal, the polygon is irregular. 
 
 38. A circle is a plane figure, 
 bounded by a cui-ved line, called the 
 circumference, which is everywhere 
 equidistant from a certain point 
 within it, called the centre. 
 
 39. The diameter of a circle is a right line dra^\n 
 througli the centre, and terminating in the circumference 
 on each side, as AB. 
 
 40. The radius of a circle is half the diameter ; or it is 
 a right line drawn from the centre to the circumference, 
 as CD. 
 
 41. An arc of a circle is anyj)art of the circumference, 
 as the arc EHF. 
 
 42. A chord is aright line joining the extremities of 
 an arc, as the line EF ; and the versed sine is part of the 
 diameter cut oiF by the chord, as GH. 
 
 43. A segment is any part of a circle, bounded by an 
 arc and its chord. 
 
 44. A semicircle is half of a circle, or a segment cut off 
 by the diameter, as ADB. 
 
 45. A quadrant is the fourth part of a circle, as AT3C. 
 
 46. A sector is any part of a circle, bounded \)y an arc 
 and two radii. 
 
 47. The circumference of every circle is supposed to be 
 divided into 360 equal parts, called degrees ; each degree 
 into 60 equal parts, called minutes ; and each minute into 
 60 equal parts, called seconds. 
 
 48. The arc of a quadrant contains 90 degrees, which 
 is the measure of a right angle. 
 
 c3 
 
6 GEOMETRICAL PAin" I. 
 
 49. An ellipse is a plane figure 
 bounded by a curved line, called the 
 circumference; but as the figure is 
 not a circle, it is described from two 
 points in the longest diameter, called 
 the foci, or focuses. 
 
 50. The longest diameter that can be drawn within an 
 ellipse is called the transverse diameter, as AB ; and the 
 shortest the conjugate diameter, as CD. Sometimes theso 
 diameters are termed axes. 
 
 51. Identical or similar figures, are such as have all the 
 sides and angles of the one equal to all the sides and 
 angles of the other, each to each ; so that if the one figure 
 were applied or laid upon the other, aU the parts would 
 coincide. 
 
 52. The perimeter of a figure is the sum of all its sides 
 taken together. 
 
 53. A proposition is something proposed to be done, or 
 to be demonstrated ; and is either a problem or a theorem. 
 
 54. A problem is something proposed to be done, and 
 requires construction. 
 
 55. A theorem is sometliing proposed to be demon- 
 strated ; or it is a truth stated, and requires proof or de- 
 monstration. 
 
 5Q. A corollary is a consequeiit truth, gained from some 
 preceding truth or demonstration. 
 
 57. Things which are equal to the same thing, are 
 equal to each other. 
 
 58. When equals are added to equals, the wholes are 
 equal. 
 
 59. When equals are taken from equals, the remains 
 are equal. 
 
 60. When equals are added to unequals, the wholes are 
 unequal. 
 
 61. "Wlien equals are taken from unequals, the remains 
 are unequal. 
 
 62. Things which are halves of the same. thing, are 
 equal. 
 
 63. The whole is equal to all its parts taken together. 
 
 64. Things vrhich coincide or fill the same space, are 
 identical or mutually equal in all their parts. 
 
PAST I. 
 
 PR0BLE3IS. 
 
 GEOMETEICAL PEOBLEMS. 
 PROBLEM I. 
 
 To bisect a sxiven line AB. 
 
 From A and B as centres, with any 
 radius, greater than half AB, in your 
 compasses, describe arcs cutting each 
 other in m and n. Draw the line mCn, 
 and it will bisect AB in C. 
 
 
 / 
 
 \ 
 
 
 
 CI 
 
 
 A 
 
 
 
 B 
 
 
 X. 
 
 
 PROBLEM IL 
 
 To bisect a given angle ABC. 
 
 From the point B, with any radius, 
 describe the arc AC. From A and C, 
 with the same, or any other radius, 
 make the intersection m. Draw the 
 line Bm, and it will bisect the angle 
 ABC, as required. 
 
 PROBLEM in. 
 To draio a line parallel to a given line AB. 
 
 Case 1. Wlien the parallel line is to pass through a 
 given point C. 
 
 From any point m, in the line AB, with the radius 7nQ 
 describe the arc Qn. From the 
 centre C, with the same radius, de- r V 
 
 scribe the arc mr. Take the dis- ~ 7 T^ 
 
 tance Cn in the compasses, and apply / / 
 
 it from 771 to r. Through C and r / / 
 
 draw the line DE, and it will be the ~ ■:^^ — ^ 
 
 parallel required. 
 
 B 4 
 
8 GEOMETRICAL PAllT I. 
 
 Case 2. Wlien the parallel line is to be at a given dis- 
 tance from AB. 
 
 From any two points m and ?2, in ^ ^ ^ ^ 
 
 the line AB, with a radius equal to '^ "^^^^^I IP" ^ 
 the given distance, describe the arcs 
 r and o. Draw the line CD, to 
 
 touch these arcs, without cutting ^ — ^ ^j — 5 
 
 them, and it will be the parallel 
 required. 
 
 Note. — This Problem may be more easily performed by means of 
 a parallel ruler, which may also be used to advantage in several 
 operations in Practical Geometry. 
 
 PROBLEM IV. 
 
 To erect a 'perpendicular from a given point C, in a given 
 line AB. 
 
 Case 1. "When the point is near the middle of the 
 line. 
 
 On each side of the point C, take 
 any two equal distances. Cm and 
 C?^. From m and ??, as centres, 
 with any radius greater than Cm or 
 C«, describe two arcs cutting each 
 
 other in r. Draw the line Cr, and i_ 
 
 it will be the perpendicular required. ^ '" ^ '' "^ 
 
 Case 2. When the point C is at or near the end of 
 the given line. 
 
 From any point m, as a centre, 
 with the radius or distance Cm, de- 
 scribe the arc 7^Cr, cutting the given 
 line in n and C. Through n and m 
 draw a line cutting the arc in r. \ 
 Draw the line Cr, and it will be the \/ 
 perpendicular. required. ^ Ai~.^ ^- c b 
 
 PROBLEM y. 
 
 From a given point C, to let fall a perpendicular iipon a 
 given line AB. 
 
 Case 1. "Wlien the point is nearly opposite to the middle 
 of the line. 
 
PART I. 
 
 PKOBLEMS. 
 
 With C as a centre, and any radius 
 a little exceeding the distance of the 
 given line, describe an arc cutting 
 AB in m and n. With the centres 
 m and n, and the same, or any radius 
 exceeding half their distance, describe 
 arcs intersecting each other in r. 
 Draw the line C;*, and CD will be 
 the perpendicular required. 
 
 Case 2. When the given point C is nearly opposite to 
 the end of the given line. 
 
 From C draw the line CM to meet 
 AB, in any point M. Bisect the line 
 CjM in the point N ; and with the 
 centre N, and radius CN or MN, de- 
 scribe an arc cutting AB in D. Draw 
 the line CD, and it will be the per- 
 pendicular required. 
 
 
 
 , ^- 
 
 c 
 
 1^^ / 
 
 \ 
 
 \ ^K 
 
 \ 
 
 ^ A \ 
 
 
 X 
 
 ' 
 
 
 
 \ ^' "-- ^- 
 
 / 
 
 
 
 D B 
 
 Note. — Perpendiculars inay be more easily raised or let fall by 
 means of a square, or of a feather-edged scale or ruler, with a line 
 across it perpendicular to its edges. 
 
 PROBLEM VI. 
 
 To find the centre of a given circle^ or one already 
 described. 
 
 Draw any chord AB ; and bisect 
 it perpendicularly with CD, which 
 will be a diameter. Bisect CD in the 
 point o, which will be the centre re- 
 quired. 
 
 PROBLEM VII. 
 
 To make a triangle with three given lines, any tico of 
 ivhich, taken together, must be greater than the third, 
 {Euclid, I. 22.) 
 
 Let the given lines be AB = 12, AC = 10, ?nd 
 BC = 8. 
 
10 
 
 GZOMETPJCAL 
 
 PART I. 
 
 From any scale of equal parts (wliicli is to be under- 
 stood as employed likervise in the 
 following Problems) lay off the base 
 
 AB. AYith the centre A, and radius 
 
 AC. describe an arc. With the cen- 
 tre B, and radius BC, describe another 
 arc, cutting the former in C. Draw 
 the lines AC and BC, and the triangle 
 will be completed. 
 
 Note. — A trapezium maybe constructed in the same manner 
 havinn; the four sides and one of the diasjonals, 
 
 PEOBLEM Yin. 
 
 Having given the base, the perpendicular, and the place of 
 the perpendicular upon the base, to construct a triangle. 
 
 Let the base AB = 12, the perpendicular CD — 6, and 
 the distance AD = 7. 
 
 Make AB equal to 12, and AD 
 equal to 7. At D erect the perpen- 
 dicular DC, which make equal to 6. 
 Join AC and BC, and the figure will 
 be completed. 
 
 Note. — A trapezium may be constructed in a similar manner, 
 by having one of the diagonals, the two perpendiculars let fall 
 thereon from the opposite angles, and the places of these perpen- 
 diculars upon the diagonal ; and a trapezoid may be constructed by 
 drawing the two parallel sides perpendicularly to their base or given. 
 distance. 
 
 PROBLEM IX. 
 
 To describe a square, whose side shall he ec/iial to a given 
 line. 
 
 Let the given line AB == 8. tv' c 
 
 Upon one extremity B, of the given I 1 
 
 line, erect the perpendicular BC, which | ! 
 
 make equal to AB. With A and C as 
 centres, and the radius AB, describe arcs 
 cutting each other in D. Join AD and 
 CD, and the square will be completed. 
 
PART 1. 
 
 PROr>LEMS. 
 
 11 
 
 PROBLEM X. 
 
 To describe a rectangular parcilldograjn, whose length 
 and breadth shall be eqaid to two given lines. 
 
 Let the length AB = 12, and the breadth BC= 6. 
 
 At B erect the perpendicular BC, which make equal 
 to 6. With A as a centre, and the 
 radius BC, describe an arc ; and 
 with C as a centre, and the radius 
 AB, describe another arc, cutting 
 the former in D. Dj-aw the lines 
 AD and CD, and the rectangle will 
 be completed. 
 
 PROBLEM XL 
 Upo7i a given right line to construct a regidar rhoj?ib2is. 
 
 Let the given line AB = 8. 
 Draw the line AB equal to 
 
 8. With A and B as 
 
 centres, and the radius AB, de- 
 scribe arcs cutting each other in D ; 
 then with B and D as centres, and 
 the same radius, make the inter- 
 section C. Draw the lines AD, 
 DC, and BC, and the rhombus will 
 be completed. 
 
 PROBLEM^ XIL 
 
 To construct an irregular rhombus^ having given its side 
 and perpendicular height. 
 
 Let the side = 8, and the perpendicular = 6. 
 
 Draw AB equal to 8 ; at A erect the perpendicular 
 AE, which make equal to 6 ; and 
 di^aw EC parallel to AB. With 
 the radius AB, and A as a centre, 
 make the intersection D ; and with 
 the same radius, and B as a centre, 
 make the intersection C. Join AD, 
 DC, and CB, and the figure v.dll be 
 completed. 
 
 B 6 
 
12 GEOMETRICAL PART I 
 
 PROBLEM XIII. 
 
 Having any two right lines given, to construct a regular 
 rhomboid. 
 
 Let the given lines be AB = 12, and BC = 6. 
 
 Draw the line AB equal to 12. Take in your com- 
 passes the line BC, and lay it from A to E. Witli A 
 and E as centres, and the radius d^ .^. 
 
 AE, make the intersection D. Then ^■p'^ y^^ 
 
 with B as a centre, and the same / / 
 
 radius, describe an arc ; and with D Z i / 
 
 as a centre, and the radius AB, ^ ^' ^^ 
 
 describe another arc, cutting the former in C. Draw the 
 lines AD, DC, and BC, and the rhomboid will be completed. 
 
 PROBLEM XIV. 
 
 Having given the base, the perpendicular, and the place 
 of the perpendicular upon the base, to construct an irre- 
 gular rhomboid. 
 
 Let the base AB = 15, the perpendicular ED = 6, 
 and the distance AE = 5. 
 
 Make AB equal to 15, and AE equal to 5. At E 
 erect the perpendicular ED, which make equal to 6 ; 
 and join AD. Yv^ith the radius AB, ^ 
 
 and D as a centre, describe an arc ; a v-P 
 
 and with B as a centre, and the / \ / ' 
 
 radius AD, describe another arc, Z — 1 / 
 
 cutting the former in C. DraAV the ^'' 
 
 lines DC and BC, and the figure will be completed. 
 
 Note. — The sum of all the interior angles of any quadrilateral 
 figure is equal to four right angles. 
 
 PROBLEM XV. 
 
 Having the transverse aiid conjugate diameters given, to 
 construct an ellipse. 
 
 Let the transverse diameter AB =14, and the con- 
 jugate diameter CD — 8. 
 
 Draw the two diameters to bisect each other perpen- 
 dicularly in the centre o. With the radius Ao, and 
 the centre D, intersect AB in F and /; these two 
 
PART I. 
 
 rilOBLEMS. 
 
 13 
 
 points will be the foci of the ellipse. Take an}' point 
 
 m, in the transverse diameter, and with F and _/' as 
 
 centres, and the radius Aw, describe the arcs G, G, 
 
 //, g. Then with the same centres, 
 
 and the radius B/«, describe arcs '^^.^K ^' 
 
 cutting the former in the points ^X' - 
 
 G, G, g, g ; thus you will have /" , , . 
 
 four points in the circumference 
 of the ellipse. Again, take a se- 
 cond point ?^, in the transverse 
 diameter, and proceeding as before, 
 you will determine other four points. By the same 
 method you may determine as many more as you please ; 
 through all of whicli, with a steady hand, draw the cir- 
 cumference of the ellipse. 
 
 PROBLEM XVI. 
 
 Upon a given line AB, to make a regular pentagon. 
 
 Make Bw^ perpendicular to AB, and equal to half 
 of it. Draw Aw, and produce it till mn be equal to 
 Bw. With the radius V>n, and A 
 and B as centres, describe arcs in- 
 tersecting each other in o, which 
 will be the centre of the circum- 
 scribing circle. From the point o, 
 with the same radius, describe the 
 circle ABODE ; and apply the line 
 AB five times to the circumference, 
 marking th(^. angular points, whicli 
 connect with right lines, and the figure will be completed. 
 
 PROBLEM XVII. 
 
 Upon a given line AB, to make a regular hexagon. 
 
 "With A and B as centres, and 
 the radius AB, describe arcs inter- 
 secting each other in o ; and with o 
 as a centre, and the same radius, 
 describe the circle ABODEF. Ap- 
 ply the line AB six times to the 
 circumference, and it vrill form the 
 hexagon required. 
 
14 
 
 GEOJIETRICAL 
 
 PAST I. 
 
 PROBLEM XVIII. 
 
 Uiyon a given line AB, to construct a regular octagon. 
 
 On the extremities of the given line AB, erect the in- 
 definite perpendiculars AF and BE. Produce the line 
 AB, both Y^^ays to m and n ; and 
 with the lines AH and BC, each 
 equal to AB, bisect tlie angles wAF 
 and ?2BE. Draw CD and HG 
 parallel to AF or BE, and each 
 equal to AB. With D and G- as 
 centres, and the radius AB, describe 
 arcs intersecting AF and BE, in the 
 points F and E. Join OF, FE, and 
 EB, and the figure will be completed. 
 
 PROBLEM XIX. 
 
 In a given triangle ABC, to inscribe a circle. 
 
 Bisect the angles A and B, vv'itli 
 the lines, Ao, Bo, and o will be the 
 centre of the required circle ; and 
 its radius will be the nearest distance 
 to any one of the sides ; hence the 
 circle may be described. 
 
 PROBLEM XX. 
 
 About a given triangle ABC, to circumscribe a circle ; or 
 to describe the circumference of a circle through three 
 given points, A, B, C. 
 
 c 
 
 Bisect the sides AB, BC, with 
 the perpendiculars mo and no ; and 
 o will be the centre of the circle, and 
 its radius will be Ao, Bo, or Co. 
 
PnOBLEMS. 
 
 15 
 
 PROBLEM XXI. 
 
 To make a triangle equal to a given trapezium ABCD. 
 
 Draw the diagonal DB, and parallel 
 to it draw CE meeting AB produced 
 in E. Join the points DE ; so shall 
 the triangle ADE be equal to the 
 trapezium ABCD. 
 
 PROBLEM XXIL 
 
 To make a right angle hy the line of chords on the plane 
 scale. 
 
 Draw the unlimited line AB ; then 
 take in your compasses 60° from the 
 line of chords, and with A as a centre, 
 describe the arc ED. Take 90° from 
 the same scale, and set off that extent 
 from D to C. Draw the line AC ; 
 and CAD will be the angle required. ^' ^ '' 
 
 PROBLEM XXin. 
 
 To make an acute angle that shall contain any number of 
 degrees ; suppose 35° 30'. 
 
 Draw the unlimited line AB ; then 
 take 60° in your compasses, and with 
 A as a centre, describe the arc ED. 
 Set off the angle 35° 30', from D to C. 
 Draw the line AC ; and CAD will 
 be the angle required. 
 
 PROBLEM XXIY. 
 
 To make an obtuse angle that shall contain any number 
 of degrees ; suppose 128° 35^ 
 
 Draw the indefinite line AJ^ ; and 
 with the chord of 60° in your com- g ,.--—- — -^^ 
 passes, describe the arc DE. Set off ^ 
 90° from D to C ; and from C to O 
 set off the excess above 90°, which 
 
 is 38° 35^ Draw the line AG- ; and .^ ^^ ^ 
 
 GAD will be the required angle. 
 
16 
 
 GEOMETRICAL 
 
 *ART I. 
 
 PROBLEM XXV. 
 
 To find the number of degrees contaiaed in any given 
 angle BAC. 
 With the chord of 60, and A as a .g 
 
 centre, describe the arc mn. Take the _ / 
 
 distance mn in your compasses, and 
 apply it to the line of chords ; and it 
 
 will show the number of degrees re- 
 
 quired. a ** c 
 
 Note 1. — The measure of an angle is an arc of any circle con- 
 tained between the two lines which form that angle, the angular 
 point being the centre of the circle ; and it is estimated by the number 
 of degrees that are contained in that arc. 
 
 Note 2. Angles may be more expeditiously laid down or measured, 
 
 bv means of a semicircle of brass, called " a Protractor," the arc of 
 which is divided into 180 degrees. 
 
 PROBLEM XXVI. 
 
 Upon a given line AB, to make a regular polygon of any 
 proposed number of sides. 
 
 Divide 360° by the number of sides, subtract the 
 quotient from 1 80°, and divide the dif- 
 ference by two. Make the angles ABC 
 and BAG each equal to the quotient 
 last found ; and the point of intersec- 
 tion C, will be the centre of the circum- 
 scribing circle. With the radius AC 
 or BC, describe the circle ; and apply 
 the chord AB to the circumference the 
 proposed number of times, and it will 
 form the polygon required. 
 
 PROBLEM XXVII. 
 
 In any given circle to inscribe a regular polygon of any 
 proposed number of. sides ; or to divide the circumference 
 into any number of equal parts. 
 
 Divide 360° by the number of sides, 
 and make the angle ACB, at the centre, 
 equal to the number of degrees con- 
 tained in the quotient ; and the arc AB 
 v^•ill be one of the equal parts of the 
 circumference ; hence the polygon may 
 be formed. 
 
PART I. 
 
 PROBLEMS. 
 
 17 
 
 Note. — The sum of all the interior angles of any polygon, 
 whether regular or irregular, is equal to twice as many right angles, 
 wauling four, as tlie figure has sides. 
 
 PROBLEM XXVni. 
 To find a mean •proportional between two given lines. 
 
 Let the given lines be AB = 32, and BC = 18. 
 
 Make AC = 50, the sum of the 
 given lines ; and with the radius 
 Ao = 25, and o as a centre, describe 
 the semicircle ADC. From the point 
 B, erect the perpendicular BD, and 
 it will measure 24, the mean propor- 
 tional sought. 
 
 Note. — A mean proportional between any two numbers may also 
 be found, by multiplying them together, and extracting the square 
 root of their product. 
 
 PROBLEM XXIX. 
 
 To raise a perpendicidar from any point D, in a given 
 line AB, hy a scale of equal parts. 
 
 Make T)m = 3 ; and from the 
 points D and m, with the distances 
 4 and 5, describe arcs intersecting 
 each other in n. From D, through 
 the point 7i, draw the line DC, and it 
 will be the perpendicular required. 
 
 Note. — This Problem may be performed by any other numbers 
 in the same proportion ; but 3, 4, and 5 are the least whole numbers 
 that will form a right-angled triangle. 
 
 PROBLEM XXX. 
 
 Given the span or chord line, arid height or versed sine of 
 the arch of a bridge or cellar, to find the radius of the 
 circle that ivill strike the arch. 
 
 Draw the unlimited line CE ; and take the versed sine 
 
18 GEO:\IETrJCAL PART I. 
 
 from a scale of equal parts, and set it from C to D. 
 
 Tlirougli the point D cbaw the line 
 
 AB perpendicularly to CE, and make 
 
 AD and BD each equal to half the 
 
 given chord ; also draw the chords 
 
 AC, BC. Bisect either of these 
 
 chords perpendicularly, with a line 
 
 meeting CE in o, which will be the 
 
 centre of the circle ; hence the arch 
 
 ACB may be described. Or, divide 
 
 the square of half the chord by the versed sine ; to the 
 
 quotient add the versed sine, and the sum will be the 
 
 diameter of the circle. 
 
 Note. — This Problem is extremely useful to Joiners in striking 
 circular arcs, forming centres for bridges, cellars, Sec. ; and also to 
 Masons and Bricklayers, in describing circular pediments and other 
 ornamental arches. 
 
 GEOMETRICAL THEOREMS, 
 
 THE DEMONSTRATIONS OF WHICH MAT BE SEEN IN THE 
 ELEMENTS OF EUCLID, SBIPSON, AND EMERSON. 
 
 THEOREM I. 
 
 When two straight lines intersect each other, the opposite 
 angles are equal. 
 
 If two straight lines AB, CD, cut (\ 
 each other in the point E, the angle \^ 
 
 AEC will be equal to the angle DEB, "^"^ ^\~ j 
 
 and CEB to AED. {Euclid 1. 15. \ 
 
 Simp. I. 3. Efn. I. 2.) ^d 
 
 THEOREM II. 
 
 The greatest side of every triangle is opposite to tlie 
 greatest angle. (Iliic. I. 18. Si?np. I. 13. Em. II. 4.) 
 
PART I. 
 
 THEOREMS. 19 
 
 theore:m III. 
 
 "Wlien a line intersects two parallel lines, it makes the 
 alternate angles equal to each other. 
 
 Let the right line EF fall upon the parallel right lines 
 AB, CD ; the alternate angles AGH, ^. 
 GHD are equal to each other ; and \ 
 
 the exterior angle EGB is equal to \ 
 
 the interior and opposite, upon the ^ \ 
 
 same side GHD ; and the two interior . \ 
 
 angles BGH, GHD, upon the same ^ ''\ "" 
 
 side, are together equal to two right ^-^ 
 
 angles. {Euc. I. 29. Swijj. I. 7. ^" 
 
 Em. I. 4.) 
 
 THEOKEM TV. 
 
 Wlien one side of a triangle is produced, the outward 
 angle is equal to both the inward opposite angles taken 
 together. 
 
 Let ABC be a triangle, and let one of its sides BC 
 be produced to D ; the exterior angle 
 ACD is equal to the two interior and 
 opposite angles CAB, ABC ; also 
 the three interior angles of every 
 triangle are together equal to two 
 ri2;ht angles. {Eiic. 1. 32. Simp. 
 V9 & 10. E7n. IL 1 & 2.) 
 
 THEOREM V. 
 
 Parallelograms standing on the same base, and bstween 
 the same parallels, are equal to each other. 
 
 Let the parallelograms ABCD, ^ _. 
 
 DBCE be upon the same base BC, 
 and between the same parallels AE, 
 BC ; the parallelogram ABCD is 
 equal to the parallelogram DBCE. 
 {Euc.LBo. Si?}?p.Il.2. Em.111.6.) 
 
 D 
 
 
 /4 
 
 / 
 
 // 1/ 
 
 
20 
 
 GEOMETRICAL 
 
 PAP.T L 
 
 THEOREM YI. 
 
 Triangles standing on the same base, and between the 
 same parallels are equal to each other. 
 
 Let the triangles ABC, DEC be 
 upon the same base BC, and between 
 the same parallels AD, BC ; the triangle 
 ABC is equal to the triangle DBC. 
 {Euc. I. 37. Simp. II. 2. Em. II. 10.) 
 
 hypo- 
 other 
 
 THEOREM VII. 
 
 In any right-angled triangle, the square of the 
 thenuse is equal to the sum^ of the squares of the 
 two sides. 
 
 Let ABC be a right-angled triangle, 
 having the right angle BAC ; the 
 square of the side BC is equal to the 
 sum of the squares of the sides AB, 
 AC. {Euc. I. 47. Simp. II. 8. Em. 
 n. 21.) 
 
 jJoTE. Pythagoras, who was born about 2400 ycai's ago, dis- 
 covered this celebrated and useful Theorem ; in consequence of 
 which, it is said, he offered a hecatomb to the gods. 
 
 THEOREM Till. 
 
 An angle at the centre of a circle is double the angle at 
 the circumference, when both stand on the same arc. 
 
 Let ABC be a circle, and BDC an 
 angle at the centre, and BAC an angle 
 at the circumference, which have the 
 same arc BC for their base ; the angle 
 BDC is double of the angle BAC. 
 {Euc. III. 20. Simp. III. 10. Em. 
 IV. 12.) 
 
 THEOREM IX. 
 
 An angle in a semicircle is a right angle. 
 
PART I. 
 
 THEOREMS. 
 
 21 
 
 Let ABC be a semicircle ; then 
 the ansxle ABC in that semicircle is 
 a right-un^le. {Euc. III. 31. Simp. 
 III. 13. Em. VI. 14.) 
 
 THEOREM X. 
 
 If a line be di-awn in a triangle parallel to one 
 sides, it will cut the two other sides proportionally. 
 
 Let DE be drawn parallel to BC, 
 one of the sides of the triangle ABC ; 
 then BD is to DA, as CE to EA. 
 {Euc. VL 2. Simp. IV. 12. Em. 
 IL 12.) 
 
 ofit> 
 
 THEOREM XI. 
 
 In tlie preceding figure, DE being parallel to BC, the 
 triangles ABC, ADE are equi-angular or similar ; there- 
 fore, AB is to BC, as AD to DE; and AB is to AC, as 
 AD to AE. {Euc. VL 4. Simp. IV. 12. Em. II. 13.) 
 
 THEOREM XII. 
 
 In a right-angled triangle, a perpendicular from the 
 right angle is a mean proportional between the segments 
 of the hypothenuse ; and each of the sides, about the right 
 angle, is a mean proportional between the hypothenuse 
 and the adjacent segments. 
 
 Let ABC be a right-angled triangle, having the right 
 angle BAC ; and from the point A let AD be drawn 
 perpendicularly to the base BC ; the triangles ABD, 
 ADC are similar to the whole tri- ^ 
 
 angle ABC, and to each other. Also 
 the perpendiculai* AD is a mean pro- 
 portional between the segments of the 
 base ; and each of the sides is a mean 
 proportional between the base and its 
 ;jegment adjacent to that side ; therefore, BD is to DA, aa 
 
22 
 
 GEOIMETRICAL 
 
 PAKT 1. 
 
 DA to DC ; BC is to B A, as B A to BD ; and BC is to CA, 
 as CA to CD. {Euc. Yl.'8. Simp. IV. 19. ^/w, VI. 17.) 
 
 THEOREM Xin. 
 
 Similar triangles are to each other as the squares of 
 their like sides. 
 
 Let ABC, ADE, be similar triangles, having the angle 
 A common to both ; then the tri- 
 angle ABC is to the triangle ADE, 
 as the squai-e of BC to the square 
 ofDE. That is, similar triangles 
 are to one another in the duplicate 
 ratio of their homologous sides. {Euc. 
 VL19. Simp. IV. 24. Em. 11. 18.) 
 
 THEOEEM XIY. 
 
 In any triangle the double of the square of a line 
 drawn from the vertex to the middle of the base, together 
 with double the square of half the base, is equal to the 
 sum of the squares of the other sides. 
 
 In any triangle ABC, double the 
 square of a line CD, drawn from the 
 vertex to the middle of the base AB, 
 together with double the square of 
 half the base AD or BD, is equal to 
 the sum of the squares of the other 
 sides AC, BC. {Simp. H. 11. Em. 
 
 n. 28.) 
 
 THEOREM XV. 
 
 In any parallelogram ABCD, the 
 sum of the squares of the two dia- 
 gonals AC, BD, is equal to the sum 
 of the squares of all the four sides of 
 the parallelogram. {Simp. 11. 12. 
 Em. HI. 9.) 
 
 THEOREM XYI. 
 All similar figures are in proportion to each otlier as 
 the souares of their homologous sides. {Simp. IV. 26. 
 Em. III. 20.) 
 
PA^RT T. THEOREMS. 23 
 
 THEOREM XVn. 
 
 The circumferences of circles, and the arcs and chords 
 of similar segments, are in proportion to each other, as the 
 radii or diameters of the circles. (Em. lY. 8 & 9.) 
 
 THEOREM XVin. 
 
 Circles are to each other as the squares of their radii, 
 diameters, or circumferences. {^Em. TV. 35.) 
 
 THEOREM XIX. 
 
 Similar polygons described, in circles are to each other, 
 as the circles in which they are inscribed ; or as the 
 squares of the diameters of those circles. {Em. IV. 36.) 
 
 THEOREM XX. 
 
 All similar solids are to each other, as the cubes of their 
 like dimensions. {Em. VI. 24.) 
 
 AN EXPLANATION 
 
 PRINCIPAL ]!kIATHEMATICAL CHARACTERS. 
 
 The sign or character = (called equality) denotes that 
 the respective quantities, between Avhicli it is placed, are 
 equal; as 4 poles =■ 22 yards = 1 chain = 100 links. 
 
 The sign -l- (called plus, or more) signifies that the 
 numbers, between which it is placed, are to be added to- 
 gether ; as 9 -}- 6 (read 9 plus 6) = 15. Geometrical 
 lines are generally represented by capital letters ; then 
 AB -}- CD sisrnifies that the line AB is to be added to the 
 line CD 
 
24 TiIATHEMATICAL CHARACTERS. PAllT I. 
 
 The sign — (called minus, or less) denotes that the 
 quantity, which it precedes, is to be subtracted ; as 15 — 6 
 (read 15 iniwis 6) = 9. In geometrical lines also, AB- 
 CD signifies that the line CD is to be subtracted from the 
 line AB. 
 
 The sign x denotes that the numbers, between which 
 it is placed, are to be multiplied together; as 5 X 3 (read 
 5 multiplied by 3) = 15. 
 
 The sio-n -=- signifies division ; as 15 -r- 3 (read 15 di- 
 vided by 3) = 5. Numbers placed like a vulgar fraction, 
 also denote division ; the upper number being the di- 
 vidend, and the lower the divisor ; as ^f = 5. 
 
 The signs : : *. : (called proportionals) denote propor- 
 tionality Tas 2 : 5 : : 6 : 15, signifying that the number 2 
 bears the same proportion to 5, as 6 does to 15 : or, in 
 other words, as 2 is to 5, so is 6 to 15. 
 
 The sign | (called vinculum) is used to 
 
 connect several quantities together ;as9H-3|— 6|X 
 2 = 12"^^"! X 2 = 6 X 2 = 12. 
 
 The sign -, placed above a quantity, represents tlie 
 
 square of that quantity ; as 5 ^- 3 1 - =: 8- = 8 x 8 = 64. 
 
 The sign ■"', placed above a quantity, denotes the cube 
 
 of that quantity ; as 9T^ -8 1^ = 12-8P = 4-' = 4 
 X 4 X 4 = 64. 
 
 The sign l'' or ^/, placed before a quantity, denotes 
 
 the square root of that quantity ; as a/9 x 4 = ^736 
 = 6. 
 
 The sign ^, placed before a quantity, represents the 
 
 cube root of that quantity ; as ^Q x 4 x 3 1 — 8 = 
 
 ^y2T^r3]-8 = ,^72-8 = ^/64 = 4. 
 
 Z. Angle; as /. A, signifies the angle A; or the angle 
 ABC, B denotes the angular point. 
 
25 
 
 PAET II. 
 MENSUEATION OF SUPERFICIES. 
 
 The area of any plane figure is its superficial content, 
 or the measurement of its surface, without any regard to 
 thickness. 
 
 The dimensions of figures are taken in lineal measure. 
 Sometimes they are taken in inches and tenths ; some- 
 times in feet, inclies, and parts ; sometimes in feet, tenths, 
 and hundredths ; and sometimes in chains and links ; and 
 the area of any figure is estimated by the number of 
 square inches, square feet, square chains, &c., contained 
 in that figure. 
 
 Xote 1 A measuring tape, (usually called a box and tape,) divided into feet and 
 
 inclies on one side, and links on the other, is admirably calculated for taking di- 
 mensions. Tapes are of various lengths ; but those of four poles, or twenty-two 
 yards, are most useful. 
 
 2. In practice, the dimensions may generally be commodiously entered upon a 
 rough sketch of the figure. 
 
 3. The following tables of lineal and square measures ought to be well under- 
 stood by the learner, before he proceeds further. 
 
 A TABLE OF LINEAL JMEASURE. 
 
 Inches. 
 
 Link. 
 
 Foot. 
 
 
 Pole or 
 Perch. 
 
 
 
 7.92 = 
 
 1 
 
 
 12 
 
 1.5151 = 
 
 1 
 
 Yard. 
 
 
 36 
 
 4.5454 
 
 3 = 
 
 1 
 
 
 198 
 
 25 
 
 16.5 
 
 5.5 = 
 
 1 
 
 Chain. 
 
 Fur- 
 longs. 
 
 
 792 
 
 100 
 
 66 
 
 22 
 
 4 = 
 
 1 
 
 
 7920 
 
 1000 660 
 
 220 
 
 40 
 
 10 = 
 
 1 
 
 Mile. 
 
 63360 
 
 8000 5280 
 
 1760 
 
 320 
 
 80 
 
 8 = 
 
 1 
 
 Note. — Two yards make one fathom ; and seven yards one rood of fencing 
 Pitching. 
 
26 
 
 MENSUnATIOX 
 
 PAUT n. 
 
 A TABLE OF SQUARE MEASUKE. 
 
 Square 
 Inches. 
 
 Square 
 Link. 
 
 
 
 
 
 Square 
 Foot. 
 
 
 62.7264 = 
 
 1 
 
 Square 
 Yard. 
 
 
 
 144 
 
 2.29.56 = 
 
 1 
 
 
 
 
 Square 
 Perch. 
 
 
 1296 
 
 20.661! 
 
 9 = 
 
 1 
 
 
 
 Square 
 Chain. 
 
 
 39204 
 
 625j 272.25 
 
 30.25= 1 
 
 
 Square 
 Rood. 
 
 
 627264 
 
 lOOOOi 4356 
 
 484i 16 = 
 
 1 
 
 
 
 Square 
 Acre. 
 
 
 15681 GO 
 
 250001 10890 
 
 1210 40 
 
 2.5 = 
 
 1 
 
 Square 
 Mile. 
 
 627264U 
 
 10000o{ 43560J 4840J 160 
 
 10 
 
 4 = 
 
 1 
 
 4014489600 64000000'27878400l 309760o! 102400 
 
 6400 
 
 2560 
 
 640 = 
 
 1 
 
 Note. — Forty-uine square j'ards make one rood of digging. 
 
 PROBLEM L 
 
 To fiyid the area of a square. 
 
 RULE. 
 
 Multiply the length of one of its sides by itself, and 
 the product will be the area. 
 
 "Note 1 The side of a square may be found by extracting the square root of its 
 
 area. 
 
 2 • When the area of a figure is found by duodecimals, it is generally said to be in 
 feet, inches, parts, &c. ; but it is evident that it is in feet, twelfths, inches, &c., be- 
 cause 144 square inches, and not 12, make a square foot. In this work I have, 
 however, conformed to the general practice. 
 
 8. The learner should carefully examine, work over, and put down, all the 
 solutions given in this book, in order that he may the better comprehend the dif- 
 ferent rules. The definitions and rules sliould also be committed to memory. 
 
 4. The diagonal of a square is equal to the square root of twice the square of the 
 side. 
 
 exa:\iples. 
 
 1. "^Yhat is the area of the square 
 ABCD, the side of which is 8 feet 
 S inches ? 
 
PART II. OF SUPERFICIES. 27 
 
 Bjj Decimals. Jjy Duodecimals. 
 
 ^^e^t- Feet. Inches. 
 
 8-5 8 6 
 
 8.5 8 6 
 
 425 
 680 
 
 68 
 4 3 
 
 r2.25feet. 
 12 
 
 3.00 inches. • 
 
 72 3 Answer. 
 Answer, 12 feet Sin 
 
 2. Required tlie area of a square floor, whose side is 
 18 feet 9 inelies. 
 
 Here 18 feei 9 inches = 18.75 ; and 18.75 X 18.75 ■=^ 
 2>o\.6Q2o feet = 2)o\ feet, 6 ew. 9 parts, the area required. 
 
 3. What is the area of a square table, the side of which 
 measures 6 feet 8 inches ? Ans. 4:4 ft. 5 in. 4 parts. 
 
 4. The side of a square court-yard measures 85 feet 
 3 inches ; what will it cost paving at 2s. 9d. per yard ? 
 
 ^^5. ^111.05. 7 id. 
 
 5. The base of the largest Egyptian pyramid is a 
 square, whose side is 693 feet ; how many acres of ground 
 does it cover ? Ans. lla. Or. 4p. 
 
 6. What is the side of a square whose area is 132496 
 square feet? Ans. S64 feet. 
 
 7. Required the side of a square garden that cost 
 ^3. 185. l^d. trenching at l^d. per square yard. 
 
 Ans. 25 yards. 
 
 Note — The pyramids of Egypt are some of the most ancient structures in the 
 •R-orld, their antiquity extending beyond tiie records of all history, and their original 
 uses being entirely unknown. Herodotus, who wrote about 2200 years ago, speaks 
 with as much uncertainty concerning the time of their building as we do at present. 
 Some have conjectured that these majestic monuments were built by the children 
 of Israel during their bondage in Egjpt ; and designed as sepulchres for the 
 Egyptian kings. Should this supposition be correct, they must have stood about 
 3300 years. The perpendicular height of the largest of these stupendous structures 
 is about .500 feet ; but if measured obliquely, it is 700 feet from the base to the ter- 
 mination at the top. 
 
 PROBLEM II. 
 
 To find the area of a rectangle. 
 
 RULE. 
 
 Multiply the lengtli by the breadth, and the product 
 will be the area. 
 
 c 2 
 
28 MENSURATION TART U. 
 
 Note 1 If the length and breadth be in inches, their product must be divided 
 
 by 144, in order to obtain the area in square feet ; but if one dimension be in feet, 
 and the other in inches, the product divided by 12 will give the area in square feet. 
 The observation, however, is not applicable when the area is found duodecimally. 
 
 2. If the area of a rectangle be divided by one of its sides, the quotient will be 
 the other side ; but the area must first be reduced to the same denomination as the 
 given side. 
 
 3. Acres, roods, and perches may be reduced into square links, by'multiplying 
 the whole quantity, in perches, by C25, the number of square links in a square 
 perch. 
 
 4. In taking dimensions with a chain, it is usual to set down the number of links. 
 In this case, five figures must be cut off, as decimals, towards the right-hand of the 
 area; and those on the left will express the number of acres. The decimals must 
 then be reduced into roods and perches, by multiplying them successively by 4 
 and by 40, cutting off five figures, as before, in each product. 
 
 5. The diagonal of a rectangle is equal to the square root of the sum of the 
 squares of any two adjacent sides. 
 
 exa:mples. 
 
 1. Required the area of the rect- 
 angle ABCD, whose length AB is 
 12.75 chains or 1275 links, and 
 breadth BC, 6.75 chains or 675 links. 
 
 Links. 
 
 1275" 
 
 675 
 
 6375 
 8925 
 7650 
 
 8.60625 
 
 4 
 
 2.42500 
 40 
 
 17.00000 
 
 8.60625 area. 
 
 An. 
 
 A71S, Sa. 2r. I7p. 
 
 2. A door measures 6 feet 3 inches, by 3 feet 6 inches ; 
 what did it cost at Is. 8d. per square foot ? 
 
 Ans. £1. I6s. o^d. 
 
 3. The length of a mahogany tea-table is 45 inches, 
 and. its breadth 42 inches ; what is its area in square 
 feet? A71S. lS.l25feef. 
 
 4. How many square feet are contained in a rectan- 
 gular fir-deal, whose length is 18 feet, and breadth 11 
 inches r A?is. 16^ feet 
 
 5. Required the area of a granary floor, whose length 
 is 45 feet 6 inches, and breadth 20 feet 10 inches. 
 
 Aiis. 947 feet, 11 in. 
 
 6. How many yards of carpet, 3 quarters wide, will 
 cover a room that measures 28 feet 6 inches, by 18 feet 
 9 inches ? A7is. 79 7/ds. 6 in. 
 
PART II. OF SUPERFICIES. 29 
 
 7. The area of a rectangular field is I4a. 2r. lip.; 
 what is its length, its breadth being 925 links ? 
 
 Ans. 157o links. 
 
 8. A rectangular allotment upon a common, cost 
 .£78. Is. lO^d. digging and levelling, at £7. lOs. per 
 acre ; what will be the expense of fencing it half round 
 at OS. 6d. per rood, its length being 1225 links ? 
 
 Ans. a^l7. 185. 8d. 
 
 PROBLEM III. 
 To find the area of a rhombus or rhomhoides 
 
 RULE. 
 
 Multiply the length of the base by the perpendicular 
 breadth, and the product will be the area. 
 
 Note 1. — If the area of a rhombus or rhomboides be divided bj- one of its dimen- 
 sions, the quotient will be the other dimension. 
 
 2. The area of a parallelogram may be found by multiplying the product of any 
 two of its sides by the natural sine of their included angle. 
 
 EXAMPLES. 
 
 1. Required the area of the irre- 
 gular rhombus ABCD, whose base 
 AB is 12 feet 10 inches, and perpen- 
 dicular DE 9 feet 8 inches. 
 
 Here 12 ft, 10 in. = 154 inches, and 9 ft. 8 in. = 116 
 inches; then (154 X 116) -r- 144 = 17864 -f- 144 = 
 124.05555/ee^ = 124 /f. in. 8 pa.y the area required. 
 
 2. Wliat is the area of the regular 
 rhomboides ABCD, whose side AB 
 is 2185, and AD 1426 links? 
 
 B^ Note 2. we have 2185 x 1426 X .8660254 (the 
 natural sine of 60°, the angle BAD) = 2698370 sq. 
 links — 26a. Sr. 37p., the area required. 
 
 3. How many square yards of paving arc there in a 
 court-yard in the form of a rhombus, whose base mea 
 sures 265 feet 9 inches, and perpendicular, which rises 
 at the distance of 100 feet from the end of the base, 
 •246 feet 3 inches ? ' Ans. 7271.21527 i/ardL 
 
 4. Measuring along the base of a field in the form of 
 
 c 3 
 
 D 
 
 / 
 
 C 
 
 / 
 
 / 
 
'30 MENSURATION PABT n. 
 
 a rhomboides, I found the perpendicular to rise at 678, 
 and its length 1264 links ; the remainder of the base 
 measured 2435 links ; what is the area of the field ? 
 
 Ajis. 39a. Ir. 15|p. 
 
 5. A window contains 36 panes, each in the form of a 
 rhombus, whose base measures 11| inches, and perpen- 
 dicular 8^ inches; what will it cost glazing at 1^. life/, 
 per square foot ? Aiis. £2. 7s. Hid. 
 
 6. A grass-plot, in a gentleman's pleasure-ground, cost 
 ^3. 14^. Id. making, at 4d. per square yard ; what is the 
 length of the' base, the perpendicular being 40 feet, and 
 the figure a rhombus ? Ajis. 50 feet. 
 
 PROBLEM IV. 
 
 To find the area of a triangle^ when the base and 'perpen- 
 dicular are given. 
 
 KULE. 
 
 Multiply the base by the perpendicular, and half the 
 product will be the area. 
 
 Or, multiply the base by half the perpendicular; or 
 the perpendicular by half the base, and the product will 
 be the area. 
 
 Jv^o/e 1 If double the area of a triangle be divided by one of its dimensions, the 
 
 quotient will be the other dimension. 
 
 2. When the base, the perpendicular, and the place of the perpendicular upon the 
 base are given, the triangle may be truly constructed ; and its other two sides 
 measured by the scale used in the construction. 
 
 3. If the area of a triangle be divided by half tlie perpendicular height, the quotient 
 u'ill be the base ; and if the area be divided by half the base, the quotient will be 
 the perpendicular height. 
 
 4. In order to find the true place of the perpendicular, in taking the dimensions of 
 a triangle, or any other figure, it will be necessary to use an instrument called a 
 *' cross, or cross-staff','' which may be made in the following manner : Procure a 
 piece of board, either of sycamore, "box, or mahogany, out cf which form the quad- 
 rant of a circle, of about four inches radius. In it, with a fine saw, make two 
 grooves at right angles to each other, parallel to the radii, and about half an inch 
 deep. Through the centre of the quadrant make a circular hole, large enough to 
 receive a staff of about three quarters of an inch in diameter. The cross or quad- 
 rant must rest and turn upon a shoulder made at a convenient distance from the 
 top of the staff; and it may be prevented from coming off, by means of a screw 
 passing through a hollow cylinder of wood or brass, made to fit the staff imme- 
 diately above the cross. The staff' must be piked with iron at the bottom, in order 
 to enter the ground readily. It may be about eight feet in length ; and if it be 
 divided into feet and inches', or into links, it will be found very useful in taking di- 
 mensions. Crosses may be constructed in dift'erent ways ; but the cross described 
 above I find to be most convenient. 
 
 EXA]\IPLES. 
 
 C 
 
 1. Eequired the area of the tri- /N. 
 
 angle ABC, whose base AB measures, /\ \. 
 
 21.68 feet, the perpendicular DC / \ ^^ 
 
 9.46 feet, and AD 8.26 feet. ^ \ ^^ 
 
PART II. OF SUPERFICIES. 31 
 
 13008 
 
 8672 
 19512 
 
 2)205.0928 
 
 102.5464 Ansiuer. 
 
 2. ^Yliat is the area of a triangular field, the base of 
 which measures 3568 links, the perpendicular 1589 links, 
 and the distance between one end of the base and the 
 place of the perpendicular 1495 links ? 
 
 A71S. 28a. Ir. I5ip. 
 
 3. Required the area of the gable end of a house, the 
 base or distance between the eaves being 22 feet 5 inches, 
 and the perpendicular from the ridge to the middle of the 
 base 9 feet 4 inches. Ans. 104 ft. 7 in, 4 parts. 
 
 4. After measuring along the base of a triangle 895 
 links, I found the place of the perpendicular, and the per- 
 pendicular itself =: 994 links ; the whole base measured 
 1958 links ; what is the area of the triangle ? 
 
 Alls. 9a. 2r. 37p. 
 
 5. How many square yards of slating are there in the 
 hipped roof of a square building ; the base or length of 
 the eaves from hip to hip being 23 feet 9 inches, and the 
 distance between the middle of the base and the vertex of 
 the roof 9 feet 6 inches ? Ans. 50 i/ds. 1 ft. 3 in. 
 
 6. The area of a triangle is 6 acres 2 roods and 8 
 perches, and its perpendicular measures 826 links ; what 
 will be the expense of making a ditch, the whole length 
 of its base, at 2^. 6d. per rood ? Ans. £6. 4^. 7^d. 
 
 PROBLEM V. 
 
 To find the area of a triangle^ the three sides only of lohich 
 
 are giveyi. 
 
 RULE. 
 
 From half the sum of the three sides subtract each side 
 severally ; multiply the half sum and the three remainders 
 continually together, and the square root of the last pro- 
 duct will be the area of the triangle. 
 c 4 
 
32 MENSURATION PART U. 
 
 2sole 1 If a triangle be accurately laid down, from a pretty large scale of equal 
 
 parts, the perpendicular maybe measured, and the area found by the last Problem. 
 
 2. If the rectangle of any two sides of a triangle be multiplied by the natural sine 
 of their included angle, the product will be double the area of the triangle ; con- 
 sequently, if double the area of a triangle be divided by the rectangle of any two of 
 its sides, the quotient will be the natural sine of their mcluded angle. 
 
 3. If the area of a triangle be divided by half the sum of the sides, the quotient 
 will be the radius of the iascribed circle. 
 
 4. Many calculations in Arithmetic, Mensuration, &c. may be greatly facilitated 
 by the assistance of logarithms ; and it is matter of surprise that their use is not 
 more generally taught in public schools. 
 
 The method of working bv logarithms is extremely simple, and may be applied 
 to almost every kind of calculation ; yet the advantages of this great discovery have 
 hitherto been chicHy enjoved by those only who have leisure and inclination to cul- 
 tivate the higher branches of the Mathematics. 
 
 The following are the most useful Rules on this subject ; they, however, apply 
 only to numbers greater than unity; for if the numbers be less than unify, the 
 indices of their logarithms will be negative, which requires a little more ingenuity 
 in the management. 
 
 Multiplication bij Logarithms. 
 RULE. 
 Add the logarithms of all the factors together, and their sum will be the logaiith-.n 
 of the product. 
 
 Division by Logarithms. 
 RULE. 
 From the logarithm of the dividend subtract the logarithm of the divisor ; an^J 
 the remainder will be the logarithm of the quotient. 
 
 Rule of Three by Logarithms. 
 
 RULE. 
 
 Add the logarithms of the second and third terms together, and from their sum 
 
 subtract the logarithm of the lirst term ; and the remainder will be the logarithm 
 
 of the fourth term. 
 
 Involution, or Raising of Potccrs. 
 RULE. 
 Multiply the logarithm of the given number by the index of the proposed power ; 
 and the product will be the logarithm of the power sought. 
 
 Evolution, or Extraction of Roots. 
 RULE. 
 
 Divide the logarithm of the given number, by the index of the proposed root ; 
 and the quotient will be the logarithm of the root required. 
 
 Further directions on this head appear to be unnecessary, as it is presumed that 
 few Teachers are without a treatise on logarithms ; it may, however, be observed, 
 that Dr. Button's Mathematical Tables are considered the most useful. 
 
 EXAMPLES. 
 
 c 
 
 1. What is the area of the triangle 
 ABC, the side AB measuring 2.5, 
 AC 20, andBC 15 chains? 
 
 Here (25 + 20 + 15) — 2 = 60 -- 2 = 30 = half the 
 sum of the sides; then 30 - 25 = 5, the first remainder j 
 30 — 20 = 10, the second remainder; and 30 —.15 = 15, 
 
PART II. 
 
 OF SUPERFICIES. 33 
 
 the third remainder; ivhence a/ (30 X 5 X 10 X lo) = 
 ^/ 22500 = 150 square chains = 16 acres^ the area re- 
 quired. 
 
 By Logarithms. 
 
 The lo2f. of . . . 
 
 30 . . . 
 
 . . = 1.4771213 
 
 5 . . 
 
 . . = 0.6989700 
 
 10 . . 
 
 . . = 1.0000000 
 
 15 . . . 
 
 . . = 1.1760913 
 
 Divide by the index of tlie root .... 2 )4.3521 826 
 The quotient is the log. of 150, the area . 2.1760913 
 
 2. Eequired the area of a grass-plot in the form of an 
 equilateral triani^le, whose side is 36 feet. 
 
 Ans. 561.18446 /ec/. 
 
 3. Wliat is the area of the gable end of a house form- 
 ing an isosceles triangle, whose base is 25, and each of its 
 other equal sides 20 feet ? Ans, 195.15618 /ee^. 
 
 4. What is the area of a triangular field, whose three 
 sides measure 2564, 2345, and 2139 links? 
 
 Ans. 23«. 2r. 0\'p. 
 
 5. The three sides of a triangular fish-pond measure 
 293, 239, and 185 yards ; what did the ground which 
 it occupies cost, at ^''185 per acre ? 
 
 Ans. ^^843. 75. U, 
 
 PROBLEM VI. 
 
 Any two sides of a right-angled triangle being given, to 
 find the third side. 
 
 RULE. 
 
 1. lyjien the two legs are given, to find the hypothenuse. 
 Add the square of one of the legs to the square of 
 the other, and the square root of the sum will be the 
 hypothenuse. 
 
 2. WJien the hypothenuse and one of the legs are given, to 
 find the other leg. 
 From the square of the hypothenuse subtract the 
 square of the given leg, and the square root of the re- 
 mainder will be the required leg. {See Theorem 7.) 
 
 Kote When the area of a right-angled triangle and the hypothenuse are given, 
 
 ti:e legs may he found bv the following General Rule : To the square of the hy- 
 riothenuse add four times" the area of the triangle, and the square root of this num- 
 
 ■ c 5 
 
34 MENSURATION PART II. 
 
 ber will be the sum of the legs. From the square of the hypothenuse take four 
 times the area of the triangle, and the square root of the remainder will be the dif- 
 ference ot the legs. Add half the difference of the legs to half their sum, and you 
 will obtain the greater leg ; but if half the difference of the legs be taken from half 
 their sum, the remainder will be the less leg. 
 
 EXAJMPLES. 
 
 1 . In the right-angled triangle ABC, 
 are given the base AB = 36, and the 
 perpendicular BC = 27, to find the 
 hypothenuse AC. 
 
 Here 362 _^ 27- = 1296 + 729 = 2025 ; and v'2025 
 = 45, the hyjjothenuse AC. 
 
 2. If the hypothenuse AC be 60, and the perpendicular 
 BC 36 ; what is th'e base AB ? 
 
 Here 60^ - 36- = 3600 - 1296 = 2304 ; and ^2304 
 = 48, the base AB. 
 
 3. Required the length of a scaling ladder to reach the 
 top of a wall whose height is 33 feet ; the breadth of the 
 moat before it being 44 feet. Ans. 56 feet. 
 
 4. What must be the length of a shore, which, strutting 
 10 feet 9 inches from the upright of a building, will sup- 
 port a jamb 18 feet 6 inches from the ground? 
 
 Ans. 21ft. 4 in. 9 parts. 
 
 5. The distance from the ridge to the eaves of a build- 
 ing is 15 feet, and the perpendicular height of the gable 
 end 9 feet ; what is the breadth of the building ? 
 
 A?is. 24 feet. 
 
 6. A ladder 46 feet in length, being pldced in a street, 
 reached a window 26 feet from the ground, on one side ; 
 and by turning it over, without removing the bottom, it 
 reached another window 35 feet high, on the other side ; 
 what is the breadth of the street ? Ans. 67.79695 feet. 
 
 7. A castle wall there was, whose height was found 
 To be 100 feet from th' top to th' ground ; 
 Against the wall a ladder stood upright, 
 
 Of the same length the castle was in height : 
 A waggish youngster did the ladder slide 
 (The bottom of it) 10 feet from the side : 
 Now I would know how far the top did fall, 
 By pulling out the ladder from the wall ? 
 
 Ans. 6 inches, nearly 
 
PART II. 
 
 OF SrrERFICIES. 
 
 35 
 
 PROBLEM VII. 
 To find the area of a trapezium. 
 
 RULE. 
 
 Multiply the sum of the two perpendiculars by the 
 diagonal upon which they fall, from the opposite angles, 
 and half the product will be the area. 
 
 Or, divide the trapezium into two triangles, in the 
 most convenient manner ; and the sum of their areas 
 found by Problem IV., or Problem V., will be the area 
 required. 
 
 Note. — If the trapezium can be inscribed in a circle, that is, if the sum of any two 
 of its opposite angles be equal to 180 degrees, its area may be found as follows : — 
 From half the sum of the four sides subtract each side severally ; then multiply the 
 four remainders continually together, and the square root of the last product will 
 be the area. 
 
 EXAMPLES. 
 
 1. It is required to lay down the 
 trapezium ABCD, and find its area ; 
 AE being 10.26 feet, AF 25.34 feet, 
 AC 40.18 feet, DF 14.32 feet, and 
 BE 12.86 feet. 
 
 Here (14.32 + 12.86) x 
 1092.0924 ; and 1092.0924 ~ 
 required. 
 
 40.18 = 27.18 X 40.18 = 
 ■ 2 =. 546.0462 feet, the area 
 
 The log. of 
 The log. of 
 
 Tlieir sum 
 The lo^. of 
 
 By Logarithms. 
 
 .... 27.18 is 
 .... 40.18 is 
 
 1.4342495 
 1.6040099 
 
 3.0382594 
 0.3010300 
 
 2.7372294 
 
 The diff. is the log. of 546.0462 the area 
 
 2. Lay down the following tra- ^ 
 
 pezium, and find its area ; AE mea- /'C^- ^\ 
 
 suring 1125, AB 3243, DE 1168, / | '^''-v:' \ 
 DF i216, DE 2418, and CF 610 /__J[__^_^^ 
 links. - . ^^ ^ 
 
 JVo^<?.— This figure is divided into two triangles, because two perpendiculars can- 
 not be taken upon either of the diagonals. 
 
 Here (3243 x 1168)-:- 2 =3787824 
 the area of the triangle ABD. 
 C 6 
 
 183912, 
 
36 siensuhation " part il 
 
 Also, (2418 X 610) -f- 2 = 1474980 ~2 - 737490, 
 the area of the triangle BCD. 
 
 Then 1893912 4- 737490 = 2631402 square links = 
 26a. Ir. lOp., the area, of the trapezium ABCD. 
 
 3. How many square yards of paving are there in a 
 trapezium whose diagonal is found to measure 126 feet 
 3 inches, and perpendiculars o8 feet 6 inches, and Qi^ feet 
 9 inches ? Ans. 871.47569 yards. 
 
 4. In taking the dimensions of a trapezium, I found 
 the first perpendicular to rise at 568, and to measure 
 835 links ; the second at 1865, and to measure 915 links ; 
 the whole diagonal measured 2543 links ; what is the 
 area of the trapezium ? Ans. 22a. \r. Op. 
 
 5. Lay down a trapezium, and find its area from the 
 following dimensions ; namely, the side AB measures 
 345, BC 156, CD 323, DA 192, and the diagonal AC 
 438 feet. Ans. 52330.33406 /ee^. 
 
 6. The sides of a trapezium, two of whose opposite 
 angles are together equal to 180 degrees, measure 30, 
 32-i-, 35, and 37^ feet ; what is its area ? 
 
 Ans. USlft. 2 in. 9 pa. 
 
 PROBLEM yni. 
 
 To find the area of a trapezoid. 
 
 RULE. 
 
 IMultiply the sum of the parallel sides by the perpen- 
 dicular distance between them, and half the product will 
 be the area. 
 
 Or, half the sum of the sides multiplied by their dis- 
 tance will give the area. 
 
 EXAMPLES. 
 
 1. What is the area of the trape- 
 zoid ABCD, the parallel sides AD 
 and BC of which are 25 and 18 ; 
 and AB, the perpendicular distance 
 between them, 38 feet ? 
 
 Here (25 + 18) X 38 = 43 x 38 = 1634 ; and 1634 
 ~ 2 = ^VJ feet, the area required. 
 
PART n. OP SUPERFICIES. 
 
 2. Kequlred the area of tlie trape- 
 zoid ABCD, whose parallel sides AB 
 and DC measure 46 feet 10 inches, 
 and 28 feet 4 inches ; DE, the per- , 
 pendicular distance between them, / 
 26 feet 9 inches ; and AE 12 feet L. 
 6 inches. 
 
 Here 46 feet 10 in. — 46.83333, and 28 feet 4 inches 
 = 28.33333 ; then (46.83333 4- 28.33333) x 26.75 = 
 ^ioA^h^m X 26.75 = 2010.708155, half of luhich is = 
 1005.o54:077 5 feet, the area required. 
 
 3. The parallel sides of a piece of ground measure 856 
 and 684 links, and their perpendicular distance 985 links ; 
 what is its area? Ans. la. 2r. 131p. 
 
 4. If the parallel sides of a garden be 66 feet 6 inches, 
 and 49 feet 3 inches, and their perpendicular distance 
 o6 feet 9 inches ; what did it cost, at ^325. 10,s. per 
 acre ? Ans. ^24. Qs. I^d. 
 
 5. The hipped roof of a square building is flat at the 
 top ; the length of the eaves, from hip to hip, is 54 feet 
 6 inches ; the side of the square at the top, is 30 feet 
 9 inches ; and the nearest distance from the top to the 
 eaves, is 18 feet 3 inches : how many square yards of 
 slating are contained in the four sides of the roof ? 
 
 Ans. 345.73611 i/ards. 
 
 PROBLEM IX. 
 
 To find the area of an irregidar polygon of any number 
 of 
 
 RULE. 
 
 Divide the figure into triangles and trapeziums in the 
 most convenient manner ; and find the area of each 
 separately; then the sum of these areas will be the area 
 of the polygon. 
 
 "iJote — In calculating the content of an irregular polygon, it is sometimes more 
 eligible to find the double area of each figure into which it is divided ; and half the 
 sum of these double areas will be the area of the whole polygon. 
 
 exa:\iples. 
 1. It is required to lay down the irregular figure 
 ABCDEFGA, and to find its area from the following 
 dimensions. 
 
38 
 
 MENSURATION 
 
 PART n. 
 
 Diagonals. 
 Feet. 
 Km = 25 1 
 AB =35.0 J 
 Ao = 12 1 
 Kn = 24.8 [ 
 AD = 36 J 
 Dp =16 1 
 Dr = 23.5 y 
 DF = 34 J 
 
 Perpe?idiculars. 
 
 Feet. 
 Cm= 8 
 
 Go 
 Cn 
 Ejo 
 Gr 
 
 = 7.5 
 = 8.6 
 = 10, 
 = 9, 
 
 i] 
 
 AB, which make = 
 at which -point erect 
 
 Construction. — Draw the line 
 35.5 ; and lay off 25 from A to m, 
 the perpendicular Cm ~ 8 ; 
 join AC and BC, and you will 
 have the triangle ABC. 
 
 With C as a centre, and the 
 radius Cji, describe an arc ; and 
 with A as a centre, and the 
 radius An, describe another 
 arc cutting the former in n. 
 Through n di'aw the diagonal 
 AD = 36, upon which lay off 
 Ao =12. At o erect the per- 
 pendicular Go = 7.5 ; join CD, DG, and GA, and the 
 trapezium ACDG will be completed. 
 
 The trapezium DEFG may be constructed in a similar 
 manner. 
 
 Calculation. — Rere 35.5 x 8 = 284, double the area 
 of the triangle ABC. 
 
 Again, (7.5 + 8.6) X 36 = 16.1 x 36 = 579.6, double 
 the area of the trapezium ACDG. 
 
 Also, (10.4 + 9.2) X 34 = 19.6 x 34 
 the area of the trapezium DEFG. 
 
 Then, (284 + 579.6 + 666.4) -f- 2 == 1530 -4- 2 = 765 
 feet, the area of the irregular polygon required. 
 
 2. It is required to lay down a pentangular field, and 
 iind its annual value at £2. 5s. per acre ; the first side 
 measuring 926, the second 536, the third 835, the fourth 
 628, and the fifth 587 links ; and the diagonal from the 
 first angle to the third 1194, and that from the third to 
 
 666.4, double 
 
 the fifth 1223 linkst 
 
 Ans. ^18. 105. I^d. 
 
 Note.— Th\s field is divided into three triangles, the areas of which may be found 
 by Problem V. 
 
PART n. 
 
 OF SUPERFICIES. 
 
 39 
 
 PROBLEM X. 
 
 Given the side of a regular jioli/gon, to find the radius of 
 its inscribed or circumscribing circle. 
 
 RULE. 
 
 l\Iultiply the given side by the number standing oppo- 
 site to the name of the polygon, in the third or fourth 
 cokunn of the following table, as the case requires ; and 
 the product will be the radius of the inscribed or circum- 
 scribing circle. 
 
 I. TABLE OF POLYGONS, &c. 
 
 No. of 
 Sides. 
 
 
 Rad. of the 
 
 Rad. of the 
 
 Names. 
 
 Inscribed 
 Circle. 
 
 Circums. 
 Circle. 
 
 3 
 
 Trigon, or equil. triangle . 
 
 0.28867510.5773503 
 
 4 
 
 Tetragon, or square . 
 
 0.5000000:0.7071068 
 
 5 
 
 Pentagon .... 
 
 0.68819100.8506508 
 
 6 
 
 Hexagon .... 
 
 0-8660254 l.OOdOOOO 
 
 7 
 
 Heptagon .... 
 
 1.0382617|1. 1523825 
 
 8 
 
 Octagon .... 
 
 1.2071068 
 
 1.3065630 
 
 9 
 
 Nonagon .... 
 
 1.3737387 
 
 1.4619022 
 
 10 
 
 Decagon .... 
 
 1.5388418 
 
 1.6186340 
 
 11 
 
 Undecagon 
 
 1.7028437 
 
 1.7747329 
 
 12 
 
 Duodecagon 
 
 1.8660254 
 
 1.9318516 
 
 l<[ote 1 If the radius of a circle be given, the side of any inscribed polygon may 
 
 be found by dividing the given radius by the number standing opposite to the name 
 of the polygon, in the fourth column of the preceding table. 
 
 2. By the assistance of this Problem, any regular polygon, whose side is given, 
 may be" easily constructed in the following manner : Find the radius of the circum- 
 scribing circle ; and to the circumferenceof this circle apply the given side the pro- 
 posed number of times, and you will have the polygon required. Or if the radius 
 of a circle be given, find the side of the inscribed polygon, with which proceed as 
 before. 
 
 3. The perpendicular let fall from the centre of a regular polygon upon one of 
 its sides is equal to the radius of the inscribed circle ; and the sum of the sides is 
 equal to the product of one side multiplied by the number of sides. 
 
 EXAIVIPLES. 
 
 1. The side of a regular pentagon 
 is 21.75 ; what are the radii of the 
 inscribed and circumscribing circles ? 
 Here .688191 X 21.75 = 
 14.96815425 = DC, the radius of the 
 inscribed circle ; and .8506508 x 
 21.75 = 18.5016549 = AC, the radi?(s 
 of the circumscribing circle. 
 
40 MENSURATION PART H. 
 
 2. If the side of a regular lieptagon be 25.25 ; what 
 is the radius of the circumscribing circle ? 
 
 Alls. 29.097658. 
 
 3. If the side of an octagonal grfiss-plot, in a gentle- 
 man's pleasure ground, measure 86 feet 10 inches ; what 
 will be the expense of making a gravel-walk from the 
 middle of one of its sides to the middle of the opposite 
 side, at 2^d. per yard, lineal measure ? Ans. 14^. 6^d. 
 
 4. If the radius of a circle be 65 feet ; what is the 
 sum of the sides of its inscribed nonagon ? 
 
 Ans. 400.1 6356 /ce^. 
 
 PROBLEM XI. 
 
 To find the area of a regular 'polygon. 
 
 RULE. 
 
 Multiply the sum of the sides, or perimeter of the 
 polygon, by the perpendicular demitted from its centre to 
 one of the sides, and half the product will be the area. 
 
 "Sote 1. — If double the area of a regular polj'gon be divided by the perper-dicular, 
 the quotient will be the sum of the sides. 
 
 2. In any figure whatever, the sum of all the inward angles is equal to twice as 
 many right angles, wanting four, as the figure has sides. 
 
 3. 'When every side of any figure is produced out, the sum of all the outward 
 angles is equal to four right angles. 
 
 EXAMPLES. 
 
 1. Required the area of the re- 
 gular hexagon ABCDEF, whose 
 side AB is 20 feet 6 inches, and 
 perpendicular Po is 17 feet 9 inches. 
 
 //ere 20.5 x 6 X 17.75 = 2183.25; 
 and 2183.25 — 2 = 1091.625 feet 
 = 1091 yif. 7 in. 6 pa., the area re- 
 quired. '^ P ^ 
 
 2. Wliat is the area of a court -yard in the form of a 
 regular pentagon, whose side measures 92 feet 6 inches, 
 and perpendicular 63 feet 8 inches ? 
 
 Ans. 14722.91666 /ee^. 
 
 3. Required the area of a heptagonal stone, whose side 
 measures 8 feet 9 inches, and perpendicular 9 feet. 
 
 Ans. 215 ft. 7 in. 6 pa. 
 
 4. 'What will the floor of an octagonal summer-house 
 cost paving with black and Avhite marble, at 45. Qd. per 
 
PART n. 
 
 OF SUPERFICIES. 
 
 41 
 
 square 'foot, the side of which measures 9 feet 6 inches, 
 and the nearest distance from one of its sides, to the 
 opposite side 22 feet 11 inches ? A?is. £91. \9s, 4-^6?. 
 
 0. A hexagonal piece of ground, in a gentleman's park, 
 cost ^29. 105. 5'^d. planting with trees, at £5. lOs. per 
 acre ; and a gravel-wallv leading from the middle of one 
 of its sides to the middle of the opposite side, cost 
 e£2. Zs. Z^d. making, at Zd. per yard, lineal measure ; 
 what was the expense of fencing the perimeter of the 
 polygon, at 6a\ 6d. per rood ? Ans. ^27. 17*. 
 
 PROBLEM XII. 
 
 To find the area of a regidar polygon, when the side only 
 is given. 
 
 RULE. 
 
 Multiply the square of the given side by the number 
 or area standing opposite to the name of the polygon, in 
 the following table, and the product will be the area. 
 (This Rule is founded on Theorem XIII.) 
 
 n. TABLE OF POLYGONS, &c. 
 
 No.of 
 
 
 Multipliers, or 
 
 Anjrle 
 
 Ancjle 
 
 Sides. 
 
 IS a. DIGS. 
 
 Areas. 
 
 ACB. 
 
 DAC. 
 
 3 
 
 Trigon . . 
 
 0.4330127 
 
 120° 
 
 30° 
 
 4 
 
 Tetragon . . 
 
 1.0000000 
 
 90° 
 
 45° 
 
 5 
 
 Pentagon . . 
 
 1.7204774 
 
 72° 
 
 54° 
 
 6 
 
 Hexagon . . 
 
 2.5980762 
 
 60° 
 
 60° 
 
 7 
 
 Heptagon . . 
 
 3.6339124 
 
 5n-° 
 
 64f° 
 
 8 
 
 Octagon . 
 
 4.8284271 
 
 45° 
 
 67i° 
 
 9 
 
 Nonagon . . 
 
 6.1818242 
 
 40° 
 
 70° 
 
 10 
 
 Decagon . . 
 
 7.6942088 
 
 36° 
 
 72° 
 
 11 
 
 Undecagon . 
 
 9.3656399 
 
 32^° 
 
 73t^° 
 
 12 
 
 Duodecagon . 
 
 11.1961524 
 
 30° 
 
 75° 
 
 Note I. — If the area of a pol.vgon be divided by the number standing opposite 
 to its name, in the foregoing table, the quotient will be the square of the polygon's 
 side. 
 
 2. The multipliers in the table of polygons, Problem X., are the radii of the 
 inscribed and circumscribing circles, when the side of the polygon is unity, or 1 ; 
 and may be found by trigonometry, in the following manner :"Divide 360 degrees 
 by the number of sides, and the quotient will be the angle ACB at the centre of 
 the polygon, half of which will be the angle A CD ; then say, as the nat. sine of 
 the angje ACD is to AD, so is the nat. co-sine of the angle ACD to CD, the radius 
 of the inscribed circle, or the perpendicular of the polygon ; and, as the nat. sine 
 of ACD is to AD, so is the radius (I) to AC, the radius of the circumscribing circle. 
 
42 menstjtjltion part n. 
 
 The multipliers in the last table are tb.e areas of their rcspectiva polygons, 
 when the side is 1, and may be found thus : Multiply the perpendicular or number 
 in column the third, Table 1, by .5 (half of AB), and the product will be the areaof 
 the triangle ACB, which being multiplied by the number of sides, we obtain the 
 area or multiplier in Table 2. 
 
 Or if the nat. tangent of the angle DAC be multiplied by the number of sides, 
 one-fourth of the product will be the multiplier. (See the figure, Problem X.) 
 
 EXAMPLES. 
 
 1. If the side of a pentagon be 8 feet 4 inches; what 
 is its area ? 
 
 Here 8 feet 4 inches = 84- feet = 25-^3; and (25 ~ 
 3)- = 625 H- 9, the square of the side ; then 1.7204774 
 X 625 -H 9 = 1075.298375 -- 9 = 119.4775972 feet, the 
 area required. 
 
 2. ^Vliat is the area of the base of a hexagonal stone 
 pillar, whose side measui-es 1 foot 6 inches ? 
 
 Ans. 5.84:567 feet. 
 
 3. If the side of an octagonal brick pillar measure 
 1 foot 5 inches ; what is the area of its base ? 
 
 Ans. 9 ft. 8 i?i. 3 pa- 
 
 4. Required the area of a decagon whose side measures 
 25 feet 9 inches. Ans. 5101 ft. Sin. 10 pa. 
 
 5. A gardener wishes to make a hexagonal grass-plot 
 that shall contain 260 square yards ; what must be the 
 length of its side ? Ans. 10 yards. 
 
 PROBLEM Xin. 
 
 Tlie diameter of a circle being given, to find the circum- 
 ference ; or, the circumference being given, to fiend the 
 diameter. 
 
 RULE I. 
 
 As 7 is to 22, so is the diameter to the circumference ; 
 or, as 22 is to 7, so is the circumference to the diameter. 
 
 RULE n. 
 
 As 113 is to 355, so is the diameter to the circum- 
 ference; or, as 355 is to 113, so is the cii'cumference to 
 the diameter. 
 
 RULE III. 
 
 Multiply the diameter by 3.1416, and the product will 
 be the circumference ; or, divide the circumference by 
 3.1416, and the quotient will be the diameter. 
 
PART IT. 
 
 OF SUPERFICIES. 
 
 43 
 
 ]^ote I There :s no figure that affords a greater variety of useful properties 
 
 than the circle ; nor is there any that contains so larcte an area within tlie same 
 perimeter. 
 
 Tlie ratio of the diameter of a circle to its circumference has never yet been 
 exactly determined ; although this celebrated Problem, called the squaring of the 
 circle, has engaged the attention and exercised the abilities of the al)lest mathe- 
 maticians, both ancient and modern. But though the relation between the diameter 
 and circumference cannot be exactly defined in known numbers ; yet approximating 
 ratios have been determined, suffic"iently correct for practical purposes. 
 
 Archimedes, a native of Syracuse, who flourished about 200 years before the 
 Christian jera, after attempting in vain to determine the true ratio of the diameter 
 to the circumference, found it to be nearly as 7 to 22. 
 
 The proportion given by Vieta, a Frenchman, and Metius, a Dutchman, about 
 the end of the IGth century, is as 113 to ."JS.d, which is rather more accurate than 
 the foi'mer ; and is a very commodious ratio, for being reduced into decimals, it 
 agrees with the truth to the sixth figure inclusively. 
 
 The first, however, who ascertained this ratio to any great degree of exactness, 
 ■was Ludolph Van Ceulen, a Dutchman. He found that if the diameter of a circle 
 be 1, the circumference will be 3.141592653589793238462643383279502884 nearly, 
 which is true to 36 places of decimals. This was thought so extraordinary a per- 
 formance, that the numbers were cut on his torub-stoiie, in St. Peter's church-yard, 
 at Ley den. 
 
 Since the invention of fluxions, by the illustrious Sir Isaac Newton, the squaring 
 of the circle has become more easy ; and the late ingenious Mr. Abraham Sharp, of 
 Little Horton, near Bradford, in Yorkshire, has not only confirmed Ceulen's ratio, 
 but extended it to 72 places of decimals. 
 
 Mr. John Machin, professor of astronomy in Grcsham College, London, has 
 also given us a quadrature of the circle, which is true to 100 places of figures ; and 
 even this has been extended, by the French mathematicians, to 128. 
 
 2. The first Rule is the proportion of Archimedes ; the second that of Vieta and 
 Metius ; and the third is an abridgment of Van Ceulen's ratio. This Rule is not 
 quite so accurate as the second ; but is most commonly used, as being most con- 
 venient, and, in most cases, correct enough for practice. 
 
 EXA3IFLES. 
 
 1. If the diameter AB of a circle 
 be 12 ; what is the circumference ? 
 
 By Rule I. 
 
 As 7 : 22 :: 12 : (264 -- 7) = 37.714285, the circum- 
 ference reqidred. 
 
 By Rule 11. 
 
 As 113 : 355 :: 12 : 37.699115, the circumference re- 
 
 By Rule III. 
 
 Here 3.1416 X 12 = 37.6992, the circumference re- 
 quired. 
 
 2. If the circumference of a circle be 45 ; what is the 
 diameter ? 
 
 By Rule I. 
 
 As 22 : 7 :: 45 : I4.3I8I8I, the diameter required. 
 
44 MENSURATION PART H. 
 
 By Rule II. 
 As 355 : 113 :: 45 : 14.323943, the diameter required. 
 
 By Ride III. 
 
 Hf^re 45 -^ 3.1416 = 14.323911, the diameter re- 
 quired. 
 
 3. If the diameter of a well be 3 feet 9 inches ; what 
 is its circumference. A7is. lift. 9 in. 4 pa. 
 
 4. The diameter of a circular plantation is 100 yards ; 
 what did it cost fencing round, at 6s. 9d. per rood ? 
 
 Ans. £16. 2s. ll^d. 
 
 5. "What is the diameter of a stone column whose cir- 
 cumference measures 9 feet 6 inches ? 
 
 Ans. Sff. iti. 3 pa. 
 
 6. The circumference of the earth is 25000 miles ; 
 what is its diameter, supposing it a perfect sphere ? 
 
 Ans. 7957.72854 miles. 
 
 7. The diameter of the sun is 883220 miles ; what is 
 his circumference ? Ans. 2774723.952 miles. 
 
 8. The circumference of the moon is 6850 miles ; what 
 is her diameter ? A7is. 2180.41762 77iiles. 
 
 9. The diameter of Venus is 7680 miles ; what is her 
 circumference ? Ans. 24127.488 77iiles. 
 
 KoW. —Those who are desirous of making themselves acquainted with the method 
 of finding the distances of the sun, moon, and planets from the earth, and also 
 their diameters, are referred to Martin's Trigonometry, vol. i. page 208. ; Fergu- 
 son's Astronomy, page 100. ; and Bonnycastle's Astronomy, page 277. 
 
 PROBLEM XIY. 
 To find the length of a7iy arc of a circle. 
 
 RULE I. 
 
 From 8 times the chord of half the arc subtract the 
 chord of the whole arc, and X of the remainder will be 
 the length of the arc, nearly. 
 
 Note 1. — Half the chord of the whole arc, the chord of half the arc, and the 
 versed sine are sides of a right-angle triangle ; any two of which being given, the 
 third mav be found by Problem VI. 
 
 2. The' difference of the diameters of any two circles, multiplied by 3.141G, will 
 give the difference of their circumference ; 'and vice versa. 
 
 3. If a line drawn through or from the centre of a circle, bisect a chord, it will be 
 perpendicular to it ; or, if it be perpendicular to the chord, it will bisect both the 
 chord and the arc of the chord. 
 
 4. If more than two equal lines can be drawn from any point within a circle to 
 the circumference, that point will be the ceTitre. 
 
 5. Any chords in a circle which are equally distant from the centre are equal to 
 each other. 
 
 6. A line perpendicular to the extremity of the radius, is a tangent to the circle. 
 
PART n. 
 
 OF SUrERFICIES. 4o 
 
 7. An angle formed by a tangent and chord is measured by half the arc of that 
 
 8. An angle at the circumference of a circle, is measured by half the arc that 
 subtends it. ,. ^, 
 
 9. All angles formed in the same segment of a circle, or standnig on the sama 
 arc, are equal. , , . ,. -t 
 
 10. An angle at the centre of a circle is double tlie angle at the circumference, if 
 botl) stand on the same base. 
 
 11. Anv angle in a semicircle is aright angle. , ,, 
 
 12. The angle formed by a tangent to a circle, and a chord drawn from the pomt 
 of contact, is equal to the angle in the alternate segment. _ 
 
 13. The sum of any two opposite angles of a quadrangle, inscribed ni a circle, 
 is equal to two right angles. , , ., . j 
 
 14. If any side of a quadrangle inscribed in a circle be produced out, the outward 
 angle will be equal to the inward opposite angle. 
 
 15. Any two parallel chords in a circle intercept equal arcs. 
 
 IG. An angle formed within a circle, by the intersection of two chords, is mea- 
 sured by half the sum of the two intercepted arcs. 
 
 17. An angle formed without a circle by two secants, is measured by half the dif- 
 ference of the intercepted arcs. 
 
 18. The angle formed by two tangents, is measured by half the diflerence ot the 
 two intercepted arcs. 
 
 EXAIklPLES. 
 
 r- 
 
 1. The chord AB of the whole 
 arc is 24, and the versed sine CD 9 ; 
 what is the length of the arc ACB ? 
 
 Here 12' + 9-' = 144 + 81 = 225 ; 
 and v/225 = 15 = AC, the chord 
 (f half the arc; then (15 X 8-24) \ j 
 
 -4- 3 = (120 - 24) -i- 3 = 96 -- 3 = ^--^^L^-^^ 
 
 32, the length of the arc required. ^ 
 
 2. The chord of the whole arc is 45, and the chord of 
 half the arc is 25.5 ; what is the length of the arc ? 
 
 Ans, 53. 
 
 3. The chord of half the arc is 21.25, and the versed 
 sine 10 ; what is the length of the arc ? Ans, 44.1666. 
 
 4. The chord of the whole arc is 30, and the versed 
 sine 8 ; what is the length of the arc ? Ans. 2>ol, 
 
 RULE II. 
 
 Divide the square of half the chord by the versed sine ; 
 to the quotient add the versed sine, and the sura will be 
 the diameter. 
 
 Subtract — of the versed sine from the diameter ; 
 50 
 
 divide - of the versed sine by the remainder ; and to the 
 
 3 '' 
 
 quotient last found, add 1 ; then this sum being multi- 
 plied by the chord of the whole arc, will give the length 
 of the arc, nearly. 
 
46 MENSURATION PAP.T H. 
 
 Xote — When great accuracy is required, the second Rule should be used, as the 
 first gives the length of the arc too little ; though near enough for most cases in 
 practice. 
 
 EXA3IPLES. 
 
 1. The cliord AB is 24, and the versed sine CD 9; 
 what is the length of the arc ACB ? 
 
 Here 12^ -f- 9 + 9 = 144 ^ 9 -|- 9 = 16 + 9 = 25 = 
 
 2 /^ 9 X 41\ 
 the diameter CE ; and 9 X -o" "^ ( 25 — — ^ — \ ^=z Q ^^ 
 
 25 - 7.38 = 6 -^ 17.62 = .34052 ; then 1 + .34052 x 
 24 = 1.34052 X 24 = 32.17248, the length of the arc 
 required. 
 
 2. The upper part or head of a window is the segment 
 of a circle, whose chord-line measures 6 feet 9 inches, 
 and versed sine 2 feet 6 inches ; what is the length of the 
 arch ? Ans. Sft. 11 in. 1 1 pa. 
 
 3. What is the length of the circular arch of a bridge, 
 the span of which is 15 feet 6 inches, and height above 
 the top of the piers 6 feet 9 inches ? 
 
 Ans. 22 ft. 4 m. 9 pa. 
 
 4. If the span of the circular roof of a cellar be 21 feet 
 9 inches ; what is the length of the arch, its height being 
 6 feet 6 inches ? Ans. 26 ft. 7 in. 5 pa. 
 
 5. If the span of a circular pediment be 18 feet 6 
 inches ; what is the length of the arch, its height above 
 the top of the entablature being 4 feet 6 inches ? 
 
 Ans. 21 ft. 3 ifi. 7 pa. 
 
 RULE m. 
 
 As 180 is to the number of degrees in the arc, so is 
 3.1416 times the radius to its length. Or, multiply to- 
 gether the number of degrees in the arc, the radius, and 
 the number .01745329; and the product will be the 
 length of the arc. 
 
 Xoie. — The length of the arc of a semicircle, a quadrant, &c. may also be found 
 by taking one-half, one-fourth, &c. of the uliole circumference. 
 
 exa:\iples. 
 
 1. Required the length of an arc of 46 degrees 35 
 minutes, the radius being 12 feet. 
 
 As 180° : 46° 35' :: 3.1416 x 12 : 9.75641, the length 
 of the arc reqidred. 
 
PART n. 
 
 OF SUPERFICIES. 47 
 
 2- VThat is the length of the semicircular arch of a 
 bridge, the span of which is 18 feet 6 inches ? 
 
 Ans. 29 Jl in. 8 pa. 
 3. The radius of a cart-wheel, from the centre of the 
 nave to the outside of the felloe, is 2 feet 3 inches ; what 
 is the length of one-sixth of the rim or circumference ? 
 
 Ans. 2 ft 4 in. 3 pa. 
 
 Noti' Those who wish to understand the i)rinciples upon which carriage wheelg 
 
 should be made, so that the carriage may be drawn by the least power, may consult 
 Marrat's Mechanics, article 788. 
 
 PEOBLEM XV. 
 
 To find the area of a circle. 
 
 RULES. 
 
 1. Multiply half the circumference by half the dia- 
 meter, and the product will be the area. Or, divide the 
 product of the whole circumference and diameter by 4, 
 and the quotient will be the area. 
 
 2. Multiply the square of the diameter by .7854, and 
 the product will be the area. 
 
 3. Multiply the square of the circumference by .07958, 
 and the product will be the area. 
 
 '^ote 1 If the area of a circle be divided by .7854, the quotient will be the square 
 
 of the diameter. 
 
 2. A circle may be considered as a regular polygon of an infinite number of sides, 
 the perimeter of which being equal to the circamference, and the perpendicular 
 equal to the radius: consequently by Problem XI. the area of a circle is equal to 
 half the circumference multiplied by half the diameter. Also, the area of a circle 
 -whose diameter is 1 is .78-54 nearly ; and by Theo. 18, circles are to each other as 
 the squares of their diameters : hence we derive the second rule. The area of a 
 circle whose circumference is 1, is 07958 ; and the areas of circles are to each other 
 as the squares of their circumferences : hence our third i-ule. 
 
 The following Rules will solve most of the useful Problems relating to the circle 
 and its equal or inscribed square, &c. 
 
 Rule 1.— The diameter of a circle multiplied by .8862269, will give the side of a 
 square equal in area. 
 
 2. The circumference of a circle multiplied by .2820948, will give the side of a 
 square equal in area, 
 
 3. The diameter of a circle multiplied by .7071068, will give the side of the 
 inscribed square. 
 
 4. The circumference of a circle multiplied by .2250791, will give the side of the 
 inscribed square. 
 
 5. The area of a circle multiplied by .6366197, and the square root of the product 
 extracted, will give the side of the ins'cribed square. 
 
 6. The side of a square multiplied by 1.414214, will give the diameter of its cir- 
 cumscribing circle. 
 
 7. The side of a square multiplied by 4.442S83, will give the circumference of its 
 circumscribing circle. 
 
 8. The side of a square multiplied by 1.128379, will give the diameter of a circle 
 equal in area. 
 
 9. The side of a square multiplied by 3.544908, will give the circumference of a 
 circle equal in area. 
 
 10. When the'diameter of a circle is 1, its area is found to be .78539816339744830- 
 961566084.581987.5721, which is true to 36 places of decimals ; but .7854 is sufficiently 
 correct for all practical purposes. 
 
 11. The areas of circles, are to each other as the squares of their diameters, or 
 their radii. 
 
48 MENSURATION PART II. 
 
 12. The area of any circle, is equal to the rectangle of half its diameter and half 
 its circumference. . 
 
 13. The area of a circle, is equal to the area of a triangle, the base being equal to 
 the circumference, and the perpendicular equal to the radius. 
 
 14. A square circumscribed about a circle is equal to four times the square of the 
 radius, and a square inscribeil in a circle to half the circumscribed square. 
 
 15. Similar figures inscribed in circles, have their like sides, and also their whole 
 perimeters, in the same ratio as the diameters of circles in which they are inscribed. 
 
 16. Similar figures inscribed in circles, are to each other as the squares of the 
 diameters of those circles. 
 
 EXAMPLES. 
 
 1. Wliat is the area of a circle whose diameter is 106, 
 and circumference 333 feet ? 
 
 Here (333 X 106) -H 4 = 35298 H- 4 = 8824i/ee^, the 
 area required. 
 
 2. Required the area of the end or base of a roller, 
 whose diameter is 2 feet 3 inches. 
 
 Here 2.25 X 2.25 = 5.0625, the square of the diameter; 
 and 5.0625 x .7854 = 3.97608 /ee^ = 3 ft. 11 in. 8 pa., 
 the area required. 
 
 3. If the circumference of a cylindrical stone column 
 be 7 feet 9 inches ; v/hat is the area of its base ? 
 
 A?is. 4 ft. 9 171. 4 pa. 
 
 4. The diameter of a cylindrical vessel is 3 feet 6 
 inches ; what is the area of its bottom ? 
 
 A71S. 9 ft. 7 in. 5 pa. 
 
 5. The diameter of a circular building at the iron 
 foundry of Messrs. Fenton, Murray, and Wood, in Leeds, 
 measures 73 feet 3 inches ; how many square yards of 
 paving are contained in the ground floor ? 
 
 Ans. 468.23475 i/ards. 
 
 6. In the midst of a meadow well stored with grass, 
 I engag'd just one acre to tether my ass ; 
 
 What length must the cord be, that he, feeding all round. 
 Hay not graze less or more than an acre of ground r 
 
 Ans. 39.25068 ijards. 
 
 PROBLEM XVI. 
 To find the area of a sector of a circle. 
 
 RULE I. 
 
 Multiply the length of the arc by the radius of the 
 sector, and half the product will be the area. 
 
 Note 1 . — The length of the arc may be found by Problem XIV. 
 
 2. If the sector be greater than a semicircle, find the arc of the remammg sector, 
 •which subtract from the circumference of the whole circle ; and the remamder will 
 be the length of the arc required. 
 
PART U. OF SLTERFICIES. 49 
 
 EXAMPLES. 
 
 1. The radius AD is 15, the chord 
 BD of the whole arc 24, and the 
 versed sine CE 6 ; what is the area 
 of the sector ABCD? 
 
 Mere v/(122 + 6'^) = ^/(lU + 36) = \/180 = 13.4164 
 = DC, the chord of half the arc; and (13.4164 x 8-24) 
 -r- 3 = (107.3312 - 24) -- 3 = 83.3312 -- 3 = 27.77706, 
 the length of the arc; then (27.77706 X 15) -H 2 = 
 416.6559 -r- 2 = 208.32795, the area required. 
 
 2. AYliat is the area of the sector of a circle whose 
 radius is 15, and the chord of the whole arc 18 feet ? 
 
 Ans. 144/^. 8 in. 10 pa. 
 
 3. If the radius of a sector be 12 feet 6 inches, and 
 the length of the arc 16 feet ; what is its area? 
 
 Alls. 100 feet. 
 
 4. The chord of half the arc is 42 feet 6 inches, and 
 the versed sine 20 feet ; what is the area of the sector ? 
 
 Ans. 1994: ft. 4: in. 9 pa. 
 
 5. AYhat is the area of a sector greater than a semi- 
 circle ; the chord of the whole arc of the remaining sector 
 being 72 feet, the chord of half the arc 45 feet, and the 
 radius 37 feet 6 inches? Ans. 2617ft. 10 in. 6 pa. 
 
 RULE II. 
 
 As 360 is to the degrees in the arc of the sector, so is 
 the area of the whole circle to the area of the sector. 
 
 Note 1. — The area of a semicircle, a qi;adrant, &c. may be most easily found by 
 taking one-half, one-fourth, £cc. of the area of the whole circle. 
 
 2. The area of a circle may be found by multiplying the square of the radius by 
 3.1416; consequently, if the area be divided by 3.1416, the quotient will be the 
 square of the radius. ^ 
 
 EXAMPLES. 
 
 1. The arc of a sector contains 35 degrees, and its 
 radius is 45 feet ; what is its area ? 
 
 Bere 452 x 3.1416 = 2025 x 3.1416 = 6361.74, the 
 area of the luhole circle; then as 360° : 35° W 6361.74 ; 
 61 8.5025 /ee^ = 618 /<f. 6 in., the area required. 
 
 2. The radius of a sector is 25 feet 9 inches, and the 
 length of its arc 218 degrees 54 minutes ; what is its 
 area? Ans. 1266 ft. 7 in. 6 pa. 
 
 3. Required the area of the semicircle, the radius of 
 which is 18 feet 3 inches ? A?is. 523 ft. 2 in. 1 jJCi. 
 
50 MENSURATION PART U. 
 
 4. What is the area of a quadrant whose radius is 
 27 feet 6 inches? A?is. 59SfL 11 in. 6 pa. 
 
 5. Wliat is the area of a sector Avhose radius is 50 feet, 
 and the length of its arc 60 degrees ? A7is. IS09 feet 
 
 PROBLEM XVII. 
 To find the area of the segment of a circle. 
 
 RULE I. 
 
 Find the area of the sector, having the same arc as the 
 segment ; also, find the area of the triangle formed by the 
 chord of the segment and the radii of the sector ; then 
 the difference of these areas, when the segment is less 
 than a semicircle, or their sum, when it is greater, will be 
 the area of the segment. 
 
 EXAMPLES. <-' 
 
 1. The radius AD is 25, and the 
 chord AB of the whole arc 40 ; what 
 is the area of the segment ABC ? 
 
 Here ^/(252 - 20-^) = v/(625-400) = ^/225 = 15 = 
 DE, therefore the versed sine CE = 10. 
 
 Agahi, ^/(20-' + 10^) = ,/(400 + 100) = ^/500 
 = 22.36068 = AC, the chord of half the arc ; and 
 (22.36068 X 8 - 40) -f- 3 = (178.88544 - 40) -f- 3 = 
 138.88544 -f- 3 = 46.295146. the length of the arc 
 ACB; then (46.295146 x 25) -^ 2 = 1157.37865 -- 2 = 
 578.689325, the area of the sector ADBp. 
 
 Now, i AB X DE = 20 x 15 = 300, the area of the 
 triangle ADB ; hence, 578.689325 - 300 - 278.689325, 
 the area of the segment required. 
 
 2. What is the area of the segment of a circle; the 
 chord of the whole arc being 60 feet, and the cliord of 
 half the arc 37 feet 6 inches ? Ans. 987 ft. 6 in. 
 
 3. The chord of the whole arc is 20, and the versed 
 sine 5 feet ; what is the area of the segment ? 
 
 Ans. 69 ft. 8 in. 
 
 4. What is the area of a segment, the arc of which is a 
 quadrant, whose radius is 24? Ans. 164.3904. 
 
TART II. OF SUPERFICIES. 1 
 
 5. "Wliat IS tlie area of a segment greater than a semi- 
 circle ; the chord of the whole arc being 102 feet 6 inches, 
 the chord of half the arc 100 feet, the chord of one 
 quarter of the arc 57 feet 6 inches, and the diameter of 
 the circle 1 16 feet 6 inches ? Ans. 8408.89399 /ee^. 
 
 6. Find the area of a segment whose arc contains 245 
 degrees 45 minutes ; the diameter of the circle being 
 48 feet 9 inches. Ans. 1 545.03853 /ee^ 
 
 RULE II. 
 
 To two-thirds of the product of the chord and height 
 of the segment, add the cube of the height divided by 
 twice the chord, and the sum will be the area of the seg- 
 ment, nearly. 
 
 Note. — When the segment is greate than a semicircle, find the area of the 
 remaining segment, wiiicli subtract from tlie area of the whole circle ; and the 
 remainder will be the area required. 
 
 EXAMPLES. 
 
 1. "WTiat is the area of a segment of a circle whose 
 chord is 32, and height or versed sine 8 ? 
 
 Here (32 x 8) X f + 8^ -- (32 x 2) = 256 x f + 512 
 -H 64 = 170.6666 -h 8 =. 178.6666, the area required. 
 
 2. The chord is 65, and versed sine 15; wliat is the 
 area of the segment? Ans. 675.96153. 
 
 3. "What is the area of a segment, greater than a semi- 
 circle, whose chord is 30, and height 20 ? 
 
 Ans. 518.26171875. 
 RULE in. 
 
 Divide the height of the segment by the diameter, and 
 find the quotient in the column of heights or versed sines, 
 in the Table at the end of Part II. 
 
 Take out the corresponding area seg., which multiply 
 by the square of the diameter, and the product will be 
 the area of the segment. 
 
 Note \ — If the quotient of the height by the diameter do not terminate in three 
 places of figures, without a fractional remainder, find the area seg. answering to 
 the first three decimals of the quotient ; subtract it from the next greater area scg. ; 
 multiply the remainder by the fractional part of the quotient, and the product will 
 be the corresponding proportional part to be added to the first area xcf;. 
 
 This method ought to be used when accuracy is required ; but for common pur- 
 poses, the fractional remainder may be omitted. 
 
 2. When the area of a segment greater than a semicircle is required, subtract the 
 quotient of the height by the diameter, from 1 : find the area seg. corresponding 
 to the remainder, which take from .785398, and the difference will be the area scg. 
 answering to the quotient. 
 
 3. The versed sines of similar segments are as the diameters of the circles to 
 which they belong, and the areas of those segments are as the squares of the 
 diameters. 
 
 D 2 . 
 
52 MENSURATION PAKT H. 
 
 EXA3IPLES. 
 
 1. What is the area of a segment whose chord is 32, 
 the versed sine 8, and the diameter of the circle 40 ? 
 
 Here 8.0 -:- 40 = .2, the quotient or tabular height; 
 and the correspond itig area seg. is .111823 ; heiice, 
 .111823 X 40- = .111823 x 1600 = 178.9168, the area 
 of the segment required. 
 
 2. What is the area of a, segment whose height is 9, 
 and the diameter of the circle 25 ? Ans. 159.09375. 
 
 3. Find the area of a segment whose height is 25, and 
 the diameter of the circle 55. Ans. 1050.60065. 
 
 4. What is the area of a segment, greater than a semi- 
 circle, whose height is QQ feet, and the chord of the whole 
 arc 60 feet 10 inches? Ans. 4435 /<f. 8 in. II pa. 
 
 5. The base of a stone column is the greater segment 
 of a circle whose chord measures 2 feet, and versed sine 
 1 foot 4 inches ; what is its area? Ans. 2.304 feet. 
 
 peoble:?^ xvin. 
 
 To find the area of a circular zone, or the space included 
 betiveen any two parallel chords and their intercepted 
 arcs. 
 
 KULE. 
 
 Find the area of that part of the zone forming the 
 trapezoid ABCD, to which add twice the area of the 
 segment AED ; and the sum will be the area of the zone 
 required. {See the next figure.) 
 
 JNV^iff ]._When great accuracy is required, the area of the segment should be 
 found bv Faile 3, Problem XVII. „,.,.„ 
 
 2. The chord AD and versed sine E?« mav be found from the parallel sides AB, 
 DC, and the perpendicular distance DF, by the help of Theorem 12, Part I. ; 
 much calculation may, however, be saved by measuring the chord and versed sine 
 in taking the dimensions of the zone. (See the Key to the Mensuration, page 22.) 
 
 3. When the parallel sides and their perpendicular distance are given, the zone 
 may be constructed in the following manner : draw the side AB ; make AF equal 
 to half the difference between AD and DC, and erect the perpendicular FD. From 
 the point D, draw DC parallel to AB, and join AD. Bisect AB and AD with the 
 perpendiculars wo, no ; and o will be the centre of the circle of which the zone is 
 a part ; thus you will determine whether the centre of the circle falls within or 
 without the zone. 
 
 EX-\:»IPLES. 
 
 1. The greater side AB measures 
 40 feet, the less DC 30 feet, the per- 
 pendicular FD 35 feet, the chord AD 
 35 feet 3 inches, and the versed sine 
 Em 7 feet 3 inches ; Vv^hat is the area 
 ofthezone ABCD? 
 
 •--^0 
 
 o 
 
 ATT 
 
PAKT n. OF SUPERFICIES. 53 
 
 By Problem VIII. tve have (40 + 30) x 35 = 70 x 
 35 = 2450; a?id2450 -- 2 = 1225, the area of the trapc 
 zoid ABCD. 
 
 Also, hij Problem XVII. Rule 2, tve have (35.25 
 X 7.25) X f + 7.253 -- (35.25 x 2) == 255.5625 x ^ ^ 
 381.078125 -- 70.5 = 170.375 -f 5.40536 = 175.78036, 
 the area of the segment XEJ) ; hence, 1225 + (175.78036 
 X 2) = 1225 + 351.56072 = 1576.56072 feet = 1576ft. 
 6 in. 8 pa., the area of the zone ABCD, as required. 
 
 2. The greater side is 120 feet, the less 75 feet, the 
 chord 39 feet 6 inches, and the versed sine 3 feet 3 
 inches ; what is the area of the zone ? 
 
 Ans. 3337/f. 4 i?i. 10 pa. 
 
 3. What is the area of a zone whose greater side mea- 
 sures 72 feet, the less 45 feet, and its breadth 19 feet 
 6 inches? Ans. 1201ft. 11 m. 1 pa. 
 
 4. The greater side is 80, the less 60, and their dis- 
 tance 70 ; what is the area of the zone ? Ans. 6326.96. 
 
 PROBLEM XIX. 
 
 To find the area of a circular ring, or the space included 
 between the circmnferences of two concentric circles. 
 
 KULE. 
 
 Multiply the sum of the diameters by their difference, 
 and this product by .7854 ; and it will give the area re- 
 quired. Or, the difference of the areas of the two circles 
 wall be the area of the ring. 
 
 l:Jote 1. — The area of a circular ring may also be found bv multiplying half the 
 sura of the circumference by half the difference of the diameters. 
 
 2. The area of part of a ring, or the segment of a sector, may be found by mul- 
 tiplying half the sura of the bounding arcs by the nearest distance between them. 
 
 exa:mples. 
 
 1. The diameters AB and CD 
 are 30 and 20 ; what is the area of 
 the circular rinj>- ? 
 
 Here (30 + 20) x (30 - 20) x .7854 == 50 x 10 x 
 .7854 -=. 500 X .7854 = 392.7, the area of the riiig re- 
 quired. 
 
 concentric' circles are 35 
 D 3 
 
54 
 
 MENSURATION 
 
 PART II. 
 
 and 23 ; wliat is the area of the ring formed by the cir- 
 cumference of those circles ? Ans. 546.6384. 
 
 3. The inner diameter of a circular building is 73 feet 
 3 inches, and the thickness of the wall 1 foot 9 inches ; 
 how many square feet of ground does the wall occupy ? 
 
 A71S. 412.335 feet. 
 
 4. What was the expense of making a moat round a 
 circular island, at 2^. 6d. per square yard ; the diameter 
 of the island being 525 feet, and the breadth of the moat 
 21 feet 6 inches ? Ans. £512. I3s. 7 id. 
 
 5. Wliat is the area of the front of a circular arch, 
 built with stones, each 3 feet 6 inches long ; the length 
 of the upper bounding arc being 35 feet 3 inches, and the 
 length of the lower 24 feet 9 inches? A?is. 105 feet. 
 
 PROBLEM XX. 
 
 To find the area of a lune^ or the space included between 
 the intersecting arcs of two eccentric circles. 
 
 RULE. 
 
 Find the areas of the two segments forming the lune, 
 and their diiFerence will be the area required. 
 
 l^ote\. — If ABCD be a square, and circles be 
 described from the points B and C. with the radii BD 
 and CD ; the area of the lune DGEFD will be equal 
 to the area of the square ABCD. 
 
 2. If ABC be a right-angled triangle, and semi- 
 circles be described on the three sides, as diameters; 
 then will the area of the said triangle be equal to the 
 sum of the areas of the two lunes D and E. 
 
 Several other curious properties or lunes may be seen in Dr. Hutton's Recrea- 
 tions, and Mathematical Dictionary. 
 
 EXAJGPLES. 
 
 1. The length of the chord AB is 
 30, the height DC 12, and DE 5 ; 
 what is the area of the lune ACBEA? 
 
PAKT II. OF SUPERFICIES. 55 
 
 By Rule 2, Problem XVII. we have 30 x 12 x f + 
 
 123 ^ (30 X 2) = 360 X f + 1728 -- 60 = 240 + 28.8 
 = 268.8, the area of the segment ACB. 
 
 Also 30 X 5 X ^ + 5^ -- (30 X 2) = 150 x | + 125 
 -^ 60 = 100 4- 2.08333 = 102.08333, the area of the 
 segment AEB ; hence, 268.8 - 102.08333 = 166.71667, 
 the area of the lime required. 
 
 2. The chord is 24, and the heights of the segments 9 
 and 4 ; what is the area of the lune ? Ans. 93.8542. 
 
 3. The length of the chord is 48, and the versed sines 
 of the segments 18 and 8 ; what is the area of the lune ? 
 
 Arts. 374.775. 
 
 'Sole. — The .inswer to this question was obtained by the assistance of Rule 2, 
 Problem XIV., and Rule 3, Problem XVII. 
 
 PROBLEM XXI. 
 To find the area of an ellipse, 
 
 RULE. 
 
 Multiply continually together the two diameters and 
 the number .7854, and the product wiU be the area of the 
 ellipse. 
 
 Note 1 The area of an elliptical ring, or the space included between the cir. 
 
 cumference of two concentric similar or dissimilar ellipses, may bo found thus : 
 From the product of the two diameters of the greater ellipse subtract the product 
 of the two diameters of the less ; multiply the remainder by .7854, and the product 
 will be the area of the ring. Or, subtract the area of the less ellipse from that of 
 the greater, and the remainder will be the area of the ring. 
 
 2. If the sum of the two diameters of an ellipse be multiplied by 1.5708, (half of 
 3.1416,) the product will be the circumference, exact enough for most practical 
 purposes. 
 
 3. To half the sum of the two diameters add the square root of half the sum of 
 their squares ; multiply the last sum by 1..5708, and the product will be the circum- 
 ference, extremely near. 
 
 4. The ellipse is equal to a circle whose diameter is a mean proportional between 
 the two axes ; hence we obtain the above Rule. 
 
 EXAMPLES. 
 
 1. What is the area of the ellipse 
 ABCD, whose transverse diameter 
 AB is 34, and conjugate CD 2^ ? 
 
 Here 34 X 25 x .7854 = 850 x .7854 = 667.59, the 
 area required. 
 
 2. If the transverse diameter be 50 feet 9 inches, and 
 the conjugate 35 feet 6 inches ; what is the area of the 
 ellipse? Ans. 1414 /f. 11 in. II pa. 
 
 D 4 
 
56 SIEXSUILITION PART II. 
 
 3. The transverse diameter of an elliptical bath mea- 
 sures 25 feet, and the conjugate 15 feet ; Tvhat was the 
 expense of paving a walk round it, with Portland-stone, 
 at 3s. 6d. per square foot ; the breadth of the walk being 
 5 feet 6 inches ? Ans. aB77. 2s. 1^(1. 
 
 4. The diameters of an elliptical piece of ground are 
 330 and 220 feet ; how many quicks will plant the fence 
 forming the circumference, supposing them to be set 
 o inches asunder ? Ans. 2073. 
 
 PROBLEM XXn. 
 
 To find the area of an elliptical segment^ Hie base of which 
 is 2)cirallel to either of the diameters of the ellipse. 
 
 KULE. 
 
 Divide the height of the segment by that diameter of 
 the ellipse of which it is a part, and find the area seg. 
 ansY/ering to the quotient, in the Table at the end of 
 Part II. Multiply the two diameters of the ellipse and 
 the area seg. thus found continually together, and the 
 product will be the area required. 
 
 EXA31PLES. 
 
 1. What is the area of the ellip- 
 tical segment ABC, cut oft by the 
 double ordinate AB ; the height CG 
 being 12, the diameter CD of the 
 
 whole ellipse 40, and EF 2o ? 
 
 Here 12 -r- 40 = .3, the tahular height; and the cor- 
 responding area seg. is .198168; the?i .198168 x 25 x 
 40 = 198.168, the area of the segment required. 
 
 2. 'What is the area of the segment ADB, the height 
 DG being 28, and the diameters of the ellipse 40 and 25 ? 
 
 A71S. 587.23. 
 
 3. TTliat are the areas of the tvv^o elliptical segments 
 made by a double ordinate parallel to the conjugate 
 diameter, at the distance of 9 from the centre of the 
 ellipse ; the diameters being 40 and 60 ? 
 
 Ans. 587.952, and 1297.008. 
 
 4. The diameters of an elliiDse are 63 feet 9 inches, and 
 
PAKT U. 
 
 OF SUPEKFICIES. 
 
 o7 
 
 95 feet 3 inclies, and the height of a segment, cut off by 
 a line parallel to the transverse, is 25 feet 6 inches ; what 
 is its area? Ans. 1781 ft. 4 in. 8 pa. 
 
 PROBLEM XXIII. 
 
 To find the area of a segment of a circle., or any other 
 curvilineal figure^ by means of equidistant ordinates. 
 
 RULE. 
 
 If a right line AN be divided into any even number of 
 equal parts AC, CE, EG, &c. ; and at the points of divi- 
 sion be erected perpendicular ordi- 
 nates AB, CD, EF, &c. ; terminated 
 by any curve BDF, &c. ; and if A be n 
 put for the sura of the extreme or ^^Xj 
 first and last ordinates AB, NO ; B ; 
 for the sum of the even ordinates ■■—r^r 
 CD, OH, LM, &c., viz. the second, 
 fourth, sixth, &c. ; and C for the sum o 
 
 IK, &c., viz. the third, fifth, &c.. 
 
 .f all the rest EF, 
 or the odd ordinates, 
 wanting the first and last : then the common distance AC, 
 or CE, &c. of the ordinates, being multiplied by the sum 
 arising from the addition of A, four times B, and two 
 times C, one-third of the product will be the area ABON, 
 very nearly ; that is, (A + 4 B + 2 C) ^ 3 x D = the 
 area, putting D = AC, the common distance of the ordi- 
 nates. 
 
 {^See Prob.o. Part 6., where the rule is expressed in 
 words.) 
 
 Note 1 — When equidistant ordinates or perpendiculars cannot conveniently be 
 taken, in consequence of the bends or corners of the boundary being at unequal 
 distances, you may proceed as in Problem 4, Fart III. 
 
 2. By the Rule given in this Problem, the contents of all solids, whether regular 
 or irregular, may be found, by using the areas of the sections perpendicular to the 
 axe, instead of the ordinates ; and it is evident that the greater the number of or- 
 dinates or sections are used, the more accurately will the area or solidity be deter- 
 mined. {See the Scholium, Frob. 1. Fart 6. Case 2.) 
 
 EXAMPLES. 
 
 1. It is required to find the area of the curved 
 ABCD, the lengths of the five equi- 
 distant ordinates being as follow ; viz. 
 the first or AD = 8, the second = 
 10, the third = 12, the fourth = 
 14, and the fifth or last BC = 15 ; 
 and the length of the base AB = 24. 
 
 Here A = 8 + 15 = 23, B = 10 -f 14 — 24, C 
 D 5 
 
 space 
 c 
 
 
 
 E 
 1 O 
 
58 
 
 MENSURATION 
 
 PART n. 
 
 24 
 
 and D == "^ =: 6 ; the7i (A + 4 13 + 2 C) x -i D = (23 
 4-96 + 24) X 4 = 143 x 2 = 286, the area required. 
 
 2. Given the lengths of seven equidistant ordinates of 
 an irregular piece of ground, as follow ; viz. 15, 19, 20, 
 23, 25, 30, and 33 feet; and the length of the base 
 72 feet : required the plan and area. 
 
 Ans. The area is 1704: feet. 
 
 A TABLE OF THE AREAS OF THE SEGMENTS 
 OF A CmCLE, 
 
 ^VHOSE DLIMETER IS UNITY, AND SUPPOSED TO BE DIVIDED 
 INTO 1000 EQUAL PARTS. 
 
 Height, 
 
 Area 
 Segment. 
 
 Height. 
 
 Area 
 Segment. 
 
 Height. 
 
 Area 
 Segment. 
 
 .001 
 
 .000042 
 
 .026 
 
 .005546 
 
 .051 
 
 .015119 
 
 .002 
 
 .000119 
 
 .027 
 
 .005867 
 
 .052 
 
 .015561 
 
 .003 
 
 .000219 
 
 .028 
 
 .006194 
 
 .053 
 
 .016007 
 
 .004 
 
 .000337 
 
 .029 
 
 .006527 
 
 .054 
 
 .016457 
 
 .005 
 
 .000470 
 
 .030 
 
 .006865 
 
 .055 
 
 .016911 
 
 .006 
 
 .000618 
 
 .031 
 
 .007209 
 
 .056 
 
 .017369 
 
 .007 
 
 .000779 
 
 .032 
 
 .007558 
 
 .057 
 
 .017831 
 
 .008 
 
 .000951 
 
 .033 
 
 .007913 
 
 .058 
 
 .018296 
 
 .009 
 
 .001135 
 
 .034 
 
 .008273 
 
 .059 
 
 .018766 
 
 .010 
 
 .001329 
 
 .035 
 
 .008638 
 
 .060 
 
 .019239 
 
 .011 
 
 .001533 
 
 .036 
 
 .009008 
 
 .061 
 
 .019716 
 
 .012 
 
 .001746 
 
 .037 
 
 .009383 
 
 .062 
 
 .020196 
 
 .013 
 
 .001968 
 
 .038 
 
 .009763 
 
 .063 
 
 .020680 
 
 .014 
 
 .002199 
 
 .039 
 
 .010148 
 
 .064 
 
 .021168 
 
 .015 
 
 .002438 
 
 .040 
 
 .010537 
 
 .065 
 
 .021659 
 
 .016 
 
 .002685 
 
 •041 
 
 .010931 
 
 .066 
 
 .022154 
 
 .017 
 
 .002940 
 
 .042 
 
 .011330 
 
 .067 
 
 .022652 
 
 .018 
 
 .003202 
 
 .043 
 
 .011734 
 
 .068 
 
 .023154 
 
 .019 
 
 .003471 
 
 .044 
 
 .012142 
 
 .069 
 
 .023659 
 
 .020 
 
 .003748 
 
 .045 
 
 .012554 
 
 .070 
 
 .024168 
 
 .021 
 
 .004031 
 
 .046 
 
 .012971 
 
 .071 
 
 .024680 
 
 •022 
 
 .004322 
 
 .047 
 
 .013392 
 
 .072 
 
 .025195 
 
 .023 
 
 .004618 
 
 .048 
 
 .013818 
 
 .073 
 
 .025714 
 
 .024 
 
 .004921 
 
 .049 
 
 .014247 
 
 .074 
 
 .026236 
 
 .025 
 
 .005230 
 
 .050 
 
 .014681 
 
 .075 
 
 .026761 
 
PART II. 
 
 OF SUPERFICIES. 
 
 59 
 
 1 
 
 Height. 
 
 Area 
 Segment. 
 
 Height. 
 
 1 
 
 Area 
 Segment. 
 
 Height. 
 
 Area 
 Segment. 
 
 .076 
 
 .027289 
 
 1 .117 
 
 .051446 
 
 .158 
 
 .079649 
 
 .077 
 
 .027821 
 
 .118 
 
 .052090 
 
 .159 
 
 .080380 
 
 .078 
 
 .028356 
 
 ' .119 
 
 .052736 
 
 .160 
 
 .081112 
 
 .079 
 
 .028894 
 
 ! .120 
 
 .053385 
 
 .161 
 
 .081846 
 
 .080 
 
 .029435 
 
 1 .121 
 
 .054036 
 
 .162 
 
 ,082582 
 
 .081 
 
 .029979 
 
 .122 
 
 .054689 
 
 .163 
 
 .083320 
 
 .082 
 
 .030526 
 
 i !l23 
 
 .055345 
 
 .164 
 
 .084059 
 
 .083 
 
 .031076 
 
 .124 
 
 .056003 
 
 .165 
 
 .084801 
 
 .084 
 
 .031629 
 
 .125 
 
 .056663 
 
 .166 
 
 .085544 
 
 .085 
 
 .032186 
 
 .126 
 
 .057326 
 
 .167 
 
 .086289 
 
 .086 
 
 .032745 
 
 .127 
 
 .057991 
 
 .168 
 
 .087036 
 
 .087 
 
 .033307 
 
 .128 
 
 .058658 
 
 .169 
 
 .087785 
 
 .088 
 
 .033872 
 
 .129 
 
 .059327 
 
 .170 
 
 .088535 
 
 .089 
 
 .034441 
 
 .130 
 
 .059999 
 
 .171 
 
 .089287 
 
 .090 
 
 .035011 
 
 .131 
 
 .060672 
 
 .172 
 
 .090041 
 
 .091 
 
 .035585 
 
 .132 
 
 .061348 
 
 .173 
 
 .090797 
 
 .092 
 
 .036162 
 
 .133 
 
 .062026 
 
 .174 
 
 .091554 
 
 .093 
 
 .036741 
 
 .134 
 
 .062707 
 
 .175 
 
 .092313 
 
 .094 
 
 .037323 
 
 .135 
 
 .063389 
 
 .176 
 
 .093074 
 
 .095 
 
 .037909 
 
 .136 
 
 .064074 
 
 .177 
 
 .093836 
 
 ) .096 
 
 .038496 
 
 .137 
 
 .064760 
 
 .178 
 
 .094601 
 
 .097 
 
 .039087 
 
 .138 
 
 .065449 
 
 .179 
 
 .095366 
 
 .098 
 
 .039680 
 
 .139 
 
 .066140 
 
 .180 
 
 .096134 
 
 .099 
 
 .040276 
 
 .140 
 
 .066833 
 
 .181 
 
 .096903 
 
 .100 
 
 .040875 
 
 .141 
 
 .067528 
 
 .182 
 
 .097674 
 
 .101 
 
 .041476 
 
 .142 
 
 .068225 
 
 .183 
 
 .098447 
 
 .102 
 
 .042080 
 
 .143 
 
 .068924 
 
 .184 
 
 .099221 
 
 .103 
 
 .042687 
 
 .144 
 
 .069625 
 
 .185 
 
 .099997 
 
 .104 
 
 .043296 
 
 .145 
 
 .070328 
 
 .186 
 
 .100774 
 
 .105 
 
 .043908 
 
 .146 
 
 .071033 
 
 .187 
 
 .101553 
 
 .106 
 
 .044522 
 
 .147 
 
 .071741 
 
 .188 
 
 .102334 
 
 .107 
 
 .045139 
 
 .148 
 
 .072450 
 
 .189 
 
 .103116 
 
 .108 
 
 .045759 
 
 .149 
 
 .073161 
 
 .190 
 
 .103900 
 
 .109 
 
 .046381 
 
 .150 
 
 .073874 
 
 .191 
 
 .104685 
 
 .110 
 
 .047005 
 
 .151 
 
 .074589 
 
 .192 
 
 .105472 
 
 .111 
 
 .047632 
 
 .152 
 
 .075306 
 
 .193 
 
 .106261 
 
 .112 
 
 .048262 
 
 .153 
 
 .076026 
 
 .194 
 
 .107051 
 
 .113 
 
 .048894 
 
 .154 
 
 .076747 
 
 .195 
 
 .107842 
 
 .114 
 
 .049528 
 
 .155 
 
 .077469 
 
 .196 
 
 .108636 
 
 .115 
 
 .050165 
 
 .156 
 
 .078194 
 
 .197 
 
 .109430 
 
 .116 
 
 .050804 
 
 .157 
 
 .078921 
 d6 
 
 .198 
 
 .110226 
 
60 
 
 MENSURATION 
 
 PAUT n. 
 
 Height. 
 
 Area 
 Segment. 
 
 Height. 
 
 Area 
 
 Segment. 
 
 Height. 
 
 Area 
 Segment. 
 
 .199 
 
 .111024 
 
 .240 
 
 .144944 
 
 .281 
 
 .180918 
 
 .200 
 
 .111823 
 
 .241 
 
 .145799 
 
 .282 
 
 .181817 
 
 .201 
 
 .112624 
 
 .242 
 
 .146655 
 
 .283 
 
 .182718 
 
 .202 
 
 .113426 
 
 .243 
 
 .147512 
 
 .284 
 
 .183619 
 
 .203 
 
 .114230 
 
 .244 
 
 .148371 
 
 .285 
 
 .184521 
 
 .204 
 
 .115035 
 
 .245 
 
 .149230 
 
 .286 
 
 .185425 
 
 .205 
 
 .115842 
 
 .246 
 
 .150091 
 
 .287 
 
 .186329 
 
 .206 
 
 .116650 
 
 .247 
 
 .150953 
 
 .288 
 
 .187234 
 
 .207 
 
 .117460 
 
 .248 
 
 .151816 
 
 .289 
 
 .188140 
 
 .208 
 
 .118271 
 
 .249 
 
 .152680 
 
 .290 
 
 .189047 
 
 .209 
 
 .119083 
 
 .250 
 
 .153546 
 
 .291 
 
 .189955 
 
 .210 
 
 .119897 
 
 .251 
 
 .154412 
 
 .292 
 
 .190864 
 
 .211 
 
 .120712 
 
 .252 
 
 .155280 
 
 .293 
 
 .191775 
 
 .212 
 
 .121529 
 
 .253 
 
 .156149 
 
 .294 
 
 .192684 
 
 .213 
 
 .122347 
 
 .254 
 
 .157019 
 
 .295 
 
 .193596 
 
 .214 
 
 .123167 
 
 .255 
 
 .157890 
 
 .296 
 
 .194509 
 
 .215 
 
 .123988 
 
 .256 
 
 .158762 
 
 .297 
 
 .195422 
 
 .216 
 
 .124810 
 
 .257 
 
 .159636 
 
 .298 
 
 .196337 
 
 .217 
 
 .125634 
 
 .258 
 
 .160510 
 
 .299 
 
 .197252 
 
 .218 
 
 .126459 
 
 .259 
 
 .161386 
 
 .300 
 
 .198168 
 
 .219 
 
 .127285 
 
 .260 
 
 .162263 
 
 .301 
 
 .199085 
 
 .220 
 
 .128113 
 
 .261 
 
 .163140 
 
 .302 
 
 .200003 
 
 .221 
 
 .128942 
 
 .262 
 
 .164019 
 
 .303 
 
 .200922 
 
 !222 
 
 .129773 
 
 .263 
 
 .164899 
 
 .304 
 
 .201841 
 
 .223 
 
 .130605 
 
 .264 
 
 .165780 i 
 
 .305 
 
 .202761 
 
 .224 
 
 .131438 
 
 .265 
 
 .166663 
 
 .306 
 
 .203683 
 
 .225 
 
 .132272 
 
 .266 
 
 .167546 
 
 .307 
 
 .204605 
 
 .226 
 
 .133108 
 
 .267 
 
 .168430 
 
 .308 
 
 .205527 
 
 .227 
 
 .133945 
 
 .268 
 
 .169315 
 
 .309 
 
 .206451 
 
 ■ .228 
 
 .134784 
 
 .269 
 
 .170202 
 
 .310 
 
 .207376 
 
 .229 
 
 .135624 
 
 .270 
 
 .171089 
 
 .311 
 
 .208301 
 
 !230 
 
 .136465 
 
 .271 
 
 .17197a 
 
 .312 
 
 .209227 
 
 .231 
 
 .137307 
 
 .272 
 
 .172867 
 
 .313 
 
 .210154 
 
 .232 
 
 .138150 
 
 .273 
 
 .173758 
 
 .314 
 
 .211082 
 
 .233 
 
 .138995 
 
 .274 
 
 .174649 
 
 .315 
 
 .212011 
 
 .234 
 
 .139841 
 
 .275 
 
 .175542 
 
 .316 
 
 .212940 
 
 .235 
 
 .140688 
 
 .276 
 
 .176435 
 
 .317 
 
 .213871 
 
 .236 
 
 .141537 
 
 .277 
 
 .177330 
 
 .318 
 
 .214802 
 
 .237 
 
 .142387 
 
 .278 
 
 .178225 
 
 .319 
 
 .215733 
 
 .238 
 
 .143238 
 
 .279 
 
 .179122 
 
 .320 
 
 .216666 
 
 1 .239 
 
 .144091 
 
 .280 
 
 .180019 
 
 .321 
 
 .217599 
 
PART n. 
 
 OF SUPERFICIES. 
 
 61 
 
 Height. 
 
 Area 
 Segment. 
 
 Height. 
 
 Area 
 Segment. 
 
 Height. 
 
 Area 
 Segment. 
 
 .322 
 
 .218533 
 
 .363 
 
 .257433 
 
 .404 
 
 .297292 
 
 .323 
 
 .219468 
 
 .364 
 
 .258395 
 
 .405 
 
 .298273 
 
 .324 
 
 .220404 
 
 .365 
 
 .259357 
 
 .406 
 
 .299255 
 
 .325 
 
 .221340 
 
 .366 
 
 .260320 
 
 .407 
 
 .300238 
 
 .326 
 
 .222277 
 
 .367 
 
 .261284 
 
 .408 
 
 .301220 
 
 .327 
 
 .223215 
 
 .368 
 
 .262248 
 
 .409 
 
 .302203 
 
 .328 
 
 .224154 
 
 .369 
 
 .263213 
 
 .410 
 
 .303187 
 
 .329 
 
 .225093 
 
 .370 
 
 .264178 
 
 .411 
 
 .304171 
 
 .330 
 
 .226033 
 
 .371 
 
 .265144 
 
 .412 
 
 .305155 
 
 .331 
 
 .226974 
 
 .372 
 
 .266111 
 
 .413 
 
 .306140 
 
 .332 
 
 .227915 
 
 .373 
 
 .267078 
 
 .414 
 
 .307125 
 
 .333 
 
 .228858 
 
 .374 
 
 .268045 
 
 .415 
 
 .308110 
 
 .334 
 
 .229801 
 
 .375 
 
 .269013 
 
 .416 
 
 .309095 
 
 .335 
 
 .230745 
 
 .376 
 
 .269982 
 
 .417 
 
 .310081 
 
 .336 
 
 .231689 
 
 .377 
 
 .270951 
 
 .418 
 
 .311068 
 
 .337 
 
 .232634 
 
 .378 
 
 .271920 
 
 .419 
 
 .312054 
 
 .338 
 
 .233580 
 
 .379 
 
 .272890 
 
 .420 
 
 .313041 
 
 .339 
 
 .234526 
 
 .380 
 
 .273861 
 
 .421 
 
 .314029 
 
 .340 
 
 .235473 
 
 .381 
 
 .274832 
 
 .422 
 
 .315016 
 
 .341 
 
 .236421 
 
 .382 
 
 .275803 
 
 .423 
 
 .316004 
 
 .342 
 
 .237369 
 
 .383 
 
 .276775 
 
 .424 
 
 .316992 
 
 .343 
 
 .238318 
 
 .384 
 
 .277748 
 
 .425 
 
 .317981 
 
 .344 
 
 .239268 
 
 .385 
 
 .278721 
 
 .426 
 
 .318970 
 
 .345 
 
 .240218 
 
 .386 
 
 .279694 
 
 .427 
 
 .319959 
 
 .346 
 
 .241169 
 
 .387 
 
 .280668 
 
 .428 
 
 .320948 
 
 .347 
 
 .242121 
 
 .388 
 
 .281642 
 
 .429 
 
 .321938 
 
 .348 
 
 .243074 
 
 .389 
 
 .282617 
 
 .430 
 
 .322928 •■ 
 
 .349 
 
 .244026 
 
 .390 
 
 .283592 
 
 .431 
 
 .323918 
 
 .350 
 
 .244980 
 
 .391 
 
 .284568 
 
 .432 
 
 .324909 
 
 .351 
 
 .245934 
 
 .392 
 
 .285544 
 
 .433 
 
 .325900 
 
 .352 
 
 .246889 
 
 .393 
 
 .286521 
 
 .434 
 
 .326892 
 
 .353 
 
 .247845 
 
 .394 
 
 .287498 
 
 .435 
 
 .327882 
 
 .354 
 
 .248801 
 
 .395 
 
 .288476 
 
 .436 
 
 .328874 
 
 .355 
 
 .249757 
 
 .396 
 
 .289453 
 
 .437 
 
 .329866 
 
 .356 
 
 .250715 
 
 .397 
 
 .290432 
 
 .438 
 
 .330858 
 
 .357 
 
 .251673 
 
 1 .398 
 
 .291411 
 
 .439 
 
 .331850 
 
 .358 
 
 .252631 
 
 ' .399 
 
 .292390 
 
 .440 
 
 .332843 
 
 .359 
 
 .253590 
 
 1 .400 
 
 .293369 
 
 .441 
 
 .333836 
 
 .360 
 
 .254550 
 
 1 .401 
 
 .294349 
 
 .442 
 
 .334829 
 
 .361 
 
 .255510 
 
 .402 
 
 .295330 
 
 .443 
 
 .335822 
 
 .362 
 
 .256471 
 
 .403 
 
 .296311 
 
 .444 
 
 .336816 i 
 
62 
 
 JIENSUEATION OF SUPERFICIES. 
 
 PART n. 
 
 ■Height. 
 
 Area 
 Segment. 
 
 Height. 
 
 Area 
 Segment. 
 
 1 
 Height. 
 
 Area 
 
 Segment. 
 
 .445 
 
 .337810 
 
 .464 
 
 .356730 
 
 .483 
 
 .375702 
 
 .446 
 
 .338804 
 
 .465 
 
 .357727 
 
 .484 
 
 .376702 
 
 .447 
 
 .339798 
 
 .466 
 
 .358725 
 
 .485 
 
 .377701 
 
 .448 
 
 .340793 
 
 .467 
 
 .359723 
 
 .486 
 
 .378701 
 
 ' .449 
 
 .341787 
 
 .468 
 
 .360721 
 
 .487 
 
 .379700 
 
 .450 
 
 .342782 
 
 .469 
 
 .361719 
 
 .488 
 
 .380700 
 
 .451 
 
 .343777 
 
 .470 
 
 .362717 
 
 .489 
 
 .381699 
 
 .452 
 
 .344772 
 
 .471 
 
 .363715 
 
 .490 
 
 .382699 
 
 .453 
 
 .345768 
 
 .472 
 
 .364713 
 
 .491 
 
 .383699 
 
 .454 
 
 .346764 
 
 .473 
 
 .365712 
 
 .492 
 
 .384699 
 
 ; .455 
 
 .347759 
 
 .474 
 
 .366710 
 
 .493 
 
 .385699 
 
 ; .456 
 
 .348755 
 
 .475 
 
 .367709 
 
 .494 
 
 .386699 
 
 ': .457 
 
 .349752 
 
 .476 
 
 .368708 
 
 .495 
 
 .387699 
 
 .458 
 
 .350748 
 
 .477 
 
 .369707 
 
 .496 
 
 .388699 
 
 .459 
 
 .351745 
 
 .478 
 
 .370706 
 
 .497 
 
 .389699 
 
 .460 
 
 .352742 ' 
 
 .479 
 
 .371705 
 
 .498 
 
 .390699 
 
 .461 
 
 .353739 i 
 
 .480 
 
 .372704 
 
 .499 
 
 .391699 
 
 .462 
 
 .354736 
 
 .481 
 
 .373703 
 
 .500 
 
 .392699 
 
 .463 
 
 .355732 
 
 .482 
 
 .374702 
 
 
 
 2^oie The use of the foregoing Table is given in Problem XVII. Rule 3; and 
 
 the method of constructing it may be seen in Moss's Gauging, page 29. 
 
PART in. LAND-SUKYEYING. 63 
 
 PART III. 
 LAND-SURVEYING, 
 
 ANn 
 
 MISCELLANEOUS QUESTIONS 
 
 IN 
 Superjicial Mensuration. 
 
 SECTION I. 
 
 LAND-SURVEYING. 
 
 LajsD is commonly measured with a chain, invented by- 
 Mr. Gunter, and known by the name of Gunters Chain. 
 It is 4 poles, or 22 yards in length, and divided into 100 
 equal parts, called links ; each link being 7.92 inches. 
 
 An acre of land is equal to 10 square chains; that is, 
 10 chains in length and 1 in breadth ; or 220 x 22 = 4840 
 square yards ; or 40 x 4 = 160 square rods, poles, or 
 perches ; or 1000 x 100 = 100000 square links. 
 
 The measurement of land is generally given in acres, 
 roods, and perches ; 4 roods being an acre, and 40 perches 
 a rood. 
 
 A statute-pole or perch is 16J feet long ; but in dif- 
 ferent parts of the kingdom tliere are, by custom, poles of 
 different lengths; as 15, 18, 21 feet, &c. 
 
 The Field-hook. 
 
 The best method of entering the field-notes, is to begin 
 at tlie bottom of the page and write upward. 
 
 Each page of the field-book must be divided into three 
 columns. In the middle column must be set down the 
 distances on the chain-line, at which any mark, offset, or 
 other observation is made ; and in the right and left-hand 
 columns respectively, those marks, offsets, and observa- 
 tions must be entered. 
 
 The crossings of fences, rivers, &c. may be denoted by 
 
64 LA^D-SrnVEYING. PART III. 
 
 lines dra-vvn across the middle column, or part of the right 
 and left-hand columns, opposite the distances on the chain- 
 line, at which they are crossed ; and the corners of fields, 
 and other remarkable turns in the fences, to which offsets 
 are taken, may be defined by lines joining or lying in the 
 same relation to the middle column, as fences, &c. do to 
 the chain-line. 
 
 Thus a tolerably accurate representation of the fences, 
 &c. may be sketched in the field, which will very much 
 assist the surveyor in drawing the plan. 
 
 With respect to the characters used to denote stations, 
 the letters of the alphabet will do very well, in small sur- 
 veys ; but in those of a large extent, nuijieral figures must 
 be used, and the sign -f (plus) placed before each figure ; 
 thus, -{- 1, or + 2, which may be read, station first, or 
 cross first, station one, or cross one, &c. Upon the plan 
 they are generally represented by this ( ) mark. 
 
 Note 1. Manr surveyors not only begin at tlie bottom of the field-book, but also 
 at the right-hand side, and write towards the left, which method I always Ibllow. 
 
 2. A straight fence may be denoted in the field-book, bj' placing S against the 
 line which represents the fence. (See the field-book to Plate III.) 
 
 3. In order to assist the memory in planning, a learner sometimes draws a rough 
 sketch of the field or estate he is about to measure, and upon it marks the station 
 in the same manner as they are put down in taking the survey. 
 
 Miscellaneous Instructions. 
 
 1. In addition to the chain, the surveyor must provide ten iron arrows, each 
 about 12 or 15 inches in length, and pointed at the bottom, to enter the ground 
 readily ; and also bent in a circular form at the top, for the convenience of holding 
 them. To the top of each arrow, a piece of red cloth should be attaclied. to make 
 it more conspicuous among the long grass, &c. Poles, likewise, generally called 
 " station staves," will be wanted as marks or objects of direction, each about 8 oi 
 10 feet in length, piked with iron at the bottom, and having a red or w^hite flag at 
 the top, that they may be better seen at a distance ; and also an offset staff, 10 or 
 12 links long, to which a cross may be fixed, as described in Prob.4, Part il. 
 
 2. In order to measure a line, let your assistant take nine arrows in his left hand, 
 and one end of the chain arid an arrow in his right ; then advancing tov/ards the 
 place to which he is directed, at the end of the chain let him put down the arrow 
 v.'hich he holds in his right hand. This the follower must take up with his chain- 
 hand, when he comes to it; the leader, at the same time, putting dovvu another at 
 the other end of the chain. In this manner he must proceed until he has put down 
 his tenth arrow ; then advancing a chain further, he must set his fcot upon the 
 end of the chain, and call out " change." The surveyor must then walk up to 
 him, if he have no oftsets to take ; and the arrows being carefully counted, one 
 must be put down at the end of the chain; then proceed as before, until the whole 
 line be measured. 
 
 3. Each change ought to be entered in the field-book, or a mistake of 10 chains 
 may happen, when the line is very long. 
 
 4. The follower may direct the' leader by the motion of his left hand ; moving it 
 to the right or left, as circumstances may require. 
 
 5. The arrows should always be put down perpendicularly, and in a right line 
 with the object of direction ; "otherwise the line will be made too long. 
 
 6. If the ditch be in the field which you are about to measure, both it and the 
 hedge usually belong to the adjoining field. In some places 5, and in others (i links, 
 from the roots of the quickwood, arc allowed for the breadth of the ditches be- 
 tween neiglibouring estates ; but for those ditches adjoinirig roads, commons, and 
 waste lands, 7 links are commonly allowed. 
 
 7. The ditches and fences must always be measured with the fields to which 
 they belong, when the whole quantity of land is required ; but in measuring crops 
 
£EGT. T. LAND-SUr.VEYING. 65 
 
 of corn, turnips, &c. only so much must bo measured as is, or has been occupied 
 by tho corn, &c. 
 
 8. When your view between two stations is obstructed by an intervening hill, 
 the line must be ranged as follows : Wit!i your assistant, go to the phice whence 
 you can distinctly see both stations ; and turning face to face, at the distance of 
 two or three chains, direct each otiier to the right or left, until you are both in a 
 right line with the stations ; then, either of you putting down a pole, the line will 
 be correctly found. 
 
 9. In surveying hill!/ ground, tho horizontal measure must always be returned, 
 except for paring, reaping, &c., in which cases the hypothenusarmeasure must 
 be given ; to obtain which, it is frequently necessary to divide a hill into various 
 figures. 
 
 10. In order to preserve the horizontal line, in ascending or descending a hill, it 
 •will be necessary to elevate the chain. If the inclination of the hill be great, the 
 chain must be elevated at several times. 
 
 King's Quadrant is adm.irably adapted for surveying hilly ground. (See my 
 Land- Surveying, page 172. ) 
 
 11. When a field is bounded by crooked fences, you must measure a line as near 
 to each as the angles or curves will permit ; in doing which, you must take the 
 oflfset perpendicularly to each corner or angle in the fence. Where the fence is 
 curved, those offsets must be so taken, that a right line drawn from the end of any 
 one to the end of the next, on each side, would neither exclude any part of the 
 land to be measured, nor include any of that which is adjacent. Terpendiculars 
 thus erected will divide the whole offset or piece of land contained between the 
 base-line and the fence into triangles and trapezoids. 
 
 12. Sometimes it is most convenient to measure a line on the outside of the 
 field, and upon it erect perpendiculars to the crooked fence. These are called 
 "insets ; " and the area thus included must be deducted from that of the whole 
 ligure. (See Ex.2, Prob. I.) 
 
 13. When the fences and ditches are to be measured with the field to which they 
 belong, it is generally most practicable to fix the stations within the fences, at a 
 little distance from the corners, and then to measure to the roots of the quick wood: 
 adding or subtracting 5 or 6 links, according to the custom of the place, for the 
 breadth of the ditch. (See Ex. 1, Prob. II.) 
 
 14. When the offsets are small, their places on the base-line may be determined 
 by laying the offset-staff at right angles across the chain; but when large, they 
 must'be found by the cross, and measured with the chain. 
 
 15. Any scale of equal parts may be used in laying down figures ; but an ivory 
 feathered-edged plotting-scale is most convenient ; as distances may be pricked off 
 by it, without using the compasses. 
 
 16. In laying down an offset, apply the scale to the base-line, and make a small 
 pencil dot at every place where a perpendicular must be erected ; then lay the 
 scale across the base, so that the line which is marked with oo, may coincide with 
 it, the edge of the scale at the same time touching one of the dots. From the dot, 
 by the edge of the scale, draw a line (which will be perpendicular to the base), and 
 upon it prick off the offset ; or it may be pricked off without drawing a line. 
 
 17. The base of each triangle and trapezoid, forming an offset, may be found by 
 subtracting the distances on the chain-line from each other. 
 
 18. In planning or laying down figures relating to surveying, the upper part of 
 the paper or book used should always, if possible, represent the north : then you 
 will have the east on your right hand, and the west on your left. 
 
 19. In taking dimensions, it is an admirable method to measure proof-linos, in 
 order that you may be able to confirm your survey, after it is planned. Some may 
 perhaps deem this tedious and superfluous ; but a person had certainly much better 
 be at the pains of discovering his own errors, than expose himself to ridicule, by 
 suffering them to be detected by some other surveyor.* 
 
 20. The line in which you have the misfortune to lose an arrow, must be re- 
 measured. 
 
 PROBLEM I. 
 
 To measure afield, of three sides. 
 
 At each corner of tlie field place a station-staff ; mea- 
 sure along the base till you arrive at tlie point from wiiich 
 you suppose a perpendicular will rise to the opposite 
 angle ; there plant your cross, and turn its index till the 
 mark at each end of the base can be seen tlirough one of 
 
66 
 
 la^"d-sur\t:yixg. 
 
 PART m. 
 
 the grooves ; then apply your eye to the other groove, and 
 if you see the mark at the opposite angle, you are in the 
 right place to measure the perpendicular ; if not, move 
 the instrument backward or forward, along the base, till 
 you can see the three marks as above directed. Enter in 
 your field-book the distance from the end of the base to 
 the cross ; measure the perpendicular, and then the re- 
 mainder of the base. 
 
 Or, measure the three sides, construct the figure, and 
 determine the length of the perpendicular by the scale ; 
 or the area may be found from the three sides, without 
 the perpendicular. 
 
 Note 1. Be especially careful, that in measuring the two parts of the base and 
 the perpendicular, no confusion of arrows take place. 
 
 2. Triangles, trapeziums, and trapezoids, may be constructed by Problems 7 
 and 8, Part I. ; and their areas found by the Rules given in Problems 4, 5, 7, and 
 8, Part II. 
 
 3. If a triangle be laid down by a scale of 1 chain or 2 chains to an inch, the 
 perpendicular may be ascertained with considerable accuracy. 
 
 4. If the examples in this Problem, or any of the following Problems, be thought 
 too few, more may be easily supplied by the teacher sketching fields, at pleasure, 
 with his pen, which the learner may measure by a scale. 
 
 This method will be found very advantageous ; as it will give the learner a good 
 idea in what manner he must take his dimensions, and enter his notes, when he 
 commences field-practice. 
 
 5. The expression, R. off B, or L. off B, &c. in the field-notes, signifies that you 
 are to turn to the right or left-hand, and measure from B, &c. 
 
 EXAMPLES. 
 
 1. Lay down a field, and find its area from the follow- 
 ing notes. 
 
 
 BC 
 
 
 
 
 846 
 
 
 64 
 
 600 
 
 
 85 
 
 400 
 
 
 50 
 
 200 
 
 
 
 
 000 
 R. offB. 
 
 
 
 AB 
 
 
 
 1253 
 
 
 
 1000 
 
 
 
 586 
 
 525 C. 
 
 Begin at 
 
 A, and 
 
 go TVest. 
 
 Per, on the left 
 
 Base line 
 
 Per. on t 
 
SECT. I. 
 
 LAND-SrRVEYJ2s"G. 
 
 67 
 
 CALCULATION. 
 
 Triangle ABC 
 
 1253 base AB. 
 
 525 perp. C«<. 
 
 2506 
 6265 
 657825 double area. 
 
 Offsets taken on the line BO. 
 
 50 
 
 2or/ 
 
 10000 
 
 85 
 
 64 
 
 l49 
 
 200 
 29800 
 
 10000" 
 
 27000 
 
 29800 
 
 15744 
 
 82544 
 
 Double areas of offsets 
 collected. 
 
 50 
 
 85 
 
 Sum. 
 
 135 
 
 200 
 27000 
 
 246 
 64 
 
 984 
 1476 
 15744 
 
 Double Areas. 
 
 657825 Trianirle ABC. 
 82544 Offsets. 
 2)740369 
 
 3.70184 
 
 4 
 
 2.80736 
 
 40 
 3 2.29440 Area 3a. 2r. Z2p. 
 
 2. Draw a plan of a field, and find its area from the fol- 
 lowing? notes. 
 
 
 CA 
 
 
 
 1252 
 
 37 
 
 1000 
 
 69 
 
 824 
 
 45 
 
 716 
 
 72 
 
 610 
 
 15 
 
 424 
 
 55 
 
 212 
 
 
 
 000 
 
 
 R. off C. 
 
LAK D - S UIl YETIXG. 
 
 PAKT ni. 
 
 
 AC 1 
 
 
 683 
 
 
 
 < 536 
 
 40 
 
 1 354 
 
 64 
 
 i 
 
 229 
 
 49 
 
 1 
 
 000 
 
 
 
 1 
 
 E. offB. 
 
 
 
 AB 
 
 
 
 
 973 
 
 
 48 
 
 745 
 
 
 76 
 
 600 
 
 
 66 
 
 495 
 
 
 25 
 
 256 
 
 
 
 
 000 
 
 
 Begin at 
 
 A, and 
 
 go N. East 
 
 A2sSlYEE. 
 
 Having constructed the figure, you will find the per- 
 pendicular falling from the angle B, upon the side CA, to 
 measiu-e 528 links. 
 
 Double areas. 
 
 661056 Triangle ABC. 
 
 68543 Offsets taken on the line AB. 
 
 95376 Ditto on CA. 
 824975 Sum. 
 
 50154 Insets on BC. 
 ?.j774821 Difference. 
 387410 =3<2. 3r. 19£p., the area required. 
 
 3. Find the area of a field from the following dimeii- 
 sionr. . 
 
SECT. I. 
 
 LAIs D-SURVE TING. 
 
 69 
 
 Fi 
 
 Easi 
 
 CALCULATION. 
 
 Here, (1946-i- 1197 + 143o;-f-2=4o78-r-2 = 22S9^1ialt 
 the smu of tlie sides; then, 2289-1946—343, the first 
 remainder; 2289- 1197 = 1092, the second remainder; 
 and 2289—1435 = 854, the third remainder; whence 
 ^/(2289 X 343 x 1092 x 854) = ^ 7^2184316136 = 855677 
 square links, = 8 «cre5, 1 roods, and 9 perches, the area 
 required. 
 
 By Logarithms. 
 
 The log. of . . . 2239 
 
 ... 343 
 
 . . . 1092 
 
 ■ . ... 854 
 
 = 3.3596458 
 = 2.5352941 
 = 3.0382226 
 = 2.9314579 
 
 Divide the sum of the log. by 2)11.8646204 
 
 The quotient is the log. of 855677, the area "5.9323102 
 
 4. Lay dov/n a field, and find its area from the follow- 
 
 ing notes. 
 
 
 DC 
 
 
 412 
 
 Return 
 
 toD. 
 
 
 GA 
 
 
 
 638 
 
 46 
 
 465 
 
 32 
 
 246 
 
 
 
 000 
 
 
 E. ofi- C. 
 
 
 BC 
 
 
 
 462 
 
 45 
 
 300 
 
 32 
 
 150 
 
 
 
 000 
 
 
 R. oitB. 
 
 proof-line. 
 
70 
 
 LA2sD-SURV£YING. 
 
 PAKT III. 
 
 
 AB 
 
 
 
 725 
 
 32 
 
 566 
 
 40 
 
 400 
 
 25 
 
 250 
 
 
 
 000 
 
 ^gin at 
 
 A, and 
 
 D, station for a proof-iine. 
 iro West. 
 
 ANSWER. 
 
 Having constructed the figure, you will find the proof- 
 line DC to measure 412 links, as in the field-book ; hence, 
 you may conclude there is no error committed in taking 
 or setting down the dimensions. You will also find the 
 perpendicular falling from the angle C, upon the ba^e 
 AB, to measure 402 links ; from which, and the offsets, 
 we obtain the area = 1«. Sr. 25p. 
 
 PROBLEM IL 
 
 To measure a field of four sides. 
 
 Measure a diagonal, and find the perpendiculars falling 
 upon it, from the opposite angles, as already directed. 
 Or, if you measure a diagonal and the four sides, the 
 figure may be constructed, and the perpendiculars deter- 
 mined by the scale. 
 
 Note 1. Always make choice of the longer diagonal, because the longer the base- 
 line of a triangle, the more obtuse is its subtending angle; consequently the per- 
 pendicular will be shorter, and its place more easily and accurately determined. If 
 you measure the four sides and both the diagonals, one of them will serve as a 
 proof-line, after you have laid down the figure. 
 
 2. If neither of the diagonals can be measured, in consequence of obstructions, 
 you must measure tie-lines across the angles of the field, which will enable you to 
 plan it : and hence the diagonal and perpendiculars may be found by the scale. 
 
 3. When two perpendiculars cannot be taken upon either of the diagonals, such 
 fields must be divided into two triangles, or into two right-angled triangles and a 
 trapezoid. 
 
 4. Unskilful surveyors affect to reduce trapeziums into rectangles, by adding each 
 two opposite sides together, and taking half their sum respectively for a mean 
 length and breadth; but this method leads to very erroneous results. (See my 
 Land- Surveying, page 124.) 
 
 EXAMPLES. 
 
 1. Draw a plan of a field, and find its area from the 
 following; notes. 
 
SECT. I. 
 
 LAND-SURVEYING. 
 
 To 
 
 the 
 
 ! Fence 
 
 48 
 
 660 
 
 
 53 
 
 630 
 
 to A. 
 
 40 
 
 500 
 
 
 25 
 
 380 
 
 
 50 
 
 200 
 
 
 62 
 
 000 
 
 
 From 
 
 Ago 
 
 Soutii. 
 
 To 
 
 the 
 
 Fence. 
 
 40 
 
 1012 
 
 
 42 
 
 950 
 
 toD. 
 
 54 
 
 600 
 
 
 33 
 
 400 
 
 
 12 
 
 260 
 
 
 30 
 
 150 
 
 
 65 
 
 000 
 
 
 From 
 
 C, go 
 
 East. 
 
 To 
 
 the 
 
 Fence. 
 
 25 
 
 615 
 
 
 35 
 
 550 
 
 to C. 
 
 30 
 
 400 
 
 
 10 
 
 240 
 
 
 22 
 
 150 
 
 
 45 
 
 000 
 
 
 From 
 
 B, go 
 
 North, 
 
 To 
 
 the 
 
 Fence. 
 
 33 
 
 1090 
 
 
 45 
 
 1045 
 
 toB. 
 
 56 
 
 800 
 
 
 40 
 
 500 
 
 
 48 
 
 300 
 
 
 30 
 
 000 
 
 
 ►egin at . 
 
 A., and go 
 
 Wec't. 
 
72 
 
 L AND- SURYE YrN'G- 
 
 PART UL 
 
 CALCULATION. 
 
 Trapeziu?n ABCD. Offsets ialicn on the line AB. 
 
 5201 
 
 495 j P^^i^- 
 1015 sum. 
 1155 diag. 
 5075 
 5075 
 1015 
 1015 
 1172325 double area. 
 
 30 
 
 48 
 
 78 
 
 300 
 
 23400 
 
 "48"" 
 
 40 
 
 88 
 
 200 
 17600 
 
 56 
 
 45 
 101 
 
 245 
 505 
 404 
 202 
 
 2"4745 
 
 40 
 
 dQ 
 
 96 
 
 300 
 
 28800 
 
 Offsets taken on the line AB. 
 
 23400 
 17600 
 28800 
 
 45 
 33 
 
 78 
 
 45 
 
 390 
 
 312 
 
 3510 
 
 24745 
 3510 
 98055 sum. 
 
 Double areas 
 collected. 
 
 45 
 22 
 67 
 
 150 
 3350 
 67 
 10050 
 
 10 
 32 
 90 
 
 2880 
 
 Offsets taken on the line BC. 
 
 30 
 35 
 
 150 
 3250 
 65 
 
 9750 
 
 "35 
 
 25 
 
 60 
 
 65 
 
 300 
 
 360 
 
 3900 
 
 10050 
 
 2880 
 
 6400 
 
 9750 
 
 390 
 32980 sum. 
 
 Double areas 
 collected. 
 
 10 
 30 
 
 40 
 160 
 
 2400 
 40 
 
 e'4oo 
 
ECT. I. 
 
 LAND- 
 
 ■SURVEYING. 
 
 
 
 Offsets taJu 
 
 '71 on 
 
 the line CD. 
 
 65 
 
 30 
 
 
 142501 
 
 
 30 
 
 12 
 
 
 4620 
 
 
 95 
 
 42 
 
 
 6300 
 
 Double areas 
 
 150 
 
 110 
 
 
 17400 
 
 collected. 
 
 4750 
 95 
 14250 
 
 420 
 42 
 4620 
 
 54 
 
 
 33600 
 
 5084 
 
 81254 
 
 sum. 
 
 12 
 
 
 33 
 
 42 
 
 
 
 45 
 
 96 
 
 
 
 140 
 
 350 
 
 
 
 1800 
 
 '4800 
 
 
 
 45 
 
 288 
 
 
 
 6300 
 
 33600 
 
 
 
 
 =rr^. 
 
 
 
 33 
 
 42 
 
 
 
 54 
 
 40 
 
 
 
 87 
 
 82 
 
 
 
 200 
 
 62 
 
 
 
 17400 
 
 164 
 
 
 
 
 492 
 
 
 
 
 73 
 
 5084 
 
 Offsets taken on the Ime DA, 
 
 53 
 _48 
 101 
 30 
 
 62 
 
 25 
 
 50 
 
 40 
 
 112 
 
 65 
 
 200 
 
 120 
 
 22400 
 
 1300 
 
 50 
 25 
 75 
 
 65 
 7800 
 
 40 
 
 180 
 
 53 
 
 6000 
 
 93 
 
 75 
 
 130 
 
 13500 
 
 2790 
 
 
 93 
 
 
 12090 
 
 8030 
 
 22400 
 13500 
 
 7800 
 12090 
 
 3030 
 58820 sum. 
 
 Double areas 
 collected. 
 
74 
 
 LAND-SURVEYING 
 
 PART HL 
 
 1172325 
 980.55 
 32980 
 81254 
 58820. 
 
 Wliole double areas 
 collected. 
 
 2) 1443434 sum. 
 7.21717 
 
 • 4 
 
 .86868 
 40 
 
 34.74720Area 7a. Or. 34|p. 
 
 To find the area of the foregoing figure by the method 
 generally called " Casting^ 
 Having constructed the figure, and taken out the chain- 
 lines, draw with a pencil by 
 the straight edge of a clear 
 piece of lantern-horn, the four 
 right lines AB, BC, CD, and 
 DA, in such a manner, that 
 the parts included may be 
 equal to those excluded, as 
 nearly as you can judge by 
 your eye; then the diagonal 
 AC will be found to measure 1274, and the perpendicu- 
 lars B«, 550, and J)a, 583 links. 
 
 Note. The use of the Parallel Ruler, in straightening crooked fences, :? given in 
 the Author's Land-Snrveying, in twelve new problems, comprising every case that 
 can possibly occur in practice. 
 
 CALCULATION. 
 
 550 I 
 
 _583P^^'P- 
 1133 sum. 
 1274 diag.. 
 4532 
 7931 
 2266 
 1133 
 
 2)1443442 
 7.21721 
 
 4 
 
 .86884 
 
 40 
 
 34.75360 Ans. 7a. Or. 34 3p. 
 
SECT. I. 
 
 land-suk\t:ying. 
 
 76 
 
 Noie. Although the method cf finding the area by casting is adopted by most 
 practical surveyors, it is certainly less correct than that by offsets, &c. ; a learner, 
 therefore, ought to practise both, until he can habitually come very near to the 
 truth by the former. 
 
 2. Lay down a field, and find its area frara the follow- 
 ing: dimensions. 
 
 
 
 953 E. 
 1142 C. 
 
 
 West. 
 
 
 AF 
 
 
 2718 
 
 D 
 
 1865 
 
 B 
 
 575 
 
 
 000 
 
 m 
 
 A, go 
 
 Area 20a. Sr. 17 ^p. 
 
 3. Required the jDlan and area of a field from the fol- 
 lowing!: notes. 
 
 
 CB 
 
 
 
 1646 
 
 Diao:. 
 
 
 720 
 
 385 D 
 
 
 L. off C. 
 
 
 
 AC 
 
 Base. 
 
 
 2030 
 
 
 B 850 
 
 620 
 
 
 From 
 
 A, go 
 
 East. 
 
 Area 11a. 3r. 7 p. 
 
 4. Draw a plan, and find the area of a field, from the 
 foUowins: dimensions. 
 
 Return 
 
 Dias:. 
 
 Dial 
 
76 
 
 LAND - SURVEYING. 
 
 TAUT m 
 
 
 ~ m^ 
 
 
 
 1244 
 
 47 
 
 1000 
 
 70 
 
 800 
 
 85 
 
 600 
 
 68 
 
 400 
 
 30 
 
 200 
 
 
 
 000 
 
 
 R.offC. 
 
 
 EC 
 
 
 
 720 
 
 85 
 
 650 
 
 112 
 
 550 
 
 88 
 
 450 
 
 
 360 
 
 
 300 
 
 
 ■ 200 
 
 
 100 
 
 
 000 
 
 
 R. off B. 
 
 
 AB 
 
 
 1350 
 
 
 1000 
 
 Begin at 
 
 A, and 
 
 Cross tlie tence. 
 
 83 
 
 130 
 
 100 
 
 
 
 sro West. 
 
 ANSWER. 
 
 Having constructed the figure by means of the four sides 
 iind the diagonal AG, you will tind the diagonal BD to 
 measure 1460 links, as in the field-book; hence. It may be 
 concluded that no error lias been made In taking or enter- 
 ing the dimensions. You will also find the perpendicular 
 from the angle A, upon the diagonal BD, to measure 618 
 links, and the perpendicular from the angle C, upon the 
 same diagonal, 613 links; hence the area is 9a. 2?\ lljo. 
 
 PROBLEM in. 
 
 To measure afield of more than four sides. 
 
 Any piece of land, consisting of more than four sides, 
 may be surveyed by reducing it into triangles and tra- 
 peziums ; thus, a field of five sides may be reduced into a 
 triangle and a trapezium ; of six sides, into two trape- 
 
SECT. I. 
 
 LAND-SUIIVEYLNG. 
 
 77 
 
 ziums ; of seven, into two trapeziums and a triangle ; of 
 eight, into three trapeziums, &c. 
 
 EXAMPLES. 
 
 1. Required the plan a^id area of a field, from the fol- 
 lowing dimensions. 
 
 ^ B'iag. 
 
 168 B. r« —h 
 
 
 CE 
 
 
 1165 
 
 
 912 
 
 D 308 
 
 415 
 
 
 RofFC. 
 
 
 AC 
 
 
 1195 
 
 
 293 
 
 From 
 
 A, go 
 
 Base. 
 
 233 B. 
 
 West. 
 
 U 
 
 \. 
 
 CAXCULATION. 
 
 Triangle ABC. 
 1195 base. 
 233 perp. 
 3585 
 3585 
 2390 
 
 278435 
 
 Trapezium BCDE. 
 3081 
 
 476 sum. 
 1165 diag. 
 23^0 
 2856 
 476 
 476 
 554540 
 
 Double Areas. 
 
 278435 
 
 554540 
 
 2 )832975 
 
 4.16487 
 
 4 
 
 .65948 
 40 
 26.37920 Area 4a. Or. 2Qp. 
 
 2. Lay down a field, and find its area from the follow- 
 ing: notes. 
 
 E. 562 
 
 £ 3 
 
LAXD-STHiVP^YING. 
 
 PAKT in. 
 
 
 AD • 
 
 
 2243 
 
 C728 
 
 1785 
 
 
 560 
 
 From 
 
 A, go 
 
 Diag. 
 
 624 B. 
 N. ^Vest. 
 Area 2Qa. Or. Sl^p. 
 
 3. Required the plan and area of a field from tlie fol- 
 lowino; dimensions. 
 
 
 CE 
 
 
 5126 
 
 
 5000 
 
 
 4040 
 
 
 4000 
 
 
 3000 
 
 
 2000 
 
 D 1695 
 
 1720 
 
 
 1000 
 
 
 R. off C. 
 
 
 AC 
 
 
 4300 
 
 
 4000 
 
 
 3000 
 
 
 2000 
 
 B 1265 
 
 1930 
 
 
 1000 
 
 From 
 
 A, go 
 
 Diao:. 
 
 Diagf. 
 
 West. 
 Area 108a. 3r. 12\p. 
 
 4. Draw a plan of a field, and find its area from the fo)- 
 lowins" notes. 
 
 BD 
 
 1238 
 
 1000 
 
 Return to 
 
 Diasr. 
 
SECT. I. 
 
 LAND-S U RVE Y1N(J. 
 
 79 
 
 DE 
 
 1500 
 
 1000 
 
 R. oif D. 
 
 Dias. 
 
 Begin at 
 
 FD 
 
 1160 
 
 1000 j 
 
 KoffF.i 
 
 EF 
 
 1260 
 
 1000 
 
 R. offE. 
 
 CE 
 
 1080 
 
 R. off C. 
 
 AC 
 
 2100 
 2000 
 1000 
 
 R. offA 
 
 DA 
 
 900 
 R. off D 
 
 CD 
 
 1340 
 
 1000 
 
 R. off C. 
 
 BC 
 
 1000 
 R. off B. 
 
 AB 
 
 1560 
 1000 
 
 A, and 
 
 Diag, 
 
 Diaff. 
 
 go West. 
 
 AJS'SWER. 
 
 Having laid down the figure, you will find the perpen- 
 diculars of the trapezium ABCD = 708 and 390, and of 
 DCEF=680 and 808 links ; hence the area of the field is 
 25 acreSy 2 roods, and 26^ perches. 
 E 4 
 
so 
 
 LAND-SURVEYIXG. 
 
 PART m. 
 
 PROBLEM IV. 
 
 To measure an irregular, narrow 'piece of land. 
 
 Divide the ground to be surveyed into triangles and 
 trapezoids, by measuring a base-line, in a convenient posi- 
 tion, and upon it erecting perpendiculars to the boundaries, 
 on each side. 
 
 'Noic\. The general method of measuring a few lands or ridges together, in 
 common fields, &c. is by taking several breadths, dividing their sum by the number 
 for a mean breadth, and multiplying this by the length, for the areaj but it is much 
 more accurate to note the place of e^ch breadth upon the base or chain-line, and 
 find the area by the rule for trapezoids. 
 
 2. In order to obtain the breadths correctly, you must measure directly across 
 the lands. 
 
 3. If the lands be curved, or longer on one side than on the other, by measuring 
 along the middle, you will obtain the mean length. 
 
 4. Paring, reaping, &c. should always be surveyed with a slack chain, in order to 
 obtain the measurement of the surface. 
 
 EXA3CPLES. 
 
 1. Draw a plan of an irregular piece of land, and find 
 
 its area from the following dimensions. 
 
 126 
 
 
 AB 
 
 
 
 1314 
 
 234 
 
 1005 
 
 
 980 
 
 
 785 
 
 312 
 
 700 
 
 
 ^^0 
 
 215 
 
 460 
 
 
 335 
 
 336 
 
 260 
 
 360 
 
 000 
 
 From 
 
 A, go 
 
 52 
 125 
 
 152 
 
 100 
 
 232 
 East. 
 
 ANSWER. 
 
 Double Areas. 
 
 324337 offsets on the right. 
 656476 ditto on the left. 
 2)980813 sum. 
 
 490406=4a. 3r. 244rp. the area required. 
 
 2. Lay down a field, and find its area from the foIlo\v 
 ing notes. 
 
SECT. I. 
 
 l.\:nd-suiiyeying. 
 
 81 
 
 
 BC 
 
 
 
 526 
 
 65 
 
 400 
 
 92 
 
 300 
 
 94 
 
 200 
 
 ^'0 
 t O 
 
 100 
 
 
 
 000 
 
 
 R.offB. 
 
 
 AB 
 
 
 
 1235 
 
 55 
 
 1100 
 
 112 
 
 900 
 
 175 
 
 650 
 
 205 
 
 500 
 
 230 
 
 350 
 
 250 
 
 200 
 
 266 
 
 000 
 
 From 
 
 A, go 
 
 526 to C. 
 
 435 
 
 328 
 
 250 i 
 
 224 
 
 245 
 
 280 
 
 365 
 
 North. 
 
 Area 6a. Ir. 2p. 
 
 3. Find the area of five lands, from the following di- 
 mensions. 
 
 2436 
 
 205 
 
 2000 
 
 210 
 
 1700 
 
 214 
 
 1400 
 
 218 
 
 1000 
 
 220 
 
 700 
 
 215 
 
 400 
 
 209 
 
 000 
 
 205 
 
 Area 5a. Or. 29p. 
 
 4. Required the area of six lands, from the following 
 notes. 
 
 3365 
 
 199 
 
 3000 
 
 
 2500 
 
 214 
 
 2000 
 
 
 1800 
 
 236 
 
 1000 
 
 
 800 
 
 201 
 
 000 
 
 175 
 
 
 Area 7a. Or. 8/7 
 
 E 
 
82 
 
 LAND-SURVEYING. 
 
 PART ni. 
 
 l^ote. The sum of the breadths in the last example is 1025, which being divdtd 
 by 5, their number, we obtain 205 for the mean breadth ; then 33G5 multiplied by- 
 SOS is equal to 6s9S25 square links, which are equal to 6 acres, 3 roods, 24 perches, 
 the area, which is too little by 24 perches, proving the inaccuracy of the common 
 method. , , j -n 
 
 If a piece of ground be narrowest about the middle, the common method will 
 give the area too much. 
 
 PROBLEM V. 
 To measure a mere or wood. 
 
 By the lielp of your cross, fix four station-staves on the 
 outside of the mere or wood, so as to form a rectangle ; 
 then measure each of its sides, taking insets to the boun- 
 daries, which must be treated as before directed. 
 
 It is, however, a more expeditious method than the 
 former, to measure the four sides of a quadi'ilateral figure, 
 having one right-angle ; then construct the figure, and 
 measure the diagonal and perpendiculars by the scale. 
 
 Note 1. If a mere or wood be of a triangular shape, its area may be found by 
 measuring the three sides of a circumscribing triangle; and proceedmg as 
 
 2. By this Problem you may survey fields into which you are not permitted to 
 enter, or which contain obstructions. 
 
 EXA3IPLE. 
 
 Let the following figure represent a wood, the area of 
 which is required. 
 
 Set up your cross at station A, and let your 
 fix the marks B and 
 D, so that the angle 
 at A may be a right 
 angle ; and measure 
 the line AB, taking 
 insets to the fence, as 
 you proceed. Then 
 fix the mark C, as 
 most convenient ; 
 measure the other 
 three lines, and you 
 will find in your field- 
 book the following 
 dimensions. 
 
 issistanl 
 
 
 DA 1 
 
 
 1550 
 
 60 
 
 1440 
 
 50 
 
 1200 
 
 
 1000 
 
 
 
 900 
 
 
 L. off" D. 
 
SECT. I. 
 
 LAND-SURVEYING. 
 
 83 
 
 
 CD 
 
 
 
 950 
 
 120 
 
 500 
 
 
 
 000 
 
 
 L. off C. 
 
 
 BC— ' 
 
 
 1340 
 
 
 
 1050 
 
 
 1000 
 
 50 
 
 700 
 
 60 
 
 400 
 
 
 
 000 
 
 
 L. oif B. 
 
 
 AB 
 
 
 
 1150 
 
 
 1000 
 
 50 
 
 900 
 
 
 
 550 
 
 100 
 
 300 
 
 110 
 
 160 
 
 
 
 000 
 
 Begin at 
 
 A, and 
 
 2;o Nortli. 
 
 ANSWER* 
 
 Having constructed the figure, you will find the diago- 
 nal BD to measure 1930, the perpendicular Aa 923, and 
 Ca 605 links ; hence the area of the ivood is 12a. 3r. 20p. 
 
 PEOBLEM YI. 
 
 To measure and plan roads, rivers, canals, ^c. 
 
 Li measuring roads, rivers, or canals, angles or tie-lines 
 must be taken at the different turns, in order to lay down 
 the chain-lines; and offsets must be taken to the boundaries, 
 as you proceed, to enable you to di^aw the plan. 
 
 Note 1. The length of a road is generally returned either in miles, furlongs, and 
 poles, or else in miles and yards. 
 
 2. A machine called a " Perambulator " is sometimes used to ascertain the 
 lengths of roads. It has a wheel of 8 feet 3 inches, or half a pole in circumference, 
 which being made to pass over the Rround, puts m motion the clockwork within ; 
 and the distance measured is pointed out by an index on the outside. 
 
 This instrument is much more expeditious for measuring the length of a road 
 than the chain ; but it is certainly less correct : for by the wheel passing over 
 stones, sinking into holes, &c. the' distance is made to appear more than it is La 
 reality. 
 
 E 6 
 
84 
 
 LiJND-SURVEYIN'G. 
 
 PART III. 
 
 EXAiEPLES. 
 
 1. Let the following figure represent a serpentine road, 
 a plan of which is required. 
 
 Begin at + 1, and measure to + 3, taking offsets on 
 both sides, as you proceed. Return to + 2, and measure 
 to ^- 4, from >yhich run a line to + 3, which will tie the 
 first and second lines. Return to + 4, and continue the 
 line to + 6. From -f 6, proceed as before, until you ar- 
 rive at + 14 ; and you will have obtained the following 
 dimensions, from which a plan may be drawn. 
 
SECT. I. 
 
 LAND-SURYEYDnG. 
 
 85 
 
 
 to -f 14. 
 
 
 58 
 
 350 
 
 60 
 
 68 
 
 200 
 
 44 
 
 
 150 
 
 + 13 is 184 from + 11 
 
 50 
 
 100 
 
 80 
 
 Go from 
 
 + 12, 
 
 Line 5. 
 
 30 
 
 720 
 
 to + 12. 
 
 70 
 
 650 
 
 
 
 600 
 
 + 11. 
 
 86 
 
 550 
 
 33 
 
 70 
 
 300 
 
 50 
 
 S is 200 from +10 
 
 200 
 
 
 120 
 
 135 
 
 Cross-fence. 
 
 Go from 
 
 + 9, 
 
 Line 4. 
 
 
 700 
 
 to + 9. 
 
 
 600 
 
 Cross-fence. 
 
 
 500 
 
 + 8. 
 
 38 
 
 480 
 
 84 
 
 40 
 
 300 
 
 60 
 
 52 
 
 180 
 
 65 
 
 
 150 
 
 + 7 is 160 from + 5. 
 
 50 
 
 100 
 
 
 Go from 
 
 + 6, 
 
 Line 3. 
 
 20 
 
 512 
 
 to + 6. 
 
 50 
 
 450 
 
 
 52 
 
 380 
 
 70 
 
 
 350 
 
 + 5. 
 
 20 
 
 300 
 
 80 
 
 
 "200 
 
 + 4 is 232 from + 3. 
 
 18 
 
 100 
 
 93 
 
 Go from 
 
 + 2, 
 
 Line 2. 
 
 
 600 
 
 to + 3. 
 
 
 480 
 
 Cross-fence. 
 
 
 400 
 
 4-2. 
 
 38 
 
 350 
 
 95 
 
 15 
 
 300 
 
 
 28 
 
 200 
 
 80 
 
 55 
 
 000 
 
 70 
 
 Begin 
 
 at+1, 
 
 Line 1. 
 
 2. Let the foregoing figure represent a river, a plan of 
 wliicli is required. 
 
86 LAXD-SUr.VEYING. PART m. 
 
 Begin at cz, and measure to c; taking offsets to the 
 river's edge, as you proceed. From c measure to d ; and 
 there take the tie or chord-line dh, which will enable you 
 to lay down the first and second lines. Continue the second 
 line to 71 ; and from m, measure to r, at which place take 
 the tie-line rn ; and thus proceed until you come to the 
 end of your survey at x. 
 
 If the breadth of the river be everyivhere nearly the 
 same, its breadth taken in different places, by the next 
 Problem, will suffice ; but if it be very irregular, dimen- 
 sions must be taken on both sides, as above. 
 
 Wlien the area is required, it must be found from the 
 plan, by dividing the river into several parts ; and taking 
 the necessary dimensions by the scale. 
 
 Ao/e 1. Any bog, marsh, mere, or wood, whatever may be its number of sides, 
 may be measured by this Problem. 
 
 2. In taking an angle with the chain, some surveyors never measure more than 
 100 links from the angular point ; but when the lines, including the angle, are of a 
 considerable length, it is much better to measure the chord-line at a greater dis- 
 tance from the angle, than one chain ; because a small inaccuracy in constructing 
 the figure, when the angular distance is short, will throw the lines, when far pro- 
 duced, considerably out of their true position. It sometimes happens, however, in 
 consequence of obstructions, that it is impossible to measure the chord-line at a 
 greater distance from the angular point, than one chain. In such cases, multiply 
 both the angular distance and chord-line by 2, 3, 4, or any larger number, as 
 circumstances may require ; and use the products resulting, in laying down the 
 angle. Or, which is the same thing, lay down the angle by a scale 2 or 3 times as 
 large as that by which you intend to draw the plan. By this means the radius or 
 angular distance be equal cd, and the chord-line bd, will be increased in length ; 
 and consequently the lines ac and cw, including the angle, will be more correctly 
 laid down in position. 
 
 PROBLEM VII. 
 
 To find the breadth of a river. 
 
 EXAIMPLE. 
 
 Let the following figure represent a river, the breadth 
 of which is required. 
 
 Fix upon any 
 object B, close b 
 
 by the edge of 
 the river, on the 
 side opposite to 
 which you stand. 
 By the help of 
 your cross, make 
 AD perpendicu- 
 lar to AB ; also make AC = CD, and erect the perpen- 
 dicular DE ; and when you have arrived at the point E, 
 in a direct line with CB, the distance DE vrill be = AB, 
 
lEGT. I. 
 
 LAND-SUKVEYING. 
 
 87 
 
 the breadth of the river ; for by Theo. 1, Part I., the angle 
 ACB = DCE, and as ACr=CD, and the angles A and 
 D right angles, it is evident that the triangles ABC, 
 CDE, are not only similar but equal. 
 
 'Note 1. The distance between A and the edge of the river must be deducted from 
 DE, when it is not convenient to fix A close by the river's edge. 
 
 2. This Problem may also be well aijplied in measuring the distance of any inac- 
 cessible object: forlet'AC equal 8. CD equal 2, and DE equal 10 chains ; then, by 
 similar trihngles, as CD : DE : : AC : AB, equal to -10 chains. 
 
 PROBLEM VIII. 
 
 To measure a line upon 'which there is an impediment, 
 
 EXA3IPLE. 
 
 Let CDEF represent the base of a building, through 
 wliich it is necessary that a straight j> 
 
 line should pass from A. 
 
 Measure from A to m ; at m, erect 
 the perpendicular 7na, which measure 
 until you are clear of the impediment, 
 as at c. Erect the perpendicular ce, 
 which measure until you are beyond 
 the building, as at b. Erect the per- 
 pendicular bd ; and make bii equal to - 
 7nc, at which point you will be in a 
 direct line with mA. Erect the per- 
 pendicular wB, which measure ; then 
 A.m added to the sum of cb and nB, 
 will give the length of the whole line 
 AB. ^ 
 
 Note 1 . This Problem is very useful when you meet with ponds, bogs, buildings 
 &c. upon a chain-line. 
 
 2. When a cross is not at hand, a perpendicular may be erected by the chain, in 
 the following manner : Measure 40 links from in to r ; and let one end of the chain 
 be kept, fast at r, and the eightieth link at m ; take hold of the fiftieth link, and 
 stretch the chain so that the two parts tow, and rii;, maybe equally tight ; then 
 will 7)nii be perpendicular to mr ; for niv), nir, and j'w, are in proportion to each 
 other as 3, 4, and 5. -(See Problem 29, Part I.) 
 
 ^--J 
 
 PROBLEM IX. 
 
 To part from a rectangular-field, any proposed quantity 
 of land, by a line parallel to one of its sides. 
 
 KULE. 
 
 Divide the proposed area, in square links, by the side 
 upon which it is to be parted off, and the quotient will 
 be the length of the other side of the figure required. 
 
88 
 
 LAI\'D-SURVEYLN-G. 
 
 PART in. 
 
 Note 1. If there be offsets nn the line upon which the proposed quantity is to be 
 parted off, deduct the area of the offsets from it, and proceed with the remainder as 
 above. 
 
 2. Acres, roods, and perches may be reduced into square links, by multiplying 
 the whole quantity, in perches, by 625, the number of square links in a perch. 
 
 EXA3IPLES. 
 
 1. From the rectangle ABCD part off 2 acres, 1 rood, 
 and 32 perches, by a line parallel to 
 AD = 700 Hnks. 
 
 Here 2a. In 2>2p.=Z92p., and 392 
 
 X 625 = 245000 square links ; then 
 
 245000 2450 or.^ v i k^ 
 
 ^ = -y- = 350 links = AE =: 
 
 DF. 
 
 2. Part off 2a. 3r. Z2p. upon the chain-line AB, so that 
 the offsets taken on that line may be included. 
 
 AB 
 
 
 1200 
 
 
 
 1000 
 
 
 900 
 
 60 
 
 600 
 
 80 
 
 300 
 
 50 
 
 000 
 
 
 
 A, go 
 
 East. 
 
 Froi 
 
 2a. 3r. 32j9.=295000 square links. 
 
 57000 the area of the offsets. 
 12,00) 2380700 the difference. 
 
 198,4 links=AE=BF. 
 
 Hence the irregular figure AGBFE, contains 2a. 2>r. 
 32p. 
 
 Note. When it is required to part from a trapezium, approaching very nearly to 
 a rectangle, any number of acres, &c. by a line parallel to one of its sides ; it may 
 be done by this Problem. 
 
 PROBLEM X. 
 
 To 'part from a trapezium or any irregular polygon what- 
 ever^ any proposed quantity of land, by a line drawn 
 parallel to any of the sides, or by a line drawn from 
 any of the angles, or from a7ty assigned point i?i one of 
 the sides, to any of the opposite sides. 
 
 RULE. 
 
 Having surveyed and laid down the whole figure, draw 
 a guess-line in the direction required, parting off, as nearly 
 
SECT. I. 
 
 LiV2sD-SURVEYlNG. 
 
 89 
 
 as can be judged, the proposed quantity; after which, by 
 the scale, measure with the greatest accuracy, the guess- 
 line, and also the quantity thus parted-ofF. Then, if the 
 guess-line or line of division be drawn from an angle, or 
 from any assigned point in a side, divide the difference 
 between the proposed quantity and the quantity parted off, 
 by half the guess-line ; and the quotient will be the per- 
 pendicular to be set off, on one side, or the other, of the 
 guess-line, accordingly as the quantity parted off is more 
 or less than the quantity proposed. To the end of this 
 perpendicular, from the point assigned, draw a new line 
 of division ; and it will part off the quantity required. 
 
 But if the guess-line be drawn parallel to any of tlie 
 sides, divide the difference before mentioned, by the whole 
 guess-line, and the quotient will be the perpendicular to 
 be set off from each end of the guess-line, on one side, or 
 the other, as above. 
 
 EXAMPLES. 
 
 1. From a trapezium, whose dimensions are contained 
 in the following notes, part off 2a. 2r. 2ip. by a line 
 parallel to the side AB. 
 
 Return 
 
 oegin 
 
 BD 
 
 
 1249 
 
 Diag 
 
 1000 
 
 
 toB.. 
 
 
 AC 
 
 
 1112 
 
 Diag 
 
 1000 
 
 
 R. off A. 
 
 
 DA 
 
 
 550 
 
 
 R. offD. 
 
 
 CD 
 
 
 979 
 
 
 R. off C. 
 
 
 BC 
 
 
 557 
 
 
 R. offB. 
 
 
 AB 
 
 
 1114 
 
 
 1000 
 
 
 at A, 
 
 and go West 
 
90 
 
 LANU-SUIIV E YIN'G. 
 
 PAKT m. 
 
 Having laid down the figure, 
 draw the guess-line m?z parallel 
 to AB ; and from n, let fall 
 the perpendicular an; then 
 suppose mn = 1058 links, a7i 
 willbe=:230, and Aa=1052 ; 
 therefore Ba = 1114 - 1052 
 =: 62 links. 
 
 /: 
 
 J 
 
 Square links. 
 Then, 1055 x230-242650(*^^ ^^'"^ f *^^ trapezoid 
 w,oQfthe area of the triano;le 
 
 And 230x31 
 
 The sum is =249780 1 
 2a. 2r. 24j9. =265^00 
 
 of 
 Ba?i. 
 
 the area of the trapezium 
 AB^^??^. 
 
 15220 the diiference between the 
 quantity proposed, and the quantity parted oif by the 
 guess-Kne, which being divided by 1058, we obtain 14.4 
 links, to be set off perpendicularly from 7n and n towards 
 D and C. Hence EF is the true line of division ; and the 
 trapezium ABEF contains 2a. 2r. 24p. 
 
 As A is very nearly a right angle, measure, in the field, 
 230 + 14.4 = 244.4 links, from A to F ; and upon any 
 part of the line AB (towards B) as at e, erect the perpen- 
 dicular er, which make = 244'4 links ; stake out the line 
 ErF, and the work will be completed. 
 
 Nntc. In order to prove the operation, find the area DCEF ; then if it be equal 
 to the ditTerence between the area of ABCD and the quantity parted oft', the work. 
 is undoubtedly right. 
 
 2. From any irregular field, whose dimensions are con- 
 tained in the following notes, part off 2a. 3r. 20p. tov.^ards 
 the line AE, by a fence made from the angle D to the 
 side AB. 
 
 EB 
 
 1398 
 
 1000 
 
 Pv. off E. 
 
 CE^ 
 
 12 
 
 10 
 
 5 
 
 11. off C. 
 
 Diasr. 
 
 Diaix. 
 
 ?7^, proof-line, goes to D, 
 and measures 324. 
 
SECT. I. 
 
 LAND-SURVEYING. 
 
 
 AC 
 
 
 1260 
 
 
 1000 
 
 
 R. offA. 
 
 
 EA 
 
 
 
 400 
 
 80 
 
 200 
 
 
 
 000 
 
 
 R. off E. 
 
 
 DE 
 
 
 
 600 
 
 25 
 
 450 1 
 
 So 
 
 300 
 
 20 
 
 150 
 
 
 
 000 
 
 
 R. off D. 
 
 
 CD 
 
 
 740 
 
 
 R. offC. 
 
 
 BC 
 
 
 550 
 
 
 R. offB. 
 
 
 AB 
 
 
 
 1250 
 
 35 
 
 1000 
 
 50 
 
 800 
 
 60 
 
 600 
 
 50 
 
 400 
 
 30 
 
 200 
 
 
 
 000 
 
 ;gin at 
 
 A, and 
 
 Diag. 
 
 Having laid down the figure, 
 draw the guess-line D?i, which 
 suppose = 766 links ; then the 
 diagonal AD will be = 824, 
 the perpendicular E« = 278, 
 and the perpendicular ar = 
 372 links ; also re will he = 
 52 links. 
 
 go West. 
 
92 LAND-SIjRYEYlXG. PART ITI. 
 
 Square links. 
 
 267800 the area of tlie trapezium Ar DE. 
 120001 ^, p ^i -r. , fDE. 
 
 16000 y^'l T'"* ^^ ^^"^ f'^'^'l EA. 
 12348 J taken on the line. |^^^._ 
 
 The sum is =308148 the area of A^iDEA. 
 2cL 3r. 20/?. = 287500 
 
 20648 the difference between the quantity 
 proposed, and the quantity parted off by the guess-line, 
 which being divided by 383, (half the guess-line,) gives 
 54 links to be set off from 7i towards A. Hence, DF is 
 the true line of division ; and the irregular figure AFDE 
 contains 2a. 3r. 20p. 
 
 Now, by the scale, Ac = 377 links. Measure therefore, 
 in the field, 377 links from A to c ; stake out the line DcF, 
 and the work will be completed. 
 
 Notel. The Rules given in this Problem, for parting-off land from irregular 
 fields, are generally adopted by Practical Land-Surveyors ; because they maybe 
 applied to any irregular figure whatever. Land, however, may sometimes be parted 
 off more directly ; for instance, the last example may be performed by finding the 
 area of the irregular figure ADE, and subtracting it from the quantity to be parted 
 off; then, if the difference be dirided by half tlie line AD, the quotient will be the 
 perpendicular of the triangle ADF ; the side AB being nearly straight from A to F. 
 
 Now, at the distance of this perpendicular, draw a line parallel to AD ; and it will 
 intersect AB in F, the point to which the division fence must be made. 
 
 2. It is not absolutely necessary to survey and plan the whole figure, in order to 
 part a portion from it, as the guess-line and portion parted off may be measured in 
 the field ; but, in my opinion, the former, in general, is a more eligible method 
 than the latter. 
 
 3. In order to divide a trapezium or an irregular polygon, among any number of 
 persons, by fences made in a given direction, proceed thus : Part off the first per- 
 son's share, then from the remainder of the figure, part off the second person's 
 share ; and thus continue, till the whole field be divided. 
 
 4. Those wiio desire to see a greater Variety of Examples in BurTeving single 
 Fields, and to make themselves fully acquainted with the Methods of Laying out. 
 Parting off, and Dividing Land ; also of Dividing a Common, &c. of variable 
 Value, among any Number of Proprietors, in the Proportion of their respective 
 Interests, may consult my Treatise on Practical Land-Surveying, Seventh Edition, 
 in which I flatter myself they will find these subjects satisfactorily treated. 
 
 PROBLEM XL 
 
 To reduce statute-measure to customary, and vice versa. 
 
 It has been before observed, that by custom the perch 
 varies in different parts of England ; and with it, conse- 
 quently, varies the acre in proportion. 
 
 In Devonshire and part of Somersetshire, 15 ; in Corn- 
 wall, 18; in Lancashire, 21 ; and in Cheshire and Staf- 
 fordshire, 24 feet in length, are accounted a customary 
 perch. 
 
SECT. I. LAND-SURVEYING. 93 
 
 GENERAL RULES. 
 
 I. To reduce statute-measure to customary, multiply 
 the number of perches, statute-measure, by the square feet 
 in a square perch, statute-measure ; divide the product by 
 the square feet in a square perch, customary measure, and 
 the quotient will be the answer in square perches. 
 
 II. To reduce customary measure to statute, multiply 
 the number of perches, customary measure, by the square 
 feet in a square perch, customary measure ; divide the 
 product by the square feet in a square perch, statute- 
 measure, and the quotient will be the answer in square 
 perches. 
 
 Notel. Tt is scarcely necessary to remark tliat the length of any perch mul- 
 tiplied by itself will give the number of square feet, in a square perch of the same 
 measure ; hence we have 1G.5 x 10.5 = 272.25, the statute perch ; 15x15 = 225, the 
 DeTonshire and Somersetshire perch; 18 x 18=3-4, the Cornwall perch ; 21x21 = 
 441, the Lancashire perch ; and 24x24=576, the Cheshire and Staffordshire perch. 
 
 2. When it is intended to find the area of an estate in customary measure only, 
 it is generally thought most convenient to take the dimensions by a chain properly 
 adapted for that purpose. The Devonshire and Somerset chain, is GO feet ; the 
 Cornwall chain, 72 feet ; the Lancashire cliain, 84 feet ; and the Cheshire and 
 Staffordshire chain, is 96 feet in length. Each of these chains is dirided into 100 
 equal links, in the same manner as the statute chain ; consequently, the customary 
 measure is found by the same Rules as the statute-measure. 
 
 3. It mav also be observed, that 4840 square yards make one statute acre ; 400f^ 
 make one Devonshire or Somersetshire acre ; 57G0 make one Cornwall acre ; 7810 
 make one Lancashire acre ; and 10240 square yards make one acre of the customary 
 measure of Chesliire and Staffordshire. (Sec the Author's Land-Surveying. Farts 
 11. III. and VI.) 
 
 EXAIVIPLES. 
 
 1. In 26a. 2r. 20p. statute, how many acres, &c. cus- 
 tomary measure, of 15 feet to a perch ? 
 
 Bi/ Rule I. 25a. 2r. 20p. 
 _4 
 102 
 
 40 
 
 4100 
 
 
 
 272.25 r 
 
 .= 16.5 
 
 X16.5 
 
 20500 
 
 
 
 8200 
 
 
 
 8200 
 
 
 
 28700 
 
 
 
 8200 
 
 4,0 
 
 
 15x 15=225)1116225.00(496,^ 
 
 4)124 r 
 
 31«. Or. Ip. Am. 
 
94 LAJ>D-SURVETIXG. PART m. 
 
 2. In Sla. Or. Ip. customary measure, of 15 feet to a 
 perch, how many acres, &c. statute-measure ? 
 
 Bt/ Rule 11. 31a. Or. Ip. 
 4 
 124 
 
 40 
 
 4961 
 
 225 
 
 24805 
 9922 
 9922 4,0 
 
 272.25)1116225.00( 410,0 
 4) 102 20 
 2oa. 2r. 20p . Ans. 
 
 3. Reduce 66a. 2>r. 36p. statute, to customary measure, 
 of 18 feet to a perch. Ans. Ala. Zr. 20p. 
 
 4. In 4.7a. 3r. 20p. customary measure, at 18 feet to a 
 perch, how many acres, &c. statute-measure ? 
 
 Ajis. 66a. 3r. o6p. 
 
 PROBLEM Xn. 
 
 To survey and plan Estates or Lordships. 
 
 Various methods are adopted by different surveyors, in 
 taking the dimensions of Estates or Lordships ; I shall, 
 however, describe only four which I conceive to be the 
 most accurate and practical. 
 
 3IETH0D I. 
 
 Having made yourself acquainted with the form of the 
 estate, either by actual examination, or by the assistance 
 of a previous plan, select two suitable places, at the greatest 
 convenient distance from each other, as grand stations ; and 
 measure a principal base, or what is generally called a 
 " main-line," from one to the other, noting every hedge, 
 brook, or other remarkable object, as you cross or pass it ; 
 taking offsets likewise to the bends or corners of the hedges 
 that are near you. 
 
 Next, fix upon some otlier suitable place, towards the 
 outside of the estate, as a third grand station ; to wliich, 
 
SECT. I. LA]S'D-SURVEYIXG. 95 
 
 from each extremity of the diagon:il or main-lhie, or from 
 two convenient points in it, lines must also be run. 
 
 These three lines being laid down, will form one large 
 triangle ; and in a similar manner, if necessary, on the 
 other side of tlie diagonal or main-line, a second triangle 
 may be formed. 
 
 The survey must then be completed by forming smaller 
 triangles, on the sides of the former ; and measuring such 
 lines as will enable you to obtain the fences of each in- 
 closure, and prove the whole of your work. 
 
 Note 1. The method of measuring estates by dividing 
 them into triangles, is particularly described in my Survey- 
 ing ; and illustrated with several rough and finished Plans, 
 and an engraved field-book. 
 
 It is also exemplified in this Work by Plate I., which is 
 a rough plan of an Estate exhibiting the chain-lines and 
 stations used in taking the survey. The field-notes are 
 not given, as they would have occupied too many pages 
 of coppei -plate ; it may, however, be observed that the 
 method of entering the notes, sketching the fences, &c. is 
 precisely the same as in the field-book belonging to Plate 
 III. ; and the directions of all the lines may be easily as- 
 certained from the following particulars. 
 
 The first or main-line leads from -{- 1 to + 8 ; the se- 
 cond hne from -}- 8 to + 10 ; and the third from + 10 to 
 + 1 ; w^hich three lines form the first large triangle. 
 
 The fourth line extends from + 2 to + 15 ; and the 
 fifth from + 15 to + 8; which two lines and part of the 
 main-line form the second large triangle. 
 
 The sixtli line leads from + 9 to -f 1 1 ; the seventh 
 from +20 to + 6 ; the eighth from + 7 to + 22 ; the 
 ninth from + 21 to + 4 ; the tenth from + 24 to + 13 ; 
 and the eleventh from + 12 to -f 23; which complete the 
 survey of the first triangle. 
 
 The twelfth line extends from + 5 to + 17; the thir- 
 teenth from + 25 to the main-line, southward of + 3 ; 
 the fourteenth from + 1 to + 14 ; the fifteenth from + 
 14 to + 26 ; the sixteenth from + 27to + 16 ; the seven- 
 teenth from + 18 to 4- 28 ; and the eighteenth from -f- 
 28 to 4- 19 ; which finish the whole survey. 
 
 2. The content of the Estate may be found in the fol- 
 lowing manner : Measure the lines upon the Plan, and take 
 the necessary offsets, by a scale of 8 chains to an inch ; and 
 
96 LAND-SURVEYING. PART III. 
 
 enter the dimensions in a field-book. From the dimen- 
 sions thus obtained, draw a plan hy a scale of 2 chains to 
 an inch ; then straighten the fences as directed in Pro- 
 blem II. ; and measure diagonals, perpendiculars, &c. from 
 which compute the content of each field. 
 
 The diagonals, perpendiculars, and contents may be 
 entered in a Book of Castings, similar to that belonging 
 to Plate III. ; and if you should not have a scale of 8 
 chains to an inch, any other scale will do just the same for 
 practice. 
 
 3. Taking the dimensions, &c. as directed in the last 
 note, will be found of infinite service to the learner ; as it 
 will tend to make him very expert in entering the field- 
 notes, laying down the lines, and casting the contents, 
 what are no small acquisitions towards becoming a com- 
 plete Land- Surveyor. 
 
SliCT. J. 
 
 LAXD-SUR^'EYING. 
 
 97 
 
 Scde S Chains to an Inch. 
 
 J 
 
9B 
 
 LAND - S URYE YING. 
 
 PART m. 
 
 PLAN IL 
 
 Srnlc S Chr.ir.:; to an Tncl 
 
SECT. I. LAXD-SURYEYIXG. 99 
 
 * 
 
 METHOD II. 
 
 Measure a main-line as nearly to one of the out-boun- 
 daries of the estate as the curves in the hedges will per- 
 mit ; noting' the crossings of fences, and taking offsets as 
 before directed. 
 
 At a convenient distance, measure another main-line 
 parallel or nearly parallel to the first line, so that a num- 
 ber of fences running in that direction may be obtained ; 
 and from any two stations in the first line measure lines 
 to any station in the second main-line, forming a triangle; 
 so will a station in the second main-line become deter- 
 mined or fixed. 
 
 From the first main-line to the. second, or from the se- 
 cond to the first, measure lines in order to obtain all those 
 fences v/hich run in that direction. The remainder of the 
 fences of the inclosures contained between the first and 
 second m.ain-lines being obtained by running lines in the 
 most convenient manner; you will have completed the 
 dimensions of a portion of the estate, which may then be 
 laid down. 
 
 Parallel or neiirly parallel to the second main-line, and 
 at a proper distance from, it, measure a third ; and pro- 
 ceed with the internal lines as before, and you will obtain 
 the dimensions of another portion of the estate, which may 
 also be laid down ; and thus go forv.^ard until you finish 
 the survey. 
 
 Note 1. This method is illustrated by Plate II., which 
 displays the chain-lines and stations used in taking the 
 survey. 
 
 The first main-line leads from n- 1 to + 6 ; the second 
 line from 4- 6 to + 7 ; and the third line, or second main- 
 line, from + 7 to + 16. The fourth line extends from -f 
 16 to -}- 1 ; the fifth, or tie-line from -h 16 to + 2 ; the 
 sixth from -f 2 to 4- 14 ; the seventh from + 17 to + 
 IS ; the eighth from + 12, through + 18, to -f 3 ; the 
 ninth from -[- 4 to M- 10 ; and the tenth line leads from 
 -f 8 to + 5 ; thus all the fences betw^eeii the first anu 
 second main-lines are obtained. 
 
 The eleventh line, or the third main-line, leads from -}- 
 19 to -f- 29 ; and the directions in which all the rest of 
 the lines were run, may be easily ascertained by inspecting 
 the plan. 
 
 2. The content of this Estate may be found in the same 
 manner as directed in Note 2, ]Method I. 
 F 2 
 
100 LAXD-SURYETIXG. PART III. 
 
 METHOD m. 
 
 An estate of four sides may frequently be conveniently 
 surveyed as follows : Measure four lines in such a manner 
 that offsets or insets may be taken to the four out-boun- 
 daries of the estate ; and tie the first and fourth lines to- 
 gether by a diagonal or tie-line measured from one to tte 
 other, at the distance of five, six, or more chains from the 
 angular point, according to the extent of the survey ; thus 
 you will be enabled to lay down the first four lines, and 
 also the out-boundaries of the estate. 
 
 Next proceed to obtain the internal fences, by measur- 
 ing lines in the most convenient manner ; some of which 
 must be run from the first to the third, or from the second 
 to the fourth line, or in some other more proper direction, 
 so that they may become proofs, and fiist-lines, into which 
 other lines may be run with propriety. 
 
 In thus proceeding, it is evident that a great deal will 
 always depend upon the dexterity and ingenuity of the 
 surveyor; as no directions can be given that will suit 
 every particular case to be met with in practice. 
 
 "Wote. — This method is illustrated in my Surveying, by Plate VII. ; and also in 
 this Work, by Plate III., which is a Rough Plan of an Estate in the Township o^ 
 Farnley, in the parish of Leeds. 
 
 METHOD IV. 
 
 The method which I here intend to describe is a com- 
 pound of all the methods of surveying with the chain ; for 
 as tliere are never two estates to be met with which are 
 exactly alike, sometimes one method claims the preference^ 
 and sometimes another ; but a skilful surveyor will always 
 adopt that by which he can take his dimensions and proofs 
 with the greatest accuracy by the fevrest lines. 
 
 If an estate be in the form of an irregular polygon of 
 five, six, or more sides, and the fences very crooked, such 
 an estate may generally be most easily surveyed by divid- 
 ing it into triangles, as in Plate I. ; but if many of the 
 fences of the different inclosures run a considerable wa}" 
 in the same direction, and the fields in general be pretty 
 neat trapeziums, it is commonly more eligible to proceed 
 as directed in Method II. 
 
 Sometimes an estate varies so much in its shape that all 
 the methods before described may be used \nX\\ propriety 
 and advantage ; and it frequently happens that an in- 
 genious surveyor adopts methods, in particular cases, en- 
 
SECT. I. LAND-SUIlV,T:Y'fNG^ 101 
 
 tire! J new to liiinself ; care, however, must always be taken 
 to make one line depend upon another, throughout the 
 whole of the survey, so tliat when you come to lay it down, 
 you may find no lines whose positions are undetermined. 
 
 Kote — For the method of measuring Hillj Ground, see my Land-Survenn^, 
 Seventh Edition, page 172. ^ 
 
 INOSCELLANEOUS INSTRUCTIONS. 
 
 1. In ranging a line, you must be extremely careful to keep the 
 poles in a direct line with each other ; which you may accurately 
 effect by always having, at least, two behind you ; and if your sight 
 be obstructed by hedges, you must cut down the intervening parts. 
 
 2. In measuring your main, or any other chain-line, put down 
 stations at every place to which you apprehend it may be necessary 
 to run lines, in order to complete the survey; and if it happen that 
 you put down more than are wanted, it will be immaterial. At each 
 station you must put down a small stake, called a " station-stake," 
 with the number of the station upon it ; and also cut a mark in the 
 ground, which may be easily done by having a smdl spade upon the 
 bottom of the ofFset-stafF. 
 
 3. In measuring your internal lines, it will give you the least 
 trouble to run them from one station to another, if you can make it 
 convenient ; if not, you must run them from, and continue them to 
 some chain-line, and measure the distance upon that line, to the 
 nearest station. The place where you run upon, or cross a chain- 
 iine, may be easily ascertained by setting up poles, at two of the 
 nearest stations in that line ; and the crossing will be at the place 
 where you are in a direct line with these poles. 
 
 4. The angle which the first line makes with the meridian line 
 must be taken with a compass ; and in doing this, an allowance must 
 be made for the variation of the needle, which is about 24 degrees 
 westerly, 
 
 5. In order to plan a large survey, provide a sheet of drawing- 
 paper, of a proper size for use ; and trace with a pencil, a meridian 
 or north and south line, in such a position, that your first station may 
 be in some convenient point in this line. Then, from your first 
 station, draw your first or main-line, making its proper angle with 
 the meridian, which you may then take out with Indian rubber. 
 Further directions appear to be unnecessary ; as any person who is 
 acquainted with the methods of laying down triangles, trapeziums, 
 and trapezoids, will find no difficulty in planning an estate. 
 
 6. The most expeditious method of laying down crooked fences, 
 is by means of an offset-scale, which must be used with the plot- 
 ting-scale in the following manner : Lay one edge of the plotting- 
 scale close by the base-line, and bring the end of the offset-scale in 
 contact with the edge of the plotting-scale, so that the edges of the 
 scales may form a right angle ; then by the edge of the offset-scale, 
 prick off, in its proper situation, the first offset, with a pencil finely 
 pointed. Keep the plotting-scale firm, and slide the offset-scale U 
 
 F 3 
 
102 LA:-ri>-sur.\^:nxG. paut in, 
 
 the place of the next perpendicular, which prick off as before ; and 
 thus proceed until all the offsets are finished. 
 
 7. In order to find the area of an estate, practical surveyors gene- 
 rally straighten the crooked fences of each field, as directed in Pro- 
 l)lem II. ; and then divide the fields into trapeziums and triangles, 
 and take such dimensions, by the scale, as are necessary to find the 
 area of each field, lliey then collect all the areas into one sum; 
 afterward find tlie area of the whole survey, as if it were a single 
 field; and if it be equal or nearly equal to the sum of the separate 
 areas, previously found, they justly infer that their survey is correct. 
 
 8. Those who do not approve of finding the area by the method of 
 casting, may make use of the offsets taken in the survey, vrhere con- 
 venient ; and if more be wanted, they may be measured by the scale; 
 for in m.easuring a number of small parts by it, some will probably be 
 taken a little too large, and others a little too small, so that, in the 
 end, they will nearly counterbalance each other. 
 
 9. Practical surveyors generally lay down their lines by a scale of 
 4 chains to an inch, when their surveys are very large ; and in com- 
 puting the contents, they measure the bases and diagonals by the 
 same scale, but the perpendiculars by a scale of 2 chains to an inch ; 
 consequently the product of the base and perpendicular of a triangle, 
 will be its area. To treat small surveys, in a similar manner, by a 
 scale of 2 chains and of 1 chain to an inch, must, of course be very 
 correct. 
 
 10. Rivers, large brooks, public roads, and common sewers, should 
 not be included in the area, but only delineated upon the plan ; and 
 marshes, bogs, heaths, rocks, &c. should also be represented or spe- 
 cified, and their measurements separately returned. 
 
 11. Sometimes the content of each field is entered within the field 
 itself; sometimes the fields are numbered, and their areas set down, 
 one after another, in some convenient part of the plan ; and some- 
 times they are entered in a book of particulars. 
 
 12. When you wish to transfer a rough plan to a fresh sheet of 
 paper, or to a skin of parchment or vellum, in order to make a finished 
 plan, proceed thus: Take a sheet of writing-paper of the same size as 
 the rough plan, and rub one side of it with black-lead powder ; then 
 lay it upon the sheet which you intend for your new plan, with the 
 black side downward ; upon both lay the rough plan ; and upon the 
 whole place weights or books, to keep them from moving. Next, run 
 your tracer gently over all the boundaries upon the rough plan, so 
 that the black lead under them may be transferred to the fresli sheet. 
 Separate the papers and trace the lines thus transferred witli a fine 
 pen and Indian ink ; as common ink ought never to be used in plan- 
 ning. You must then proceed to enter such names, remarks, or ex- 
 planations, as may be judged necessary ; drawing by the pen, the 
 representations of hedges, bushes, trees, woods, hills, gates, stiles, 
 bridges, roads, the bases of buildings, &c. in their proper places ; run- 
 ning a single dotted line for a foot-path, a double one for a carriage- 
 road, Src. Rivers, brooks, lakes, &c. should be shaded with crooked 
 or waved lines, bold at the edges, and fainter towards the middle. 
 Hills may also be represented by crooked lines, bold about the middle 
 of the hill, but fainter towards the top and bottom; and the bases of 
 
SECT. I. la:nd-sukveying. 103 
 
 buildings must be shaded by straight, diagonal lines. Roads should 
 be shaded M'ith a brownish colour, laid on with a camel-hair pencil ; 
 and fields in a variety of forms, with a line pen and Indian ink. In 
 some convenient part of the Plan, you may write, in consjncuous cha- 
 racters, the title of the Estate, ornamented with a compartment. In 
 another vacancy introduce tlie Scale by which the ])lan lias been laid 
 down ; and also a meridian-line, with the compass or fiOv.'Gr-de-luct> 
 pointing north. 
 
 Tlie whole may then be bordered with black lines at a convenien 
 distance from each otiicr ; and the space between them shaded by a 
 hair-pencil, with Indian ink. 
 
 1:3. A plan well finished with Indian ink, as directed above, has a. 
 very elegant appearance, and is considered by most persons to excel 
 those done in colours ; but the work is very tedious, and requires 
 much time to do it well, in consequence of which most surveyors 
 prefer finishing their plans with colours. Some, however, not only 
 embellish them with Indian ink, but also wash the different fields 
 with various shades of colouring. 
 
 14. In colouring a plan, meadow and pasture ground should be 
 washed with a transparent green, the pasture ratb.er lighter than th(» 
 meadow ; arable land with a mixture of red and yellow, or of red, 
 brown, and yellow, of various shades, so that too many fields may not 
 api)ear exactly alike ; and some surveyors use both light blue and 
 lake, in colouring plans. 
 
 An excellent greeii, of various shades, may be made by mixing light 
 blue and gamboge. 
 
 Rivers, brooks, lakes, &c. should be coloured with Prussian blue 
 mixed with a little Indian ink. The hedges must be done with a 
 strong shade, which should be softened off towards the middle with a 
 lighter one. If the quickwood hedges be not done with a pen and 
 Indian ink, they may be represented by running strong narrow 
 shades, of various colours, round the boundaries of the different fields. 
 
 15. Reeves's and Newman's water-colours are entitled to recom- 
 mendation. They must be ]n-epared for use in the following manner: 
 Dip one end of the cake in clear water, and rub a little of it upon a 
 clean earthen plate ; then mix it with water, by your Iiair-pencil, 
 until you have brought it to any consistence you please. Indian ink 
 must be prepared for use in the same manner. 
 
 16. When parchment is used in planning, it must first be rubbed 
 ■with clean flannel dipped in the best Paris v.'hiting. This operation 
 clears its surface from grease, and makes the pen slide more freely. 
 
 1 7. The learner will fully comprehend what has been said on the sub- 
 ject of embellishing plans, by examining a well-finished coloured map. 
 
 IS. The method of transferring plans by means of a slieet cf paper, 
 rubbed with black lead, will do very well for small surveys ; but 
 when the survey is large, it is necessary, not only to transfer the 
 original plan, but also to reduce it to a smaller scale, in order that 
 the finished plan may be of a convenient size. This may be effected 
 in different ways ; as by squares, by proportional compasses, &c, ; Init 
 the most expeditious and accurate method is by an instrument called 
 a Pentagraph. 
 
 Those who desire to make themselves ac(]uainted with the method 
 F 4 
 
104 
 
 LAND- SURVEYING. 
 
 PART ni. 
 
 of using this instrument, are referred to my Land- Surveying, Seventh 
 Edition ; also to Adam's Geometrical and Graphical Essays, in 
 which valuahle work they will likewise find the description and use 
 of the Plane- Table, Theodolite, Spirit-Level, and every other Mathe- 
 matical Instrument. 
 
 Note. — The area of the following Estate, Plate III., was found from a Plan laid 
 down by a scale of 2 chains to an inch. The crooked fences were straightened as 
 directed in Problem II., and the diagonals measured by the scale used m plannmg ; 
 but the perpendiculars by a scale of 1 chain to an inch. 
 
 The diagonals, perpendiculars, &c. are contained in the following Book of 
 Castings. 
 
 A Book of Dimensions, Castings, 
 
 Belonging to Plate III. 
 
 
 «; 
 
 1 
 
 1 
 
 c 
 o 
 
 1 
 
 Quantity 
 
 Quantity In 
 
 
 2 
 
 ^ 
 
 
 «*- 
 
 m 
 
 A. R. P. 
 
 ° 
 
 §^ 
 
 "to 
 
 c 
 o 
 
 s 
 
 A. Dec. 
 
 
 ;^ 
 
 5 
 
 e 
 
 o 
 t/2 
 
 c^ 
 
 
 
 1 
 
 1375 
 
 143 
 
 227 
 
 370 
 
 5.08750 
 
 5 
 
 
 
 14 
 
 2 
 
 1542 
 
 280 
 
 298 
 
 578 
 
 8.91276 
 
 8 
 
 3 
 
 26 
 
 3 
 
 1120 
 
 264 
 
 208 
 
 472 
 
 5.28640 
 
 5 
 
 1 
 
 5f 
 
 4 
 
 1226 
 
 217 
 
 213 
 
 430 
 
 5.27180 
 
 5 
 
 1 
 
 3-^ 
 
 5 
 
 1225 
 
 220 
 
 194 
 
 414 
 
 5.07150 
 
 5 
 
 
 
 lU 
 
 6 
 
 1552 
 
 374 
 
 252 
 
 626 
 
 9.71552 
 
 9 
 
 2 
 
 34^ 
 
 7 
 
 1088 
 
 210 
 
 168 
 
 378 
 
 4.11264 
 
 4 
 
 
 
 18 
 
 8 
 
 874 
 
 180 
 
 — 
 
 180 
 
 1.57320 
 
 1 
 
 2 
 
 W, 
 
 »{ 
 
 1262 
 
 254 
 
 76 
 
 330 
 
 4.16460 
 
 V 
 
 
 
 H 
 
 1030 
 
 84 
 
 — ■ 
 
 84 
 
 .86520 
 
 10 
 
 790 
 
 136 
 
 190 
 
 326 
 
 2.57540 
 
 2 
 
 2 
 
 12 
 21$ 
 
 WTiole Quantity - 52.63652 
 
 52 
 
 2 
 
SECT. I. 
 
 LAND-SURVEYING. 
 
 1 05 
 
 y 5 
 
106 
 
 LAND-SURVEYING. 
 
 PART ni 
 
 ^ <f G O "/ 
 
 9^: 
 
 O I 
 
BECT. I. 
 
 LAND-SURVEYING. 
 
 107 
 
 + 3 
 
 <:i/7^i07A 
 
 /. ao 
 
 ^j 6 o o 
 
 ^200 
 if 
 
 & 
 
 .10 
 
 COG 
 
 ■JiO 
 OS 
 
 63 
 
 /. 
 
 April IStli, 1814. The Notes of an Estate situate in 
 \ the Township of Farnley, in the Parish of Leeds. 
 
 FIELD BOOK. 
 
 F 6 
 
108 
 
 LAND-SURVEYING. 
 
 TART in. 
 
SECT. I. LAND-SURVEYING. 109 
 
 THE METHOD OF SURVEYING WITHOUT 
 THE CROSS-STAFF. 
 
 It has been suggested, and deemed advisable in revising 
 tliis work, to add a few more examples on the subject of 
 land-surveying, where it will be observed that we have 
 varied the method in a slight degree from the former, 
 cliiefly in that of forming and keeping the field-book, as 
 also without the aid of the cross-staff; in taking dimen- 
 sions, the stations being represented by the numbers 1, 2, 
 3, 4, &c., which are read from station first to station 
 second, from second to third, from third to fourth, &c. ; 
 inserting three cyphers at the commencement of each 
 station, and by keeping the fence on the right hand, the 
 offsets will always appear on the right-hand side of the 
 field-book, unless you cross the fence, when they will be- 
 come insets : the learner is, however, strongly recom- 
 mended to make himself fully acquainted with the former 
 part on this subject. 
 
 PROBLEM XIII. 
 To measure a field of four sides. 
 
 EXA3IPLES. 
 
 1. Let the following figure represent a field to be 
 measured without the aid of the cross-staff. 
 
 Begin at station 1, and measure the first side to station 
 2, from 2 to 3, and so on until you get round the field ; 
 taking special care to take and enter all the oftsets in a 
 book, as the irregularity of the fence may require ; then 
 measure the diagonal from 1 to 3, and the field-book will 
 be completed. 
 
 CONSTRUCTION. 
 
 Draw the first side, and set off from 1 to 2, 680 from a 
 scale of equal parts, then take the second side 720, and 
 the diagonal 1080 in your compasses, and the point of in 
 tersection will be the third station ; in like manner, 800 
 and 515 taken from the same scale, and the point of inter- 
 section will be the fourth station; the diagonal being 
 drawn, and the perpendiculars measured on the same 
 scale, will be 7w2, 450, and w4, 360 
 
no 
 
 LAND-SURVEYING. 
 
 PAET m 
 
 / FrH 
 
 
 
 
 
 J- / J- 
 
 / 
 
 UO 
 
 80 
 
 
 2 S 
 
 W 
 
 
 1 6 C 
 
 '^U 
 
 
 
 
 
 
 
 SCO 
 
 
 
 z. 
 
 7 2 
 
 
 
 
 SCO 
 
 5U 
 
 
 260 
 
 
 
 
 
 
 
 
 
 / Z 
 
 
 
 5 
 
 6 
 
 Z2 
 
 
 A S 
 
 20 
 
 
 2 AC 
 
 
 
 
 
 
 2>0 
 
 V 
 
 
 r tn£^ 
 
 OZC^^ 
 
 / i 
 6 H 
 
 32 
 ZA 
 
 2 
 
 ^0 
 
 
 
 
 2> i 
 
 
 
 
 
 
 so 
 
 
 FIELD BOOK. 
 
SECT. 1. 
 
 LANJ)-SUiiYETmG. 
 
 Ill 
 
 CALCULATIONS. 
 
 Ai'ea of the trapezium - 874800 
 
 Area of the olfsets, lirst side - 27000 
 Area of the offsets, second side 24480 
 Area of the offsets, third side - 24840 
 Area of the olfsets, fourth side 30420 
 Double areas 2)98l540 
 
 4,90770=4^. 3r. 25 p. 
 
 2. Let the following figure represent a field of live sides, 
 it is required to form the lield-book, and find its area. 
 
 /. 
 
 S /, 
 
 1 
 
 / 
 
 6 
 
 3 
 
 
 Z Z 
 
 1 
 
 6 
 
 
 
 
 
 
 zoo 
 
 s 
 
 
 2 \20 
 
 
 1 6 0\0 
 
 
 ■10 
 
 22 
 
 /' 
 
 
 
 
 
 
 3 (j 
 
 4. 
 
 
 3 U 
 
 w 
 
 
 ■2> 
 
 2Z 
 
 
 z i 
 
 SO 
 
 
 i &?^ 
 
 
 
 3 
 
 0,^0 
 
 
 
 
 3^0 
 
 3 
 
 
 2 2 
 
 uo 
 
 
 4 U 
 
 10 
 
 
 J 0\Z(> 
 
 2 
 
 
 
 
 
 
 2> S 
 
 z 
 
 
 2 / 
 
 ^0 
 
 
 '12 
 
 20 
 
 i 
 
 
 
 
 
 Having measured the field, fci^med the field-book, and 
 constructed the figure, we find the perpendiculars tc be, 
 W22, 160, n^, 322, and oo, 146 links. 
 
112 
 
 LAXD-SUR^-EYIXG. 
 
 4 
 
 PAET m. 
 
 Area of the trapezium 
 Area of the triangle 
 Area of the first side offsets 
 Area of the second side offsets 
 Ai-ea of the third side offsets 
 Ai-ea of the fourth side offsets 
 
 Double areas 
 
 - 13400 
 
 - 13840 
 
 - 5220 
 
 - 6320 
 2) 406820 
 
 2,03410 
 
 A?is. 2a. Or. r,lp. 
 Lnv down a Held, and fird \ii area, from the follow- 
 
SECT. I. 
 
 LAXD-SURVEYING. 
 
 113 
 
 FIELD BOOK 
 
li 
 
 L AN B- SUli Vi: V iXG. 
 
 PART in. 
 
 Having laid down the figure from a scale of equal parts, 
 we find the perpendiculars as follows : w2, 184, n4:, 296, 
 and 00, 200, consequently the respective areas will be as 
 below : — 
 
 
 Square links. 
 
 Area of the trapezium 
 
 - 240000 
 
 Ai-ea of the triangle 
 
 - 96000 
 
 Ai-ea of the first side 
 
 - 2340 
 
 Area of the second side - 
 
 - 6040 
 
 Ai-ea of the third side 
 
 - 7040 
 
 Area of the loui-th side 
 
 - 4472 
 
 Area of the fifth jiide 
 
 - 3720 
 
 
 359612 
 
 Deduct inset 
 
 4000 
 
 Double niees 
 
 2)355612 
 
 
 1,77806 
 4 
 
 
 3,11224 
 
 
 40 
 
 
 4,48960 
 
 
 Ans. la. or. 
 
 ip- 
 
SECT. I. 
 
 LAND-SURYE i' IXG. 
 
 115 
 
 4. Let tlic following figure represent a field of six sides 
 it is required to be measured and planed. 
 
 CONSTRUCTION. 
 
 Suppose tlie field-book to be formed as below, draw the 
 first side, and set oif 700 from your scale of equal parts : 
 take 600, the second side in your compasses, which inter- 
 sect with the diagonal 1, o, 1000, and the point of inter- 
 section will be the third station ; with the side 1, 6, and 
 the diagonal 3, 6, intersect again, and you have the sixth 
 station. Then take the side 3, 4, and the diagonal 6, 4, and 
 the point of intersection is the fourtli station ; again, take 
 the two sides 4, o, and 6, 5, and the point of intersection 
 is the fifth station : thus the field v/ill be divided into two 
 trapeziums, and the perpendiculars will be 7i2, 340, 7/i6, 
 220, pS, 310, o5, 470. 
 
 \7C 
 
 I, UO 6 
 
 6 
 3 
 1 
 
 e 8 o\o i 
 /-el — iF~§-e 
 
 1 C 
 
 3 10 
 
 -f-^F-e-0 
 
 (:• 
 
 A S 
 c 
 
 c 
 
 6 
 
 6^0 
 
 UT 
 
 .5 2 
 
 2 ; 
 
 
 
 i) 
 
 
 
 
 5 
 
 3 /; 
 
 18 
 
 
 
 C h 
 30 
 
 
 
 6 
 
 3 
 
 2) 2 
 
 
 
 
 
 50 
 
 \ 7 D 
 
 2 
 
 \ 5 5 C 
 
 ' !■ 
 
 \ 5 
 
 
 
 1 ISO 
 
 0-9 
 
 ;| c c g\o 
 
116 
 
 LAND-SUPtVEYING. 
 
 PART nL 
 
 Area of the trapeziums . _ . 1043600 
 
 Area of offsets on the 1st, 2cl, 3d. and 4th sides 66360 
 
 ri09960 
 Area of the inset on the 6th side - 22400 
 
 Double areas 2)1087560 
 5,43780 
 
 4 
 
 1,75120 
 
 40 
 
 30,04800 
 
 Ans. 5a. Ir. 30p. 
 
 Note 1. It must always be rdToembered that when the cross-staff is not used in the 
 field in measuring the perpendiculars Irom the diagonals &c., great care should be 
 taken in raying down the field accurately from the scale of equal parts; and, for this 
 purpose, it will be necessary that you use as large a scale as possible in order to get 
 the true dimensions, and when due attention is paid to these observations, it is a 
 matter of doubt, whether the perpendiculars can be measured truer by using the 
 cross in the field, or by measuring them on the figure by the same scale with which 
 the field was constructed. 
 
 2. The plan of a field may be very readily taken by laying the rough draft upon 
 a sheet of paper, then with a fine pricker or needle beginat one corner of the figure, 
 and prick holes through the paper so as to leave an impression on the paper below : 
 this must be continued at all angles, at every offset, and where every offset begins 
 and ends until you get round the field, then draw the lines with a fine pencil ; it 
 may then be coloured if required. 
 
SECT. ir. MISCELLANEOUS QUESTIONS. 117 
 
 SECTION IT. 
 
 MISCELLANEOUS QUESTIONS 
 
 CONCERNIXG 
 
 SUPERFICIAL MENSURATION. 
 
 1. The length of a floor is 35 feet, and its breatltli 24 
 feet ; what is the area of three such floors, deducting for 
 two well-holes, one of which measures 10 feet 6 inches bv 
 4 feet 3 inches, and the other 8 feet 9 inches by 3 feet 6 
 inches ? Ans. 2444^ feet. 
 
 2. From a mahogany plank 16 inches in breadth, 6 
 square feet are to be sawn ofl' ; at what distance from the 
 end must the line be struck ? A7is. 4^ feet. 
 
 3. What must be the length of a chord which will 
 strike the circumference of a circular plantation that shall 
 contain just an acre and a half of ground ? 
 
 Ans. 48.072 yards. 
 
 4. A gentleman has ordered a rectangular court-yard 
 to be paved, which measures 4-5 feet 9 inches in front, and 
 32 feet 6 inches broad. A foot-path 5 feet 6 inches in 
 breadth, leading to the front door of his house, is to be 
 laid with Portland-stone, at 3^. 4d. per square foot, and 
 the rest with Purbeck-stone, at 2s. 2d. per square foot ; 
 what will be the expense of the whole ? 
 
 Ans. £176. \9s. l^d, 
 
 5. The annual rent of a triangular field is .€43. 155.; 
 its base measures 25, and perpendicular 14 chains ; what 
 is it let for per acre ? A?is. £2. 10s. 
 
 6. The transverse diameter of the ellipse in Grosvenor 
 Square measures 840 links, and the conjugate 612, within 
 the wall ; the wall is 14 inches thick ; what quantity of 
 ground does it enclose, and how much does it occupy ? 
 
 Afis. The icall cjicloses 4a. Or. 6j)., and occupies 
 1760.531 square feet. 
 
 7. The area of a right-angled triangle is 60, and the 
 hypothenuse 17; required the two legs. Ans. 15 and 8. 
 
 8. Two sides of an obtuse-angled triangle are 5 and 10 
 chains ; what must be the length of the third side, that the 
 triangle may contain just 2 acres of ground ? 
 
 Ans. 8.06225 or 13.60147 chains. 
 
118 MISCELLANEOUS QUESTIONS PART III, 
 
 9. Two travellers, A and B, departed from an inn, at 
 the hour of eight in the morning ; A proceeded north-west, 
 at the rate of 6 miles an hour, and B north-east, at the 
 rate of 8 miles an hour ; hov/ far were they distant from 
 each other, at tvrelve o'clock of the same daj ? 
 
 Ans. 40 miles. 
 
 10. Two boys amusing themselves at a game called 
 snatch-apple, in a room 13 feet high, find that by standing 
 12 feet from each other, the apple, which is suspended 
 from the ceiling by a string, and in a right line between 
 them, when put in motion, just touches each of their 
 mouths. Eequired the area of the sector described by the 
 string and apple ; the perpendicular height of each boy's 
 mouth, from the ground, being 5 feet. Ans. 64.3o726/ee^. 
 
 11. What is the area of an isosceles triangle inscribed 
 in a circle whose diameter is 24 ; the angle included by 
 the equal sides of the triangle being 30 degrees ? 
 
 'Ans. 134.3538. 
 
 12. A maltster has a kiln that is 18 feet square ; but he 
 intends to pull it dov/n, and build a new one, that may dry 
 three times as much at once as the old one ; what must be 
 its length, if its breadth be 24 feet ? Ans. 4.0^ feet. 
 
 13. The side AB of a triangular iield is 40, BC 30, and 
 CA 25 chains ; required tlie sides of a triangle parted off 
 by a division-fence made parallel to AB, and proceeding 
 from a point in CA, at the distance of 9 chains from the 
 angle A. A^ns. 16, 19.2, and 25.6 chains. 
 
 14. A field in the form of a right-angled triangle is to 
 be divided between two persons, by a fence made from the 
 right angle meeting the hypotlienuse perpendicularly, at 
 the distance of 880 links from one end ; required the area 
 of each person's share, the length of the division -fence 
 being 660 links. Ans. 2a. 3r. 241^. and \a. 2r. 2l^p. 
 
 15. It is required to part from a triangular field whose 
 three sides measure 1200, 1000, and 800 links respectively, 
 1 acre, 2 roods, and 16 perches, by a line parallel to the 
 longest side. 
 
 Ans. Hie sides of the remaining triangle are 927, 
 7721, and 618 links respectively. 
 
 16. The perambulator, or surveying wheel, is so con- 
 trived, as to turn just twice in the length of a pole, or 16^ 
 feet ; what is its diameter? Ans. 2.Q2Qfeet. 
 
 17. Required the side of an equilateral triangle whose 
 area is just two acres. A.ns. 6.79617 chai?is. 
 
SECT. II. IN SUPERFICIES. 119' 
 
 18. The sides of a triangle are 20, 30, and 40 respec- 
 tively ; what is the area of its inscribed circle ? 
 
 Ans. 130.8999. 
 
 19. In an isosceles trianc^lc, two circles are inscribed 
 touching each other and the sides of the triangle ; the 
 diameters of the circles are 9 and 25 ; required the sides of 
 the triangle. Ans. 44.27083, 44.27083, mid 41.66666. 
 
 20. The base of a field, in the form of a trapezoid, is 
 30, and the two perpendiculars are 28 and 16 chains re- 
 spectivelj ; it is required to divide it equally between two 
 persons, by a fence parallel to the perpendiculars. 
 
 Ans. The division-fence is 22.8035 chains, and it di- 
 vides the base into tico parts, whose lengths are 17.0087, 
 and 12.9913 chains respectively. 
 
 21. A field in the form of an equilateral triangle, con- 
 tains just half an acre; what must be the length of a 
 tether, fixed at one of its angles, and to a horse's nose, to 
 enable him to graze exactly half of it ? Ans. 48.072 yards. 
 
 22. The diameter of a circular estate is 25 chains ; 
 what is the length of the chord which divides it into two 
 segments whose areas are to each other as 2 to 1 ? 
 
 Ans. 24.1062 chains. 
 
 23. In turning a one-horse chaise within a ring of a 
 certain diameter, it was observed that the outer wheel 
 made three turns while the inner wheel made two. The 
 wheels were of equal height ; and supposing them fixed at 
 the statutable distance of 5 feet asunder on the axle-tree; 
 what was the circumference of tlie track described by each 
 vrheel ?. 
 
 Ans. The length of the track described by the outer 
 icheel is 94.248 /ee^, and tJiat described by the inner wheel 
 62.Smfeet. 
 
 24. If the frustrum of a cone whose diameters are 8 and 
 12 feet respectively, be made to revolve with its slailt side 
 upon a horizontal plane, until it returns to its first posi- 
 tion ; what will be the area of the space passed over by 
 the slant side, the length of which is 10 feet? 
 
 Ans. 1570.S feet. 
 
 25. Being desirous of finding the height of a steeple, I 
 placed a looking-glass at the distance of 100 feet from its 
 base, on the horizontal plane, and walking backward 5 
 feet, I saw the top of the steeple nppear in the centre of 
 the glass ; required the steeple's height, my eye being 5 
 feet 6 inches from the ground. A?is. IIO feet. 
 
120 MISCELLANEOUS QUESTIONS. PART in. 
 
 26. Three men bought a grinding-stone of 50 inches 
 diameter, each paying -1- part of the expense ; what part 
 of the diameter must each person grind down for his share? 
 
 Ans. The first must grind doiun 9.1752, the second 
 11.9537, and the third 28.8675 inches. 
 
 27. "Wanting to know the height of the cathedral at 
 York, I measured the length of its shadow, and found it 
 to be 200 feet. At the same time, a staff 5 feet long, cast 
 a shadow of 4 feet ; required hence the height of that 
 elegant and magnificent structure. Ans. 2o0 feet. 
 
 28. At Matlock, near the Peak, in Derbyshire, where 
 there are many surprising curiosities of nature, is a rock 
 by the side of the river Derwent, rising perpendicularly 
 to a wonderful height, which, being inaccessible, I endea- 
 voured to measure in the following manner : At the dis- 
 tance of 340 feet from the bottom of the rock, I fixed a 
 staff, 8 feet in length, perpendicularly to the horizontal 
 plane ; and at a convenient distance from this, I fixed an- 
 other, 3 feet long, so that by looking over both their tops, 
 I could just see the summit of the rock. The distance 
 between the staves I found to be 4 feet 6 inches. From 
 these data I determined the perpendicular altitude of the 
 rock ; and you are requested to repeat the process. 
 
 Ans. The height of the rock is 128.5925 ya ds. 
 
 29. Tliere are two columns in the ruins of Persepolis, 
 left standing upright ; one of which is 64 feet above the 
 horizontal plane, and the other 50. Between these, in a 
 right line, stands a small statue, the head of which is 97 
 feet from the summit of the higher, and 86 feet from the 
 top of the lower column, whose base is just 76 feet from 
 the centre of the figure's base ; required hence the distance 
 between the tops of the two columns. Ans. \ol .OZQ^l feet. 
 
 30. A gentleman a garden had. 
 
 Five score feet long and four score broad ; 
 
 A walk of equal width half round 
 
 He made, that took up half the ground : 
 
 Ye skilful in geometry, 
 
 Tell us how wide the walk must be. 
 
 Ans. 25.96S76feet 
 
 Note 1. Persepolis, mentioned in the 29th question, was the ancient capital of 
 the Persian empire. This immense and renowned city was taken by Alexander the 
 Great, about 300 years before the Christian era, who at the instigation of the de- 
 praved courtezan Thais, ordered it to be set on fire. Its magnificent ruins are 
 about .50 miles N. E. of Schiras, the present capital of Persia. 
 
 2. The 30th question was first proposed in the Ladies' Diarj- for the year 1708. 
 The poetry is not very elegant ; but the question is more original m its present 
 form than "it would be in prose. 
 
SECT. L 
 
 MENSURATION OF SOLIDS. 
 
 121 
 
 PAET IV. 
 
 MENSUEATION OF SOLIDS. 
 
 SECTION I. 
 
 DEFINITIONS. 
 
 1. A SOLID is a figure which generally consists of three 
 dimensions ; viz. length, breadth, and thickness. 
 
 2. The measurement of a solid is called its solidity, 
 capacity, or content. 
 
 3. The contents of solids are estimated by a cube, whose 
 side is one inch, one foot, one yard, &c. ; hence the solidity 
 of a body is said to be so many cubic inches, feet, yards, 
 &c. as are contained in that body. 
 
 4. A' cube is a solid having six 
 equal square sides. 
 
 5. A 
 
 parallelopipedon is a solid 
 
 having six rectangular sides, every |j||i 
 opposite two of which are equal and '' "'' 
 parallel. -^^ 
 
 6. A prism is a solid whose ends are tvv'o equal, parallel, 
 
 and similar plane figures; and its ^ _ 
 
 sides rectangles. \^ ^ 
 
 It is called a triangular prism \\ 
 when its ends are triangles ; a square 
 prism, when its ends are squares ; a pentagonal prism, 
 when its ends are pentagons, &c. 
 
 G 
 
122 
 
 MENSURATION 
 
 PAKT IV. 
 
 7. A cylinder is a solid conceived to be described 
 tlie revolution of a riglit-angled 
 parallelogram about one of its sides, 
 which remains fixed, and is called 
 the axis of the C}dinder ; or it is a 
 solid whose ends are parallel circles, and its sides right 
 lines. 
 
 8. A pyramid is a solid the base 
 of which is any plane figure what- 
 ever, and its sides are triangles, 
 meeting in a point, called the vertex 
 of the pyramid. 
 
 9. A cone is a solid conceived to 
 be described by the revolution of a 
 right-angled triangle about one of its 
 leo^s, which remains fixed, and is 
 called the axis of the cone ; or it is 
 a pyramid of an infinite number of 
 sides, having a circle for its base. 
 
 10. The frustum of a pyramid or cone is that part 
 which remains, when the top is cut off by a plane parallel 
 to the base. The part cut off the top is called a segment. 
 
 11. A ivedge is a solid whose base is a rectangle, its two 
 ends plane triangles, and its two opposite sides terminate 
 in an edge. 
 
 12. A prismoid is a solid whose 
 bases or ends are two right-angled 
 parallelograms, being parallel but 
 not similar to each other ; and its 
 sides four plane trapezoids. 
 
 13. A sphere or glohe is a solid conceived to be formed 
 
SECT. I. 
 
 OF SOLIDS. 
 
 123 
 
 by the rotation of a scmi-circlc about 
 its diameter, wliich remains fixed, 
 and is called the axis or diameter of 
 the sphere ; or it is a solid bounded 
 by one continued convex surface, 
 every part of which is equally distant 
 from a point within, called the centre. 
 
 14. The segment of a sphere is any part of it cut oif by 
 a plane. If the plane pass through the centre, it will 
 divide the sphere into two equal parts called hemispheres. 
 
 15. The zone of a sphere is a part intercepted between 
 two parallel planes ; and if these planes be equally distant 
 from the centre, it is called the middle zone of the sphere^ 
 
 16. A circular spindle is a solid 
 conceived to be formed by the revolu- 
 tion of a circular segment about its 
 chord, which remains fixed. 
 
 17. A cylindrical ring is a round solid, or a cylinder 
 bent into a ring. 
 
 18. A regidar body is a solid contained under a certain 
 number of similar and equal plane figures. 
 
 19. The faces of the solid are the plane figures under 
 wbich it is contained ; and the linear sides or edges are 
 the sides of the plane faces. 
 
 20. The greatest number of regular bodies which can 
 be formed is five; viz. 1st, the tetraedron, or regular 
 pjTamid, which has four triangular faces ; 2nd, the hex- 
 aedron, or cube, which has six square faces ; 3rd, the 
 octaedron, which has eight triangular faces ; 4th, the do- 
 decaedron, which has twelve pentagonal faces ; and 5th, 
 the icosaedron^ which has twenty triangular faces. 
 
 ifote. If the following figures be made of pasteboard, and the lines cut half 
 through, so that the parts may be turned up and glued together, they will represent 
 the five regular bodies ; namely, figure 1 the tetraedron, figure 2 the hexaedron, 
 figure 3 the octaedron, figure 4 the dodecaedron, and figure 5 the icosaedron. 
 
 G 2 
 
124 
 
 MENSUKATIOX 
 
 PART IV. 
 
 A Table of Solid Measure, 
 
 Cubic Inches. 
 
 Cubic Foot. 
 
 
 
 
 1728 
 
 1 
 
 Cubic Yard. 
 
 
 
 A6656 
 
 27 
 
 = 1 
 
 C. Pole. 
 
 
 
 7762392 
 
 44921 
 
 1663 
 
 
 C. Fur. 
 
 
 496793088000 
 
 287496000 I064S000 
 
 64000 
 
 = 1 
 
 C.Mile.j 
 
 254358061056000 
 
 147197952000i54517760O0 
 
 1 
 
 32768000 
 
 512 
 
 1 
 
 PEOBLEM I. • 
 
 To find the solidity of a cube. 
 
 RULE. 
 
 Multiply tlie side of the cube by itself, and that product 
 again by the side ; and the last product will be the solidity 
 required. 
 
 Note 1. The surface of a cube may be found by multiplying the square of its 
 side by six. 
 
 ■ 2. The side of a cvibe may be found by extracting the cube root of its solidity. 
 
 . 3. The best method of giving boys a competent knowledge of the different Prob- 
 lems in Mensuratiun of Solids, is to have the figures formed of wood. Tliis will be 
 found very advantageous ; as the learner may be taught both the theory and the 
 practice at the same time. 
 
SECT I. 
 
 OF SOLIDS. 
 
 125 
 
 EXAlirLES. 
 
 1. The side AB or BC of tlie cube 
 ABCDEFGH is 5 feet 6 inches; 
 what is its solidity ? 
 
 3ij Decimals. 
 
 By 
 
 Duodecimals. 
 
 Feet, 
 5.5 
 
 5.5 
 
 275 
 275 
 
 
 Ft. In. 
 5 6 
 5 6 
 
 27 6 
 2 9 
 
 30.25 
 5.5 
 
 
 30 3 
 5 6 
 
 15125 
 15125 
 
 
 151 ^pa. 
 15 1 6 
 
 166.375 
 
 
 166 4 6 ^;z5. 
 
 12 
 
 4.500 
 
 12 
 
 6.000 
 
 Ans. 166 ft 4 in. 6 pa. 
 
 By Logarithms. 
 
 The log. of .... 5.5 is 0.7403627 
 
 Multiply by the index of the power ... 3 
 
 The product is the log. of 166.375, the ans. . 2:2210881 
 
 2. Wliat is the solidity of a cube, whose side measures 
 3 feet 6 inches ? ' 
 
 Fiere Z feet 6 inches=^Z.5 ; and 3.5'' = 3.5 x3.5 x 3.5 = 
 12.25x3.5=42.875 /6'ej^=:42/<f. 10 m. 6pa. {or 6" se 
 conds) the solidity required. 
 
 3. K the side of a cubical piece of timber measures 
 2 feet 9 inches ; what is its solid content ? 
 
 Ans. 20ft. 9 in. 6^ pa, 
 
 4. Wliat is the solidity of a cubical stone, whose side 
 measures 2 feet 10 inches ? Ajis. 22ft. 8 i?i. IV 4'"'. 
 
 G 3 
 
126 
 
 MENSURATION 
 
 5. TTliat did the inside of a cubical bin, that liolds 
 exactly 5 Winchester bushels, cost painting, at l^d. per 
 square foot ? Ans. 2s. 6^d. 
 
 PROBLEM n. 
 To find the solidity of a i:)arallelopipedon. 
 
 RULE, 
 
 Multiply the length by the breadth, and that product by 
 the depth or altitude, and it will give the solidity required. 
 
 'Note 1. In order to find the surface of a parallelopipedon, multiply the jjeriphery 
 of its end by its length, and the product will be the area of the sides ; of which add 
 twice the area of its end, and the sum will be the whole surface. 
 
 2. The surfaces of all similar solids are to each other as the squares of their like 
 lineal dimensions. This is erident from Tiieorem IG. Part I. 
 
 3. If the content of a parallelopipedon be divided by the product of any two of 
 its dimensions, the quotient will be the other dimension. 
 
 4. Parallelopipedons of equal altitudes are to each other as their bases ; and if 
 equal bases they are to each other as their altitudes. 
 
 EXAMPLES. 
 
 1. "What is the solidity of the pa- 
 rallelopipedon ABCDEFGH, whose 
 length AB is 12 feet 9 inches, breadth 
 BC 8 feet 6 inches, and depth or alti- 
 tude AE 6 feet 3 inches ? 
 
 By Decimals » 
 Feet. 
 
 12.75 
 
 8.5 
 6375 
 10200 
 
 108-375 
 
 6.25 
 
 541875 
 
 216750 
 
 650250 
 
 677.34375 
 
 12 
 
 4.12500 
 12 
 
 1.50000 
 
 12 
 
 6.00000 
 
 By Duodecimals. 
 
 Feet. In. 
 
 12 9 
 _ 8_6 
 102 
 
 6 4 6 ^^ 
 108 4 6 
 6 3 
 
 650 3 
 2 7 1 1 6^"^ 
 677 4 1 GAju 
 
 Ans. 677/e. 4 in. V 6' 
 
SECT I. 
 
 OF SOLIDS. 12' 
 
 ' By Lofjar'ithms. 
 
 The Ion. of , . . . 12.75 Is .... 1. 10.55102 
 
 , .... 8.5 25 ... . 0.9294189 
 
 .... 6.25 Is .... 0.79588^0 
 
 Their sum is the log. o/ 677.34375, the cms. 2.8308091 
 
 2. The lengtli of a piece of timber is 24.65 feet, the 
 breadth 2.82 feet, and the depth 1.56 feet; what is its 
 solidity ? Ans. 108.44028 feet. 
 
 3. In a quarry near Balbec, is a stone whose length is 
 70 feet, its breadth 14 feet, and its depth 14 feet 5 inches; 
 what is its solidity ? Ans. 14128/^. 4 in. 
 
 4.. The length of a cellar is 25 feet, its breadt4i 15 feet, 
 and its depth^ 7 feet ; what did it cost digging, at 3 jc?. per 
 solid yard? Ans. £1. lOs. ^d. 
 
 5. A brick-kiln measures 27 feet 6 inches in length, 18 
 ftjet 4 inches in breadth, and 10 feet 9 inches in height ; 
 how many bricks, each 10 inches long, 5 inches broad, 
 and 3 inches thick, will it contain at once, admitting them 
 to be close piled ? ^ Ans. 62436. 
 
 6. A parallelopipedon measures 18 feet in length, 12 
 feet in breadth, and 8 feet 6 inches in depth ; what did it 
 cost painting, at 9^d. per square yard? A?is. £4. 2s. lO^d. 
 
 PROBLEM III. 
 
 To find the solidity of a prism. 
 
 EULE. 
 
 Multiply the area of the base by the perpendicular 
 height of the prism, and the product will be the solidity. 
 
 'Sof£ 1. If the area of the base in inches, be multiplied by the height in inches, 
 the product must be divided by 1728, in order to obtain the content in cubic feet ; 
 but when the area of the base in inches, is multiplied by the height in feet, the 
 product divided by 144, will give the jsolidity in feet. 
 
 2. The surface of a prism may be found thus : Multiply the area of one side by 
 the number of sides ; to the product add twice the area of the base, and the sum 
 will be the turface required. Or, if the circumference of a prism be multiplied by 
 its length, the product will be the surface of all the sides; to which add twice the 
 area of its base, and the sum will be the whole surface. 
 
 3. The altitude of a prism may be found by dividing the solidity by the area of 
 the base. 
 
 4. The area of the base may be found by Problems 11 and 12, Varl 11. 
 
 5. All prisms and cylinders of equal bases and altitudes are equal to each other. 
 
 EXAMPLES. 
 
 1, Eequired the solidity of the triangular prism ABC 
 DEF, whose length AB is 14 feet 8 inches, and each of 
 G 4 
 
128 
 
 MENSURATION 
 
 TART IV. 
 
 the equal sides of the base ADE 6 feet 
 4 inches. 
 
 Here 14: ft. 8 in. = 176 inches, and Q ft. 
 4in.=:7 6 inches; then by Prob. 12, Part 11.^ 
 we have .4330127 X 76- = .4330127 X 5776 
 = 2501.0813552, the area of the base 
 ADE. 
 
 Or, by Prob. 5, Part II., tee have C76 
 + 76 4- 76) -f- 2 = 228 -2 = 114, half the 
 sum of the sides; and 114— 76= 38 ~ each 
 remainder ; whence J (1 14 x 38 X 38 X 38) 
 = ^6255408 = 2501.08136, the area of 
 the base as before; then (2501-08136 X 
 176) ^ 1728 = 440190.31936 -- 1728 = 254.73976 /eei? 
 = 254 Z'^. 8 in. 10 pa., the solidity required. 
 
 2. The perpendicular altitude of a stone pillar is 8 feet 
 9 inches, and the side of its square base 2 feet 3 inches ; 
 what is its solidity? ' Ans. 44 ft. 3 in. 6^ pa. 
 
 3. The length of a square beam of timber is 32 feet 
 6 inches, and its circumference or girt 5 feet ; what is its 
 solidity? Ans. 50ft. 9 in. 4^ pa, 
 
 4. The side of a pentagonal stone measures 10 inches, 
 and its length 9 feet 4 inches ; what is its solidity ? 
 
 A72S. U. 1 5124 feet. 
 
 5. A hexagonal prism measures 2 feet 6 inches across 
 tlie centre of its end, from corner to corner, and 25 feet 
 3 inches in length ; what is its solidity ? 
 
 Ans. 102.502225 feet. 
 
 6. In the walls of Balbec is a stone in the form of a 
 square prism, whose length measures 63 feet, and its 
 breadth and depth each 12 feet ; what is its solid content? 
 
 Ans. 9072 feet. 
 
 7. The gallery of an assembly-room is supported by 16 
 octagonal prisms of wood, each of which measures 4 feet 
 8 inches in circumference, and 12 feet in height; required 
 the solidity of the whole, and what they cost painting at 
 9d. per square yard ? 
 
 Ans. The solidity of the 16 prisms is SI 5. 457 2368 fee f^ 
 and they cost £S. \4s. 8d. painting. 
 
 Kofe. T>albec, mentioned in the sixth question, is a town of Syria, situated about 
 37 miles north of Damascus. Its remains of antiquity display, according to the best 
 judges, the boldest plan that ever was attempted. 
 
 The inhabitants of Asia consider Solomon as its founder ; and the nobleness of 
 the architecture, the beauty of the ornaments, and the stupendous execution of t!ie 
 whole, seem to fix its foundation to a period before the Christian era; but, in all 
 probability, the Jews knew little of the Grecian style of building and ornamenting 
 in the time of Solomon. 
 
SECT I. OF SOLIDS. 129 
 
 proble:vi IV. 
 
 To find the soIkUtj/ of a cylinder, 
 
 RULE. 
 
 Multiply the area of the base by the perpendicular alti- 
 tude of the cylinder, and the product will be the solidity. 
 
 Note \. If the circumference of a cylinder be multiplied by its height, the product 
 will be the convex surface ; to which" add twice the area of the base, and the sum 
 •will be the whole surface of the cylinder. 
 
 2. Tlie altitude of a cylinder may be found by dividing the solidity by the area of 
 the base. 
 
 3. The area of the base of a cylinder may be found by Problem 15, Part II. 
 
 4. Similar prisms and cylinders are to each other, as the cubes of their altitudes, 
 or of any other like dimensions. 
 
 EXAJIPLES. 
 
 1. Required the solidity of the cylinder ^ -v,^c 
 
 ABCD, the diameter of the base AB of ^^^^llllllIZ^ 
 which is 2.76 feet, and the altitude BC Hljii 1 1{ 
 
 4.58 feet. m I 
 
 Here 2.76^ X .7854 = 2.76 x 2.76 x i||||i 
 .7854= 7.6176 x .7854 = 5.98286304, the || : 
 area of the base; arid 5.98286304 x 4.58 i|| 
 = 27.4015127232 /ee^, the solidity re- ^fe ||r 
 quired. ^ "^ 
 
 2. The altitude of a cylindrical stone column is 8 feet 
 3 inches, and its circumference 5 feet 6 inches ; what is 
 its solidity? Ans. I'^.m^l'^ feet. 
 
 3. The circumference of a cylindrical piece of timber is 
 6 feet 8 inches, and its length 24 feet ; what is its solid 
 content? ^725. 84.88533 /ee^. 
 
 4. The diameter of a rolling-stone is 18.7 inches, and 
 its length 4 feet 9 inches ; required its solidity. 
 
 Ans. 9.05952 /ee/^. 
 
 5. The diameter of a well is 3 feet 9 inches, and its 
 depth 45 feet ; what did it cost sinking at 7^. Zd. per cubic 
 yard? Ans. ^6. 13^. ^\d. 
 
 6. The greater diameter of a hollow iron roller is 1 foot 
 9 inches, the thickness of the metal 1^ inch, and the length 
 of the roller 5 feet ; now supposing a cubic foot of cast iron 
 to weigh 464 pounds avoirdupois; what did the roller 
 cost at 195. 9r/. per hundred weight, and how many times 
 will it turn round in rolling 5 acres of land ? 
 
 Ans. The roller cost £13. Is. O^d., and it ivill turn 
 round 7923.16926 times in rolling 5 acres of land. 
 G 5 
 
ISO MENSURATION PART IV. 
 
 PROBLEM y. 
 
 To find the solidity of a pyramid. 
 
 liULE. 
 
 Multiply the area of the base by 4 of the perpendicular 
 height, and the product will be the solidity. 
 
 Uote 1. The surface of a pyramid may be found thus : Multiply the perimeter of 
 the base by the length of the side, or slant height, and half the product will be tlii? 
 surface of the sides ; to which add the area of the base, and the sum will be thu 
 whole surface. 
 
 2. The perpendicular height of a pyramid or cone is a line drawn from the vertex 
 to the middle or centre of the base ;' and the slant height is the distance between 
 the middle of one side of the base and the vertex. 
 
 3. In order to find the slant height of a pyramid from the perpendicular height, 
 or the perpendicular height from the slant height, it will be necessary to find th<s 
 distance between the centre of the base and the middle of one of its sides, which 
 may be done bv the following Rule: Multiply the number answering to the poly- 
 gonal base, in the third column of the table of polygons, Problem 10, Part 11., by 
 the side of the base ; and the product will be the distance required. 
 
 4. If three times the solidity of a pyramid be divided by the area of the base, the 
 quotient will be the perpendicular altitude. 
 
 5. The area of the base of a pyramid may be found by Problems 11 and 12, 
 Part II. 
 
 6. p:very triangular pyramid, is the third part of a prism of the same base and 
 altitude. 
 
 7. In any pyramid, a section parallel to the base is similar to the base ; and those 
 two sections are to each other as the squares of their distances from the vertex. 
 
 EXAIVLPLES. 
 
 1. Required the solidity of the hexa- 
 gonal pyramid ABODE, each side of 
 its base being 3 feet 9 inches, and its per- 
 pendicular altitude EF 15 feet 9 inches. 
 
 By Prob. 12, Part II., we have 
 2.5980762 x 3.75'-^ = 2.5980762 x 14.0625 
 = 36.5354465625, the area of the base; 
 and 36.5354465625 x 15.75 -- 3 = 36. 
 5354465625 x 5.25 = 191.811094453125 
 feet, the solidity required. 
 
 2. The three sides of a triangular 
 pyramid are 6, 7, and 8 feet, and its per- 
 pendicular altitude 18 feet ; what is its 
 solidity? Ans. 12 1.998975 /ee^. 
 
 3. The altitude of a square pyramid is 15 feet 9 inches, 
 and the side of its base 2 feet 6 inches ; required its so- 
 lidity. Ans. ^2 ft. 9i?i. 9 pa. 
 
 4. The spire of a church is a regular hexagonal py- 
 ramid, whose perpendicular height measures 48 feet, the 
 side of its base 10 feet, the perpendicular height of the 
 cavity or hollow part 42 feet, and its side at the base 8 
 feet ; how many cubic feet of stone are contained in the 
 spire? Ans. 1829.0456448 /ee^. 
 
SECT. I. 
 
 OF SOLIDS. 
 
 131 
 
 5. The slant height of a regular, pentagonal, stone py- 
 ramid, measures 18 feet 6 inclies, and the side of its base 
 3 feet ; what was its whole expense ; 6s. 9d. being paid 
 for each cubic foot, and 6Ui- per square foot, for polishing 
 the sides ? 
 
 A71S. £35. I5s. l^d. 
 
 PROBLEJil VI. 
 
 To find the solidity of the frustum of a jyyramid. 
 
 GENEKAX EULE. 
 
 To the areas of the two ends of the frustum add the 
 square root of their product ; and their sum being mul- 
 tiplied by 4- of the perpendicular height, will give th^ 
 solidity. 
 
 Note 1. This rule will also give the solidit.v of the frustum of a rigiit cone; riiid 
 likewise that of the frustum of an elliptical cone, generally called a cylindroid. 
 {Sec Proh. 4. Fart VIII. Gauging.) 
 
 2. To find tlie perpendicular altitude of the frustum of a pyramid : To the arens 
 of the two ends add the square root of their product, for a divisor ; then, divide 
 three times the solidity or content, by this divisor, and the quotient will be tlis 
 
 altitude required This rule will hold good, whatever may be the form of the 
 
 ends of the frustum. 
 
 RULE n. 
 
 If the ends he regular 'polygons. 
 
 Add together the square of a side of each end, and 
 the product of those sides ; multiply the sum by 4- of the 
 height, and this product by the tabular area belonging to 
 the polygon, (Prob. 12,. Part II.) and the last product will 
 be the solidity. 
 
 Note. The surface of the frustum of a pyramid may be found thus : Multiply 
 half the sum of the perimeters of the two ends, by the slant height, and the product 
 will be the surface of the sides ; to which add the areas of the ends, and the sum 
 will be the whole surface of the frustum. 
 
 exa:\iples. 
 1. What is the solidity of the frustum e 
 
 ABCDEF, of a square pyramid ; the side 
 AB of the greater end being o feet, the side 
 DE of the less end 2 feet, and the perpen- 
 dicular height GH 12 feet ? 
 
 By the General Rule. 
 
 Here -5x5=: 25, the area of the greater 
 end ; and 2 x 2 = 4, the area of the less ; 
 also, ^/ (25 X 4) = ^/ 100 = 10, the square 
 root of their product; then (25 -f 4 + 
 
 10) X 4 =;, 39x4; 
 quired. 
 
 156 feet, the solidity re- 
 G 6 
 
132 . MENSUKATIOX PART IV. 
 
 By Rule 11. 
 
 Here 5-' + 2^ + (5 x 2) = 25 + 4 + 10 == 39 ; then 
 39 X 4 X 1 = XZQfeet^ the solidity as before. 
 
 2. Each side of the greater end of a piece of squared 
 timber is 28 inches, each side of the less end 14 inches, 
 and its length 18 feet 9 inches ; how many solid feet does 
 it contain ? Ans. 59.54861 1 feet. 
 
 3. The length of a piece of timber is 17 feet 3 inches, 
 and its ends are similar rectangles; the length of the greater 
 being 36, and its breadth 20 inches ; the length of the less 
 18, and its breadth 10 inches, how many solid feet are 
 contained in the frustum? Ans. 50.3125 /ee^. 
 
 4. "VYhat is the solidity of an octagonal stone pillar ; 
 each side of the greater- end being 12 inches, each side of 
 the less end 8 inches, and the perpendicular altitude 
 9 feet ? Ans. 30.58003 feet. 
 
 5. The roof of a portico is supported by four regular, 
 hexagonal, marble pillars, whose perpendicular altitude is 
 15 feet, the side of the greater end 8 inches, and the 
 side of the less end 5 inches ; what did they cost at £4. lOs-. 
 per solid foot ? A71S. £209. 9s. 4|^. 
 
 6. The perpendicular altitude of a square chim^iey, in 
 Leeds, is 120 feet 9 inches, the side of its base 10 feet 9 
 inches, and the side of its top 5 feet 9 inches. The cavity 
 or hollow part is a square prism, whose side is 2 feet 6 
 inches ; how many soHd feet are contained in the chimney, 
 and what is the area of its four sides ? 
 
 Ans. The solidify of the chimney is 7715.421875 cubic 
 feet, and the area of the four sides is 3985.603941 square 
 yeet. 
 
 PROBLEM VIL 
 
 To find the solidity of a cone. 
 
 RULE. 
 
 Multiply the area of the base by the perpendicular 
 height, and 4- of the product will be the solidity. 
 
 2^ote 1. The altitude of an oblique cone, or one whose axis does not make a right 
 angle with the plane of a base, may be found bydemitting a perpendicular from the 
 vertex of the cone, upon the plane. 
 
 2. If the circumference of the base of a cone be multiplied by half its slant 
 height, the product will be the convex surface ; to which add the area of the base, 
 and the sum will be the whole surface of the cone. 
 
 3. The solidity of every cone is equal to one third of the solidity of its circum- 
 scribing cylinder. 
 
 4. All prisms, cylinders, pyramids, and cones, of the same altitude, are to each 
 
SECT. I. 
 
 OF SOLIDS. 
 
 133 
 
 other as their bases ; consequently, prisms, cylinders, pyramids, and cones of equal 
 bases, are to each other as their altitudes. 
 
 5. The i)erpendicijl ir altitude of a cone may be found by dividing three times the 
 solidity by the area of the base. 
 
 G. 'ihearea of the base of a cone may be found by Problem 15, Part II. 
 
 7. The surfaces of similar right cones areas the squares of their axes, and their 
 solid contents as the cubes of their axes. 
 
 exa:mples. 
 
 1. Required the solidity of the cone 
 ABC; the diameter AB of the base 
 being 12 feet, and the perpendicular 
 altitude DC 18 feet 6 inches. 
 
 Here .7854 x 12-' = .7854 x 144 = 
 113.0976, the area of the base; and 
 (113.0976 X 18.5) -- 3 = 2092.3056 -- 
 3 = 697.4352 feet, the solidity re- 
 quired. 
 
 2. The diameter of the base of a 
 conical piece of timber is 5 feet 6 
 
 inches, and the length of its slant side 25 feet 9 inches ; 
 what is its solidity? Ans. 202.75954/eet 
 
 3. The circumference of the base of a conical stone is 
 12 feet 6 inches, and its perpendicular altitude 8 feet 
 9 inches ; what is its solidity? Ans. 36.266927 /ee^. 
 
 4. ^Yhat wiU be the expense of tooling or dressing a 
 conical spire, at %\d. per square foot ; the circumference 
 of the base being 30, and the slant height 45 feet ? 
 
 ^Ans. ^18. 55. 1\d. 
 
 PROBLEM VIII. 
 
 To find the solidity of the frustum of a cone. 
 
 EULE I. 
 
 To the product of the diameters of the two ends, add 
 the sum of their squares ; then this sum being multiplied 
 by tlie height, and again by .2618, (one-third of .7854,) 
 will give the solidity. 
 
 RULE n. 
 
 Divide the difference of the cubes of the diameters of 
 the two ends, by the difference of the diameters ; then this 
 quotient being multiplied by .7854, and the product by \ 
 of the height, will give the solidity. 
 
 cone may also be found by the general 
 
 Note 1." The solidity of the frustum o 
 rule, given in Problem G. 
 
134 MENSURATION PART IV. 
 
 •> The surface of the frustum of a cone mav be found thus : Multiply half the 
 Mun of the perimeters of the two ends, by the slant height, and the product willTie 
 the convex surface ; to which add the areas of the ends, and the sum will be tho 
 whole surface of the frustum. ,i,„ „,„/.„ot 
 
 3 To find the perpendicular altitude of the frustum of a cone : 1 o the prodiint 
 of the diameters of the two ends add the sum of their squares, and multiply t\ur^ 
 result by .2018 for a divisor ; then, divide the solidity of the frustum by this diviKor, 
 and the'quotient will be the perpendicular altitude. 
 
 EXAMPLES. 
 
 1. ^Vhat is the solidity of the frustum ABCD of a cone ; 
 the diameter AB of the greater end ^ <::::4=I:::Nq 
 being 7 feet, the diameter CD of the |--~-Xl^ 
 less end 5 feet, and the perpendicular f|| i 'M 
 altitude EF 8 feet 3 inches ? 1 Nlil 
 
 By Rule 1. i | |1 
 
 Here (7 X 5) + 7^ + 5^=r:35 + 49 + 25 ||--''''i'"'''|!i „ 
 = 109 ; mid 109 x 8.25 X .2618=899.25 ^ %---"-^--l# 
 
 X .2618 — 235.42365 feet, the soliditij ^ -"^ 
 
 rermired. 
 
 By Ride 11. 
 
 Here {V - 5^)-^ (7 - 5) x .7854 x 8.25 -4-3 = (343 - 125) 
 --2 X .7854 X 2.75 = 218-r-2 x .7854 x 2.75=109 x .7854 
 X 2.75 =:85.6086x 2.75 = 235.42365 /ee^, the solidity as 
 before. 
 
 2. The greater diameter of a piece of timber is 2 feet 
 6 inches, the less 1 foot 3 inches, and the length 25 feet 
 9 inches ; what is its soUdity ? Aiis. 73^3351 feet. 
 
 3. The circumference of a piece of round timber is 65 
 inches at the " greater end, 35 inches at the less end. 
 and its len"-th 33 feet 9 inches ; what is its solid content? 
 
 A?is. 48.027773 /ee^. 
 
 4. The pediment in the front of the court-house, in 
 Leeds, is supported by four stone columns, whose perpen- 
 dicular altitude is 22 feet 9 inches, the diameter at tho 
 bottom 2 feet 9 inches, and at the top 2 feet 3 inches ; re- 
 quired their solidity. Ajis. 448.1 85237 /^e.'. 
 
 5. The slant height of the hovel or chimney of a pot- 
 kiln is 45 feet, the circumference at the bottom 113 feet 9 
 inches, and at the top 31 feet 3 inches ; required its convex 
 surface. A7is. 3262.5 feet, 
 
 6. The top diameter of a lime-kiln is 15 feet 10 inches, 
 the bottom diameter 3 feet 8 inches, and the perpendicu- 
 lar depth 21 feet 4 inches ; how many bushels of lime wiB 
 it burn at once ; allowing live pe(;ks, or 2688 cubic inches 
 to a bushel? Ans. 1156.80693 bushels. 
 
SECT. T. OF SOLIDS. 13.J 
 
 7. The sluift of Pompey's pillar is a single stone of granite 
 whose ])erpendicular altitude is 90 feet, the diameter at the 
 base 9 feet, and at the top 7 feet 6 inches ; required its 
 solidity. A?is. 4824.3195/m^ 
 
 Note. Pompey's Pillar is situated near Alexandria, in Egypt. It is a fine 
 rcfnilar column, of the Corinthian order ; and its whole height, including the 
 capital and pedestal, is 114 feet. Upon its top has been a statue, which must have 
 been of a gigantic stature, to hare appeared of the ordinary size of a man, to a 
 spectator at the bottom. 
 
 The remains of this statue, consisting of a foot and an ancle, were only dis- 
 covered about 30 years ago, by some Jollt/ sons of Neptune, that ascended the 
 pillar, and drank abowl of punch upon the top of it, amidst the shouts and ac- 
 clamations of the natives, who were astonished at the address and boldness oi the 
 Jiritish tars. 
 
 This object thev accomplished by means of a rope, which they contrived to draw 
 orex the top of the pillar, bv the assistance of a paper kite. One of the company then 
 ascended ; and a kind of s'hroud being constructed, seven persons more went up ; 
 .-vnd the whole descended again, without the least accident, except the falling of a 
 portion of the capital, which they carried off in triumph. 
 
 Learned men and travellers liave made many fruitless attempts to discover in 
 honour of whom this majestic monument was erected. The best informed have 
 (included that it could not be in honour of Pompey, since neither Strabo nor 
 Diodorus Siculus has mentioned it. The Arabian historian Abulfeda, in his de- 
 scription of Egypt, calls it " The Pillar of Severus ; " hence Mr. Savary conjectures 
 that it was erected to the honour of the emperor Severus by the inhabitants of 
 Alexandria, to whom, as history informs us. he granted several favours. 
 
 Severus died at York, a.d. 211 ; and Pompey's Pillar did not receive that name 
 until the fifteenth century. 
 
 PROBLEM IX. 
 
 To find the solidity of a cunens or ivedge. 
 
 KULE. 
 
 To twice the length of the base, add the length of the 
 edge, multiply the sum by the breadth of the base, and 
 this product by the perpendicular height of the wedge ; 
 and I of the last product will be the solidity. 
 
 Ntite 1. The surface of a wedge may be obtained by finding the area of each of its 
 sides or parts separately, and adding them into one sum. 
 
 2. When the length of the edge is equal to the length of the base, the solidity 
 of the wedge is evidently equal to half the solidity of a prism of the same base and 
 altitude ; but if the edge be longer or shorter than the base, the wedge is manifestly 
 greater, or less than half a prism. 
 
 EXAJIPLES. 
 
 1. Required the solidity of the wedge ABCDEF, the 
 length AB of its base being 26 inches, 
 
 the breadth BC 18 inches, the length ^;| "■' 
 
 of the edge EF 15 inches, and the per- ^| 
 
 pendicular height PE 28 inches. ,^]! 
 
 Here (26 x 2) + 15 = 52 + 15=67 ; M| 
 
 and{Ql X 18 X 28) -6 = 67 X 3x28 ^j 
 
 = 5628 cithic inches, the solidity re- ^^ ^^^ 
 
 quired. a 
 
136 mensuhation part it. 
 
 2. Required the solidity of a wedge, the length and 
 breadth of the base being 82 and 35 inches, the length cf 
 the ed'^e 115 inches, and the perpendicular height 8 feet. 
 
 Ans, 90Al666feef. 
 
 3. What is the soldity of a wedge, the length and 
 breadth of whose base is 5 feet 9 inches and 3 feet 6 
 inches, the length of the edge 4 feet 3 inches, and the 
 perpendicular height 9 feet? A?is. 82.687 5 feet. 
 
 PROBLEM X. 
 To find the solidity of a prismoid. 
 
 EULE. 
 
 To the sum of the areas of the two ends add four times 
 the area of a section parallel to and equally distant from 
 both ends ; multiply this sum by the perpendicular height, 
 and i of the product will be the solidity. 
 
 Note 1. The length of the middle section is equal to half the sum of the lengths 
 of the two ends ; and its breadth is equal to half the sum of their breadths. 
 
 2. The surface of a prismoid may be obtained by adding the areas of its four sides 
 and two ends into one sum. , ,, . x. r a -^ -o 
 
 3. As the ends of a prismoid are rectangles, their areas may be found by Pro- 
 blem 2, Part II. , ^ . .J m .u r.u c 
 
 4 To find the perpendicular height of a prismoid : To the sum of the areas of 
 the two ends add four times the area of the middle section, for a divisor; then, 
 divide six times the content bv this divisor, and the quotient will be the perpen- 
 dicular altitude. This Rule wUl hold good for cylindroids, whatever may be the 
 form of their ends ; and also for all frustums and solids whose parallel sections are 
 similar figures. 
 
 REIVIAEK. 
 
 The last Rule, and the similar Rules given in the preceding Problems, will be 
 found useful when the content or capacity of a solid, and all the horizontal dimen- 
 sions are given, to find the perpendicular altitude. This is often the case in giving 
 orders for the making of cisterns, couches, tuns, vats, tubs, and other vessels oi 
 various descriptions. {See Gauging, Part VIII.) 
 
 SCHOLIUM, 
 
 The Rule given in this Problem for a prismoid, is very clearly 
 demonstrated in Simpson's Fluxions, page 178, first edition; and also 
 in a similar manner at page 302, of HoUiday's Fluxions. 
 
 There is likewise a very elegant dem.onstration given in Proposi- 
 tion III., Section I., Part'lV., of Dr. Hutton's Mensuration, third 
 edition, in which it is shown, to be true for all frustums whatever, 
 and for all solids whose parallel sections are similar ligures ; and 
 Mr. Fletcher, in his second edition of his Universal Measurer, Part 
 III., page 54, and Mr. Moss, in his Gauging, page 175, third edition, 
 say that it is nearly true for any other solid, whatever may be its 
 form. 
 
 exa:mples. 
 ]. TTliat is the solidity of the prism.oid ABCDEFGH, 
 the length AB of the greater end being 18, and its 
 
SECT. I. OF SOLIDS. 137 
 
 breadth EC 8 inclies ; the lenjrth EF 
 of the less end 12, and its breadth FG- 
 6 inches ; and the perpendicular height 
 PE 18 inches? 
 
 Here 18x8 + 12x6= 144 -h 72 
 = 216, the sum of the areas of the 
 two ends. 
 
 Also, (18 + 12)-+2=30--2 = lo, 
 the length of the jniddle section ; and 
 (8 + 6-4-2 = 14-^-2 = 7, the breadth of the middle sec- 
 tion ; then lox7x4 = l()5x 4 = 420, four times the 
 area of the middle section ; whence (216 + 420) x 18-^6 
 = 636 X 3 = 1908 cid)ic inches, the solidity required. 
 
 2. How many solid feet of timber are contained in a 
 beam whose ends are rectangles ; the length and breadth 
 of the greater being 32 and 20 inches; the length and 
 breadth of the less 16 and 8 inches ; and the perpendicular 
 length 25 feet ? Ans. 61 ^ feet. 
 
 3. The perpendicular altitude of a stone pillar is 8 feet 
 6 inches ; the length and breadth of the greater end are 
 26 and 16 inches ; and the length and breadth of the less 
 end 18 and 10 inches ; required the solidity of the stone. 
 
 A71S. 17.118055 feet. 
 
 4. The length and breadth of a fish-pond at the top are 
 1 32 and 64 yards ; the length and breadth at the bottom 
 116 and 48 yards; and the perpendicular depth 12 feet 
 9 inches ; what did it cost digging, at 6^d. per cubic yard ? 
 
 Ans. £832. Us. 6d. 
 
 5. What is the capacity of a coal waggon, the inside 
 dimensions of which are as follow : at the top, the length 
 is 801, and breadth 56 inches ; at the bottom, the length 
 is 41-|, and the breadth 32 inches ; and the perpendicular 
 depth 47 inches? Ans. 129814 ctibic inches. 
 
 Xofe. Coals are frequently carried from pits in prismoldal waggons, having four 
 cast iron wlieels, which run upon a railway. 
 
 They art; prevented from leaving the railway by a particular construction of the 
 wheels ; the inner edge of the rim of eacli wheel being made to project. Coals are 
 conveyed, in waggons of the above description, from tiie pits in the vicinity of New- 
 castle, to the river Tyne ; from those near Sunderland, to the jriver Weir ; and from 
 Middleton Colliery, to the town of Leeds. 
 
 In the last question are given the dimensions of a Leeds coal waggon, which 
 liffer s but little from those of other places. 
 
 ■ Instead of rails, the waggon-road between Middleton and Leeds, is made with 
 bars of cast-iron ; and the waggons are drawn by a machine which moves along the 
 road before them, by the force of steam. This machine was invented by Mr.Blen- 
 kinsop, of Middleton ; and is called " Blenkinsop's. Patent Steam Carriage." It is 
 considered to be one ,of the most curious pieces of mechanism that ever appeared 
 in England, or perhaps in the world. 
 
 It moves at the rate of about three miles and a half in an hour ; and is capable of 
 drawing 30 waggons at once, upon level ground, which are computed to weigh 105 
 tons 
 
lo3 mexsl'hatiox vxjit ly. 
 
 PROBLEM XL 
 To find the solidity of a sphere or globe. 
 
 RULE. 
 
 Multiply tlie cube of tlie diameter by .5236, or the cube 
 of the circumference by .016887, and the product will be 
 the solidity. 
 
 Note 1. The convex surface of a sphere may be found b}' multiplying the circum- 
 ference by the diameter. Or, multiply the square of the diameter by 3.1416, and 
 the product will be the convex surface. 
 
 The solidity of a sphere is equal to two thirds of the solidity of its circumscribing 
 cylinder. 
 
 3. The surface of a sphere is equal to 4 times the area of a circle of the sama 
 diameter as tlie sphere ; or to the area of a circle whose diameter is double that of 
 the sphere; or to the coutcx surface of the circumscribing cylinder. 
 
 EXA5IPLES. 
 
 L What is the solidity of a globe -^^^^^^ 
 
 whose diameter AB is 25 inches ? ,.^S^^^^^ 
 
 Here 2o^ X .5236 =25 X 25 x ^p ^^^ 
 
 25 X .5236 = 625 x 25 x .5236 = ^b ^p^ 
 
 15625 X .5236 =: 8181.25 cubic in- ^~ ~^^ 
 
 ches, the solidity required. ^fcj ' ~^'^ 
 
 2. Wnat is the solidity of a ^^^^ „. .. . / 
 spherical stone whose circumfer- -—-.-- 
 
 ence measures 5 feet 6 inches ? Ayis. 2.80957/ce^. 
 
 3. If the circumference of a cannon-ball be 18.6 in- 
 ches ; what is its solidity ? Ans. 108.66541 inches. 
 
 4. The diameter of the moon is 2180 miles ; what is 
 her solidity in cubic miles ? Ans. 5424617475.2 miles. 
 
 5. The diameter of the earth is 7957J miles ; what is 
 its solidity in cubic miles, and its convex surface in square 
 miles ? 
 
 A71S. The solidify is 263858149120 ciibic miles, and the 
 cojivex surface is 198944286 square miles. 
 
 PROBLEM XII. 
 
 To find the solidity of a segment of a sphere. 
 
 To three times the square of the radius of the segment's 
 base, add the square of its height ; and this sum multi- 
 plied by the height, and tlie product again by .5236, wiU 
 give the solidity. 
 
SKCT. T. OF SOLIDS. 139 
 
 RULE n. 
 
 From three times the diameter of tlie sphere subtract 
 twice the height of the segment ; multiply the remainder 
 by the square of the height, and that product by .52361 
 for the solidity. 
 
 Note 1. The surface of the segment of a sphere may he found thus : Multiply the 
 circumference of the whole gphore by the height of "the segment, and the product 
 will he the convex surface ; to which add the area of the base, and the sum will be 
 the whole surface. 
 
 2. Tiie area of a spherical triangle, or the spherical surface included by the arcs 
 of three great circle? of the sphere, intersecting each other, may be found by the 
 following rul« . 
 
 As two right angles, or IRO degrees. 
 
 Are to the area of a great circle of the sphere ; 
 
 So is the excess of the three angles above two right angles. 
 
 To the area of the triangle. 
 
 EXAMPLES. 
 
 1. The radius An of the base of 
 the segment ABC is 9 inches, and 
 the height Cn 7 inches ; what is the 
 solidity of the segment ? 
 
 Bf/ Rule /., we have 9- X 3 + 
 72 = 81x3 + 49 = 243+49 = 292; \ 
 and 292 X 7 X .5236=2044 X .5236 
 
 =: 1070.2384 inches, the solidity re- '-- --''' 
 
 quired. 
 
 2. K the diameter of a sphere be 3 feet 6 inches ; what 
 is the solidity of a segment whose height is 1 foot 3 
 inches? Ans. 6.6 '^S feet. 
 
 3. "^"^Hiat is the solidity of a spherical segment whose 
 height is 24.8 inches, and the radius of its base 30.6 
 inches ? Ans. 25.73099 /ee^. 
 
 4. Required the solidity and convex surface of each of 
 the frigid zones, whose height is 330.0075 miles, and the 
 diameter of the earth 7957f miles. 
 
 A.ns. The soUdity is 1323679753.4398 cubic wiles, and 
 the convex surface is 8250209.7425 square miles. 
 
 PROBLEM XIII. 
 
 To find the solidity of the frustum or zone of a sphere, 
 
 RULE. 
 
 To the sum of the squares of tlie radii of the two ends, 
 add J of the square of their distance, or the height of the 
 
140 MENSUK^^^TION PART IV. 
 
 zone ; and this sum being multiplied by the said height, 
 and the product again by 1.5708, will give the solidity. 
 
 Note. The convex surface of a zone may be found by multiplying the circumfer- 
 ence of the whole sphere by the height of the zone. 
 
 EXAMPLES. 
 
 1. Required the solidity of the zone ABCD; the greater 
 
 diameter AB being 28, the less dia- 
 
 meter CD 20, and the height mn 15 %rir~|IZ±^c 
 
 Here 14-' + 10-' + 15- -- 3 = 196 S \ ^B 
 
 H- 100 -h 225 -H 3 = 296 + 75 == t- 1 ^H 
 
 371 ; and 371 x 15 x 1.5708 = A^:3^r^^pB 
 
 5o65 X 1.5708 = 8741.502 inches, 
 
 the solidity required. '' "' 
 
 2. What is the solidity of a zone whose greater diameter 
 is 9 feet 3 inches, less diameter 6 feet 9 inches, and height 
 5 feet 6 inches ? Ans. 370.32425 /eeif. 
 
 3. Required the solidity and convex surface of the 
 torrid zone, the top and bottom diameters of v/hich are 
 each 7297.735 miles, its height 3173.14565 miles, and the 
 circumference of the earth 25000 miles? 
 
 Ans. The soliditij is 149455316338.46698 cubic miles, 
 and the convex surface is 79328641.25 square miles. 
 
 PROBLEM XIV. 
 
 To find the solidity of a circidar spindle. 
 
 RULE. 
 
 Find the area of the revolving segment ACBEA, which 
 multiply by half the central distance OE. Subtract the 
 product from 4- of the cube of AE, half the length of the 
 spindle , and multiply the remainder by 12.5664, and the 
 product will be the solidity. 
 
 Kote 1. The surface of a circular spindle may be found by the following rule: 
 Multiply the length AB of the spindle by the radius OC of the revolving arc ACB. 
 INIultipiy also the length of the said arc by the central distance OE, or distance 
 between the centre of the spindle and the centre of the revolving arc. Subtract 
 the latter product from the former ; and the remainder being multiplied by 5 2832, 
 will give the surface required. 
 
 2. The convex surface of any segment or zone of a circular spindle, cut off per- 
 pendicularly to the chord of the revolving arc, may be found in the same ma.nner ; 
 using the length of the solid, and the part of th.e arc which describes it, instead of 
 the length of the w hole spindle and whole arc. 
 
A/: 
 
 SECT. I. OF SOLIDS. 141 
 
 EXAMPLES. 
 
 1. Required the solidity of the cir- 
 cuhir spindle ABCD, whose length 
 AB is 30, and middle diameter CD 
 16 inches. / '^ 
 
 Bu Rule 2, Prob. U, Part IL, ive I -^ \ 
 
 y^f/d-e I0---8 -h 8=1 225 --8 + 8 = \ P / 
 
 28.125 -{- 8 = 36.125, the diameter CF; \ \ / 
 
 a?id 8 X I -+ (36.125 -(8 x 41) )-4-50 \ | 
 = 5.3333 -- (36.125-6.56) = 5.3333 ''"-^y-^" 
 
 -~ 29.565 = .18039 ; then 1 + .18039 
 X 30 = 1.18039 X 30 = 35.41 17, the length of the revolv- 
 ing arc ACB ; and by Ridel. Prob. 16, lue have 35.4117 
 X 18.0625 = 639.6238, half of which is 319.8119, the 
 area of the sector AOBC. 
 
 Again, CO-CE = 18.0625-8 = 10.0625 = OE; and 
 OE X AE =: 10.0625 x 15 = 150.9375, the area of the 
 triangle AOB ; then 319.81 19 - 150.9375 = 168.8744, the 
 area of the revolving segment ACB. 
 
 Now, by the Rule, ive have (15---3 — 168.8744 x 10.0625 
 -- 2) X 12.5664 =: (3375 -- 3-168.8744 x 5.03125) x 
 12.5664 = (1125 - 849.6493) x 12.5664 = 275.3507 X 
 12.5664 = 3460.16703648 inches, the solidify required. 
 
 2. YrHiat is the solidity and superficial content of a cir- 
 cular spindle, whose length is 96, and middle diameter 72 
 inches ? 
 
 . f Solidity, 138.51593 feet. 
 ^^^*- \ Superficies, 130.82161 feet. 
 
 PROBLEM Xy. 
 
 To find the solidity of the middle frustum or zone of a 
 circular spindle. 
 
 RULE. 
 
 From the square of half the length of the whole spindle, 
 take 4- of the square of half the length of the frustum; and 
 multiply the remainder by the said half length of the 
 frustum. Multiply the revolving area which generates 
 the frustum, by the central distance, and subtract this pro- 
 duct from the former ; then the remainder beinc: multi- 
 plied by 6.2832, v/ill give the solidity required. 
 
142 
 
 MENSURATION 
 
 PART IV. 
 
 EXAMPLES. 
 
 1. Wliixt is tlie solidity of the frustum ABCD, vrliose 
 length mn is 40, its greater 
 cliameter EF 32, and its least 
 diameter AD or BC 24 inches ? 
 
 Draw DG jmrallel to mn, 
 tlten ice have DG- = ^1 7Jm=20. 
 Also, i EF - -i AD = 16 - 12 
 = 4 = EG ; and DG-' + EG^ 
 = 202 + 42 = 400 +- 16 = 416, 
 the square of the chord DE ; 
 then hy Theo. 12, Part I., DE-' -- EG = 416 -- 4 = 104, 
 the diameter of the generating circle, half of which is 52, 
 the radius OE or OH ; consequently OE~EI = 52-16 
 = 36, tJte central distance 01. 
 
 Again OW - OP = 52-' - 36-' = 2704 - 1296 = 1408, 
 the square of HI, half the length of the ivhole spindle. 
 
 By Rule 3, Prob. 17, Part II., ^oe have 4 -r- 104 = 
 .03846, the tabular height ; and the corresponding Area 
 Seg. 25.00994; then .00994 X 104-' = .00994 x 10816 = 
 107.51104, the area of the segment DEO. Also, mn x 
 D-'/z = 40 X 12 = 480, the area of the rectangle jnDCn ; 
 and 107.51104 -]- 480 = 587.51104 the generating area 
 QvJyECn. 
 
 Noiv, hy the Rule, we have (1408-202 — 3) x 20 = 
 (1408 - 400 -- 3) X 20 = (1408 - 133.33333) x 20 = 
 1274.66667 X 20 = 25493.33334, the first product ; and 
 587.51 104 X 36 = 21 150.39744, the second product, ivhich 
 taken from the first, leaves 4342.9359; then 4342.9359 X 
 6.2832 = 27287.53484688 inches, the solidity required. 
 
 2. "Wliat is the solidity of the middle frustum of a cir- 
 cular spindle, whose length is 50, greatest diameter 40, 
 and least diameter 30 inches ? Ans. 30.84257 feet. 
 
 PROBLEM XVI. 
 To find the solidity of a cylindrical ring. 
 
 RULE. 
 
 To the thickness of the ring, add the inner diameter 
 and this sum bein 
 
 ness, and 
 solidity. 
 
 multiplied by the square of the thick- 
 the product again by 2.4674, will give the 
 
SECT. I. 
 
 OF SOLIDS. 
 
 143 
 
 Xofe. The surface of a cylindrical rinp may bo found by the following rule : 'lo 
 tlie tliickness of the ring, and the inner diameter ; and tliis sum being multiplied 
 by the tliickness, and the product again by 'J.seOG, will give the surface required. 
 
 EXAJIPLES. 
 
 1. Wliat is the solidity of a cylin- 
 drical ring, whose thickness AB or 
 CD is 6, and the inner diameter 
 BC 20 inches ? 
 
 Here (20 + 6) X 6"- x 2.4674 =26 
 X 36 X 2.4674 = 936 x 2.4674 = 
 2309.4864 inches, the soUditij re- 
 quired. 
 
 2. What is the solidity of an anchor-ring, the inner dia- 
 meter of which is 24.6 inches, and its thickness 6.4 
 inches ? Ans. 3133.005824 inches. 
 
 3. Required the solid and superficial contents of a cy- 
 lindrical ring, whose thickness is 9, and inner diameter 32 
 inches. 
 
 Ans I '^^^'^^Z' 8194.2354 inches. 
 ' \ Sifperjicies, 3641.8824 inches. 
 
 PROBLEM XYII. 
 
 To find the solidity or superfii,cies of any regular body. 
 
 RULES. 
 
 1. Multiply the tabular solidity, in the following table, 
 by the cube of the linear edge of the body, and the pro- 
 duct will be the solidity. 
 
 2. Multiply the tabular area, by the square of the linear 
 edge, and the product will be the superficies. 
 
 Solidities and Surfaces of Regular Bodies. 
 
 No. of 
 Sides. 
 
 Names, 
 
 Solidities. 
 
 Surfaces. 
 
 4 
 
 6 
 
 8 
 
 12 
 
 20 
 
 Tetraedron . 
 Hexaedron . 
 Octaedron . 
 Dodecaedron 
 Icosaedron . 
 
 .1178511 
 1.0000000 
 
 .4714045 
 7.6631189 
 2.1816949 
 
 1.7320508 
 
 6.0000000 
 
 3.4641016 ' 
 
 20.6457288 i 
 
 8.6602540 ; 
 
 "Kote. The above table shows the solidity and superficies of the five Begular 
 Bodies, when the linear edge of each is 1, or uniiv. 
 
144 
 
 MENSURATION 
 
 PART IV. 
 
 EXA3IPLES. 
 
 1. Wliat is the solidity and super- 
 ficies of the tetraedi'on ABCD, whose 
 linear edge is 8 inches ? 
 
 Here .1178511 x 8-^= .1178511 
 X 512 = 60.3397632 inches, the 
 solidity ; and 1.7320508 x 8- = 
 1.7320508 X 64 = 110.8512512 
 inches, the superficies. 
 
 2. "What is the solidity of the 
 hexaedron ABCD, whose side AB 
 is 6 feet ? 
 
 A71S. 216 feet. 
 
 3. Required the solidity of the 
 octaedi'on ABODE, whose linear 
 side is 22 inches. a 
 
 Ans. 5019.515116 i7iches. 
 
 4. The linear side ot the dodecae- 
 dron ABCDi: is 5.68 feet ; what is 
 its solidity ? 
 
 A?is 1404.26984/ee^. 
 
 5. What is the solidity and super- 
 ficies of the icosaedron ABCDEF, 
 whose linear edsce is 15 ? 
 
 ^^^ r Soliditi/, 7363.22028. 
 ' ■ \ Superficies, 1948.55715. 
 
 ,m 
 
SECT. I. 
 
 OF SOLIDS. 145 
 
 Notr. The five Regular Bodies are sometimes called Platonic Bodies. 
 
 It appears from an ancient Greek Epif^rani, quoted bv Scarburgh in his " Eng- 
 Bsh Euclid," (Oxford, 1705, ) that "the five Platonic Bodies, which the wise 
 Pythagoras found out, were indeed discovered by him ; but Plato elucidated and 
 taught them in the clearest manner ; and Euclid took them as the foundation of 
 his own imperishable renown." 
 
 Pythagoras was a native of the island of Samos, and was born about 590 years 
 before Christ; Plato was born at Athens B.C. 429; and Euclid flourished at 
 Alexandria, about 300 years before the Christian a^ra. 
 
 PROBLEM XYIIL 
 
 To find the solidity of an irregular solid. 
 
 RULES. 
 
 1. Divide the irregular solid into different figures; and 
 the sum of their solidities, found by the preceding Pro- 
 blems, will be the solidity required. 
 
 2. If the figure be a compound solid, whose two ends 
 are equal plane figures ; the solidity may be found by 
 multiplying the area of one end by the length. 
 
 3. To find the solidity of a piece of wood or stone, that 
 is craggy or uneven, put it into a tub or cistern, and pour 
 in as much water as will just cover it; then take it out, 
 and find the content of that part of the vessel through 
 which the water has descended, and it will be the solidity 
 required. 
 
 4. If a solid be large and very irregular, so that it can- 
 not be measured by any of the above Rules, the general 
 method is to take lengths, in tvro or three different places ; 
 and their sum divided by their number, is considered as a 
 mean length. 
 
 A mean breadth and a mean depth are found by similar 
 processes. 
 
 Sometimes the length, breadth, and depth, taken in the 
 middle, are considered as mean dimensions. 
 
 yotc. The unhewn blocks, in the freestone quarries, in the vicinity of Leeds, are 
 generally measured by the method desL-ribed in the last Pule. The dimensions, 
 however, are not, in general, taken to the extremities of the stones, in order to 
 make an allowance for the waste in hewing. 
 
 exa:\iples. 
 
 1. The lower part of a stone is a paralleloplpedon, the 
 breadth of whose end is 7, and its depth 5 feet. The 
 upper part is a triangular prism, the perpendicular of 
 whose end is 4 feet ; required the solidity of the stone, its 
 length being 18 feet. 
 
146 
 
 MENSURATION. 
 
 PART IV 
 
 By Rule I. 
 Here 7x5x18 = 35 xl8=: 630, the solidity of the 
 lower part; and 7x2x18=14x18 =252, the solidity 
 of the upper part ; then 630 + 252 = SS2feet, the solidity 
 of the tvhole stone. 
 
 By Ride II. 
 Here 7 X 5 + 7 X 2 = So + U = 49, the area of the 
 end; and 49 X 18 = 882 feet, the solidity as before. 
 
 2. Being desirous of finding the solidity of an irregular 
 piece of wood, I immersed it in a cubical vessel of water ; 
 and when it was taken out, the water descended 6 inches ; 
 required the solidity of the wood, the side of the vessel 
 being 30 inches. 
 
 By Rule III 
 Here 30 X 30 x 6 = 900 x 6 = 5400 inches, the solidity 
 required. 
 
 3. The lengths of an irregular block of marble, taken in 
 different places, are 8 feet 6 inches, 8 feet 10 inches, and 
 9 feet 2 inches ; the breadths 4 feet 7 inches, 4 feet 2 
 inches, and 4 feet 9 inches ; the depths 3 feet 2 inches, o 
 feet 5 inches, and 3 feet 1 1 inches ; 'required its solidity. 
 
 
 By 
 
 Rule IV. 
 
 
 I.ENGTHS. 
 
 feet in. 
 
 8 6 
 
 8 10 
 
 9 2 
 
 BREADTHS. DEPTHS. 
 
 feet in. feet in. 
 
 4 7 3 2 
 4 2 3 5 
 4 9 3 11 
 
 3)26 6 
 
 3)13 
 
 6 3)10 
 
 6 
 
 8 10 mean. 
 
 4 
 
 6 mean. 3 
 
 6 mean. 
 
 
 feet 
 
 8 ; 
 
 4 
 
 in. 
 
 10 mean length. 
 6 mean breadth. 
 
 
 
 35 
 4 
 
 4 
 5 
 
 
 
 39 
 3 
 
 9 
 
 6 mean denth. 
 
 
 
 119 
 19 
 
 3 
 10 6" 
 
 
 
 139 
 
 1 &' Ans. 
 
 
SECT. T. 
 
 OF SOLIDS. 
 
 147 
 
 4. Wanting to know the solidity of an irregular block 
 of marble, I immersed it in a cylindrical tub of water, 
 whose diameter was 34.8 inches ; and on taking it out, I 
 found the fall of the water to be 12.6 inches ; what was 
 the solidity of the marble ? Aiis. U984:.5002S i7iches. 
 
 5. Required the solidity of an irregular block of York- 
 shire-stone, Avhose dimensions are as follow : viz. lengths 
 taken in different places, 1 1 feet 3 inches, and 1 1 feet 9 
 inches ; breadths, 5 feet 5 inches, 5 feet 9 inches, and 6 
 feet 4 inches; depths, 4 feet 5 inches, 4 feet 8 inches, 
 and 5 feet 2 inches. Ans. 318 ft. 7 in. 9 pa. 
 
 PROBLEM XIX. 
 
 To find the magnitude or solidity of a body from its lueight. 
 
 RULE. 
 
 As the tabular specific gravity of the body. 
 Is to its weight in avoirdupois ounces. 
 So is one cubic foot, or 1728 cubic inches, 
 To its content in feet, or inches, respectively. 
 
 A TABLE 
 
 OF THE SPECIFIC GRAVITIES OF BODIES, 
 
 EineGold. *. 
 
 . 
 
 
 19640 
 
 Standard Gold . 
 
 
 18888 
 
 Quick Silver . . 
 
 
 14000 1 
 
 Lead . . . 
 
 
 
 11325 
 
 Fine Silver 
 
 
 
 11091 
 
 Standard Silvc 
 
 ir . 
 
 
 10535 
 
 Copper . . 
 
 
 
 9000 
 
 Gun Metal 
 
 
 
 8784 
 
 Cast Brass 
 
 
 
 8000 
 
 Steel . . 
 
 
 
 7850 
 
 Iron . . . 
 
 
 
 7645 
 
 Cast Iron . 
 
 
 
 7425 
 
 Tin . . . 
 
 
 
 7320 
 
 Load- stone 
 
 
 
 . 4930 
 
 Granite . . 
 
 
 
 . 3000 
 
 Marble . 
 
 
 
 . 2700 
 
 Glass . . 
 
 
 
 . 2642 
 
 Flint . . 
 
 
 
 . 2570 
 
 Bristol-stone . . 
 
 . 2510 
 
 Portland- stone . . 
 
 . 2496 
 
 Mill-stone . . . 
 
 . 2484 
 
 Yorkshire-stone . 
 
 . 2442 
 
 Clay .... 
 Grind- stone . . 
 
 . 2160 
 . 2143 
 
 Burford-stone . 
 
 . 2049 
 
 Brick .... 
 
 . 2000 
 
 Light Earth 
 
 . . 1984 
 
 Ivory .... 
 Chalk . . . 
 
 . 1825 
 . 1793 
 
 Sand .... 
 
 . 1520 
 
 Lignum-vitce . 
 
 . 1333 
 
 Ebony . . . 
 Coal .... 
 
 . 1331 
 . . 1250 
 
 Pitch . . . , 
 
 , . 1150 
 
 Mahogany . . 
 Box-wood . . 
 
 . . 1063 
 
 . . 1030 
 
 2 
 
 
148 MEXSUKATIOX OF SOLIDS. TART lY. 
 
 Sea-water . 1026 or 1030 i Maple Too 
 
 Common water . . 1000 , Cherry-tree .... 715 
 
 Oak 925 Pear-tree 661 
 
 Gunpowder, shaken . 922 | Cedar of Lebanon . .613 
 
 Logwood .... 913 Ehn 600 
 
 Beech 852 I Willow 585 
 
 Ash 800 Fir 550 
 
 Yew 797 Poplar 383 
 
 Apple-tree .... 793 Cork 240 
 
 Plum-tree .... 785 Atmospheric Air . .1.2 
 
 Note 1. The specific gravities of bodies are their relative weights contained under 
 the same given magnitude, as a cubic foot, a cubic inch, .Jtc. 
 
 2. As a cubic foot of water weighs just lOfO ounces avoirdupois, the numbers in 
 the foregoing Table express not only the specific gravities of the several bodies, 
 hut also the weight of a cubic foot of each, in avoirdupois ounces. 
 
 3. The several sorts of wood, mentioned in the preceding Table, are supposed to 
 be dry. 
 
 4. in some tables, sea-water is 102G, and in others 1C30. 
 
 exa:\iples. 
 1. What is the solidity of a marble chimney-piece, 
 whose weight is 210 /Z». 15 oz. avoirdupois ? 
 
 ft. ft. 
 
 : 1 : 14. Ans 
 
 2700)3375(1.25 
 
 2. What is the solidity of an irregular Yorkshire-stone, 
 whose weight is 228 Ih. 15 oz. avoirdupois ? Ans. l\foot. 
 
 3. A piece of carved mahogany weighs 49 Ih. l-ioz. 
 avoirdupois ; what is its solidity ? 
 
 Ans. 1297.21919 inches. 
 
 PROBLEM XX. 
 
 I'o find the icelght of a body from its magnitude 01 
 solidity. 
 
 RULE. 
 
 As one cubic foot, or 1728 cubic inches, 
 Is to the content of the body, 
 So is its tabular specific gravity, 
 To the weight of the body. 
 
SECT. II. CAEPEXTEU'S KUL=^.. 149 
 
 EXAMPLES. 
 
 1. The solidity of a grindstone is 3 feet ; what is its 
 weight ? 
 
 ft. ft. oz. lb. oz. 
 
 As 1:3:: 2143 : 401 13. Ans. 
 ^3 
 
 1) 6429 
 16)6429(401 
 
 2. The solidity of a beam of dry oak is 25 feet ; what is 
 its weight ? A?is. 1445 lb. 5 oz. 
 
 3. Required the weight of a block of marble, whose 
 length is 63 feet, and its breadth and thickness each 12 
 feet ; these being the dimensions of one of the stones in 
 the walls of Balbec. 
 
 Ans. 683.4375 tons, ivhich is nearly equal to the bur- 
 then of an East hidia ship. 
 
 SECTION II. 
 
 DESCRIPTION AND USE OF THE 
 CARPENTER'S RULE. 
 
 This instrument is commonly called CogeshalVs Slid- 
 ing Ride, and is much used in measuring timber and arti- 
 ficers' work ; not only in taking the dimensions, but also 
 in casting up the contents. 
 
 It consists of two pieces of box, each one foot in length ; 
 and coimected together by a folding joint. 
 
 One side or face of the rule is divided into inches and 
 half-quarters, or eighths ; and on the same face there are 
 also several plane scales, divided into twelfth parts, which 
 are designed for planning such dimensions as are taken in 
 feet and inches. 
 
 On one part of the other face are four lines marked A, 
 B, C, D ; the two middle ones, B and C, being upon a 
 slider. 
 
 Three of these lines, viz. A, B, C, are exactly alike, and 
 are called double lines ; because they proceed from 1 to 
 10, twice over. The fourth line, D, is a single one, pro- 
 ceeding from 4 to 40, and is called the girt line. 
 H 3 
 
150 carpenter's rule. part IV. 
 
 The use of the double lines, A and B, is for working 
 proportions, and finding the areas of plane figures ; and the 
 use of the girt-line D, and the other double line C, for 
 finding the contents of solids. 
 
 "^Ylien 1 at the beginning of any line is accounted 1, 
 then the 1 in the middle. will be 10, and the 10 at the end 
 100 ; and when 1 at the end is accounted 10, then the 1 
 in the middle is 100, and the 10 at the end 1000, &c. ; 
 and all the smaller divisions are altered in value accord- 
 ingly. 
 
 Upon the girt-line are also marked "W" G at 17.15, and 
 A G at 18.95, the wine and ale gauge-points, to make the 
 instrument serve the purpose of a Gauging Rule. 
 • On the other part of this face there is commonly either 
 a table of the value of a load, or ^0 cubic feet of timber, 
 at all prices, from 6d. to 24d per foot ; or else several plane 
 scales divided into twelfth parts, and marked 1, |, -J, and 
 :J, signifying that the inch, | inch, &c. are each divided 
 into 12 equal parts. 
 
 The edge of the rule is generally divided decimally, or 
 into tenths ; namely, each foot into 10 equal parts, and 
 each of those into ten other equal parts. By this scale 
 dimensions may be taken in feet,tenths, and hundi'edths of 
 a foot, which is a very commodious method of taking di- 
 mensions, when the contents are to be cast up decimally. 
 
 THE USE OF THE SLIDING RULE. 
 
 PROBLEM I. 
 
 Multiplication by the sliding rule. 
 
 rule. 
 
 Set 1 upon A, to the multiplier upon B ; then against 
 the multiplicand on A, wiU be found the product upon B. 
 
 2Jote. When the third term runs beyond the end of the line, seek it on the other 
 radius or part of the line ; and increase the product 10 times, or ICO times, &c. as 
 the case requires. 
 
 EXAMPLES. 
 
 1. What is the product of 24 multiplied by 12 ? 
 
 As Ion A: 12 on B : : 24 on A : 288 on B, the Ans. 
 
 2. Required the product of 36 multiplied by 18. 
 
 Ans. 648. 
 
SECT. II. 
 
 carpenter's rule. 151 
 
 3. Multiply 12.8 by 6,5. Arts. 83.2. 
 
 4. What is the product of 68 multiplied by 35 ? 
 
 A?is. 2380. 
 
 PROBLEM 11. 
 Division hy the sliding rule. 
 
 rule. 
 
 Set the divisor on A, to 1 upon B ; then against the 
 dividend on A, is the quotient on B. 
 
 Koie. When the dividend runs beyond the end of the line, diminish it by 10 or 100 
 times, in order to make it fall upon A ; and increase the quotient in the same pro- 
 portion. 
 
 EXAMPLES. 
 
 1. What is the quotient of 432 divided by 12 ? 
 
 As 12 on A: 1 on B : *. 432 on A .* 36 on B, the Aiis. 
 
 2. Wliat is the quotient of 9752 divided by 46 ? 
 
 Ans. 212. 
 
 3. If a board be 8 inches broad ; what length must be 
 sawn off, to make a foot square ? Ans. 18 inches. 
 
 PROBLEM in. 
 To square any number. 
 
 RULE. 
 
 Set 1 upon D, to 1 upon C ; then against the number 
 upon D, will be found the square upon C. 
 
 EXAMPLES. 
 
 1. Wliat is the square of 25 ? 
 
 As I on J) : \ onC\'.26 onT> : 625 on C, the Ans. 
 
 2. Required the square of 36. Ans. 1296. 
 
 3. What is the square of 58 ? Ans. 3364. 
 
 PROBLEM IV. 
 
 To extract the square root of any number. 
 
 RULE. 
 
 Set 1 upon C, to 1 upon D ; then against the number 
 on C, is the root on D. 
 
 H 4 
 
152 CARPENTER'S RULE. PART IV. 
 
 Note. In order to value this rightly, you must suppose the 1 on C to be some of 
 these squares, 1, 100, 10000, &c. which is nearest to the given number ; and then 
 the root corresponding will be the value of the 1 upon D. 
 
 EXASIPLES. 
 
 1. What is the square root of 144 ? 
 
 Asl onQ :l onT>:\ 144 on C : 12 on D, the A?is. 
 
 2. Required the square root of 900. Ans. 30. 
 
 3. What is the square root of 95S6 ? Ans. 97.9. 
 
 PROBLEM V. 
 To find a mean proportioiial between two numbers. 
 
 RULE. 
 
 Set one of the numbers on C, to the same on D ; then 
 against the other number on C, will be the mean on D. 
 
 EXAMPLES. 
 
 1. What is the mean proportional between 9 and 16? 
 ^5 9 o?z C : 9 o?^ D : : 16 o/^ C ; 12 on D, the Ans. 
 
 2. Required a mean proportional between 15 and 27. 
 
 A71S. 20.1. 
 
 3. The segments of the hypothenuse of a right-angled 
 triangle made by a perpendicular from the right-angle, are 
 18 and 32 ; what is the perpendicular ? Ans. 24. 
 
 PROBLEM VI. 
 
 To find a fourth proportional to three numbers ; or to 
 perform the Rule of Three. 
 
 RULE. 
 
 Set the first term on A, to the second on B ; then against 
 the third term on A, stands the fourth on B. 
 
 'Note. The finding of a third proportional is exactly the same ; the second num- 
 ber being twice repeated. 
 
 Thus, suppose a third proportional was required to 80 and 60. 
 
 As 80 on A : GO on B : : 60 on A : 45 on B, the third proportional sought. 
 
 EXAJUPLES. 
 
 1. What is the fourth proportional to the three num- 
 bers, 12, 24, 36 ? 
 
 As\2 on A: 24 on B: : 36 on A : 72 on B, the Ans. 
 
SECT. II. CAKrLXTKli's RULE. 163 
 
 2. If 1 foot of timber cost Ss. ; what will 180 feet cost? 
 
 Ans. 540s. 
 
 3. If 50 feet of timber cost 71. ', what will 2500 feet 
 cost ? A?is. £o50. 
 
 PROBLEM VU. 
 To find the areas of plane figures by the sliding rule. 
 
 RULES. 
 
 1. To find the area of a recta7igle or rhomhoides. 
 
 As 1 upon A, is to the perpendicular breadth ujion B ; 
 so is the length upon A, to the area upon B. 
 
 2. To find the area of a triangle. 
 
 As 2 upon A, is to the perpendicular upon B ; so is the 
 base upon A, to the area upon B. 
 
 3. To find the area of a trapezium. 
 
 As 2 upon A, is to the sum of the two perpendiculars 
 upon B ; so is the diagonal upon A, to the area upon B. 
 
 4. To find the area of a regular polygon. 
 
 As 2 upon A, is to the perpendicular upon B ; so is the 
 sum of the sides upon A, to the area upon B. 
 
 5. To find the diameter aiid circumference of a circle, 
 the one from the other. 
 
 As 7 upon A, is to 22 upon B ; so is the diameter upon 
 A, to the circumference upon B ; and vice versa, 
 
 6. To find the area of a circle. 
 
 As 4 upon A, is to the diameter upon B ; so is the cir- 
 cumference upon A, to the area upon B. 
 
 Or, as 1 upon D, is to .7854 upon C ; so is the diameter 
 upon D,,to the area upon C. 
 
 'Note. In order to exemplify the foregoing Rules, the learner may work the ques- 
 tions in the respective Problems of Fart II. 
 
 PROBLEM VIIL 
 To find the contents of solids by the sliding rule. 
 
 RULES. 
 
 1. To find the solidity of a cube. 
 
 As 1 upon D, is to the side upon C ; so is the side upon 
 D, to the solidity upon C. 
 
 2. To find the solidity of a parallelopipedon. 
 
 As 1 upon A, is to the breadth of the base or end upon 
 H 5 
 
154 .TI3IF,ER MEASURE. PAET IV 
 
 B ; SO is the length of the base upon A, to the area of the 
 base upon B ; and, as 1 upon A, is to the length of the 
 solid upon B ; so is the area of the base upon A, to the 
 soliditj upon B. 
 
 3. To find the solidify of a prism, or a cylinder. 
 Find the area of the base by the last Problem, with 
 
 which proceed as in the last Eule. 
 
 4. To find the solidity of a cone, or a pyramid. 
 
 Find the area of the base by the last Problem ; then, as 
 3 upon A, is to the length of the solid upon B ; so is the 
 area of the base upon A, to the solidity upon B. 
 
 Note. Examples for practice may be found in Section I. 
 
 SECTION III. 
 
 TIMBER MEASURE. 
 
 PROBLEM I. 
 To find the superficial content of a hoard or plank. 
 
 RULE. 
 
 Multiply the length by the breadth, and the product will 
 be the superficial content. 
 
 Note. If the board taper, add the breadth of the two ends together ; and half their 
 sum will be a mean breadth. 
 
 By the Sliding Rule. 
 As 12 upon B, is to the breadth in inches upon A; so 
 is the length in feet upon B, to the content upon A, in feet 
 and fractional parts. 
 
 EXAMPLES. 
 
 1. If the length of a plank be 12 feet 6 inches, and its 
 breadth 1 foot 3 inches ; what is its superficial content ? 
 
 By Decimals. By Cross 3Iultiplication. 
 
 feet. feet in. 
 
 12.5 12 6 
 
 1.25 1 3 
 
 625 12 6 
 
 250 3 1 6 ^ 
 
 ^^5 15 7 6^ Am, 
 
 15-625 Ans. ' 
 
SECT. 111. TIMBER MEASURE. 155 
 
 By the Sliding Rule. 
 As 12 ow B : lo on A : : 12.5 ow B • • 15.6 on A. 
 
 2. Required the content of a board whose length is 25 
 feet 8 inches, and breadth 1 foot 7 inches. 
 
 Ans. 40 ft. 7 hi. 8 pa. 
 
 3. What is the content of a board whose length is 18 
 feet 10 inches, and breadth at the broader end 2 feet 3 
 inches, and at the narrower end 1 foot 7 inches ? 
 
 Ans. S6ft. 1 i7i. 2 pa. 
 
 4. If the length of a mahogany plank be 35 feet 9 
 inches, and its breadth 3 feet 6 inches ; what is the value 
 of three such planks at 25. 9d. per square foot ? 
 
 Ans. £51. 125. S^d. 
 
 Note. Mahogany is a production of the warmest parts of America. It is alsc 
 foimd plentifully in the islands of Cuba, Jamaica, and Hispaniola. 
 
 This tree grows tall and straight, rising often sixty, and sometimes a hundred 
 feet from the ground to the arms ; and is about 4 feet in diameter. The foliage is 
 a beautiful deep green ; and the appearance of the whole tree is very elegant. 
 
 The mahogany brought from Jamaica is most valued, in consequence of its 
 firmness, durabiiity, and beauty of colour. 
 
 This wood has been used in medicine with the same effect as Peruvian bark. 
 
 PROBLEM II. 
 To find the solidity of squared or four-sided timber. 
 
 RULE. 
 
 Multiply the mean breadth by the mean thickness, and 
 this product by the length, and it will give the solidity, 
 according to the customary method. 
 
 By the Sliding Rule. 
 As 1 upon A, is to the mean breadth upon B ; so is the 
 mean thickness upon A, to the mean area upon B, or the 
 area of the middle section ; and, as 1 upon A, is to the 
 length upon B ; so is the mean area upon A, to the solidity 
 upon B. 
 
 Koic 1. If the tree be equally broad and thick throughout, the breadth and thick- 
 ness, taken in any part, will be the mean breadth and thickness ; but if it taper re- 
 gularly from one end to the other, the breadth and thickness must be taken in the 
 middle. Or, take the dimensions of the two ends, and half their sum respectively 
 will be the mean breadth and thickness. 
 
 2. Some measurers multiply the square of one-fourth of the circumference, or 
 quarter girt, taken in the rniddle, by the length, for the solidity; but when the 
 ends are not equal squares, this method always gives too much ; and the more the 
 breadth and thickness differ, the greater will be the error. 
 
 3. The Rule given in this Problem, is generally used \n practice ; and when the 
 tree is equally broad and thick throughout, it gives the true content ; but in taper- 
 ing timber, it always gives too little ; consequently, the method mentioned in the 
 last note is sometimes nearer to the truth. 
 
 When the true content of a tapering tree is wanted, it must be found by the 
 General Rule, for the frustumof a pyramid, given in Problem VI., Section I., or by 
 the Rule for a prismoid. given in Probi*-;;!! .K. 
 H CS 
 
156 . TIMBER MEASURE. FART IV. 
 
 exa:mples. 
 
 1. If tlie length of a piece of timber be 9.8 feet, its 
 breadtli 2.6 feet, and its thickness 1.5 foot ; what is its 
 solidity ? 
 
 Here 2.6 X 1.5 X 9.8 = 3.90 x 9.8 ^ 38.220. Ans. 
 
 By the Sliding Rule. 
 
 As \ on A '. 2.6 on B : : 1.6 on X'. 3.9 on B, the mean 
 area. 
 
 As 1 on A : 9.8 o;i B : : 3.9 on A : 38.22 on B, the 
 solidity. 
 
 2. The mean breadth of a piece of timber is 2.86 feet, 
 its mean thickness 1.93 foot, and its length 18.64 feet ; 
 what is its soKdity ? Ans. 102.889072 feet. 
 
 3. Each side of the greater end of a piece of squared 
 timber is 28 inches, each side of the less end 14 inches, 
 and its length 18 feet 9 inches ; how many solid feet does 
 it contain ? Ans. 57.421 875 /ee?. 
 
 Note. The true content measured as the frustum of a pyramid, is 59.548G feet. 
 See Example 2, Problem VI., Section I. 
 
 4. The length of a piece of timber is 17 feet 3 inches ; 
 at the greater end, the breadth is 36 inches, and the 
 thickness 20 inches ; and at the less end, the breadth is 
 18 inches, and the thickness 10 inches; what is the 
 solidity ? Aiis. 48.5 1 5625 feet. 
 
 Note. The true content, measured as the frustum of a pyramid, is 50.3125 feet. 
 See Example 3, Problem VI., Section I. 
 
 5. A piece of timber is 32 inches broad, and 20 inches 
 deep, at the greater end ; 16 inches broad, and 8 inches 
 deep, at the less end ; and its length is 25 feet ; what is 
 its solidity? Afis. 58-1 feet. 
 
 Note. The true content, measured as a prismoid, is 61^ feet. See Example 2, 
 Problem X., Section I. 
 
 PROBLEM in. , 
 
 To find the solidity of round or U7isquared timber. 
 
 RULE I. 
 
 Multiply the square of \ of the circumference, or 
 quarter girt, by the length, and the product will be the 
 content, according to the common practice. 
 
SECT. III. • TIMBER 3IEASURE. 157 
 
 By the Sliding Rale. 
 
 As the length upon C, i? to 12 upon D ; so is | of the 
 girt in inehes, upon D, to the content upon C. 
 
 RULE II. 
 
 Multiply the square of -I of the girt by twice the length, 
 and the product will be the solidity nearly. 
 
 By the Sliding Rule. 
 
 As twice the length upon C, is to 12 upon D; so is -1- 
 of the girt upon D, to the content upon C. 
 
 Note 1. The girt must be taken in the middlo of the tree, if it taper regularly ; if 
 not, take several girts, and their sum, divided by their number, will give something 
 near a mean girt. If the tree be very irregular, divide it into two or moie parts, 
 and find the content of each separately. 
 
 2. The girt of a tree is generally taken with a string, which being folded into 
 four equal parts, and applied to a carpenter's rule, gives the quarter girt ; but it is 
 more expeditious to take the girt with a tape divided into equal parts of 4 inches 
 each ; and numbered at 4 inches from the end with 1, at 8 inches with 2, at 12 
 inches with 3, 6cc., by which means the quarter girt can be obtained by merely 
 girting the tree. 
 
 If the second Rule be used, the girt may be taken with a tape divided into equal 
 p-arts of 5 inches each. 
 
 Each divijion of 4 or 5 inches, may also be subdivided into halves and quarters, 
 and numbered successively with i, h, f ; hence the quarter girt may be obtained 
 with accuracy. The other side of the tape may be divided into feet and inches. 
 
 3. In measuring a tree which has its bark on, an allowance is generally made, by 
 deducting so much from the girt as is judged sufficient to reduce the tree to sucli 
 a circumference as it would have without its bark. This deduction may be most 
 easily made from the quarter girt; and should be proportioned according to the 
 thickness of the bark. 
 
 For ash, beech, elm, young oak, &c. i or f of an inch for every foot of the quarter 
 girt, will generally be sufficient ; but for old oak, whose bark is much thicker, it 
 will be found necessary to allow an inch, an inch and a half, and sometimes two 
 inches, for every foot o'f the quarter girt. 
 
 4. That part of the boughs, or of the trunk of a tree, which is less than 24 inches 
 in circumference, or C inches quarter girt, is generally cut off, and sold at an inferior 
 price ; not being considered timber. 
 
 5. The first Rule in this Problem, gives the solidity about \ part less than the 
 true quantity, or nearly what the quantity would be if the tree were squared ; so 
 that it seems intended to make an allowance for the squaring of the tree. 
 
 6. The second Rule gives nearly the true content, whether the tree be a cylinder, 
 or the frustum of a cone; and as it is full as easy in practice as the first, it ought 
 always to be used when accuracy is required; if the Rules for a cylinder, the frus- 
 Unn of a cone, &c. be thought too prolix. ( See Examples 3 and 4.) 
 
 7. If a piece of round tapering timber be cut through exactly in the middle, the 
 two parts will measure to the most possible ; and to more than the whole by either 
 of the Rules. 
 
 8. To find where a round tapering tree must be cut, so that the part next the 
 greater end may m.easure the most possible : From the greatest girt take 3 times 
 the least ; then, as the difference of'these girts, is to the remainder, so is one-third 
 of the whole length, to the length to be cut off from the less end. Or, cut the tree 
 where the girt is one-third of the greatest girt. 
 
 When the greatest girt does not exceed 3 times the least, this Rule is imprac- 
 ticable. 
 
 9. Forty feet of round timber, measured by the quarter-girt method, are called 
 a load ; but when the content of timber is correctly found, fifty feet are accounted 
 a load. s 
 
 Fifty feet of dry oak weighs rather more than a ton and a quarter. 
 
 10. The content of a piece of timber, according to the quarter-girt method, may 
 be readily found from the following Table, thus : Multiply the area corresponding 
 to the quarter girt, in inches, by the length of the timber, in feet ; and the product 
 will be the solidity in feet and decimal parts. 
 
158 
 
 TBIBER MEASURE. 
 
 PART rv. 
 
 A TABLE FOR I^IEASURma TIMBER. 
 
 Quarter 
 Girt. 
 
 Area. 
 
 Quarter 
 Girt. 
 
 Area. 
 
 Quarter 
 Girt. 
 
 Area. 
 
 Inches. 
 6 
 
 6| 
 
 Feet. 
 
 .250 
 .272 
 .294 
 .317 
 
 Inches. 
 12 
 
 121 
 12| 
 
 Feet. 
 
 1.000 
 1.042 
 1.085 
 1.129 
 
 Inches. 
 
 18 
 181 
 19" 
 191 
 
 Feet. 
 
 2.250 
 2.376 
 2.506 
 2.640 
 
 7 
 7| 
 
 .340 
 .364 
 .390 
 .417 
 
 13 
 
 131 
 13i 
 
 1.174 
 1.219 
 1.265 
 1.313 
 
 20 
 201 
 21 
 21i 
 
 2.777 
 2.917 
 3.062 
 3.209 
 
 8 
 
 81 
 81 
 8| 
 
 .444 
 .472 
 .501 
 .531 
 
 14 
 
 141 
 141 
 14| 
 
 1.361 
 1.410 
 1.460 
 1.511 
 
 22 
 221 
 23 
 231 
 
 3.362 
 3.516 
 3.673 ! 
 3.835 
 
 9 
 
 91 
 
 91 
 
 .562 
 .594 
 .626 
 .659 
 
 15 
 151 
 151 
 15| 
 
 1.562 
 1.615 
 1.668 
 1.722 
 
 24 
 241 
 25 
 251 
 
 4.000 
 4.168 
 4.340 
 4.516 
 
 10 
 
 , 101 
 
 io| 
 
 .694 
 .730 
 .766 
 .803 
 
 16 
 
 16i 
 
 161 
 
 m 
 
 1.777 
 1.833 
 1.890 
 1.948 
 
 26 
 
 261 
 
 27 
 
 27i 
 
 4.694 
 4.876 
 5.062 
 5.252 
 
 11 
 
 111 
 lU 
 11| 
 
 .840 
 .878 
 .918 
 .959 
 
 17 
 
 IH 
 
 17| 
 
 2.006 
 2.066 
 2.126 
 2.187 
 
 28 
 
 281 
 
 29 
 
 291 
 
 30 
 
 5.444 
 5.640 i 
 5.840 i 
 6.044 
 6.250 
 
 EXAMPLES. 
 
 1. "VYliat is tlie solidity of a tree "vvliose girt in the 
 middle is 100 inches, and length 18 feet ? 
 
SECT. in. TISIBER 
 
 MEASURE. 
 
 By Rule I. 
 
 
 In. In. 
 
 
 Ft. In. 
 
 25=i of 100. 
 
 
 2 1 
 
 25 
 
 
 2 1 
 
 125 
 
 
 4 2 
 
 50 
 
 
 2 r 
 
 625 
 
 
 ■1 4 V 
 
 18 
 
 
 18 
 
 5000 
 
 
 78 1 6'^ws. 
 
 625 
 
 
 
 
 144)11250(78.125/^.^^^5. 
 
 
 By the Table, 
 
 4.340 
 18 
 34720 
 4340 
 78.120 Ans. 
 
 159 
 
 By the Sliding Rule. 
 As 18 upon C : 12 upon D : : 25 z^^otz D 
 
 By Rule II. 
 
 In. In. 
 
 20=1 o/lOO. 
 
 20 
 400 
 36 
 
 78 upon C. 
 
 144)14400(100 /if. A71S. 
 144 
 00 
 
 Ft. 
 
 In. 
 
 
 
 8 
 
 
 
 8 
 
 
 
 1 
 
 4' 
 
 
 8 
 
 
 2 
 
 9 
 
 4' 
 
 36 
 
 
 
 100 
 
 
 
 
 
 By the Sliding Rule. 
 As 36 upon C ,* 12 upon D ! : 20 upon D : 100 upon C. 
 
 2. The length of a tree is 32 feet 6 inches, and its girt 
 in the middle, after allowing for the bark, is 60 inches ; 
 what is its content ? 
 
 . ^By the first Rule, 50.78125 feet. 
 ^^'\By the second Rule, Q5 feet. 
 
 3. The circumference of a cylindrical piece of timber 
 is 6 feet 3 inches, and its length 24 feet ; what is its 
 solidity ? 
 
 J S By the first Ride, 66 ft. 8 in. 
 L % ^^^^ second Ride, S5ft. 4 in. 
 
150 TDIBER MEASURE. I'AKT IV. 
 
 Note. The true content measured as a cylinder, is 84 feet 10 inches. See Ex- 
 ample 3, Problem IV., Section I. 
 
 4. The circumference of a piece of round tapering 
 timber is 65 inches at the greater end, 35 inches at the 
 less end, and its length 33 feet 9 inches j what is its 
 solidity ? 
 
 . r % thejirst Rule, 36.621 feet 
 ^^^- I By the second Rule, 46.875 feet 
 
 Note. The true content measured as the frustum of a cone, is 84 feet. See Ex- 
 ample 3, Problem VIII., Section I. 
 
 5. Wliat is the content of an oak tree whose dimensions 
 are as follow : the length of the trunk is 45 feet, and its 
 quarter girt, in the middle, 30 inches ; the length of a 
 large bough 16 feet, and its quarter girt 14 inches; and 
 thelength of another bough 12 feet, and its quarter girt 
 10 inches? Ans. 311.3611 1/ee^. 
 
 PROBLEM lY. 
 To measure and value standing timber. 
 
 To find the content. 
 
 RULE. 
 
 Obtain the dimensions by some of the methods described 
 in the following pages, and find the solidity by the Rules 
 given in the last Problem. The first Rule is generally 
 adopted, and the content found by the Sliding Rule. 
 
 MISCELLAKEOUS INSTRUCTIONS 
 
 FOR FINDING THE 
 
 DIMENSIONS OF STANDING TIMBER, 
 
 AND FOR SETTING OUT WOODS AND VALUING THEil 
 PREVIOUSLY TO A FALL. 
 
 I. Various methods are used in order to obtain th; dimensions of 
 standing trees ; but their girt and altitudes may be found most cor- 
 rectly by means of a ladder, and a long staff divided into feet and 
 inches, and numbered from the top to the bottom, on one side, and 
 from the bottom to the top on the opposite side, for the convenience 
 of taking dimensions. 
 
SECT. ni. 
 
 TniBKR MEASURE. 
 
 161 
 
 The ladder will enable you to take the girt at or near the middle 
 of the tree; and with the statF you may measure the boughs and the 
 upper part of the trunk. The lower part of the trunk may generally 
 be measured by a tape. 
 
 2. Divide the arc of the quadrant of a circle of any convenient 
 radius, into 90 equal parts, or degrees; and figure them at every 
 10 degrees thus; 10, 20, 30, &e. to 90. Upon that radius which is 
 contiguous to 90 degrees, place two small brass sights ; and from the 
 centre or angular point suspend a ])lummet. Fix the quadrant to a 
 square staff, of a convenient length for use, by means of a nail passing 
 through both ; and upon the end of this nail screw a small nut, so 
 that the quadrant may be made fast to the staff at pleasure. 
 
 The neck of the nail should be round, so that the quadrant may be 
 readily turned upon it ; but that part which is contained within the 
 staff should be square, to prevent the nail from turning 
 round with the quadrant. ■* 
 
 Then, in order to find the height of a tree, screw the 
 quadrant fast to the staff, so that the plummet may hang 
 exactly at 45 degrees, when the staff is perpendicular to 
 the horizon ; move the staff backward or forward, always 
 keep it perpendicular, until you can see the top of the 
 tree through both the sights ; measure the distance be- 
 tween the bottom of the staff and the bottom of the tree, 
 to which add the height of your eye ; and the sum will 
 be the height of the tree, supposing the ground to be 
 horizontal. 
 
 If, by reason of impediments, you cannot retire so far 
 from the tree as directed above, make the quadrant fast, 
 so that the plummet may hang at 631 degrees, when you 
 view the top of the tree through both the sights ; then 
 twice your distance from the tree added to the height of 
 your eye, will give the height of the tree. 
 
 If this also be impracticable, take any angle of altitude, 
 and measure the distance to the bottom of the tree; then, 
 by a scale of equal parts, draw a line equal to the mea- 
 sured distance; and at one end of this line erect the per- 
 pendicular, and at the other end, by Problem XXIII., 
 Part I., make an angle equal to the angle of altitude. 
 Measure the perpendicular of the right-angled triangle 
 thus formed, by the same scale from which the base was 
 taken, to which add the height of your eye ; and you 
 will obtain the height of the tree. 
 
 3. Divide a square staff, AB, of about 7 or 8 feet in 
 length, into feet and inches, for the convenience of mea- 
 suring the distance between the place of observation and 
 the tree, or taking any other dimensions. 
 
 Upon one side of this staff, at a commodious distance 
 from the bottom, fix a rectangular board, CDEF, whose 
 length DE is exactly equal to twice its breadth CD, 
 which breadth may be about 4 or 5 inches. At C and 
 D fix sights, or small iron pins ; and also at G and E ; 
 making DG and GE each equal to CD. 
 
 Then, when the top of a tree is seen through the 
 
 B 
 
162 TIMBER HIEASURE. PART IV. 
 
 sights at C and G, the tree's height is equal to your distance from Its 
 bottom added to the height of your eye ; but if seen through the 
 sifThts at C and E, its height is equal to twice your distance from its 
 bottom, adding the same height as before. 
 
 In making an observation with this instrument, it ought to be fixed 
 perpendicularly to the horizon, which may be done by means of a 
 plummet suspended from n. 
 
 In taking the altitude of a tree growing upon an inclined plane, 
 you must endeavour to make your observations from a place upon a 
 level with the bottom of the tree. If this cannot be done, direct the 
 horizontal sights at C and D, towards the lower part of the tree, and 
 let your assistant make a mark upon it ; then find the height of the 
 tree above this mark, as before, to which add the distance of the mark 
 from the ground, which must, in this case, be considered the height 
 of the eye ; and the sum will be the height of the tree. 
 
 The simplicity of this instrument seems to give it a preference to 
 the quadrant ; practice will, however, soon enable a person to judge 
 pretty correctly of the heights of trees in general, independently or 
 any instrument. 
 
 4. In order to obtain the girt of a tree, at or near the middle, pro- 
 ceed thus : If the tree taper pretty regularly, take the girt about two 
 feet from the bottom, to which add 24 inches, the supposed girt at 
 the top, and half the sum will be nearly the girt at the middle ; as 
 the height of the trunk ought only to be taken to that part which 
 will measure 24 inches in circumference. Or, take the quarter girt 
 about 2 feet from the bottom, to which add (5 Inches ; and half the 
 sum will be the quarter girt at the middle nearly. 
 
 If the upoer part of the trunk be nearly as thick as the bottom, 
 which is sometimes the case ; take the girt as far from the ground as 
 you can reach, from which girt m.ake a small deduction, at discretion, 
 in order to reduce it as nearly as you can to the girt at the middle. 
 
 5. In measuring timber in a wood, you will frequently meet with 
 8, 10, 12, or more trees, nearly of the same dimensions, so that by 
 measuring one of a medium size, you may, by it, estimate the contents 
 of the rest, without measuring them ; and thus be enabled to proceed 
 with considerable expedition. 
 
 6. When a fall of wood takes place, it is often necessary to cut 
 down many young trees, some of which will not contain more than a 
 foot of wood, in order to make room for others to come to a greater 
 matuuty. If the quarter girt of these be less than 4 inches, their 
 contents cannot be found by the Sliding Rule, consequently they 
 must be computed by the Pen ; practice will, however, soon enable a 
 person to estimate tlie contents of small trees with considerable ac- 
 curacy, independently of calculation. 
 
 Young trees are called pole-icood, and valued at a less price than 
 larger trees, or such as are called timber. 
 
 7. In valuing a wood, the branches, the bark of oak trees, and the 
 under-wood must be taken into consideration, as well as the timber 
 and pole-wood. 
 
 The strongest of the under-v/ood, and the boughs or branches 
 which are not timber, are called cord-wood, and valued in Yorkshire 
 and other northern counties, by the statute cord of 128 cubic feet; 
 viz. 4 feet in breadth, 4 feet in height, and S feet in length. 
 
SECT. ni. TIMBER MEASURE. 163 
 
 Cord-wood is sometimes piled up and sold in parcels of those dimen- 
 sions ; and sometimes it is made into piles much larger, and the num- 
 ber of cords they contain found by admeasurement. 
 
 In Sussex, and several other southern counties, a pile of wood con- 
 taining 126 cubic feet, viz. 3 feet broad, 3 feet high, and 14 feet long, 
 is accounted a cord of wood. 
 
 The strongest of the cord-wood is made into props for the roofs of 
 coal-mines, posts, rails, &c. ; and the other into charcoal, which is 
 used in manufacturing gunpowder, polishing copper or brass, making 
 powerful fires for melting metals, &c. Tlie price of cord-wood is at 
 present from nine to fifteen shillings per cord. 
 
 Bark is valued by the ton ; and the price fluctuates in proportion 
 as we have supplies from foreign countries. The price is, at present, 
 from 1 •:/. to 1 51. , according to the quality of the article. 
 
 Bark is a very strong astringent, and is chiefly used in tanning the 
 hides of cattle into leather. 
 
 Under-wood is valued by the acre, if the number of acres contained 
 in the wood be known ; if not, it must be estimated by the great, at 
 the discretion of the valuer. The price of under-wood is regulated 
 by the quantity, quality, and demand there is for it. In some woods 
 it is worth 51. per acre, and in others it is scarcely worth 5s. 
 
 Part of the under- wood is used for wicker-work, such as panniers, 
 coal-baskets, &c. ; and some of it is only fit for hedging materials. 
 
 8. When a. fall of wood is about to take place, the Valuer or Wood- 
 man must first mark all the trees which are intended to stand, by 
 causing a ring of white or red paint to be made quite round the trunk 
 of each tree. This is called " setting-out." 
 
 The trees that are permitted to stand should be at proper distances 
 from each other ; of different ages ; and such as are likely to make 
 the greatest improvement. Those which are straight and elegant, 
 and have few branches, are generally preferred ; this part of the busi- 
 ness, however, depends very much upon the skill and experience of 
 the woodman. 
 
 WTien the wood has all been set out, those trees which are not marked 
 must be valued ; and in doing this, attention must be paid to the 
 sizes, qualities, and species of the trees. Large trees, if sound, are 
 always worth more per foot, than small ones ; and if the crooked oak 
 trees, and large crooked boughs, be particularly adapted for ship 
 timbers, they are worth more per foot than straight trees. 
 
 In measuring and valuing, som.e persons number every tree, by a 
 scribe-iron, or with paint ; others only count them, and mark each 
 tree with a cross, to show that it has been measured; and others 
 number the large trees, and count the small ones, setting dov/n the 
 contents of 10, 15, or 20 of these in one sum. 
 
 9. No regular rules can be given by which the quantity of bark, 
 and the cord-wood a tree contains, may be computed ; as the bark of 
 trees differs very much in thickness, and the cord-wood depends 
 entirely upon the quantity of top. Sometimes there is a great deal 
 more bark upon the top of a tree, than upon the bole. 
 
 If the bole of a tree measures 30 feet or 360 inches in length, 60 
 inches in circumference in the middle, and the bark be half an inch 
 thick ; we have 3G0 multiplied by 60 multiplied by 1-, equal to 
 10800 cubic inches, the solidity of the bark, which being divided 
 
164 TIMBER MEASURE. TART lY. 
 
 by 2218, the cubic Inches in an imperial busbel, we obtain 5 bushels 
 nearly, the quantity of bark upon the trunk, to which must be added 
 the quantity estimated to be on the top ; practitioners, however, are 
 never at the trouble of calculating the quantity of bark upon the 
 trunk, but make an estimate of what they suppose to be upon the 
 whole tree. Besides, when bark is chopped, it will measure to a 
 great deal more than in a solid state, as found by the above calcula- 
 tion. 
 
 In most places in the West Riding of Yorkshire, 6 pecks are ac- 
 counted a bushel of bark, and 9 such bushels a quarter ; and it has 
 been found by experiment, that about 8 quarters of the bark of pretty 
 large trees, or about 12 of that of small trees, or pole-wood, will weigh 
 a ton; consequently, we may reckon about 10 quarters, upon a me- 
 dium, to make a ton. 
 
 10. When an estate, containing a wood, or a great number of trees 
 in the hedge-rows, is intended to be sold, all the trees should be 
 valued previously to the sale ; as there have been instances where 
 Estates have been sold, without considering the woods ; and the pur- 
 chaser has immediately disposed of the timber for nearly the price of 
 the Estate. 
 
 11. When a rough calculation of a large wood is wanted, or time 
 will not permit you to be very accurate, let an acre be measured off 
 in that part of the wood which appears to be of a mean value, and 
 measure and estimate the timber, &c. which it contains ; then this 
 estimation being multiplied by the number of acres in the wood, will 
 give the value of the wood nearly. Or measure and estimate an acre 
 of tlie best and an acre of the worst; and multiply half the sum of 
 the two estimations by the number of acres in the wood, for the whole 
 value. 
 
 1 2. In measuring and valuing standing timber, various methods are 
 adopted, by different persons, in setting down the dimensions, &c. 
 Some enter the oaks, elms, &c. separately in different parts of the 
 book ; and others enter them promiscuously, having a column in 
 which they specify the name of each tree. Some valuers estimate the 
 quantity of cord- wood in the top of each tree, or in the tops of two or 
 three trees taken together ; and others value the top of each tree. 
 
 The two following examples will serve to illustrate what has 
 been said on the subject of measuring and valuing standing timber. 
 The content of each tree was found by the Sliding Rule ; and the 
 learner ought to repeat the process, in order to make himself expert 
 at casting by the Rule. 
 
 EXAMPLES. 
 
 1. Find the contents of six oak trees, and the value of 
 the timber contained in each tree, from the following 
 dimensions and values. 
 
SECT. III. 
 
 TIMBER IVIEASUKE. 
 
 165 
 
 Number Bush. 
 
 Cords 
 
 Length 
 
 Quarter 
 
 Contents 
 
 1 
 
 Values 
 
 
 of of 
 
 of 
 
 in 
 
 Girts in 
 
 in 
 
 per 
 
 
 Tree?. | Bark. 
 
 Wood. 
 
 Feet. 
 
 Inches. 
 
 Feet. 
 
 Foot. 
 
 Tree. 
 
 
 
 
 
 
 A^ d. 
 
 £ 5. d. 
 
 1 1 li- 
 
 2 1 31 
 
 
 
 12 
 
 81 
 
 6 
 
 2 
 
 12 
 
 
 
 16f 
 
 12^ 
 
 171 
 
 2 6 
 
 2 3 9 
 
 3 32 
 
 1 
 
 391 
 
 18 
 
 89 
 
 4 3 
 
 18 18 3 
 
 4 15 
 
 1 
 
 23 
 
 15? 
 
 39f 
 
 3 
 
 5 19 3 1 
 
 5 1 21 
 
 
 
 15 
 
 91 
 
 91 
 
 2 3 
 
 1 1 41 
 
 6 1 80 
 
 3 
 
 4 
 
 44 
 
 281 
 
 248 
 
 6 
 
 74 8 0" 
 
 Total 1341 
 
 4 
 
 — 
 
 
 — 
 
 — 
 
 103 2 71 
 
 2. Required the contents of ten trees, and the value of 
 the timber in each tree, from the following dimensions 
 and values. 
 
 1 
 
 
 
 
 'Lon<7ths 
 
 Quarterl Contents 
 
 Value 
 
 
 1 
 
 Xo. of 
 
 Names of the 
 
 1 Value of 1 
 
 in 
 
 Girts in 
 
 in 
 
 per 
 Foot. 
 
 Value of 
 
 Trees. 
 
 Trees. 
 
 1 the Tops, j 
 
 Feet. 
 
 Inches. 
 
 Feet. 
 
 each Tree. 
 
 
 \£. 
 
 s. 
 
 ^.I 
 
 
 
 
 s. d. 
 
 £ 
 
 s. d. 
 
 1 
 
 Ehn . . 
 
 
 
 5 
 
 6 
 
 421 
 
 121 
 
 441 
 
 3 Oj 6 
 
 12 9 
 
 2 
 
 Ash . . 
 
 
 
 5 
 
 
 
 381 
 
 llf 
 
 361 
 
 3 
 
 5 
 
 9 6 
 
 3 
 
 Pine . 
 
 
 
 3 
 
 oi 
 
 36 
 
 141 
 
 521 
 
 3 6 
 
 9 
 
 3 9 
 
 4 
 
 Fir . . 
 
 'o 
 
 1 
 
 6 
 
 28 
 
 10 
 
 19| 
 
 1 9 
 
 1 
 
 13 81 
 7 6 
 
 1 o 
 
 Beech . 
 
 
 
 4 
 
 6 
 
 39f 
 
 161 
 
 75 
 
 2 6 
 
 9 
 
 6 
 
 Sycamore 
 
 3 
 
 6 
 
 261 
 
 481 
 
 9f 
 
 17^ 
 
 2 6 
 
 2 
 
 3 11 
 
 4 3 
 
 / 
 
 Poplar . 
 
 
 
 2 
 
 6' 
 
 12;i 
 
 o4| 
 
 3 
 
 8 
 
 8 
 
 Yew . 
 
 1 
 
 2 
 
 6 
 
 91 
 
 381 
 
 961 
 
 4 9|22 
 
 18 41| 
 
 19 
 
 9 
 
 Fir . . 
 
 
 
 2 
 
 
 
 321 
 
 I'H 
 
 
 2 o! 3 
 
 10 
 
 Larch . 
 
 
 
 2 
 
 el 
 
 1 
 
 30| 
 
 151 
 
 51i 
 
 3 6 
 
 8 
 
 19 41 
 
 
 Total 
 
 2 
 
 12 
 
 6' 
 
 I 
 
 
 — 
 
 — 
 
 1 
 - j78 
 
 11 3| 
 
 yote. Having described the methods of measuring and valuing standing timber; 
 and, in the last Example, mentioned a variety of trees, it is presumed that this 
 section cr.nnot be better concluded than by a short descrii'^ion of the natures, pro- 
 perties, and uses of these trees. 
 
 Besides, it is absolutely necessary, in order to become a valuer of timber, to be 
 made acquainted with a few of the leading properties of trees, and their comparatire 
 usefulness. 
 
166 A DESCRIPTION PART IV 
 
 A DESCRIPTION 
 
 OF THE NATUEE AND PROPEETIES OF SOME OF 
 
 THE MOST USEFUL TIMBER TREES. 
 
 THE OAK. 
 
 The Oak stands at the head 'of all British timber trees, as well for 
 its utility, as for its majestic appearance. It is slow of growth ; but 
 if permitted to stand, it arrives at a size equal, if not superior, to 
 that of any other tree of the forest ; and by the vast arms which it 
 throws out on every side, it forms a mass which fills the eye of the 
 spectator, and impresses him with gigantic ideas of its masculine 
 strength. The oak delights most in a rich, strong soil, in which it 
 strikes its root to a great depth. It loves hilly, better than boggy 
 ground, and thrives best in large plantations. 
 
 The uses to which oak is applied, are numerous. It is a firm 
 wood, and will endure all weathers, climates, and seasons ; hence it is 
 used for posts, rails, window-frames, casks, water-pails, carts, wag- 
 gons, wheel-spokes, &c. In machinery, no other wood is equal to 
 it, where a great stress is to be borne, as in mill-work, &c. ; and in 
 water- works it is inferior to none. It is also used for household furni- 
 ture, such as tables, bedsteads, chests of drawers, &c. ; but it has ac- 
 quired its chief fame, in this country, by its use in ship-building, 
 being much superior to foreign oak ; and has, no doubt, contributed 
 very materially to the naval glory of Old England. 
 
 Some oak trees have arrived at an enormous size. In Dr. Hunter's 
 edition of Evelyn's Sylva, is given a figure of the old oak of Cow- 
 thorpe, in Yorkshire, which measures 48 feet in circumference, at a 
 yard from the ground. About a mile and a half from Shrewsbury, 
 there is an oak whose girt is 44 feet at the bottom, 27 at the distance 
 of 8 feet from the ground ; and it is 41 i feet in height. In Hainault 
 Forest, near Barking, in Essex, there is an oak which measures 36 
 feet in circumference. The tree has been known through many cen- 
 turies by the name of Faijilop. Mr. Gilpin, in his " Remarks on 
 Forest Scenery and other Woodland Views," says that the tradition 
 of the country traces this tree half-way up the Christian sera. 
 
 Some of our best poets have noticed the usefulness and remarkable 
 longevity of the oak. 
 
 " Let India boast her plants, nor envy we , 
 
 The weeping amber and the balmy tree. 
 While by our oaks the precious loads are borne, 
 And realms commanded which those trees adorn." — Pope. 
 
 " The monarch oak, the patriarch of the trees. 
 Shoots rising up, and spreads by slow degrees : 
 Three centuries he grows, and «hree he stays 
 Supreme in state, and in three more decays." — Dryden 
 
SECT. m. OF TIMBER TREES 167 
 
 THE ASH. 
 
 This tree generally grows tall and elegant, and makes a graceful 
 appearance, when contrasted with trees of greater bulk. It flourishes 
 most in woods, but will also thrive well in good soils, upon open 
 grounds. 
 
 There are few trees which excel the ash in utility ; for its wood, 
 noxt to that of oak, is employed for the greatest variety of purposes. 
 It may be peculiarly termed the husbandman'' s tree ; for it is one of 
 the principal materials in making ploughs, harrows, carts, waggons, 
 spokes and felloes for wheels, and various other implements for rustic 
 use. It is also employed by the turner for dairy utensils ; and at sea 
 It is used for oars and hand-spikes. 
 
 The toughness of its wood rendered it a favourite with the heroes 
 of old, for the shafts of their potent spears ; hence it is poetically 
 termed " the martial ash." Homer arms his heroes with spears of 
 ash. 
 
 " From Pelion's cloudy top an ash entire. 
 Old Chiron lell'd, and sharp'd it for liis sire." — Pope's Homer. 
 
 THE BEECH. 
 
 This is one of the most stately of timber trees ; large woods are 
 wholly composed of it in some parts of this country. It particularly 
 delights in a chalky soil, where it will flourish and arrive at a great 
 size, although the land may have all the appearance of barrenness. 
 
 The wood of the beech is brittle, and apt to decay ; but being of a 
 flne grain, and easily wrought, it is used for a variety of domestic 
 purposes. It is employed by the turner and cabinet-maker ; the 
 former using it for his larger ware, and the latter for common chairs 
 and other articles of furniture. It can be split so thin, that it is used 
 for band-boxes, hat-cases, book-covers, and scabbards of swords. 
 
 THE ELM. 
 
 The common elm is a large timber tree, of great beauty and utility. 
 It grows to a great height, and at the same time, if permitted, ex- 
 pands some massy arms. It loves an open situation, and a black 
 clayey soil. 
 
 The wood of the elm is hard and tough ; and is vised for a great 
 variety of purposes. Is is particularly serviceable in situations where 
 it is kept constantly wet ; and is therefore used for ship-planks, water- 
 pipes, pumps, mill-wheels, &c. It is also employed for axle-trees, 
 wheel- naves, gate-posts, chopping-blocks, &c. 
 
 THE SYCAMORE. 
 
 This tree is of quick growth, arrives at a large size, and flourishes 
 best in open places and sandy ground. 
 
 Its wood is soft and very white ; hence it is proper for the use of 
 the turner, who makes it into bowls, trenchers, and other domestic 
 utensils. In consequence of its lightness, it is also occasionally wrought 
 up as cart and plough timber. 
 
168 A DESCRIPTION PART IT. 
 
 THE BLACK POPLAR. 
 
 The name of black seems given to this tree, in order to distinguish 
 it from the nhiie poplar ; for its leaves are a heautiful green, and the 
 tree has nothing dark in its appearance. It loves a rich and moist 
 soil; it arrives at a great size; and is one of the tallest and most 
 stately to be seen when full grown 
 
 The wood of the poplar is tougher and harder than fir ; and is 
 frequently used instead of it, for laths, packing-boxes, roofing, floor- 
 ing, &c. 
 
 The white poplar does not arrive at so large a size as the black. It 
 grows best in moist situations ; and is very conspicuous from the 
 whiteness of its foliage. 
 
 THE ALDER. 
 
 The alder flourishes best in boggy situations, and by the sides of 
 rivers, &c. It sometimes arrives at a large size ; it is, however, chiefly 
 used in this country for poles. Virgil, in his Georgics, tells us that, 
 by hollowing its trunk, it was anciently made into boats or canoes. 
 
 The wood of the alder will remain long sound under v/ater. It is 
 employed for pumps, water-pipes, piles, &c. It is likewise used for 
 shoe-heels, closes, and turners' work. 
 
 THE WEYMOUTH PINE. 
 
 This tree is a native of North America, where it is called the 
 " white pine." It frequently arrives at the height of ICO feet ; and 
 is therefore preferred to the rest of the pine tribe for the masts of 
 large ships. Our men of war are generally furnished with masts of 
 this species of tree, from the timber-yards of Nova Scotia. 
 
 Lord Weymouth introduced it into this country ; on which account 
 it is generally known by his name ; and has, for some time, been a 
 great favourite with planters, 
 
 THE FIR. 
 
 Linnaeus considers all the fir tribe to be only different species of 
 the pine. Some of them grow best on mountainous situations ; others 
 prefer bogs and swamps ; and they alone often compose extensive 
 woods, clothing barren and desolate regions, unfit for human culture. 
 
 INIany species of this tribe, are now become'common in our planta- 
 tions. Of these I shall only mention a few of the principal. 
 
 The Scotch Fir grows naturally in some parts of the Highlands of 
 Scotland; and also upon the mountainous parts of Norway, Sweden, 
 and Russia. Tlie fir grown in the native forests of Scotland is good, 
 but affords a very scanty supply ; consequently, the greatest part of what 
 is consumed by us, is brought from Norway, and the countries border- 
 ing upon the 13altic. 
 
 No wood is at present used amongst us in such quantities as the 
 fir, which, under the name of deal, is employed about buildings, for 
 boards, planks, beams, rafters, joists, &c. 
 
 Fir is also used for the masts and yards of ships ; and from it we 
 obtain the important productions of turpentine, tar, and pitch. It 
 
SECT. III. OF TIlVrBER TREES. 169 
 
 may, therefore, be called the sailors' tree, with as much propriety as 
 the oak. 
 
 The Spruce Fir is a native of Norway ; and is said to afford the 
 white deal. From the green tops of this tree is made the antiscor- 
 butic beverage called " Spruce-beer." 
 
 The Silver Fir grows in Norway and Germany, and from its trunk 
 the yellow deal is said to be procured. It also yields great abundance 
 of tar. 
 
 The Larch Fir is a native of the Alps and Apennines ; and for 
 beauty and durability of wood, it greatly surpasses the Scotch fir. It 
 is now become a great favourite in this country, and thrives well on 
 barren and sandy soils. In some countries ships are built of this wood, 
 which are represented as proving very durable. The larch is remark- 
 ably resinous, and from it is obtained Venice turpentine. 
 
 THE BIRCH. 
 
 There are four species of birch. The common birch-tree may be 
 cultivated upon barren land, where better trees will not grow ; for 
 there is no ground so bad as not to allow this to thrive in it. It will 
 grow in moist springy land, or in dry gravel, or sand, where there is 
 little surface ; hence persons who are in possession of poor land, can- 
 not employ it better than by planting it with these trees. They have 
 sometimes been planted upon ground that produced nothing but 
 moss ; and in nine or ten years after planting, have been sold for 10/. 
 per acre ; the after-produce greatly increased. 
 
 ITie wood of this tree is used by broom-makers, hoop-benders, and 
 turners. 
 
 THE YEW. 
 
 The yew is a native of this country, and is found in rocky and 
 mountainous situations. It is frequently planted in church-yard?, 
 probably on account of its being an ever-green. The leaves are of a 
 poisonous nature ; both horses, cows, and children have died in con- 
 sequence of eating them. 
 
 This tree is remarkable for its toughness and elasticity ; and is 
 celebrated for the purposes to which its wood was anciently applied 
 in making that most formidable weapon of our ancestors, — the long- 
 bow. 
 
 " Th' elastic yew, whose distant wound 
 With England's rivals heap'd the ground." — William's Miscellanies. 
 
 • The wood of the yew is at present valued by cabinet-makers and 
 iulayers on account of its beautiful red veins ; and it is also a good 
 material for flood-gates, axles, cogs for mill-wheels, and other works 
 where strength and durability are required. 
 
 The yew seldom grows to a great height ; but sometimes arrives 
 at an amazing thickness of trunk. 
 
 In Darnley church-yard, near Matlock, in Derbyshire, is a yew- 
 tree whose circumference is 33 feet. 
 
 THE HOLLY. 
 
 The holly is an ever-green, and a native of the woods in this coun- 
 try, where it sometimes arrives to the height of 20 or 30 feet. It ii, 
 
 I 
 
170 A DESCRIPTION OF TLMBEK TREES. PART IV. 
 
 however, more usually seen in gardens and hedgerows, in the state of 
 a shrub. 
 
 The wood of a full-grown holly is valuable. It is the whitest of 
 all our hard woods, and is therefore used by inlayers ; and it is some- 
 times stained black, to imitate ebony. It is also excellent for the uses 
 of the turner, carver, and mill-wright ; being extremely firm and 
 durable. 
 
 THE BOX. 
 
 The box is another ever-green tree or shrub, which is sometimes 
 met with here in a wild state ; but more commonly in gardens and 
 plantations. A strong shallow soil, of the limestone kind, seems to 
 suit it best. 
 
 The wood of the box is of a pale-yellow colour ; and being very 
 hard, smooth, and solid, it is much used for various purposes. Mathe- 
 matical and musical instruments are made of it ; also knife-handles, 
 combs, shuttles, &c. It is the only tree whose wood is serviceable to 
 the engraver. 
 
 Having described the natures, properties, and uses of most of the 
 trees found in our woods and plantations, I shall conclude with the 
 following extract from Mr. John Tuke's Survey of the Agriculture 
 of the North Riding of Yorkshire ; a work that abounds with useful 
 and just observations : — 
 
 " Most people, I think, concur in this point, that for the last half 
 century, the wobd in this kingdom has been terribly on the decline. 
 
 " That glooiuy prospect is now become tremendous, and sufficiently 
 visible to awaken the fears of every thinking person. The axe is 
 often heard, but the planter is seldom seen. Let us cast our thoughts 
 towards the future support and welfare of our navy — our sole pro- 
 tection ! — and we must tremble at the continual disappearance of our 
 oak. Some speedy method must be adopted to remedy this great 
 national evil ; or, besides the danger from fierce external foes, we 
 must determine to go barefoot : we should never think of looking to 
 foreign countries for a constant supply of oak-bark to tan our leather. 
 Let Britain help herself! 
 
 " Each nobleman and gentleman should insert, in the agreement 
 with his tenants, a clause to compel them to plant and protect, in the 
 corners of their fields, and upon pieces of waste ground, a certain 
 number of good oak, elm, and ash trees, annually. These trees 
 should be found by the landlord ; and he should enforce the perfornn- 
 ance of this clause as rigidly as the payment of the rent ; then will 
 the rising generation have cause to bless the wisdom and policy of 
 the present age." 
 
 Note. It has, however, been found within the last few years, that England can 
 have a sufficient supply from her own colonies. It is now ascertained that the 
 timber in the fertile islands of New Zealand far exceeds that of the English, both 
 in size and variety ; the cowdie tree is a fine specimen ; it possesses a high degree 
 of flexibility, as well as strength : the wood is fine grained, and is of sufficient 
 length and strength to serve for the main and fore top masts of the largest three- 
 decker ; the Board of .Admiralty being now procuring supplies of this timber for 
 the use of the Royal Navy. 
 
SECT. IV. mSCELLANEOUS QUESTIONS. 171 
 
 SECTION IV. 
 MISCELLANEOUS QUESTIONS 
 
 CONCERNING 
 
 SOLIDS. 
 
 1. TViiAT is the solidity of a cubical stone, whose dia- 
 gonal is 30 inches? Ajis. 9545.94154 inches. 
 
 2. The length of a paralleiopipedon is 10, its breadth 
 6, and its depth 4 feet ; what is the length of a similar 
 paralleiopipedon, whose solidity is three times as much ? 
 
 Ans. UA2249feet. 
 
 3. The altitude of a cylinder is 20, and its diameter 10 
 inches ; what is the altitude of another cylinder whose 
 solidity is twice as much, its diameter being 30 inches ? 
 
 Ans. 4A inches. 
 
 4. The diameter of the old Winchester bushel is 18^ 
 inches, and its depth 8 inches ; what is the diameter of 
 that bushel whose depth is 7 inches ? 
 
 A71S. 19.77733 inches. 
 
 5. Two men bought a conical piece of timber, which is 
 to be equally divided between them, by a plane parallel to 
 the base ; what will be the altitude of each part, the height 
 of the cone being 21 feet, and the diameter of its base 3 
 feet 6 inches ? 
 
 Ans. The altitude of the upper part is 16.66771 /ee^; 
 and that of the loiver part 4.33229 feet. 
 
 6. Each side of the greater end of a piece of timber, in 
 the form of the frustum of a square pyramid, is 5 feet, 
 each side of the less end 2 feet, and its perpendicular alti- 
 tude 15 feet;. it is required to divide it equally between 
 two jTflrsons, by a plane parallel to the ends. 
 
 Ans. The height of the part next the less end is 10.25709, 
 and the height of the part next the greater end is 
 A.I A291 feet. 
 
 7. A gentleman has a bowling-green, 500 feet in length, 
 and 300 feet in breadth, which he would raise 1 foot 
 higher, by means of the earth to be dug out of a ditch 
 with which he intends to surround it ; what will be the 
 depth of the ditch, if its breadth be every where 9 feet ? 
 
 Ans. 10.18744/ee^. 
 1 2 
 
172 mSCELLAlfEOUS QUESTIONS PART IV 
 
 8. What must be tlie diameter of tlie bore of a cannon, 
 which is cast for an iron ball of 36 lb. in weight, so that 
 the diameter of the bore may be ^ of an inch more than 
 that of the ball ? A??s. 6.4747 inches. 
 
 9. If a heavy sphere whose diameter is 8 inches, be put 
 into a conical vessel, full of water, whose diameter is 10, 
 and altitude 12 inches; it is required to determine how 
 many cubic inches of water will run over. 
 
 Ans. 210.17722 cubic inches. 
 
 10. A cubical foot of brass is to be di'awn into wire J■^^ 
 of an inch in diameter ; what Vvdll be the length of the 
 wire, allowing no waste in the metal ? 
 
 A?is. 31.25217 ?mles. 
 
 11. How many 4-inch cubes can be cut out of a 12- 
 inch cube ? Ans. 27. 
 
 12. The length of a piece of square timber is 12 feet, 
 and its solidity 48 feet ; what wiU be its solidity when it 
 is made into the greatest cylinder possible ? 
 
 Ans. 37.6992 feet. 
 
 13. A farmer borrowed part of a haystack of his neigh- 
 bour, which measured 8 feet every way, and paid him 
 back by two equal cubical pieces, each side of Avhich mea- 
 sured 4 feet : Query, was the lender fully paid ? 
 
 A}is. He was paid only 4- j9«;*/. 
 
 14. A piece of timber is 20 inches broad, and 16 inches 
 thick ; at wliat distance from the end must a section be 
 made, so that the part sawn oif may measure exactly o 
 cubic feet ? Ans. 27 inches. 
 
 15. A ship's hold measures 120 feet in length, 33 in 
 breadth, and 6 in depth ; how many bales of goods, 6 feet 
 long, 4 feet broad, and 3 feet deep, may be stowed therein, 
 leaving a gangway, 3 feet broad, the whole lenath of the 
 hold? ^W5. 300. 
 
 16. The diameter of a circular island is 300 feet, and 
 it is surrounded by an oblique-sided moat; how many 
 cubic yards of earth were dug out of it, the breadth at the 
 top being 10 feet, the breadth at the bottom 4 feet, and the 
 perpendicular depth 8 feet? Ans. 2019.93244 cubic yards. 
 
 17. Suppose it be found, by measurement, that a man 
 of war, with its ordnance, rigging, and appointments, 
 draws so much water as to displace 55000 cubic feet oi' 
 sea-water ; required the weight of the vessel. 
 
 Ans. 1574 ^% tons. 
 
 18. The length of a cylindrical piece of timber is 15 
 
SECT. IV. CONCERNING SOLIDS. 173 
 
 feet, and its diameter 4 feet ; what will be its solidity 
 when hewn into a square prism? Ans. 120 feet. 
 
 19. How many bricks, each 8 inches long, 4 inches 
 broad, and 3 inches thick, will build a wall 200 feet long, 
 10 leet high, and 1 foot thick? Ans. 36000. 
 
 20. What is the weight of a cast iron cannon-ball 
 whose diameter is 4 inches ? A7is. 9 pounds. 
 
 21. The ball upon the top of St. Paul's church, in Leeds, 
 is 3 feet in diameter, and it is covered with copper ■:^^ of 
 an inch thick ; required the value of the copper, at 2s. i\d. 
 per pound avoirdupois. Ans. £1 \2s. Z\d. 
 
 22. What is the weight of a bomb-shell, or hollow sphere 
 of cast iron, whose inside diameter is 9 inches and the 
 thickness of the metal an inch and a half ? 
 
 Ans. 140.4746 lb. 
 
 23. The length and breadth of the greater end of a 
 niill-hopper, in the form of a prismoid, are 40 and 30 
 inches ; the length and breadth of the less end 10 and 6 
 inches ; and its perpendicular altitude 36 inches how 
 many imperial bushels will it hold ? 
 
 Ans. 8.277 bushels. 
 
 24. A tree whose length is 28 feet, girt 45 inches in 
 the middle, with a rope 1 inch in diameter ; required its 
 true girt, and likewise its solidity, according to the quarter- 
 girt method of measuring timber ; the length of the rope, 
 Avhen extended in a right-line, being equal to the circum- 
 ference of its centre, when formed into a circle, in taking 
 the girt of the tree. 
 
 Ans. The true girt is 41.858364 inches, and the solidity 
 21.293157 feet ; hence appears the propriety of 
 girting trees ivith a small string, or with a mea- 
 suring-tape. 
 
 25. A leaden pipe, 12 feet in length, weighs 100 lb. 
 avoirdupois ; required the thickness of the lead, the dia- 
 meter of the bore being 2 inches. Aiis. .2408 of an inch. 
 
 26. Required the thickness of the shell of a hollow 
 sphere of copper, weighing 3 lb. avoirdupois, so that being 
 put into common water, it may be just immersed by its 
 own weight ? Ans. .1042 of an inch. 
 
 27. The bore of a syringe which holds one pint, wine 
 measure, is H inch in diameter ; what is the length of the 
 piston? " Ans. \6.2>^9m i7iches. 
 
 28. If a balloon contains 1000 yards of silk, ^,- of a yard 
 
 I 3 
 
174 jnSCELLANEOUS QUESTIONS. PART IV. 
 
 wide ; what is its diameter, admitting it to be a perfect 
 sphere ? Ans. 15.45095 yards. 
 
 29. A hollow sphere of ebony, whose diameter is 18 
 inches, is found to sink just 6 inches in sea-water ; re- 
 quired the thickness of the shell. Ans. .647333 of an inch, 
 
 30. One ev'ning I chanc'd with a tinker to sit. 
 Whose tongue ran a great deal too fast for his wit ; 
 He talk'd of his art with abundance of mettle ; 
 
 So I ask'd him to make me a flat-bottom'd kettle : 
 Let the top and the bottom diameters be. 
 In just such proportion as five is to three ; 
 Twelve inches the depth I proposed, and no more ; 
 And to hold in ale-gallons seven less than a score. . 
 He promis'd to do it, and straight to work went ; 
 But when he had done it, he found it too scant ; 
 He alter'd it then, but too big he now made it ; 
 For though it held right, the diameters fail'd it : 
 Thus making it often too big, and too little, 
 The tinker at last quite spoiled his kettle ; 
 But he vows he will bring his said promise to pass, 
 Or he'U utterly spoil every ounce of his brass : 
 Now, to save him from ruin, I pray find him out 
 The diameter's length, for he'll ne'er do't I doubt. 
 
 Ans. The bottom diameter is 14.640098, and the top 
 diameter, 24.400163 inches. 
 
 Note. This ingenious question was, I believe, first proposed in the Ladies' Diary, 
 for the year 1711. 
 
PAKT V. artificers' WORK. 175 
 
 PART V. 
 ARTIFICERS' WORK. 
 
 The Artificers loliose Works are here to he treated of^ are 
 Bricklayers, 3Iasons, Carpenters and Joiners, Slaters 
 and Tilers, Plasterers, Painters, Glaziers, Plumbers, 
 and Pavers, 
 
 The contents of the works of all artificers, whetlier super- 
 ficial or solid, must be found \>j the Rules given in the 
 foregoing Problems, for the respective figures. This 
 ought to be particularly attended to in taking the dimen- 
 sions. 
 
 The following Rule, which should be got by heart, is 
 better adapted for Cross 3Iultiplication, than any other 
 I have seen ; and as 12 fourths make 1 third, 12 thirds 1 
 second, 12 seconds 1 inch, and 12 inches 1 foot, these 
 numbers, when not too large, may be more expeditiously 
 tui'ned into the higher denominations by the Pence Table, 
 than by dividing them by 12. 
 
 RULE. 
 
 Feet multiplied into feet give feet. 
 Feet multiplied into inches give inches. 
 Feet multiplied into seconds give seconds. 
 Inches multiplied into inches give seconds. 
 Inches multiplied into seconds give thirds. 
 Seconds multiplied into seconds give fourths. 
 
 The measures chiefly used by Artificers, are contained 
 in the following table : — 
 
 12 inches 
 
 ->! 
 
 
 - 1 lineal foot. 
 
 144 square inches 
 
 
 
 1 square foot. 
 
 9 square feet 
 
 
 
 1 square yard. 
 
 63 square feet 
 
 
 - make - 
 
 1 square rood. 
 
 100 square feet 
 
 
 
 1 square. 
 
 272i square feet, or \ 
 
 
 1 square rod, pole, or 
 
 30^ square yards 
 
 JJ 
 
 
 perch. 
 
 2\ote. As the number 27'2i is rather troublesome to divide by, it is customary, in 
 
 practice, to divide by 272, omitting the ^ ; but if this be not thought correct 
 
 enough, convert the \ into decimals ; or reduce the divisor to the improper frac- 
 
 I 4 
 
176 bricklayers' work. part v. 
 
 tion lz?r, and multiply the content in feet by 4, and divide the product by 1089 ; 
 
 4 
 and the quotient will be the number of rods sought. Divide the remainder, if anj-, 
 by 4, and the quotient will be the feet. 
 
 TJie rod of 272i square feet, is the square of 16i feet ; but in some places the cus- 
 tomary rod is the square of 18 feet; viz. 324 square feet. 
 
 BRICKLAYERS WORK. 
 
 Bricklayers generally compute their work by the rod of 
 2721- square feet, and at the rate of a brick and a half 
 thick ; consequently, if a wall be more or less in thickness 
 than this standard, it must be reduced to it, in the follow- 
 ing manner : Multiply the superficial content of the wall, 
 in feet, by the number of half bricks which it is in thick- 
 ness ; and ^- of the product will be the content sought. 
 
 In some places, however, brickwork is measured by the 
 rod of 63 square feet; that is, 21 feet in length, and 3 
 feet high ; and then no regard is paid to the thickness of 
 the wall, in measuring ; as the price of the workmanship 
 is regulated according to the thickness. 
 
 Note 1. In taking the dimensions of a building, measure half round the outside 
 and half round the inside ; and the sum of these two will be the true compass of 
 the building. Or, 4 times the thickness of the wall taken from the whole compass, 
 on the outside, or added to the whole compass within, will give the true compass ; 
 which multiply by the height, and the product will be the superficial content of the 
 walls ; and if the breadth of the building be multiplied by the height of one gable- 
 end, you will obtain the content of both the gable-ends. 
 
 In some places it is customary to take the whole compass, on the outside, in 
 order to make an allowance to the workmen, for the trouble of turning the corners ; 
 but this should not be done for both workmanship and materials, except specified 
 in the agreement ; but for workmanship only. 
 
 If the workmen be allowed so much per yard, lineal measure, for the corners, 
 which is the custom in some places, the true compass of the building ought then to 
 be taken for both workmanship and materials. 
 
 2. When the height of a building is unequal, measure a piece round the bottom, 
 so as to make the upper part all of one height : and in doing this, holes must be 
 dug in the ground, to enable you to take your dimensions to the foundation. 
 Several altitudes of the bottom p'art must be taken ; and their sum, divided by their 
 number, may be considered as a mean altitude. 
 
 3. In most buildings of two or more stories, the walls decrease in thickness, to- 
 wards the top ; and this diminution generally consists of half a brick, in each story. 
 The thickness is set off on the inside, and commonly in a place where the floor will 
 be laid; a contrivance by which the set-off \% concealed. 
 
 The stories that are of different thicknesses must be measured separately ; except 
 by agreement, one price be allowed for the whole, which is not often the case. 
 
 When the walls of a building are of different thicknesses, the upper rooms are 
 broader than the lower ones ; hence a set-off may be discovered, although the walls 
 are plastered and the floors laid. 
 
 4. Doors and windows must always be deducted for materials, if there be no 
 stipulation to the contrary ; but for workmanship, these deductions are seldom 
 made, except the doors and windows be very numerous, as in workshops, &c. ; or 
 larger than the usual size, as in shop-fronts, &c. 
 
 In these cases the Surveyor must exercise a discretionary judgment; for the 
 windows of buildings vary so much in size and number, that no special Rule can 
 be given. 
 
 5. If a chimney stand by itself, without any party wall being joined to it, take 
 the girt, in the middle, for the length, and the height of the story for the breadth ; 
 but if the chimney-back be a parly-wall, and the wall be measured by itself, you 
 must girt the chimney round, to the wall, on each side, for the length ; and take 
 the breadth the same as before. 
 
 When a chimney is wrought upright from the mantel-tree to the ceiling, the 
 
PART V. bricklayers' WORK. 177 
 
 thickness of the whole, is penorally considc-rcd the same as that of the jar^i/js ,- and 
 no deduction is ever made f(ir the vacancy between the floor and the rnnntcl-trrc, 
 because of the gathering of the breast and wings, to make room for tlie hearth in 
 the next story. . . . ., -j ., r 
 
 Chituney shafts above the roof, are measured by girtmg them, m the miaaic, lor 
 the length, and taking the height for the breadth. 
 
 Their thickness is generally accounted half a brick more than it is m reality, in 
 consideration of the plastering and scafTolding. 
 
 G. In some places, double measure, for workmanship, is allowed for dinnnci/s, in 
 consequence of their being more troublesome to be made than the other parts of 
 the building; and in others they are done at so much per yard, lineal measure, or 
 at so much per piece. 
 
 It is also customary, in most places, for bricklayers, to charge so much extra for 
 every arcli thcv turn ; and this charge is regulated by the size of the arch. They 
 also make a difference in the price between an inside and an outside arch, charging 
 less for the former than the latter. 
 
 EXAMPLES. 
 
 1. The length of a wall is 86 feet 9 inches, its height 
 12 feet 6 inches, and its thickness 3 bricks ; how many 
 standard rods of brick-work does it contain ? 
 
 B?/ Decimals. By Cross 3TultipUcation. 
 
 Feet. Feet. Inches, 
 
 86.75 length. 86 9 
 
 12.5 height. 12 6 
 
 43375 1041 
 
 17350 43 4 6 '' 
 
 8675 _ 1084 4 6 
 
 1084.375 Q 
 
 6 3 )6506 3 r.ft in. 
 
 3)6506^250 r. ft. in. 272) 2168 9 0( 7 264 9 
 272)2168.750(7 264 9 Ans. 
 
 2. In taking the dimensions of a cottage, I find that 
 half the compass, on the outside, is 40 feet, half the com- 
 pass within 37 feet, the height from the foundation to the 
 eaves 9 feet, the height of the gable-end 5 feet, and its 
 breadth 18 feet ; how many rods of brickwork are con- 
 tained in the building, deducting for a door, which mea- 
 sures 6 feet by 3, and a window whose height is 5, and 
 breadth 4 feet? Ans. 11 roods, 52 feet. 
 
 3. The circumference of a circular building at the iron 
 foundry of Messrs. Fenton, Murray, and Wood, in Leeds, 
 is 241 feet, and its height, from the bottom to the eaves, 
 49 feet 6 inches ; how many rods are contained in the 
 wall; deducting for a door whose breadth is 12, and 
 height 14 feet ; for 96 windows, each of which measures 
 
 I 5 
 
178 MASOXS' WORK. PART V. 
 
 6 feet 7 inches bj 4 feet ; and for 14 windows, whose 
 heights are 9 feet 6 inches, and breadths 5 feet 4 inches ? 
 
 Alls. 135 rods, 19 feef. 
 4. The true compass of a building, two stories high, is 
 120 feet 6 inches ; the height of the lower story is 10 feet 
 6 inches, and the thickness of the wall 2 bricks; the 
 height of the upper story is 8 feet 3 inches, and the thick- 
 ness of the wall 1^ brick ; the gable-end measures 20 feet 
 in breadth, and 8 feet in height, and is 1-i brick thick ; 
 what did the brick-work cost at 14/. 125. per standard 
 rod ; deducting for 6 windows in the lower story, each of 
 which measures 5 feet 6 inches by 4 feet 3 inches, and a 
 door whose height is 6 feet 6 inches, and breadth 3 feet 
 9 inches, and four windows in the upper story, whose 
 dimensions are 4 feet 9 inches by 4 feet 3 inches ? 
 
 Ans. £136 6s. 9d. 
 
 ISIASONS' WORK. 
 
 All kinds of stone-work belong to Masonry ; and thc^ 
 measures generally used are the lineal foot, the square 
 foot, the square yard, the square rod of 63 feet, and the 
 cubic foot. 
 
 Carved mouldings, &c. are generally measured by the 
 lineal foot ; and ornamental work, such as arches, archi- 
 traves, friezes, cornices, chimney-pieces, &c. by the square 
 foot. Also all tooled or cleansed work is measured by 
 the square foot ; viz. door-posts, window-jambs, flags, 
 steps, &c. ; but rough flagging is generally measured by 
 the square yard. 
 
 Walls are sometimes measured by the square yard, and 
 sometimes by the rod of 63 square feet. 
 
 Columns, pillars, blocks of marble or stone, &c. are 
 measured by the solid foot ; and sometimes the contents 
 of walls are found in the same measure. 
 
 Solid measure is chiefly used for materials, and super- 
 ficial for workmanship ; in some places, however, masons 
 are paid so much per rod, for workmanship and ma- 
 terials, and the price is regulated by the thickness of the 
 wall. 
 
 Note 1. The dimensions of stone bm'ldings are taken in the same manner as in 
 Bricklayers' Work ; and deductions must be made for doors, truidotvs, Sec, except 
 the agi'eement prohibits it. These deductions, however, ought only to be made 
 for materials ; as the workmen are fully entitled to receive pay for the whole as ^ 
 walling, in consequence of the trouble of fixing the wiadow-jambs, &c. 
 
PART V. 
 
 JLASONS' WORK. 179 
 
 In measuring tooled or cleansed fronts, the doors and windows must be deducted ; 
 as the price of the workmanship, in these cases, is too considerable for them to be 
 included. 
 
 2. The walls of the upper stories of buildings are, In general, not so thick as 
 those of the lower stories ; but the price for the workmanship is commonly the 
 same, in the consideration of tlie trouble of scaffolding, and the labour of carrying 
 up the materials. 
 
 3. In some places it is customary to measure door-posts, windoiv -jambs, slops, 
 &c. by the cubic foot for the materials, and the superficial foot for the workman- 
 ship ; and in others, so much per superficial foot, is charged for workmanship and 
 materials ; and in taking the breadth, or girt, the tape is made to ply close over 
 every i)art of the stone that has been tooled, except it appear that the workmen have 
 inte7itionalli/ tooled more than is necessary. 
 
 The length of a circular window-head, or door-head, is found by taking half the 
 gum of the greater and less arches. 
 
 4. In making the notch in the window-jambs, for the frame of the window, some 
 workmen are in the habit of tooling or chiseling the jambs further on the inside, 
 than the window-frame requires, in order to make the work measure as much as 
 possible : few architects will, however, allow more than 3 or 4 inches for both sides 
 of the notch. 
 
 5. In 7iieasuring a flight of steps make the tape ply close over them, m the 
 middle, from the top to the bottom, for the length ; and take the length of a step 
 for the breadth . Or, if the steps be all of one size, multiply the area of one step by 
 the number of steps. 
 
 The etids and the landings ought always to be measured by themselves. 
 
 6. All the parts of ornamental frontispieces must be measured separately : viz. 
 plinths, dados, columns, pilasters, architraves, friezes, cornices, pediments, &c. &c. 
 
 7. It is customary, in some places, to allow double measure for all kinds of 
 circular work, such as cylindrical or conical columns, circular pediments, arched 
 door or ivindow-heads, &c. ; and also for cornices, feathered gables, &c. ; in other 
 places, only the area and half the area are allowed; it is much better, however, to 
 proportion the price to the workmanship, and take the true measurement. 
 
 It is also customary, in most places, to pay the workmen so much per yard lineal, 
 for the trouble of turning the corners of buildings, and hewing the stones in a 
 proper manner to form these corners. 
 
 If the corners be formed by tooled or cleansed coins, they must be measured 
 separately, and added to the tooled or cleansed work. In this case no other 
 charge ought to be made for the corners. 
 
 It may likewise be observed, that masons are generally allowed something extra 
 for the trouble of turning arches. 
 
 EXAMPLES. 
 
 1. The length of a wall is 86 feet 9 inches, its height 
 10 feet 6 inches, and its thickness 2 feet 3 inches; re- 
 quired its superficies and solidity. 
 
 By Decimals. B// Cross Maltiplication. 
 
 Feet. Feet. Inches. 
 
 86.75 length. 
 10.5 height. 
 43375 
 8675 
 910.875 superficies. 
 
 2.25 thickness. 
 4554375 1821 9 
 
 1821750 227 8 7 
 
 86 
 10 
 
 9 length. 
 6 height. 
 
 867 
 
 43 
 
 910 
 
 2 
 
 6 
 
 4 6" 
 
 10 6 superficies. 
 3 thickness. 
 
 1821750 2049 5 7 6 solidity. 
 
 204 9.46875 solidity. 
 
 I 6 
 
180 JIASONS WORK. PAKT V 
 
 2. The length of a flight of Yorkshire-stone steps is 
 10 feet 6 inches, and their breadth 3 feet 9 inches ; the 
 landing measures o feet 3 inches hy 4 feet 6 inches ; to 
 what do they amount, at Is. 9d. per square foot, for work- 
 manship and materials? A?is. £5 10s. Sd. 
 
 3. The side of a square pillar, of Portland- stone, mea- 
 sures 1 foot 9 inches, and its height 7 feet 6 inches ; Avhat 
 is the value of 6 pillars of the same dimensions, at 5s. 9d. 
 per cubic foot? A7is. £'S9 I2s. od. 
 
 4. The circumference of the base of a stone column 
 measures 6 feet 10 inches, the circumference at the top 3 
 feet 8 inches, and the slant height 9 feet 3 inches ; re- 
 quired the expense of cleansing or polishing the convex 
 surfaces of 4 such columns, at 1^. 9d. per square foot. 
 
 Ans. £\Q I9s. U^^d. 
 
 5. The threshold of a door measures 4 feet 6 inches in 
 length, 1 1 inches in breadth, and 5 inches in thickness ; 
 the head is 4 feet 11 inches long, 10 inches broad, and 7 
 inches thick ; each jamb is 6 feet 5 inches in height, 10 
 inches broad, and 7 inches thick ; how many cubic feet of 
 stone do they all contain ? Ans. 10 ft. 4 in. 2 pa. 
 
 6. A window-sole measures o feet 10 inches in length, 
 and 15 inches in girt ; the head is 5 feet 8 inches long, 
 and girts 11 inches; the height of each jamb is 6 feet 4 
 inches, and the girt, including 3 inches for the notch, 1 
 foot 2 inches ; what will be the expense of tooling the soles, 
 heads, and jambs of 8 such windows, at 7d. per square 
 foot? Atis. £6 7s. 2id. 
 
 7. The mantel-tree of a fireplace measures 6 feet 3 
 inches in length, 9 inches in breadth, and 6 inches in 
 thickness ; the height of each jamb is 4 feet 9 inches, the 
 breadth 9 inches, and the thickness 6 inches ; the length 
 of the slab is 6 feet 3 inches, and its breadth 2 feet 6 
 inches ; the two coves are each 4 feet 9 inches high, and 
 9 inches broad ; the chimnej'-piece measures 6 feet 9 inches 
 by 6 inches ; what did the marble cost, the jambs and 
 mantel being at 4/. 10^. per cubic foot, and the coves, 
 slab, and chimney-piece at 125. 6d. per square foot ? 
 
 Ans. ^42 18^. Ud. 
 
 8. Farnley Chapel, in the parish of Leeds, measures 66 
 feet 8 inches in length, and 32 feet 7 inches in breadth, on 
 the outside ; its perpendicular height from the foundation 
 to the ridge is 32 feet, and from the foundation to the 
 eaves, 22 feet 6 inches ; and the thickness of the wall 
 is 2 feet ; how many rods of materials are contained 
 
PART V. A DESCRIPTION, ETC. 181 
 
 in the building ; taking its true compass, and deducting 
 for 7 arched Avindows, whose heights, from the sole to the 
 crown or middle of the arch are 1 1 feet 81 inches, from 
 the sole to the spring of the arch, or where it begins to 
 turn, 9 feet 4 inches, and breadths 4 feet 9 inches ; and 
 for an arched door, whose height, from the threshold to 
 the crown of the arch, is 9 feet 4^ inches, from the thresh- - 
 old to the spring of the arch, 7 feet, and breadth 4 feet 
 9 inches ? ^^s. 66 roods, 23 feet. 
 
 9. A cloth-mill at Armley, in the parish of Leeds, mea- 
 sures 192 feet in length, and 32 in breadth, on the out- 
 side ; its perpendicular height, from the foundation to the 
 eaves, is 35 feet, the height of the gable-end 8 feet, and 
 the thickness of the wall 2 feet : how many roods does the 
 building contain; taking the true compass, and making 
 deductions for 2 doors, one of which measures 9 feet 6 
 inches by 6 feet, and the other 9 feet 6 inches by 5 feet ; 
 for 69 windows, whose dimensions are 7 feet 6 inches by 
 4 feet, and 78 windows, each of which is 7 feet in height,, 
 and 4 feet in breadth ? Ans. 179 roods, 20 k feet 
 
 A DESCRIPTION 
 
 OF 
 
 PORTLAND, YORKSHIRE, AND PURBECK 
 STONE ; 
 
 AND ALSO OP VARIOUS KINDS OF SIARBLE. 
 
 The learned and ingenious Mr, Robert Boyle observes, that a com- 
 petent knowledge of the stones used in building, is of the greatest 
 importance ; one stone dug out of a quarrj' being found to moulder 
 away in a few winters, while another will brave the weather for many 
 ages'. The same author adds, that although some stones, will decay in 
 afcw years, others will not have attained their full hardness in half a 
 century. 
 
 Free-stone is very much used in building. It is of a sandy nature, 
 and mav be cut freely in any direction ; hence it derives its name. 
 It is generally of reddish, yellowish, whitish, or greyish colour ; and 
 is sometimes mixed with small particles of mica, or vestiges of shells. 
 
 There are many excellent quarries of free-stone in England. Those 
 in the peninsula of Portland, in Dorsetshire, and in the vicinity of 
 Leeds, in Yorkshire, are the most celebrated. 
 
182 CARPENTERS AND PART V. 
 
 Portland-stone is of a soft nature, when dug out of the quarry ; 
 but in process of time becomes harder. Some of the finest structures 
 in London are built with this stone, as the piers and arches of "West- 
 minster Bridge, the magnificent cathedral of St. Paul's, &c. &c. 
 
 Yorkshire-stone is harder when dug out of the quarry, than Port- 
 land ; and it becomes still more hard by being exposed to the weather. 
 Large quantities of this stone are sent to different parts of England. 
 It is noted not only for its use in building, but also for its durability 
 Avhen laid under water ; hence it is much used for bridges, docks, &c. 
 
 KirkstaU Ahhey, situated about three miles from Leeds, is built of 
 this stone ; and most of the stones of which the outer walls are com- 
 posed, are not only perfectly sound, but extremely hard. A few are 
 in a state of decay ; which prove the justness of Mr. Boyle's observ- 
 ations before quoted. 
 
 It is 667 years since this Abbey was built. Part of it was un- 
 roofed, at the dissolution of the house, in the reign of Henry VIII. ; 
 and the whole is now completely in ruins. 
 
 PuRBECK-STONE is dug out of quarrics in the isle of Purbeck, in 
 Dorsetshire ; and is of a much harder nature than either Portland 
 or Yorkshire-stone. 
 
 There are numerous sorts of this stone, the finest of which take a 
 good polish ; and are used for chimney-pieces, hearths, grave-stones, 
 &c. The coarser kinds make excellent paving-stones. 
 
 Marble is a fine valuable stone, extremely hard and compact ; and 
 capable of receiving a very beautiful polish. It is chiefly used in 
 works of ornament, as columns, statues, altars, tombs, chimney-pieces, 
 &c. &c. 
 
 Of this stone there are many varieties, as white, black, purple, &c. 
 
 English white marble is generally veined with red. Derbyshire 
 marble is variously clouded and diversified with brown, red, and yel- 
 low. That of Devonshire is either black with white veins, or red 
 variegated with grey and orange. Marble of Auveryne, in France, is 
 of a pale red, mingled with violet, green, and yellow. 
 
 Various other kinds are denominated by the places from which they 
 are brought, as Liege and Namur, in the Netherlands ; Languedoc, 
 in France ; Savoy, Sicily, Spain, &c. &c. 
 
 CARPENTERS' AND JOINERS' WORK. 
 
 To tliis branch belongs all the wood-work of a house ; 
 viz. flooring, partitioning, roofing, wainscotting, &c. &c. 
 
 Flooring,' partitioning, and roofing are generally com- 
 puted by the square of 100 feet. In some places, how- 
 ever, naked flooring and roofing are measured by the 
 cubic foot for materials, and by the square of 100 feet, for 
 workmanship. 
 
 Wainscotting, doors, window-shutters, &c. are com- 
 monly measured by the square foot ; enriched mouldings 
 
PART V. joiners' WORK. 183 
 
 and several other articles, by the lineal foot ; and some 
 things are done at so much per piece. 
 
 Note 1. In measuring naked floorinii, take the length of the room added to the 
 bearin<' ofihe joists, or what they are let into the wall, at each end, for one dimen- 
 sion ; and the breadth of the room added to the bearing of i\vQ girders, for the other 
 dimension. , , j- • 
 
 If there be no girders, the length of the room must be taken for one dimension ; 
 and its breadth added to the bearing of the joists for the other. 
 
 Thcsirders and joists of floors intended to bear great weights, ought to bo let 
 into the wall, at each end, § of the wall's thickness ; but in common floormg, the 
 girders have seldom more than 9 or 10 inches bearing, and the joists about 5 or 6 
 
 For boarded flooring, take the length and breadth of the room ; and deductions 
 must always be made for hearths and well-holes. 
 
 In naked flooring, no deductions are made for hearths, in consequence of the 
 additional trouble, and waste of materials; but well-holes must always be de- 
 
 If the materials of the naked flooring be charged by the cubic foot, multiply the 
 solidity of one joist by the number of joists of the same dimensions ; and when the 
 eirders are all of the same size, proceed with them in a similar manner ; if not, 
 find the solidity of each separately. 
 
 2. Partitions are measured from wall to wall for one dimension, and from floor to 
 floor, or as far as they extend, for the other dimension ; and deductions must be 
 madefor doors and windows, except the agreement includes them. 
 
 Strong partitions, made with framed timber, are generally measured in the same 
 manner as flooring. o,««/. . ^i 
 
 Weather-boarding is sometimes computed by the square of 100 feet, the same as 
 partitioning ; sometimes by the square foot, the same as wainscottmg ; and some- 
 times it is measured by the lineal foot. ,j J .„<..- .1 • 1 
 
 3. In roofing, the length of the house in the inside, added to I of the thickness 
 of each gable, is to be taken for the length ; because the purlins ought to be let 
 into the wall, at each end, f of its thickness ; and the breadth or girt will be equal 
 to twico the distance measured from the ridge, down the principal rajter, over the 
 end of the i?(:-Z»crtTO, to the wall. , ,,r , ,. 
 
 If the roof be covered with tiles or slates before the Joiners Work be measured 
 off, which is very often the case, you must endeavour to ascertain tlie true girt as 
 nearly as possible. In general twice the distance between the top of the ridge and 
 the extremity of the eaves, will be rather too much ; this, however, depends very 
 aiuch upon the depth of th(- tie-beam. , , ■ 
 
 In measuring roofing, v hether for workmanship or m.atenals, no deductions 
 ought to be made for the holes of chimney-shafts, sky-lights, or luthern-lights, on 
 account of their additional trouble, and the waste of materials. 
 
 When angles, running from the ridge to the eaves, are formed m a roof, that 
 angle which bends outward is called a hip, and that which bends inward a valley; 
 and in roofing, it is customary to allow the Joiner some additional measurement, 
 in consideration of the waste of timber, and extra trouble. 
 
 In most places the length of the hip or valley is multiplied by 2 feet, and the 
 product added to the content of the roof. 
 
 The sides of the hipped roof of a rectangular building are trapezoids, and the 
 ends triangles ; hence, it is evident that if the contents of these figures be found 
 separately, their sum will be the content of the whole roof. 
 
 Or, if the length of one side and the breadth of one end, taken half way between 
 the ridge and the eaves, be multiplied by the girt of the roof, the product will be the 
 content. ,. „ . ., c 
 
 When the solidity of the tie-beams, rafters, king-posts, purlins. Sec. in the root 
 of a building, are required ; they must be found in the same manner as directed for 
 girders and joists, in naked flooring. r u j 
 
 4. The pitches of roofs are frequently made according to the fancy ol the de- 
 signer. Formerly they were made much higher than Architects, m general, make 
 them at present ; some regard, however, ought always to be paid to the covering 
 they are to receive. ^ , , .,■,. -.u- ..i 
 
 When the length of the rafters is f of the breadth of the building, within the 
 walls, the roof is said to be of a true pitch. In this case, the angle formed at the 
 ridge, by the rafters, is 83° 37' ; and the perpendicular height of the gable 5 of the 
 breadth'ofthe building, in the inside. 
 
 Some Architects use this pitch when the covering is pantiles ; and others take no 
 more than half the breadth of the building. 
 
 When the covering is slate, some persons take i of the breadth of the building 
 for the height of the gable ; and others consider this pitch too low, and take i of 
 the breadth. . , 
 
 When the height of the gable is equal to i the breadth of the building, the angle 
 
184 carpenters' and part v 
 
 V the ri(l"e is 90°; when it is equal to i, the angle is 112° 37' ; and when equal to 
 i, th" angle is 1260 52'. 
 
 5. IVainscotting is measured by taking the compass of the room, as it is upon 
 the floor, for the length, and the height from the floor to the ceiling, or as far as 
 the wainscot extends, for the breadth ; and in doing this, you must gird over the 
 swelling panels, &c. ; making the tape ply close into all the mouldings. 
 
 Cornices are sometimes done by tlie lineal foot, and sometimes by the square 
 foot ; they, however, must always be measured separately from the wainscotting. 
 
 Chimneys, window-seats, cheek-boards, soffits, casings, architraves, &c. must be 
 measured by themselves ; and deductions ought always to be made for doors, win- 
 dows, fire-places, and other openings. 
 
 6. Inmeasuring a door, it is usual to take its length, measured along the straight 
 edge, and once its thickness for the length ; and its breadth, pressing the tape into 
 all the mouldings of the panels, and once its thickness for the breadth. 
 
 If a door be panelled on both sides, it is customary, in some places, to allow 
 double measure for workmanship ; but if one side only be panelled, the area and 
 half the area are taken for workmanship. It is much preferable, however, to take 
 the real measurement, and regulate the price according to the thickness of the door, 
 the number of the panels, &c. 
 
 This is the method followed in London, and ought to be adopted in every other 
 place. (See Crosby's Builder's Price Book, by Mr. John Phillips, Surveyor, 
 page 178.) 
 
 For the architrave of a door, measure round the outermost edge for the length ; 
 and take the girt of its front and two edges, for the breadth. 
 
 Architraves are sometimes executed by the lineal foot. 
 
 7. Windows are measured by taking the distance between the under side of the 
 sill and the upper side of the top-rail, for the height ; and the distance between out- 
 side and outside of the jambs, for the breadth. 
 
 Sometimes windows are made at so much per piece. 
 
 WiJidow-shulters and the architraves of windows are measured in the same 
 manner as doors, and their architraves. 
 
 8. In measuring stair-cases, make the tape ply close over the steps, from the top 
 to the bottom, for the length ; and take the length of a step for the breadth. Or, 
 if the steps be all of the same size, multiply the area of the riser and tread by the 
 number of steps. 
 
 The ends and landings must be taken by themselves. 
 
 The strings are measured either by the lineal or cubic foot ; the string boards by 
 the square foot ; and the brackets are generally charged at so much per piece. 
 
 For the balustrade, take the whole length of the hand-rail for the length ; and the 
 height of the baluster and hand-rail, upon the landing, for the breadth. 
 
 Formerly there was much more carved work in balusters than there is at present, 
 and it was 'then customary to allow double measure ; but now the true measure is 
 taken, and the price is proportioned according to the workmanship. 
 
 Hand-rails are sometimes executed by the lineal foot, and the balusters charged 
 at so much a piece ; and in some places both hand-rails and balusters are computed 
 bv lineal measure. In this case they must be taken separately. 
 
 '9. Centring for vaults \& TLie&%\iveiX by tiiking its length for one dimension, and 
 girting over the arch for the other ; but in groin-centring, it is customary, in some 
 places, to allow double measure, on account of the additional trouble. It is a more 
 proper method, however, to proportion the price to the workmanship. (See 
 Crosby's Builder's Price Book, page 1G6.) 
 
 10. The following articles are generally done by lineal measure; viz. boxing 
 to windows, skirting boai'ds, beads, fillets, stops, cuppings, astragals, and water 
 trunks. 
 
 Examples in Flooring. 
 
 1. The length of a floor is 45 feet 7 inches, and the 
 bearing of the joists is 6 inches at each end ; its breadth 
 is 22 feet o inches, and the bearing of the girders is 9 
 inches at each end ; required the number of squares of 
 naked and boarded flooringr. 
 
PART V. 
 
 JOINERS WORK. 
 . Bij Cross Multiplicatio7i. 
 
 1 
 
 
 NAKED 
 
 FLOORING. 
 
 
 feet. 
 
 22 
 
 1 
 
 inches. 
 
 5 
 6 
 
 feet. 
 
 45 
 
 1 
 
 46 
 23 
 
 inches. 
 
 7 
 
 
 23 
 
 11 breadth. 
 
 7 length. 
 11 breadth. 
 
 
 
 
 
 151 
 92 
 
 42 
 
 5 
 
 
 8 5^' 
 
 
 1,00)11,14 
 
 1 5 
 
 
 
 Ans. 11 squares and 14/ 
 
 
 BOARDEI 
 
 ) FLOORING. 
 
 
 
 feet. 
 
 45 
 22 
 
 inches. 
 
 7 length. 
 5 breadth. 
 
 
 
 102 
 90 
 18 
 
 10 
 
 11 W 
 
 
 
 1,00)10,21 
 
 9 11 
 
 
 185 
 
 yiw5. 10 squares and 2\ fecL 
 
 2. In a common naked floor, there are 1 1 girders, whose 
 lengths are 20 feet 9 inches, and scantlings (viz. breadths 
 and depths), 1 foot by 7 inches ; 144 joists, each 8 feet 
 6 inches in length, and their scantlings 5^ inches by 
 4 inches ; required the solidity of the whole. 
 
 Alls. ^20 ft. 1 in. 9 pa. 
 
 3. In a naked floor, the girder is 1 foot 2 inchcis deep, 
 1 foot broad, and 25 feet long ; there are 10. bridging 
 joists, whose scanthngs are 5 inches by 3 inches, and 
 lengths 24 feet ; 10 binding joists, whose scantlings are 
 10 inches by 51 inches, and lengths 9 feet 6 inches. The 
 ceihng joists are 24 in number, each 5 feet 9 inches in 
 length, and their scantlings 3 inches by 2-^- inches ; what 
 is the value of the whole, at 4^. 9d. per cubic foot ? 
 
 Ans. £23. 3^. 9]f/. 
 
 4. A house of two stories, besides the ground floor, 
 measures 45 feet 9 inches in length, and 21 feet 6 inches 
 in breadth. There are 6 hearths whose dimensions are, 
 
186 CARPENTEKS' AlvD PAKT Y. 
 
 two of 6 feet by 4 feet 6 inches ; two of 6 feet hj 4 feet 
 3 inches ; and two of 5 feet 9 inches bj 4 feet. The well- 
 hole measures 12 feet 8 inches bj 4 feet 3 inches ; re- 
 quired the expense of the boarded flooring, at 61. 5s. per 
 square. Ans. ^168. 5s. 
 
 Examples in Partitioning. 
 
 1. The length of a partition between two rooms is 25 
 feet 6 inches, and its height 8 feet 10 inches ; how many- 
 squares does it contain ? Ans. 2 squares, 25 feet. 
 
 2. A partition measures 36 feet 4 inches in length, and 
 12 feet 3 inches in height ; what did it cost, at 61. 1 5s 
 per square ? Ans. £30 Os. 9d. 
 
 Examples in Roofing. 
 
 1. How many squares are in the roof of a building 
 whose length is 86 feet 10 inches, and girt 28 feet 6 
 inches ? A^is. 24 squares, 14:^ feet. 
 
 2. The length of a house, within the walls, is 32 feet 
 3 inches, and the purlins have 9 inches bearing at each 
 end ; the girt of the roof is 23 feet 4 inches ; how many 
 squares does it contain ? An^. 7 squares, SI ^ feet. 
 
 3. The length of a hipped roof, at the eaves, is 46 feet 
 
 8 inches, and the breadth of the end 25 feet ; the length 
 of the ridge is 21 feet 8 inches, and its distance from the 
 eaves, 1 8 feet 9 inches ; each of the 4 hips measures 22 
 feet 6 inches in length ; how many squares are contained 
 in the roof, allowing 2 feet in breadth, for the hips ? 
 
 Ans. 19 squares, oO feet. 
 
 4. Let the following figure represent a truss for the 
 roof of a building ; the tie-beam AB, is 28 feet long, 7 
 inches broad, and 12 inches deep ; the principal rafters 
 AC, BC, measure 16 feet 9 inches in length, 6 inches in 
 breadth, and 10 inches in depth ; the king-post CD, is 
 
 9 feet 3 inches in height, and its scantling at the bottom, 
 1 foot 4 inches by 6 inches ; the braces DE, DF, are 5 
 feet 3 inches long, and their scantlings 6 inches by 4 
 inches ; the punchins GH, KL, are 3 feet 3 inches in 
 height, and their scantlings 6 inches by 4 inches ; how 
 many cubic feet are contained in the whole, deducting 
 the two pieces cut out of the king -post, whose lengths are 
 6 feet, and breadths ol inches? Ans. Zl ft. 6 in. 6'\ 
 
PAIIT V 
 
 joiners' work. 
 
 187 
 
 Note. If the pieces, sawn out of the king-post, in order to give the ends of the 
 braces what Joiners call a " square butment," be 2i inches or more, in breadth, 
 and exceed 2 feet in length, thej' must be deducted from the solidity of the king- 
 post ; but if they be of smaller dimensions than these, no deduction must be made, 
 as the pieces cut out are considered to be of little or no value. 
 
 In measuring these pieces, the shortest lengths ought always to be taken ; because 
 the ends, in general, must be cut square, to render the pieces fit for use. 
 
 Examples in Wainscotting, S^c. 
 
 1. A room of wainscot, being girt downwards over the 
 mouldings, is 12 feet 10 inches in height, and 98 feet 6 
 inches in compass, as upon the floor ; what did it cost, at 
 \s. Zd. per square foot ? Ans. £79 0^. \\d. 
 
 2. A door measures 7 feet 2|- inches by 3 feet 10 inches ; 
 there are two architraves, one on each side of the door, 
 whose lengths are 19 feet 7-J inches, and breadths 9^ 
 inches ; the linijig-boards or casings round the door-way 
 measure 17 feet 7 inches in length, and 11 inches in 
 breadth. Now there are five doors with architraves and 
 linings of the above dimensions, in the Court-House, in 
 Leeds ; required the content of these doors, in square 
 feet, and also the contents of their architraves and linings. 
 
 A?is. The contents of the doors are \ZS ft. 1 in. W ; the 
 architraves, 156 ft. 4 in. 4'^ &'' ; and the linings 
 SO feet, 7 iriches, l'\ 
 
 3. The pulpit of a church meas.ures 12 feet 6 inches by 
 4 feet 10 inches; the minister's reading-desk 12 feet 4 
 inches by 4 feet 3 inches ; and the clerk's reading-desk 
 10 feet 8 inches by 3 feet 6 inches. The divisions between 
 the pews are 72 in number, each of which is 10 feet 6 
 inches in length, and 3 feet 8 inches in height ; the doors 
 and ends, taken together, measure 332 feet 3 inches in 
 length, and 3 feet 8 inches in height ; Avhat did the whole 
 cost, at \s. 6d. per square foot ? A?is. ^^310 10s. 7^d. 
 
 4. A room of wainscot is 135 feet 6 inches in compass, 
 and 14 feet 10 inches in height ; the cornice girts 10 
 inches, and its length is equal to the compass of the room ; 
 
188 slaters' a^d tilers' work. part v. 
 
 the door is mahogany, and measures 7 feet 4 inches by 
 3 feet 9 inches ; the length of the surrounding architrave 
 is 20 feet 3 inches, and its girts 9 inches; the casings 
 round the door- way measure 17 feet 11 inches in length, 
 and 1 foot 2 inches in breadth ; there are three pair of 
 window-shutters, each of which measures 7 feet 3 inches 
 by 4 feet 9 inches ; the soffit and cheek-boards of each 
 window, are 21 feet 9 inches in length, and 15 inches in 
 breadth ; the fire-place measures 6 feet 3 inches by 4 feet 
 9 inches ; what did the whole cost, the door being charged 
 ISs. 6d. the cornice 2s. 3d. the architrave Is. 9d. the 
 window-shutters Is. 6d. and all the other articles 1*. 2c?. 
 per square foot? A7is. £161 Is. 6ld. 
 
 SLATERS' AND TILERS' WORK. 
 
 Slating and Tiling are generally computed either by 
 the square yard, or by the square of 100 feet ; and the 
 length of the ridge is taken for one dimension, and the 
 girt of the roof, from eaves to eaves, for the other dimen- 
 sion. 
 
 Note 1. In measuring slating, the tape is made to ply over the eaves, and returned 
 Kp the under side till it meets the wall or eaves-board, in order to make an allow- 
 ance for the double row of slates at the bottom ; but in tiling, the distance betweep 
 tSie extremities of the eaves is taken for the girt of the roof. 
 
 2. It is customary, in some places, instead of returning the tape to the wall or 
 eaves-board, to add 3 feet to the girt of the roof, viz. 18 inches for each side ; and 
 when hips or valleys occur, either in slating or tiling, their length is generally 
 added to the content of the roof; that is, an allowance of one foot in breadth, the 
 whole length of the hips and valleys, is made for the waste of materials. 
 
 3. Sky-lights and chimney-shafts are generally deducted, if they be large ; but no 
 deductions ought to be made for common luthern-lights, garret windows on the 
 roof, and small chimney-shafts. 
 
 EXA5IPLES. 
 
 1. How many square yards of slating are there in a 
 roof whose length is 65 feet 10 inches, and breadth or girt 
 29 feet 6 inches ? 
 
 Bf/ Cross Multiplication. 
 feet, inches. 
 
 65 10 
 
 29 6 
 
 609 2 
 
 130 
 
 32 11 
 
 9)1942 1 
 215 7 1 
 
 Ans. 215 yards^ and 7 feet. 
 
PART V. PLASTEREKS' WORK. 189 
 
 2. What Avas th(? expense of covering the roof of a barn 
 v/ith pantiles, at 21. 'Ss. 6d. per square ; the length being 
 37 feet 8 inches, and the girt 25 feet 6 inches ? 
 
 Ans. £20 175. 9|c?. 
 
 3. The roof a school measures 56 feet 9 inches in lengtli, 
 and 32 feet 4 inches in girt ; what did it cost covering 
 with Tavistock slates, at 21. I5s. per square? 
 
 Ans. £50 9s. 2\d. 
 
 4. A roof measures 28 feet 4 inches in length, and 24 
 feet 6 inches in girt, exclusive of the allowance at the 
 eaves, which is 18 inches on each side; what did it cost 
 covering with Yorkshire slates, at 25. 9c?. per square yard ; 
 deducting for two sky-lights, one of which measures 10 
 feet 8 inches by 6 feet 9 inches, and the other 6 feet 10 
 inches by 5 feet 6 inches ? Ans. £10 4^. "Jd. 
 
 PLASTERERS' WORK. 
 
 Plasterers' work is principally of two kinds ; namely, 
 plastering upon laths, called ceiling ; and plastering upon 
 walls or partitions made of framed timber, called render- 
 ing. 
 
 These different kinds must be measured separately, ex- 
 cept they be done at the same price, which is not often 
 the case. 
 
 The contents of ceiling and rendering are estimated 
 either by the square yard, or by the square of 100 feet ; 
 and if cornices do not exceed 9 inches in girt, they are 
 rated at so much per foot, lineal measure ; but all above 
 tliis girt are computed by the square foot. 
 
 Enriched mouldings are measured by the lineal foot. 
 
 'Sole 1. When Plasterers find materials, deductions must always be made for 
 fire-places, doors, and windows ; and the returns at the tops and sides of doors 
 and windows must be measured separately ; but for workmanship only, these de- 
 ductions are generally omitted, the plastered returns being allowed to counter- 
 balance them. 
 
 2. Deductions are never made for cornices, enriched mouldiugs, festoons, or other 
 ornaments ; because the spaces occupied by them are always plastered previously 
 to such ornaments being made ; and the length of a coriiice, running round the toi> 
 of a room, is always taken equal to the compass of the room, as u])on the floor. 
 
 It is likewise customary to allow so much a piece extra, for every corner above 
 four, in the cornice. 
 
 3. In measuring plastered timber partitions, in large warehouses, &c. where the 
 quai-tcrs and braces project from the plastering, i part of the whole area is gene • 
 rally deducted ; because the projecting timbers are not plastered. 
 
 This work is commonly called " rendering between quarters." 
 
 4. Whitewashing and colouring are measured in the same manner as plastering; 
 and in timber partitions, \ of the whole area is commonly added, for the sides of 
 the quarters and braces. 
 
190 TALN'TEKS WORK. PART V. 
 
 EXAMPLES. 
 
 1. A ceiling measures 43 feet 10 inches in length, and 
 25 feet 6 inches in breadth ; how many square yards does 
 it contain ? 
 
 By Cross Multiplication. 
 feet, inches. 
 
 43 10 
 25 6 
 
 235 
 86 
 21 
 
 10 
 
 
 
 11 
 
 
 9)1117 
 
 9 
 
 
 124 
 
 1 
 
 9 
 
 Alls. 124 yards, and l^foot. 
 
 2. The compass of a room is 138 feet 8 inches, and its 
 height 10 feet 3 inches ; what will be the expense of 
 plastering the wall with stucco, at 9d. per square yard, 
 workmanship only? A7is. £5 ISs. 5^d. 
 
 3. A partition which measures 85 feet 10 inches in 
 length, and 15 feet 6 inches in height, is rendered be- 
 tween quarters on both sides, and whitewashed ; what did 
 the whole cost, workmanship and materials ; the lathing 
 and plastering being charged Is. 6d. per yard, and the 
 whitewashing 2^d. per yard? Ans. £21 11 5. 9d. 
 
 4. A room measures 32 feet 10 inches in length, 21 feet 
 6 inches in breadth, and 1 2 feet 3 inches in height ; and 
 the girt of the cornice is 11 inches. The ceiling cost 
 Is. 9d. per yard, the rendering lOd. per yard, and the 
 cornice Is. id. per foot ; what was the expense of the 
 whole? A?is. £19 135. 3^^ 
 
 PAINTERS' WORK. 
 
 Painters generally compute the contents of all large 
 articles, such as wainscotting, doors, window-shutters, &c. 
 by the square yard ; and every part is measured ujDon 
 which the colour is laid. 
 
 Cornices and enriched mouldings are estimated by the 
 lineal foot ; window-frames at so much a piece ; and window - 
 squares at so much per dozen, according to their size. 
 
PART V. 
 
 painters' work. 191 
 
 Deductions must always be made for fire-places and 
 other openings. 
 
 Note\. In measuring trainscottincr, doors, window-shutters, ^c. painters always 
 gird over the swelling panels, in taking both the length and breadth ; and press the 
 tape into all the mouldings. 
 
 2. Balustrades are generally measured by taking the length of the hand-rail, for 
 one dimension, and twice the height of the"baluster, upon the landing added to the 
 girt of the hand-rail, for the other dimension. 
 
 3. For lattice-work, double the area of one side is generally taken for the mea- 
 surement of both sides ; and the area and lialf the area of one side, for palisading ; 
 but no general rules can be given for these works, as they vary so much in the dis- 
 tance of their perpendiculars and horizontal parts. 
 
 4. Painters proportion their prices to the nature of the colouring, the number of 
 coats the work receives, &c. &c. 
 
 EXAaiPLES. 
 
 1. If a room be painted, wliose lieiglit is 15 feet 6 
 inches, and compass 98 feet 9 inches ; how many yards 
 does it contain ? 
 
 By Decimals. By Cross Multiplication. 
 
 feet. 
 
 98.75 
 15.5 
 
 49375 
 49375 
 
 9875 
 
 9) 1530.625 
 " 170.069 Ans. 
 
 feet. 
 
 inches. 
 
 98 
 
 9 
 
 .15 
 
 6 
 
 501 
 
 3 
 
 98 
 
 
 
 49 
 
 6 6'' 
 
 ))1530 7 6 
 
 170 
 
 7 6 
 
 2. A door measures 7 feet 4^ inches by 3 feet 8 J inches ; 
 what did 4 such doors cost painting on both sides, with 
 3 coats, at 10c?. per yard? Ans. £\ Os. A\d. 
 
 3. The length of the hand-rail of a staircase is 21 feet 
 9 inches, and the girt of the hand-rail, and twice the height 
 of the baluster, upon the landing, 7 feet 10 inches ; how 
 many yards of painting are contained in the balustrade ? 
 
 Ans. 18 yards, S feet, 4^ in. 
 
 4. The lattice window of a dairy measures 4 feel 10 
 inches by 3 feet 6 inches; how many square yards of 
 painting does it contain, taking the area of both sides ? 
 
 Ans. 3 yards, Q feet, 10 in. 
 
 5. The compass of a wainscotted room is 118 feet 6 
 inches, and its height, to the under side of the cornice, 12 
 feet 8 inches ; the door measures 7 feet 2 inches by 3 feet 
 6 inches; the Hning round the door-way is 17 feet 4 
 inches long, and 9 inches broad ; there are four pair of 
 window-shutters, each of which measures 6 feet 10 inches 
 
192 GLAZIERS WORK. PART Y. 
 
 by 4 feet 6 inches ; the soffit and cheek -boards of each 
 window, are 19 feet 6 inches in length, and 14 inches in 
 breadth ; the fire-j)lace, which is to be deducted, measures 
 6 feet 6 inches, by 5 feet 2 inches ; the door and window- 
 shutters are painted on both sides ; required the expense 
 of giving the whole 4 coats ; the cornice, which is 117 feet 
 in length, being charged Sd. per lineal foot, and aU the 
 other articles Is. lOd. per square yard. Ans. £21 8s. 3d. 
 
 GLAZIERS' WORK. 
 
 Glaziers take their dimensions either in feet, inches, 
 and parts, or in feet, tenths, and hundredths ; and estimate 
 their work by the square foot. 
 
 Note \. The most general method of measuring a window is by taking the length 
 and breadth, without making any deduction for the cross-bars between the panes ; 
 sometimes, however, the measurement of a window is found by multiplying the 
 area of one pane by the number of panes. 
 
 2. If windows be circular or elliptical, they must be measured as if they were 
 squares or rectangles ; the greatest lengths and breadths always being taken, in 
 order to make a compensation for the waste of glass in cutting the panes into proper 
 shapes. 
 
 EXA3IPLES. 
 
 1. How many feet of glass are contained in a window 
 which measures 7 feet 10| inches by 4 feet 7^ inches? 
 
 JBt/ Cross Multiplication. 
 
 ft. 
 
 in. 
 
 pa. 
 
 
 7 
 
 10 
 
 9 
 
 
 4 
 
 7 
 
 6 
 
 
 31 
 
 7 
 
 
 
 
 4 
 
 7 
 
 3 
 
 3/// 
 
 
 
 3 
 
 11 
 
 4 6'"' 
 
 36 
 
 6 
 
 2 
 
 7 6A 
 
 2. A pane of plate glass measures 2 feet 8 inches by 1 
 foot 6 inches; what did 8 such panes cost, at 14^. 6d. 
 per square foot ? Ans. £23 As. 
 
 3. The diameter of a circular window is 2 feet 9^ inches ; 
 for how many feet must the glazier be paid, taking the 
 window as a square? Ans. 1 ft. 9 in. 6^ pa. 
 
 4. The base of a triangular sky-light measures 10 feet 
 6 inches, and the perpendicular 5 feet 8 inches ; what did 
 it cost glazing, at 1*. 10c?. per square foot ? 
 
 Ans. £2 145. G^e/. 
 
PART Y. 
 
 PLUBIBERS WORK. 
 
 193 
 
 5. There is a house with three tiers of windows, four in 
 a tier ; the height of the first tier is 7 feet 2 inches, of the 
 second 6 feet 4 inches, and of the third 5 feet 8 inches ; 
 and the breadth of each window is 4 feet 6 inches. The 
 hei^-ht of a semi-circular window, above the door, is 2 
 feet ; what did the whole cost glazing, at 2^. 6d. per square 
 foot Ans. £44 2^. 6d. 
 
 PLUMBERS' WORK. 
 
 pLmiBERs' work is generally done at so much per pound, 
 or else by the hundred weight of 112 pounds ; and the 
 price is regulated according to the value of the lead at the 
 time when the work is performed. 
 
 Note. Sheet lead, used in roofing, guttering, &c. commonly weighs from 7 to 12 
 pounds per square foot ; and leaden pipe varies in weight, per yard, according to 
 the diameter of its bore in inches. 
 
 The first of the following Tables shows the weight of a square foot of sheet lead, 
 in pounds, to ea«ch of the subjoined thicknesses, in tenths and hundredths of an 
 inch ; and the second exhibits the general weight of a yard of leaden pipe, accord- 
 ing to the diameter of its bore. 
 
 TABLE L 
 
 TABLE IL 
 
 Thickness 
 
 of 
 Sheet Lead. 
 
 Pounds 
 
 to a 
 
 Square Foot. 
 
 Thickness 
 
 Qf 
 
 Sheet Lead. 
 
 Pounds 
 
 to a 
 
 Square Foot. 
 
 tV 
 
 5.899 
 
 .15 
 
 8.848 
 
 .11 
 
 6.489 
 
 .16 
 
 9.438 
 
 1 
 
 6.554 
 
 } 
 
 9.831 
 
 .12 
 
 7.078 
 
 .17 
 
 10.028 
 
 ^ 
 
 7.373 
 
 .18 
 
 10.618 
 
 .13 
 
 7.668 
 
 .19 
 
 11.207 
 
 .14 
 
 8.258 
 
 X 
 
 11.797 
 
 i 
 
 8.427 
 
 .21 
 
 1 
 
 12.387 
 
 o 
 
 1 
 
 Bore of 
 eaden Pip 
 in Inches. 
 
 Pounds 
 
 per 
 
 Yard. 
 
 v-1 
 
 
 f 
 
 10 
 
 1 
 
 12 
 
 li- 
 
 16 
 
 u 
 
 18 
 
 H 
 
 21 
 
 2 
 
 24 
 
 EXAMPLES. 
 
 1. A sheet of lead measures 18 feet 10 inches in length, 
 and 5 feet 6 inches in breadth; required its weight, at 
 9-J- lb. to a square foot. 
 
194 TAVERS' WORK. PART V 
 
 By Cross Multiplication. 
 Ft. In. 
 
 18 
 
 10 
 
 5 
 
 6 
 
 94 
 
 2 
 
 9 
 
 5 
 
 103 
 
 7 
 
 7 content. 
 
 As I ft. : 9 /3. 8 or. : : 103/f. 7 i?i. : S cwt. 3 qu. 4 /i. 
 
 02;. 10 dr., the iveight required. 
 
 •2. What is the weight of a sheet of lead, whose length is 
 15 feet 10 inches, breadth 4 feet 6 inches, and thickness 4- 
 or .2 of an inch ? A?is. 840.53625 pounds. 
 
 3. If I buy 150 yards of leaden pipe, whose bore is If 
 inch ; what will it cost me at S^d. per pound, admitting 
 each yard to weigh 21 pounds? Ans. £49 4s. 4^d. 
 
 4. What cost the covering and guttering of the roof of 
 a church with lead, at £1 18^. per cwt. ; the length of the 
 roof being 82 feet 9 inches, and its girt 65 feet 3 inches ; 
 the length of the guttering 165 feet 6 inches, and its breadth 
 
 1 foot 9 inches ; admitting the thickness of the lead to be I 
 of an inch? Ans. £948 15^. 11^^. 
 
 PAVERS' WORK. 
 
 Pavers estimate their work by the square yard ; and such 
 dimensions must always be taken as will give the true 
 area. 
 
 examples. 
 
 1. A rectangle measures 85 feet 9 inches in length, and 
 43 feet 6 inches in breadth ; how many square yards of 
 paving does it contain ? 
 
 By Decimals. By Cross Multiplication. 
 
 Feet Feet. In. 
 
 85.75 length. 85 9 length. 
 
 43.5 breadth. 43 6 breadth. 
 
 42875 287 3 
 
 25725 340 
 
 34300 42 10 6'' 
 
 9 )3730.125 9 )3730 1 6 
 
 414.4583 yds., the Ans. 414yds. 4 ft. Ans. 
 
PAET V. A'AULTED .VND AKCIIED ROOFS. 195 
 
 2. The base of a triangle measures 76 feet 8 inches, and 
 the perpendicular 42 feet 3 inches ; what did it cost paving 
 with Aberdeen granite, at 10^. 6d. per square yard? 
 
 A?is. £94 9s. 6d. 
 
 3. If the parallel sides of a trapezoid be 68 feet 7 inches, 
 and 45 feet 3 inches, and their perpendicular distance 98 
 feet 6 inches ; what will it cost paving with Guernsey 
 pebbles, at' 65. 6d. per square yard ? A?is. £202 8s. II ^d, 
 
 4. How many yards of paving does the trapezium con- 
 tain, whose diagonal measures 136 feet 8 inches, and per- 
 pendicular 68 feet 2 inches, and 06 feet 4 inches ? 
 
 Ans. 945 7/ds. 2^ ft 
 
 5. The length of a street is 538 feet 6 inches, and its 
 breadth 65 feet 8 inches; what did it cost paving with 
 Purbeck-stone, at 5s. 6d. per square yard ? 
 
 A?is. £1080 95. 9^d. 
 
 6. A rectangular court-yard measures 96 feet 9 inches 
 in length, and 74 feet 6 inches in breadth. Across the 
 middle and round the extremities of the yard, is a foot- 
 way, 5 feet 3 inches broad; and paved with Guernsey 
 granite, at 9^. 6d. per square yard. The rest is paved 
 with Jersey pebbles, at 5s. 9d. per square yard ; required 
 the expense of the whole. Ans. £272 8s. Zld. 
 
 VAULTED AND ARCHED ROOFS. 
 
 Arched Roofs are either vaults, domes, saloons, or groins. 
 
 Vaidted roofs are formed by arches springing from the 
 opposite walls, and meeting in a line at the top. 
 
 Domes are made by arches springing from a circular 
 or polygonal base, and meeting in a point at the top. 
 
 Saloons are formed by arches connecting the side walls 
 to a flat roof, or ceiling, in the middle. 
 
 Groins are formed by the intersection of vaults with 
 each other. 
 
 Vaulted roofs are commonly of the three following 
 sorts : 
 
 1. Circular roofs, are those whose arch is some part of 
 the circumference of a circle. 
 
 2. Elliptical or oval roofs, or those whose arch is an 
 oval, or some part of the circumference of an ellipsis. 
 
 3. Gothic roofs, or those which are formed by two cir- 
 
 K 2 
 
196 VAULTED AND AECHED KOOFS. PART Y. 
 
 cular arcs, struck from different centres, and meeting in a 
 point over the middle of the breadth or span of the arch. 
 
 Kote. Domes and saloons are of various figures ; they, however, seldom occur in 
 the pr.ictice of measuring; but most cellars are covered either with vaults ov 
 groins. 
 
 PROBLEM I. 
 
 To find the content of the vacuity of a circulai^, an 
 elliptic, or a Gothic vaulted roof 
 
 RULE. 
 
 Multiply the area of one end by the length of the roof 
 or vault, and the product will be the content required. 
 
 'Note 1. If the arch be the segment of a circle, the area of the end may be found 
 by Problem 17, Part II. ; if it be elliptical, multiply the span by the height, and the 
 product by .7854, for the area of the end ; but if it be a Gothic arch, the area of the 
 end must be obtained by finding the areas of the two circular segments and the 
 triangle of whicli the end is composed. 
 
 2. The upper sides of all arches, whether vaults or groins, are built up solid, 
 above the haunches, to the same height as the crown of the arch. 
 
 3. The solidity of the materials in any arched roof, may be found thus : find the 
 content of the whole, considered as solid, from the spring of the arch to the upper 
 side of the crown ; find also the contents of the vacuity; tlien the difference of these 
 two contents will be the solidity required. 
 
 4. The whole arch, considered as a solid, will be a parallelopipedon, the content 
 of which may be found by Problem 2, Part I V. 
 
 EXAMPLES. 
 
 1. Required the content of the vacuity of a semi-circular 
 vault, the span or diameter of which is 20 feet, and its 
 length 60 feet. 
 
 .7854 
 
 400= the square of 20. 
 2 )314.1600 
 
 157.08 the area of the end. 
 60 the length. 
 
 9424.80 Ans. 
 
 2. The span of an elliptical vault is 30 feet, its height 
 10 feet, and its length 50 feet ; what is the content of the 
 vacuity? Ans. 11781 feet, 
 
 3. Required the content of the vacuity of a Gothic vault, 
 whose span is 30 feet, the chord of each arch 32 feet, the 
 versed sine, or distance of each arch from the middle of 
 these chords, 8 feet, and the length of the vault 35 feet 6 
 inches. Ans. 27755.0502 feet. 
 
PAET V. VAULTED AKD ARCHED HOOFS. 197 
 
 4. Let ABCD denote the end or ^ g c 
 
 upright section of a semi-circular [f^----^— ;::^i ,___ 
 
 roof. The span EF is 18 feet, the i W'^^^^m 
 
 thickness of the wall AE or FB, at i /,,- ^-; 
 
 the spring of the arch, 3 feet, the ^i"]' 
 
 thickness GH, at the crown of the [J 
 
 arch, 2 feet, and the length of the 
 
 vault 56 feet 9 inches ; how many solid feet are contained 
 
 in the roof? Ans. 77 61 A25 feet. 
 
 PROBLEM IL 
 
 To find the concave or convex surface of a circular^ an 
 elliptic^ or a Gothic vaulted roof 
 
 RULE. 
 
 Multiply the length of the arch by the length of the 
 vault, and the product will be the superficies required. 
 
 I\otc. The convex length of an arch may be easily found by making a line ply 
 close over it ; but for the concave length, this method is not quite so applicable ; 
 for if care be not taken, the dimension will be made too short. 
 
 If the arch be the segment of a circle, its true length may be found by Problem 
 14, Part H. 
 
 EXASIPLES. 
 
 L The span of a semi-circular vault is 30 feet, and its 
 length 40 feet ; what is its concave surface ? 
 
 3.1416 
 30 
 
 2 )94.24 80 
 47X24"= length of the arch. 
 40 
 Ans, 1884.960 square feet. 
 
 2. The length of a vault is 62 feet 9 inches, and that of • 
 the arch 54 feet 6 inches ; how many square yards are 
 contained in the roof? Ans. 379.986 yards. 
 
 3. Required the concave surface of a bridge consisting 
 of 5 circular arches ; the span of each arch being 96 feet, 
 the height, above the top of the piers, 36 feet, and the 
 length 45 feet. Ans. 28955,232 /eeif. 
 
 "Note. Those who desire to make themselves acquainted with tlie essential pro- 
 perties, dimensions, proportions, and other relations of the various parts of a 
 bridge, are referred to Dr. Hutton's Principles of Bridges. 
 
 In this valuable little work, the learned Doctor proves, that the equilibrial arch, 
 described in Problem V., is the most proper for a bridge of several arches. Next 
 to it, the elliptical arch claims the preference ; after it the cycloidal arch ; and 
 lastly the arch of a circk 
 
 K 3 
 
198 VAULTED AND ARCHED EOOFS. PART V. 
 
 As for parabolic, hyperbolic, and catanarian arches, they ought never to be ad- 
 mitted into a bridge consisting of several arches ; but may, in some cases, be used 
 for a bridge of one arch, which is to rise an unusual height. 
 
 PROBLEM in. 
 
 To find the solid content of a dome ; its height, and the 
 dimensions of its base being given. 
 
 RULE. 
 
 Multiply the area of the base by the height, and | of 
 the product will be the solidity. 
 
 EXAIvn'LES. 
 
 1. What is the solid content of a hemispherical dome; 
 the diameter of the base being 40 feet ? 
 
 .7854 
 
 1600 = square of 40. 
 4712400 
 7854 
 
 1256.6400 = area of the base. 
 
 20 = height. 
 
 25132.8000 
 2 
 
 3) 50265.6000 
 
 16755.2000 = solidity. 
 
 2. Required the solidity of an octagonal dome ; each 
 side of the base being 20 feet, and the height 21 feet. 
 
 Alls. 27039.19176/ec^. 
 
 PROBLEM IV. 
 To find the superficial content of a spherical dome. 
 
 RULE. 
 
 Multiply the area of the base by 2, and the product will 
 be the superficies required. 
 
 l^ote. If the dome be elliptical, the product of the two diameters multiplied by 
 1.5708, will give the superficial content, sufficiently near for practical purposes. 
 
 EXAiVIFLES. 
 
 1. Required the superficies of a hexagonal spherical 
 dome ; each side of the base being 20 feet. 
 
PART V. VAULTED AND ARCHED ROOFS. 199 
 
 2.5980762 = tabular area, Prob. 12, Part 11. 
 400 = square of 20. 
 
 1039.2304800 = area of the base. 
 2 
 
 2078.4609600 = superficies required. 
 
 2. What will an octagonal spherical dome cost painting, 
 at Is. 3d. per yard ; each side of the base being 10 feet ? 
 
 Ans. £6 Us. V^d. 
 
 PROBLEM V. 
 To find the solid content or vacuity of a saloon. 
 
 RULE. 
 
 Multiply the area of a transverse section by the mean 
 compass of the solid part of the saloon ; subtract this pro- 
 duct from the whole vacuity of the room, supposing tlie 
 walls to go upright, all the height, to the flat ceiling, and 
 the difference will be the answer 
 
 Note 1. If the base of the saloon be a rectangle, the vacuity of the room, or the 
 whole upright space, will be a parallelopipedon, the content of which may be found 
 by Problem 2 : if the base be a regular polygon, the vacuity will be a prism, the 
 content of which may be found by Problem 3 ; and if the base be a circle, the 
 vacuity will be a cylinder, the content of which may be obtained by Problem 4, 
 Part IV. 
 
 2. For practical purposes, the mean compass of the solid part may be found by 
 adding the compass taken at the middle of the arch to the compass of the room, 
 taken within the walls, and dividing the sum by 2 ; but in solving the second and 
 third examples, the mean compass of the solid part was found with mathematical 
 accuracy. 
 
 EXAIUPLES. 
 
 1. If the height AB of a saloon be 3.2 feet, the chord 
 ADC of its face 4.5 feet, and the distance DE of its mid- 
 dle part from the arch be 9 inches ; required the solidity, 
 supposing the mean compass of the saloon to be 50 feet. 
 
 By Rule 2, Problem 17, Part II., 
 we have (4.5 X .75) xf + .753_-(4.5 x 
 2) = 3.375 X f X .421875 -v- 9 = 2.25 -h 
 .046875 =2.296875, the area of the seg- 
 ment ADCEA. 
 
 Again, by Problem 6, Part II., we 
 have AC- -AB-=:.4.5- -3.2- =20.25 
 — 10.24 = 10.01 ; and ^/ 10.01 = 
 3.163858 = EC; then EC x ^ AB = 
 3.163858 x 1.6=5.0621728, the area of the triangle ABC\ 
 K 4 
 
 c 
 
 
 B 
 
 "\~~---^ 
 
 
 
 \ 
 
 
 ^ 
 
 -C 
 
200 VAULTED AND ARCHED KOOFS. PART V. 
 
 consequently 5.0621728-2.296875=2.7652978, the area 
 of the transverse section, AECBA. 
 
 Noio, 2.7652978 X 50^138.26489 /ee#, the content of 
 the solid part, which being taken from the whole upright 
 space, will leave the content of the vacuity within the 
 room. 
 
 2. What is the solid content of a saloon with a circular 
 quaclrantal arch of 2 feet radius, springing over a rectan- 
 gular room of 20 feet long, and 16 feet broad ? 
 
 Ans. 580.2065 cubic feet. 
 
 3. A circular building of 40 feet diameter, and 25 feet 
 high to the ceiling, is covered with a saloon, the circular 
 quadrantal arch of which is 5 feet radius ; required the 
 capacity of the room in cubic feet ? 
 
 Ans. 30766.496 cubic feet, 
 
 PROBLEM yi. 
 To find the superficial content of a saloon. 
 
 RULE. 
 
 Find its breadth by applying a line close to it across 
 the surface ; and its length by measuring along the middle 
 of it, quite round the room ; then the product of these two 
 dimensions will be the surface required. 
 
 Is^ote. The area of the flat ceiling must be added to the area found by the above 
 Rule, in order to obtain the ichole surface of the saloon. 
 
 EXAIIPLES. 
 
 1. The girt across the face of a saloon is 5 feet 3 inches, 
 and its mean compass 94 feet 6 inches ; what is the area 
 of its surface ? 
 
 Here 94.5 X 5.25 = A9Q.126 feet, the area required. 
 
 2. The mean compass of a saloon is 126 feet 10 inches, 
 and the girt across its face 8 feet 6 inches ; what is the 
 area of its surface ? Ans. lOlSft. 1 in. 
 
 PROBLEM Vn. 
 
 To find the solid content of the vacuity formed by a groin 
 arch, either circular or elliptical. 
 
 RULE. 
 
 Multiply the area of the base by the height, and the 
 product by .904, and it will give the solidity required. 
 
PART V VAULTED AND ARCHED ROOFS. 201 
 
 Note. Groins are sometimes measured as if they were solid, in consideration 
 of the great trouble and waste of materials in forming the arches and mterscctions. 
 
 EXAMPLES. 
 
 1. "Wliat is the content of the vacuity formed by a cir- 
 cular groin, springing from the sides of a square base, 
 each side of which is 14 feet ? 
 
 14 
 14 
 
 _14 
 
 196= area of the base. 
 7= height, or radius. 
 I372 
 .904 
 5488 
 12348 
 
 1240.288 = solidity required. 
 
 2. Wliat is the solid content of the vacuity formed by 
 an elliptical groin ; the side of its square base being 24 
 feet 6 inches, and its height 8 feet 3 inches ? 
 
 Ans. 4476.6645 /ee^. 
 
 PROBLEM Vin. 
 
 To find the concave surface of a circular or an elliptical 
 groin. 
 
 RULE. 
 
 Multiply the area of the base by 1.1416, and the pro- 
 duct will be the superficies required. 
 
 Note. In measuring works where there are many groins in a range, the cylin- 
 drical pieces between the groins, and on their sides, must be taken separately. 
 
 EXABIPLES. 
 
 1. What is the concave surface of a circular groin 
 arch ; the side of its square base being 15 feet 6 inches ? 
 
 Here 15.5 X 15.5 = 240.25, area of the base; and 
 240.25 X 1.1416 = 274.2694 /ee?, the answer required. 
 
 2. The base of a groin is a rectangle whose sides are 20 
 and 26 feet ; required the concave surface of the arch. 
 
 Ans. 593.632 /ee^. 
 K 5 
 
202 
 
 GENERAL ILLUSTRATION. 
 
 PART V 
 
 GENERAL ILLUSTRATION. 
 
 Haying gone through the Works of Artificers, and 
 noted the methods of measuring buildings, and computing 
 their contents, I now proceed to give a general illustra- 
 tion of the whole, by assigning the dimensions of a house, 
 and from thence computing the contents of the works of 
 the different Ai'tificers employed in building it. 
 
 In performing this task, are shown the methods of 
 ruling the book, entering the dimensions, with the con- 
 tents ; then the method of abstracting the contents ; and 
 lastly, of forming the bills of expenses of the work and 
 materials. 
 
 The building of which I have made choice, for the 
 general illustration, consists of two stories, beside the 
 cellars, and of two rooms upon each floor ; wliich will be 
 found quite sufficient to exemplify the methods of mea- 
 suring the works of Artificers. 
 
 The whole length of the building, on the outside, is 53 
 feet 6 inches, and its breadth 24 feet. The walls of the 
 cellars are 6 half bricks, or 24 inches in thickness ; those 
 of the lower story 20 inches ; and those of the upper 
 story 16 inches. 
 
 One of the lower rooms is considered to be the kitchen, 
 and the other the parlour ; and they are frequently dis- 
 tinguished by these denominations, in the following notes. 
 
 The plans of the different stories, and the elevation, 
 could not be given without a folding plate ; but it is pre- 
 sumed that the reader will find no difficulty in compre- 
 hending the dimensions without them. 
 
 Note. The columns of numbers, in the following forms, are sufficiently explained 
 by the titles at the tops of them ; excepting the figures 2, 3, tVc. in the first column ; 
 which figures signify that there are more than one article of the same dimensions ; 
 consequently the contents arising from the dimensions to which these figures are 
 prefixed, must be multiplied by 2, 3, 8ic. ; and the products entered in the column 
 of contents. 
 
 THE BRICKLAYER'S WORK. 
 
 
 Dimensions. 
 
 Half 
 Bricks 
 thick. 
 
 Contents. 
 
 Titles. 
 
 
 Ft. In. 
 
 147 
 10 6 
 
 6 
 6 
 
 Ft. In. 
 1543 6 
 
 420 
 
 The cellar walls. 
 
 2 
 
 20 
 10 6 
 
 IMiddle walls of ditto. 
 
PAKT V. 
 
 GENERAL ILLUSTRATION. 
 
 203 
 
 
 Dimensions. ] 
 
 Half- 
 Bricks 
 thick. 
 
 Contents. 
 
 Titles. 
 
 2 
 
 Ft. 
 20 
 19 
 
 In. 
 
 
 
 Ft. 
 
 5 760 
 
 In. 
 
 
 
 
 
 1 
 
 
 
 6 
 
 
 
 
 
 9 
 
 
 
 2 
 
 
 
 10 
 
 
 
 
 
 
 
 6 
 
 2 
 
 4 
 
 Common arclies over the two 
 cellars. 
 
 
 20 
 10 
 
 
 
 
 5 
 
 5 
 
 4 
 
 3 
 
 3 
 
 3 
 
 5 
 
 4 
 
 3 
 
 6 
 
 5 
 
 5 
 
 5 
 
 4 
 
 4 
 
 3 
 
 200 
 1911 
 1837 
 162 
 1071 
 141 
 247 
 204 
 90 
 47 
 44 
 35 
 65 
 76 
 34 
 93 
 
 Ditto under the passage. 
 
 
 147 
 13 
 
 
 
 
 Outer walls of the ground 
 stoiy. 
 
 
 147 
 12 
 
 
 6 
 
 Ditto of the upper story. 
 
 
 24 
 6 
 
 
 9 
 
 Gable-ends of the outer walls. 
 
 2 
 
 21 
 25 
 
 
 6 
 
 Partition walls. 
 
 
 21 
 6 
 
 
 9 
 
 Gable-ends of ditto. 
 
 2 
 
 13 
 9 
 
 
 6 
 
 Chimney shafts of the ground 
 story. 
 
 2 
 
 ~12~ 
 
 8 
 
 2 
 
 Ditto of the upper story. 
 
 2 
 
 6 
 6 
 
 9 
 
 8 
 
 Ditto of the gable-ends. 
 
 2 
 
 6 
 3 
 
 10 
 6 
 
 The deductions are as follow : viz. 
 
 Cellar doors. 
 
 
 8 
 5 
 
 
 6 
 
 Front door. 
 
 
 7 
 5 
 
 
 
 
 Back door. 
 
 2 
 
 6 
 5 
 
 6 
 
 
 Windows of the lower rooms. 
 
 3 
 
 ^5- 
 
 4 
 
 8 
 6 
 
 Ditto of the upper rooms. 
 
 
 6 
 1 ^ 
 
 10 
 
 
 Staircase window. 
 
 •1 
 
 6 
 3 
 
 8 
 6 
 
 Doors in partition walls. 
 
 
 
 
 
 
 
 K 6 
 
204 
 
 GENERAL ILLUSTKATION. 
 
 PART V. 
 
 In order to abstract tlie foregoing contents ; that is, to 
 collect them into one sum, make the deductions, and re- 
 duce the neat contents to the standard thickness of one 
 brick and a half, proceed thus : Make only two columns 
 for the whole contents, and two for the deductions of the 
 same thickness ; viz. one column for the contents that are 
 one brick in thickness, and the other for the contents that 
 are one brick and a half in thickness ; and dispose of the 
 superior denominations in one or both of these columns, 
 by entering them more than once ; thus, the contents that 
 are 2 bricks in thickness must be set down twice in the 
 one-brick column ; those that are 21 bricks in thickness, 
 once in the one-brick column, and once in the brick-and- 
 half column ; and those contents whose tliickness is 3 
 bricks, must be entered twice in the brick-and-half 
 column; then add up all the columns, and reduce the 
 sums in the one brick columns to the standard thickness 
 of one brick and a half, which add to the respective sums 
 in the brick-and-half columns. Lastly, take one of the 
 sums thus obtained, from the other, and the remainder 
 will be the whole reduced content of the brick -work. 
 
 ' Abstract of the Brick- Work. 
 
 Contents. 
 
 Contents. 
 
 Deducti 
 
 ons. 
 
 Deductions. 
 
 11 brick thick. 
 
 1 brick thick. 
 
 li brick thick. 
 
 1 brick thick. 
 
 Ft. In. 
 
 Ft. In. 
 
 Ft. 
 
 In. 
 
 Ft. In. 
 
 1543 6 
 
 760 
 
 47 
 
 10 
 
 44 
 
 1543 6 
 
 200 
 
 47 
 
 10 
 
 35 
 
 420 
 
 1911 
 
 44 
 
 
 
 65 
 
 420 
 
 1837 6 
 
 35 
 
 
 
 76 6 
 
 760 
 
 1837 6 
 
 66 
 
 
 
 76 6 
 
 200 
 1911 
 
 162 
 1071 
 
 141 9 
 
 247 
 90 
 
 247 
 204 2 
 204 2 
 
 93 
 
 4 
 
 34 2 
 
 34 2 
 
 333 
 243 
 
 
 6 
 
 365 4 
 2 
 
 7201 4 
 2 
 
 576 
 
 6 
 
 
 
 3)730 8 
 
 3)14402 8 
 
 243 6 
 
 4800 10 
 
 
 8509 9 
 
 
 
 
 
 
 
 4800 10 
 
 
 
 
 
 13310 7 
 
 
 576 6 
 
 
 
 
 
 12734 1 
 
 
PART V 
 
 GENERAL ILLUSTRATION. 
 
 205 
 
 Then 12734 feet being clividecl by 272, we obtain 46 
 rods 222 feet of brick-and-lialf waD, for workmanship and 
 materials. 
 
 Next find the contents of the corners or coins of the 
 different stories of the honse, to which add the deductions 
 for doors and windows ; and you will obtain the quantity 
 to be charged for workmanship only. 
 
 Cellar walls. 
 
 Feet. In. 
 
 2 thickness. 
 4 multiply. 
 
 8 breadth. 
 10 6 height. 
 80 
 
 4 
 
 84 
 
 content. 
 
 6 half bricks thick. 
 
 Walls of the ground story. 
 Feet. In. 
 
 1 8 thickness. 
 
 4 multiply. 
 6 8 breadth. 
 
 13 height. 
 78 ' 
 8 8 
 86 8 content. 
 
 5 half bricks thick. 
 
 3)504 
 
 3)433 4 
 
 168 reduced content. 144 5 reduced content. 
 
 Walls of the upper story. 
 
 Feet. In. 
 1 4 thickness. 
 4 multiply. 
 5 4 breadth. 
 126 height. 
 
 64 
 
 2 
 
 66 8 content. 
 
 Feet. In. 
 QQ 8 content. 
 4 multiply. 
 
 3)266 
 
 88 10 reduced content. 
 168 ditto. 
 144 5 ditto. 
 401 3 total of the coins. 
 576 6 deductions. 
 272)977 9(3 rods 161 feet for 
 816 workmanship only. 
 
 161 rem. 
 
 THE BRICKLAYER S BILL. 
 Rods. Feet. £ 
 
 46 222 of brick-and-half wall, workmanship 
 
 and materials, at per rod . — 
 
 3 161 of ditto, for workmanship only, at 
 
 per rod — 
 
206 
 
 GENERAL ILLUSTRATION. 
 
 PART V. 
 
 THE IVIASON'S WORK 
 
 
 Dimensions. 
 
 Contents. Titles. 
 
 
 Ft. 
 
 155 
 
 2 
 
 In. 
 
 3 
 
 Ft. 
 
 348 
 
 40 
 
 29 
 
 11 
 
 6 
 
 10 
 
 15 
 
 36 
 
 37 
 
 26 
 
 32 
 
 18 
 
 13 
 
 35 
 
 22 
 
 13 
 
 In. 
 9 
 
 1 
 
 
 
 
 
 10 
 
 
 
 
 
 8 
 
 4 
 
 3 
 
 
 
 4 
 
 4 
 
 
 
 6 
 
 6 
 
 Stone base or plinth. 
 
 
 53 
 
 
 6 
 9 
 
 Facia to the front of the house 
 
 
 14 
 2 
 
 6 
 
 
 Jambs of the front door. 
 
 
 5 
 
 2 
 
 6 
 
 
 Head of ditto. 
 
 
 5 
 
 1 
 
 6 
 3 
 
 Frize. 
 
 
 6 
 
 1 
 
 8 
 6 
 
 Cornice. 
 
 
 6 
 2 
 
 8 
 3 
 
 Threshold. 
 
 4 
 
 5 
 
 1 
 
 6 
 
 8 
 
 Steps to the front door. 
 
 2 
 
 8 
 2 
 
 
 
 4 
 
 Ends of ditto. 
 
 
 7 
 3 
 
 "6" 
 6 
 
 Landing of ditto. 
 
 2 
 
 10 
 
 1 
 
 8 
 6 
 
 Front window jambs. 
 
 2 
 
 5 
 1 
 
 6 
 
 8 
 
 Soles of ditto. 
 
 2 
 
 5 
 1 
 
 
 
 4 
 
 Heads of ditto. 
 
 3 
 
 9 
 
 1 
 
 4 
 3 
 
 Upper window jambs 
 
 3 
 
 5 
 
 1 
 
 
 6 
 
 Soles of ditto. 
 
 3 
 
 4 
 
 1 
 
 6 
 
 
 Heads of ditto. 
 
PART Y. 
 
 GENERAL ILLUSTRATION. 
 
 207 
 
 
 Dimensions. 
 
 Contents. 
 
 Titles. 
 
 
 
 Ft. 
 
 18 
 
 1 
 
 In. 
 
 
 8 
 
 Ft. 
 30 
 
 8 
 14 
 
 7 
 
 5 
 28 
 16 
 25 
 22 
 18 
 11 
 21 
 12 
 
 8 
 18 
 18 
 12 
 
 In. 
 
 
 3 
 
 7 
 4 
 10 
 4 
 6 
 8 
 8 
 4 
 8 
 4 
 
 
 
 8 
 
 
 Back door jambs and head. 
 Threshold of ditto. 
 Staircase window jambs. 
 Sole of ditto. 
 
 Head of ditto. 
 
 Jambs to the fire-places in 
 lower rooms. 
 
 Mantels to ditto. 
 
 Slabs or hearths to ditto. 
 
 Coves to ditto. 
 
 Back cover to ditto. 
 
 Chimney-pieces or cornices 
 ditto. 
 
 Jambs to the fire-places in 
 upper rooms. 
 
 Mantels to ditto. 
 
 Chimney-pieces or cornices 
 ditto. 
 
 Slabs to ditto. 
 Coves to ditto. 
 Back coves to ditto. 
 
 
 
 5 
 
 1 
 
 6 
 6 
 
 
 
 11 
 
 1 
 
 8 
 3 
 
 
 
 5 
 1 
 
 6 
 
 4 
 
 
 
 5 
 1 
 
 
 2 
 
 
 2 
 
 8 
 
 1 
 
 6 
 
 8 
 
 the 
 
 2 
 
 5 
 
 1 
 
 6 
 6 
 
 
 2 
 
 5 
 2 
 
 6 
 4 
 
 
 4 
 
 4 
 
 1 
 
 3 
 
 4 
 
 
 4 
 
 2 
 
 1 
 
 9 
 
 8 
 
 
 2 
 
 5 
 
 1 
 
 
 2 
 
 to 
 
 2 
 
 8 
 
 1 
 
 
 
 4 
 
 the 
 
 2 
 
 4 
 
 1 
 
 6 
 
 4 
 
 
 2 
 
 4 
 
 1 
 
 
 
 
 to 
 
 2 
 
 4 
 2 
 
 6 
 
 
 
 4 
 
 4 
 1 
 
 
 2 
 
 
 4 
 
 2 
 
 1 
 
 3 
 
 
208 
 
 GENEK^U. ILLUSTRATION. 
 
 P.VRT V. 
 
 
 Dimensions. 
 
 Contents. 
 
 Titles. 
 
 2 
 
 Ft. 
 14 
 3 
 
 In. 
 
 6 
 
 4 
 
 Ft. 
 96 
 
 47 
 
 106 
 
 83 
 
 82 
 
 78 
 
 8 
 
 720 
 
 190 
 
 36 
 
 385 
 
 196 
 
 23 
 
 In. 
 
 8 
 6 
 4 
 3 
 
 9 
 9 
 
 
 9 
 9 
 4 
 10 
 
 Chimney -tops of stone. 
 
 2 
 
 15 
 
 1 
 
 10 
 6 
 
 Bases and facias to ditto. 
 
 
 58 
 1 
 
 
 10 
 
 Cornice to the front of the house. 
 
 
 55 
 
 1 
 
 6 
 6 
 
 Blocking course to ditto. 
 
 4 
 
 10 
 2 
 
 3 
 
 
 Tabling. 
 
 15 
 
 3 
 
 1 
 
 6 
 6 
 
 Steps down to the cellars. 
 
 
 3 
 2 
 
 6 
 6 
 
 Landing of ditto. 
 
 2 
 
 20 
 
 18 
 
 
 
 
 Tooled flagging in the cellars. 
 
 
 20 
 9 
 
 
 6 
 
 Ditto between the cellars. 
 
 
 10 
 3 
 
 6 
 6 
 
 Well-hole of the cellar-stairs to be 
 deducted. 
 
 
 20 
 
 18 
 
 8 
 8 
 
 Polished flagging in the kitchen. 
 
 
 20 
 9 
 
 8 
 6 
 
 Ditto in the passage. 
 
 
 5 
 
 4 
 
 6 
 
 4 
 
 Kitchen fire-place to be deducted. 
 
 ABSTRACT OF THE JIASONRT. 
 
 In making this abstract, aU the articles of the same 
 price must be collected together : thus, the polished work 
 must be brought into one sum ; the tooled work into 
 another ; the cornices into another, &c. &c. 
 
PART • 
 
 V^. GENERAL ILLUSTRATION. 
 
 
 Polished Work, 
 
 Feet. 
 
 In. 
 
 348 
 
 9 stone base or plinth. 
 
 40 
 
 1 facia to the front. 
 
 29 
 
 jambs to the front door. 
 
 11 
 
 head of ditto. 
 
 6 
 
 10 frize. 
 
 15 
 
 threshold. 
 
 36 
 
 8 steps to the front door. 
 
 37 
 
 4 ends of ditto. 
 
 26 
 
 3 landing of ditto. 
 
 32 
 
 front window jambs. 
 
 18 
 
 4 soles of ditto. 
 
 13 
 
 4 heads of ditto. 
 
 35 
 
 upper window jambs. 
 
 22 
 
 6 soles of ditto. 
 
 13 
 
 6 heads of ditto. 
 
 28 
 
 4 jambs to the fire-places in the lower rooms. 
 
 16 
 
 6 mantels to ditto. 
 
 21 
 
 4 jambs to the fire-places in the upper rooms. 
 
 12 
 
 mantels to ditto. 
 
 96 
 
 8 chimney tops. 
 
 47 
 
 6 bases and facias to ditto. 
 
 83 
 
 3 blocking course to the front. 
 
 82 
 
 tabling. 
 
 1073 
 
 2 sum. 
 
 
 Tooled Work. 
 
 Feet. 
 
 In. 
 
 30 
 
 back door jambs and head. 
 
 8 
 
 3 threshold of ditto. 
 
 14 
 
 7 staircase window jambs. 
 
 7 
 
 4 sole of ditto. 
 
 5 
 
 10 head of ditto. 
 
 78 
 
 9 steps down to the cellars. 
 
 8 
 
 9 landing of ditto 
 
 153 
 
 6 sum. 
 
 
 Cornices. 
 
 Feet, 
 
 . In. 
 
 10 
 
 cornice of front door. 
 
 11 
 
 8 chimney-pieces in the low rooms. 
 
 3 
 
 ditto in the upper rooms. 
 
 106 
 
 4 cornice to the front of the house. 
 
 136 
 
 sum. 
 
 209 
 
210 GENERAL ILLUSTRATION. PART V. 
 
 
 Slabs and Coves 
 
 Feet 
 
 In. 
 
 25 
 
 8 slabs of the low room fire-piaca-;. 
 
 22 
 
 8 coves of ditto. 
 
 18 
 
 4 back coves of ditto. 
 
 18 
 
 slabs of the upper room lire-places. 
 
 18 
 
 8 coves of ditto. 
 
 12 
 
 back coves of ditto. 
 
 115 
 
 4 sum. 
 
 Tooled Flagging 
 Feet. In. 
 
 720 flagging in the cellars. 
 190 ditto between the cellars. 
 910 sum. 
 36 9 deduction. 
 
 9 )873 3 
 
 97 square yards 
 
 Polished Flagging. 
 Feet. In. 
 
 385 9 flagging in the kitchen. 
 196 4 ditto in the passage. 
 582 1 sum. . 
 23 10 deduction. 
 
 9 )558 3 
 
 62 square yards. 
 
 THE ]\IASON S BILL. 
 
 Feet. In. ^ «• d. 
 
 1073 2 of polished work . . at per foot, 
 
 153 6 of tooled ditto . . . at— — per foot, 
 
 136 of cornices . . . .at per foot, 
 
 115 4 of slabs and coves . . at per foot, 
 
 97 yards of tooled flagging . at per yard, 
 
 62 ditto of polished ditto . . at per yard, --_ 
 
 £ 
 
PAKT V. 
 
 GENERAL ILLUSTRATIOX. 
 
 211 
 
 THE CARPENTER'S AND JOINER'S WORK 
 
 
 Dimensions, 
 
 Contents. 
 
 Titles. 
 
 
 Ft. In. 
 22 8 
 19 8 
 
 Ft. 
 445 
 
 385 
 
 23 
 
 948 
 
 824 
 
 34 
 
 90 
 
 74 
 
 60 
 
 4 
 
 38 
 
 5 
 
 60 
 
 4 
 
 92 
 
 26 
 
 In. 
 9 
 
 9 
 
 10 
 
 10 
 
 10 
 
 6 
 
 1 
 
 5 
 
 
 
 10 
 
 
 
 6 
 
 
 
 10 
 
 9 
 
 4 
 
 Naked flooring of the parlour ; 
 
 viz. girders, and binding and bridging joists. 
 
 
 20 8 
 18 8 
 
 Boarded flooring of ditto. 
 
 
 5 6 
 
 4 4 
 
 Hearth to be deducted from the 
 boarded flooring. 
 
 2 
 
 23 4 
 20 4 
 
 Naked flooring of the upper rooms. 
 
 2 
 
 21 4 
 19 4 
 
 Boarded flooring of ditto. 
 
 2 
 
 4 6 
 3 10 
 
 Hearths to be deducted from the 
 boarded flooring. 
 
 
 11 6 
 
 7 10 
 
 Naked flooring of the staircase. 
 
 
 9 6 
 
 7 10 
 
 Boarded flooring of ditto. 
 
 10 
 
 4 
 1 6 
 
 Deal steps to staircase, first flight. 
 
 10 
 
 10 
 
 7 
 
 End of ditto. 
 
 
 9 6 
 
 4 
 
 Foot-pace. 
 
 
 9 6 
 
 7 
 
 Face of ditto. 
 
 10 
 
 4 
 1 6 
 
 Steps of the second flight. 
 
 10 
 
 10 
 
 7 
 
 Ends of ditto. 
 
 
 26 6 
 3 6 
 
 Balustrade of the staircase. 
 
 
 39 6 
 
 8 
 
 
 
 String-board of ditto. 
 
212 
 
 GENEEAL ILLUSTEATIOX. 
 
 TJ-JXT y. 
 
 
 Dimensions. 
 
 Contents. 
 
 Tines. 
 
 2 
 
 Ft. In. 
 23 4 
 20 4 
 
 Ft. 
 948 
 
 224 
 
 1441 
 
 35 
 
 14 
 
 20 
 
 30 
 
 11 
 
 21 
 
 20 
 
 12 
 
 12 
 
 20 
 
 12 
 
 25 
 
 41 
 
 25 
 
 In. 
 
 10 
 
 
 10 
 
 
 
 4 
 5 
 4 
 1 
 
 7 
 7 
 6 
 7 
 7 
 1 
 2 
 3 
 
 Ceiling-joists over the upper 
 rooms. 
 
 
 21 4 
 10 6 
 
 Ceiling-joists over the staircase. 
 
 
 52 9 
 
 27 4 
 
 Roofing. 
 
 
 7 6 
 4 8 
 
 Front door, 6 panelled. 
 
 
 21 6 
 8 
 
 Casings to ditto. 
 
 
 24 6 
 10 
 
 Architrave to ditto. 
 
 
 6 9 
 4 6 
 
 Back door, 4 panelled. 
 
 
 19 
 
 7 
 
 Casing to ditto. 
 
 
 6 
 3 6 
 
 Cellar door, 4 panelled. 
 
 
 6 6 
 3 2 
 
 Kitchen door, 6 panelled. 
 
 
 16 10 
 9 
 
 Casings to ditto. 
 
 
 18 10 
 8 
 
 Architrave to ditto. 
 
 
 6 6 
 3 2 
 
 Parlour door, 6 panelled. 
 
 
 16 10 
 9 
 
 Casings to parlour door. 
 
 2 
 
 18 10 
 8 
 
 Ai'chitraves to ditto. 
 
 2 
 
 6 6 
 3 2 
 
 Upper room doors, 6 panelled. 
 
 2 
 
 16 10 
 9 
 
 Casings to ditto. 
 
PAUT Y. 
 
 GENERAL ILLUSTRATION. 
 
 113 
 
 
 Dimensions. 
 
 Conten 
 
 ts. j Titles. 
 
 1 
 
 4 
 
 Ft. 
 
 18 
 
 
 
 In. 
 10 
 
 8 
 
 Ft. 
 
 50 
 50 
 46 
 15 
 23 
 16 
 61 
 42 
 47 
 26 
 57 
 
 in. 
 2 
 
 6 
 
 
 9 
 
 8 
 1 
 
 7 
 2 
 
 8 
 6 
 
 Arcliitraves to ditto. 
 
 2 
 
 
 
 4 
 
 10' 
 
 4 
 
 Window shutters to lower win- 
 dows. 
 
 2 
 
 23 
 1 
 
 
 
 
 Casings to ditto, viz. cheek-boards, 
 window -boardOj and soffits. 
 
 
 27 
 
 
 
 
 7 
 
 Architrave to parlour window. 
 
 
 23 
 
 1 
 
 8 
 
 
 Casings to staircase window. 
 
 
 27 
 
 
 8 
 
 7 
 
 Architrave to ditto. 
 
 3 
 
 20 
 
 1 
 
 4 
 
 
 Casings to upper windows. 
 
 3 
 
 24 
 
 
 4 
 
 7 
 
 Architraves to ditto. 
 
 2 
 
 5 
 
 4 
 
 8 
 2 
 
 Sashes of the lower windows. 
 
 
 6 
 
 4 
 
 2 
 
 4 
 
 Ditto of the staircase window. 
 
 3 
 
 5 
 3 
 
 
 10 
 
 Ditto of the upper windows. 
 
 ABSTRACT OP THE CARPENTER's AND JOINER'S WORK. 
 
 This abstract must be made in the same manner as that 
 of the Masonry ; viz. by collecting all the articles of the 
 same price into one sum, making the proper d(3ductions, 
 &c. &c. 
 
 Naked Flooring. 
 
 Feet. In. 
 
 445 9 parlour floor. 
 948 10 upper floors. 
 90 1 staircase. 
 
 1,00 )14,84 8 sum. 
 
 14. squares, 84 feet, 8 inches. 
 
214 
 
 GENERAL ILLUSTRATION. 
 
 PART V. 
 
 Boarded Flooring. 
 
 Feetc In. 
 
 385 9 parlour floor. 
 624 10 upper floors. 
 74 staircase. 
 
 1285 
 
 58 
 
 sum. 
 
 4 hearths to be deducted. 
 
 J, 00) 1 2,26 8 
 
 12 squares. 26 feet, 8 inches. 
 
 Ceiling Joists. 
 
 Feet. In. 
 
 948 10 upper rooms. 
 
 224 staircase. 
 
 1,00 )11,72 10 sum. 
 
 11 squares. 72 feet, 10 inches. 
 
 Doors, 4: panelled. 
 
 Feet. 
 
 In. 
 
 30 
 
 4 back door. 
 
 21 
 
 cellar door. 
 
 51 
 
 4 sum. 
 
 
 Stairs Steps. Doors, 6 panelled. 
 
 Feet. 
 
 In. Feet. In. 
 
 60 
 
 first flight 35 front door. 
 
 4 
 
 10 ends of ditto. 20 7 kitchen door. 
 
 38 
 
 foot-pace. 20 7 parlour door. 
 
 5 
 
 6 face of ditto. 41 2 upper room door; 
 
 60 
 
 second flight. 117 4 sum. 
 
 4 
 
 
 i.\j enus 01 uiito. 
 
 26 
 
 4 string-board. 
 
 199" 
 
 6 sum. 
 
 
 Door and Wijidoio Casings. 
 
 
 Feet. In. 
 
 
 14 4 front door. 
 
 
 11 1 back door. 
 
 
 12 7 kitchen door. 
 
 
 12 7 parlour door. 
 
 
 25 3 upper room doors. 
 
 
 46 lower room windows. 
 
 
 23 8 staircase window. 
 
 
 61 upper room windows. 
 
 
 206 6 sum. 
 
V. 
 
 GENERAL ILLUSTKATI 
 
 
 Architraves. 
 
 Peet. 
 
 In. 
 
 20 
 
 5 front door. 
 
 12 
 
 6 kitchen dooi- 
 
 25 
 
 1 parlour door. 
 
 50 
 
 2 upper room doors. 
 
 15 
 
 9 parlour window. 
 
 16 
 
 1 staircase window. 
 
 42 
 
 7 upper windows. 
 
 182 
 
 - 
 
 7 sum. 
 
 
 Window Sashes. 
 
 Feet. 
 
 In. 
 
 47 
 
 2 lower windows. 
 
 26 
 
 8 staircase window. 
 
 57 
 
 6 upper windows. 
 
 131 
 
 4 sum. 
 
 115 
 
 THE CARPENTER S AND JOINERS BILL. 
 Sqrs. Ft. In. £ s. d. 
 
 14 84 8 of naked flooring, . at — per square, 
 
 12 23 8 of boarded flooring, at — per square, 
 
 11 72 10 of ceiling joists, . . at — per square, 
 
 14 41 10 of roofing, . . .at — per square, 
 
 199 6 of stair-steps, . .at — per foot, 
 
 92 9 of balustrade, . .at — per foot, 
 
 117 4 of doors 6 panelled, at — per foot, 
 
 51 4 of doors 4 panelled, at — per foot, 
 
 206 6 of doors and window 
 
 casings, . . .at — per foot, 
 
 182 7 of architraves, . . at — per foot, 
 
 131 4 of window sashes, . at — per foot, 
 
 THE SLATER'S WORK. 
 
 The length of the roof, for the slating, is 52 feet 6' 
 inches, and its girt, allowing 18 inches at each of the- 
 eaves, is 29 feet 10 inches; hence the content is 1566|- 
 square feet =15 squares 66^ feet, at per square . . ► 
 
216 
 
 GENERAX ILLUSTRATION. PART V* 
 
 THE PLASTERER'S WORK. 
 
 ■ 1 
 
 Dimensions. 1 
 
 1 
 
 Contents. 
 
 Titles. 
 
 2 
 
 Ft. 
 18 
 20 
 
 In. 
 
 8 
 8 
 
 Ft. 
 
 771 
 
 137 
 82 
 74 
 28 
 
 868 
 23 
 32 
 28 
 
 868 
 23 
 35 
 32 
 28 
 
 633 
 
 558 
 
 In. 
 
 6 
 9 
 8 
 5 
 
 10 
 
 4 
 6 
 
 10 
 
 4 
 
 6 
 
 10 
 6 
 1 
 
 Ceilings of the lower roomc?, 
 3 coats. 
 
 2 
 
 82 
 
 
 8 
 10 
 
 Cornices of ditto. 
 
 
 82 
 
 8 
 
 Enriched mouldings in the parlour, 
 lineal measure. 
 
 
 9 
 
 7 
 
 6 
 10 
 
 Ceiling in the lower part of the 
 staircase, 3 coats. 
 
 
 34 
 
 
 8 
 10 
 
 Cornice of ditto. 
 
 
 82 
 10 
 
 8 
 6 
 
 Walls of the parlour, hard-finish- 
 ing. 
 
 
 6 
 3 
 
 8 
 6 
 
 Door of ditto to be deducted. 
 
 
 6 
 5 
 
 6 
 
 
 Window to be deducted. 
 
 
 5 
 5 
 
 6 
 3 
 
 Fire-place to be deducted. 
 
 
 82 
 10 
 
 8 
 6 
 
 Walls of the kitchen, 2 coats. 
 
 
 6 
 3 
 
 8 
 6 
 
 Door to be deducted. 
 
 
 7 
 5 
 
 
 
 
 Back door to be deducted. 
 
 
 6 
 5 
 
 6 
 
 
 Window to be deducted. 
 
 
 5 
 5 
 
 6 
 3 
 
 Fire-place to be deducted. 
 
 
 60 
 10 
 
 4 
 6 
 
 Walls of the lower part of the 
 staircase, hard-finishing. 
 
 
 60 
 9 
 
 4 
 3 
 
 Ditto of the upper part. 
 
PAIJT V 
 
 GEXEllxVL ILLUSTRATION. 
 
 217 
 
 
 Dimensions. 
 
 Contents. 
 
 Titles. 
 
 
 Ft. In. 
 8 
 5 6 
 
 Ft. In. 
 
 44 
 
 46 8 
 39 
 34 2 
 25 6 
 46 8 
 573 2 
 
 50 3 
 824 10 
 141 8 
 
 1572 6 
 46 8 
 
 51 
 45 
 
 Front door to be deducted. 
 
 2 
 
 6 8 
 3 6 
 
 Kitchen and parlour doors to be 
 deducted. 
 
 
 13 
 3 
 
 Space occupied by the staircase, 
 to be deducted. 
 
 
 6 10 
 5 
 
 Staircase window to be deducted. 
 
 
 5 8 
 4 6 
 
 Window at tlie top of the stair- 
 case, to be deducted. 
 
 2 
 
 6 8 
 3 6 
 
 Upper room doors to be deducted. 
 
 
 60 4 
 9 6 
 
 Ceiling over the staircase, 3 coats. 
 
 
 60 4 
 10 
 
 Cornice of ditto. 
 
 2 
 
 21 4 
 19 4 
 
 Ceiling of the upper rooms, 3 coats. 
 
 2 
 
 85 
 10 
 
 Cornices of ditto. 
 
 2 
 
 .85 
 9 3 
 
 Walls of the upper rooms, 2 coats. 
 
 2 
 
 6 8 
 3 6 
 
 Doors to be deducted. 
 
 2 
 
 5 8 
 4 6 
 
 Windows to be deducted. 
 
 2 
 
 5 
 
 4 6 
 
 Fire-places to be deducted. 
 
 ABSTRACT OF THE PLASTERING. 
 
 This abstract must be made by collecting all the ceiling 
 of 3 coats, into one sum ; the hard-finishing into another, 
 &c. &c. ; and by making the proper deductions. 
 
 L 
 
218 
 
 GENERAL IL 
 
 LUSTHATION. 
 
 
 PAllT V 
 
 Ceiling. 
 
 Rendering. 
 
 Deductions. 
 
 
 Cornices. 
 
 3 Coats. 
 
 Hard 
 
 Finishing. 
 
 2 Coats. 
 
 Hard 
 Finishing. 
 
 Ft. In. 
 23 4 
 32 6 
 28 10 
 44 
 46 8 
 39 
 34 2 
 25 6 
 46 8 
 
 2 Coats. 
 
 Ft. In. 
 771 6 
 
 74 5 
 573 2 
 824 10 
 
 Ft. In. 
 868 
 633 6 
 558 1 
 
 Ft. 
 
 868 
 
 1572 
 
 In. 
 
 6 
 
 Ft. 
 23 
 35 
 32 
 28 
 46 
 51 
 45 
 
 '4 
 
 
 6 
 10 
 8 
 
 
 
 Ft. In. 
 
 137 9 
 28 10 
 50 3 
 
 141 8 
 
 2440 
 262 
 
 6 
 
 4 
 
 2059 7 
 320 8 
 
 9)2243 11 
 
 9)2178 
 
 2 
 
 358 6 
 
 Yds. Ft. In. 
 
 9)1738 11 
 
 Yds. Ft. 
 242 
 
 In. 
 2 
 
 
 249 2 11 
 
 Yds. Ft. In. 
 193 1 11 
 
 
 
 262 
 
 4 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 320 8 
 
 
 THE PLASTERERS BILL. 
 
 Yds. Ft. In. £ s. d. 
 
 249 2 1 1 of ceiling, 3 coat?, at — per yard, . 
 
 193 1 11 of rendering, liard-iinishing, at — 
 
 per yard, 
 
 242 2 of ditto, 2 coats, at — per yard, . . 
 
 358 6 of cornices, at — per foot, . . . 
 
 82 8 of enriched mouldings, at — per 
 
 foot, 
 
 J? 
 
 Note. The followiiig Works will be found extremely useful to 
 Bricklayers, Masons, Joiners, or any other persons who are desirous 
 of obtaining a knowledge of Architecture. 
 
 Miller's Designs for Cottages, Farm-houses, Country-houses, Villas, 
 Lodges for Park or Garden Entrances, with Plans of the Offices 
 belonging to each Design, on 32 Qur.rto Plates, Price, sewed, lOs. 6d. 
 
 Rawlin's Designs of Houses for Gentlemen and Tradesmen, Par- 
 sonages and Summer Retreats ; with Back-fronts, Sections, &c. ; 
 together with Banquetting Rooms and Churches ; to which is added 
 the Masonry of semicircular and elliptical Arches, on 51 Royal 
 Quarto Plates, Price 1/. Is. 
 
 The Student's Instructor, in drawing and -working the Five Orders 
 of Architecture ; fully explaining the best Methods of striking re- 
 gular and quirked mouldings, for diminishing and glueing of Columns 
 and Capitals, for finding the true diameter of an Order to any given 
 Height, for striking the Ionic Volute circular and elliptical, with 
 finished Examples, on a large Scale, of the Orders, their Planceers, 
 &c. on 33 Octavo Plates, by Fcter NicI}olso?i, Price, bound, lOs. 6d. 
 
PART V. GENERAL ILLUSTRATION. 219 
 
 Paints Euihlcr^s Assistant, demonstrating in the most easy and prac- 
 tical manner, all the principal Rules of Architecture, from the Ground 
 Plan to the ornamental Finish, illustrated with several new and use- 
 ful Designs of Houses, witli their Plans, Elevations, and Sections, on 
 42 Folio Plates, Price, bound, \Gs. 
 
 Pain's Practical House Carpenter, containing a great Variety of use- 
 ful Designs in Carpentry and Architecture ; as Centering for Groins, 
 Niches, &c. Examples for Roofs, Sky-lights, &c. The Five Orders 
 laid down by a New Scale, IMouldings, &c. at large, with their Enrich- 
 ments. Plans, Elevations, and Sections of Houses for Town and 
 Country ; Lodges, Hot-houses, Green-houses, Stables, &c. Design 
 for a Church, with Plan, Elevation, and two Sections; an Altar- 
 piece, and Pulpit. Designs for Chimney-pieces, Shop Fronts, and 
 Door Cases. Section of a Dining-room, and Library. Variety of 
 Staircases, with many other important Articles, and useful Embel- 
 lishments, on 14G Quarto Plates, Price, bound, ISs. This is Pain's 
 last Work. 
 
 L 2 
 
220 jyiENSUKATJON OF HAT-STACKS. PART VI. 
 
 PAET VI. 
 
 THE METHOD OF 
 
 MEASURING HAY-STACKS, 
 
 DEAIXS, CANALS, SIAEL-PITS, PONDS, ^HLL-DAZMS, EMBANK- 
 MENTS, QUARRIES, COAL-HEAPS, AND CLAY-HEAPS. 
 
 HAY-STACKS. 
 
 « 
 
 The contents of hay-stacks are found in order to ascer- 
 tain their weights ; which must, of course, varj according 
 to the density of the hay. 
 
 Some stacks will not weigh more than 8 or 10 stones 
 l^er cubic yard, and others will weigh 15 or 16 stones ; it 
 is not, however, the measurer's province to determine the 
 weight, but only the number of cubic yards which the 
 stack contains ; and leave the buyer and seller to settle 
 about the weight as they think proper. 
 
 PROBLEM I. 
 To measure a hay-stack^ having a circular base. 
 
 CASE I. 
 
 When the stack is straight from the bottom to the eaves, 
 and from the eaves to the top, as in the following figure ; 
 the upper part may be taken as a cone, and the lower 
 part as a conical frustum. 
 
 RULE. 
 
 Multiply the square of the circumference at the bottom 
 AB, by .07958, or for general practice, by .08 ; and the 
 product will be the area of the base. 
 
 Find the area of a section at the eaves DC, in the same 
 manner. To the sum of these areas, add the square root 
 of their product ; multiply this sum by the perpendicular 
 height GF = HD, and \ ot tne product will be the soli- 
 dity of the frustum ABCD. 
 
PAUT VI. 
 
 MENSURATION OF HAY- STACKS. 
 
 221 
 
 Multiply the area of the section at the eaves, by tlie 
 perpendicular height FE = DK, and -} of the product will 
 be the solidity of the cone DCE. 
 
 To the solidity of the frustum add that of the cone ; 
 and the sum will be the content of the whole solid 
 ABCEDA. 
 
 Note 1. Some measurers take the dimensions of hay-stacks, canals, marl-pits, &c. 
 with a tape divided into yards, tenths, and hundredths ; but one divided into feet 
 and tenths, is considered, by most practitioners, to be much preferable. 
 
 2. When the dimensions are taken in feet, the content must be divided by 27, in 
 order to reduce it to cubic yards. 
 
 EXA3IPLES. 
 
 1. The circumference at the base 
 AB, of the following figure, is 40, 
 the circumference at the eaves DC 60, 
 the perpendiculiw height GF=HD 
 15, and FE = DK 16 feet ; how 
 many solid yards does the stack con- 
 tain ? 
 
 CALCULATION. 
 
 Here 40 x 40 x .08=128, the area 
 of the base ; and 60 X 60 X .08 = 288, 
 the area of a section at the eaves ; 
 also, V(128x288)=^/36864=192, the square root of 
 their product; then (128+288-f 192) x 15--3 = 608 x5 
 = 3040 cubic feet, the content of the frustum ABCD. 
 
 ^^«m, (288xl6)-f-3=4608--3=1536 cubic feet, the 
 content of the cone CDE. 
 
 Lastly, (3040-1- 1536)--27=4576--27 = 169?/«r6?5, 13 
 feet, the ansiver required. 
 
 2. The circumference of the base of a hay-stack is 52 
 feet 9 inches, the circumference at the eaves 75 feet 6 
 inches, the perpendicular height of the lower part 16 feet 
 3 inches, and that of the upper part 18 feet 6 inches ; how 
 many tons are contained in the stack ; admitting each 
 cubic yard to weigh 15 stones or 210 pounds ? 
 
 Ans. 28 tons, 10 cwt. 1 qr. 17 lb. 
 
 CASE n. 
 
 When the stack is bulged from the bottom to the fop, as 
 in the folloioing figure ; the equidistant ordinate methods 
 described in Problem 23, Part II., must be adopted. 
 L 3 
 
222 IMENSURATION OF HAY-STACKS. TAUT \1. 
 
 RULE, 
 
 1. Find the areas of as many circular sections, taken at 
 equal perpendicular distances from the bottom, as you 
 judge sufficient, by multiplying the square of the circum- 
 ference of each section by .08. Proceed v>ath these areas 
 in the same manner as if they were equidistant ordinates ; 
 and the result avIU be the solidity of the stack, from the 
 bottom to the uppermost or last section. 
 
 2. Multiply the area of the base of the remaining part, 
 at the top, which may be considered as a cone, by its per- 
 pendicular height ; and l of the product will be the 
 solidity. 
 
 3. Add these two solidities together, and the sum will 
 be the content of the whole stack. 
 
 Note 1. Always make choice of an odd number of sections, in order that the num- 
 ber of parts into which the solid is divided, may be equal. Five or seven will, in 
 general, be sutficient ; of which one must be at the bottom, and another at the 
 eaves, or as near to them as possible. 
 
 2. Great care must be taken to obtain the dimensions of the sections at equal 
 perpendicular distances ; for if the slanting distances be taken, it is evident that 
 the content will be made too much. 
 
 SCHOLIUM. 
 
 The method of finduig the areas of curvilineal figures, by means of 
 equi-distant, perpendicular ordinates, was first demonstrated by the 
 illustrious Sir Isaac Newton. 
 
 Mr. Robert ShirtclifFe, in his Tlieory and Practice of Gauging, ap- 
 pears to have been the first who applied it to finding the areas of 
 curvilineal vessels used by Brewers, Distillers, &c. ; and after him 
 Mr. Samuel Farrer, in the Appendix to Overley's Gauging. Their 
 Rules, however, Avere extremely tedious; and, though true to de- 
 monstration, were not general, but particular, according to the num- 
 ber of ordinates used. 
 
 To obviate this inconvenience, the general rule given in Problem 23, 
 Part II. of this Work, was deduced from Simpson's Dissertations, 
 page 109, by Mr. Thomas Moss; and demonstrated in his valuable 
 Treatise of Gauging, page 235. 
 
 Dr. Hutton, in his Mensuration, Proposition I., Section II., Part 
 IV., has also given an elegant demonstration of the same Rule; and 
 adds, in a Corollary, that it will obtain for the contents of all solids, 
 by using the areas of the sections perpendicular to the axe, instead of 
 the ordinates. 
 
 The Doctor particularly recommerids it for the purpose of gauging 
 and ullaging casks ; hence, it is evident that it may be applied with 
 propriety and success to the Mensuration of Hay-stacks, Canals, 
 Marl-pits, and other irregular figures, as being the best approxima- 
 tion that has yet been, or perhaps ever can be given ; for by taking 
 an indefinite number of sections, the content of an irregular solid 
 may be obtained to any degree of accuracy. 
 
PAIIT VI. SIENSURATIOX OF IIAY-STACKS. 223 
 
 EXAMPLES. 
 
 1. Let the annexed figure re- 
 present a hay-stack, whose di- 
 mensions are as follow ; viz. the 
 cirt at the bottom AB=36, at 
 CD=:o4, at EF=66, at GH= 
 58, and at KL = 37 feet. The 
 perpendicular distance between 
 each section is 5 feet, and the 
 perpendicular height of the co- 
 nical part iniiM 4 feet 6 inches ; 
 how many cubic yards are con- 
 tained in the stack ? 
 
 CALCULATION. 
 
 Here 36 X 36 X .08 = 103.68 the area of the bottom or 
 first section; 54 X 54 X .08=233.28, the area of the second 
 section; QQ X QQ x .08 = 348.48, theareaofthe thirdsection; 
 58 X 58 X. 08 =269. 12, the area of the fourth section ; and 
 37 X 37 X .08=109.52, the area of the fifth or last section; 
 then hy 'proceeding according to the Ride for equi- distant 
 ordinates, lue have A= 103.68 + 109.52=213.2, B = 233.28 
 + 269.12 = 502.4, 0=348.48, f«2^D=5; consequenthj 
 (A+4B4 2C)-4-3xD = (213.24- 2009.6 + 696.96)--3 x5 
 = (2919.76 x5)^3 = 14598.8--3=4866.26 feet, the so- 
 liditjj of the part ABLK. 
 
 Again, (109.52 x4.5)--3=492.84-r-3 = 164.28/eeif, the 
 solidity of the conical part KLM. 
 
 i:«5%, (4866.26+ 164.28)--27=5030.54-f-27=186.31 
 cubic yards, the content required. 
 
 2. The dimensions of a hay-stack are as follow ; viz. 
 the girt of the bottom or first section = 127.2, the girt of 
 the second = 145.4, the girt of the third = 156.5, the girt 
 of the fourth = 168.7, the girt of the fifth = 148.3, "the 
 girt of the sixth = 121.8, and the girt of the seventh or 
 last section =: 68.6 feet. The perpendicular distance be- 
 tween each section is 8 feet, and the perpendicular height 
 of the conical part, at the top, 7.4 feet ; how ma?iy cubic 
 yards are contained in the stack ? 
 
 Ans. 2970.49641 yards. 
 
 L 4 
 
224 MENSURATION OF HAT-STACKS. FART Vl. 
 
 PROBLEM II. 
 
 To measure a liay-stack, having a rectangular base. 
 
 CASE I. 
 When the stack is straight from the bottom to the eaves, 
 and from the eaves to the top, as in the following figure ; 
 the lower part may be taken as a prismoid, and the upper 
 part as a triangular prism. 
 
 RULE. 
 
 1. Multiply the mean lengtliof the bottom by themear 
 breadth, and the product will be the area of the bottom. 
 Find the area of a section at the eaves in the same manner. 
 Multiply half the sum of the lengths of the bottom and 
 eaves, by half the sum of the breadths ; and the product 
 will be the area of a section equally distant from the 
 bottom and eaves. To the area of the bottom add the 
 area of the section at the eaves, and four times the area of 
 the middle section ; multiply this sum by the perpendiculan 
 height, from the bottom to the eaves ; and -i of the pro- 
 duct will be the solidity of the lower part. 
 
 2. Multiply the breadth at the eaves by the perpen- 
 dicular height from the eaves to the top ; and half the 
 product will be the area of the end ; which being multi- 
 plied by the mean length, will give the solidity of the 
 upper part. 
 
 3. Add these two solidities together, and the sum will 
 be the content of the whole stack. 
 
 Note 1 . Sometimes stacks are longer on one side than the other, and broader 
 at one end than the other ; in such cases, take half the sum of the lengths for a 
 mean length, and half the sum of the breadths for a mean breadth. 
 
 2. Some stacks are higher and broader at the ends than in the middle ; when 
 this is the case, a proper allowance must be made in taking the dimensions. 
 Allowance also ought to be made for the thatch. 
 
 3.' If the ends of the upper part be not equal, find the area of both ends; and 
 take half their sum for a mean area, which multiply by the length for the solidity. 
 
 4. If the length of the top or ridge GH, be more or less than the length of the 
 base or eave EF, it is evident that the upper part of the stack is in the form of a 
 cuneus or wedge ; hence its true content may be found by Problem 'J, Section I. 
 Part IV. 
 
 EXAMPLES. 
 
 1. Let the annexed figure represent a hay-stack, the 
 dimensions of which are as follow ; viz. the mean length 
 at the bottom 36-8, and the mean breadth 18,3 ; the m^ean. 
 length at the eaves 44.6, and the mean breadth 25.9 ; 
 the perpendicular height from the bottom to the eaves 
 18.6, and from the eaves to the top 15.5 feet; how many 
 cubic yards does the stack contain; the mean length of 
 the upper part being 43.7 feet ? 
 
PART YI. MENSURATION OF HAY-STACKS. 225 
 
 CALCULATION. 
 
 Here 36.8x18.3 = 673.44, the area of the base; and 
 44.6 X 25.9=1155.14, the area of the section at the eaves. 
 
 Also, (36.8 + 44.6)^2 = 81.4^2=40.7, the length of 
 the middle section; and (18.3 + 25.9)--2=44.2-f-2 = 22.1, 
 the breadth of the middle section; then 40.7 X 22.1 x 4 = 
 3597.88, /owr times the area of the middle section; ichence 
 (673.44+1155.14 + 3597.88) x 18.6--6 = 5426.46 x 3.1 = 
 16822.026 /ee^, the solidity of the prismoid ABCDEF. 
 
 Again, (25.9 x 15.5)--2=401.45-h2 = 200.725, the area 
 of the ewtZEDG; a7id 200.725 X 43.7 =877 1. 6825/6^6^, the 
 soliditT/ of the prism FEDGH. 
 
 Z«5%(16822.026 + 8771.6825)-4-27=25593.7085--27 
 = 947.9151 cubic yards, the answer required. 
 
 2. How many cubic yards are contained in a liay-stack 
 of the following dimensions ; viz. the length at the bottom 
 49.4, and the breadth 24.6 ; the length at the eaves 58.8, 
 and the breadth 34.2 ; the perpendicular height of the 
 lower part 25.8 ; the perpendicular height of the upper 
 part 21.7 feet, and its length the same as that at the 
 eaves? Ans. 2335.143 yards. 
 
 CASE II. 
 
 JVlien a stack is bulged from the bottom to the eaves, 
 and from the eaves to the top, as in the following figure ; 
 recourse must be had to the equi-distant ordinate method. 
 
 RULE. 
 
 1. Find the areas of as many equi-distant, parallel sec- 
 tions as you think sufficient, with which proceed as if 
 they were the equi-distant ordinates ; and the result will 
 L 5 
 
226 MENSURATION OF HAY-STACKS. PAET Yl. 
 
 be the solidity of the stack, from the bottom to the upper- 
 most section. 
 
 2. Multiply the area of the end of the remaining piece, 
 at the top, which may be considered as a triangular prism, 
 by its length ; and the product wiU be its solidity. 
 
 3. Add these two solidities together, and the sum will 
 be the content of the v/hole stack. 
 
 Notel. The v.'eight of a stack maybe ascertained to a considerable degree of 
 accuracy, in the following manner : Cut out a portion extending to the centre of 
 the stack, from the top to the bottom, and weigh it : and also measure the vacuity 
 from which it is taken : then say, as the content of this part is to its weight, so is 
 the content of the whole stack, to its weight. 
 
 2. If the stack have a rectangular base, the portion to be weighed should not be 
 cut off the end, but taken out about half way between the end and the middle j 
 where it may be supposed the hay is of a medium density. 
 ♦ 
 
 EXAJIPLES. 
 
 1. Let the annexed figure represent a hay-stack, the 
 dimensions of wliich are as follow ; viz. the length AB 
 38 feet, and breadth BC 13 feet ; the length DE 41 feet, 
 and breadth EF 18 feet ; the length GH 45 feet, and 
 breadth HI 22 feet ; the length lO. 44 feet, and breadth 
 LM 19 feet ; the length NO 42 feet, and breadth OP 
 12 feet ; the mean length of the triangular prism, at the 
 top, 41 feet, and the perpendicular height of its end 
 5 feet ; how many cubic yards are contained in the stack ; 
 the perpendicular distance betr/een each parallel section 
 being 6 feet ? 
 
 
 A 
 
 CALCULATION. 
 
 ITe?'e 38 X 13=494, the area of the first section; 41 x 
 18 = 738, the area of the second section; 45x22=990, 
 the area of the third section; 44 x 19—836, the area of 
 the fourth sectio7i ; and 4:2 X 12=504, the area of the fifth 
 or last section ; then by proceeding according to the Ride 
 
PART YI. MENSDRATIOX OF HAT-STACKS. 227 
 
 for cqui-distant ordinatcs, we have A=494 4-504=998, 
 B = 738 + 836 = 1574, C=990, a7id D=6; cooiscquently, 
 (A + 4 B + 2C)4-3xD=(998H- 6296+ 1980) x{] =9274 X 
 2 = \S64:Sfeet, the solidity of the part ABCPON. 
 
 Again, (12 x 5)^2 =60+2 = 30, the area of the etid 
 OPR; a?id 41 X SO = 12S0 feet, the solidity of the prism 
 NOPRS. 
 
 Lastly, (18548 + 1230)--27=19778h-27=:732 yards, 
 14 feet, the ansiver required. 
 
 2. The dimensions of a luiy-stack are as follow ; viz. tlis 
 length of the bottom or first section 70.8, and its breadth 
 20.2 ; the length of the second 75.4, and its breadth 28.7 ; 
 the length of the third 81.2, and its breadth 32.4; the 
 leno-th of the fom-th 86.7, and its breadth 36.8 ; the length 
 of the fifth 84.9, and its breadth 34.6 ; the length of the 
 sixth 83.4, and its breadth 30.9 ; the length of the seventh 
 82.7, and its breadth 16.5 ; the mean length of the trian- 
 gular prism, at the top, 81.5, and the perpendicular height 
 of its end 6.4 feet ; how many cubic yards are contained 
 in the stack; the perpendicular distance between each 
 section being 8 feet ? Ans. 4668.8 yards. 
 
 3. Wliat is the weight of a hay-stack which measures 
 425 cubic yards 18 feet; the weight of a piece cut out of 
 the stack from the top to the bottom, being 1 ton, 3 cwt. 
 2 qrs. 15 lb., and the vacuity from which it is taken, 
 16 cubic yards 12 feet ? Ans. 30 tons, 11 ciut. 3 qrs. 1 lb. 
 
 REMARK. 
 
 Sometunes hay is stored up in hams or shades. "When this is the 
 case, the lower part of the mow, from the bottom to the eaves, will 
 be a parallelopipedon, the content of which may be found by Prob- 
 lem 2 ; and the upper part, from the eaves to the ridge of the build- 
 ing, will be a triangular prism, the content of v/hich may be obtained 
 by Problem 3, Part IV. ; then the sum of these two contents will be 
 the whole content of the mow. 
 
 If the hay docs not extend to the ridga, the upper part will be a 
 trapezoidal prism, the content of which may be obtained by finding 
 the area of the end by Problem 8, Part II. ; and multiplying this 
 area by the length of the prism. (See Problem 18, Part IV., on the 
 method of measuring compound and irregular solids.) 
 
 Note. "VSTien hay-mows are irregular, mean dimensions must be taken. 
 
 L 6 
 
228 MENSURATION OF PART \1. 
 
 DRAINS AND CANALS. 
 
 A Drain is an artificial channel made to convey water 
 from marshes, bogs, and other low grounds, for the pur- 
 pose of making improvements in agriculture, &c. 
 
 A Ciis^AL is an artificial, navigable river, most commonly- 
 made for the purpose of transporting goods from one 
 place to another, by water-carriage. 
 
 Note 1. Drains and Canals are made with sloping sides, and are almost invariably 
 dug by the cubic yard ; hence, it is of the utmost moment to ascertain their con- 
 tents with accuracy. . , , , , , - i, >. j 
 
 The common method of measuring them is to take the breadths of the top and 
 bottom, in different places, and their sum being divided by their number, the 
 quotient is considered as a mean breadth. Several depths are likewise measured, 
 in various places, and their sum divided by their number, is taken for a mean 
 depth • then the length, breadth, and depth being multiplied continually together, 
 the last product is taken for the content ; but it is evident that this process must 
 lead to very erroneous results. ■, r u e 
 
 2 In this country, many thousands of acres of bogs, marshes, and fens have ot 
 late been made fit for the purposes of agriculture, by means of drains, particularly 
 in the counties of York and Lincoln ; and improvements of this kind are still car- 
 ried on in the latter county, with great spirit. 
 
 PROBLEM IIL 
 
 To find the number of cubic yards lohich have been dug 
 out of a drain or canal. 
 
 RULE. 
 
 Find the areas of as many equi-distant perpendicular 
 transverse sections, as you judge sufficient, with which 
 proceed as if they were equi-distant ordinates ; and the 
 result will be the content between the first and last sec- 
 tions. (See Problem 23, Part 11.) 
 
 Note 1. The distance between the sections must be nneasured along 
 the middle of the bottom of the drain or canal ; and when the ground 
 is very uneven, it will be necessary to take the sections nearer to each 
 other, than when it is pretty level. 
 
 2. When a drain or canal is very long, divide it into several parts 
 or kngths, and find the content of each separately by the above Rulej 
 and the sum of these contents will be the content of the whole. 
 
 3. Let the annexed figure repre- 
 sent a perpendicular, transverse sec- 
 tion of a drain or canal, where the 
 ground is even ; then it is evident 
 that if a line AB, be stretched across 
 the top, it will be parallel to the 
 bottom DC, admitting it to be level ; • ^ n 
 and the perpendicular Dm will be equal to the perpendicular Ln, 
 which perpendiculars may be easily found by erecting a straight start 
 perpendicularly to the bottom DC, and so as to touch the line AB, 
 at m and n. 
 
PART VI. 
 
 DRAINS AND C.VNALS. 
 
 229 
 
 In this case, the section ABCD is a trapezoid, tlic area of which 
 may be found by Problems, Part II. 
 
 4. Let the subjoined figure de- 
 note a perpendicular, transverse sec- 
 tion of a drain or canal, where the 
 ground is uneven ; then it is evident 
 that the section is a trapezium, be- 
 cause AB is not parallel to DC. " '• 
 
 Now, as two perpendiculars cannot be taken upon either of the 
 diagonals, the figure must be divided into two triangles in the follow- 
 ing manner: Measure the diagonal BD, and at right-angles to it, the 
 perpendicular C?i ; measure also the line AB, and at right-angles to 
 it, the perpendicular Dm ; hence, the area of the section may be found 
 by Problem?, Part II. 
 
 Or, the section may be divided 
 into two right-angled triangles, and 
 a trapezoid, as in the annexed figure, 
 by measuring the line AB, and at 
 right-angles to it, the two perpen- 
 diculars Dwi and Cn ; and hence 
 the area may be found. 
 
 Or, the area of the section may 
 be obtained by dividing it into the 
 two triangles A Dm, BCn, and the 
 trapezoid DmnC. 
 
 In this case. Dm and Cm, which 
 are at right-angles to DC, are the 
 parallel sides of the trapezoid, and 
 also the bases of the two triangles; and their perpendiculars are Ar 
 and Ba. 
 
 5. If the bottom of a drain or 
 canal be a curve line, as in the sub- 
 joined section ; then the area of the 
 quadrilateral figure ABCD, must 
 l>e found by some of the foregoing 
 methods ; and the area of the part 
 DEC, by the method of equi-distant ordlnates, described in Pro- 
 blem 23, Part II. 
 
 Note. The Rule given in Problem 23, Part IT., being expressed in an algebraic 
 form, is seldom perfectly comprehended by learners ; but the following one may 
 be easily understood, and committed to memory. 
 
 RULE. 
 
 To the sum of the first and last ordinatcs, add four 
 times the sum of all the even ordinates, and twice the sum 
 of all the odd ordinates, not including the first and last ; 
 multiply this sum by the common distance of the ordinates, 
 divide the product by 3, and the quotient will be the area 
 required. 
 
230 
 
 MEKSUIiATlON OP 
 
 PART VI. 
 
 EXAMPLES. 
 
 1. Let the subsequent figure represent part of a drain 
 or canal ; required the number of cubic yards that were 
 dug out of it ; the distance between each perpendicular, 
 transverse section being 60 feet, and the dimensions of 
 these sections, as follow : 
 
 First Section ABCD. 
 
 Feet. 
 ABrr:60.6 
 
 J)m=iCn =15.4 
 
 Second Section EFGH. 
 Feet. 
 HF=:52.3 
 G?^=I3.5 
 EF=63.4 
 H?7i=14.8 
 
 Third Section KLMN. 
 Feet. 
 
 ]Srm=:13.5 
 mn —TlA 
 M?i=18.3 
 \in =23.8 
 
 Fourth Section PQES. 
 
 Feet. 
 
 Pr=15.2 
 SR=29.7 
 E?i =20.5 
 Qa = l7.3 
 
 Fifth Section TUYXTr. 
 Feet. • 
 TU =61.5 
 ■WT=38.4 
 "Wm=V?^ =10.2 
 qa = 5.3 
 X;s = 7.4 
 pr = 5.8 
 
 ^ xP 
 
PART VI. DRAINS AKD C.VNALS, 231 
 
 CALCULATION. 
 
 First Section. 
 Here (60.6 + 28.7) X 15.4=89.3x15.4=1375.22; and 
 1375,22 -r-2=687.61, the area of the trapezoid ABCD. 
 
 Second Section. 
 
 Here 52.3 x 13.5=706.05, double the area of the 
 triangle FGH ; and 63.4 x 14.8 = 938.32, double the 
 area of the triangle EFH ; then (706.05 ~\- 938.32) --2 = 
 1644.37-^2=822.185, the area of the trapezium EFGH. 
 
 Third Section. 
 Here 12.2 x 13.5 = 164.7, double the area of the triangle 
 KmN; also (13.5 + 18.3) x 27.4 = 31.8 x 27.4 = 871.32, 
 double the area of the trapezoid Nm^zM; and 23.8 x 18.3 
 =435.54, double the area of the triangle JAhi ; then 
 (164.7 ^ 871.32+435.54)^2 = 1471.56--2 = 735.78, the 
 area of the trapezium KLMN. 
 
 Fourth Section. 
 Here 14.6x15.2=221.92, double the area of the tri- 
 angle ^Vm ; also (14.6 + 20.5) x 29.7= 35.1 x 29.7 = 
 1042.47, double the area of the trapezoid ^mn'R ; and 20.5 
 X 17.3=354.65, double the area of the triangle QR/^; then 
 (221.92 + 1042.47 + 354.65)-2=i619.04--2=809.52,^/ie 
 area of the trapezium PQRS. 
 
 Fifth Section. 
 Here (61.5 + 38.4) x 10.2=99.9 x 10.2 = 1018.98 ; and 
 1018.98--2=509.49, the area of the trapezoid TUVW; 
 and by the Rule for equi-distant ordinates, we have A=0, 
 B = 5.3 + 5.8 = ll.l, C=7.4, and D=38.4--4=9.6 ; then 
 (A-f 4B + 2C)x 4- D=(44.4 + 14.S)x9.6--3 = 59.2x3.2 
 = 189.44, the area of the part VXW; then 509.49 + 
 189.44=698.93, the area of the whole section TUVXW. 
 
 To find the solid content. 
 
 Here A = 687.61 + 698.93 = 1386.54, B = 822.185 + 
 809.52 = 1631.705, C = 735.78, and'D=QOi then, (A + 
 4B +2C) X J,D=(1386.54 + 6526.82 + 1471.56) x 60-3 = 
 9384.92x20=187698.4 cubic feet; and 187698.4-27= 
 6951.792 cubic yards, the answer required. 
 
 2. Required the number of cubic yards dug out of part 
 of a drain or canal, from the follov.dng dimensions ; each 
 
232 
 
 MENSUKATION OF MARLPITS. 
 
 PART YI. 
 
 perpendicular, transverse section being divided into two 
 triangles in the same manner as the second section EFGH, 
 in the foregoing figure ; and the common distance of the 
 sections 100 feet: 
 
 First Section. 
 Feet. 
 HF = 35.9 
 G?^= 8.2 
 EF=:41.7 
 Hm= 9.6 
 
 Third Section. 
 
 Feet. 
 
 HF=37.1 
 Gn= 9.8 
 EF =42.2 
 llni^ 9.4 
 
 Fifth Section. 
 
 Feet. 
 
 HF = 37.1 
 Gn = 9.5 
 EF =42.8 
 Hw^ = 10.2 
 
 Second Section. 
 
 Feet. 
 
 HF=36.1 
 Gn = 8.9 
 EF =39.8 
 Hm= 9.1 
 
 Fourth Section. 
 
 Feet. 
 HF=36.5 
 Gn r= 9.2 
 EF =41.5 
 Hwi=10.2 
 
 Sixth Section. 
 
 Feet. 
 
 HF =38.6 
 Gn= 9.4 
 EF =43.2 
 Hm= 9.8 
 
 Seventh Scctioji. 
 Feet. 
 HF=37.6 
 Gn— 9.8 
 EF=42.6 
 H;?2 = 10.3 
 
 Ans. The area of the first sectlo7i=34:7 .S5 ; the area 
 ofthesecond=S4:l.7S5i the area of the thlrd=SSO.lS ; 
 the area of the fourth = 379.55 ; the area of the fifth— 
 394.505 ; the area of the slxth=Z9Z.\ ; the area of the 
 seve?ith=40S.6S ; and the co7itent=225259} cubic feet= 
 8342 cubic yards, 25^ feet, the answer required. 
 
 MAELPITS. 
 
 Marl is a kind of rich clay, and is used as manure for 
 land, in Lancashire, Cheshire, Derbyshire, and other 
 counties in England ; and is commonly dug by the cubic 
 yard. 
 
PART. VI. MENSURATIOX OF SIAKLPITS. 233 
 
 Marlpits are of various forms. Sometimes they are 
 laid out in the shape of a rectangle, sometimes in that of 
 a trapezium, and sometimes as an irregular polygon. 
 
 In digging a marlpit the sides are ahvays sloped, in 
 order to prevent the upper edges from slipping in ; and as 
 the bottom is seldom perfectly level, there is generally a 
 little variation in the depths. 
 
 PROBLEM IV. 
 
 To find how many Cubic Yards have been dug out of a 
 Marlpit. 
 
 CASE I. 
 
 When the top and bottom of a marlpit are rectangles, the 
 pit may be considered as a prismoid, 
 
 KULE. 
 
 To the sum of the areas of the top and bottom, add 
 four times the area of a section half-way between them ; 
 multiply this sum by the mean perpencUcular depth, and 
 |- of the product will be the solidity. 
 
 Note 1. When the sides and ends are regularly sloped, the length of the middle 
 section will be equal to half the sum of the lengths of the top and bottom, and its 
 breadth equal to half the sum of their breadths ; but if the inclination of the sides 
 and ends be not regular, the length and breadth of the middle section must be 
 found by actual admeasurement. 
 
 2. A mean depth must be found by taking several depths, at equal distances from 
 each other, and dividing their sum by their number. 
 
 EXAJIPLES. 
 
 1. Let the subsequent figure represent a marlpit, the 
 dimensions of which are as 
 follow : viz. the length AB 
 of the top = 58.6 feet, and the 
 breadth BC = 36.2 feet ; the 
 length EF of the middle sec- 
 tion =55.7 feet, and its breadth 
 FG — 33.4 feet ; the length 
 KX of the bottom = 52.8 feet, 
 and breadth LM = 30.4 feet ; 
 how many cubic yards of marl were dug out of the pit, 
 its mean perpendicular depth being 8.6 feet ? 
 
234 SIENSURATIOX OF SIAKLPITS. PART ^^.. 
 
 CALCULATION. 
 
 Here 58.6x36.2=2121.32, the area of the top; 55.7 X 
 33.4x4 = 1860.38x4=7441.52, /owr times the area of 
 the middle section; and 52.8 X 30.4 = 1605.12, the area of 
 the bottom; then (2121.32 + 7441.52 + 1605.12) x 8.6 = 
 11167.96x8.6=96044.456, which being divided by 6, we 
 obtain 16007.409 cubic feet=:592.867 cubic yards, the 
 answer required. 
 
 2. A marlpit measures 86.4 feet in length, and 36.8 
 feet in breadth at the top ; the length of the middle sec- 
 tion is 82.2 feet, and its breadth 32.7 feet ; the bottom is 
 78.3 feet in length, and 28.6 feet in breadth ; how manr 
 cubic yards of marl were dug out of the pit, its mean 
 perpendicular depth being 10.5 feet ? 
 
 A?is. 1048.098 7/ards. 
 
 CASE 11. 
 
 To find the number of cubic yards luhich have been dug 
 out of a marlpit, the top and bottom ofivhich are triangles, 
 trapeziums, polygons, or any other figure whatever. 
 
 KULE. 
 
 Take such dimensions as will give you the area of the 
 top, the area of the middle section, and the area of the 
 bottom ; proceed with these areas in the same manner as 
 directed for a prismoid, and the result will be the content, 
 nearly. {Seethe Scholium, Prob. 10, Part IV. Sect. I.) 
 
 Note ]. The above Rule will give the content of the generality of marlpits very 
 near the truth ; if, however, a pit exceed 10 or 12 feet in depth, and the sides be 
 not pretty regularly sloped, it will be more accurate to find the areas of 6 equi- 
 distant, horizontal sections ; and proceed with them in the same manner as if they 
 were equi-distant ordinates. (See the SCHOLIUM, Problem 1, Case 2, Part VI.) 
 
 2. When the top and bottom of a marlpit are trapeziums approaching nearly to 
 rectangles, and the inclination of the bottom from the plane of the l-.orizon very 
 considerable, so as to make the pit much deeper on one side than on the other, the 
 content may be most correctly obtained by finding the areas of several equi-distant, 
 perpendicular sections ; and proceeding with them as directed in the last Problem. 
 
 The contents of the two pieces at the ends, resembling wedges, may be found by 
 Problem 9, Part IV., Section L 
 
 3. The content of the space or road leading into the pit, may be obtained by the 
 last Problem. 
 
 The area of the first section, on that where the road commences, will evidently 
 be nothing ; and the distance between the first section and the middle of the slant 
 height of the pit must be taken for the length of the space. 
 
 4. When the edges of a pit give way, and slide in, these slips must be measured, 
 if they be thrown out or removed bv the labourers . The general method of doing 
 this is by multiplving the mean length, the mean breadth, and the mean depth 
 continually together, for the content. This method is certainly inaccurate ; yet in 
 such cases as these, it may, in general, suffice ; and sometimes it is the only one 
 that can be adopted with any probability of success. 
 
FAT.T Tl. 
 
 MENSURATION OF JtLVRLPITS. 
 
 235 
 
 EXAIMPLES. 
 
 1. Let the following figure denote the uppermost, 
 
 'y\ 
 
 horizontal section of a marlpit, 
 the diagonal AC of which mea- 
 sui-es 86.0, the perpendicular 
 Da 26.8, and the perpendicular 
 B« 32.6 feet. The diagonal of 
 tlie middle section is found to 
 be 80.6, and one of the perpen- 
 diculars 23.9, and the other 29.3 feet. The diagonal of 
 the bottom measures 74.3, and the perpendiculars 20.7, 
 and 26.5 feet ; how many cubic yards were dug out of the 
 pit, its mean perpendicular depth being 10.8 feet ? 
 
 CALCULATION. 
 
 Here (26.8 + 32.6) x86.5==59.4 x 86.5 = 5138.1 ; cmd 
 5138.1 -=-2 = 2569.05, the area of the top. 
 
 Also, (23.9 + 29.3) X 80.6=53.2x80.6=4287.92; and 
 4287.92 ^2x4 = 2143.96 x 4 = 8575.84, four times the 
 area of the middle section. 
 
 Again, (20.7 + 26.5) x 74.3=47.2 x 74.3 = 3506.96; and 
 3506.96-5-2=1753.48, the area of the bottom. 
 
 Lastly, (2569.05 + 8575.84 + 1753.48) x 10.8 -^ 6 = 
 12898.37 X 1.8 = 23217.066 cubic feet= 859.8913 cubic 
 yards, the answer required. ' 
 
 2. Admit the following d 
 
 figure ABODE to denote three 
 horizontal sections of a marl- 
 pit ; also the figure FLMG- to 
 represent the space or road 
 leading into the pit ; required 
 the number of cubic yards 
 that were dug out of each ; the 
 mean depth of the pit being 
 11.4 feet; the equi-distant, 
 perpendicular sections at HK, 
 LM, trapezoids ; the whole 
 length of the space 43.8 feet ; 
 as below. 
 
 \ 
 
 a 
 
 \ 
 
 ^.^'' \ 
 
 -. — ' "' 
 
 
 A 
 
 L 
 
 
 M B 
 
 
 H 
 
 F 
 
 
 K. 
 G 
 
 
 
 and the other dimensions 
 
236 JVIENSUKATION OF MARLPITS. PART VI. 
 
 Horizontal Sections of the Pit. 
 
 Uppermost Section. Middle Section. Bottom Section. 
 Feet. Feet. Feet. 
 
 AC = 118.61 AC=111.71 AC=104.^ 
 
 = 118.61 AC=111.7"1 AC=104.81 
 
 = 43.4 \ E« = 39.8 \ E« = 36.4 \ 
 
 = 35.2 J B«= 31.6 J Ba= 28.3 J 
 
 CE = 102.81 CE= 95.91 CE=: 88.61 
 
 D«= 26.4 J I)a= 22.6 j Da = 19.5 J 
 
 Ea 
 Ba 
 
 Perpendicular Sections of the Space, 
 
 Section at HK. Section at LM. 
 
 Feet. Feet. 
 
 AB=12.4 AB=12.8 
 
 DC= 8.5 DC= 7.9 
 
 Dm = C?z = 5.7 D;?? = Cw =11.2 
 
 {^See the figure in the last Problem, Note S.) 
 
 CALCULATION OF THE PIT. 
 
 Uppermost Section. 
 
 Here (43.4 + 35.2) x 118.6 = 78.6 x 118.6 = 9321.96, 
 double the area of the trapezium ABCE; and 102.8 x 26.4 
 =2713.92, double the area of the triangle CDE; then 
 (9321.96 + 2713.92)-^2=rl2035.88-T-2=60l7.94, the area 
 vf the whole section ABODE. 
 
 Middle Section. 
 
 Here (39.8 + 31.6) x 111.7 = 71.4 x 111.7 = 7975.38, 
 double the area of the trapezium ABCE ; and 95.9 X 22.6 
 =:2 167.34, double the area of the triangle CDE; then 
 (7975.38 + 2167.34) --2x4 ='l0142.72 -- 2 x 4 =5071.36 
 X 4=20285.44, ybwr times the area of the whole section 
 ABCDE. 
 
 Bottom Section. 
 
 Here (36.4 + 28.3) x 104.8 = 64.7 x 104.8 = 6780.56, 
 double the area of the trapezium ABCE ; and 88.6 x 19.5 
 = 1727.7, double the area of the triangle CDE; then 
 (6780.56 + 1727.7)-- 2 = 8508.26 -r- 2 = 4254.13, the area 
 of the whole section ABCDE. 
 
PAKT VI. MENSURATION OF PONDS^ ETC. 237 
 
 To find the ^olid content. 
 
 Here (6017.94 + 20285.44 + 4254.13) x 11.4 -- 6 = 
 30557.51 X 1.9 = 58059.269 cubic yee^ = 2150.343 cubic 
 yards, the content of the pit. 
 
 CALCULATION OF THE SPACE. 
 
 Section at IIK. 
 
 Here (12.4 + 8.5) x 5.7 =20.9 x 5.7 = 119.13; and 
 119.13-^2 = 59.565, the area of the trapezoid ABCD. 
 
 Section at LM. 
 
 Here {12.S +7.9) x 11.2 = 20.7 X 11.2 = 231.84 ; a?id 
 231.84--2 = il5.92, the area of the trapezoid ABCD. 
 
 To find the solid content. 
 Here A= 1 15.92, B = 59.565, C = 0, and D=43.8 -r- 2 
 = 21.9; ^/ie?i(A + 4B4-2C)x4-D=(115.92 + 238.26)x 
 21.9 -- 3 = 354.18 x 7.3 = 3585.514 cubic jeet= 95.759 
 cubic yards, the content of the space. 
 
 3. The diagonal of the first or uppermost horizontal 
 section of a marlpit is 118.6, and the perpendiculars are 
 64.8, and 43.5 feet ; the diagonal of the second section 
 112.4, and the perpendiculars 61.7, and 40.6 feet; the 
 diagonal of the third section 106.3, and the perpendiculars 
 58.5, and 37.7 feet ; the diagonal of the fourth section 
 100.8, and the perpendiculars 55.9, and 34.6 feet ; the 
 diagonal of the fifth or bottom section 94.3, and the per- 
 pendiculars 52.4, and 31.5 feet; required the number of 
 cubic yards that were dug out of the pit, the sections 
 being taken at equal distances from each other, and the 
 mean perpendicular depth of the pit 16.8 feet. 
 
 Ans. The area of the first section — 6422.19 ; the area 
 of the second = 51 49. 2Q ; the area of the third=5llS.0S ; 
 the area of the fourth = 4561.2 ; the area of the fifth = 
 3955.885 ; and the content = 86584.365 cubic feet = 
 3206.828 cubic yards, the ansiuer required. 
 
 DIRECTIONS rOR 
 
 MEASURING PONDS AND MILL-DAMS. 
 
 Ponds and mill-dams are commonlj dug by the cubic 
 yard ; and assume a variety of shapes. 
 
238 MENSURATION OF EMBANlOkEENTS. P.iP.T VI 
 
 If the top and bottom of a pond or dam be rectangles, 
 it may be treated as directed in the last Problem, Case I. 
 if the top and bottom be triangles, trapeziums, or polygons, 
 the content may generally be found by Case 11., of the 
 last Problem ; but if both these methods fail, you must 
 proceed as directed in Problem 3, for drains and canals. 
 In this case, the contents of the pieces at the ends, re- 
 sembling wedges, must be found by Problem 9, Part IV., 
 Sect. I. 
 
 ]Mr. TV. Putsey, Teacher of the Mathematics, at Picker- 
 ing, informs me that the ponds made upon the wolds in 
 Yorkshire, are generally of a conical shape ; hence their 
 contents may be found by Problem 7, Part IV, Sect. I. 
 
 Mr. P. who is a practical measurer, has also very 
 kindly communicated the following method of taking the 
 dimensions of a conical pond in which there is water. 
 
 EXA^IFLE. 
 
 Let ABC denote a perpen- 
 dicular section of a conical pond, 
 whose dimensions are required. 
 
 Extend a cord over the pond, 
 with which take the diameter c 
 
 AB ; make a ring fast to the middle of the cord or 
 diameter, as at D, through which put the end of the 
 plumb-line ADC ; let your assistant keep one end of the 
 diameter at B ; while you hold the other, and the plumb- 
 line at A ; permit the plummet to descend to the bottom 
 of the pond, as at C ; then di'aw back both the cords, and 
 measure CD, which will be the perpendicular depth of the 
 pond. 
 
 Note 1. ^^^len the top of a pond is not a perfect circle, measure two diameters 
 at right angles to each other ; and take half their sum for a mean diameter. 
 
 2. Cellars generally form parallelopipedons ; and when they are dug by the cubic 
 yard, their contents may be found by Problem 2, Section I., Part IV. 
 
 DIRECTIONS FOR 
 
 3MEASURING ElMBANKMENTS. 
 
 The most correct method of measuring embankments, 
 is to proceed in the same manner as directed in Pro- 
 blem 3, for drains and canals ; and if the following ob- 
 servations be well understood, no difficulties will arise, in 
 
PART VI. 
 
 MENSURATION OF QUATIPJES. 
 
 taking the dimensions, raid finding the areas of the 
 sections. 
 
 Let ABCD represent a trans- 
 verse perpendicukir section of 
 an embankment, made upon 
 level ground ; then if DC be 
 parallel to AB, it is evident 
 that the section is a trapezoid, 
 the area of which may be found 
 by Problem 8, Part 11. 
 
 The distance of the parallel sides, or height of the sec- 
 tion, may be obtained by placing a staff perpendicular to 
 the horizon, at A or B, and producing DC to E or F ; 
 then will AE or BE be the perpendicular distance of the 
 parallel sides AB, DC ; and AB, the breadth of the bottom 
 of the embankment, is evidently equal to ED +DC +CF. 
 
 Again, let G-HKL, denote a perpendicular, transverse 
 section of an embankment, 
 made upon uneven ground ; 
 tlien it is evident that the sec- 
 tion is a trapezium ; and as the 
 diagonals and perpendiculars 
 cannot be measured, the con- 
 tent must be found in the fol- 
 lowing manner : Produce LK, 
 the horizontal Kne of the top, both ways, to M and N, 
 and let fall the perpendiculars iMG and NH ; then by 
 Problem 8, Part II., half the sum of these parallel sides 
 multiplied by their perpendicular distance MN, will 
 give the area of the trapezoid GHNM, from which sub- 
 tract the sum of the areas of the tAvo right-angled triangles 
 GjNIL, HNK, and the remainder will be the area of the 
 section GHKL. 
 
 Note. Unskilful measurers affect to determine the contents of embankments by 
 finding what they call mean breadths and thicknesses ; but no person who has a 
 scientific knowledge of mensuration, will have recourse to such an erroneous pro- 
 cess, if the foregoing method can by any means be adopted. See Drains and 
 Canals, Note 1. 
 
 DIRECTIONS FOB 
 
 IMEASURING QUAERIES, 
 
 The baring of quarries is generally done by the cubic 
 yard ; and sometimes tlie stones themselves are got by the 
 same measurement. 
 
1540 MENSUHATION OF COAL-HEAPS. PART VI. 
 
 Quarries, in general, are very irregular ; they may 
 however, commonly be measured by some of the methods 
 already described for canals, marlpits, ponds, &c. 
 
 If a quarry be so irregular that none of these methods 
 can be adopted, the general method is to take such dimen- 
 sions as will give the area of a mean horizontal section ; 
 then this area being multiplied by the mean depth of the 
 quarry ; the product is taken for the content. 
 
 Note 1 . Sometimes a quarry may be most correctly measured by dividing it 
 into several parts ; and taking the dimensions of each part separately. 
 
 2. When the stones that are left jutting out of the sides of the quarry, are mea- 
 sured with the vacuity, mean dimensions of these stones must be taken, and their 
 contents subtracted, in order to obtain the true content of the vacuity. 
 
 DIRECTIONS FOB 
 
 MEASURING COAL-HEAPS. 
 
 "When the stock of a colliery is to be taken, or an ex- 
 change of tenants takes place, it becomes necessary to 
 ascertain the quantity of coals which are laid unsold at 
 the pits ; and as pit-heaps are generally very irregular, 
 both in extent and thickness, it is no easy task to find 
 their contents with a tolerable degree of accuracy. 
 
 The method generally adopted, is to take such dimen- 
 sions as will give the area of a mean horizontal section of 
 the heap ; and to multiply this area by the mean thick- 
 ness, for the content. 
 
 Note I. The perpendicular height of any point at the extremity of a coal-heap, 
 may be obtained in the following manner: Place a spirit or water-level upon the 
 top of the heap, close by the edge ; let your assistant hold a pole upon the ground. 
 In a perpendicular direction, at the bottom of the heap ; direct the level towards 
 the pole, and note that point in it which is seen through the level ; measure the 
 distance between the bottom of the pole and this point ; subtract the height of the 
 level from the said distance ; and the remainder will be the height of the heap, at 
 the place where the observation is made. 
 
 2. When a coal-heap is upon level ground, the height or thickness of any place 
 between the pit and the extremity of the heap, may be found thus : By Note 1, 
 find the height of the heap where it is most elevated ; then let your assistant place 
 the pole upon that part of the heap the height of which you wish to obtain ; and 
 without removing the level from the highest point of the heap, direct it towards 
 the pole ; measure the distance between the bottom of the pole and the point seen 
 through the level ; from this distance take the height of the level ; and if the re- 
 mainder be subtracted from the greatest height of the heap, before found, you will 
 obtain the height required. 
 
 3. If the ground on which the coal-heap is laid be level, and the upper surface 
 of the heap a regular inclined plane, rising gradually from the pit to the utmost 
 extremity of the heap, which is sometimes the case, it is evident that the mean 
 thickness of the heap will be equal to half the sum of the two heights taken at that 
 part of the heap adjoining the pit, and at the utmost extremity or highest point of 
 the heap. 
 
 4. When a coal-heap is extremely irregular, it fs generally necessary to divide it 
 into several parts, and take the dimensions of each part separately ; in this case, 
 heights must be taken in such places as are most likely to give the mean thickness 
 of each part ; and when the ground is not level, a proper allowance must be made 
 ft>r this circumstan-je. 
 
TAUT \J. 
 
 w i:x-i i;a' 
 
 X OF CLAY-nEAPS. 
 
 241 
 
 ri. In some part'- oi J-liiftl'iiid, .'i !)>• 
 make a bushel of co-il*. and .".((i bu-ln 
 in a coal-heap be divided by 'ifiss. •. 
 the number of bushel j contiiinod in 
 every person whoinfusincs .t eoal-ii 
 
 customary measure of tTie ])l;uv ' 
 
 '2815 cubic inches. (.SV<; ///<• A"liiu, 
 penal Measures, see 2'art VIII. <■/ 
 
 0. Notwithstandinic Avliat lias b>'. ; 
 mid coal-heaps, a grr.it disil will alv. 
 I'or it is impossible to -ive direction- 
 ■.vith in the practic.: ijl'iiK-;i«nriiig- tii 
 
 cks, M'inchester measure, or 2083 cubic inches, 
 :1s a chaldron ; therefore, if the cubic 'inches 
 r the cubic feet by 1 .555, the quotient will be 
 the heap ; hut as tliis measure is not generul, 
 .^ap ou;;iiL to make! himself acquainted witli the 
 Hie content of the new imperial coal-bushel is 
 •'.V Arithmetic, pp. 63. 353. For the New Im- 
 'iis Work.) 
 
 1 advanced on the subject of measuring quarries 
 ys d,?pend uymn the ingenuity of the measurer ; 
 that will suit every particular case to be met 
 ■se irregular figures. 
 
 Clay is frc-i'.'-ml 
 surfiice of the iivk-m 
 ;incl tliea sold !'v- i 
 purpose, of m:iki I rj- i 
 
 When this i- rip 
 t^iin its solid conirii 
 be approximaii-'! '..y 
 of the base of i lie ii. 
 ]nost convenicMir 
 direeticns giv.'ji 
 ?md then takv h; 
 area. — Also, i;sk 
 their nuniber, !'<.•. 
 by the mean <!.;> 
 rent ; and, \i ;•' n 
 reduce it to cii'-i- 
 
 il,l- 
 
 < r.Ai -HEAPS. 
 
 y ur- out of pits, and laid upon the 
 i<;, -• nerally in very irregular lieaps ; 
 li'- .-iibic yard or by tlie ton, for the 
 
 • it becomes necessary to ascer- 
 
 .early as possible ; and this may 
 allowing methods : Find the area 
 
 iid also the area of its top, in the 
 , according to the rules and 
 liiferent Problems of Part II. ; 
 
 I im of these areas, for the mean 
 depths, and divide their sum by 
 depth. Multiply the mean area 
 le product will be the solid con- 
 n cubic feet, divide it by 27, to 
 
 >i;mi 
 
 I r! 
 rin 
 
 \ ur ■ 
 
 Note 1. When tl)« b.i^f- oi i c!; 
 !*roblem 5, Part 111. 
 
 li. When the dhufn.-ion- iir- 
 t«»itent will be in i-ii'iic y:u(U. 
 taken in feet, tenth-, nna Inm.ii t- 
 oi' a foot. 
 
 3. When a clay-ii.- •i)',!'; n-r 
 found by taking the dinH'i.sion- ■ 
 .•<ion, as'you take it, bo a.- to J'. • 
 middle horizontal section. — 'I'ii 
 4:/i(l iaf,/ii}^j" as thv di)!!»!>,ioi'i 
 ..f the hea]), but lar^f r ;han tlms 
 
 4. When a clay-heap ;-.Mini' ■ 
 blern '2, Part IV.), >oi i' m. .I'-ii 
 and divide their sum !>y r'n 'ii- m 
 mean depth are fonuil' in tw- ^; 
 three dimensions is taken fi.r <hi 
 
 5. When a heap i - very large y 
 into two or three p;iJ ti ; "and tak 
 separately 
 
 -;i ap is very irregular, Its area may be found by 
 
 ■■;^..;n in yards, tenths, and hundredths, the solid 
 nu decimals of a yard : but if the dimensions be 
 ')i>, the content will be in cubic feet, and decimals 
 
 ■ ■. irregular, the mean area may sometimes be 
 uis top of the heap ; and increasing each dimcn- 
 :>- nearly as you can judge) the dimensions of the 
 li.ethodof measuringis generally called '' givi/nr 
 lin.s obtained are smaller than those of the base 
 • ■rthe top. 
 
 II.- thing of the form of a parailelopipedon (Pro- 
 /> ;ake lengths, in two or three difi'erent places ; 
 3r, for a mean length. A mean breadth and a 
 11- manner: and the continued product of these 
 •"! d content. 
 
 id irregular, it is frequently advisable to divide it 
 th'^ dimensions, and Und the content of each part 
 
 When a clay-bf ;>» resembles a n-ismoid, or the frustum of a cone, the rule 
 given in Problem lO, I'arr l.V., m,.y be applied with propriety. For other useful 
 ilules and ObservcHu.'i>. tlie r^-adf r is referred to Problem 18, Part IV. ; and also 
 to tiie methods of inei<-uiing n?:irl-pits and coal-heaps. 
 
 7. A cubic foot of ci.ay. jucording to the table of specific gravities, page 147., 
 weighs 2160 ounces a-, oiruiijtois : c<.asequently, a cubic yard will weigh 1.627 tons ; 
 therefore, if we multi|>iy iIk- mnnbt r of cubic yards, in a day-heap, by 1.C27, the 
 
 )roduct will be Ihv 
 
 ,'hT 
 
 roj. 
 
 M 
 
242 
 
 JIENSURATION OF CLAY- 
 
 :aps. 
 
 PAIiT TI. 
 
 8. B}^ experience, it has been found that when clay is very compact, a cubic yard 
 •will weigh 1| tons ; consequently, the multiplier will be 1.75 ; but when it is less 
 dense, a cubic yard will not weigh more than 1^ or 1^ tons ; in these cases, the 
 multiplier will be 1.25 or 1.5, to bring cubic yards into tons. 
 
 9. A measuring-tape divided into I'ect, tenths, and half-tenths, is best adapted for 
 taking the dimensions ; as we then liave no trouble in reducing the inches to the 
 decimals of a foot ; for, in this case, the tenths are given by the tape ; and the 
 half-tenths are 5 hundredths of a foot ; hence we have alv.ays the dimensions in 
 feet and tenths, or in feet, tenths, and hundredths of a foot ; thus, 28.(i^ feet s=28.65 
 feet, &c. 
 
 10. The rules and directions given for clay-heaps will apply to all compost and 
 manure heaps ; when it is required to determine their solid contents for the pur- 
 poses of sale, &c. 
 
 EXAIIPLE. 
 
 Eequired tlie number of cubic yards, in two clay -heaps, 
 resembling parallelopipedons, from the following dimen- 
 sions ; and also the weight of the clay, supposing each 
 cubic yard to v/eigh 1.627 tons. 
 
 1 
 No. 
 
 Lengths. 
 
 Ereadths. 
 
 1 
 
 Depths. 
 
 1 
 
 Feet. 
 
 93.85 
 86.75 
 74.65 
 
 Feet. 
 
 56.55 
 48.45 
 39.35 
 
 Feet. 
 
 9.65 
 
 8.85 
 7.70 
 
 1 
 
 2 
 
 62.25 
 58.30 
 46.45 
 
 43.25 
 39.70 
 
 8.35 
 
 1 
 
 Content by the third and fourth NotcL 
 
 
 No. I. 
 
 
 Length. 
 
 Breadth. 
 
 Depth. 
 
 Feet. 
 
 Feet. 
 
 Feet. 
 
 93.85 
 
 56.55 
 
 9.65 
 
 86.75 
 
 48.45 
 
 8.85 
 
 74.65 
 
 89.35 
 
 7.70 
 
 3)255.25 
 
 3)144.35 
 
 3)26.20 
 
 85.083 mean. 
 
 48.116 mea7i. 
 
 8.666 7nean 
 
 Then, 85.083 X 48.116 — 4093.853628 square feet, the 
 mean area; and 4093.853628 x 8.666=35477.335540248 
 cubic f est, the solid content of No. I. 
 
PART VI. A DESCRIPTION OF C.\:NALS. 243 
 
 Ko. II. 
 
 Length. Breadth. Depth. 
 
 Feet. Feet. Feet. 
 
 62.25 43.25 8.35 
 
 58.30 39.70 "=^ 
 
 ^6.45 2)82:95 
 3)167^ 41:4 75 mean. 
 
 65.6Q6 mean. 
 
 Then, 55.666 x 41.475 =.-2308.74735 square feet, the 
 mean area; and 2308.74735 x 8.35 = 19278.0403725 
 cubic feet, the solid content of No. IT. 
 
 Lastly, 35477.335 + 19278.040 = 54755.375 cubic feet, 
 the solid content of both heaps ; then, 54755.375 -^-27 = 
 2027.9768 cubic yards, the lohole content. 
 
 Weight by the seventh Note. 
 
 Here, 2027.9768 x 1.627 = 3299.5182536 tons-Z2^% 
 tons, the ID eight required. 
 
 KEStARK. 
 
 Having, in the preceding pages, treated of the Mensu- 
 ration of Drains and Canals, it is presumed that a short 
 account of a fe^v of the principal ones, will not be un- 
 acceptable to the young reader ; as it will tend to give 
 him some idea of the great Improvements that Agriculture 
 and Commerce have received, and are daily receiving by 
 the use of them ; and will inform him what stupendous 
 Works have been effected by the Ingenuity, Perseverance, 
 and united Efforts of men. 
 
 A DESCRIPTION 
 
 OF SOME OF 
 
 THE PRINCIPAL CANALS IN ENGLAND, 
 
 SCOTLAND, FRANCE, AND CHINA. 
 
 Canals are to be met with in every civilised country; and perhaps it 
 will not be going too far to say, that the internal commerce of no 
 nation has received greater improvements by them, than that of Great 
 Britain. 
 
 The Duke of Bridgewater" s Canal, is a work that begins at Worsley, 
 seven miles from Manchester ; v/here, at the foot of a mountain, com- 
 M 2 
 
244 A DESCr.lPTION OF CANALS. PAKT VI. 
 
 posed Of coal, a basin is cut, containing a great body of water, wiiich 
 serves as a reservoir to the navigation. The canal runs under a hill, 
 by a subterraneous passage from this basin, nearly three quarters of a 
 mile, to the Duke's coal-works. In some places the passage is cut 
 through the solid rock, and in others arched over witli brick ; and at 
 certain distances, air funnels, some of which are thirty-seven yards in 
 perpendicular height, are cut through the rock to the top of the hill. 
 At Barton-bridge, three miles from the basin, is an aqueduct, which, 
 for upwards of 200 yards, conveys the canal across a valley and the 
 navigable river Irwell. There are three arches over this river ; the 
 middle one is 60 feet wide, and 3vS feet above the water in the Irwell ; 
 and will admit the largest barges that navigate that river, to pass 
 under with masts and sails standing. At Longford bridge, the canal 
 turns to the right, and crossing the Mersey, passes near Altringham, 
 Dunham, Grappenhall, and Kaulton, into the tideway of the Mersey, 
 at Runcorn Gap, where barges can come into the canal from Liver- 
 pool at low water. 
 
 This canal is 29 miles in length. It was begun in 1758, and 
 finished in 1763, under the direction of the great mechanic and en- 
 gineer, Brindley. 
 
 Note. From this canal, at Worsley, there is a branch cut to tlio 
 town of Leigh ; and thence to Wigan, where it communicates v.ith 
 the Leeds and Liverpool canal. — From the Duke's canal, there is also 
 another branch which terminates at Manchester ; and thus unites this 
 great ehiporiwn with the maritime town of Liverpool. 
 
 The Grand Junctioji Canal, is a woi'k that joins several other canals 
 in the centre of the country, which thence form a communication 
 between the rivers Thames, Severn, Mersey, and Trent; and, conse- 
 quently, an inland navigation to the four principal sea-ports, London, 
 Bristol, Liverpool, and Hull. 
 
 This canal commences at Braunston, on the west-borders of Nor- 
 thamptonshire, passes by Daventry, to Stony Stratford, in Buckiiig- 
 iiamshire, thence on the confines of Bedfordshire, west of Leighton 
 Buzzard, to Tring, Berkhamstead, and Rickmansworth, in Herts, and 
 through Middlesex, by Uxbridge, to Brentford, where it enters the 
 Thames, 12 miles, by that river, above London. Its whole length is 
 iqjwards of 90 miles. 
 
 The Grand Trunk Canal, is a work that forms a communication 
 between the rivers Mei'sey and Trent, and, of course, between the 
 Irish sea and the German ocean. Its length is 92 miles, from the 
 Duke of Bridgewater's canal at Preston on the Hill, in Cheshire, to 
 Wildon-ferry, in Derbyshire, where it communicates with the Trent. 
 'I'iie canal is carried over the river Dove, in an aqueduct of 23 arches, 
 and over the Trent, by an aqueduct of 6 arches. At Preston on the 
 Hill, it passes under ground 1241 yards; at Barton, and in the 
 neighbourhood, it has two subterraneous passages ; and at Hardcastle 
 Hil-l, in Staftordshire, it is conveyed under ground 2880 yards. In 
 the neighbourhood of Stafford, a brancli is made from this canal, to 
 run near Wolverhampton, and to join the Severn near Bewdley : from 
 this again other branches cross Warwickshire to Braunston, where 
 commences the Grand Junction Canal to the Thames at Brentford. 
 
PAP.T VI. A DESCRIPTION OF CAN^\XS. 245 
 
 This canal was begun in 17 66, by Brindlcy, who died before its 
 completion ; but it was finished by IMr. Ileushall, his brother-in-law, 
 in 1777 
 
 TJie Leeds and Liverpool Canal. " The first act for this canal 
 passed in 1770, by which the undertakers, a body corporate, by the 
 name of " Tlie Company of Proprietors of the Canal Naviyntion from 
 Leeds to Liverpool,'^ were empowered to raise the sum of 320,000/. in 
 shares, deemed personal estate. 
 
 The work was prosecuted with spirit ; and in a few years there 
 were finished 33^ miles on the Yorkshire side, viz. from Leeds to 
 Holme Bridge, near Gargrave ; and 28 miles on the Lancashire side, 
 viz. from Liverpool to Newbrough ; but the money subscribed being 
 all expended, the works were discontinued for several years. 
 
 In 1790 another act was obtained, to enable the com})any to vary 
 the line from the north to the south side of the river Calder ; and to 
 raise a further sum of money : in consequence of which the works 
 Tvere again resumed. 
 
 In 1794 another act passed, to enable the company to make a con- 
 siderable alteration in the line, so as to pass by the market-towns of 
 Burnley, Blackburn, Chorley, and Wigan, where it joins the head of 
 the Douglas Navigation ; by means of which it unites the Lancashire 
 side of this canal at Newbrough. 
 
 The whole length of the canal, according to this arrangement, is 
 129 miles. 
 
 The summit of this canal is at a village, called Foulridge, near 
 Colne, where it passes under ground 1630 yards, by a tunnel, whicii 
 is 18 feet high, and 17 feet wide. ITie fall, eastward, from the sum- 
 mit to the navigable river Aire, at Leeds, is 409^ feet ; and westward 
 to the basin, at Liverpool, 431 feet; which basin is about 52 feet 
 above the river Mersey, at low water. Between the summit and 
 Leeds are 44 locks, viz. 15 to Holme Bridge, (from which place to 
 near Bingley, there is a level pool of 17 miles in length,) 11 from 
 Bingley to the Junction with the Bradford canal, and 1 8 from thenco 
 to Leeds. Between the summit and Liverpool are 47 locks. 
 
 There are several large aqueduct bridges on this canal, particularly 
 those over the river Aire below Bingley, and above Gargrave. 
 
 It has a commimication with the town of Bradford, by means of 
 a branch, about three miles in length, called " Bradford Canal." 
 
 The general advantages which the public will derive from this 
 canal, are more than the limits of this paper will allow to be ex- 
 plained. It will tend greatly to encourage trade and manufactures ; 
 will bring into extensive use inexhaustible rocks of lime-stone, slate, 
 flags, and free-stone ; excellent mines of coal ; and great quantities of 
 timber for building ships, houses, &c. 
 
 This canal is 42 feet wide at the top, with 5 feet depth of water ; 
 and the dimensions of the locks upon it, are about 70 feet in length, 
 and 152- feet in width. 
 
 It was completed in the latter part of the year 1816; and now 
 
 forms a most eligible communication between the port of Liverpool 
 
 and the town of Leeds; and from tlicnce, by the river Aire, and other 
 
 canals and river navigations, to the port of llull. Thus the object of 
 
 M 3 
 
246 A DESCRIPTION OP C^VNALS. PAET VI. 
 
 the original promoters of this noble undertaking is obtained, viz. " A 
 2v'avigable Communication between the East and West Seas. 
 
 Xote. Mr. Thomas Beeston, of the Cr.nal office, Leeds, and Mr. John Eottomiey, 
 of the Canal office, Bradford, very kuidly favoured me with the above description 
 of the Leeds and Liverpool Canal. 
 
 TVie Great Canal, in Scotland, is a v.-ork that forms a junction be- 
 tween the Forth and Clyde. Its length is 35 miles from the influx 
 of the Carron, at Grangemouth, to the Junction with the Clyde, six 
 miles above Dumbarton. In the course of this navigation, the vessels 
 are raised to the height of 155 feet above the level of the sea; and 
 passing afterward upon the summit of the country for IS miles, they 
 descend into the river Clyde, and have from thence free access to the 
 Atlantic ocenn. This canal is carried over 36 rivers and rivulets, and 
 2 great roads, by 38 aqueducts of hewn stone. In some places it 
 passes through mossy ground, and in others through solid rock. 
 
 The road from Edinburgh to Glasgovv- passes under it near Falkirk, 
 and over it, by means of a drawbridge, six m.iles from Glasgow. In 
 the course of this inland navigation are many striking scenes ; par- 
 ticularly the romantic situation of the stupendous aqueduct over the 
 Kelvin, near Glasgow, 420 feet in length, carrying a great artificial 
 river over a natural one, where large vessels sail at the height of 65 
 feet above the bed of the river below. The utility of this communi- 
 cation between the German and Atlantic oceans, to the commerce of 
 Great Britain and Ireland, in their trade -^vith Norway, Sweden, and 
 the Baltic, must be strikingly evident ; as it shortens the nautical 
 distance in some instances 800, and in others 1000 miles. 
 
 This canal cost about 300,000/. It was begun in 1768, imder the 
 direction of Mr. Smeaton, and finished in 1790, by Mr. Whitworth ; 
 Mr. Smeaton having resigned, in consequence of the bad state of his 
 health. 
 
 TJie Royal Canal of Languedoc, in France, is a work that effects 
 an inland communication between the r^Iediterranean and Atlantic. 
 From the port of Cette, in the Mediterranean, it crosses the lake of 
 Thau ; and, below Toulouse, is conveyed by three sluices into the 
 Garonne. At St. Ferreol, near Revel, between two rocky hills, is a 
 grand basin, above 1000 feet in diameter, into which the rivulet 
 Laudot is received ; and hence three large cocks of cast brass open, 
 and discharge the water, which then goes under the name of the river 
 L.audot, and continues its course io the canal called Rigole de la 
 Plaine. Thence it is conveyed to another reservoir near Naurouse, 
 out of vv'hich it is conveyed by sluices, both to the Mediterranean and 
 Atlantic, as the canal requires it ; this being the highest point be- 
 tween the two seas. Near Beziers are eight sluices, whieh form a 
 regular and grand cascade, 936 feet long, and 66 fefet high, by which 
 vessels cross the river Orb, and continue their voyage on the canal. 
 Above it, between Beziers and Capestan, is the Mai- Pas, where the 
 canal is conveyed, for the length of 720 feet, under a mountain. At 
 Adge is a round sluice, with three openings, three different depths of 
 water meeting here ; and the gates are so contrived, that vessels may 
 pass through by opening which sluice the master pleases. The canal 
 has 37 aqueducts, and its length from Toulouse to Beziers, where it 
 joins the river Orb, is 152 miles. 
 
PAKT VI. A^ ACCOUNT OF DUAI.S'S. 247 
 
 This canal was begun in 1 G66, and completed in 1 68 1 , under the 
 direction of the engineer lliquet. 
 
 Tlie Imperial Canal, in China, is a stupendous work, extending 
 nearly through the whole empire, from north to south ; and serves to 
 convey goods between Pekin and Canton ; being only interrupted by 
 a mountain, in the province of Kiang-si, of about one day's joiarney. 
 This canal was finished about the year 980 ; and 30,000 men were 
 employed 43 years in completing it ; its whole length being 825 miles. 
 The traffic upon it is exceedingly great ; and it is, in various other 
 respects, an object of wonder and admiration to Europeans. 
 
 There are in England, as well as in other nations, many more 
 Canals far too numerous to be mentioned here ; those who desire to 
 see a more circumstantial account, may consult Mr. Phillips's General 
 History of Inland Navigation, which is a work replete with useful in- 
 formation. 
 
 AN ACCOUNT OF SOME OF THE 
 
 PRINCIPAL DRAINS IN THE COUNTY OF YORK, 
 
 AND THE 
 
 AGRICULTURAL . LIIPROVEMENTS MADE BY THEM. 
 
 Extensive Drainages have of late years been executed in this county; 
 and little land, which is capable of improvement by such means, now 
 remains undrained. 
 
 The Drainage known by the name of the " Holderness Drainage," 
 lies chiefly adjoining to and on the east side of the river Hull. It 
 extends from north to south about eleven miles ; and contains, by 
 survey, 11211 acres. Previously to the execution of this undertaking, 
 much of the level was usually covered with water for above half of 
 the year ; and on some parts were extensive cars or lakes. 
 
 This Drainage, though still imperfect, has made a great improve- 
 ment in the adjoining lands, many of which from having been of 
 little or no value, are now let from fifteen to thirty shillings per acre ; 
 and this has been accomplished at an expense of somewhat less than 
 5/. per acre. 
 
 The Act of Parliament for executing this Drainage, was obtainea 
 in 1762 ; but it was not until the year 1775, that an assessment was 
 levied for 24,000Z. to carry it into effect. Since that time, cngineei-s 
 have been consulted on the further improvement of this Drainage ; 
 and the consent of the proprietors has been obtained for raising fur- 
 ther supplies; in consequence of which the sum of 30,000Z. has been 
 advanced, in order to complete the Avork. 
 
 The winter of 1763-4 being unusually wet, the water on the west- 
 side of the river Hull was higher than common, and occasioned a 
 breacii in the bank of the river, near the upper end of this level, by 
 which the whole of it was overflowed, together with part of the town- 
 ships of Stoneferry, Marfleet, Bilton, Preston, Southcoats, and Dry- 
 pool. The turnpike road between Hull and Headon, for nearly 
 M 4 
 
248 AN ACCOUNT OF DRAINS. PART VI. 
 
 four miles, stood from two to four feet deep in water, as did likewise 
 that leading from Whitecross to Beverly ; and persons going to 
 market were obliged to pass in boats ; but it is now believed that 
 these low-lands will never be flooded again, if the banks of the river 
 Hull can be preserved strong enough to resist the water of floods. 
 
 The Beverley and Barmston Drainage lies parallel to the Holderness 
 Drainage, but on the contrary side of the river Hull ; and extends 
 from Barmston, along the course of that river, nearly to Kingston- 
 upon-Hull, a distance of about twenty-four miles. 
 
 The Act of Parliament for executing this Drainage, was obtained 
 about the year 1792. It is divided into two parts, the upper or more 
 northern of which, containing 2,136 acres, extends from the township 
 of Foston to Barmston, where it has an outfall into the German 
 Ocean. The lower or southern part commences at the same point, 
 and has its outlet into the river Hull, at a place called Wincolmlee, 
 near the town of Hull. It contains 10,432 acres. 
 
 This Drainage is very effectual, as the water in the drains com- 
 monly stands three or four feet below the surface of the land. 
 
 The expenses hitherto incurred, amount to 135,000/. or about ten 
 pounds fifteen shillings per acre ; and the average improvement of tlie 
 land is about fourteen shillings per acre per annum ; but as the an- 
 nual expense of keeping the drains and works in repair, amounts to 
 between twelve and thirteen hundred pounds, the clear interest upon 
 the capital expended, is reduced to little more than 51. per cent, per 
 annum ; hence it appears not to have been a very profitable under- 
 taking to the proprietors. The public, however, has been benefited 
 by land being brought into cultivation, which before was of little 
 value. 
 
 Tlie Keyingham Drainage is divided into two parts, having distinct 
 outlets. The eastern part commences at Sand-le-Mere, aiid dis- 
 charges itself into the Humber, at a place called " Stone- Creek ;" 
 and is in length about fourteen miles. The other part lying a little 
 to the north and west of the former, commences at Aldborough, and 
 empties itself also into the Humber at a place called '* Headon 
 Haven ; " being in length about eleven miles. 
 
 The Act of Parliament for executing this Drainage was obtained 
 in 1802 ; and the level contains 5,505 acres of land. 
 
 The whole expenses of the act and the completion of the works, 
 amounted to 30,000/. or about five pounds nine shillings per acre. 
 Prior to the commencement of this Drainage, the Commissioners 
 made a valuation of the low grounds; and subsequently to the com- 
 pletion of it, they made an estimate of the improvements which had 
 been effected, amounting to 2,475/. per annum, which is rather more 
 than 8/. per cent. ; hence it appears that this has been a profitable 
 undertaking. 
 
 The Harford and Derwent Drainage contains upwards of 10,500 
 acres, of which nearly 4,500 are situated in the East Riding, and the 
 remainder in the North Riding. 
 
 The Act of Parliament for this Drainage, was obtained in 1800; 
 jtnd it has been completed at a total expense of 41,612/. which is 
 
PART VI. AN ACCOUXT OF DIl^VI^?. 249 
 
 nearly at the rate of 4Z. per acre ; but as the land has been benefited 
 to the amount of 17s. per acre ; after paying the annual asse-smotit, 
 the disbursements are paying an interest, to the proprietors of about 
 21 Z. per cent, per annum ; hence it appears that this has been a ver^- 
 advantageous undertaking. 
 
 Spalding -Moor and iraUhtg-Fcn Drainacjo. was completed about 
 thirty years ago ; and has greatly improved about 20,000 acres of 
 land which were before subject to be flooded. It commences at 
 ZMarket Weighton, and enters the Humhcr near a place called " Foss- 
 dyke Clough," situated upon the banks of that river. 
 
 ' This Drain also serves as a nnvirrable Canal; and turns out groatly 
 to the advantage of the land-owners and proprietors, not only in 
 draining the lands, but by facilitating the carriage of the produce and 
 manufactures of a large extent of country, which before were con- 
 veved at a very great expense by land-carriage. 
 
 Note. Those who desire to see more on this subject are referred to 
 the excellent Agricultural Survey of the East Riding of Yorkshire, 
 by H. E. Strickland, Esq. 
 
 AN ACCOUNT OF 
 
 AGRICULTURAL IMPROVEMENTS 
 
 MADE IN THE COUNTY OF LINCOLN, 
 BY 3IEANS OF DRAINS. 
 
 The extensive Drainages which have been executed in diflTereat parts 
 of the county of Lincoln are far too many to be enumerated here ; 
 but the following extract from Mr. Young's Agricultural Survey of 
 that County, will give the reader some idea of the great improve- 
 ments that have been made by them : " It is very probable tliat no 
 county in the kingdom has made equal exertions in this very im- 
 portant work of draining. The quantity of land thus added to the 
 kingdom has been great ; fens of water, mud, wild fowl, frogs, and 
 agues, have been converted to rich pasture and arable ground, v.'orth 
 from 20s. to 40s. an acre : health improved, morals corrected, and the 
 community enriched. These, when carried to such an extent, arc 
 great works ; and reflect the highest credit on the good, sense and 
 energy of the proprietors. 
 
 " Without going back to very remote periods, there cannot have 
 been less than 150,000 acres drained and improved, on an average, 
 from 5s. an acre to 25s. ; or a rental created of 150,000?. a year. But 
 suppose it only 100,000/. and that the profit has, on ?n average, been 
 received during the period of thirty years ; the rental has in that time 
 amounted to three millions, and the produce to nearly ten ; and 
 when with the views of a political arithmetician, we reflect on the 
 circulation that has attended this creation of wealth through industry ; 
 the number of people sujiportcd ; tlie consumption of manufactures ; 
 the shipping employed ; the taxes levied by the State ; and all the 
 classes of the community benefited ; the magnitude and importaiiot? 
 M 
 
250 elktngton's mode op draining. PAET \1, 
 
 of such works will be seen ; and the propriety well understood of 
 giving all imaginable encouragement and facility to their execution. 
 
 " Early in the days of republican France, decrees were issued for 
 draining marshes : I do not ask what progress has been made ; but I 
 would demand, if any drainages equal to this have been executed in 
 that kingdom during a century? From Bourdeaux to Bayonne, in 
 one of the finest climates in Europe, nearly all is marsh. What 
 Frenchman has been so actuated by the blessings of republican 
 security, as to lay out one louis on that or any other marsh or bog. 
 
 " Undertakings of this kind prove the reliance of the people on 
 the secure possession of what their industry creates ; and an English- 
 man must examine his native land with a cold heart, who does not 
 pray for the continuance of a system of legislation which has tended 
 so powerfully to adorn, improve, and cultivate the country, and to 
 diffuse prosperity and happiness through the whole of society." 
 
 MR. JOSEPH ELKINGTON'S 
 METHOD OF DRAINING LAND, 
 
 APPIiOVED OF, AND RECOiEIENDED BY THE BOARD OF 
 AGRICULTURE. 
 
 Several gentlemen have favoured the public with Treatises en the 
 subject of draining land ; but Mr. Elkington's Method appears to be 
 now almost universally adopted. Mr. Elkington was a Warwickshire 
 farmer, and practised the art of draining land with unexampled suc- 
 cess. The publication of this art was represented, to the Board of 
 Agriculture, to be one of the greatest means of promoting the im- 
 provement of this country that could be suggested. In consequence 
 of this, the House of Commons, on the 10th of June, 1795, voted an 
 Address, " That His Majesty v.-ould be graciously pleased to give 
 directions for issuing to Mr. Joseph Elkington, as an inducement to 
 discover his Mode of Draining, such sum as his Majesty, in his Avis- 
 dom, shall think proper, not exceeding the sum of lOCOZ. sterling." 
 
 Mr. Elkington's health being extremely precarious, there was a 
 risk that the public might lose the benefit of the knowledge he had 
 acquired, by the experience of above thirty years, in a species of im- 
 provement, which, in these kingdoms, ought to be considered as the 
 basis of every other. To prevent so unfortunate a circumstance, the 
 Board resolved to send Mr. John Johnstone, to visit, in company 
 with iMr. Elkington, the principal Drainages he was then executing ; 
 and to take Drawings and Sections of the same. 
 
 The result of that journey was draM-n up by Mr. Johnstone ; and 
 has been published under the title of " An Account of the Mode of 
 Draining Eand, according to the System practised by Mr. Joseph 
 Elkington." This Work has been well received by the public; and 
 has already been of great service in promoting agricultural improve- 
 ments. It not only points out the best mode of draining hoc/s and 
 marshes ; but also treats very extensively of the most approved me- 
 
PAET Yl. ELKINGTON'S MODE OF DK.VHimG. 251 
 
 thods of making under-drains ; hence it will be found of infinite ser- 
 vice to every person who is troubled with springy grounds. 
 
 The following extracts from some of the Agricultural Reports of 
 those counties in England, where Mr. Elkington executed the most 
 remarkable Drainages, will serve to shew the superiority of his method 
 over every other. 
 
 Countf of Wartcich, hy Mr. John Wedge. 
 " Draining is, without doubt, the first step towards the improve- 
 ment of all wet land. It has been practised with much success in 
 this county for several years; but more particularly so, since Mr. 
 Elkington, a farmer of this district, introduced a method of draining 
 boggy" lands, by making deep drains, and boring at the bottom or 
 sides of them, through the different under- strata, so as to tap the 
 springs, and thereby, in many instances, cure large tracts of land with 
 very iew drains." 
 
 County of Leicester, hy Mr. John Monk. 
 " The most capital improvements have been made under the direc- 
 tion of a Mr, Elkington, who is supposed to be the first in that line in 
 the world. After forming the drain, by beginning at the fell, and 
 working upwards, he makes use of a hover to find the spring, with 
 which he generally succeeds, which has a wonderful efifect in draining 
 the land." 
 
 County of Worcester, hy Mr. W. J. Pomeroy. 
 '• In speaking of under-drains, it may be thought right to mention, 
 that various experiments have been made at Ev/ell- Grange, the seat 
 of the Earl of Plymouth, and in that neighbourhood ; but that by 
 boring, after Mr. Elkington's method, deserves to be most particularly 
 noticed." 
 
 County of Derhy, hy Mr. Thomas Brown. 
 
 " Every other method seems to bend to that practised by Mr. 
 Elkington, whose practice is becoming every day more extensive, and 
 seems to me the most effectual of all others, for carrying off subter- 
 raneous waters. He lays a stone drain from three to six feet below 
 the surface, in such a direction as to cut the source of the spring, and 
 with such a declivity as to scour itself. Wherever he finds the source 
 of the spring below the level of his drain, he bores, and with such 
 judgment, that, to a stranger, his auger seems possessed of the virtue 
 of that rod with which Moses struck the rock ; for the water imme- 
 diately gushes out, and perhaps lays land that before was too wet to 
 carry a sheep, sufficiently dry to carry the heaviest ox. This method 
 is certainly effectual against springs." 
 
 Some readers may perhaps think that I have expatiated too much 
 on the excellency of Mr. Elkington's mode of Draining ; but I am of 
 opinion that this science, which tends so powerfully to promote the 
 general welfare, cannot be too much recommended. 
 
 SI 6 
 
252 
 
 COXIC SECTl-'V.-i 
 
 PART TTT. 
 
 PAET ^W. 
 CONIC SECTIONS AXL Till: IE SOLIDS. 
 
 DEFIXITIO-. 
 
 1. Conic Sections are plar-- li-ur- 
 formed by cutting a cone. 
 
 According to the different pc-iii-i!: . 
 the cutting plane, there will ariF--' ji\< 'lii 
 ferent figures or sections. 
 
 2. If the cutting plane pass til i'<M-ii '1. 
 vertex of the cone, and anypr.n "T ili 
 base, the section will be a tj-iavrh . 
 
 3. If a cone be cut into two p:ii.-i 
 plane parallel to the base, the ser-ii- 
 be a circle. 
 
 4. If a cone be cut by a plai- ]'; 
 through its two slant sides, in on -'i 
 direction, the section will be an > llij, 
 ellipsis. 
 
 5. If a cone be cut by a plan 
 parallel to either of its slant 
 section will be a parabola. 
 
PAKT VII. 
 
 /. \ II 
 
 ?OLrDS. 
 
 253 
 
 iio nvo parrs, 
 It i 11110(1, would 
 tlir srctioR is 
 
 i<<ilil\ 111- foimod h)'' 
 111'.- iircceding de!i- 
 , viz. :i frianglc, ;i 
 I :i lii/i)crholn ; but 
 noininnted CONIC 
 
 «'i-.-(' liiameter of an ellipse or :i 
 ' AK between the vertice.s ; and 
 a riglit line drawn from the A'er- 
 tiLiiiro into two eqnal parts, as 
 
 6. If a cone be eii 
 by a plane, Avhich, if <• 
 meet the opposite coii< 
 (Milled a hyperbola. 
 
 Nofr. All the figures that r:in 
 ctirting a rone, are mcntioncii 
 tiitious ; and are five in nunih 
 c/'/c/f, an c/lipse, a parabola. ; 
 the last tiirec onlv, arc usua'.lv 
 .SECTION'S. 
 
 7. The vertex oi^ ••(•nlo sretion, is 
 that point where the ciitdiig plane 
 meets the opposite sides of the cone, ^ " 
 as at A and B ; hen«'f^ tlie <'Uipse and the hyperbola have, 
 each of them, two von lees : hnv. tiio parabola has only one 
 vertex. 
 
 8. The axis or tr:iii~\ 
 hyperbola, is the distaiic 
 the axis of a parabola i- 
 tex, so as to divide tin' 
 
 9. The conjnriatt n'unndrr of an ellipse or hyperbola 
 is a right line dra%\ ii throMgh the centre of the figure, 
 perpendicularly to tli«' Transverse; hence the semi-con- 
 jugate diameter of a liypt'ibola is the perpendicular dis- 
 tance from the cenir.- of tlie figure, or middle of the 
 transverse diameter A !*>. t«» the vertices of the opposite 
 cx)nes. 
 
 10. An ordinate i- ;i right line drawn from any point 
 in the curve, perpend! <-iilarlv to the axis. 
 
 11. An absciss is that part of the axis which is con- 
 tained between the vertex and the ordinate. 
 
 12. The jmjrunctcr of any diameter is a third propor 
 tional to that diamelor aii'l its conjugate. 
 
 The parameter is -ouKtiiiios called the latifs rectum. 
 }o. ThefocTfS is ; 
 i.s equal to half the y 
 
 14. Tlie ellipse and li\ -porl-ola have each two /be/; but 
 ()]\c focf/s. 
 
 f// ipso id is a 
 «' r<'\<ihjtion of 
 >f iis diameters, 
 made about the 
 
 >iiit in the axis where the ordinate 
 
 IlllOtCI'. 
 
 the parabola has onl\ 
 lo. A spheroid (» 
 solid generated by t 
 an ellipse about one 
 If the revolution be 
 transverse diameter, the solid is called 
 Ti prolate spheroid ; but U" about the 
 conjugate diameter, ;:ii nhhitr spheroid 
 
 
254 
 
 COI^IC SECTIONS 
 
 PAUT vn. 
 
 16. A coJioid is a solid formed by the revolution of a 
 parabola, or hyperbola, about its axis ; and is accordingly 
 called parabolic, or hyperbolic. 
 
 The parabolic conoid is also called a 
 paraboloid ; and the hyperbolic conoid, 
 a hyperboloid. 
 
 17. An elliptic, a parabolic, or a 
 hyperbolic spindle, is a solid formed by 
 the revolution of a segment of an ellipse, 
 a parabola, or a hyperbola, about its 
 double ordinate, which remains fixed. 
 
 PROBLEM I. 
 
 In an ellipse, to find the transverse, or conjugate, or 
 ordinate, or absciss ; having the other three given. 
 
 CASE I. 
 
 When the transverse, conjugate, and absciss are given, to 
 Jind the ordinate. 
 
 RULE. 
 
 As the transverse diameter is to the conjugate, so is 
 the square root of the product of the tvv^o abscisses, to the 
 ordinate. 
 
 Note 1. An ellipse may be constructed by Problem 15, Part I. ; its area found by 
 Problem 21, Part II. ; and its circumference by Note 3, of the same Problem. 
 
 Also, the area of an elliptical-segment may be obtained by Problem 22, Part II. 
 
 2. An ellipse may also be constructed in the following manner : Having found 
 the foci F, t as directed in Problem 15, Part I., take a thread equal in length to the 
 transverse diameter AB, and fasten its ends, with two pins, in the points F, f ; then 
 stretch the thread to its greatest extent ; and by moving a pencil round, within the 
 thread, keeping it always tight, you will trace out the curve of the ellipse. 
 
 This construction is founded on the second property of the ellipse, given at the 
 end of this Problem. 
 
 EXAMPLES. 
 
 1. In the ellipse ABCD, the trans- 
 verse diameter AB is 60, the con- 
 jugate diameter CD 20, the absciss 
 BE 12, and the absciss AE 48 ; 
 what is the length of the ordinate ^ 
 EF? 
 
 ^5 60 : 20:: v'(48xl2) : 20-^60V(48x 12) ==4-^/576 
 =4-x24=8=EF, the ordinate required. 
 
 2. If the transverse diameter be 100, the conjugate 7o, 
 and the less absciss 20 ; what is the ordinate ? Ans. 30. 
 
 c F 
 
PART yn. AlsD TUEIK SOLIDS. 255 
 
 CASE II. 
 
 VPlieyi the transverse, conjugate, and ordinate are given, 
 to find the absciss. 
 
 RULE. 
 
 As the conjugate diameter is to the transverse, so is 
 the square root of the difference of the squares of the 
 semi-conjugate and ordinate, to the distance of that ordi- 
 nate from the centre. Then this distance being added to 
 the semi-transverse, will give the greater absciss ; but if 
 subtracted from it, the remainder will be the less absciss. 
 
 EXAMPLES. 
 
 i. The transverse diameter AB is 60, the conjugate 
 diameter CD 20, and the ordinate EF 8 ; what are the 
 abscisses AE and BE ? 
 
 As 20 : 60 :: v (lO'^- 8^0 : 60 -- 20 v(io-'-8)-^= 3 
 
 ,/ (100-64) =3 -v/ 36 = 3 X 6 = 18 =: EG-, the distance 
 from the centre. 
 
 Then 30 + 18=48 = AE, the greater absciss; and SO 
 — 18 = 12 = BE, the less absciss. 
 
 2. What are the two abscisses to the ordinate 30, the 
 diameters being 100 and 75 ? Aiis. 20 and 80. 
 
 CASE III. 
 
 When the conjugate, ordinate, and absciss are given, to 
 find the transverse. 
 
 RULE. 
 
 From the square of the semi-conjugate subtract the 
 square of the ordinate; and extract the square root of 
 the remainder. Add this root to the semi-conjugate, if 
 the less absciss be given ; but when the greater absciss is 
 given, subtract this root from the semi-conjugate ; and 
 reserve the sum or dilference. Then say, as the square 
 of the ordinate is to the rectangle of the conjugate and 
 absciss, so is the reserved sum or difference to the trans- 
 verse diameter. 
 
 EX^lMPLES. 
 
 1. The conjugate diameter CD is 20, the ordinate EF 8, 
 and the absciss BE 12 ; required the transverse diameter 
 AB. 
 
Job COXIC SECTION'S TART \U. 
 
 Here ^/(10'^-8-') = v/(100 -64)= \/36=6, and 10 + 
 6= 16 ; then, as 82 : 20 x 12 : : 16 : (20 x 12 x 16) -- 8-' = 
 (20 X 12 X 2)--8 = (5 X 12 X 2)--2=5 x 12r=60, the trans- 
 verse diameter required. 
 
 2. If the conjugate dicameter be 75, the ordinate 30, 
 and the greater absciss 80 ; what is the transverse ? 
 
 Ans. 100. 
 
 CASE IV. 
 
 J^Hien the transverse, ordinate, and absciss are given, to 
 find the conjugate. 
 
 RULE. 
 
 As the square root of the product of the two abscisses 
 is to the ordinate, so is the transverse diameter to the 
 conjugate. 
 
 'Sole. The transverse diameter may be found from the conjugate, in the same 
 maniier ; using tlie abscisses of the conjugate, and their ordinate perpendicular to 
 the conjugate. 
 
 EXAMPLES. 
 
 1. The transverse diameter AB is 60, the ordinate 
 EF 8, and the absciss BE 12 ; required the conjugate 
 diameter CD. 
 
 j^ere60 — 12=48 =AE, the greater absciss, and ^/(48 
 xl2)=,/576=24; then, as 24, : 8::60: (8x60)--24r= 
 60 + 3=: 20, the conjugate diameter required. 
 
 2. If the transverse diameter be 100, the ordinate 30, 
 and the absciss 20 ; what is the conjugate diameter ? 
 
 Ans. 75. 
 
 PROPERTIES OF THE ELLIPSE. 
 
 1. The square of the distance of the focus from the centre, is equal to the differ- 
 ence of the square of the semi-axes. 
 
 2. The sum of two lines drawn from the foci, to meet at any point at tlie curve, 
 is equal to the transverse axis. 
 
 3. The parameter, or double ordinate at the focus, is equal to the square of the 
 conjugate diameter divided by the transverse. 
 
 4. The rectangles of the segments of any diameter, are as the squares of their or- 
 dinates. 
 
 5. All the parallelograms circumscribed about an ellipse, are equal to one an- 
 other, and each equal to the rectangles of the two axes. 
 
 fi. An ellipse is to the rectangle of the two axes, as any circle is to the square of 
 Its diameter. 
 
 7. The areas of ellipses are to one another, as the rectangles of their transverse 
 and conjugate axes. 
 
 8. As the transverse axis of an ellipse is to the conjugate, so is the area of a circle 
 whose diameter is the transverse, to the area of the ellipse. 
 
 9. The square of the transverse diameter is to the square of the conjugate, as tlie 
 rectangle of the abscissa is to the square of the ordinate. 
 
 10. The area of an ellipse is a geometrical mean proportional number between 
 the area of two circles whose diameters are equal to the transverse and conjugate 
 diameters of the ellipse. 
 
PAKT vn. 
 
 AND THEIR SOLIDS. 
 
 257 
 
 PROBLEM n. 
 
 To describe a parabola, the absciss CD, ajid the ordinate 
 BD being given. 
 
 Construction. Bisect the ordinate 
 BC in «; join Ca, and draw a n per- v\^'- 
 
 pendiciilar to it, meeting CD pro- /fi^pX 
 
 duced in n. .»•)/-. -.I;.. ..V,. 
 
 Make CE and CF each equal to V./^./.Ll.lV 
 D;^ ; and F will be the focus of the . / ~" '' j \ " \ , 
 parabola. *j •,"" j V'"." V 
 
 Take any number of points, r, r, l_ » I \ \ 
 
 &c. in the axis, through which draw a <.l- - a w 
 
 the indefinite, double ordinates, srs, 
 srs, &c. perpendicularly to CD. 
 
 With the radii EF, Er, &c. and the centre F, describe 
 arcs cutting the corresponding ordinates in the points, s, s, 
 &c. ; then draw the curve through all the points of inter- 
 section, and ACB will be the parabola required. 
 
 Note. Various methods of constructing parabolas and liyperbolas, may be seen in 
 Emerson's Conic Sections. 
 
 PROBLEM IIL 
 To find any parabolic absciss or ordinate. 
 
 RULE. 
 
 As any absciss is to the square of its ordinate, so is any 
 other absciss to the square of its ordinate. Or, as the 
 square root of any absciss is to its ordinate, so is the square 
 root of any other absciss to its ordinate ; and vice versa. 
 
 1. 
 
 EXAMPLES. 
 
 The absciss CE is 9, and its 
 
 ordinate DE 6 ; what is the ordinate 
 AG, whose absciss CG- is 16? 
 
 As 9 : 62 :: 16 : (36xl6)--9 = 
 576-i-9=:64 ; and^/Q\=S=AO, the 
 ordinate required. 
 
 Or, «5,/9 : 6:: yi6 : (6x^/i6) 
 
 -HV9=(6x4)-f-3 = 24-H3 = 8 = 
 AG, the same as before. 
 
258 CONIC SECTIONS PAET vn. 
 
 2. The ordinate AG- is 8, tlie ordinate DE 6, and the 
 absciss CG 16; what is the absciss CE? 
 
 As 82: 16::62: (16x6-)--82=.(16x36)-f-64=:o76-- 
 64=9=:CE, the absciss required. 
 
 Or, as 8: V16::6 : (a/16x6)-^8==(4 x6) --8=24 
 ►1-8=3 ; and 3 x 3=:9=CE, the same as before. 
 
 3. If an absciss be 12, and its ordinate 15 ; required the 
 ordinate whose absciss is 27. A?is. 22^. 
 
 4. If an absciss be 18, and its ordinate 12, what is the 
 absciss whose ordinate is 16? Aiis. 32. 
 
 PROBLEM IV. 
 
 To find the length of the arc of a parabola, cut off by a 
 double ordinate. 
 
 RULE. 
 
 To the square of the ordinate add ^ of the square of 
 the absciss ; multiply the square root of the sum bj 2, and 
 the product wiU be the length of the curve required. 
 
 Note. This rule will not hold good, when the absciss is greater tlian half the or 
 dinate ; that is, when the ordinate falls below the focus. 
 
 EXAMPLES. 
 
 1. If the absciss CE be 6, and the ordinate DE 12 ; what 
 is the length of the curve DCF ? 
 
 Here 12^ + 62x4 =144 + (36 x4)~3 = 144 + 144--3 = 
 144 + 48 = 192; and ^ 192 x 2 = 13.8564 x2 = 27.7128, 
 the length of the curve required. 
 
 2. "Wliat is the length of a parabolic curve, wliose 
 absciss is 7, and ordinate 15? Ans. 34.0782. 
 
 PROBLEM V. 
 
 To find the area of a parabola, its base and height being 
 
 given. 
 
 RULE. 
 
 Multiply the base by the height, p. 
 
 and. -3- of the product will be the area. 
 
 EXATvEPLES. 
 
 1. TVliat is the area of the para- 
 bola, ABC, whose height BD is 18, 
 and base or double ordinate AB 24 ? 
 
 ./ 
 
 ^p 
 
 \l.- 
 
 / 
 
 1 >- 
 
 1 
 
 \ 
 
 / 
 
 1 
 
 \ 
 
 
 JJ 
 
 
 
PAKTVn. AND THEIR SOLIDS. 259 
 
 Here 24 x 18 x ^. = (24 x 18 x 2)-r-3= 8 x 18 x 2= 288, 
 the area required. 
 
 2. The absciss is 24, and the double ordinate 72 ; what 
 is the area of the parabola ? Ans. 1152. 
 
 PROBLEM VI. 
 
 To find the area of a •parabolic frustum 
 
 EUI.E. 
 
 Divide the difference of the cubes of the two ends of the 
 frustum, by the difference of their squares ; multiply the 
 quotient by ^ of the altitude, and the product will be the 
 area • 
 
 EXAMPLES. 
 
 1. Wliat is the area of a parabolic frustum AEFC, the 
 end AC being 24, EF 16, and the altitude DG 10? 
 
 Here (243-163) -- (242- iq2) ==( 13824 -4096) -r- (576 
 -256) = 9728 --320=30.4; and 30.4 x (10x2)-- 3 = 
 (30.4 X 10x2)^3=608-4-3=2022, the area required. 
 
 2. Required the area of a parabolic frustum, the greater 
 end of which is 20, the less end 12, and the height 6. 
 
 Ans. 98. 
 
 PROPEKTIES OF TPIE PARABOLA. 
 
 1. The distance of the focus from the vertex is equal to the square of any ordi- 
 nate divided by 4 times its absciss. 
 
 2. The parameter, latus rectum, or double ordinate at the focus, is equal to the 
 square of .any ordinate divided by its absciss. 
 
 3. The dis'tance between the focus and tlie vertex is equal to \ of the parameter, 
 or i of the ordinate at the focus. 
 
 4". As the parameter is to the sum of any two ordinates, so is the difference of 
 those ordinates to the difference of their abscisses. 
 
 5. All parabolas are similar to each other. 
 
 G. The area of a parabola is equal to \ of the area of the circumscribing parallel- 
 ogram. 
 
 7. The square of any two ordinates are to each other as their corresponding ab- 
 scisses. 
 
 8. Abscisses are to each other as the squares of their ordinates. 
 
 PROBLEM Vn. 
 
 To co7istruct a hyperbola, having the transverse and con- 
 jugate diameter given. 
 
 Construction. Draw CD equal to the conjugate dia- 
 meter ; and bisect it perpendicularly with a transverse 
 diameter AB. From E, the middle of the transverse 
 
260 
 
 CONIC SECTIONS 
 
 PAET vn 
 
 /H 
 
 diameter, witli the radius EC or 
 ED, describe the circle C/DF 
 cutting AB produced in F and/, 
 which points will be the foci of the 
 hyperbola. In AE produced, take 
 any number of points m, on, &c. ; 
 and from F and/, as centres, with 
 tlie distances Bw?, Aw, as radii, de- 
 scribe arcs cutting each other in the 
 points s, s, &c. Through these points 
 draw the curve sJjs, and it will be 
 the hyperbola required. 
 
 No/i'. If the straight lines EC, ED, be drawn from the point E, through the ex- 
 tremities of the conjugate axis CD, they will be the asymptotes of the hyperbola ; 
 the proj)erty of which is to approach continually to the curve, but not to meet 
 it, imtil they be infinitely produced. (See Emersoti's Conic Sectio7is, Bool; II. 
 Prob.lb.) 
 
 PROBLEM A^in. 
 
 In a hijperboJa, to find the transverse, or conjugate, or 
 ordinate, or absciss. 
 
 CASE I. 
 
 The transverse and conjugate diameters, and the two 
 abscisses being given, to find the ordinate. 
 
 KULE. 
 
 As the transverse diameter is to the conjugate, so is 
 the square root of the product of the two abscisses to the 
 ordinate. 
 
 Note. The less absciss added to the transverse diameter, will give the greater 
 absciss. 
 
 EXAMPLES. 
 
 1. In the hyperbola ABC, the 
 transverse diameter is 30, the con- 
 jugate 18, and the absciss BD 10; 
 what is the ordinate AD ? 
 
 As 30: 18 ::,/ (40xio) : is -- 
 
 30 v/ (40 X 10) = (3 X v/ 400) -- o = 
 
 (3 X 20)-7-5=:3 X 4=12, the ordinate required. 
 
 2. The transverse diameter is 48, the conjugate 42, and 
 the less absciss 16 ; what is the ordinate? Ans. 28 
 
PART VII. AND THEIR SOLIDS. 261 
 
 CASE ir. 
 
 The transverse and conjugate diameters, and an ordinate 
 being given, to Jind the abscisses. 
 
 RULE. 
 
 To the square of half the conjugate, add the square of 
 the ordinate ; and extract the square root of the sum. 
 Then say, as the conjugate diameter is to the transverse, 
 so is the said square root to half the sum of the abscisses, 
 or the distance between the ordinate and the centre. 
 
 Then this distance being added to half the transverse 
 diameter, will give the greater absciss ; and their diiier- 
 ence will be the less absciss. 
 
 EXAMPLES. 
 
 1. The transverse diameter is 30, the conjugate 18, and 
 the ordinate 12 ; what are the two abscisses? 
 
 Here ,/(92 + 122)r= ,/(81 + 144)= ^/22o = 15 ; the7i, a^ 
 18 : 30:: 15 : (30xlo) -- ]8 = (5xl5)---3=5x5=:2o, 
 half the sum of the abscisses; hence ^ 25 + 15=40, the 
 greater absciss ; and 26 — 16=^10, the less absciss. 
 
 2. If the transverse diameter be 48, the conjugate 42, 
 and the ordinate 28 ; what are the two abscisses ? 
 
 Ans. 64 aiid 16. 
 
 CASE III. 
 
 The transverse diameter, the two abscisses, and the ordi- 
 nate being given, to find the conjugate. 
 
 RULE. 
 
 As the square root of the product of the two abscisses 
 is to the ordinate, so is the transverse diameter to the 
 conjugate. 
 
 EXAMPLES. 
 
 1. The transverse diameter is 30, the ordinate is 12, 
 and the two abscisses are 40 and 10 ; required the con- 
 jugate diameter. 
 
 ^erey(40xl0)= V400=20, the square root of the 
 product of the two abscisses; then, C5 20 : 12: :30 : (12 
 
262 CONIC SECTIONS PART VII. 
 
 X 30) -4- 20= (12 X 6) -f- 4 = 3x6=: 18, the conjugate re- 
 
 2. The transverse diameter is 48, the ordinate 28, and 
 the abscisses are 64 and 16 ; required the conjugate 
 diJimeter. Ans. 42. 
 
 CASE IV. 
 
 The conjugate diameter^ the ordinate^ and the two abscisses 
 being given., to find the transverse. 
 
 RULE. 
 
 To the square of half the conjugate add the square of 
 the ordinate, and extract the square root of the sum. ^ To 
 this root add half the conjugate, when the less absciss is 
 used, but subtract it when the greater absciss is used ; 
 and reserve the sum or difference. Then say, as the 
 square of the ordinate is to the product of the absciss and 
 conjugate, so is the sum or difference, above found, to the 
 transverse required. 
 
 EXAMPLES. 
 
 1. The conjugate diameter is 18, the ordinate is 12, 
 and the less absciss 10 ; required the transverse diameter. 
 
 iJere -v/ (92 + 122)= ^/(8H- 144)= v225 = 15; and 15 
 + 9=24. Also, 12x12 = 144, the square of the ordinate; 
 and 18x10=180, the product of the absciss and con- 
 jugate; then, as 144 : 180::24 : (180x24)--144=(15 x 
 24)-^ 12= 15 X 2 = 30, the transverse required. 
 
 2. The conjugate diameter is 42, the greater absciss is 
 64, and the ordinate 28 ; vf hat is the transverse diameter ? 
 
 Ans. 48. 
 
 PROBLEM IX. 
 To fond the length of a hyperbolic curve. 
 
 KULE. 
 
 To 21 times the square of the conjugate, add 9 times 
 the square of the transverse ; and to the said 21 times the 
 square of the conjugate, add 19 times the square of the 
 transverse ; and multiply each sum by the absciss. To 
 
PAET \^I. AND THEIR SOLIDS. 263 
 
 each of these two products add 15 times the product of 
 the transverse multiplied by the square of the conjugate ; 
 then say, as the less of these sums is to the greater, so is 
 the double ordinate to the length of the curve, nearly, 
 
 EXAJEPLES. 
 
 1. The transverse diameter of the 
 h}^erbola ABC, is 30, the conjugate 
 18, the absciss BD 10, and the double 
 ordinate AC 24 ; what is the length 
 of the curve ABC ? 
 
 Here (18- X 21) + (30- x 9) r= (324 
 x21) + (900x9)-6804-h8100r=14904r=21 times the 
 square of the conjugate, added to 9 times the square of 
 the transverse; and 6804 + (900 x 19) =6804+ 17100== 
 23904 = 21 times the square of the conjugate, added to 19 
 times the square of (he transverse; then 14904x10= 
 149040, and 23904 x 10=239040. 
 
 Again, 324 x 30 x 1 " = 145800=15 times the product 
 of the transverse midtiplied by the square of the conjugate; 
 and 149040+145800=294840, and 239040+145800= 
 384840; then, as 294840 : 384840:: 24 : 31.326007, the 
 length of the curve required. 
 
 2. The transverse diameter of a hyperbola is 45, the 
 conjugate 27, tl\e absciss 15, and the double ordinate 36; 
 what is the leno-th of the curve ? Ans. 46.989. 
 
 PROBLEM X. 
 
 To find the area of a hyperbola, the transverse, conjugate, 
 and absciss being given. 
 
 RULE. 
 
 Multiply the transverse by the absciss ; to the product 
 add 4 of ^lie square of the absciss ; and multiply the 
 square root of the sum by 21. Add the product, last 
 found, to 4 times the square root of the product of the 
 transverse and absciss ; and divide the sum by 75. Di- 
 vide 4 times the product of the conjugate and absciss hj 
 the transverse ; then this quotient being multiplied by 
 the former, will give the area, nearly. 
 
264 CONIC 5i:t Tioxs part ^'Jl. 
 
 EXA>ll"Li:S. 
 
 1. Required the area of a li^perbola, wliose traiisverae 
 diameter is 30, the conjugate 18, and the absciss 10. 
 
 Here 21 V(30 x 10 -r 10- x ••) = 21 ,/ (300 + 100 x i) 
 = 21 .x/(300 f 500 -^ 7) = 21 ^^(300 + 71.428571) = 
 21,/ 371.428571 =21x19.272=404.712. 
 
 Again, (4 V (30 x 10) -1- 404.712 ) -- 75 = (4 ^/ 300 -^ 
 404.712) -^75=(4x 17.3205 -^ 404.712) -75= (69.282 -f- 
 404.712)-r-75=473.994--75=6.ai99. 
 
 Lastit/, (18 X 10 X 4) -- 30 x (S.3I99 = (18 x 4) -^ 3 x 
 6.3199=6x4x6.3199=24x6.3199 = 151.6776, the area 
 required. 
 
 2. The transverse diamet(;'r of a liyperbola is 50, tlie 
 coijju2:ate 30, and the absciss 25 : what is its area? 
 
 Ans. 805.0902. 
 
 PKOPEKTIES OP TIU: IllPEllBOLA. 
 
 1. 'Ihe square root of the sum of the squares ol' the seiai-axes, is equal to the 
 Jistance of the focus from the middle of the transverse axis. 
 
 2. If half the transverse axis be subtracted from the distance between the focus 
 and the middle of the said axis, the remainder will be the distance of the focus 
 from the vertex. 
 
 3. The difference of two lines drawn from the twu :oci to any point in the curve, 
 is equal to the transverse axis. 
 
 4. The parameter, latus rectum, or double ordiiure at the focus, is equal to tlve 
 square of the conjugate divided by the transverse. 
 
 Note. Those who desire to obtain a comj.Iet<- inowledge of the properties of 
 Conic Sections, are referred to the works of Emerson, Hutton, and Simson, on 
 tliat subject. 
 
 PROBLlv.\i; XL. 
 
 To find the solldJ!;i <•/ n spheroid. 
 
 IIMLW. 
 
 Multiply the square of the re\ ohhig axis by the fixed 
 axis, and the product thence arising by .5236 ; and tJie 
 last product will be the solidity. 
 
 Sole 1. 'J'he multiplier .5236 is * of 3.141?;. 
 
 •2. A semi-s))heroid is equal to 5 of a cy,:;iM.;-, >i to double a cone of the same 
 base and altitude. 
 
 EXAIMi'J.E^. 
 
 1. What is the solidity of th.e pro- 
 late spheroid ABCD, whose trans- 
 verse or fixed axis AC is 50, and the 
 conjugate or revolvina- axisBD 30? 
 
 ilere 30-' x 50 X .5236 = 900 X 50 
 X .5236=45000 x .5236=23562, ihc 
 soVidity req a Ired. 
 
PART VII. AND THEIR SOLIDS. 265 
 
 2. What is the content of an oblate spheroid, whose 
 axes are 50 and 30 ? Ans. 39270. 
 
 3. Eequired the solidity of a prolate spheroid, whose 
 axes are 45 and 95. Ans. 100727.55. 
 
 PROBLEM Xn. 
 
 To find the solidity of the segment of a spheroid. 
 
 CASE I. 
 JVIien the base is circular, or parallel to the revolving axis. 
 
 RULE. 
 
 From triple the fixed axe, take double the height of 
 the segment ; multiply the cliiference by the square of the 
 height, and the product thence arising by .5236 ; then 
 say, as the square of the fixed axe is to the square of the 
 revolving axe, so is the last product to the solidity re- 
 quired. 
 
 EXAMPLES. 
 
 1. Required the solidity of the 
 segment ABC of a prolate spheroid, 
 the height Bwi being 5, the fixed 
 axe, BD 50, and the revolving axe / ' 
 EF30. / I \ 
 
 Here (50 x 3) — (5 X 2) x 5^ x [ 1 J 
 
 .5236 = (150- 10) X 25 X .5236= e d f 
 
 140 X 25 X .5236 = 3500 x .5236 = 1832.6 ; then, as 2500 : 
 900 :: 1832.6 : 659.736, the soliditg required. 
 
 2. The axes of a prolate spheroid are 100 and 60 ; what 
 is the solidity of that segment whose height is 10, and its 
 base parallel to the revolving axe .^ Ans. 5277.888. 
 
 3. The axes of an oblate spheroid are 50 and 30 ; what 
 is the solidity of that segment whose height is 6, and its 
 base parallel to the revolving axe ? Ans. 4084.08. 
 
 CASE IL 
 
 Wlien the base is elliptical, or perpendicular to the re- 
 volving axe. 
 
 RULE. 
 
 From triple the revolving axe, take double the height 
 of the segment \ multiply tlie difference by the squai-e of 
 
 N 
 
266 CONIC SECTIONS PART VII. 
 
 the heiglit, and the product thence arising by .5236 ; then 
 say, as the revolving axe is to the fixed axe, so is the last 
 product to the solidity required. 
 
 EXASIPLES. 
 
 1. What is the solidity of the n 
 segment ABC of a prolate spheroid, Ay^^^^^^^^^ c 
 
 the height Bm being 6, the fixed /''^^— — ^ ^ '■•.. 
 
 axe EF 50, and the revolving axe ^i; "|' ;'^' 
 
 BD30? V^ i ,./ 
 
 Here (30 X 3) -(6 X 2) x 6^ x ~^~"ir" 
 
 ,5236 = (90-12) X 36 X .5236 = 78 x 36 x .5236 = 2808 
 x .5236 = 1470.2688 ; then, as 30 : 50 : : 1470.2688 : 
 2450.448, the solidity required. 
 
 2. The axes of an oblate spheroid are 50 and 30 ; what 
 is the solidity of that segment whose height is 6, and its 
 base perpendicular to the revolving axe ? 
 
 Ans. 1560.74688, 
 
 PROBLEM XIIL 
 
 To find the content of the middle frustum of a spheroid, 
 
 CASE I. 
 
 When the ends are circidar, or jjcirallel to the revolving 
 
 axe. 
 
 KULE, 
 
 To twice the square of the middle diameter, add the 
 square of the diameter of one end ; multiply the sum by 
 the length of the frustum, and the product thence arising 
 by ,26 f8, for the solidity. 
 
 Note 1. A cask in the form of the middle frustum of a prolate spheroid is by 
 Gangers called a cask of theirs/ varirty ,- hence this Rule is useful in cask gauging. 
 
 2. The old ale gallon contains 282, the old wine gallon 231, and the old Win- 
 chester bushel 2150.4 cubic inches The new imperial galloii contains 277.274, and 
 
 the new imperial bushel 2218.192 cubic inches. {Sec the New Imperial Tables, 
 Fart VIII.) 
 
 3. If the content of anj' vessel, in cubic inches, be dividedby 277.274, and 2218.192; 
 the respective quotients will be the content in imperial gallons and bushels. {See 
 Gauging, P rob. I., Part VIII.) 
 
 EXAJMPLES. 
 
 1. What is the solidity of the middle frustum EGHF 
 of a prolate spheroid ; the middle diameter CD being 30, 
 
PAKT YII. 
 
 AND THEIR SOLIDS. 
 
 267 
 
 the diameter of eacli end EF or GH 
 18, and the length AB 40 ? 
 
 Here ( (30-' X 2) + 18-) X 40 X 
 .2618 = ( (900 X 2) + 324) x 40 x 
 .2618 = (1800 + 324) x 40 x .2618 = 
 2124 X 40 x .2618 = 84960 x .2618= 
 22242.528, the solidity required. 
 
 2. "Wliat is the solidity of the middle frustum of an 
 oblate spheroid ; the diameter of each end being 40, the 
 middle diameter 50, and the length of the frustum 18 ? 
 
 Atis. 31101.84. 
 
 3. The length of a spheroidal cask is 30, the bung dia- 
 meter 24, and the head diameter 18 inches; what is its 
 content in imperial gallons ? Ans. 41.088 gallons. 
 
 CASE 11. 
 
 JVheii the ends are elliptical, or 2^(^rpe7idicular to the 
 revolfing axe. 
 
 EULE. 
 
 To twice tlie product of the transverse and conjugate 
 diameters of the middle section, and the product of the 
 transverse and conjugate of one end; multiply the sum 
 by the height of the frustum, and the product thence 
 arising by .2618, for the solidity. 
 
 exa:mples. 
 
 1. In the middle frustum EFGH, 
 of an oblate spheroid, the diameters 
 of the middle section AB are 50 
 and 30 ; those of the end EF or GH 
 40 and 24; and its height 7nn 18; 
 what is the solidity ? 
 
 Here (50 x 30 x 2) + (40 x 24) x 18 x .2618 = (3000 + 
 960) X 18 X. 2618 = 3960x18 x. 2618= 71280 x. 2618 = 
 18661.104, the solidity required. 
 
 2. What is the solidity of the middle frustum of a pro- 
 late spheroid ; the diameters of the middle section being 
 100 and 60, those of each end 80 and 48, and its height 
 36? Ans. 149288.832. 
 
268 CONIC SECTIONS PART ^'U. 
 
 PROBLEM XIY. 
 To find the solidity of an elliptic spindle. 
 
 RULE. 
 
 From 3 times the square of the middle or greatest dia- 
 meter take 4 times the square of the diameter at i of the 
 length, or equally distant between the middle and one 
 end ; and from 4 times this last diameter take 3 times the 
 middle diameter ; and \ of the quotient arising from 
 dividing the former difference bj the latter, will give the 
 central distance. 
 
 Find the axes of the ellipse ; and also the area of the 
 generating segment. 
 
 Divide 3 times this area by the length of the spindle ; 
 from the quotient subtract the middle diameter ; and 
 multiply the remainder by 4 times the central distance, 
 before found. 
 
 Subtract this product from the square of the middle 
 diameter; multiply the remainder by the length of the 
 spindle, and the product thus obtained by .5236, for the 
 solidity. 
 
 EXA3IPLES. 
 
 1. What is the solidity of the ^ 
 
 elliptic spindle A CBD A, the length --^' iV^"^^^j'> 
 
 AB being 40, the middle diameter . "^-^I^^^^-- 
 
 CD 12, and the diameter EF, at i ^.L ^-_- ^l 
 
 of the length, 9.49546 ? "x,^ i 
 
 Here (12^ x 3-9.49546- x 4) -- "" i""""" 
 
 (9.49546 x4-12x3)x-i-=: (432 - 
 
 360.0546) -T- (37.98184-36) x ^=. (71.3454--1.98184) x 
 
 ■1=36 x^=9, the central distance 0G-. 
 
 Also, {9 + 12^2) x2 = {9-^Q)x2 = lox2^m, the con- 
 jugate diameter CH. 
 
 By Problem 1, Case 4, Part VIL, 2ve have 30-6=24 
 =GH, the greater absciss ; and a/(24x6)— ^7144= 12; 
 then, as 12 : 20 (AG) : 1 30 (CH) : 50, the transverse 
 diameter KL. 
 
 Again, by Problem 22, Part IL, loe have 6-=-30=.2, 
 the tabular height ; and the corresponding Area Seg, is 
 .111823;Me;i.lll823x 50x30-^5.59115x30 = 167.7345, 
 the area of the generating segment ACB. 
 
 Now, (167.7345 x 3-f-40- 12) x 9 x 4=:(503.2035-t- 40 
 
PAKT Tn. 
 
 AND TIIEIll SOLIDS. 
 
 2G9 
 
 - 12) X 36 = (12.5800875-12) x 36 = .5800875 x 36 
 =20.88315 ; then (12-' -20.88315) x 40 x.5236= (144 
 
 — 20.88315) X 40 X .5236 = 123.11685 x 40 x .5236 = 
 4928.674 X. 5236=2578.5593064, the solidity required. 
 
 2. The length of an elliptic spindle is 80, the middle 
 diameter 24, and the diameter at \ of the length 18.99094; 
 what is the solidity ? Ans. 20628.4744512. 
 
 PROBLEM XY. 
 
 To find the soliditij of a 'parabolic conoid. 
 
 RULE. 
 
 Multiply the square of the diameter of the base by the 
 altitude, and the product thence arising by .3927, (one 
 half of .7854 ;) and the last product wiU be the solidity. 
 
 "Note 1. This Ilule will give the solidity of any segment of a paraboloid, wliose 
 base is a circle. 
 2. A paraboloid is equal to half of its circumscribing cylinder. 
 
 EXA3IPLES. 
 
 1. "VYhat is the solidity of the para- 
 boloid ADB ; the height Dm being 42, 
 and the diameter of the base AB 24 ? 
 
 Here 24- x 42 x .3927 = 576 x 42 x 
 .3927 = 24192 X. 3927=9500.1984, the 
 solidity required. 
 
 2. Wliat is the solidity of a para- 
 boloid, the height of which is 30, and 
 the diameter of its base 50 ? 
 
 Ans. 29452.5. 
 
 3. Required the solidity of the segment of a paraboloid, 
 the height of which is 9, and the diameter of its base 12. 
 
 Ans. 508.9392. 
 
 PROBLEM XYL 
 
 To find the solidity of the frustum of a paraboloid, tohen 
 its ends are perpendicular to the axis of the solid. 
 
 KULE. 
 
 Multiply the sum of the squares of the diameters of tho 
 two ends by the height of the frustum, and the product 
 N 3 
 
270 
 
 CONIC SECTIONS 
 
 PAUT vn. 
 
 and tlie last product will be the 
 
 thus obtained by .3927 
 solidity. 
 
 Xofe. If the lower frustums of two equal paraboloids be joined together at their 
 preater ends, they will form a figure which, by Gaugers, is called a cask of the 
 third variety ; hence this Rule may be applied in Gauging. 
 
 EXAMPLES. 
 
 / 
 
 1. "Wliat is the content of the 
 parabolic frustum ABCD ; the dia- 
 meter AB of the greater end being 
 30, that of the less end DC 24, and 
 the height mn9? 
 
 Here (30^ + 242) x9 x .3927 = (900 a^ 
 + 576) X 9 X .3927 = 1476 x 9 x. 3927 
 = 13284 X .3927 = 5216.6268, the content of the frustum. 
 
 2. ^Vhat is the solidity of the frustum of a paraboloid ; 
 the diameter of the greater end being 80, that of the less 
 end 40, and the altitude 45 ? Airs. 141372. 
 
 3. The length of a cask, in the form of tAvo equal 
 frustums of a parabolic conoid, is 30, the bung diameter 
 24, and the head diameter 18 inches; what is its content 
 in imperial gallons ? Aijs. 38.2397 gallons. 
 
 PROBLEM XVII. 
 To find the solidity of a parabolic spindle. 
 
 RULE. 
 
 Multiply the square of the middle diameter by the 
 length of the spindle, and the product thence arising by 
 .41888, and it will give the solidity. 
 
 2^ote. A parabolic spindle is equal to -^ of its circumscribing cylinder ; hence we 
 obtain the multiplier .41888, which is -^ of .7854. 
 
 EXA5IPLES. 
 
 1. The length AB of the parabolic 
 spindle ACBD i-s 81, and the middle 
 diameter CD 27; what is its solidity? 
 
 Here 21"- X 81 X .41888 = 729 X 
 81 X .41888 = 59049 x .41888 = 
 24734.44512, the solidity required. 
 
 2. What is the solidity of a parabolic spindle, wliose 
 length is 20, and middle diameter 8 ? Ans. 536.1664. 
 
PART \'II. Am) THEIR SOLIDS. 271 
 
 PROBLEM XVIII. 
 
 To find the solidity of the middle frustum of a parabolic 
 spindle. 
 
 RULE. 
 
 Add 8 times the square of the middle diameter, 3 
 times the square of the end diameter, and 4 times the pro- 
 duct of those diameters into one sum ; then this sum 
 being multiplied by the length, and the product thence 
 arising by .05236, (one-fifteenth of .7854) will give the 
 solidity. 
 
 Note. A cjisk in the form of the middle frustum of a parabolic spindle, is called a 
 cask of the second variety ,• and is the most common of any of the varieties ; hence 
 the above Rule. is very useful in Cask Gaujjing. 
 
 EXAMPLES. 
 
 1. "What is the solidity of the 
 middle frustum ABCD, of a para- 
 bolic spindle ; the diameter of the 
 end AB being 12, the middle dia- »«! 
 meter EF 16, and the length m?i 
 20? 
 
 Here (16-^ x 8) + (12-' x 3) + (16 X 
 12x4) = 256 X 8 + 144x3 + 192x4=2048+432 + 768 
 = 3248 ; a7id 3248 x 20 x .05236 = 64960 x .05236 = 
 3401.3056, the solidity required. 
 
 2. What is the solidity of the frustum of a parabolic 
 spindle, whose length is 25, middle diameter 20, and end 
 diameter 15 ? Ans. 6643.175. 
 
 3. The length of a cask, in the form of the middle 
 frustum of a parabolic spindle, is 30, the bung diameter 
 24, and the head diameter 18 inches ; what is its content 
 in imperial gallons? A71S. 41.4009 gallons. 
 
 PROBLEM XIX. 
 
 To find the solidity of a hyperholoid. 
 
 RULE. 
 
 To the square of the radius of the base, and the square 
 of the diameter in the middle between the base and the 
 vertex; then this sum being multiplied by the altitude, 
 
 N 4 
 
272 
 
 CONIC SECTIONS 
 
 PART YIl. 
 
 and the product thus obtained bj .5236, will give the 
 solidity. 
 
 EXAMPLES. 
 
 1. '\\niat is the solidity of the 
 hjperboloid ACB, the altitude Cr 
 being 25, the radius Ar of the base 
 26, and the middle diameter m7i 34? 
 
 Here 26'^ + 34^ = 676 + 1156 = 
 1832; a?id 1832 x 25 x .5236 = 
 45800 X .5236=23980.88, the solidity required. 
 
 2. The altitude of a hyperboloid is 75, the radius of the 
 base 78, and the middle diameter 102 ; what is the soli- 
 dity? Ans. 647483.76 
 
 PROBLEM XX. 
 To find the solidity of the frustum of a hyperboloid. 
 
 KULE. 
 
 To the sum of the squares of the greatest and least 
 semi-diameters, add the square of the middle diameter ; 
 then this sum being multiplied by the altitude, and the 
 product thence arising by .5236, wiU give the solidity. 
 
 EXAJIPLES. 
 
 1. "Wliat is the solidity of the /"; \ 
 hyperbolic frustum ABCD, the dia- 
 meter AB of the greater end being 
 20, the diameter CD of the less end 
 12, the middle diameter mn 17, and 
 the altitude rs 24 ? 
 
 Here 102 + q2 + 172:= loo + 36 + 
 289=425 ; then 425 x 24 x .5236 = 
 10200 X .5236 = 5340.72, the solidity 
 of the frustum. 
 
 2. The diameter of the greater end of a hyperbolic 
 frustum is 40, the diameter of the less end 24, the middle 
 diameter 34, and the altitude 48 ; what is the solidity ? 
 
 Ans. 42725.76. 
 
PART VII. AND THEIR SOLIDS. 273 
 
 PROBLEM XXI. • 
 
 To find the soUdity of a hyperbolic spindle. 
 
 RULE. 
 
 To tlie square of the greatest diameter, add the square 
 of double the diameter taken at ^ of the length ; then this 
 sum being multiplied by the length, and the product thus 
 obtained by .1309, will give the solidity, nearly. 
 
 Nufe. When great accuracy is not required, this Rule may be used for any 
 spindle formed by the revolution of a conic section, or part of a conic section, about 
 Its axis ; as it will always give nearly the true soliditj', 
 
 EXA3IPLES. 
 
 1. "What is the solidity of a hyperbolic spindle, whose 
 length is 36, the greatest diameter 24, and the diameter at 
 i of the length 16.73338? 
 
 Here 24^ + (16.73338 x 2)^= 24^+ 33.46676-'=o76 4- 
 1120.0240249 = 1696.0240249; then 1696.0240249x36 
 X .1309 = 61056.8648964 x .1309=7992.34361493876, 
 the answer required. 
 
 2. The length of a hyperbolic spindle is 30, the 
 greatest diameter 20, and the diameter at -^ of the length 
 13.94449 ; what is the solidity? Ans. 4625.20177. 
 
 PROBLEM XXn. 
 
 To find the solidity of the middle frustum of an elliptic 
 or a hyperbolic spindle. 
 
 RULE. 
 
 To the sum of the squares of the greatest and least dia- 
 meters, add the square of double the diameter taken 
 exactly in the middle between them ; and this sum being 
 multiplied by the length, and the product again by .1309, 
 will give the solidity. 
 
 "Kote. This Rule will also give nearly the content of any frustum or segment 
 formed by the revolution of a conic section, or part of a conic section, cither about 
 the axis of the section, or about any other line. 
 
 EXAZ^n^LES. 
 
 1. The length of the middle frustum of an elliptic 
 spindle is 60, the greatest diameter 48, the least dia- 
 meter 36, and the diameter in the middle between them 
 45.23635 ; what is the solidity ? 
 N 5 
 
274 CONIC SECTIONS AND TIIELK SOLIDS. PART Vn. 
 
 Here 48-'+ 36^+ (45.23635 x 2)- = 48- + 36-' + 90.4727- 
 = 2304 + 1296 + 8185.30944529=11785.30944529; then 
 11785.30944529 x 60 x .1309 = 707118.5667174 x .1309 
 =92561.82038330766, the solidity required. 
 
 2. The length of the middle frustum of a hjiDerbolic 
 spindle is 40, the greatest diameter 32, the least diameter 
 24, and the diameter in the ixdddle between them 29; 
 what is the solidity ? Ajis. 25991.504. 
 
 3. What is the content of the middle frustum of any 
 spindle ; the length being 50, the greatest diameter 40, 
 the least diameter 30, and the diameter in the middle be- 
 tween them 37.69696 ? Ans. 53565.871567. 
 
 4. What is the content of the segment of any spindle ; 
 the length being 15, the greatest diameter 12, and the 
 middle diameter 9 ? Ans. 918.918. 
 
PART \TII. GAUGING. 1 i O 
 
 PAET YIII. 
 GAUGING. 
 
 Gauging is the art of finding the capacities or contents ot 
 all kinds of vessels used by Maltsters, Brewers, Distillers, 
 Wine-Merchants, Victuallers, &c. &c. ; such as cisterns, 
 couches, coppers, coolers, tuns, vats, stills, casks, &c. &c. 
 
 KE^LIRK. 
 
 The Art of Gauging is of such general utility, in the common affairs of life, that 
 
 there are few persons who do not frequently want its assistance To Candidates 
 
 for the Flxcise and Customs, it is indispensable ; and to Victuallers, Common 
 Brewers, Maltsters, Distillers, Rectifiers, Wine-Merchants, Spirit-Merchants, OiU 
 TMerchants. Cider and Perry Makers, Vinegar Makers, &c.&c., it is of considerable 
 moment ; for without its aid they could not ascertain, with accuracy, the quantity 
 of those articles which they buy, 'sell, or manufacture. 
 
 NEW IMPERIAL TABLES. 
 
 Few Acts of Parliament immediately interest so many 
 individuals, as that passed on June 17th, 1824, for esta- 
 blishing uniformity of Weights and Measures. Every 
 person that has any dealings, either in buying or selling, 
 particularly in commodities that are measured, must feel 
 desirous to know how he is affected by the change that 
 has taken place ; in order that he may neither be the ob- 
 ject of intentional fraud himself, nor take, inadvertently, 
 any undue advantage in his transactions with others. 
 
 The content of the New Lnperial Standard Gallon is 
 fixed at 277.274 cubic inches ; and this is now the stan- 
 dard measure of capacity, to be used througliout the 
 United Kingdom of Great Britain and Ireland, for both 
 Spirits, Wine, Ale, Beer, and all other liquids ; and also 
 for every kind of grain, and all other dry commodities. 
 But as the capacity of the gallon is changed, so likewise 
 is the capacity of the tierce, the hogshead, the puncheon, 
 the pipe, and the tun, in Wine and Spirit Measure ; the 
 firkin, the kilderkin, the barrel, the hogshead, the pun- 
 cheon, the butt, and the tun, in Ale and Beer Measure ; 
 and the peck, the busliel, the strike, the coom, the quarter, 
 the wey, and the last, in Dry Measure. This being the 
 N 6 
 
276 
 
 GAUGING. 
 
 PART Mil. 
 
 case, the author has calculated Tables, Avhich exhibit the 
 capacities of each of these denominations, in cubic inches. 
 
 TABLES OF WINE AND SPIRIT JVIEASURE. 
 
 Table I. Exhibiting the different denominations i?i this 
 measure. 
 
 4 gills make 1 pint, p^. 
 
 2 pints 1 quart, qt. 
 
 4 quarts 1 gallon, gal. 
 
 42 gallons ....'.... 1 tierce, tier. 
 
 63 gallons 1 hogshead, hhd. 
 
 84 gallons 1 puncheon, ptm. 
 
 2 hogsheads, or 126 gallons ... 1 pipe or hutt, j^ipe* 
 
 2 pipes, 4 hogsheads, or 252 gal. . 1 tun, tun. 
 
 Table II. Exhibiting the number of cubic inches con- 
 tained in each denomination according to the new 
 standard gallon. 
 
 Cubic Inches. 
 
 277.274= 
 11645.508 = 
 17468.262 = 
 23291.016^ 
 34936.524 = 126 = 3 
 69873.048=252=6 
 
 1 gallon. 
 42 = 1 tierce. 
 63 = H = 1 hogshead. 
 84=2 =:1-V=1 puncheon. 
 
 2 =1^=1 pipe. 
 
 4 =3*'=2 = 1 tun. 
 
 A'b/e. As 277.274 cubic inches make one imperial gallon ; 
 34.65925 make one pint ; and 8.G648I25 make one gilL 
 
 .3185 make one quart ; 
 
 TABLES OF ALE AND BEER IVIEASURE. 
 
 Table I. Exhibiting the different denominations in this 
 
 measure. 
 
 2 pints 
 
 4 quarts 
 
 9 gallons 
 
 2 firkins, or 18 gallons . . 
 2 kilderkins, or 36 gallons 
 1-J- barrel, or 54 gallons 
 
 2 barrels, or 72 gallons 
 
 3 barrels, 2 hogsheads, or 108 gallons 
 2 butts, 4 hogsheads, or 216 gallons 
 
 make 1 quart, qf. 
 1 gallon, gal. 
 1 firkin, ^r. 
 1 kilderkin, kil. 
 1 barrel, bar. 
 1 hogshead, hhd. 
 1 puncheon, pun. 
 1 butt, butt. 
 1 tun, tun. 
 
PAKT \TII. 
 
 GAUGING. 
 
 277 
 
 Table IL Exhibiting the numher of cubic inches con- 
 tained in each denomination, according to the new 
 standard gallon. 
 Cubic Inches. 
 
 1 gallon. 
 
 277.274== 
 
 2495.466= 
 
 4990.932 = 
 
 9981.864= 
 
 14972.796 = 
 
 19963.728= 
 
 9= 1 firkin. 
 18= 2= 1 kilderkin. 
 36= 4= 2=1 barrel. 
 54= 6= 3=1^=1 hogshead. 
 72= 8= 4 = 2" =1-1 = 1 puncheon. 
 6=3 =2"'=U = 1 butt. 
 
 29945.592=108 = 12 
 59891.184=216=24=12=6 =4 = 3 "^ =2=1 tun. 
 
 TABLES OF CORN OR DRY MEASURE. 
 
 Table I. Exhibiting the different 
 measure. 
 
 2 pints 
 
 2 quarts, or 4 pints 
 2 pottles, or 4 quarts 
 2 gallons, or 8 quarts 
 
 4 pecks, or 8 gallons 
 2 bushels, or 8 pecks 
 
 ■4 bushels, or 2 strikes 
 2 cooms, or 8 bushels 
 
 5 quarters, or 40 bushels 
 2 weys, or 10 quarters 
 
 de?iominations in this 
 
 make 1 quart, qt. 
 1 pottle, pot. 
 1 gallon, gal. 
 1 peck, ph. 
 1 bushel, bu. 
 1 strike, strike. 
 1 coom, cooin. 
 1 quarter, qr. 
 1 wey, wey. 
 1 last, last. 
 
 Table II. Exhibiting the number of cubic inches con- 
 tained in each denomination, according to the new 
 standard gallon. 
 
 Cubic Inches. 
 
 277.274 = 
 
 554.548= 
 
 2218.192 = 
 
 4436.384= 
 
 8872.768 = 
 
 1 gallon. 
 
 2= 1 peck. 
 8= 4= 1 bushel. 
 16= 8= 2= 1 strike. 
 32= 16= 4= 2= 1 coom. 
 17745.536= 64= 32= 8= 4= 2= 1 quarter. 
 88727.680=320=160=40=20=10= 5 = 1 wej. 
 177455.360=640=320=80=40 = 20=10=2=1 last. 
 
 Note I. The immbcr of cubic inches in the old wine gallon is 231 ; in the old ale 
 gallon, 282 ; in the old corn gallon, 2G8.8 ; and in the old corn bushel, 2150.4. 
 
 2. The old Scotch boll for wheat, beans, j)eas, rye, salt, and grass-seeds, contains 
 878934 cubic inches, equal to 4 bushels, 2 quarts, 1^ pint of our old measure. 
 
278 GAUGING. PAKT VHI. 
 
 3. The old Scotch boll for oats, barley, and malt, contains 12822. 09G cubic inches, 
 equal to 5 bushels, 3 pecks, 1 gallon, 2 quarts, li pint of our old measure. 
 
 4. If the Scotch boll for wheat, &c. be still continued, it will contain, according 
 to the new standard, 90GG.342 cubic inches, equal to 4 bushels, 2 quarts, 1^ pint of 
 our new measure, the same proportion as before. 
 
 .5. If the Scotch boll for oats, &c. be still continued, it will contain, according to 
 the new standard, 13225.728 cubic inches, equal to 5 bushels. Specks, 1 gallon, 2 
 quarts, li pint of our new measure, the same as before. 
 
 6- In Ireland the old malt and corn gallon contains 272^ cubic inches ; conse- 
 quently, the content of the bushel is 2178 cubic inches The old Irish gallon, for 
 
 all liquids, contains 217.6 cubic inches. 
 
 7. The old Scotch pint, for liquids, contains 103.4, the quart 206.8, and the gallon 
 827.23 cubic inches ; consequently, 1 Scotch gallon is equal to 2.98^43 imperial gal- 
 lons = 2 gal. 3 qt. 1 pt. 3i gills ; or 3 gallons, very nearly. 
 
 8. In Scotland, 4 pecks make 1 firlot, and 4 firlots 1 boll ; consequently, the 
 wheat peck contains 549.33375, and the oat peck 801.381 cubic inches ; the wheat 
 firlot 2197.335, and the oat firlot 3205.524 cubic inches. {See the 2nd and 3rd 
 Notes.) 
 
 KULES 
 
 For reducing any number of old gallons of wine, to the 
 new standard ; and vice versa. 
 
 1. As 277.274 cubic inclies, are to 231 cubic inches; 
 so are any number of old gallons, to the required number 
 of new gallons. 
 
 2. As 231 cubic inches, are to 277.274 cubic inches; 
 so are any number of new gallons, to the required number 
 of old gallons. 
 
 Note. The preceding rules will serve for Ale Measure and Dry Measure ; but 
 282 must be used for the old ale gallon ; 268.8 for the old corn gallon ; 2150.4 for 
 the old corn bushel, and 2218.192 cubic inches, for the new corn bushel. 
 
 EXAMPLES. 
 
 1. How many new imperial gallons are contained in 
 1 hogshead, or 63 old gallons of wine ? 
 
 As 277.274 : 231: :63 : 52.48598 new gallons — 62 gal. 
 1 qt. 1 Jit. 3^ gills, very nearly. 
 
 2. How many old gallons of wine are contained in 
 
 1 hogshead, or 63 gallons, imperial measure ? 
 
 As 231 : 277.274; :63 : 75.62018 old gallons = 16 gal 
 
 2 qt. Opt. 3f gills, very nearly. 
 
 3. How many new gallons are contained in 1 hogshead, 
 or 54 old gallons of ale ? 
 
 As 277.274 : 282:: 54 : 54.92040 new gallojis=6A:gal. 
 
 3 qt. 1 pt. li gill, very nearly. 
 
 4. How many old gallons of ale are there in 1 hogs- 
 head, or 54 gallons of imperial measure ? 
 
 As 282 : 277.274! :54 : 53.09502 old gallons = 6Z gal 
 qt. Opt. 3 gills, very nearly. 
 
 5. How many new bushels of corn are there in 1 last, 
 or 80 old bushels ? 
 
PART VIII. GAUGING. 279 
 
 As 2218.192 : 2150.4 :: 80 : 77.55505 7ieiv bushels = 
 77 bush. 2pk. gal. 1 qf. l^ jjinf, very nearhj. 
 
 6. HoAv many old bushels of corn are there in 1 last, 
 or 80 new bushels ? 
 
 As 2150.4: 2218.192:: 80 : 82.52202 ohl bushels = 
 82 bush. Ii^k. Ogal. qt. I jjt. Ih f/ill, very nearly. 
 
 2d. JIaliipUers, by ivhich old gallons and bushels may be 
 reduced to imperial gallons and bushels. 
 
 Multiply .83311 by the content of any vessel, in old 
 wine gallons ; and the product will be imperial gallons. 
 Multiply 1.01705 by old ale gallons ; and the product 
 will be imperial gallons. Multiply .96945 by old corn 
 bushels ; and the product will be imperial bushels. 
 
 EXAMPLES. 
 
 1. If a cask contain 1 hogshead, or 63 old gallons of 
 wine ; how many imperial gallons will it hold ? 
 
 Here .83311 X 63 = 52.48593 imperial gallons, the same 
 as obtained in the first example, by the preceding rules. 
 
 2. If the content of a vessel be 1 hogshead, or 54 old 
 gallons of ale ; what is its content in imperial gallons ? 
 
 i Jlere 1.01705x54 = 54.9207 i?nperial gallons, the same 
 CIS obtained in the third example, by the preceding rules. 
 
 3. How many imperial bushels are contained in 1 last, 
 or 80 old bushels ? 
 
 Here .96945 x 80 = 77.556 imperial bushels, the same 
 as found in the fifth example, by the preceding rides. 
 
 3d. Multipliers, by which imperial gallons and bushels 
 may be reduced to old gallons and bushels. 
 
 Multiply 1.20033 by the content of any vessel, in im- 
 perial gallons ; and the product will be old wine gallons. 
 Multiply .98325 by imperial gallons ; and the product 
 will be old ale gallons. Multiply 1.03153 by imperial 
 bushels ; and the product will be old Winchester bushels. 
 
 EXAMPLES. 
 
 1. If a cask contain 52.48593 imperial gallons ; how 
 many old wine gallons will it hold ? 
 
 Here 52.48593 x 1.20033 = 63 old ivine gallons. 
 
 2. If the content of a vessel be 54.9207 imperial gal- 
 lons ; Mdiat is the content in old ale gallons ? 
 
 Here 54.9207 x .98325=54 old ale gallojis. 
 
280 GAUGING. PART Vin. 
 
 3. How many old Wincliester bushels are contained in 
 77.556 imperial bushels ? 
 
 Here 77.556 x 1.03153=:80 old bushels. {See the pre- 
 ceding Examples.) 
 
 4th. Multipliers, for reducing Irish bushels and Scotch 
 bolls to imperial bushels ; and vice versa. 
 
 Multiply .98188 by Irish bushels, and the product will 
 be imperial bushels ; or multiply 1.01846 by imperial 
 bushels, and the product will be Irish bushels. 
 
 Multiply 3.96239 by Scotch wheat bolls, and the pro- 
 duct will be imperial bushels ; or multiply .25238 by im- 
 perial bushels, and the product will be Scotch wheat bolls. 
 
 Multiply 5.78043 by Scotch oat bolls, and the product 
 will be imperial bufehels ; or multiply imperial bushels by 
 .173, and the product will be Scotch oat bolls. 
 
 EXAMPLES. 
 
 1. How many imperial bushels are there in 1 last, or 
 80 Irish bushels ? 
 
 Here .98188 X 80=78.5504 imperial bushels. 
 
 2. How many Irish bushels are there in 78.5504 im- 
 perial bushels ? a 
 
 Here 78.5504 X 1.01846 = 80 Irish bushels. 
 
 3. How many imperial bushels are there in 10 Scotch 
 bolls of wheat ? 
 
 Here 3.96239 x 100 = 396.239 imperial bushels. 
 
 4. How many Scotch bolls of wheat are there in 
 396.239 imperial bushels ? 
 
 Here 396.239 X .25238 = 100 Scotch bolls of wheat. 
 
 5. How many imperial bushels are there in 100 Scotch 
 bolls of oats ? 
 
 Here 5.78043 X 100=578.043 imperial bushels. 
 
 6. How many Scotch bolls of oats are there in 578.043 
 imperial bushels ? 
 
 Here 578.043 X. 173 = 100 Scotch bolls of oats. 
 
 Irish liquid gallons may be reduced to imperial gallons, by multi- 
 plying by .78478 ; and imperial gallons may be reduced to Irish 
 liquid gallons, by multiplying by 1.27425. 
 
 1. The new imperial gallon is greater than the old wine gallon, by 
 46.274 cubic inches, which is nearly \ of 277.274, or I of 231 ; hence, 
 6 old gallons are equal to 5 new gallons. 
 
 2. The new gallon is less than the old ale gallon, by 4.726 cubic 
 inches, which is nearly gL of 282, or ^L of 277.274; hence 60 new 
 gallons are nearly equal to 59 old gallons. 
 
PART Vni. GAUGING. 281 
 
 3. The new gallon is greater than the old corn gallon, by 8.474 
 cubic inches, which is nearly equal to r,'., of 2f)8.8 ; and the new bushel 
 is greater than the old bushel, by ()7.792 cubic inches; hence 32 
 new bushels are nearly equal to 33 old bushels. 
 
 4. Old wine gallons may be reduced into new gallons, by subtract- 
 ing \', and new gallons may be brought into old wine gallons, by 
 adding^. 
 
 5. Old ale gallons may be reduced into new gallons, by adding ^'g ; 
 and new gallons may be brought into old ale gallons, by subtract- 
 ing 3S- 
 
 6. Old bushels may be reduced into new bushels, by subtracting 
 5'^ ; and new bushels may be brought into old bushels, by adding ^^. 
 
 OBSERVATION. 
 
 Those who wish to see more on the subject of the New 
 Imperial Weights and Measures^ are referred to the 
 Author's Treatise on Practical Arithmetic, Third Edi- 
 tion, in which he has given extensive Tables, comparing 
 the old and new measures ; and also a copious abstract 
 of the Act of Parliament, for establishing uniformity of 
 Weights and Measures throughout the united kingdom of 
 Great Britain and Ireland. 
 
 A DESCRIPTION OF THE SLIDING RUXE. 
 
 This instrument is in the form of a paralleloplpedon ; 
 has four sliding pieces, which run in grooves, and is com- 
 monly made of box. 
 
 It was invented by Mr. Thomas Everand, about the 
 year 1683, and is generally called Everand' s Sliding 
 Rule; but it has since been much improved by Mr. 
 "Verie ; and is now adapted to the new imperial measure. 
 
 It is of various lengths, as 6, 9, 12, 18 inches, &c. ; but 
 12 inches is the most common length. 
 
 This rule is much used in Gauging, in consequence of 
 the ease and expedition with which calculations may be 
 made by it. 
 
 Upon the first face of the Rule, is a line marked A, 
 which is called Gunter's Line, from its inventor jNIr. 
 Edmund Gunter ; and is numbered from the left to the 
 right with the figures 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ; and the 
 space between each of these figures are graduated into 
 subdivisions. At 2218.2, is a brass pin, marked I]\IB, 
 signifying the cubic inches in the imperial bushel ; and ac 
 277.3, is another brass pin, marked IMG, denoting tlie 
 number of cubic inches in the imperial gallon. 
 
282 GAUGING. PAKT \^^. 
 
 The second line on this face is upon the slide, and 
 marked B. It is divided exactly in the same manner as 
 tliat marked A. There is also another slide B, which is 
 used with the former ; the two brass ends are then placed 
 together, in which position they form a double radius. 
 At 277.3, on the second radius, is a brass pin marked G 
 or IG-, denoting the number of cubic inches in the im- 
 perial gallon. 
 
 By these lines, multiplication, division, proportion, &c. 
 may be performed ; and the manner of reading and using 
 them is precisely the same as the lines A and B, upon the 
 Carpenter's Rule. (See Section 2, Part IV.) 
 
 Upon the same face of the Kule, is another line marked 
 MD, signifying malt- depth ; and is numbered from the 
 right to the left, with the figures, 3, 4, 5, 6, 7, 8, 9, 10, 2. 
 
 The 1, or 10 on this line is placed directly opposite to 
 MB, on the line A ; and if it be called 1000, the last di- 
 vision at the left end of the Rule, will denote 2218.2, the 
 cubic inches in the imperial bushel. — This line is used 
 with the lines A and B, in malt gauging. 
 
 Upon the second face of the Rule, or that opposite to 
 the one already described, is a line marked D. This line 
 begins on the upper edge of the Rule, and is numbered 
 from the left to the right with the figures 1, 2, 3, 31, 32, 
 which is at the right end of the Rule. The line is then 
 continued from the left end of the other edge, 32, 4, 5, 6, 
 7, 8, 9, 10. ^ 
 
 At 18.79 is a brass pin, marked IG, signifying the cir- 
 cular gauge-point for imperial gallons ; at 47.1 is another 
 brass pin, marked MS, which is the square gauge-point 
 for imperial malt bushels; and at 53.1, is a third pin, 
 marked MR, denoting the round or cii'cular gauge-point 
 for imperial malt bushels. 
 
 Upon this face of the Rule, is a line marked C. This 
 line is upon the slide ; and is numbered and divided in 
 the same way as the lines A and B. Belonging to the 
 Rule, is also another slide C, which is used with the former, 
 in the same manner as the two slides marked B. 
 
 The line D, on this Rule, is similar to that marked D, 
 on the Carpenter's Rule ; and is used with the two slides 
 C, just described ; in finding the contents of vessels 
 whose form is that of a cube, a parallelopipedon, a cy- 
 linder, &c. &c. 
 
 By these lines the square root of any number may also 
 
PART Vni. GAUGING. 283 
 
 be readily extracted ; for if 1 on C be set to 1 on D, we 
 have against any proposed number on C, its root on D. 
 
 The back of the first slide, marked C, is divided, next 
 the upper edge, into inches and tenths ; and numbered 
 from the left to the right, with 1, 2, 3, 4, 5, &c. ; the se- 
 cond line is marked spheroid, and the third second variety; 
 and both are numbered from the left to the right with the 
 figures 1, 2, 3, 4, &c. ; and the spaces between each of 
 these figures are divided into ten equal parts. 
 
 These three lines are used for finding a mean diameter 
 between the bung and the head of casks of the first and 
 second varieties ; which is performed in the following 
 manner : Find the difference between the bung and head 
 diameters, on the first line, or line of inches ; then against 
 it, for each variety, is a number, which being added to 
 the head, will give the mean diameter sought ; hence the 
 cask is reduced to a cylinder, whose content maybe found 
 by Problem III. 
 
 The back of the second slide C, contains the gauge- 
 points, divisors, and factors or multipliers, for imperial 
 gallons and bushels. 
 
 The third face of the Rule contains a line marked Seg. 
 St. or SS, signifying segments standing. This line is 
 numbered from the left to the right, on the upper edge of 
 the Rule, with 1, 2, 3, 4, 5, 6, 7, 8. It is then continued 
 on the other edge, 8, 9, 10, 20, &c. to 100. 
 
 This line is used with the two slides, marked C, in 
 finding the ullage of a standing cask, or the quantity of 
 liquor it contains when it is not full. 
 
 Upon the fourth face of the Rule, or the one opposite 
 to that last described, is a line marked Seg. Ly. or SL, 
 denoting segments lying. This line is numbered nearly 
 in the same manner as the last ; and is used with the 
 slides C, in ullaging lying casks. 
 
 "Sole 1. There are various kinds of Sliding Rules, some of which have a line 
 marked E, for extracting the cube roots of numbers; and others a line for the 
 third variety of casks ; they are, however, all upon the same principle, and may be 
 easily understood from the foregoing description. 
 
 2. Very nf.it and commodious Slidin? Rules, have lately been made flat, with 
 only two slides. These slides are the lull thickness of the rule ; and being divided 
 on each side, answer every purpose of the four sides; and the rule is considerately 
 reduced both in size and weight. 
 
 3. It may not be improper to observe, that in the preceding description of the 
 Sliding Rule, tlie figures mentioned, refer to the larger figures placed at the 
 primary divisions of the rule ; most of the subdivisions being marked with smaller 
 figures. — For the method of estimating the values of the divisions and subdivisions, 
 the reader is referred to the Description of the Carpenter's Rule, Sectioa 2, 
 Part IV. 
 
284 GAUGING. PAUT vm. 
 
 THE METHODS OF FINDING THE 
 
 MULTIPLIEES, DIVISOES, AND GAUGE- 
 POINTS, 
 
 CONTAINED IN" THE SUBSEQUENT TABLE. 
 
 The multipliers and diYisors are cliiefly obtained by 
 division, in the following manner : 
 
 Divisors for Multipliers for 
 
 squares. circles. 
 
 277.274)1.0000 &c. (.003607 imperial gaUotis. 
 2218.192)1.0000 &c. {.000451 imperial bushels. 
 
 Divisors for IMultipiiars for 
 
 squares. circles. 
 
 277.274).785398 &c. (.002833 imperial gallons. 
 2218.192).785398 &c. (.000354 imperial bushels. 
 
 Divisors for circles. 
 
 .785398) 277.274(353.04 imperial gallons. 
 .785398)2218.192(2824.29 imperial bushels. 
 
 GAUGE-POINTS. 
 
 The gauge-points are found by extracting the square 
 roots of the divisors. 
 
 Divisors for Gauge-points for 
 
 squares. squares. 
 
 277.274) their square (16.65 imperial gallons. 
 2218.192) roots are (47.10 iinperial bushels. 
 
 Divisors for Gauge-points for 
 
 circles. circles. 
 
 353.04) their square (18.79 imperial gallons. 
 2824.29) roots are (53.14 imperial bushels. 
 
 Thus all the numbers in the following table were ob- 
 tained. 
 
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 Note. The use of the Multipliers, Divisors, and Gauge-points, : 
 following Problems. 
 
 shown in the 
 
286 GAUGING. PART YUI- 
 
 PROBLEM I. 
 
 The side of a vessel in the for jn of a cichc, being given in 
 inches ; to find its content in imperial gallons and 
 bushels. 
 
 RULE. 
 
 Multiply the side by itself, and that product again by 
 the side ; and the last product will be the content of the 
 vessel in cubic inches ; which being multiplied by .003607, 
 and .000451 ; or divided by 277.3, and 2218.2, the respec- 
 tive products or quotients will be the content of the 
 vessel in imperial gallons and bushels. 
 
 Note 1. The definitions of the cube, the jiarallelopipedon, the cylinder, &c. &c. 
 may be seen in l^art IV.; and if the content of any vessel be found'in cubic inches, 
 by the rules given in that Part, and then divided by 282, 231, and 2150.4, the respec- 
 tive quotients will be the content of the vessel in old ale and wine gallons, and 
 malt bushels ; but the Rules and Examples given in the following Problems will 
 be found more particularly adap.'td to Gauging, than those in Part IV. 
 
 2. If the content of any vessel, in cubic inches, be divided by 277.3, and 2218.2 ; 
 the respective quotients will be the content in imperial gallons and bushels. 
 
 3. Gaugers always take their dimensions in inches or in inches and tenths ; and 
 when we say the side or diameter of a vessel measures so many Indies, we mean 
 the internal, not the external dimensions. 
 
 EXAMPLES. 
 
 *• 1. The side of a cubical vessel measures 46 inches ; 
 what is its content in imperial gallons and bushels ? 
 
 Here 46x46x46=2116x46=97336, the content in 
 cubic inches; then 97336 X .003607=351.0909 imjjerial 
 gallons ; and 97336 X .000451 =43.8985 imperial bushels. 
 
 Or, 97336-^277.3 = 351.0133 imperial gallons ; and 
 97336-T-2218.2=43.8806 imperial bushels. 
 
 Note. The divisors are used in all the following solutions ; the work, howeverj 
 may be proved by the multipliers given in the preceding Table- 
 
 Bf/ the Sliding Mule. 
 
 In this operation, the square gauge-points 16.65, and 
 47.10, upon the line D, must be used. 
 
 On D. On C. On D. On C. 
 
 J r 16.65 "I . AC • ' AC { 2>ol.O imperial gallons. 
 
 ^^ 1 47. 10 J • ^t» .. 4b . \^4^4^^Q i.j.perial bushels. 
 
 Note. New Sliding Rules have not always the square guagc-points marked upon 
 them ; but small brass pins may be easily inserted at those points. 
 
 2. The side of a cubical cistern is 134 inches; what is 
 its content in imperial gallons and bushels ? 
 
 Ans. 8676.8986 imperial galloi^s, and 1084.710-1 im- 
 perial bushels. 
 
PAET Vni. GAUGING. 287 
 
 3. The side of a cubical wine-vat measures 135.6 in- 
 ches ; how much wine will it contain at once ? 
 
 Ans. 8991.4389 (/allo?is = 8991 gal 1 qt. \\pt. 
 
 PROBLEM II. 
 
 The lengthy breadth, and depth of a vessel, iii the form of 
 a parallelopipedon, being given in inches ; to find its 
 content in imperial gallons and bushels. 
 
 RULE. 
 
 Multiply the length by the breadth, and the product 
 thence arising by the depth ; and the last product will be 
 the content in cubic inches ; which being divided by 
 277.3 and 2218.2, will give the content in imperial gal- 
 lons and bushels. 
 
 Note, As the sides of vessels, in the form of a parallelopipedon, are seldom per- 
 fectly regular and parallel, it is best to take several lengths, in different places ; 
 and divide their sum by their number for a mean length. A mean breadth and 
 depth may be found in the same manner 
 
 EXAMPLES. 
 
 1. A cistern in the form of a parallelopipedon, mea- 
 sures 96 inches in length, 66 in breadth, and 48 in depth ; 
 what is its content in imperial gallons and bushels ? 
 
 Here 96 X 65 X 48=299520, the content in cubic inches; 
 ^Ae?z 299520-^277.3=1080.1298 gallons; aw</ 299520-t- 
 2218.2 = 135.0284 bushels. 
 
 By the Sliding Rule. 
 
 As 96 on C : 96 on J) \ \ 65 on Q,'.19 on D, which is 
 a mean proportional between the length and breadth. ( See 
 Problem 5, page 152.) Then, 
 
 On D. On C. On D. On C 
 
 .r 16.651 .. _ . ^lOm.O gallons. 
 
 ^^147.10/ • ^« •• '^ • I \Z5.0 bushels. 
 
 2. The length of a vessel in the form of a parallelo- 
 pipedon, is 136, its breadth 94, and its depth 62 inches ^ 
 what is its content in imperial gallons and bushels ? 
 
 Ans. 2858.3050 gallotis, and 357.3203 bushels.. 
 
 3. A water-trough measures 85.3 inches in length, 54.7 
 in breadth, and 32.9 in depth ; how many imperial gallons* 
 will it contain ? 
 
 Ans. 553.5825 gallons = 553 gal. 2 gt. OJ pf. 
 
288 GAUGING. PART Ym. 
 
 4. A maltster has a cistern vv^hose length is 132, breadth 
 
 118, and depth 46 inches ; how much barley can he steep 
 
 at a time, admitting the water to occupy ^ of the cistern ? 
 
 A?is. 193.8047 bushels = 24 qr. 1 bush. 3 ^ pk. 
 
 PROBLEM III. 
 
 The diameter and depth of a vessel, in the form of a 
 cylinder being given ; to find its content in imperial 
 gallons and bushels. 
 
 RULE. 
 
 Multiply the square of the diameter by the deptli ; and 
 divide the product by 353, for imperial gallons ; and by 
 2824.3, for imperial bushels. 
 
 liote 1. As cylindrical vessels are seldom perfectly round, it is best to measure 
 cross diameters, in different parts ; and divide their sum by tiieir number, lor a 
 mean diameter. 
 
 2. If a cvlindrical vessel be placed in an inclining position, so that the hquor 
 forms an elliptical surface, the content mav be found by the following Rule : Mul- 
 tiply the square of the diameter of the vessel by half the sum of the greatest and 
 least depths of the liquor ; and divide the product by 353, for imperial gallons. 
 \See Notes, Prob. 5.) 
 
 EXiOIPLES. 
 
 1. The diameter of a cylindrical vessel is 34, and its 
 depth 45 inches ; what is its content in gallons and 
 bushels ? 
 
 Here 34 X 34 X 45 = 52020 ; then 52020 --353 = 
 U1.2>Q64. gallons; a?^c^52020--2824.3= 18.4187 bushels. 
 
 By the Sliding Rule. 
 Here the circular gauge-points 18.79, and 53.14, upon 
 the line D. mu.'^t be used. 
 
 On D. On C. On D. On C. 
 
 18.791 . ,r .. o, . f 147.4 <7aZ^o?z5. 
 
 ^^l53'.14| • "^^ •• ^^ ' \ IS.4: bushels. 
 
 2. The diameter of a cylindrical vessel is 46.7, and its 
 depth 68.4 inches ; what is its content in gallons and 
 bushels? A71S. 422.5860 gallons, and 52.8176 bushels. 
 
 3. A cylindrical mash-tun measures 94 inches in dia- 
 meter, and 82 in depth ; how many bushels of malt will it 
 contain at once ? A?is. 256.5421 bushels. 
 
 4. At Heidelberg, in Germany, is a cylindrical wine cask, 
 
PART VIII. GAUGING. 289 
 
 whose depth is 27, and diameter 21 feet ; how many gal- 
 lons Avill it contain, imperial measure ? 
 
 A?is, 58286.9575 gallons. 
 
 Xole. The convivial monum.^nt of ancient hospitality, mentioned in the 4th Ex- 
 ample, was formerly kept full of tlic best Rhenish wine, and the electors have 
 given many entertainments on its platform ; but it now onlv serves as a melancholy 
 instance of the extinction of that hospitality ; for it is suffered to moulder in a damp 
 vault, quite empty. 
 
 Although this vessel is of an extraordinary magnitude, yet it is much inferior in 
 capacity, to many of the London i)orter-vats" 
 
 PROBLEM IV. 
 
 Give?} the dimensions of a vessel in the form of a prismoid, 
 or the frustum of a square pyramid, or a cylindroid ; 
 to find its content in imperial gallons and bushels. 
 
 To the sum of the areas of the two ends, add four times 
 the area of the middle section parallel to them; multiply 
 this sum by the perpendicular depth, and I of the product 
 will be the content in cubic inches ; which divide by 
 277.3 for imperial gallons, and 2218.2 for imperial bushels. 
 {See the Scholium, Prob. 10. Sect. I. Part IV.) 
 
 Noiel. A cylindroid is a figure resembling the frustum of a cone; but having 
 elliptical instead of circular ends. Sometimes one end is circular, and the other 
 elliptical. 
 
 2. When the vessel is a prismoid, the length of the middle section is equal to 
 half the sum of the lengths of the two ends ; and its breadth is equal to half the 
 sum of their breadths. 
 
 .3. If the ends be elliptical, the transverse diameter of the middle section will be 
 equal to half the sum of the transverse diameters of the two ends ; and the conjugate 
 diameter equal to half the sum of the conjugate diameters of tlie two ends. 
 
 4. If one end be an ellipse and the other a circle, add the transverse diameter of 
 the elliptical end to the diameter of the circular end ; and take half the sum for the 
 transverse diameter of the middle section. The conjugate diameter of the middle 
 section may be found in a similar manner ; it is better, however, in all cases of 
 practice, to take the real dimensions of the sections. 
 
 .5. When the ends are rectangles, their areas may be found by Problem 2; when 
 they are circles, we may obtain their areas by Problem 15 ; and when they are 
 ellipses, we can find their areas by Problem 21, Part II. 
 
 EXAMPLES. 
 
 1. The length and breadth of the bottom of a vessel in 
 the form of a prismoid measures 72 and 64, the length 
 and breadth of the top 96 and 82, and the perpendicular 
 depth 65 inches ; what is its content in gallons and 
 bushels ? 
 
 Here (72 v 64) + (96 X 82) =4608 + 7872 = 12480, the 
 area of the tuo ends. 
 
 Also, (72-f 96) --2 = 168-^2 = 84, the length of the 
 middle section; and (64 + 82) -- 2 = 146 -r-2=73, the 
 
 o 
 
290 GAUGING. PAET ■^.^II. 
 
 breadth of the middle section ; then 84 X 73 x 4=6132 x 4 
 = 24528, four times the area of the middle section ; whence 
 (12480 + 24528) X 65 --6=(37008x65)--6==240o520--6 
 =400920, the content in cubic inches ; then 400920-7-277.3 
 = 1445.7987 gallons; and 400920 --2218.2 = 180.7411 
 bushels. 
 
 Note. This and some of the following Problems may be performed by the Sliding 
 Rule ; but as the operations are too tedious for practice, they are omitted. 
 
 2. Each side of the bottom of a cistern, in the form of 
 the frustum of a square pyramid, measures 86, each side 
 of the top 78, and the perpendicular depth 74 inches ; 
 what is its content in gallons and bushels ? 
 
 Ans. 1795.7831 gallons, and 224:A9Zl bushels. 
 
 3. The perpendicular depth of a vessel, in the form of 
 the frustum of an elliptical cone, is 46.8 inches ; the 
 transverse and conjugate diameters of the bottom measure 
 49.6 and 37.8 inches ; the transverse and conjugate dia- 
 meters of the top, 67.2 and 50.4 inches ; required the 
 content in gallons and bushels. 
 
 Ans. 343.8297 gallons, and 42.9825 bifshels. 
 
 4. The perpendicular depth of a vessel with an elliptical 
 base and a circular top, is 52.6 inches ; the transverse and 
 conjugate diameters of the bottom measure 61.6 and 46.2 
 inches ; and the diameter of the top measures 42.6 inches ; 
 required the content in gallons and bushels. 
 
 Ans. 345.4747 gallons, and 43.1882 bushels. 
 
 PROBLEM Y. 
 
 To find the content of a vessel in the form of the frustum 
 of a cone, in imperial gallons and bushels. 
 
 RULE. 
 
 To three times the product of the top and bottom di- 
 ameters, add the square of their difference ; multiply the 
 sum by the depth, and divide the product by 1059 for im- 
 perial gallons, and 8472.9 for imperial bushels. 
 
 Note 1. The general rule given in Problem G, Section I., Part IV. will give the 
 content of any frustum, whatever may be the form of the two similar ends ; that is, 
 whether they be polygons, circles, or ellipses. 
 
 2. In taking the dimensions of circular vessels, it is best, in all cases, to measure 
 cross diameters, as directed in the second Note, in Case I. of the next Problem. 
 
 3. When the frustum of a cone is cut by a diagonal plane passing through the 
 opposite extremities of the ends ; or whpn a vessel of that shape is placed in an in- 
 clining position, so that the liquor just touches the opposite extremities of the top 
 and bottom, the two parts or sections thus formed are called elliptic hoofs ; and 
 their contents may be found by the following Rule: Multiply the product of the 
 diameters of the ends of the frustum by a mean proportional between them ; and 
 cube the diameter of the hoof's base. 
 
PART ^^II. GAUGING. 291 
 
 From the greater of the numbers thus found subtract the less ; and divide the 
 remainder by the difference of tlie diameters ; then the quotient being multiplied 
 by the perpendicular height of the hoof, and the product by .2018, will give the 
 content in cubic inches ; which divide by 277.3, and you will obtain the content in 
 gallons. 
 
 4. If the greater end of the frustum be considered as the base of the hoof, the 
 product of the diameters by a mean proportional between them, will be less than 
 the cube of the diameter of tlie base; but the contrary will be the case, when the 
 less end of the frustum represents the base of the hoof. 
 
 5. If a vessel be placed in such a position that tiie liquor just covers the bot- 
 tom, and part of one side, measure the diameter at the bottom, the diameter at the 
 upper extremity of the liquor, and also the liquor's perpendicular depth, or height 
 of the hoof; then proceed with these dimensions as directed in Note 3, and you 
 will obtain the content. 
 
 6. In Nesbit's and Little's Practical Gauping, there are Rules given for finding 
 the contents of all cylindrical, pyramidical, and conical hoofs, or ungulas, that 
 can possibly be formed, by placing in various positions vessels containing fluids. 
 
 exa:mples. 
 
 • 1 . The bottom diameter of a vessel, in the form of the 
 frustum of a cone, is 46, the top diameter 62, and the 
 depth 60 inches ; what is its content in gallons and 
 bushels ? 
 
 Here 62 x 46 X 3 = 2852 x 3 = 8556, three thnes the pro- 
 duct of the top and bottom diameters. Also, (62—46)-=: 
 16 X 16=256, the square of their difference ; then (8556 
 + 256) X 60=8812 x 60=528720 ; and 528720 -- 1059 = 
 A99.2QZ^ gallo7is ; likeivise, 528720-1-8472.9 = 62.4013 
 bushels. 
 
 2. The bottom diameter of a guile-tun is 115, the top 
 diameter 98, and the depth 75 inches ; how many impe- 
 rial gallons will it contain ? Ajis. 2414.9433 gallons. 
 
 3. The bottom diameter of a wine-vat is 78.6, the top 
 diameter 64.3, and the depth 72.8 inches; what quantity 
 of wine will it contain? Ans. 1056.3513 gallons 
 
 PROBLEM VI. 
 
 To find the content of a circular vessel, ivith curved sides. 
 CASE I. 
 
 When the sides are not much curved, o.s in the following 
 figure. 
 
 KULE. 
 
 To the sum of the squares of the top and bottom diam- 
 eters, add four times the square of the middle diameter ; 
 multiply this sum by the depth ; divide the product by 
 2118.24, and 16945.74 ; and the respective quotients will 
 be the content in imperial gallons and bushels. 
 o 2 
 
292 GAUGING. TAKT Vni, 
 
 Note 1. The foregoing divisors are found by multiplying the circular divisors 
 by G. 
 
 2. As vessels of this kind are seldom perfectly round, it is best to measure two 
 di;;meters at right angles to each other ; and take half their sum for a mean dia 
 meter. (Sec Problem 7.) 
 
 3. The Rule given in this Problem is the same, in substance, as that given in 
 Problem 10, Section I., Part IV. {Hee the Scholium in that Probleift.) 
 
 EXAIMPLES. 
 
 1. The diamete'f AB, of the fol- .^- -^^. 
 
 lowing vessel, measures 50, the dia- ',^>1_ _ ;" 
 
 meter CD 56, tlie diameter EF 54, ^ ; 
 
 and the depth w?i 46 inches ; what ^y| ■ 
 
 is its content in gallons and bushels ? W. \ 
 
 Here 50^ +54^^=2500 + 2916= | _,..— ^-.. 
 5416, the sum of the squares of the r^^--- — -^ -- 
 top and bottom diameters; and. 
 56- X 4=3136 x4= 12544, four times the square of the 
 middle diameter. Then, (54 1 6 -h 1 2544) x 46 = 1 7960 x 46 
 = 826160; and 826160 -- 2118.24= 390.0219 gallons; 
 also, 826160-r-16945.74 = 48.7532 bushels. 
 
 2. The bottom diameter of a mash-tun is 75.3, the top 
 diameter 81.6, the diameter taken in the middle between 
 them 84.7, and the depth 56.9 inches ; what is its content 
 in imperial bushels? Ans. 137.7527 bushels. 
 
 3. If the length of a cask be 30, the bung dianieter 24, 
 and the head diameter 18 inches ; what is its content in 
 imperial gallons ? Ans. 41. S082 gallons. 
 
 CASE II. 
 
 To find the content of a circular vessel, ^chen its sides 
 are much curved, by taking five diameters, at equal 
 distarices from each other, as in the following figure. 
 
 RULE. 
 
 Add into one sum, the squares of the top and bottom 
 diameters, twice the square of the middle diameter, and 
 four times the sura of the squares of the diameters taken 
 at one-fourth and at three-fourths of the depth ; multiply 
 this sura by the depth ; divide the product by 4236.48, 
 and 33891.48 ; and the respective quotients M'ill be the 
 content in imperial gallons and bushels. 
 
 yote. The foregoing divisors are found by multiplying the circular divisors by 
 twelve. 
 
PART Mil. GAUGING. ' 293 
 
 REMARK. 
 
 The Author has deduced the preceding Rule, from the 
 method of equidistant ordinates, described in Problem 23, 
 Part II, ; and it may be used for all vessels of an ordinary- 
 size ; but if a vessel be very deep, its sides very much 
 curved, and great accuracy required, the content must be 
 found by the following Rule : Find the areas of as many 
 equidistant, parallel sections as you think necessary, by 
 multiplying the square of the mean diameter of each sec- 
 tion by .7854. Then to the sum of the areas of the two 
 end sections, add four times the sum of the areas of all the 
 even sections, and twice the sum of the areas of all the 
 odd sections, not including the sections at the ends ; mul- 
 tiply the sum by the common distance of the sections ; 
 divide the product by 3 ; and the quotient will be the 
 content in cubic inches. — Divide the content thus found 
 by 277.274, and 2218.192 ; and the respective quotients 
 will be the content in imperial gallons and bushels. 
 
 Note 1 . Always inake choice of an odd number of sections, in order that the num- 
 ber of parts into wliich the vessel is divided may be equal. Seven or nine will, in 
 general, be sufficient, except when tlie vessel is very deep, in which case it may be 
 necessary to take eleven, thirteen, &c. as the case may require. 
 
 2. The perpendicular depth of the vessel must first be taken, in order to deter- 
 mine the common distance of the ordinates, which may be found by dividing the 
 whole depth by the number of sections minus one. 
 
 .3. Great care must be taken to obtain the diameters of the sections at equal per- 
 pendicular distances from each other; for if their distance be measured upon the 
 side of the vessel, it is evident that the process will be incorrect. {See Pro- 
 blem 1, Case 11., Part yi.) 
 
 EXAMPLES. 
 
 1. The diameter AB, of the 
 
 following vessel, measures 49, CD ^^C^r^'"^" ~ '^^^1' 
 
 62, EF 68, GH 65, KL 56, and GiL^ZlrTTT^pH 
 the depth mn 48 inches ; what « ; :i|]i| 
 
 is its content in imperial gallons ^||i j" %^ 
 
 and bushels ? ^Jii, ; |S) 
 
 Here 49^ + 56^ = 2401 + 3 1 36 ^^::i:::::^m 
 
 = 5537, the sum of the squares — '^-""^ 
 
 of the top and bottom diameters; 68- x 2 = 4624 x 2 ~ 
 9248, twice the square of the middle diameter ; and (62- -f- 
 GS^) X 4=(3844-|-4225) x 4^ 8069 x 4 = 32276,/oMr times 
 the sum of the squares of the diameters taken at one- 
 fourth arid at three-fourths of the depth. Then, (5537 + 
 9248 + 32276) r 48=47061 x 48 = 2258928 ; awc^ 2258928 
 *7- 4236.48=533.2087, the content in imperial gallons; 
 o 3 
 
294 GAUGIXG. PART VIII. 
 
 also 2258928--33891.48 = 66.6ol7, the content in impe- 
 rial bushels. 
 
 2. The bottom diameter of a gnile-tmi measures 74.2, 
 and that of the top 84.3 inches ; the diameter taken at i 
 of the depth from the bottom, is 93.6, and that at | of the 
 depth 98.7 inches ; the middle diameter 102.4, and the 
 depth 60.8 inches ; how m.uch ale will it contain at once ? 
 
 A77S. 1544.1428 gallons. 
 
 3. What is the content of a vessel in imperial gallons 
 and bushels, the depth of which is 120 inches, and the 
 mean diameters of seven equidistant, parallel sections, as 
 follow : the diameter of the bottom or first section 124, 
 the second 146, the third 161, the fourth 164, the fifth 166, 
 the sixth 157, and the diameter of the top, or seventh 
 section 144 inches ? 
 
 Ans. The area of the first section w =12076.3104, the 
 5eco;iJ= 16741.5864, the third =20358.3534, the fourth = 
 21124.1184, Me//^A = 21642.4824, Me 5^0:^/^ = 19359.3246, 
 and the seventh =:l6286.054i4 ; then by the method of equi- 
 distant ordinates, or parallel sections, we find the content 
 = 2275094.36 cubic inches = 8205.2206 gallons, and 
 1025.6525 bushels, the answer required. 
 
 'Note. In order to find the content of a copper with a convex or a concave bottom, 
 generally called a rising or a. falling crown, pour in as much water as will just 
 cover the bottom ; and make marks with chalk upon the sides of the copper, at the 
 surface of the water; then draw it into a vessel of a known measure, or one that 
 may be easily gauged ; after which find the content of the remainder of the copper, 
 as directed in this Problem, which being added to the number of gallons requisite 
 to cover the bottom, will give the whole content. — The content of a still may also 
 be found in the same manner. 
 
 PROBLEM YII. 
 
 To gauge and inch a round guile-tun according to the 
 method practised in the Excise. 
 
 RULE. 
 
 Take cross diameters of the guile-tun, in the middle of 
 every ten inches from the bottom upwards ; that is, mea- 
 sure the first diameters at five inches from the bottom, 
 the second at fifteen, the third at tAventy-five, &c. ; and 
 take half the sum of each two for mean diameters. Di- 
 vide the square of the first mean diameter by 353.04, 
 multiply the quotient by ? ; and the product will be the 
 content of the first ten inches of the vessel. Find the 
 content of every ten inches of the perpendicular depth in 
 
PAKT VIII. GAUGING. 295 
 
 tlie same manner, wliicli being added togetlier, will give 
 the content of the whole tun in imperial gallons. 
 
 Xoir. It must here be observed that the square of the diainetcr of any section 
 divided by 3')3.04, will give the area of that section in imperial gallons ; for in finding 
 the arcas'of figures, Gangers always consider their depth to be one inch. 
 
 2. Mash-tuns are used for the purpose o'" macerating malt in hot water, to ex- 
 tract from it the saccharine substance ; and guile-tuns, sometimes called working- 
 vats, are vessels in which wort is fermented, in order to convert it into ale or beer. 
 
 REMARK. 
 
 The mG5t eligible method of taking the dimensions of 
 circular vessels, is to quarter them, in order to obtain 
 cross diameters as correctly as possible. — First, take the 
 diameter of the top of the vessel, which multiply by 
 .707, or .7 ; and the product will be the side of the in- 
 scribed square. {See Part IL Froh. 15, Rule III. in the 
 Notes,) 
 
 Apply the side of the square thus found, four times to 
 the top of the vessel, by your dimension cane, or tape ; 
 and mark, with chalk, each angle of the inscribed square ; 
 and thus will the top of the vessel be quartered. — With a 
 steady hand, draw a chalk line, from each quartering 
 point, down the inside of the vessel, to the bottom ; taking 
 care that these lines be equally distant from each other, 
 not only at the bottom of the vessel, but also at every 
 horizontal section ; otherwise the vessel will not be truly 
 quartered in every part. 
 
 Place your dimension cane, or other instrument, in a 
 perpendicular direction, with its end resting upon the 
 bottom of the vessel, about half Avay between the centre 
 of the bottom and the side ; at the same time laying a, 
 rod, or holding a cord diametrically across the top, so as 
 to come in contact with the dimension cane ; and thus 
 you will have the perpendicular depth of the vessel, upon 
 the dimension cane, where it intersects the rod or cord at 
 the top. 
 
 Nov/, in order to take cross diameters in the middle of 
 every ten inches, mark the dimension cane with chalk, at 
 5, 15, 25, 35 inches &c. ; and place that instrument in 
 the same manner in which it stood in taking the depth ; 
 then bring a rod horizontally in contact with the cane, at 
 5 inches from the bottom, so that the rod and cane may 
 form right angles ; and let the end of the rod, at the same 
 time, touch one of the quartering lines ; and tliere make a 
 mark with chalk, on the inside of the vessel ; and you will 
 have one point at which a diameter must be taken, 
 o 4 
 
296 GAUGING. PAKT VID. 
 
 Do the same at 15, and 25 inches, &c. from the bot- 
 tom ; and you will obtain all the points in one of the 
 quartering lines, at which diameters must be taken. 
 These points may then be transferred to the other three 
 quartering lines, by a pair of compasses or a dipping 
 piece ; and hence you may proceed to take cross diameters, 
 as before directed ; always beginning at the bottom. 
 
 Note When vessels are small, cross diameters may be most easily taken by an 
 instrument called the Diameter Rod, or Rule. When they are large, they must be 
 measured by the Dimension Cane. 
 
 EXAMPLES. 
 
 1. Let the following figure ^ 
 
 ABCD represent a guile-tun, d^^^---^^ ^^Z^^ 
 
 in the form of the frustum wfe^^ —Ir-. ^0g'' 
 
 of a cone, whose perpendicular % [_ ,i-Jg 
 
 depth m?i is 30, the mean dia- ''m i ||«* 
 
 meters pq 42.2, rs 45.3, and jM^^-^^^^^^^^^^J^m^ 
 
 wx 48.6 inches ; what is its jj| ^---t— --j |^b 
 content in imperial gallons ? 
 
 Here (42.2 x 42.2) -- 353.04 = 1780.84 -- 353.04 = 
 5.044 gallons, the area of the section at pq ; and 5.044 X 
 10=50.44 gallons, the content of the first 10 inches of 
 the perpendicular depth. 
 
 Again, (45.3 x 45.3) -- 353.04 = 2052.09 -- 353.04 = 
 5.812 gallons, the area of the section at rs ; and 5.812 X 
 10:^58.12 gallons, the content of the second 10 inches of 
 the depth. 
 
 Also, (48.6 X 48.6) -r- 353.04 = 2361.96 -4- 353.04 = 
 6.690 gallons, the area of the section at ivx ; and 6.690 X 
 10 = 66.90 gallons, the content of the third and last 10 
 inches of the depth. 
 
 Lasthj, 50.44 + 58.12 + 66.90 = 175.46 5ra/Zow5 =4 bar- 
 rels, 3 firkins, and 4.46 gallons, the whole content of the 
 guile-tun. 
 
 Note. In Practice, -when the diameters are taken in the middle of every ten 
 inches, it Is not necessary to multiply by 10 ; for if the decimal dot, m the area, be 
 removed one figure towards the right-hand you will obtain the content the sanie as 
 if you actually multiplied the area by 10; but in the foregoing Example I have 
 multiplied the area of each section by 10. in order to show the learner how to pro- 
 ceed when diameters are taken in the middle of every 6 or 8 inches ; or when the 
 vessel is not full of liquor. 
 
 2. If the depth of the liquor in the foregoing guile-tun 
 be 26.8 inches ; how many gallons does it contain ? 
 
 Here 50.44 + 58.12 = 108.56 gallons, the content of the 
 first 20 inches of the depth ; and 6.69 X 6.8=45.492 gal- 
 
PAKT VII r. 
 
 GAUGING. 
 
 297 
 
 Io)/s, the content of the remaining G.S inches; then 108.56 
 + 45.492=154.052 gaUons, the content required. 
 
 3. If the dopth of tlie liquor in the preceding guile-tim 
 be 18.4 inches ; how many gallons does it contain ? 
 
 Ans. 99.2608 gallons. 
 
 Ki:^IARKS. 
 
 1. In order to facilitate the Practice of Gauging, it is 
 necessary to be furnished with a measuring-tape, divided 
 into inches, a gauging-rod, divided into inches and tenths, 
 a chalk-line, plummet, &c. as the case may require ; and 
 also a dimension-book ruled in columns proper for the 
 dimensions intended to be taken. 
 
 2. The following Table, in which are entered the di- 
 mensions, areas, and contents of the foregoing guile-tun, 
 may serve as a specimen for mash-tuns, guile-tuns, cop- 
 pers, stills, &c. 
 
 3. By the solution to the first Example, we find the 
 content of the first 10 inches from the bottom to be 50.44 
 gallons ; and as 9 gallons make 1 firkin, and 4 firkins 1 
 barrel, we have only to divide 50.44 gallons by 9 and 4, 
 successively ; and we obtain 1 barrel, 1 firkin, and 5.44 
 gallons, for the reduced content of the first 10 inches. — 
 In the same manner we have 58.12 gallons = 1 barrel, 2 
 firkins, and 4.12 gallons, for the reduced content of the 
 second 10 inches ; and likewise 66.90 gallons = 1 barrel, 
 3 firkins, and 3.90 gallons, for the reduced content of the 
 third 10 inches of the depth ; the whole being equal to 4 
 barrels, 3 firkins, and 4.46 gallons, as in the preceding 
 Solution and following Table. 
 
 A. B.'s Guile-tun, gauged March 10th, 1827. 
 
 Depths. 
 
 I 
 
 Cross 
 Diameters. 
 
 Mean 
 Diam. 
 
 Areas in 
 Gallons. 
 
 Areas. 
 
 Contents. 
 
 10 
 10 
 10 
 
 41.8 
 44.7 
 48.5 
 
 42.6 
 45.9 
 
 48.7 
 
 42.2 
 45.3 
 48.6 
 
 5.044 
 5.812 
 6.690 
 
 bar. 
 
 
 
 
 
 firk. 
 
 
 
 
 
 gal. 
 
 5.044 
 5.812 
 6.690 
 
 bar. 
 1 
 1 
 1 
 
 4 
 
 firk. 
 1 
 
 2 
 3 
 
 3 
 
 gal. 
 
 5.44 
 4.12 
 3.90 
 
 4.46 
 
 1 
 
 30 
 
 Whole depth and content . . 
 
 RE^IARK. 
 
 Li order to find the content of the foregoing guile- tun, 
 at every incli from the bottom, proceed thus : Add the 
 
 o o 
 
298 
 
 GAUGING. 
 
 PART -^in. 
 
 area of tlie first section to itself, and the sum will be the 
 content at two inches of the depth ; to this add the same 
 area, and the sum will be the content at three inches ; 
 and thus continue adding the area of the first section, 
 until you come to the tenth inch ; and you will have the 
 content at every inch of the first ten inches. To this 
 content add the area of the second section, and you will 
 obtain the content at eleven inches from the bottom ; and 
 thus proceed until you come to the 30th inch, changing 
 the area at every ten inches ; and you will have the con- 
 tent at every inch of the depth, as in the following Table. 
 
 A Table of the foregoing Guile-tun. 
 
 Wet 
 In. 
 
 Contents. 
 
 Wet 
 In. 
 
 Contents. 
 
 Wet 
 In. 
 
 Contents. 
 
 1 
 
 bar. 
 
 
 
 firk. 
 
 
 
 gall. 
 
 5.044 
 5.044 
 
 11 
 
 bar. 
 1 
 
 firk. 
 
 2 
 
 gall. 
 
 2.252 
 5.812 
 
 21 
 
 bar. 
 
 3 
 
 flrk. 
 
 
 
 gall. 
 
 7.250 
 6.690 
 
 2 
 
 
 
 1 
 
 1.088 
 5.044 
 
 12 
 
 1 
 
 2 
 
 8.064 
 5.812 
 
 22 
 
 3 
 
 1 
 
 4.940 
 6.690 
 
 3 
 
 4 
 
 5 
 
 ~6" 
 
 
 
 1 
 
 6.132 
 5.044 
 
 13 
 
 1 
 
 3 
 
 4.876 
 5.812 
 
 23 
 
 3 
 
 2 
 
 2.630 
 6.690 
 
 
 
 2 
 
 2.176 
 5.044 
 
 14 
 
 2 
 
 
 
 1.688 
 5.812 
 
 24 
 
 3 
 
 3 
 
 0.320 
 6.690 
 
 
 
 
 2 
 
 7.220 
 5.044 
 
 15 
 16 
 
 2 
 
 
 
 7.500 
 5.812 
 
 25 
 
 3 
 
 3 
 
 7.010 
 6.690 
 
 3 
 
 3.264 
 5.044 
 
 2 
 
 1 
 
 4.312 
 5.812 
 
 2% 
 
 4 
 
 
 
 4.700 
 6.690 
 
 7 
 8 
 
 
 
 1 
 
 3 
 
 8.308 
 5.044^ 
 
 17 
 
 T8~ 
 
 2 
 
 2 
 
 1.124 
 5.812 
 
 27 
 
 4 
 
 1 i2.390 
 6.690 
 
 
 
 4.352 
 5.044' 
 
 2 
 
 2 
 
 6.936 
 5.812 
 
 28 
 
 4 
 
 2 |0.080 
 6.690 
 
 9 
 
 1 
 
 1 
 
 0.396 
 5.044 
 
 19 
 
 2 
 
 3 
 
 3.748 
 5.812 
 
 29 
 
 4 
 
 2 16.770 
 6.690 
 
 10 
 
 * 
 
 1 
 
 1 
 
 5.440 
 5.812 
 
 20 
 
 3 
 
 ^ 
 
 0.560 
 6.690 
 
 30 
 
 4 
 
 3 
 
 4.460 
 
 Note 1. The stars or asterisks, in the preceding Table, denote the changes of the 
 areas; and their use vvUl tend to prevent mistakes, in tabtc/ating other vessels. 
 
 2. The use of the foregoing Table is evident ; for when a vessel is once gauged 
 a.ndhiched, we have only to take the depth of the liquor, and against it we find the 
 content required: thus if the depth be 2.3 inches, we have 3 B. 2 F. 2.63 G. for the 
 content. If the depth be 23.5 inches, we take one-half of the area of the third 
 section; namelv, one-half of G.Cf* G. = 3.345 G., which being added to the content 
 at 23 inches, we' obtain 3 B. 2 F. 5.D75 G. for the content required. Again if the 
 
PART Tin. GAUGma. 299 
 
 depth be 23.8 inches, we take fjj of fi.G9 G = 5.352 G ; and this being added to the 
 content at 23 inches, we have 3 B. 2 ¥. 7.982 G. for the content at 23.8 inches of the 
 deptli. 
 
 3. Any inash-tun, guile-tun, copper, or still, whose sides are curved, may be 
 gauged and inched in the same manner ; but if the sides be much curved, cross- 
 diameters must be taken in the middle of every 4, 6, or 8 inches of the perpendicular 
 depth, as the case may require ; and the areas of the different sections must be 
 used, as before directed, in tabling the vessels. 
 
 4. In order to gauge and inch a mash-tun, or a gui!e-tun, in the form of a pris- 
 moid, or the frustum of a square pyramid, take lengths and breadths in the middle 
 of every G, 8, or 10 inches of the perpendicular depth, as the case may require ; and 
 divide the product of each length and breadth by 277.3, and the quotients will be 
 the areas of the sections in imperial gallons ; then enter the dimensions, areas, and 
 contents, in Tables, as before directed. 
 
 5. Large mash-tuns, and guile-tuns are generally fixed in an inclining position, 
 called tlie drip or fall of the tun, in order that the liquor may be conveniently 
 drawn off; wlien this is the case, pour in as many gallons of water as will just 
 cover the bottom ; and make marks upon the sides of the tun, at the surface of thfl 
 water ; then draw it off, and proceed with the remainder of the tun, as before di- 
 rected. 
 
 6. The water that is used to cover the bottoms of guile-tuns, and of coppers with 
 rising or falling crowns, must either be measured with a gallon measure, or drawn 
 qffl into a vessel of a known capacity, or into one that may be easily gauged ; and 
 the quantity must be added to the content of the other part of the guile-tun or 
 copper. 
 
 • PROBLEM Vin. 
 
 To gauge a back or cooler. 
 
 Backs or coolers are vessels which receive the wort, 
 when let out of the copper, in order to be cooled. They 
 are commonly of a rectangular form, and seldom exceed 
 nine or ten inches in depth. 
 
 Multiply the mean length by the mean breadth ; divide 
 the product by 277.3, and the quotient will be the area in 
 imperial gallons ; which being multiplied by the mean 
 depth, will give the quantity of wort contained in the 
 cooler. 
 
 2^ote 1. In taking the dimensions of a cooler, find the middle of both ends, and 
 also that of the sides, at which take the length and breadth ; but as coolers are 
 never fixed in a horizontal position, but inclined a little, in order that the wort 
 may run off, and as their bottoms are I'requently warped ; it is necessary to take the 
 depth of the liquor in several places, and divide the sum of these depths by their 
 number, for a mean depth. 
 
 2. In order to find a constant dipping-place, which will save much trouble, pro- 
 ceed thus: Find a mean depth, as directed in the last note ; then try in different 
 parts of the cooler, until you find a phice of the same depth as your mean depth ; 
 at which make a mark upon the side of the cooler, for a constant dijjping-place. 
 
 3. It lias been found by experiments that 10 gallons of hot wort wid only mea- 
 sure to 9 gallons when the wort is cold; therefore, an allowance of one gallon in 
 every ten must be made, when wort is gauged hot. In order to do this, multiply 
 the number of warm gallons by .9, and the product will be the number of gallons 
 when cold. 
 
 EXA3IPLES.- 
 
 1. The length AB of a cooler is 125.6 inches, the 
 breadth CD 73.4 inches, and the depth of the wort taken 
 o 6 
 
300 
 
 GAUGmG. 
 
 PART ATII. 
 
 in ten different places, as below ; how many gallons are 
 contained in the cooler ? 
 
 Depths. 
 Inches. 
 
 At a =4.3 
 b=:4.2 
 c =4.2 
 c?=:4.5 
 e=4.4 
 g=4.6 
 A =4.0 
 A =4.9 
 ?n = 4.8 
 n=4.6 
 10)45.0 
 
 4.5 mean. 
 
 Inches. 
 
 125.6 length, 
 73.4 breadth. 
 5024 
 3768 
 8792 
 277.3)9219.04(33.24 gallons, the area for one inch deep. 
 
 Gallons. 
 33.24 area. 
 
 4.5 mean depth. 
 16620 
 13296 
 
 149.580 the content of the wort in the cooler. 
 By the Sliding Rule. 
 
 Divisor. Length. Breadth. Area. 
 
 As 277.3 on A : 125.6 onB\\ 73.4 on A : 33.24 on B. 
 And, 
 
 Unity. Area. Depth. Content. 
 
 As Ion A: 33.24 onl^\\ 4.5 on A : 149.58 on B. 
 
 Note. The constant dipping-place maj* be either at dor h; but it is preferable to 
 find one nearer the side of the cooler, as at r. 
 
 2, If the mean depth of warm wort, in the foregoing 
 cooler, be 5.8 inches ; how many gallons will there be 
 when the wort is cold ? 
 
PART Vm. GAUGING. 301 
 
 Gallons. 
 
 33.24 area. 
 0.8 depth. 
 26o92 
 16 620 
 
 192.792 gallons of warm ivort. 
 .9 multiplier. 
 
 173.5128 gallons wlieii cold. 
 
 3. If the mean depth of warm wort in the preceding 
 cooler be 6.4 inches ; how many gallons when cold ? 
 
 Ans. 191. ^62A gallons. 
 
 Note 1. The depth of the liquor in a cooler is always taken to the tenth of an 
 inch ; and in order to facilitate the practical part of Gauging, Excise officers gene- 
 rally make a Table, exhibiting the content in barrels, firkins, and gallons, at every 
 tenth of an inch. This is called tcnthivg a cooler ; and maybe performed in the 
 following manner : Find ttie area of the cooler, and reduce it to barrels, firkins, 
 and gallons ; which being divided by ten, will give the content at one-tenth of an 
 inch in flepth. Add this content to itself, and the sum will be the content at two- 
 tenth?. Again, to tiiis content add that of the first tenth, and you will obtain the 
 content at three-tenths of the depth. Continue this operation until you arrive at 
 six rir seven inches of the depth, which will generally be found sufficient ; as the 
 wort in coolers is not often deeper. 
 
 2. It has been before observed that coolers are generally rectangular ; but should 
 Tou meet with one of any other shape, its area may be found by the Rules given in 
 tart II. 
 
 PROBLEM IX. 
 
 To gauge a cistern, couch, or floor of malt. 
 
 RULE. 
 
 Take the dimensions as directed in the last Problem ; 
 then multiply the mean length by the mean breadth ; di- 
 vide the product by 2218.2, and the quotient will be the 
 area ; which being multiplied by the mean depth, will 
 give the content in bushels. 
 
 'Note 1. According to Act of Parliament, barley must lie under water in the 
 cistern, forty hours; in which time it is supposed to swell or increase to one-fourth 
 more ; so that four bushels in twenty are allowed for this increase. 
 
 From the cistern the barley is removed to the couch ; and after having lain there 
 twentv-four hours, it is deemed a floor. The same allowance is made in the couch 
 as in tlie cistern ; but when the corn has been thrown out of the couch into the 
 floor, and there grown according to the usual custom, it is supposed to increase 
 one-half ; consequently an allowance is made of ten bushels in every twenty. 
 
 2. If cistern or couch -bushels be multiplied by .8, the product v.ill be neat 
 bushels ; but floor-bushels must be multiplied by .5, in order to reduce them to 
 neat bushels. 
 
 3. The duty is always charged upon the best gauge of the cistern, couch, or 
 floor ; and in order to find from which the charge will arise, without reducing 
 them to neat bushels, proceed thus : Multiply the btst gauge of the cistern or 
 couch by 1.6; and if the product exceed the floor bushels, the charge must be 
 made from the cistern or couch ; but if not, tlie charge must be made from the 
 floor. 
 
 This multiplier is found by dividing eight-tenths by five-ienths. (i'.v ike last 
 Note.) 
 
302 GAUGING. PART Vni. 
 
 EXAMPLES. 
 
 1. Tlie mean lengtli of a cistern is 96, the mean breadth 
 64, and the mean depth 32 inches ; what is the area and 
 content in malt bushels ? 
 
 To find the area. 
 Inches. 
 96 length. 
 64 breadth. 
 384 
 576 
 2218.2)6144.0(2.769 area in bushels. 
 
 By the Sliding Rule. 
 M. B. Length. Breadth. Area. 
 
 As 2218.2 on A : 96 ow B : ! 64 on A ; 2.77 on B. 
 
 To find the content. 
 Bushels. 
 2.769 area. 
 32 depth. 
 5538 
 8307 
 
 88.608 content. 
 
 By the Sliding Ride. 
 
 Unitv. Area. Depth. Content. 
 
 As 1 'on A : 2.77 ow B : : 32 on A : 88.6 on B. 
 Or, 
 The content may be readily found by means of the line 
 M D, on the Sliding Rule, without knowing the area : thus, 
 Lenj^th. Depth. Breadth. Content, 
 
 As 96 on B : 32 o?z M D : : 64 on A : 88.6 on B. 
 
 2. The mean length of a floor of malt is 115, the mean 
 breadth 112, and the mean depth 4.6 inches; what is its 
 content in Hoor -bushels ? 
 
 Inches. 
 115 length. 
 112 breadth. 
 230 
 115 
 115 _ 
 2218.2 )1288 0.0(5.806 area in bushels. 
 
PART Ylir. GAUGIXG. 303 
 
 GAUGIXG. 
 
 Bushels. 
 
 5.806 area. 
 4.6 dej^th. 
 34836 
 23224 
 
 26.7076 content. 
 
 By the Slidhnj Rule. 
 
 Length. Depth. Breadth. Content. 
 
 As 115 on B : 4.6 on MD :: 112 on A ; 26.7 on B. 
 
 3. The mean length of a cistern is 126.4, the mean 
 breadth 62.6, and the mean depth of the barley 32.8 in- 
 ches ; hoAY many neat bushels are contained in the cistern? 
 
 Inches. 
 
 126.4 length. 
 62.6 breadth. 
 
 7584 
 2528 
 7584 
 
 2218.2)7 912.64 (3.567 area in bushels. 
 
 Bushels, 
 
 3.567 area. 
 32.8 depth. 
 28536 
 7134 
 10701 
 
 116.9976 content in cistern bushels. 
 
 .^multiplier in Note 2. 
 
 93.59808 content in neat bushels. 
 
 ^ 4. If the mean depth of the barley in the foregoing 
 cistern be 38.6 inches; hov/ many neat bushels does it 
 contain ? ^^^^^ 1 2 o. 14896 bushels. 
 
 5. The length of a couch is 136.2, the breadth 72.6, 
 and the depth of the barley 42.8 inches ; what is its con- 
 tent in neat bushels ? Ans. 152.60768 bushels. 
 
 6. If the depth of the barley in the foregoing couch be 
 46.3 inches ; how many neat bushels does it contain ? 
 
 Ans. 165.08728 bushels. 
 
 . 7. The length of a floor of malt is 236, the breadth 212, 
 
 and the depth 5.2 inches ; what is its content in neat 
 
 bushels .^ Ans. 58.643 bushels. 
 
304 
 
 GAUGIXG. 
 
 PART vni. 
 
 8. If the best cistern-gauge be 68.4, the best couch- 
 gauge G9.8, and the best floor-gauge 109.o bushels ; from 
 which will the charge of the duty arise ? 
 
 A?is. From the couch, 
 
 Note. In this Problem, we have supposed the cistern, couch, and floor to be in 
 the form of a parallelopipedon, which is most commonly their shape ; but their 
 contents maybe obtained by the Rules given in Part IV., Section I., whatever form 
 they may assume. 
 
 PROBLEM X. 
 Cask Gauging. 
 
 The performance of this part of Gauging is the most 
 difficult that occurs, as no Rules can be given by wliicli 
 the exact form of casks may be ascertained. 
 
 There are commonly reckoned four forms or varieties 
 of casks, viz. 
 
 1. The middle frustum of a spheroid. 
 
 2. The middle frustum of a parabolic spindle. 
 
 3. The lower frustums of two equal paraboloids. 
 
 4. The lower frustums of two equal cones. 
 
 The Rule for finding the content of the 1st variety, is 
 given in Prob. 13. Part VII. ; that for the 2d in Prob. 18. 
 Part VII. ; that for the 3d in Prob. 16. Part VII. ; and 
 the content of a cask of the 4th variety may be obtained 
 by Prob. 8. Sect. I. Part IV. ; but it is very probable 
 that there never was a cask that agreed exactly Avith any 
 of the varieties ; for very few casks are to be met with 
 that will contain so much as the first form, or so littla as 
 the third or fourth ; so that the second variety is the 
 most general form of casks. 
 
 Note. Excise officers generally consider all casks as belonging to the first variety, 
 and gauge them as such ; but this practice ought to be abolished, as being injurious 
 to the Trader. {See Ncsbifs and Little's Practical Gauging.) 
 
 To take the dimensions of a standing cask. 
 
 Measure the distance between 
 the inside of the chimb, close to 
 the head, and the outermost sloped 
 edge of the opposite staff, which 
 will be the head diameter within 
 the cask, very nearly. 
 
 In order to find the bung dia- 
 meter, lay a straight rod AB 
 across the centre of the head, and 
 perpendicular to it, place another 
 straight rod am, so as to touch 
 the bulge of the cask at C ; mea- 
 
TART VIII. 
 
 GAUGING. 
 
 305 
 
 sure the distance between the outer edge of cacli chimb 
 at b and c ; also measure ah, which should be equal to 
 inn ; then he being added to twice ab, will give the bung 
 diameter CD, including the thickness of the staff on each 
 side of the cask. From this take twice the thickness of 
 the staff, at the bulge, as nearly as your judgment directs, 
 having regard to the size of the cask ; and you will obtain 
 the internal bung diameter. 
 
 It is unnecessary to give any directions for taking the 
 length of a cask in this position, admitting a hole to be in 
 the upper head, which is most commonly the case. 
 
 Kote\. The external bung diameter of a standing cask may also be found by 
 dividing the circumference by 3.1416. {See Prob. XIII. Part II.) 
 
 2. The staves of casks, in general, are thicker at the bulge than at the head; 
 London-made casks, however, have their staves commonly much thicker at the 
 head than at the bulge. The best method of forming a correct judgment is to 
 examine empty casks of the same size and make as those you are about to 
 gauge. By this means, you w ill come to tolerably correct conclusions relating to 
 the deductions necessary to be made for the thickness of the heads, staves, &c. 
 {See the Notes in the next Article.) 
 
 To take the dimensions of a hjing eask. 
 
 Measure the head diameter in the same manner as 
 directed for a standing 
 cask ; and in taking the 
 bung diameter CD, make 
 an allowance for the 
 thickness of the staff at 
 the bung-hole. 
 
 The length may be 
 most expeditiously found 
 by a pair of callipers, al- 
 lowing for the thickness 
 of the heads, according 
 
 to the size of the cask ; but as it cannot be expected that 
 every person concerned in Gauging is furnished with 
 this instrument, the length may be obtained in the follow- 
 ing manner . — Apply a straight rod AB, to the bulge of 
 the cask ; and, at right angles to it, place two others, ma 
 and na, touching the chimbs at each end, and making mr 
 equal to nr ; then measure the distance mn, from which 
 subtract the depths of the chimbs, together with the thick- 
 ness of the heads, as nearly as you can judge, and the 
 remainder will be the internal len^rth of the cask. 
 
 yotel. In taking the dimensions of a cask, the ganger ought carefully to ob- 
 serve that the bung-hole be in the middle ; tiiat tlie bung-statl' be regular and even 
 
30G 
 
 GAUGING. 
 
 TART YUI. 
 
 within ; and that the staff opposite the bung-hole be neitlier thicker nor thinner 
 than the rest, which he may easily ascertain bv the gauging-rod ; and if any impro- 
 priety be discovered, a proper allowance must be made for it, in the dimensions. 
 
 2. It is also necessary to observe that the heads of the casks be equal, and truly 
 circular ; if not, take cross diameters of each head, and divide their sum by foui 
 for a mean diameter. 
 
 To determine the content of a cask of any of the four 
 varieties, hy finding a mean diameter, or reducing it to 
 a cylinder. 
 
 RULE. 
 
 Multiply the difference between the bung and head 
 diameters, when it is 6 inches, or less, 
 
 By 
 
 .68 
 
 m 
 
 .55 
 .5 
 
 for the 
 
 variety. 
 
 Or, if the difference between the bung and head diame- 
 ters exceed 6 inches, 
 
 By 
 
 7 n 
 
 64 
 
 57 
 
 -for the^ 
 
 "1 
 
 2 
 
 3 
 
 52 
 
 
 4 
 
 variety. 
 
 Add the product to the head diameter, and the sum 
 will be a mean diameter. Square the mean diameter, and 
 multiply that • square by the length of the cask ; then, if 
 the product thence arising be divided by 353.04, you will 
 obtain the content of the cask, in imperial gallons. 
 
 By the Sliding Rule. 
 
 Find the difference between the bung and the head 
 diameters, on the inside of the slide marked C ; and oppo- 
 site to it, for each variety, is a number, which being added 
 to the head, will give the mean diameter. Then, as the 
 gauge-point on D, is to the length of the cask on C, so is 
 the mean diameter on D, to the content on C. 
 
 Note. Neither of the above methods of finding a mean diameter, is strictly true ; 
 but in consequence of their simplicity, they are generally adopted by officers of the 
 Excise. {See Moss's Gauging, Section X.) 
 
PAKT Vm. GAUGINCi^ 307 
 
 EX/UIPLF.S. 
 
 1. If tlie length of n cask be 30, the bung diameter 24, 
 and the head diameter 18 inches; what is its content in 
 imperial gallons, for the first, second, and third varieties ? 
 
 For the middle frustum of a spheroid, or the first variety. 
 
 Bung diameter 24 inches. 
 Head diameter 18 inches. 
 
 Difference 6 
 
 V Multiplier ^ 
 
 48 
 36 
 
 Product 4.08 
 
 Head diameter 18. 
 Mean diameter 22.08 
 
 Ditto 22.08 
 
 T7664 
 4416 
 4416 
 
 487.5264 square. 
 30 length. 
 Bi.Yisor 3o3.04) T4625?7920( 4 1.428 gallons. 
 
 By the Sliding Rule. 
 
 The difference between the bung and head diameters is 
 6 inches, against which, on the line of inches, we find 
 4.16 on the line marked spheroid^ which being added to 
 18, the head diameter, gives 22.16 inches, for the mean 
 diameter. Then, 
 
 OnVi. OnC. OnJ). OnC. 
 As 1S.79 : 30 :: 22.16 : 41.5 gallons 
 
 Note. The content given by the Kule on page 267. is 41.808R gaP.ons. 
 
30S GAUGING. PART VIH. 
 
 For the middle frustum of a parabolic spindle, or the 
 second variety. 
 
 Bunp: diameter 24 inches. 
 Head diameter 18 inches. 
 
 Difference 6 
 
 Multiplier .62 
 
 12 
 36 
 
 Product 3.72 
 
 Head diameter. 18. 
 Mean diameter 21.72 
 
 Ditto 2^72 
 
 471.7584 square. 
 30 length. 
 353.04)141 52.7520(40.088 gallons. 
 
 By the Sliding Rule. 
 
 Against 6, the difference of the diameters, on the line 
 of inches, we find 3.8 on the line marked second variety, 
 which being added to the head diameter 18, we obtain 
 21.8 for the mean diameter. Then, 
 
 OnJy. OnC. OnT>. OnC. 
 .-45 18.79 : 30 :: 21.8 : 40.15 gallons. 
 
 Note. The content found by the Rule on page 271. is 41.4009 gail^^ns. 
 
 For the loicer frustums of two equal paraboloids, or the 
 third variety. 
 
 Bung diameter 24 inches. 
 Head diameter 18 inches. 
 
 Difference G 
 
 Multiplier 55 
 
 "30 
 30 
 
 Product 3.30 
 
 Head diameter. 18. 
 Mean diameter 21.30 
 
 Ditto .^^1.30 
 
 453.6900 square. 
 30 length. 
 353.04)13610.7060(38.552 galloyis. 
 
PART Vm. GAUGING. 309 
 
 By the Sliding Rule. 
 
 The mean diameter, found as above, is 21.3 inches. 
 Then, 
 
 OnT). OuQ. Onl). On C. 
 As 18.79 : 30 :: 21.3 : 3S.6 gallons. 
 
 Note. Tht! content found by the Kule on page 270. is 38.2397 gallons. 
 
 2. The length of a cask is 45, the bung diameter 36, 
 and the liead diameter 27 inches ; what is its content in 
 imperial gallons, for the first, second, and tliird varieties ? 
 
 Ans. For the first variety^ 141.343; for the second^ 
 136.796; and for the third, 131.586 im2)erial gallons. 
 
 Koie. By the followinor General Rules, the content of this cask is 1.33.188 gal- 
 lons, which is between the second and third varieties. (See the Second Exampl:.) 
 
 GENERAL RULES 
 
 For finding the contents of casks from the head diameter, 
 hung diameter, and length; without paying any regard 
 to their variety. 
 
 RULK I. 
 
 Add into one sum, 39 times the square of the bung 
 diameter, 25 times the squr.re of the head diameter, and 
 26 times the product of those diameters ; and multiply the 
 sum by the length of the cask. Then, if the product thus 
 obtained, be multiplied by .000031473, or divided by 
 31773.244; the result will be the content of the cask, in 
 imperial gallons. 
 
 'Note. This Rule is taken from Dr. Hutton's excellent Mathematical and Philo- 
 sophical Dictionary, vol. i. page 528., and has now been adapted, by the author, 
 to the new imperial gallon. It gives the contents of casks less than the Kule 
 for the second variety, and more than that for the third, being nearly in the 
 middle between them. But the Dr. o!)serves that, " It agrees woJl with the real 
 contents of cas.ks ; as hath been proved by several casks which have actually been 
 filled with a true gallon measure, after their contents had been computed by this 
 method." 
 
 KULK II. 
 
 Divide the head diameter by the bung diameter, to two 
 places of decimals ; and find that number in the column 
 of quotients, in the following Table, against which we 
 have a multiplier or factor, for imperial gallons. Multiply 
 this factor by the square of tlie bung diameter, and the 
 product thence arising by the length of the cask, and 
 the last result will be the content of the cask in iin[)crial 
 gallons. 
 
310 
 
 GAUGING. 
 
 PAKT Tin. 
 
 Xoiel. If the quotient of the head by the bung diameter do not terminate in 
 tvro places of figures, without a fractional remainder, find the multiplier answering 
 to the first two decimals of the quotient, and subtract it from the next greater 
 multiplier ; then if the remainder be multiplied by the fractional part of the 
 quotient, the product will be the corresponding proportional part to be added to 
 the first multiplier. This method ought always to be used when the fractional 
 remainder is large, or accuracy required. 
 
 2. This Rule gives the content very nearly the same as Rule I., and will be 
 found much easier in practice, as it requires a great deal fewer figures in the 
 operation, particularly when the fractional remainder is rejected. 
 
 A NEW TABLE 
 
 Of multipliers or factors, for Ji?iding the content of any 
 cask, in imperial gallons. 
 
 Quotient of 
 the head 
 
 Multipliers 
 
 for 
 
 Imperial 
 
 Gallons. 
 
 Quotient of 
 the head 
 
 Multipliers 
 
 divided by 
 the bung 
 diameter. 
 
 divided by 
 the bung 
 diameter. 
 
 Imperial 
 Gallons. 
 
 .50 
 
 .0018333 
 
 .76 
 
 .0023038 
 
 .51 
 
 .0018494 
 
 .77 
 
 .0023240 
 
 .52 
 
 .0018657 
 
 .78 
 
 .0023443 
 
 .53 
 
 .0018822 
 
 .79 
 
 .0023649 
 
 .54 
 
 .0018988 
 
 .80 
 
 .0023856 
 
 .66 
 
 .0019155 
 
 .81 
 
 .0024064 
 
 .56 
 
 .0019324 
 
 ■ .82 
 
 .0024274 
 
 .57 
 
 .0019495 
 
 i .83 
 
 .0024486 
 
 .58 
 
 .0019667 
 
 j .84 
 
 .0024698 
 
 .59 
 
 .0019842 
 
 : .85 
 
 .0024913 
 
 .60 
 
 .0020017 
 
 i .86 
 
 .0025131 
 
 .61 
 
 .0020193 
 
 .87 
 
 .0025348 
 
 .62 
 
 .0020372 
 
 .88 
 
 .0025568 
 
 .63 
 
 .0020552 
 
 .89 
 
 .0025788 
 
 .64 
 
 .0020734 
 
 i .90 
 
 .0026012 
 
 .65 
 
 .0020917 
 
 .91 
 
 .0026236 
 
 .66 
 
 .0021102 
 
 .92 
 
 .0026462 
 
 .67 
 
 .0021288 
 
 1 .93 
 
 .0026689 
 
 .68 
 
 .0021476 
 
 .94 
 
 .0026918 
 
 .69 
 
 .0021666 
 
 .95 
 
 .0027149 
 
 .70 
 
 .0021858 
 
 .96 
 
 .0027381 
 
 .71 
 
 .0022050 
 
 .97 
 
 .0027615 
 
 .72 
 
 .0022244 
 
 .98 
 
 .0027850 
 
 .73 
 
 .0022440 
 
 .99 
 
 .0028086 
 
 .74 
 
 .0022638 
 
 1.00 
 
 .0028325 
 
 .75 
 
 .0022837 
 
 
 
FAKT VIIL GAUGING. 31 1 
 
 EXAMPLES. 
 
 1. The length of a cask is 30, the bung diameter 24, 
 and the head diameter 18 inches ; what is its content in 
 imperial gallons ? 
 
 By Rule I. 
 
 Here 24 X 24 X 39 :r^ 576 X 39 = 22464, thirty -nine 
 times the square of the bung diameter ; 18 X 18 x 25 = 
 324 X 26 ^=^ 8100, twenty times the square of the head 
 diameter ; and 24 X 18 x 26 = 432 x 26= 1 1232, twenty 
 six times the product of the diameters. Then, (22464 + 
 8100 4- 11232) X 30 X .000031473 = 41796 x 30 x 
 .000031473 = 1253880 x .000031473 = 39.46336 im- 
 perial gallons. Or, 1253880 -- 31773.244 = 39.46339 
 imperial gallons, the same as before. 
 
 By Rule 11. 
 Here 18 -4- 24 = .75, the quotient of the head divided 
 by the bung diameter. Opposite to this in the preceding 
 Table, ive have the mtdtiplier .0022837 ; then, .0022837 
 x 24-^ X 30 = .0022837 x 576 x 30 =: 1.3154112 x 30 
 = 39.462336 imperial gallons, the same as by the first 
 Rule. 
 
 Noie. The content of this cask, found by the foregoing Rules, for the first variety, 
 is 41.428 gallons ; for the second variety, 40.088 gallons; and for the third variety, 
 38..5.52 gallons: hence we see that the content given by the General Rules is be 
 tween the second and third varieties. 
 
 2. The lengtt of a cask is 45, the bung diameter 36, 
 and the head diameter 27 inches ; what is its content in 
 imperial gallons ? 
 
 Ans. By the first Rule, 133.188 ; and by the second 
 Rule, 133.185 imperial gallons. 
 
 3. What is the content of a cask whose bung and head 
 diameters are 32 and 24, and length 40 inches ? 
 
 Ans. By the first Rule, 93.5427 gallons. 
 
 4. The bung diameter of a cask is 45, the head diameter 
 34.2, and the length 56.4 inches ; ivhat is its content in 
 imperial gallons ? 
 
 Ans. By the second Rule, 263.1169 gallons. 
 
 5. Wliat is the content of a cask whose bung and head 
 diameters are 37.3 and 28.6, and length 52.8 inches ? 
 
 Ans. By the secoiid Ride, 170.2364 gallons 
 
312 GAUGING. PART "STIT. 
 
 DIAGONAL ROD. 
 
 To find the cvntent of a cask by the diagonal or gauging 
 rod. 
 
 This rod has two lines of numbers graduated upon it. 
 One of them is divided into inches and tenths, for the 
 purpose of taking dimensions ; and the other is called a 
 diagonal line, and expresses the content of any cask, in 
 imperial gallons, corresponding to the cask's diagonal in 
 inches and tenths. 
 
 "WHien the rod is put into the bung-hole of a cask, so as 
 to meet the head, where it intersects the staff opposite the 
 bung-hole, the content of the cask is exliibited in imperial 
 gallons on the diagonal line, reckoning from the end of 
 the rod to the centre of the bung-hole. 
 
 The diagonal rod is very much used in gauging, in con- 
 sequence of the ease and expedition with which the con- 
 tents of casks may be obtained by it, and it commonly 
 gives nearly the same as the General Rules. 
 
 Its construction is founded upon Theo. 20. Part I. ; 
 viz. that similar solids are to each other as the cubes of 
 their like dimensions. 
 
 T^ote 1. Those who have not a diagonal rod, may nevertheless find the content of 
 a cask bv tliat method in the following manner : _ Measure the diagonal of the 
 cask in inches and tenths ; then multiply the cube of this diagonal by .002266, and 
 the product will be the content of the cask in imperial gallons. 
 
 2. If the bung and head diameters, and length of a cask be given, the diagonal 
 may be found by the following Rule: — To the square of half the sum of the 
 diameters, add the square of half the length ; and the square root of the sum thence 
 arising will be the diagonal, or distance between the centre of the bung-hole and 
 the point where the middle of the opposite staff and head intersect each other. 
 
 EXAJIPLES. 
 
 1. The diagonal of a cask is found to be 20 inches; 
 what is its content in imperial gallons ? 
 
 By the Diagonal Line. 
 Opposite to 20 inches, the given diagonal of the cask, 
 we find on the diagonal line, 18 gallons, the content re- 
 quired. 
 
 By Note I. 
 
 Here 20 X 20 x 20 = 8000, the cube of the diagonal; 
 and .0022m X 8000= 18.128 gallons, the same as by 
 the diago7ial line. 
 
 2. The length of a cask is 30, the bung diameter 24, 
 and the head diameter 18 inches; required the diagonal 
 and content. 
 
PART VTII. GAUGING. 313 
 
 Bij Note 2, ive have (24+18) -^ 2 = 42 •- 2 = 21, half 
 the sum of the diameters; and 30-^2= 15, half the 
 length of the cask ; then A/(2r- + 15-) = V(441 + 225) = 
 ^/666 = 25.8 inches, the diagonal. 
 
 Again, by Note 1, ice have 666 x 25.8 x .002266 = 
 17182.8 X .002266 = 38.9362248 gallons; which is 
 nearly the same as given by the General Rules. ( See the 
 first Example?) 
 
 3. The length of a cask is 45, the bung diameter 36, 
 and the head diameter 27 inches ; required the diagonal 
 and the content. 
 
 Ans. The diagonal is 38.71 inches, and the content 
 131.4437 gallons ; which is the same as found for the third 
 variety; and differs only 1.14: gallons from the content 
 given by the General Rules. ( See the second Example.) 
 
 PROBLEM XI. 
 
 To ullage a standing cask. 
 
 The liquor contained in a cask when it is not full, is 
 called the wet ullage ; the vacuity or space not occupied 
 by the liquor, is termed the dry ullage ; and the method 
 of finding the content of the liquor, is called uUaging a 
 cask. 
 
 RULE. 
 
 Divide the wet inches by the length of the cask, to three 
 places of decimals ; and if the quotient exceed .500, add 
 to the said quotient one-tenth of the excess ; but if it be 
 less than .500, subtract one-tenth of the deficiency : then 
 the whole content of the cask being multiplied by the sum 
 in the first case, or the remainder in the second, will give 
 the ullage required. 
 
 Note. When the whole content of the cask is not known, it must be found from 
 proper dimensions, before the above Rule can be applied. 
 
 By the Sliding Rule. 
 
 Set the length of the cask upon C to 100 on the lino 
 marked Segt. St. or SS ; then against the wet inches on 
 C, you will have a segment on the line SS, which call a 
 fourth number. 
 
 Set 100 on the line marked A, to the content of the cask 
 upon B ; then against the fourth number, before found, on 
 A, is the quantity of liquor in the cask, upon B. 
 
 P 
 
314 GAUGING. PAIIT VIH. 
 
 EXAMPLES. 
 
 1. The length of a cask is 30, the bung diameter 24, 
 the head diameter 1 8, and the wet inches 22 ; what is the 
 uUage, in imperial gallons ? 
 
 Here 22 ~ 30 = .733, lohich exceeds .500, hj .233, 
 one-tenth part of ivhich is .0233 ; then .733 + .0233 = 
 .7563, the multiplier. 
 
 The ivhole content of the cask, as found in the Gejieral 
 Rules, in the last Problem, is 39.46 imperial gallons'; 
 then, 39.46 X .7563 = 29.843598 imperial gallons, the 
 ullage required. 
 
 By the Sliding Rule. 
 
 OnC. 0?zSS. OnC. On SS. 
 ^5 30 : 100 : : 22 : 75.5, a fourth number. 
 And, 
 On A. On B. On A. On B. 
 
 As 100 : 39.46 : : 75.5 : 29.6 gallons. 
 
 2. Let the dimensions and content be the same as in 
 the last example : what is the ullage in imperial gallons, 
 for 8 wet inches ? 
 
 Here 8 -4- 30 = .2m, ivhich is less than .500 by .234, 
 one-tenth part of which is .0234 ; then .266 — .0234 = 
 .2426, the multip>lier ; hence, 39.46 x .2426 = 9.572996 
 gallons, the ullage required. 
 
 Note. If we add 29.84, the ullage of the first example, to 9.57. the ullage of the 
 second ; we obtain 39.41 gallons, the whole content of the cask, nearly. 
 
 By the Sliding Ride. 
 
 On C. On SS. On C. On SS. 
 As^O : 100 :: 8 : 24.5, a fourth ?iumber. 
 
 And, 
 
 On A. On B. On A. On B. 
 
 ^s 100 : 39.46 : : 24.5 : 9.7 gallons. 
 
 3. The length of a cask is 50, the bung diameter 40, 
 the head diameter 30, and the wet inches 20 ; what are 
 the wet and dry ullages, in imperial gallons ? 
 
 Ans. By Ride II. Problem X. the content is found to 
 be 182.696 ; and hence the Wet ullage 71.25144, and the 
 dry ullage 111.44456 imperial gallon? 
 
PART Vm. GAUGING. 313 
 
 PROBLEM Xn. 
 
 To ullage a lying cash, 
 
 RULE. 
 
 Divide tlie wet indies hj the bimg diameter, to three 
 places of decimals ; and if the quotient exceed .500, add 
 to the said quotient one-fourth of the excess ; but if it be 
 less than .500, subtract one-quarter of the deficiency: 
 then multiply the whole content of the cask by the sum 
 in the first case, or the remainder in the second, and the 
 product will be the ullage required. 
 
 By the Sliding Rule. 
 
 Set the bung diameter upon C to 100 on the line 
 marked Seg. Ly. or SL ; then against the wet inches on 
 C, you will have a segment on the line SL, which call a 
 fourth number. 
 
 Set 100 on the line marked A, to the content of the 
 cask upon B ; then against the fourth number on A, is 
 the quantity of liquor in the cask, upon B. 
 
 EXA3IPLES. 
 
 1. The length of a cask is 30, the bung diameter 24, 
 the head diameter 18, and the wet inches 15 ; what is the 
 ullage, in imperial gallons ? 
 
 Here 15 -^ 24 = .^2^, ivhich exceeds .500 by .125, 
 one-fourth part of which is .03125 ; then .625 -f .03125 
 ^ .65625, the multiplier. 
 
 The whole content of the cash, as found by the General 
 Hides, in the tenth Problem, is 39.46 imperial gallons ; 
 then, 39.46 x .6oQ26 = 25.895625 imperial gallons, the 
 idlage required. 
 
 By the Slidijig Bide. 
 
 On C. On SL. On C. On SL. 
 ^5 24 : 100 :: 15 : Q1 , a fourth number. 
 And, 
 On A. OwB. On A. 0?z B. 
 
 ^5 100 : 39.46 :: 67 : 26.4 gallons. 
 
 2. Let the dimensions and content be the same as in 
 the last example ; what is the ullage, in imperial gallons, 
 for 9 wet inches ? 
 
 Here 9 -^ 24 = .375, which is less than .500, by .125. 
 p 2 
 
316 mSCELLANEOUS QUESTIONS PART Vin. 
 
 one-fourth part of lohich is .03125 ; then .375 — .03125 
 = .34375, the multiplier; hence, 39.46 X .34375 = 
 13.564375 imperial gallons, the ullage required. 
 
 Note. If we add 25.89, the ullage of the first example, to 13.56, the ullage of the 
 second ; we obtain 39.45 gallons, the whole content of the cask, nearly. 
 
 By the Sliding Rule. 
 
 On C. On SL. On C. On SL. 
 ^5 24 : 100 :: 9 : ^2.5, a fourth number. 
 And, 
 On A. On B. On A. On B. 
 
 As 100 : 39.46 : : 32.5 : 12.8 gallons. 
 
 3. The length of a cask is 32, the bung diameter 25.6, 
 the head diameter 19.2, and the wet inches 15.8 ; what 
 are the wet and dry ullages, in imperial gallons ? 
 
 Ans. By Rule II. Problem X. the content is found to 
 be 47.892660224 : and hence the wet ullage 30.9506, and 
 the dry ullage 16.942 imperial gallons. 
 
 IVnSCELLANEOUS QUESTIONS 
 
 CONCERNING 
 
 GAUGING. 
 
 1. The perpendicular depth of a vessel in the form of 
 a parallelopipedon is 52, its breadth 75, and the diagonal 
 of its bottom 125 inches ; what is its content in imperial 
 gallons? Ans. \A0QA19 gallons. 
 
 2. A vessel in the form of a parallelopipedon contains 
 675 imperial gallons ; its length is 85, and its breadth 64 
 inches ; what is its depth ? A?is. 34.4044 inches. 
 
 3. The diagonal of a cylindrical vessel is 45, and its 
 diameter 27 inches ; what is its content in imperial 
 gallons ? Ans. 74.3456 gallons. 
 
 4. The greatest diameter of a vessel in the form of the 
 frustum of a cone, is 96, the least diameter 48, and its 
 slant height 5 1 inches ; how many imperial gallons will it 
 contain ? Ans. 685.3257 gallons. 
 
 5. A reservoir measures 144 inches in length, 122 in 
 breadth, and 85 in depth ; how long will a person be in 
 filling it with water, by means of a pump ; supposing he 
 makes 30 strokes in a minute, and lifts 3 pints of water 
 at each stroke ? A?is. 7 hours, 58.717 minutes. 
 
 6. The slant heijrht of a cistern in the form of the 
 
PAET Vni. IN GAUGING. 317 
 
 frustum of a square pyramid is 153, the perpendicular 
 height 1 35, and the side of the least end 92 inches ; what iS 
 its content in imperial gallons? Ans. 13935.2326 gallons. 
 
 7. The altitude of a vessel in the form of a hexagonal 
 prism is 60, and the side of its base 30 inches ; what is its 
 content in imperial bushels ? A7is. 63.2477 bushels. 
 
 8. The greatest depth of the liquor in a cylindrical 
 vessel, placed upon an inclined plane, is 38 inches, the 
 least depth 32 inches, and the diameter of the vessel 36 
 inches ; how many imperial gallons does it contain ? 
 
 Ans. 128.4985 gallons. 
 
 9. The transverse diameter of an elliptical bath mea- 
 sures 144, the conjugate 112, and the depth 60 inches ; 
 how many imperial gallons will it contain ? 
 
 Ans. 2740.7712 gallons. 
 
 10. If the internal diameter of a hollow sphere be 100 
 inches ; how many bushels of corn, imperial measure, will 
 it hold ? Ans. 236.0472 bushels. 
 
 11. The top diameter of a conical vessel measures 32, 
 and its slant height 34 inches ; how many imperial gallons 
 of wine will it contain ? Ans. 29.002 gallons. 
 
 12. The top diameter and depth of a vessel, in the 
 form of the greater segment of a globe, are 32 inches 
 each ; how many imperial gallons will it hold ? 
 
 A71S. 108.2773 gallons. 
 
 13. If the linear side of each Platonic body be 30 
 inches ; required their respective contents in imperial 
 gallons. 
 
 Ans. The content of the tetraedron is 11 "4748, the 
 hexaedron 97'3674, the octaedron 45.8994, dodecaedron 
 746.1385, and the icosaedroji 212.4261 imperial gallons. 
 
 14. Two porters agreed to drink off a pot of strong 
 beer at two pulls, or a draught each; now, the first 
 having given it a black eye, as it is called, or drunk till 
 the surface of the liquor just touched the opposite edge of 
 the bottom, gave the remaining part to the other ; what 
 was the difference of their shares, supposing the pot was 
 the frustum of a cone, whose top diameter was 3.7, bottom 
 diameter 4.23, and perpendicular depth 5.7 inches ? 
 
 Ans. 7.06511 cubic inches, 
 
 15. At Konigstein, near Dresden, in Germany, is a 
 cask whose head diameter is 25, bung diameter 26, and 
 perpendicular altitude 28 feet ; how many gallons of wine, 
 imperial measure, will it contain ? 
 
 p 3 
 
318 CO:iIPUTING DISTANCES BY PART YIU. 
 
 Ans. The content found by Rule IL, Problem X., is 
 89674.383 gallons, which exceeds the content of the cask 
 at Heidelberg, by 31387.4255 imperial gallons. {See 
 Example 4, Problem III.) 
 
 The Konijrstein cask was begun in the j'ear 1722, and finished in 1725, under the 
 direction of General Kj-aw ; and is considered to be the largest cask in the world. 
 
 It consists of 157 staves, each 8 inches in thickness ; and one of its heads is com- 
 posed of 26, and the other of 28 boards. 
 
 The top or upper head of tins enormous cask is railed round, and affords suffi- 
 cient room for twenty persons to regale themselves ; and there are several sorts of 
 large goblets, called"" Welcome Cups," offered to strangers, who are invited to 
 drink by a Latin inscription, which in English is as follows : — " Welcome, Tra- 
 veller, and admire this Monument, dedicatfd to Festivity, in order to exhilarate the 
 Mind with a cheerful Glass, in the year 1725. by Frederic Augustus, King of 
 Poland, and Elector of Saxoni/, the Father of his Country, the TiTUS of his Age, 
 the Delight of Mankind: Drink, therefore, to the Health of the Sovereign, the 
 Country, the Electoral Family, and Baron Kyaw, Governor of Konigstein ; and if 
 thou be able, according to the Dignity of this Cask, the most capacious of all Casks, 
 drink to the Prosperitj- of the whole Universe ; so farewell." 
 
 'Note. Those who wish to see this subject more fully treated, are referred to 
 Nesbit's and Little's Treatise on Practical Gauging. 
 
 THE 
 
 METHOD OF COJilPUTING DISTANCES 
 
 BY THE 
 
 VELOCITY OF SOUND. 
 
 The velocity of sound, or the space through which it 
 is propagated in a given time, has been very differently 
 estimated by authors who have wntten upon this subject. 
 Koberval states it at the rate of 560 feet in a second of 
 time ; Gassendus at 1473 ; Musenne at 1474 ; Duhamel, 
 in the History of the Academy of Sciences at Paris, at 
 1338 ; Newton at 968 ; Derham, in whose measure Flam- 
 stead and Halley acquiesce, at 1142. 
 
 The reason of this variety is ascribed, by Derham, 
 partly to some of those gentlemen using strings and 
 plummets instead of regular pendulums ; partly to the 
 too small distance between the sonorous bod}^ and the 
 place of observation ; and partly to no regard being paid 
 to the winds. 
 
 By the account since published by M. Cassini de Thury, 
 in the Memoirs of the Royal Academy of Science at 
 Paris, 1738, where cannon were fired at various, as well 
 as great distances, under many varieties of wind, weather, 
 and other circumstances, and where the measm-es of the 
 different places had been settled with the utmost exact- 
 ness, it was found that sound was propagated, on a 
 
PART ^III. THE VELOCITY OF SOi;:sD. 319 
 
 medium, at tlie rate of 1038 French feet, in a second of 
 time ; and as the French foot is to the English, in the 
 proportion of 15 to 16, it follows that 1038 French feet 
 are equal to 1107 English feet. Therefore, the difference 
 of the measures of Dcrham and Cassini is 35 English 
 feet, or 33 French feet in a second. 
 
 The medium velocity of sound, therefore, is nearly at 
 the rate of 1 mile, or 5280 feet, in 42- seconds, or 1 league 
 in 14 seconds, or 13 miles in 1 minute. But sea miles 
 are to land miles nearly as 7 to 6; and, therefore, sound 
 moves over 1 sea mile in nearly 5^- seconds, or 1 sea 
 league in 16 seconds. 
 
 It is also commonly observed, that persons in health, 
 have about 75 pulsations, or beats of the artery at the 
 wrist, in 1 minute ; consequently, in 75 pulsations, sound 
 flies about 13 land miles, or U-i sea miles, which is nearly 
 1 land mile in 6 pulsations, and 1 sea mile in 7 pulsations, 
 or 1 league in about 20 pulsations ; and hence the distance 
 of objects may be found by knowing the time occupied by 
 sound, in passing from those objects to an observer. 
 
 Having given the time, in seconds or pulsatio7is, that 
 sound is in passing from an object to an observer, to find 
 the distance of the object. 
 
 RULES. 
 
 1. As 1 second is to 1107 feet, so is the time in 
 seconds, to the distance of the object in feet, vfhich being 
 divided by 3, will give the distance in yards ; and tliis 
 again divided by 1760, will give the distance in miles. 
 
 2. As 75 pulsations is to 22140 yards, so is the time, in 
 pulsations, to the distance in yards. 
 
 3. Wlien accuracy is not required, say, as 6 pulsations 
 is to 1 mile, or 1760 yards; so is the time in pulsations, 
 to the distance in miles, or yards. Or take -{^^ of the time 
 in seconds, for the distance in miles. 
 
 'Note. In \\ pulsations, sound flies a quarter of a mile ; in 3 pulsatiors, half a 
 mile ; in 4v'- pulsations, three quarters of a mile ; in G pulsations, one mile, &c. &c.: 
 hence the distance of an object from which sound proceeds, may be easily ascer- 
 tained without the trouble of calculation. 
 
 exa:mples. 
 
 . 1. After observing a flash of lightning, I found by 
 my watch that it was 16 seconds before I heard the 
 thunder ; how far was I from the cloud whence it came ? 
 p 4 
 
320 ^ TONNAGE OF SHIPS. PAET Vin. 
 
 As 1 : 1107 :: 16 
 
 16 
 6642 
 1107 
 
 3) 17712 yards. 
 1760)5904(3 miles, 624 yards. A7is. 
 
 2. If the report of the Tower- guns be heard at 
 Shooter's Hill, 38 seconds after they are fired ; what dis- 
 tance are the two places from each other ? 
 
 Here (38 X 3) — 14 = 114 -f- 14 = 8.142 miles. 
 
 3. After seeing the flash of a gun at sea, I counted 58 
 pulsations, at the wrist ; what was the distance between 
 the ship from which the gun was fired, and the place 
 where I stood ? A?is. 9.728 77iiles. 
 
 4. Perceiving a man at a distance felling a tree, I 
 remarked that 5 of my pulsations passed between seeing 
 him strike with the axe, and hearing the report of the 
 blow ; what was the distance between us ? 
 
 A?is. 1476 yards. 
 
 5. The report of a clap of thunder was heard 4 pulsa- 
 tions after seeing the flash of lightning ; what was the 
 distance of the cloud from which the thunder issued ? 
 
 A71S. U80.8 yards. 
 
 Note 1. Here it may not be improper to remark, that when the report of thunder 
 is heard nearly at the same instant the lightning is seen, the observer is in great dan- 
 ger ; as it is evident that the thunder cloud is very near him. 
 
 2. It may also be remarked that persons should never take shelter, in a thunder- 
 storm, under a tree ; as the branches will attract the lightning, if it comes within 
 the sphere of their attraction. Many have met premature deaths by taking shelter 
 under trees, during ■thunder-storms. 
 
 METHODS 
 
 OF 
 
 ASCERTAINING THE TONNAGE OF SHIPS. 
 
 PKELianNAKY OBSERVATIONS. 
 
 If the number of cubic feet of water which a ship dis- 
 places, in sinking from the light water-rTiark to the load 
 water-markj be divided by 35, the number of cubic feet 
 
PART vin. 
 
 TONNAGE OF SHIPS. 321 
 
 of sea-water in 1 ton, the quotient will be the number of 
 tons, which the vessel is capable of carrying, or the 
 vessel's true burthen. 
 
 Now the number of cubic feet of water so displaced, is 
 exactly equal to the solid content of so much of the body 
 of the ship, as is contained between the two water-lines or 
 marks ; but in consequence of the great variety of forms 
 given to this part of the vessel, no exact practical Rule 
 can be given that will apply in all cases. If, indeed, the 
 areas of three or five horizontal sections of that part of 
 the hull contained between the said two water-lines, could 
 be obtained by means of equi-distant ordinates, then the 
 content might be correctly found by Problem VI., Part 
 VIII. ; but as taking of the dimensions of these sections 
 would be attended with great difficulty, we shall give the 
 following Parliamentary Rules for ascertaining the ton- 
 nage of Merchants' and King's ships. 
 
 CASE I. 
 When the vessel is laid dry, 
 
 RULE. 
 
 Measure the length on a straight line along the rabbit 
 of the keel of the ship, from the back of the main stern- 
 post, to a perpendicular line let fall from the fore-part of 
 the main-stem, under the bow-sprit; from this length 
 subtract f of the extreme breadth; and the remainder 
 will be the length of the keel, for tonnage. The breadth 
 must be taken from outside to outside of the plank, in the 
 broadest part of the ship, whether above or below the 
 main-wales, exclusive of all manner of doubling planks, 
 or sheathing, that may be wrought upon the sides of the 
 vessel ; then multiply the length of the keel, in feet, by 
 the breadth ; divide the last product by 94 ; and the quo- 
 tient will be the tonnage required. 
 
 EXAMPLES. 
 
 1. The length from the back of the stern-post to a line 
 let fall from the fore-part of the main-stem, is 88 feet 
 6 inches ; the extreme breadth from outside to outside of 
 the plank, 26 feet 6 inches ; required the tonnage of the 
 ship. 
 
 p 5 
 
322 TONNAGE OF SHIPS. PAKT VIU. 
 
 feet. 
 
 Gross length 88.5 of the heel. 
 26.5 X f = 15.9 the deduction. 
 True length 72.6 difference. 
 
 26.5 breadth of the beam. 
 3630 
 4356 
 1452 
 1923.90 first product. 
 
 Then 1923.90 x 13.25, half the breadth — 25491.6750 
 second product. 
 
 And 25491-675 -4- 94 = 271.188 tons, the tonnage re- 
 quired. 
 
 2. The lengtli from the back of the stern-post to a line 
 let fall from the fore-part of the main-stem, is 108 feet 
 9 inches ; the extreme breadth from outside to outside of 
 the plank, 29 feet 6 inches ; required the tonnage of the 
 vessel. Ans. 421.469 to7is. 
 
 CASE II. 
 When the vessel is afloat, 
 
 RULE. 
 
 Let fall a plumb-line over the stern of the ship, and 
 measure the distance between this line and the aft-part of 
 the stern-post, at the land water-mark; then measure 
 from the top of the said plumb-line, in a parallel direction 
 with the water, to a perpendicular point immediately over 
 the load water-mark, at the fore-part of the main-stem ; 
 from the last measured distance subtract the former ; and 
 the remainder will be the ship's extreme length. From 
 this length deduct 3 inches for every foot of load-di'aught 
 of water, for the rake abaft ; and also f of the ship's ex- 
 treme breadth, for the rake of the stem ; and the remainder 
 will be the true length of the keel, for tonnage. The 
 extreme breadth must be measured, and the tonnage 
 found as directed in the first Case. 
 
 EXA3IPLE. 
 
 The true length of an eighty-gun ship, after all deduc- 
 tions are made, in taking the dimensions, is 150 feet 9 
 inches ; and the extreme breadth 50 feet 6 inches ; re- 
 quired the tonnage of the vessel. Ayis, 2044.9478 tons. 
 
 'Note 1 . It is found, by experience, that ships of war carry less ; and most mer- 
 chant ships carry cousiclerably more tonnage than they are rated at, by the pre- 
 ceding Rules. 
 
 2. Some writers on this subject, divide by 100, instead of 94, for King's ships. 
 On the same principles, the divisors for Merchants' ships should be decreased, 
 perhaps to 00 or 92. 
 
 3. Solutions to the last two Questions may be found in the Key to Nesbit's and 
 Little's Practical Gauging. 
 
PAKT IX. TRIGONOMETRY. 323 
 
 PAKT IX. 
 TRIGONOMETRY. 
 
 Trigonometry is an important branch of the mathe- 
 matical sciences ; by it we determine the magnitudes of 
 the earth and planets, their mutual distances and motions r. 
 it is that part of science which teaches how to measure 
 the sides and angles of triangles, plane and spherical. 
 
 Plane trigonometry is the art of measuring plane tri- 
 angles : and of determining the sides and angles which 
 principally depend on the properties of the circle, and 
 circular arcs. 
 
 Every circle is divided into 360 equal parts called de- 
 grees, each degree into 60 equal parts called minutes, each 
 minute into 60 equal parts called seconds, and so on by a 
 sexigesimal division into thirds, fourths, fifths, &c. 
 
 An angle is spoken of as containing as many degrees, 
 minutes, seconds, &c. as are contained in the arc, or part 
 of the circumference hj which it is measured. 
 
 Degrees, minutes, seconds, &c. are marked at the top 
 of the figures of their quantity in the order °, ', ^\ "^, ^^^ 
 &c. as 20° 34' 40'' ZT" 48^^ are 20 degrees, 34 minutes, 
 40 seconds, 37 thirds, 48 fourths ; but it is not usual is 
 modern practice to extend the division by 60 beyond 
 seconds, lower denominations, if there be any, being ex 
 pressed in decimal parts of a second ; thus, in the above 
 number, 37'" 48"'^ = -f^'^ of a second ; hence it would be 
 written 20° 34' 40'^63. 
 
 A right angle is measured by the fourth part of the 
 circumference, or 90° ; an obtuse angle is greater than 
 90°, and an acute angle is less than 90°. 
 
 The complement of an angle is what it wants of a right 
 angle or 90°, and the supplement is what it wants of two 
 right angles, or 180°. 
 
 The complement of an arc is what it wants of a quad- 
 rant, or 90°, And the supplement is what it wants of a 
 semicircle, or 180". 
 
 The chord of an arc is a straight line drawn from one 
 extremity of the arc to the other, as AB is a chord to the 
 p 6 
 
324 TRIGONOIMETRY. PART IX. 
 
 arc ADB ; it is also tlie chord of the arc AEB, the re- 
 mainder of the circle. 
 
 The sine of an arc, is a straight 
 line drawn from one extremity of 
 the arc, perpendicular to, and ter- 
 minated by, the diameter drawn 
 from the other extremity of the arc. 
 As AF perpendicular to, and ter- 
 minated by the diameter DE, is the 
 sine of the arc AD; it is also the 
 sine of the arc AE, the supplement 
 of the arc AD. 
 
 The tangent of an arc, is a straight line drawn per- 
 pendicularly to the extremity of the diameter, just touch- 
 ing the arc, in a point at one extremity of that arc, and 
 continued from thence to meet a line drawn from the 
 centre of the circle through the other extremity, which 
 line is called 
 
 The secant of the arc : thus DG is the tangent of the 
 arc DA, and CG is the secant of the same arc ; DG is also 
 the tangent of the arc AE, and CG the secant of the arc 
 AE, the supplement of AD. 
 
 The versed sine of an arc is that part of the diameter 
 contained between the sine and the arc ; thus FD is the 
 versed sine of the arc AD, and FE is the versed sine of 
 the supplemental arc AE. 
 
 If the arc DH be taken equal to a quadrant, and the 
 diameter HI be drawn perj^endicular to the former, DE, 
 then the arc AH is the complement of the arc AD, and 
 AK, HL, CL are the sine, tangent, and secant respect- 
 ively of the arc AH ; we have, in other words, AK the 
 sine of the complement of the arc AD, HL the tangent of 
 the complement of the same arc ; or, more briefly, AK is 
 the co-sine of the arc AD, HL is the co-tangent, CL the 
 co-secant, and HK is the co-versed sine, of the arc AD. 
 
 In place of the arc we may put the angle at the centre, 
 which the arc subtends and measures : thus, AF, DG, 
 CG is the sine, tangent, and secant respectively of the 
 angle ACD, or angle ACE, and AK, HL, CL is the sine, 
 tangent, and secant respectively of the angle ACH, or 
 the co-sine, co-tangent, and co-secant of the angle ACD : 
 ACH being the complement of ACD. Conversely, ACD 
 is the complement of ACH, and AF, the sine of ACD or 
 ACE, is the co-sine of ACH, and similarly of the tangent 
 
PAKT IX. TRIGONOMETRY. 325 
 
 and secant ; hence, when the sine, cosine, tangent, co- 
 tangent, secant, cosecant of any obtuse angle is required, 
 we may take the sine, cosine, &c. of the supplemental 
 acute angle : or, for the sine of an obtuse angle take the 
 cosine of its excess above a right angle ; for the cosine 
 of an obtuse angle take the sine of its excess above a 
 right angle : in place of the tangent, cotangent, secant, 
 cosecant of an obtuse angle, we may take the cotangent, 
 tangent, cosecant, secant of the excess of the given angle 
 above a right angle. 
 
 The arc ADB is double of the arc AD, and its chord 
 AB is double of the sine AF ; hence the sine of any arc is 
 half the chord of twice the arc. 
 
 The versed sine of an arc and the versed sine of its- 
 supplement are together equal to the diameter ; thus FD, 
 FE are together equal to DE. 
 
 The figure KACF is a parallelogram, and KA and CF 
 are equal; taking CF, therefore, in place of AK for the 
 cosine of the angle ACD, or of the arc AD, it is manifest 
 when the arc is less than a quadrant that its versed sine and 
 cosine are together equal to the radius, and that when the 
 arc is greater than a quadrant the excess of its versed 
 sine above its cosine is equal to the radius : thus FD and 
 CF are together equal to the radius CD, and the excess 
 of EF above CF is equal to the radius CE. 
 
 In the right-angled triangle ACF, whereof the radius 
 CA is the hypothenuse, and AF, CF the sine and cosine 
 of the same angle, viz. of the angle ACF, the squares 
 of the sine and cosine of any angle are together equal to 
 the square of the radius. 
 
 In the right-angled triangle CDG, we find the squares 
 of the radius and tangent are together equal to the square 
 of the secant. 
 
 From the right-angled triangle LCH, the squares of 
 the radius and cotangent are together equal to the square 
 of the cosecant. 
 
 From the similar triangles CFA, CDG, we have CF : 
 FA: : CD : DG ; or the rectangle under CF and DG equal 
 to the rectangle under FA and CD, or the rectangle under 
 the cosine and tangent equal to the rectangle under the 
 sine and the radius. 
 
 Again, CF : CA : : CD : CG ; or the rectangle under 
 CF and CG is equal to the rectangle under CA and CD, 
 
326 TRIGONOMETRY. PART IX. 
 
 or the square of CA or of CD ; that is, the rectangle under 
 the cosine and secant is equal to the square of the radius. 
 
 In like manner from the similar triangles ACK, LCH, 
 we find the rectangle under the sine and cotangent equal 
 to the rectangle under the cosine and radius ; and the 
 rectangle under the sine and cosecant equal to the square 
 of the radius. 
 
 Comparing the triangles CDG-, CHL together, they are 
 similar, and DG- : CD : : CH : HL ; whence the rectangle 
 under the tangent and cotangent (i. e. under DG and 
 TTTi) is equal to the square of the radius (i. e. to the rect- 
 angle under CD and CH). 
 
 By the known property of the circle, the rectangle 
 under CF and, FD is equal to the square of AF ; or the 
 square of the sine is equal to the rectangle under the 
 versed sine, and the versed sine of the supplement (which 
 Mendoza in his Tables calls the suversed sine, as HK is 
 the coversed sine). 
 
 These are the most important and useful properties of 
 the lines di-awn in and about the circle. 
 
 It is plain, from an inspection of the diagram, that the 
 sine and cosine can never exceed the radius, the secant 
 and cosecant can never be less than the radius, and the 
 tangent and cotangent may have any degree of magnitude 
 whatever. 
 
 And in the same proportion that the tangent is less or 
 greater than the radius, so is the cotangent greater or 
 less. 
 
 "Wlien the angle is 90°, its sine becomes the radius 
 CH. 
 
 The side of a hexagon which is the chord of 60° is equal 
 to the radius, whence the sine of 30° is half the radius. 
 
 Wlien the angle ACD is 45°, the triangle CDG has the 
 angle CGD also 45°, and therefore DG equal to CD; 
 whence the tangent of 45° is equal to the radius. 
 
 In every plane triangle the greater side is opposite the 
 greater angle, and conversely the greater angle is opposite 
 the greater side. 
 
 And when two sides are equal, the angles opposite are 
 equal; and when two angles are equal, the sides opposite 
 to them are equal. 
 
 In every plane triangle the tln-ee angles are together 
 equal to two right angles, or 180°; hence if one of the 
 angles be a right or obtuse angle, the rest are acute. 
 
PART IX. TRIGONOMETRY. 32 < 
 
 In every plane triangle, if one angle be known tlie 
 sum of the other two is found by subtracting the known 
 one from 180° ; and if two of the angles are known, the 
 third is found by subtracting the sum of the given angles 
 from 180°. 
 
 Hence whenever two angles of a plane triangle are 
 known, the whole three may be considered as known. 
 
 And in aright-angled plane triangle, if one of the acute 
 angles be known the other is found by subtracting the 
 known one from 90°, the acute angles being complements 
 one of the other. 
 
 Every plane triangle consists of six parts, viz. three 
 sides and three angles ; and it is necessary that three of 
 these be known, in order to determine the other three, 
 and consequently the triangle itself. 
 
 The combinations of six things taken three together are 
 twenty, whence it would appear that there are twenty cases 
 in general in plane triangles; but several of these cases fall 
 under one general case, so that there are in reality but 
 three, the case of the three angles being given being ex- 
 cluded, for the triangle cannot be determined from these 
 data alone : since an infinite number of triangles have 
 their three angles respectively equal, it is therefore neces- 
 sary that a side be included among the given quantities. 
 
 The tlieee general cases are : 
 
 1. Given — A side and angle, opposite each other, to- 
 gether with another side or another angle. 
 
 2. Given — Two sides and the angle included. 
 
 3. Given. — The three sides. 
 
 These include right-angled triangles of course; but 
 these latter are of sufficient importance to form a distinct 
 class, and be treated of separately : and by sides we shall 
 understand not any indiiFerently, but only those about the 
 right angle, that opposite being always called the hypo- 
 thenuse ; and here are three cases, viz. 
 
 1. Given — An acute angle, and with the hypothenuse 
 or side. 
 
 2. Given — The hypothenuse and a side. 
 
 3. Given — The two sides. 
 
 Now the Trigonometrical Tables of Sines, Tangents, &c. 
 are a register of 5400 right-angled triangles, in which the 
 hypothenuse or a side is made radius, and expressed by 
 a power of 10 ; and in this table a right-angled triangle 
 may always be found similar to any proposed right- - 
 
328 TRIGONOMETRY. PART IX. 
 
 anoied triangle, whence the latter can be determined by 
 proportion from the former : the actual operations being 
 performed bj logarithms, whose uses are copiously ex 
 plained in directions which accompany the tables. And, 
 note — in the following pages no subtraction by loga- 
 rithms is admitted in the operations, except a preliminary 
 subtraction for obtaining a certain result called arithme- 
 tical complement, and the subtraction of 10 or its multi- 
 ples, 100 and its multiples, &c. when necessary. 
 
 KIGHT-ANGLED PLANE TEIANGLES. 
 
 PROBLEM L 
 
 Given an acute angle, with the hypothenuse or a side, to 
 determine the triangle. 
 
 RULE. 
 
 As sine of any angle : side opposite : : sine of any other 
 angle : its opposite side. 
 
 DEMONSTRATION. 
 
 Let ABC be the proposed triangle right-angled at G, 
 andADE the tabular triangle similar to it; let AD be 
 made radius : then DE will be the sine ^ 
 
 of the opposite angle A, and AE the 
 cosine of the same angle, or sine of the 
 complement of B ; that is, the sine of the 
 opposite angle D, and AD the radius, is 
 equal to the sine of 90°, the opposite 
 angle E. Thus each side of ADE will 
 be the sine of its opposite angle, and 
 the sides of ABC being as the similar 
 sides of ADE, are also as the sines of the opposite and 
 equal angles A, B, and C ; whence in the triangle ABC 
 any one side : sine of angle opposite : *. any other side 
 : sine of its opposite angle. And conversely, 
 
 Sine of any one angle : side opposite : *. sine of any 
 other angle I its opposite side. 
 
 Li using logarithms, the logarithm of the first term is 
 subtracted from the sum of the logarithms of the second 
 and third; and when the first term is radius, its logarithm, 
 which is lOj will be written after the logarithm of the 
 
PAKT IX. TRIGONOMETRY. 329 
 
 second term witli the sign of subtraction, thus taking up 
 only one line instead of two. 
 
 And when radius occurs in the second or third 
 terms, its logarithm 10 will be added at once to the 
 logarithm of the third or second term, thus saving a line. 
 
 And when the subtractive term is other than radius, 
 we take its arithmetical complement in the following 
 manner : — 
 
 When the number is less than 1, its arithmetical com- 
 plement is its difference to 1 ; when the number is greater 
 than 1 and less than 10, the arithmetical complement of 
 it is its diiference to 10 ; if it be greater than 10 and less 
 than 100, then it is taken from 100, and so on. 
 
 Now supposing the number decimally expressed, the 
 arithmetical complement is found at once by subtracting 
 each digit, beginning at the highest or left hand, from 9, 
 and the last significant figure on the right from 10. 
 This avoids carriage in the subtraction, and may be effect- 
 ed at sight ; so that while any number is being read its 
 arithmetical complement may at the same time be written. 
 Thus, of .375 the arithmetical complement is .625, of 
 .3750 it is .6250, of .03074990 it is .96925010, of 3.75 
 it is 6.25, of 37.5 it is 62.5, and so on ; and the arithme- 
 tical complement of the arithmetical complement of any 
 number being taken, reproduces the original number. Thus 
 the arithmetical complement of 375 being 625, the arith- 
 metical complement of this reproduces the original num- 
 ber 375. 
 
 This is likewise true in angles and arcs. The comple- 
 ment of the complement of an arc reproduces the original 
 angle or arc; and the supplement of the supplement 
 reproduces the original angle or arc ; and thus the cosine 
 of the complement of an angle or arc is the sine of the angle 
 or arc itself : similarly the cotangent of the complement is 
 the tangent, the cosecant of the complement is the secant. 
 
 That number from which a lesser number is taken to 
 obtain its arithmetical complement, may be called the con- 
 stant, being the sum of the number and its arithmetical 
 complement. 
 
 The constant of two acute angles or arcs less than 90°, 
 which are complements of each other, is 90°. 
 
 The constant of two arcs or angles, which are sup- 
 plements of each other, is 180°. 
 
 When instead of subtracting a number its arithmetical 
 
330 TKIGONOJklETKi. PAET IX. 
 
 complement is added, the constant must be subtracted in 
 place of the original number. 
 
 Thus, 1463-358==1463 + 642-1000; 642 being the 
 arithmetical complement of 358, and 1000 the constant. 
 
 When several numbers are to be subtracted, their 
 arithmetical complements may be added instead, and the 
 sum of the constants subtracted from the sum of the 
 added quantities. 
 
 This method is made use of in the modern astrono- 
 mical tables, wherein it frequently happens that some 
 quantities are to be added and some subtracted ; but by 
 increasing each quantity by a constant, these all become 
 additive, and the sum of the constants which is known is 
 subtracted once for all from the sum, the computer having 
 nothing to do but to add previous to this. Single sub- 
 traction avoids the mistake of adding instead of subtract- 
 ing, and subtracting instead of adding, — mistakes which 
 might be said to be unavoidable in the former case, when 
 the number of quantities are very numerous (as they 
 are in Damoiseau's Tables of the Moon, wherein 46 cor- 
 rections are required to obtain the moon's true place in the 
 heavens). Thus, for instance, instead of performing the 
 diversified operations 90° -43° + 76° -32° -15'^ let the 
 constant 100° be applied to each subtractive quantity, 
 which are three, and we have the uniform operations 
 90° -f 57° + 76° -f 68° -i- 85° - 300°. 
 
 EXASIPLES. 
 
 1. The hypothenuse is 420, and one of the acute angles 
 is 41° 30^: to find the sides, and the other acute angle. 
 
 The other acute angle is 90° -41° 30^=48° 30^ 
 As sine 90° : hypothenuse 420 .. . 2.62325-10 
 :: sine 41° 30'. . . 9.82126 
 
 : opp. side 278.3 . . . 2.44451. 
 
 As sine 90 ; hypothenuse 420 .. . 2.62325-10 
 :: sine 48° 30' . . . 9.87446 
 
 : opp. side 314.6 . . . 2.49771 
 
 By the Gunter''s Scale. 
 
 1st. Extend the compasses from 90° to 41° 30' on the 
 line of sines ; that extent will reach from 420 to 278.3 on 
 the line of numbers. 
 
 2d. Extend the compasses from 90° to 48° 30' on the 
 
PART rx. TRIGONOMETEr. 331 
 
 line of sines ; that extent will reach from 420 to 314.6 on 
 the line of numbers. 
 
 2. Let the greater side be 276, and one of the acute 
 angles 40° 10' : to find the hypothenuse, the remaining 
 side, and the other acute angle. 
 
 Since 40° 10' is less than 45°, it is the lesser of the 
 two acute angles ; the greater is 49° 50% and therefore 
 opposite the greater side 276. 
 
 As sine 49° 50' ar. co. 0.11681-10 
 
 : 276 : : sine 90° 12.44091 
 
 : hypothenuse 361.2 . • . . . 2.55772 
 
 Assine49°50 ar. co. 0.11681 — 10 
 
 : 276 2.44091 
 
 : : sine 40° 10' 9.80957 
 
 opposite side 233 2.36729 
 
 JB^ the Gunter''s Scale. 
 
 1st. Extend the compasses from 49° 50' to 90° on the 
 line of sines ; that extent will reach from 276 to 361.2 on 
 the line of numbers. 
 
 2d. Extend the compasses from 49° 50' to 40° 10' on the 
 line of sines ; that extent will reach from 276 to 233 on 
 the line of numbers. 
 
 3. The hypothenuse is 744, one of the acute angles 
 30° 44' ; required the other acute angle, and the sides. 
 
 Ans. Angle 59° 16', the sides 380.2 and 639.5. 
 
 4. One of the sides is 470 links, the angle adjacent to 
 it 40° ; required the opposite angle, the other side, and 
 the hypothenuse, and also the area of the triangle. 
 
 Ans. Angle 50°, side 394.4 links, hypothenuse 613.5 
 links ; area 3 roods, 28.3 per^ches. 
 
 5. The hypothenuse of a right-angled triangle is 924 
 yards, one of its angles 50° 10' ; required the other acute 
 angle, and the sides. 
 
 Ans. Angle 39° 50', sides 709.5 and 591.9 ijards. 
 
 6. Standing on the bank of a river, and wanting to 
 know the distance of an object on the opposite side, I 
 measured 250 yards in a straight line at right angles to 
 the direction of the object, and then found the angle be- 
 tween this line and the direction of the object at the 
 second station to be 6Q° 30' ; required the breadth of the 
 
332 TRIGONOMETRY. PART IX. 
 
 river, and the distance of the object from the second 
 station. 
 
 Ans. Breadth of the river 575 yards, the distance of 
 
 the object from the first station ; and 621 yards the 
 
 distance from the second station. 
 
 PROBLEM II. 
 
 Given the hypothenuse and a side, to find the other side 
 and the angles. 
 
 RULE I. 
 
 By the proportions in page 328., we have hypothenuse 
 ; sine 90^ : *. given side : sine of the opposite angle. 
 
 Then find the other acute angle, which is opposite the 
 required side ; and sine 90° : hypothenuse : : sine of angle 
 opposite the required side : the required side. 
 
 When the angles are not required, the side may thus 
 be found : — 
 
 The square of the hypothenuse is equal to the squares 
 of the sides, whence the square of one side is equal to the 
 difference of the squares of the hypothenuse and of the 
 other side, which is equal to the rectangle of the sum 
 and the difference of the hypothenuse and side. 
 
 Whence, in logarithms. 
 
 The logarithm of the square of a side, equal to twice 
 the logarithm of the side, is equal to the logarithm of the 
 sum of the hypothenuse and other side, added to the loga- 
 rithm of the difference between the hypothenuse and this 
 other side. Consequently, 
 
 RULE II. 
 
 Add to the logarithm of the sum of the h}^othenuse 
 and given side the logarithm of the difference between 
 the hypothenuse and given side, and half the sum of the 
 two logarithms will be the logarithm of the side required. 
 
 EXA3IPLES. 
 
 1. The hypothenuse is 420, one of the sides is 310; 
 required the angles, and the other side. 
 
PAKT TX. TRIGONOMETKY, 333 
 
 As liypothenuse 420 .... ar. co. 7.37675 — 10 
 
 : sine 90° :: side 310 12.49136 
 
 : sine 47° 34' 9.86811 
 
 Tlie other angle 42° 26' . 
 
 As sine 90° : liypothenuse .... 2.62325 — 10 
 
 :: sine 42° 26' 9.82913 
 
 : other side 283.4 2.45238 
 
 By the Gunters Scale. 
 
 1st. Extend the compasses from 420 to 310 on the line 
 of numbers ; that extent will reach from 90° to 47° 34^ on 
 the line of sines. 
 
 2d. Extend the compasses from 90' to 42° 26' on the 
 line of sines ; that extent wiU reach from 420 to 283.4 on 
 the line of numbers. 
 
 2. The hypothenuse is 420, one of the sides 283 ; re- 
 quired the other side. 
 
 . 420 
 
 . 283 
 
 Hypothenuse 
 Side . . 
 Sum . . 
 Diiference 
 Sum . . 
 Other side 
 
 . 703 . . 2.84696 
 
 . 137 . . 2.13672 
 
 . . . . 4.98368 
 
 . 310.3 . 2.49184 
 
 3. The hypothenuse is 564, one of the sides 372 ; re- 
 quired the angles, and the other side. 
 
 Ans. Angles 41° 16' and 48° 44', side 423.9. 
 
 4. Given the hypothenuse 196, one side 142, to find the 
 other side. Ans. 135.1. 
 
 5. The hypothenuse of a right-angled triangle is 670 
 yards, one of its sides 526.2 yards ; required the angles, 
 the other side, and the area of the triangle. 
 
 Ans. Angles SS° 15' and 51° 45, side 414.8, area 
 109133.88 2/ards. 
 
 PROBLEM III. 
 
 Given the two sides, to determine the angles and 
 hypothenuse. 
 
 RULE. 
 
 As one of the sides : other side ! ! radius r tangent of 
 the angle adjacent to the first side, or opposite the second. 
 
334 TMGONOMETRY. PAKT IX. 
 
 Aiid, — 'As sine of tliis angle : second side : *. sine 90° : 
 hypotlienuse. 
 
 DEMONSTRATION. 
 
 Bj the proportion, page 328., transposed, 
 
 As one of the sides : other side : : sine of angle opposite 
 the first side ; sine of angle opposite second side. 
 
 : : cosine of angle adjacent to first side : sine of same 
 angle. 
 
 Multiply the last two terms of this proportion by the 
 tangent of the angle ; then, 
 
 As one of the sides .* other side : I cosine of angle ad- 
 jacent to the first side x tangent : : sine X tangent. 
 
 But cosine x tangent = sine x radius ; by page 326. 
 
 Therefore, as one of the sides : other side : : sine of 
 angle adjacent to first sidex radius : sine x tangent. 
 
 Dividing out the sine from the last two terms, 
 
 As one of the sides : the other or second side : : radius 
 : tangent of the angle adjacent to the first side, or oppo- 
 site the second. 
 
 And as sine of this angle : second side : *. sine 90° : hy- 
 pothenuse. 
 
 EXA3IPLES. 
 
 1 . The sides are 560 and 444 links ; required the angles 
 and hypothenuse. 
 
 As 560 ar. CO. 7.25181-10 
 
 : 444 : : rad 12.64738 
 
 : tan. 38° 25 9.89919 
 
 Other angle 51 °35^ 
 
 As sine 38° 25 ... . . ar. co. 0.20665 — 10 
 
 : 444 :: sine 90° 12.64738 
 
 : hyp. 714.6 2.85403 
 
 B^/ the Gimter's Scale. 
 
 1st. Extend from 444 to 560 on the line of numbers ; 
 this extent on -the line of tangents will reach from 45° to 
 38° 25' in one direction, and to 51° 35' in the opposite 
 direction. 
 
 2d. Extend the compasses from 38° 25' to 90° on the 
 line of sines ; that extent will reach from 444 to 714.6 on 
 the line of numbers. 
 
FART IX. TRIGONOarETRY. 335 
 
 2. The sides of a right-angled triangle are 375 and 
 428 yards ; required tlie angles and lijpothenuse. 
 
 Ans. Angles 41° 13' and 48° 47', hypothenuse 569. 
 
 3. The sides of a triangle are 300 and 400 ; required 
 the hypothenuse. Ans. 500. 
 
 OTHER EXAMPLES. 
 
 1. The side of a triangle is 125, the angle adjacent to 
 it 26° 35' ; required the other angle, the other side, and 
 the hypothenuse. 
 
 Ans. Angle 63° 25', side 62.55, hypothenuse 139.8 
 
 2. The hypothenuse is 637, and one of the angles 
 47° 37^ ; required the other angle and the side. 
 
 Ans. Angle 42° 23', sides 470.5 and 429.4, 
 
 3. The sides of a right-angled triangle are 25 and 125 
 required the angles and hypothenuse. 
 
 Ans. Angles 11° 19' and 78° 41', hypothenuse 127.5 
 
 4. The hypothenuse is 1023, and one of the sides 473 
 required the angles, and the other side. 
 
 Ans. Ajigles 27° 33' and 62° 27', side 907 
 
 5. The hypothenuse is 813, and one of the sides 312 
 required the other side only. Ans. 750.8 
 
 OBLIQUE-ANGLED PLANE TEIANGLES. 
 
 PROBLEM I. 
 
 Given a side and its opjyosite angle, together ivith anothei 
 side or angle, to determine the triangle. 
 
 RULE. 
 
 As the sine of any one angle : side opposite : '. sine of 
 any other angle ; its opposite side. 
 
 And, Any one side .* sine of angle opposite : *. any other 
 side : sine of its opposite angle. 
 
 DEMONSTRATION. 
 
 Let ABC be the proposed triangle, and DEF a circle 
 described with the radius of the tables, in which is in- 
 scribed the triangle DEF, similar to the triangle ABC ; 
 so that DE : DF : EF: : AB : AC : BC. Now the side 
 DE is the chord of the arc DE, or of the angle at the 
 centre which this arc subtends, and which is double the 
 
336 TRIGONOMETRY. PART IX. 
 
 imgle DFE at the circumference ; lience we may gay DE 
 is the chord of twice the opposite c 
 
 angle F, DF the chord of twice 
 the opposite angle E, and EF the 
 chord of twice the opposite angle 
 
 D. But the chord of an arc is ^^ ' ^' 
 twice the sine of half the arc 
 
 (page 325.) ; consequently DE is / ^^ ^ 
 
 equal to 2 sine F, DF to 2 sine / ^^ \\ 
 
 E, and EF to 2 sine D ; whence dL^_ \! ^ 
 
 we have 2 sin. F : 2 sin.' E : 2 \ / ^ 
 
 sin. D: : AB : AC : BC. Trans- 
 posing, dividing out by 2, and 
 putting the equal angles, viz. C for F, B for E, and A 
 for D, there results AB : AC .* BC : : sin. C : sin. B : 
 sin. A ; that is, the sides of plane triangles are as the sines 
 of their opposite angles, and conversely; hence, in the 
 same triangle, the sine of any given angle ; its opposite 
 side : *. sine of any other given angle \ its opposite side. 
 
 'Note. If the sum of two angles, being known, be less than 90'^, its sine may be 
 used for that of the third angle ; and if greater, the cosine of the excess may be 
 used for the sine of the third angle, these quantities being equal. See page 325. 
 
 EXAMPLES. 
 
 1. The two angles of a triangle being 50° 30' and 68° 
 20', and the side opposite this second angle 500, it is 
 required to determine the triangle. 
 
 First angle 50° 30' 
 
 Second angle 68° 20 
 
 Sum 118° 50' 
 
 Excess 28° 50' 
 
 As sine 68° 20' . . . . ar. co. 0.03182-10 
 : opposite side 500 .... 2.69897 
 
 :: sine 50° 30' 9.88741 
 
 : opposite side 415.1 .... 2.61820 
 
 As sine 68° 20' 0.03183-10 
 
 : opposite side 500 .... 2.69897 
 
 : : cosine excess 28° 50' . . . 9.94252 
 
 ; third side 471.3 2.67332 
 
 By the Gunter's Scale. 
 
 Knowing the three angles 50^ 30', 68° 20^ and 180° — 
 (50° 30'-t-68° 20') =180°-118° 50' =60° 10'. 
 
i-AIcT iX. riilGONOJDETRY. 337 
 
 1st. Extend the compasses from 68° 20' to 50° 30' oa 
 the line of sines; this extent will reach from 500 to 415.1 
 on the line of numbers. 
 
 2d. Extend the compasses from 68° 20' to 61° 10' on 
 the line of sines ; this extent will reach from 500 to 471.3 
 on the line of numbers. 
 
 When one angle and two sides are given, the second 
 angle found by the proportion of p. 335 is sometimes doubt- 
 ful ; that isj it may be either acute or obtuse, since the 
 same sine answers to two angles supplements of each 
 other. This happens when the given angle is opposite the 
 lesser of the two given sides ; for if opposite to the greater, 
 it must be acute. See the following example. 
 
 2. Two sides of a triangle are 710 and 406, and the 
 angle opposite the less is 34° ; required the other side and 
 angles. 
 
 As the side 406 . . . ar. co. 7.39147 — 10 
 : sine of its opposite angle 34° . 9.74756 
 
 :: side 710 2.85126 
 
 : sine of its opposite angle . . 9.99029 
 
 Second angle is 77° 66', or 102° 4'; and the third is 
 68° 4', or 43° 56\ 
 
 As sine 34° ar. co. 0.25244-10 ar. co. 0.25244-10 
 : opposite side 406 2.60853 .... 2.60853 
 : : sine 68° 4' . . 9.96737 or 43° 56 . . 9.84125 
 : third side 673.5 . 2.82834 or 503.8 . . 2.70221 
 
 JB?/ the Guntej-^s Scale. 
 
 1st. Extend the compasses from 406 to 710 on the line 
 of numbers; this extent will reach from 34° to 77° 56' on 
 the line of sines. 
 
 Then 180° -(34° + 77° 560 =180°- 111° 56'^68° 4', 
 the third angle. 
 
 2d. Extend the compasses from 34° to 68° 4' on the 
 line of sines ; this extent will reach from 406 to 673.5 on 
 the line of numbers. And this is one solution. 
 
 Taking now 180° -77° 56'= 102° 4' in place of 77° 56', 
 the third an^le is 180° — (102° 4' + 34°) = 180° -136° 4' 
 = 43° 56'=^ 77° 56'- 34°. 
 
 And extending the compasses from 34° to 43° 56' on 
 Q 
 
338 TRIGOXOJIETHr. PART IX. 
 
 the line of sines, the same extent will reach from 406 to 
 503.8 on the line of numbers ; which is the other solu- 
 tion. 
 
 Note. When the question admits of two solutions, the second angle in the second 
 solution is the supplement of the second angle in the first solution, and the third 
 angle of the second solution is the difference between ihe second angle in the first 
 solution and the given angle. Thus, in the above example, first solution, given 
 angle is 34°, second angle 77*^ 5G', third angle 68° 4' ; in the second solution, second 
 angle is 180°— 77° 56'=102O 4', and the third angle is 77° 56'— 34°=430 56'. 
 
 3. Two angles of a triangle are 42° and 98°, the side 
 opposite the lesser is 124 ; required the third angle, and 
 the remaining sides. 
 
 Ans. Angle 40°, sides 183.5 and 119.1. 
 
 4. Two sid^s of a triangle are 300 and 480, the angle 
 opposite the lesser is 38° 40' ; required the other angles, 
 and remaining side. 
 
 Ans. 88° 30' and 52° 50', side 382.6. Or, 
 91° 30' cmd 49° 50', side 367. 
 
 5. Two sides of a triangle are 300 and 480, the angle 
 opposite the greater is 38° 40' ; required the other angles, 
 and the remaining side. 
 
 Ans. 22° 59' and 1 18° 2 1 ', sic?e 676.1 
 
 PROBLf:M 11. 
 Given two sides and the included angle, to find the rest. 
 
 RULK. 
 
 As the sum of the sides : their difference: '.cotangent of 
 half the given angle : tangent of an angle E. 
 
 Then the lesser of the unknov>^n angles of the triangle 
 is the complement of the sum of the angle E and half the 
 given angle. 
 
 And the greater of the unkuowm angles is the sum of 
 the complement of half the given angle, and the angle E. 
 
 The other side is found by Problem I. 
 
 DEMONSTUATIOX. 
 
 Let ABC be the proposed triangle; let DE be the 
 radius of the tables and FG {)assing through E at right 
 angles to DE ; let DF, DG meet FG in the points F and 
 G, making equal angles EDF, EDG, each equal to half 
 the supplement of the angle C, /. e. half the exterior angle 
 
PAP.T IX. TRlGOXOMETlir. 339 
 
 at C ; then will tlie angle FDG equal the supplement of 
 C, and if FD be produced, the 
 angle GDH will equal the angle C. 
 Let Gil be drawn meeting FH in 
 H, and making the angle DGH 
 equal to one of the angles A, B, 
 equal to A the greater for in- 
 stance ; then angle DHG equal to 
 the angle B, whence the triangles 
 GHD, ABC are similar ; bisect 
 FH in I, and join EI; and since 
 FG, FH are bisected in E and I, 
 EI will be parallel to GH, and 
 angle EID equal to the anj^le 
 GHD ; and drawing from D, DK 
 parallel to IE the angle KDF, equal to the angle GHD, 
 equal to the angle B ; the angle EDF, equal to half the 
 supplement of C, is equal to half the sum of the angles 
 A and B, and the angle EDK is equal to this half sum 
 diminished by B, and therefore equal to half the differ- 
 ence of the angles A and B ; and DE being radius, EF is 
 the tangent of EDF the half sum, and EK is tiie tangent 
 of EDK the half difference. Now, FH being the sum of 
 HD, DF, or HD, DG, and FI half that sum, and DI being 
 the difference between FI, FD, or their equals HD, DG, 
 we have by similar triangles FI ; DI: lEF : EK, or twice 
 FI : twice DI: : EF : EK; that, is the sum of the sides DH, 
 DG : their difference : : tangent of half the sum of their 
 opposite angles : tangent of half their difference ; and 
 since the sides BC, AC are proportional to DH and DG, 
 the sums and differences will be proportional. Whence 
 by substitution, 
 
 Sum of the given sides (AC, BC) : difference of these 
 sides : : tangent of half the sum of their opposite angles 
 (A, B) : tangent of half their difference. 
 
 Adding the half difference to tlie half sum gives the 
 greater angle, and subtracting the half difference from 
 the half sum gives the lesser angle. 
 
 Note. Instead of taking the tangent of half the sum of the unknown angles, we 
 may take the cotangent of half the given angle ; and adding the half difference of the 
 unknown angles to half the known angle gives the complement of the lesser angle, 
 and adding the half difference to the complement of half the given angle gives the 
 greater angle. 
 
 All the angles being known, the third side may be found by the 1st Problem. 
 
 Q 2 
 
340 TRIGONOMETRY. PART IX, 
 
 EXAjMPLES. 
 
 1. Two sides of a triangle are 90 and 120, the included 
 angle 104°; required tlic other two angles, and the re- 
 maining side. 
 
 Here 210 is the sum of the sides, and 30 the diiference, 
 half angle 52°. 
 
 Sum 210 ar. CO. 7.67778-10 
 
 Difference 30 1.47712 
 
 Cotangent 52° 9.89281 
 
 Tangent i difference 6° 22' . . 9.04771 
 
 52° 38° 
 
 6° 22' 6° 22' 
 
 Sum 58° 22' 44° 22' 
 
 31° 38' lesser angle. 
 44° 22' greater angle. 
 
 Since sine 104° equal cosine 14°, 
 
 As sine 31° 38' .... ar. co. 0.28027-10 
 : opposite side 90 1.95425 
 
 :: cosine 14° 9.98690 
 
 : third side 166.5 2.22142 
 
 By the Gunter''s Scale. 
 
 1st. Extend the compasses from 210 (the sum) to 30 
 (tlie diff.) on the line of numbers ; the same extent will 
 reach from 38° (half sum of the unknown angles) to 6° 22' 
 (half their diff.) on the line of tangents. 
 
 Then 38° + 6° 22'=-44° 22^ greater angle, 
 38°-6° 22'=31° 38' lesser angle. 
 
 2d. Extend the compasses from 31° 38' to 76° (the 
 supplement of 104°) on the line of sines, and this extent 
 will reach from 90 to 166.5 on the line of numbers. 
 
 2. Two sides of a triangle being 180 and 212, and the 
 Included angle 72° 30'; required the other two angles, and 
 the remaining side. Ans. 47° 24' and 60° 6', side 233.2. 
 
 3. The distance of the earth from the sun 100 millions of 
 miles, the distance of Jupiter from the sun 520 millions of 
 miles, and the angle of these distances 20° ; required the 
 distance of Jupiter from the earth in millions of miles. 
 
 Ayis. 428 millions of miles : 
 
PART IX TRIGONOMETRY. 341 
 
 PROBLEM III. 
 Given the three sides, to find the angles 
 
 RULE. 
 
 As the base (for wliich the longest side is taken) : 
 sum of the other two sides *. *. difference of these two sides 
 : diiference of the segments of the base made by a per- 
 pendicular from its opposite angle. Add this number to, 
 and subtract it from, the base ; then 
 
 Half sum = greater segment 
 Half diiference = lesser segment. 
 
 Then as the greater of the other two sides : greater 
 segment : '. sine 90° ; sine of angle opposite the greater 
 segment = cosine of angle adjacent to the greater seg- 
 ment. 
 
 And as lesser side : lesser segment : ; sine 90° : sine of 
 angle opposite the lesser segment = cosine of the angle 
 adjacent to the lesser segment. 
 
 Hence, the angles at the base being known, the angle 
 opposite is known ; for angle opposite = 180° — (angles at 
 the base) = sum of the complements of the angles at the 
 base. 
 
 DEMONSTRATION. 
 
 Let ABC be the proposed triangle, with either angular 
 point, say C, as a centre, and with either of the two sides 
 AC,BC, — the larger, AC for instance, 
 — as a radius, describe the circle AFD ; 
 produce AB to meet the circle in 
 D, and BC both ways to meet the 
 circle in E and F; join CD, and let fall 
 the perpendicular CG w^iich bisects 
 AD, and the rectangle under FB-, BE 
 will be equal to the rectangle under AB -BD ; or AB : FB 
 : : BE : BD. Now CA and CF being equal, FB is the 
 sum of the sides AC, BC ; and C A, CF, CE being equal, BE 
 is the difference of the sides AC, BC. The side AB is divided 
 in G into the segments AG, GB ; and since AG is equal to 
 GD, therefore BD is equal to the difference of the seg- 
 ments AG, GB, the side AB being the sum. Hence, call- 
 ing the greatest side the base (for then the perpendicular 
 Q 3 
 
342 TRIGONOMETRY. TART IX. 
 
 drawn from the opposite angle will fall within the triangle), 
 there results base : sura of the other two sides : : difierence 
 of these two sides ; difference of the segments of the base. 
 Adding this difference to the whole base, and halving the 
 sum, gives the greater segment, which subtracted from 
 t]ie whole base leaves the lesser segment ; and the trianglp 
 ABC is thus divided into two riijht-angled triangles, 
 whose liypothenuses AC, BC, and bases AG-, GB, are 
 known ; thence determining the angles ACG, BCG, their 
 complements give the angles CAB, CBA, and their sum 
 the angle ACB. Thus AC .* radius : : AG- : sine ACG ; or 
 side of the given triangle ABC : adjacent segment of the 
 base : I radius : sine of angle opposite the segment. 
 
 Xotr. The greater segment is adjacent to the greater side. 
 EXAMPLES. 
 
 I. The sides of a triangle are 184, 140, and 92 ; re- 
 quired the angles. 
 
 Here 184 is made base. 
 
 As base 184 ar. co. 7.73518 — 10 
 
 : sum of sides 232 2.36549 
 
 :: difference of sides 48 . . . 1.68124 
 : difference of segments 60 . . 1.78191 
 
 Base 184 
 
 Diff. seg 60 
 
 Sum 244 
 
 Greater seg 122 
 
 Lesser seg 62 
 
 As side 140 ar. co. 7.85387 — 10 
 
 Adj. segment 122 2.08636+10 
 
 Cosine 29° 22' 9.94023 
 
 As side 92 8.03621-10 
 
 Adj. segment 62 1.79239+10 
 
 Cosine 47° 38' 9.82860 
 
 One angle 29° 22' 60° 38' 
 
 Another angle 47° 38 42° 22' 
 
 Third angle 103° 0' 
 
TAKT IX. TKIGONOMETKY. 343 
 
 By the Gunter^s Scale. 
 
 1st. Extend the compasses from 184 the base, to 232 
 the sum of the sides on the line of numbers ; this extent 
 will reach from 48 the difference of the sides, to 60 the 
 dilfcrence of the segments of the base. Then add this to, 
 and subtract it from, the base ; half the sum and difierence 
 gives 122 the greater, and 62 the lesser segment. 
 
 2d. Extend the compasses from 140 the greater of the 
 two sides, to 122 the greater of the segments on the line 
 of numbers ; this extent will reach from 90° to 60° 38' on 
 the line of sines; then 90°- 60° 38'= 29° 22' is one of 
 the angles. 
 
 3d. Extend the compasses from 92 the lesser side, to 
 62 the lesser segment on the line of numbers; this extent 
 vrill reach from 90° to 42° 22', the complement of which, 
 47° 38^ is another angle; and 60° 38^ + 42° 22'= 103° is 
 the third angle. 
 
 2. The sides of a triangle are 76, 148, and 110; re- 
 quired the angles. Ans. 29° 54', 103° 56', 46° 10' 
 
 PART X. 
 
 THE APPLICATION OF TRIGONOMETRY 
 TO HEIGHTS AND DISTANCES. 
 
 The mensuration of heights and distances, and of inac- 
 cessible objects, depends on plane and oblique trigo- 
 nometrj, as also on oblique sailing. 
 
 The most proper instruments for taking the horizontal 
 and vertical angles are the theodolite, the sextant, the 
 quadrant, and the compass ; the former being furnished 
 with one or two telescopes and a vertical arc. 
 
 The base lines are generally measured by a tape-line of 
 50 or 100 feet, or by the Gunter's chain. 
 Q 4 
 
S44 
 
 TRIG0N03IETRY. 
 
 THE COIMPASS. 
 
 PART X. 
 
 EXAMPLES. 
 
 1. Wanting to know the height of a tower whi 
 inaccessible, or intercepted bj a 
 moat, I found by the quadrant 
 the angle at the side of the moat 
 and the top of the tower to be 
 38° 30' ; I then measured 200 yards 
 in a straight line from the tower, 
 and found the angle to be 27° 20' : 
 required the height of the tower, 
 and the width of the moat. 
 
 By Theorem 6, Parti., ang. CDB-ang. CAB: 
 Act) =38° 30' -27° 20' ==11° 10^ 
 
 Then, as sine ang. ACD 11° 10' ar. co. 0.71295- 
 
 : side AD 200 2.30103 
 
 : : sine ang. CAB 27° 20' . . 9.66197 
 : side DC 474.2 . . . ' . . 2.67595 
 
 10 
 
PAKT X. TRIGONOMETRY. 345 
 
 And, as sine 90° : hvp. DC 474.2 . .2.67595-10 
 : : sine ang. CDB SS'^ 30' . . 9.79415 
 : side BC 295.2 2.47010 
 
 And, as sine 90° : hyp. DC 474.2 . 2.67595-10 
 :: sine DCB = cos. 38° 30' . . 9.89354 
 
 : side DB 371.1 2.56949 
 
 Hence 371.1 yards is the width of the moat, and 295.2 
 yards the height of the tower. 
 
 2. Wanting to know the height of a cloud, I found it 
 formed an angle of 23° 30' ; I then moved 300 yards in a 
 straight line towards the cloud, and found the angle to be 
 40° 30' : required the height of the cloud, and its distance 
 from my first and latter stations. See Fig. of Ex. 1. 
 
 Here, ang. CDB -ang. CAB ^ ang. ACD = 40° 30'- 
 
 side DC. 
 As sine ACD : side AD : : sine ADC = sine CDB (its 
 
 supplt.) : AC 
 As sine 90° : hyp. DC : : sine CDB 40° 30' : BC. 
 
 A?is. BC is 265.7 ^ards, DC 409.1, and AC 666.4. 
 
 3. Two observers, at the distance of 880 yards from 
 each other, took at the same instant the angle of elevation 
 of a balloon ; the angle at the nearest station was 34° 30', 
 and at the other 26° : required the height of the balloon. 
 See Fig. of Ex. 1. 
 
 Here, ang. CDB - ang. CAB = ang. ACD = 34° 30' - 26° 
 
 = 8° 30'. 
 As sine ACD 8° 30' : side AD 880 : : sine CAB 26° : 
 
 side DC. 
 As sine 90° : hyp. DC T: sine CDB 34° 30' : perp. BC. 
 Ans. BC 1478 t/ards, the height of the baUoon. 
 
 4. Being on the bank of 
 a river, and observing a 
 tower standing upon a hill 
 on the opposite side of the 
 river at station A, I took 
 my observation, or angles C 
 and D, at the top and 
 bottom of the tower, and 
 found the amrle CAB to be 
 56°, and DAB 28° 30'. 
 
 Q 5 
 
346 TRIG0N03IETRY. PART X. 
 
 There being a moat on my other side, I measured 240 
 yards on the bank of the river to station E ; I then found 
 the angle CAE to be 78° and CEA 76° : required the 
 height of the tower and hill, and the distance from 
 station A, to the top and bottom of the tower. 
 
 Here, ang. ACE = 180°-(ang. CAE +ang. CEA) = 180° 
 
 -154° = 26°. 
 ang. ADB=complt. of DAB =90° - 28°30'=61°30'. 
 
 = supplt. of ang. ADC. 
 ang. CAD = ang. CAB-ang. DAB = 56°-28°30' 
 
 =27° 30\ 
 ang. ACD=complt. of CAB = 90° -56° =34°, and 
 
 AE=240. 
 
 As sine ang. ACE : side AE : : sine ang. CEA : side AC. 
 As sine ang. ADC = sine ang. ADB ; side AC : : sine ang. 
 
 ACD : side AD. 
 As sine ang. ADC = sine ang. ADB : side AC : : sine ang. 
 
 CAD : side CD. And, 
 As sine 90° : hyp. AD : : sine ang. DAB ; side BD. 
 
 Ans. CD the height of the toiver 279.1 yards, 
 BD height of the ^hill 161.3. The two dis- 
 tances AC, AD, 531.2 and 338 yards. 
 
 5. Sailing from Nelson Haven to New Plymouth, New 
 Zealand, on the Taranaki coast, I observed an object on 
 the summit of Mount Egmont, the angle of elevation 
 being 9° 50' ; I then sailed six miles upon the same 
 point of the compass, and found the angle of the same 
 object to be 18° 34' : required the height of the mountain 
 above the level of the sea. See Fig. Ex. 1. 
 
 Ans. 2.15 miles, or 3784 yards, or 11352 feet. 
 
 6. Sailing along the shore of Port Nicholson, New 
 Zealand, I saw a cape of land which bore from meN.N.E. ; 
 I then sailed N.W. by W. 24 miles, when the same cape 
 bore E.N.E. : required the distance of the cape at both 
 stations. 
 
 Let A represent the place of observation, and AB the 
 N.N.E. direction, which is 2 points, and AC the N.W. by 
 W. course, being 5 points ; then the angle BAC is 7 points, 
 or 78° 45'. Draw the dotted line a b parallel to the N.S. 
 line, and take angle aCB := 67° 30' =6 points for the 
 E.N.E. course, then angle ^CA = angle CAN = 5 points. 
 
TART X. 
 
 TRIGONOMETRY. 
 
 34*; 
 
 and aCB = 6 points ; 
 whence angle BCA= 
 (16 - 11) points = 6 
 points, or 56° 15', and 
 angle B=:(16— 5 — 7) 
 points = 4 points, or 
 45°. Then sine ang. B 
 45° : side AC 24 miles 
 :: sine ang. BAG 78° 
 45' : side BC ; and sine 
 ang. B : side AC : : sine 
 ang. BCA : side AB. 
 Ans. BC is 33.29 miles, 
 
 a?id AB 28.22 miles. 
 
 7. From a station at 
 A an object at B, the 
 distance being 8 miles, 
 bore N.E. bjN. ; another 
 object at C, distance 51 
 miles, bore S.E. : re- 
 quired the distance of 
 the two objects. 
 
 Ans. 10.56 miles. 
 
 8. Suppose two ships sail from the same port at the 
 same time ; one sails N.IC. by E. 28 miles, the other sails 
 S.E. i S. 34 miles : I demand the distance of the two 
 ships. Ans. 41.86 miles. 
 
 ^. Saint Paul's Cathedral at London bore from me 
 KN.E., and travelling N.W. by W. 12 miles it then bore 
 N.E. ; required the distance of Saint Paul's from both 
 stations. 
 
 Ans. Tfie distances are the two equal sides of an 
 isoceles triangle, being 30^ miles. 
 
 10. 1 observed a tower, 
 CD, on the top of an 
 inaccessible hill BD ; the 
 angle of elevation of the 
 top of the hill is BAD 
 36°, and the top of the 
 tower BAC 49° 15^ I 
 then measured in a di- 
 rect line AE 120 yards, 
 and found the anirle of 
 
348 TKIGONO^IETRY. PAliT X. 
 
 the top of the tower to be EEC 30" 30' : required the 
 height of tower and the hill, and its distance from my 
 first station, allowing 5 feet for the height of the instru- 
 ment. 
 
 A?is. Height of toicer 53.7 yards, height of hill 94.87 
 yards, distance from station A 123.7 yards. 
 
 11. From a window near London Bridge, which ap- 
 peared to he on a level with the 
 bottom of the Monument on the 
 opposite side of the river, the 
 angle of elevation of the top 
 is 10° 02", and from another 
 window in the same building 24 
 feet above the former the angle 
 
 is 8°56 ; required the height and distance of the Monu- 
 ment. 
 
 Ans. Height of Monument 2\o.\ feet, distance 1216 feet. 
 
 Note. ThR erection of the Monument was commenced in the year 1671, and 
 finished in the year 1677, in remembrance of the dreadful fire of London in 1G66, 
 and the rebuilding of the City in the reign of Charles II., under the inspection of 
 Sir Christopher Wren. Leigh, in his " Picture of London," says this noble pile is 
 undoubtedly the first modern column in the world : it is of the Doric order, fluted, 
 and exceeds in height the pillars of Trajan and Antoninus. 
 
 12. A celebrated navigator sailing on the Caspian Sea, 
 observed through his glass an object on the summit of 
 Mount Ararat, which iDore S.W. by W. ; he then sailed 
 36 miles N.W., when the same object bore S.W. by S. ; 
 required the distance of the object at each station. 
 
 Ans. Distance 92.26 miles each station. 
 
 Note. IMount Ararat is a celebrated mountain of Armenia, on whic^h Noah's 
 ark rested at the time of the flood, where it is supposed still to remain on the 
 summit. The capacity of the ark was 2.730782 cubic feet, which is about twenty 
 times larger than a first-rate man of war, and may be considered the object ob- 
 served. A Russian traveller says this mountain is 17260 feet above the level of the 
 sea, and is doubtless volcanic. Several attempts have been made to reach the top 
 of the mountain, but no one has got much beyond the snow limit. Sir R. K. Porter 
 visited this region, and is of opinion the inaccessible summits have never been 
 trodden by the^foot of man since the days of Noah. 
 
 13. Being at sea I observed two headlands, whose bear- 
 ings from each other I found by the chart to be W.N.W. 
 and E.S.E., and distance 20 miles ; the first bore from me 
 S.W. by S., and the second S.E. : required my distance 
 from each headland. 
 
 Ans. The S.W. by S. headland 7.8, a7id S.E. 20 miles. 
 
 14. Two ships sail from the sam.e port or station A ; 
 one sails N.AT. by N. 28 miles to station B, and the other 
 
PART X. 
 
 TRIGONOMETRY. 
 
 49 
 
 sails N.E. by N. 38 miles to station C ; required the dis- 
 tance and bearings from each other. 
 
 Ans. T/te hearing o/'B from C, is W. hy S., and C 
 from B, E. hy N. ; distance 37.6 wz7e5. 
 
 15. Wanting to know the height of a tower CD, on the 
 opposite side of a river, which appeared to be on a level 
 where I stood at A, I 
 there fixed a station, and 
 then ascended the acclivity 
 of a hill AB, in a straight 
 line from the tower 250 
 yards, when I found the 
 angles of depression to be - 
 as follows, viz. that of my 
 
 first station EBA 44°, of the bottom of tower EBC 28°, 
 and the angle EBD of depression with the top of tower 
 was 18°. Required the height of the tower, distance 
 from my first station to its bottom, as also the horizontal 
 distance ; and my height above the top of the tower. 
 
 Alls. Horizontal distance 326.6 ; distance from sta- 
 tion to toiver 146.8; height of the tozver 67.53; 
 and my height above the top 106.16 yards. 
 
 16. The distance between a tower T, and a church C, 
 was known to be 20 miles ; 
 from a ship at anchor A, I 
 saw a windmill at M, in a 
 right line with the tower ; 
 the bearing of both was 
 Is.W. by W.: the church at 
 the same time bore N.N. W. 
 •JW. After which we weigh- 
 ed anchor, and sailed upon 
 a W. by N. course AH = 
 25 miles, and the mill and the church were in a straight 
 line HC, and bore N.E. by N. Required the distance 
 between the church, tower, and windmill, at each station, 
 or place of observation, A and H ? 
 
 1st. In the triangle ACH the angles are known, and 
 the side AH is known. 
 
 For AC bears N.N.W. -JW., or 2|- points, 
 AH — W. by N.' or 7 points, 
 
 whence angle CAH is 4^ points, or 50° 37^ 
 
 The bearing of HC is N.E. by N., or 3 points ; the com- 
 
350 TRIGONOMETRY. PART X. 
 
 plement of this added to the angle HAW, wiiic;i is one 
 point, the sum is 6 points, or 67° 30', the aiiglc AHC, 
 whence angle ACH is 5^ points, or 61° 52". 
 
 As sine ACH : AH: : sine AHC : AC. 
 sine ACH : AH: -.sine CAH : HC. 
 
 2d. In the triangle ACT, the angle TAC and the oppo- 
 site side TC are known, together with AC, wiience AT 
 may be found. 
 
 For AC bears N.N.W. IW., or 21 points, 
 AT — N.W. by"W. or 5 points. 
 
 Hence angle TAC is 2i points, or 28° 7". 
 As TC : sine TAC:': AC : sine ATC. 
 Angle ACTrr: 180° -(angle TAC + angle ATC). 
 As sine TAC : TC: : sine ACT : AT. 
 
 3d. In the triangle AMH the angles are known, and 
 the side AH is known. 
 
 For AM bears N. W. by W., or 5 points, 
 AH — W. by N. or 7 points ; 
 whence angle MAH is 2 points, or 22° 30'. 
 
 AHM or AHC is 6 points, or 67° 30', whence angle 
 AMH is 8 points, or a right angle. 
 
 As sine 90° : AH: : sine AHM : AM, 
 sine 90° : AH:: sine IMAH : HM. 
 4th. In the triangle TMH, which is right-angled at M 
 (for since AMH is a right angle, TMH is likewise one), 
 TM=:AT — AM is known, and HM is known. 
 
 Then, as HM : radius ::TM : tangent THM, 
 sine THM : TM::sine 90° : TH. 
 
 A?is. AM=23.09, Ans. HM=: 9.57, 
 AC=26.19, HC=21.91, 
 
 ATr=38.84, HT=z 18.43. 
 
 1 7. Coasting along the sea shore, 
 I observed at A, two headlands, B 
 and C ; the first bore N.N.W., the 
 second N.N.E. jE. Then steering 
 E.N.E. IE. 16 miles to D, the 
 first liea'dland bore W.X.W., the -^1 
 
 secondN.W.by N. iW. Required [ 
 
 the bearing and distance of the ( 
 
 two headlands from each other ? 1 
 
TART X. TRIGONOMETRY. 351 
 
 1st. In the triangle BAD the angles, and the side AD 
 are known, 
 
 For AB bears N.N.W., or 2 points, 
 And AD — E.N.Pl IE., or 6i points, 
 
 whence angle BAD is 8^ points, or 95° 37^ : 
 
 And DB bears W.N.W., or 6 points, 
 AD — E.N.E. ^E. or 61 points ; 
 
 whence angle ADB is (16—121) points =31 points, or 
 39° 22', and the angle ABD 4 points, or 45°." 
 
 As sine ABD : AD: '.sine BAD : BD, 
 sine ABD : AD: : sine ADB : AB. 
 
 2d. In the triangle ADC the angles, and the side AD 
 are known, 
 
 For DB bears W.N.W., or 6 points, 
 
 DC -. N.W.byN.iW.or3ipoints; 
 
 whence angle BDC is 2| points, and angle ADB is 31 
 points, whence angle ADC is 61 points, or 70° 19'. 
 
 AC bears N.N.E. |E., or 2| points, 
 AD — E.N.E.lE., or 61 points; 
 
 whence angle DAC is 3 J points, or 42° 11', 
 and angle ACD is 6 points, or 67° 30. 
 
 As sine ACD : AD: : sine ADC : AC, 
 sine ACD : AD:: sine DAC : CD. 
 
 3d. In the triangle BCD the sides CD, BD are known, 
 and also the included angle BDC, which is 2| points, or 
 30° 56'. 
 
 Then BD + CD : BD - CD : : cotang. l BDC : tang, of 
 half the difference of the angles DBC, DCB : 
 
 And sine DBC : CD:: sine BDC : BC, 
 
 the distance of the two headlands B and C. 
 
 The bearing of BD is W.N.W., and of BD, it is E.S.R, 
 which is 6 points, or 67° 30' ; to this add the angle DBC, 
 which is 25° 29', the supplement of the sum is 87° 1', the 
 bearing of BC, that is, C bears N. 87° V E. from B, and 
 B bears S. 87° V W. from C, and their distance is 13.81 
 miles. 
 
 Note. — The determination of the distances and the bearings of any number of 
 points is effected in a similar manner by taking observations of each in siiccession ; 
 if this be applied to buildings, we may, in a similar manner, determine the ground 
 plan of the outline, and the elevation of the building, whether it be perpendicular, 
 
352 
 
 TRIGONOMETRY. 
 
 PART X. 
 
 of the object AB, 
 
 or inclining as the leaning tower of Pisa, or combine vertical and inclining liuej 
 as the Pyramids of Egypt. 
 
 18. Wishing to know 
 which is oblique to the 
 ground plane, and its per- 
 pendicular altitude BC, I 
 took a first station at D, 
 from which the object 
 appeared upright, and ob- 
 served the angle ADB, 
 63° 20'; fixing a station stafi' at D, I measured DE, 150 
 feet, and at the second station E, I took the angles DEB, 
 55°, DEA, 33° ; leaving a station staff at E, I returned to 
 my former station D, and took the angles : BDE 72° 10' ; 
 ADE, 102° 50'. Required the length AB, the vertical 
 altitude BC, and the inclination BAG. 
 
 1st. In the triangle BDE the angles, and the side DE 
 are known, for angle DBErr.l80°-(BDE + DEB). 
 Then, as sine DBE : DEr'.sine DEB : BD. 
 2d. In the triangle ADE the angles, and the side DE 
 are known, for angle DAE = 180°-(ADE-hDEA). 
 Then, as sine DAE : DE: : sine DEA : AD. 
 3d. In the triangle ABD the sides BD, AD are known, 
 and also the included angle ADB, whence AB and angles 
 BAD, ABD are determined bj Prob.. II. of Oblique-angled 
 Triangles. 
 
 The angle BAD is the angle of inclination of the object 
 AB. 
 
 4th. In the right-angled triangle BCD, 
 
 As sine 90° : BD: : sine ADB : BC, the altitude of AB. 
 Ans. Length AB^ 14:5.7 feet, i7icU?i. BAD, 71° 4', 
 perp. BC = IS7.S feet. 
 
 19. I wanted to know a ___B 
 
 the distance between two 
 places A and B, but could 
 not meet with any station 
 from whence I could see 
 both objects. I measured 
 a line CD ^200 yards; 
 from C the object A was 
 visible, and from D the 
 object B was visible, at 
 each of which places I set 
 
PART X. TUIGOXOMETRY. 353 
 
 up a pole. I also measured FC = 200 yards, and DE=r200 
 yards, and at F and E set up poles. I then took the 
 andes : AFC, 83° ; ACF, 54° 3V ; ACD, 53° 30^ BDC, 
 156° 2o ; BDE, 54° 30^ BED, 88° 30'. Required the 
 distance AB. 
 
 1st. In the triangle AFC angle FAC = 180°-(AFC-h 
 ACF). 
 
 Then, as sine FAC : EC: '.sine AFC : AC. 
 
 2d. In the triangle ACD there are known AC, CD, 
 and the included angle ACD, whence the angle CDA and 
 the side AD may be found. 
 
 3d. In the triangle BDE angle DBE=180°-(BDE 
 + BED). 
 
 Then, as sine DBF ; DEnsine BED : BD. 
 
 4th. In the triangle ABD there are known AD, BD, 
 and the angle BD A = angle BDC — angle ADC ; whence 
 AB is determined := 345.4 yards. 
 
 c 
 
 20. From a station at D, I perceived 
 three ob'ects A, B, C, whose distances 
 from each other I knew to be as follow, 
 viz. AB = 12 miles, BC=9 miles, and 
 AC = 6 miles. At D I took the angles : 
 ADC =22° 30', and BDC = 33° 45'. 
 Hence it is required to find my distance 
 from the objects. 
 
 Note.— In the triangles ADC, BDC, one side and an angle opposite only is given 
 in each, whence it would not seem possible to determine the other sides : but be- 
 cause AB is given, we may, from a property of the circle, determine what is re-, 
 quired. 
 
 1st. Let the triangle ADB be circumscribed by the 
 circle ADBE, and CD joined, cutting the circle in E, and 
 AE, BE be joined. 
 
 The angles ADE, ABE, which stand on the same seg- 
 ment AE are equal, as also the angles BDE, BAE ; 
 therefore in the triangle ABE all the angles, and the side 
 AB are known ; whence AE, BE may be found by Prob. 
 I. Oblique-angled Triangles. 
 
 2d. In the triangle ABC, whereof the sides are given, 
 the angles may be found by Prob. III. Triangles. 
 
 Hence the angles CAE, CBE become known. 
 
 [If the angles ADC, BDC taken together are less than 
 the angles CAB, CBA taken together, then the angle 
 
354 TRIG0X03IETRY. PART X. 
 
 AEB, the supplement of ADB, is greater than ACB, the 
 supplement of the angles CAB, CBA, and then C lies 
 beyond E, out of the triangle AEB : and the angles CAE, 
 CBE are the differences of the an2;les at the base in the 
 triangles ABC, ABE. 
 
 If the point C lie within the triangle ABE, in this case 
 the angles at the station, viz. ADC, BDC taken together 
 are greater than the angles CAB, CBA taken together ; 
 and the angles CAE, CBE are the differences of the angles 
 CAB, EAB ; CBA, EBA as before. 
 
 If the point C be within the triangle ABD, the angles 
 CAE, CBE will be the sum of the angles at the base in 
 the two triangles ABC, ABE, instead of the difference, as 
 in the former case.] 
 
 In the triangle ACE, having AC, AE, and the angle 
 CAJ^^, find the angle ACE ; whence the angle BCE is 
 known, since BCA is known. 
 
 3d. Now, in the triangles ADC, BDC, all the angles 
 are known in each, and likewise a side ; whence the sides 
 AD, CD, BD can be found. 
 
 Ans. AD = 14, CD=15.6, BD=10.7. 
 
 21. Tliree objects A, B, C, forming q 
 
 a triangle, were visible from one sta- 
 tion, at D within the triangle ; the 
 angle ADB was observed to be 123° 
 45', CDB = 132° 22", and consequently 
 ADC=:103° 53'; the distance AB was 
 known to be 12 miles, from B to C 9 
 miles, and from A to C 6 miles. Re- 
 quired the distance of each object from 
 the station at D. 
 
 1st. Here angle ABE wangle ADE=suppl. of ADC 
 angle BAE wangle BDE=:supph of CDB. 
 
 Hence, in the triangle BAE, all the angles are known, 
 and also the side AB, whence AE, BE are found by 
 Prob. I. of Oblique-angled Triangles. 
 
 2d. Finding the angles of the triangle ABC from the 
 sides being given, we have, by adding, the angles CAE, 
 CBE ; and the sides about the angle CAE being known, 
 viz. AC, AE, all the angles of the triangle ACE may be 
 found: and angle BCE becomes known, as also angle 
 CBE. 
 
 3d. In the triangles ADC, BDC all the angles are 
 
PART X. 
 
 TRIGONOAIKTKY. 
 
 .355 
 
 known, and a side AC, BC in each is known ; whence the 
 other sides are found, viz. AD 5.5, DC 1.4, DB 8. 
 
 22. From a station at D, I perceived three objects 
 A, B, C, whose distances were as follow : AB 12 miles, 
 AC 13 miles, BC 7 miles ; at D the angle ADB was 20°. 
 D, was in CA produced. Required the distances DA, DB. 
 and DC. 
 
 In the triangle CAB find 
 the angle CAB, from the sides 
 given ; the supplement of this 
 angle is angle BAD; hence, 
 in the triangle ABD, all the 
 angles are known, and the side 
 AB is known ; whence AD, 
 BD may be found, and CD = AD + AC. 
 
 Ans. AD 7.4 miles, BD 18.6, CD 20.4. 
 
 In a similar manner the distances may be found when 
 D is on one of the sides of the triangle ABC. 
 
 23. Three objects A, B, C, 
 in a straight line whose dis- 
 tances were, AC 36 yards, 
 BC 84 yards, AB 120 yards, 
 were visible from one station 
 at D, from whence the angle 
 ADC was 18° 56', and BDC 
 25° 21. Required the dis- 
 tances DA, DC, DB. 
 
 1st. The triangle ABD being circumscribed by the 
 circle, and DC produced to E, angle ADC — angle ABE, 
 and angle BDC wrangle BAE. 
 
 Hence, in the triangle ABE, all the angles, and the side 
 AB are known ; whence the sides AE, BE are determined. 
 
 2d. In the triangle CAE, the sides AE, AC, and the 
 included angle CAE, being known, the angles AEC, BEC 
 are known, for BEA is known. 
 
 3d. In the triangles AED, BED all the angles are 
 known, and a side AE, BE in each ; whence AD is found 
 =94.8, CD=108.5, and BD=168. 
 
 24. Wishing to know the dimensions of the pyramid 
 ABCDE, which was inaccessible on all sides, I chose a 
 station G, such that the edge AE appeared perpendicular 
 to the ground plane ; I measured the angles : AGB, 
 20° 4^ AGD, 43° 52^; DGE, o8°29'j AGE, 43° 49'; 
 
356 
 
 TRIGONOMETRY. 
 
 PART X. 
 
 BGE, 47° 22^ I then took a second station at H, 
 measuring GH 36 feet, AE appearing vertical from H, as 
 at G; I measured the angles: AHB, 17° 2r ; AHD, 
 37° 12'; AHE, 39° 18". I then took a third station at 
 I, from which the edge BE appeared vertical, and measured 
 the angles : BIA, 49° 37' ; BIG, 40° 32'; BIE, 58° 58'; 
 EIA, -70° 32' ; EIC, -66° 55'. I then moved from I to K, 
 measuring IK 57 feet, where, as at I, BE appeared ver- 
 tical : the angles observed were, BKA, 39° ; BKC, 
 29° 43' ; BIflE, 48° 27'. And at station L, from whence 
 the edge CE appeared vertical, I observed the angles: 
 CLE, 57° 31'; CLD, 31° 48'; at M, in the direction of 
 CL produced, I measured LM 60 feet, and the angle 
 CMD 25° 35'. From these data it is required to find the 
 dimensions of the pyramid, its vertical height, the area of 
 the ground plan, and of each of its faces, and its solid con- 
 tent, su|)posing it to be solid. 
 
 1st. In the triangle BGH the side GH is known, and 
 all the angles ; then, 
 
 As sine GBH : GH: : sine AHB : BG. 
 
 In a similar manner DG may be found in the triangle 
 DGH ; and EG in the triangle EGH. 
 
 2d. In the triangle BGE, the sides BG, EG are known, 
 and the included angle BGE; whence the edge EB is 
 determined by Prob. 11. Oblique-angled Triangles. 
 
 In a similar manner the edge ED is determined from 
 the triangle EGD. 
 
 3d. In the triangles AIK, OIK we may find lA, IC, 
 just as BG, DG were found in the triangles BGH, DGH. 
 
rART X. TRIGONOMETHY. 357 
 
 Aiid from the triangle lEK, EI may be found in q 
 similar manner as EG was found in the triangle EGIi. 
 
 And in the triangles EIA, EIC, the edges EA, EC 
 are determined in the same manner as E13, ED, in the 
 triangles EGB, EGD, were determined. 
 
 4th. In the triangle EAG the sides EG, EA are known, 
 and the opposite angle AGE ; then 
 
 As EA : sine AGE: : EG : sine EAG; 
 
 and the supplement of this is the inclination of EA to the 
 ground plan, which is known, as also the angle AEG ; the 
 vertical altitude EF is thus found: As sine 90° : EA: I sine 
 of the inclination of EA : elevation EF. 
 
 And sine AGE : AE: : sine AEG : AG. 
 
 In a similar manner may BI, CL be found in the tri- 
 angles BEI, CEL. 
 
 5th. In the triangle DAG the sides AG, DG are known, 
 with the included angle AGD ; wdience the base AD is 
 known, and the angles GDA, DAG. 
 
 Similarly, in the triangle BGA, the base line AB may 
 be found, and the angles BAG, GBA ; and in the triangle 
 CBI the base BC may be found, and in the triangle CLD 
 the base CD. 
 
 Hence all the base lines of the pyramid ai'e known, like- 
 wise all its edges, and its altitude. 
 
 6th. In the ground plan ABCD let the diagonal BD be 
 drawn ; since the angles BAG, DAG are known, the angle 
 BAD is known ; whence the diagonal BD may be found 
 by the second Problem of Oblique-angled Triangles : thus 
 the triangles ABD, CBD have a common base BD, which 
 is known, and likewise their sides known; whence the 
 ground plan may be drawn, and its area found, as in 
 p. 130. 
 
 7th. In each of the faces ABE, ADE, CDE, BCE the 
 sides are known; whence their areas are determined as 
 in p. 130. 
 
 8th. The area of the ground plan being known, and the 
 altitude EF, the solid content is found, as in p. 132. 
 
 A)is. BG = 226.5, DG= 187.5, EG=289.6, BE = 
 215.2, DE=249.4, IA=: 194.7, 10=150.6, EI= 
 233.7, EA=249.4, EC=222.9, a?ig. EAF = 
 53^29', EFr:r 200.5, AG = 60.5, ajir/. BAF= 
 27°3; iU3=i71, ang. DAF=60° 7^ a?ig. BAD 
 
358 TKI(310>;0.METIiY. PART X. 
 
 =87° lO; AD = loO, BI=i:41.6, BC=:122, ang. 
 ECF = 64° 3; CL = 30.1, LD =239.3, CD= 
 213.8, BD=221.8. Area of ground plan = 
 25670 square feet ; area o/'y«ce ABE =18094 
 square feet; area of face BCE— 12810 square 
 feet ; area of face CDE = 22344 square feet; 
 area of face ADE= 17839 square feet ; solid 
 content =17 15612 cubic feet. 
 
 MENSURATION OF ALTITUDES BY THE 
 BAR03IETER AND THERMOMETER. 
 
 The following method for determining difference of 
 levels by the btirometer and thermometer is the reduction 
 of a formida marked tt in p. 64. of the translation of Fran- 
 cceur's Hydrostatics. 
 
 1. At each of the stations whose difference of level is 
 required an observer is placed, provided with a barometer 
 and two thermometers, one thermometer being attached 
 to the barometer to ascertain the temperature of the mer- 
 cury of the barometer, the other being detached and 
 placed in the open air to ascertain the temperature of the 
 air ; each observer at tlie same time, previously agreed 
 upon, notes the states of his barometer and thermometers. 
 
 Both barometers, it is almost needless to say, must be 
 similar in their construction, viz. if one be graduated 
 by inches and parts, such as is generally used in England, 
 the other must also be of the same construction, or if one 
 be a metrical barometer*, such as is generally used in 
 France, the other must also be of the metrical construc- 
 tion : a similar remark applies to the four thermometers, 
 viz. if one be of Fahrenheit's graduation, all must be so ; 
 if one be of the centigrade graduation, all four must be so. 
 
 2. Whatever kind of barometers be used, the difference 
 of the logarithms of their indications, omitting the index. 
 
 * The barometer used in England is graduated by inches and decimal parts ; the 
 metrical barometer is sradiiated by millemetres or 'thousandths of a metre, and 1 
 inch English=25.39 millemetres. In Fahrenheit's thermometer, the freezing 
 point is marked 32°, the boiling point by 212°, the interval being 180°: in the cen- 
 tigrade till rmometer, the freezing point is marked 0°, and the boiling point is 
 marked 100°; hence 180° of Fahrenheit's thermometer=100O of the centigrade 
 thermometer. And note, 1 metre=3. 280899 English feet. 
 
PART X. TlilGONOMETRY. 359 
 
 being taken, is to be increased or diminished according to 
 tiie states of the attached thermometers ; thus, take the 
 difference of the heights of the attached thermometers, 
 and multiply it by .00008, if they are of the centigrade 
 construction, or divide four times the difference by 90000, 
 if they be of Fahrenheit's construction ; then, if the degrees 
 of the lower thermometer are less than those of the upper 
 one, this product or quotient is to be added to the ditier- 
 ence of the barometer logarithms, but added in the con- 
 trary case. The result is the first factor. 
 
 3. Add together the degrees of the detached thermo- 
 meters, and multiply the sum by .002, if they are centi- 
 grade, but for Fahrenheit's, diminish the sum by 64°, and 
 divide the remainder by 900 ; the result added to 1 gives 
 the second factor. 
 
 4. If the first factor be less than .07, multiply together 
 the number 18393 and the two factors; the product is the 
 difference of the heights of the upper and lower station, 
 nearly, in French metres ; or multiply the number 
 60345.6 and the factors together, the result is the height, 
 nearly, in English feet. 
 
 When the difference of the barometer logarithms is 0.7 
 or above, 18336 is to be used instead of 18393 for the 
 metres, and 60158.6, instead of 60345.6, for the feet. 
 
 5. A correction for the rotation of the earth may be. 
 found as follows, — 
 
 Take the difference between 90° and twice the latitude^, 
 and then, As radius : sine of the difference : ! .002837 X 
 approximate height : correction for latitude. This cor- 
 rection is to be added to the approximate height when the 
 latitude is less than 45°, but subtracted when the latitude- 
 is greater than 45°. 
 
 6. A second correction, when the difference of the baro- 
 meter logarithms is not less than .07, may be found as 
 follows, — 
 
 Add to the approximate height, if in metres, the num- 
 ber 15926.6, if in feet, the number 52253.8 ; divide for 
 metres by 6366198, for feet by 20898000, and multiply 
 by the approximate height; the result is the second cor- 
 rection, which is always to be added. 
 
 A part of this correction, viz. -jVoVit ^^ i^ is due to the 
 diminution of the weight of the mercury at the upper 
 station. 
 
 Tlie whole of this second correction may he omitted 
 
360 trigoxo:u:etp.y. part x. 
 
 when the difference of the barometer logarithms is less 
 than 0.7. 
 
 The approximate altitude corrected gives the altitude 
 very near to that determined by geometrical measure- 
 ment. 
 
 It is hardly necessary to say that the instruments should 
 be very carefully constructed, similar in their indications 
 and furnished with verniers, that we may be enabled to 
 estimate the smallest fractions of the scale. The time se- 
 lected should be such as to afford the circumstances most 
 favourable for observations, viz. calm weather, the middle 
 of the day, &c. 
 
 Note. — The reader may consult on this subject a very elegant paper by M. Ra- 
 mond, from whom the above numbers are taken. See als'o Mecaiiique Celeste, *om. 
 iv. p. 289. 
 
 Log. of 18393=4.26465 
 Log. of 18336=4.26324 
 Log. of .002837=7.45287-10 
 Log. of g^g 1-3-^=3.19613 -10 
 Log. of 60345.6=4.78065 
 Log. of 60158.6=4.77923 
 Log. of 1.8686=0.27151 
 Log. of .-0-^0 =2.67990 - 10 
 
 EXA3irLES. 
 
 1. We shall take as an example, the measurement of 
 Mont Blanc, the highest mountain in Europe. The first 
 who ascended to the summit of it was Saussure ; he made 
 there observations stated in his Voyage aux Alpes, 
 No. 2003. ; and at the same time other observers were 
 stationed at Geneva in the valley of Chamouny. Let us 
 confine ourselves to the observations made at mid-day, as 
 they are attended with more certainty. 
 
 At the distance of one metre below the summit of 
 Mont Elanc, the barometer indicated 434.2""" and the 
 centigrade thermometer 2.87 below the freezing point ; at 
 an elevation of 35.55™ above the Lake of Geneva, the alti- 
 tude of the barometer was 73S.5""*j and that of the ther- 
 mometer 28.25. 
 
 (No mention is made here of thermometers attached 
 and detached; we may suppose them to agree at each sta- 
 tion,) 
 
PART X. TlilGONOMETIlY. 361 
 
 Barometers. 
 
 Lower station . . 738.5 . Log. .86835 
 Upper station . . 434.2 . Log. .63769 
 
 Dilierence =.23066 
 
 Subtract .00249 = 1 
 lstf\ictor =.22817 
 
 
 Thennomcters 
 
 Above freezing . 
 
 . 28.25 
 
 , 
 
 . . 28.25 above freezing. 
 
 Below freezing . , 
 
 . 2.87 
 
 , 
 
 . . 2.87 below freezing. 
 
 Difference . . . 
 
 , 31.12X 
 
 } 
 
 Sum 25.38 x 
 
 
 .00008 
 
 .002 
 
 Subtract 
 
 .00249 
 
 
 =.05076 
 
 2d factor . . . 
 
 
 
 = 1.05076 
 
 Log. 18336 . =4.26324 
 Log. 1st factor =9.35826-10 
 Log. 2d foctor =0.02150 
 
 Sum .... =3.64300=Log. 4395.6 metres, approxi- 
 mate height. 
 
 1st Correction. Latitude of Chamouny 45° 58'. 
 
 Log =3.64300 
 
 Log. .002837 . =7.45286-10 
 Log. sine 1° 56' =8.52810-10 
 
 Log. sum . . =9.62396 — 10=log. .4207 ; hence, .4 is 
 the first correction, and is subtractive. 
 
 2d Correction. 
 
 
 
 15926.6 
 
 
 
 4395.6 
 
 
 
 Sum 20322.2 . 
 
 . Log. 
 
 = 4.30797 
 
 3 6<>ly» 
 
 • Log. 
 
 = 3.19613-10 
 
 4395.6 . 
 
 • Log. 
 
 = 3.64300 
 
 
 Sum 
 
 = 1.14710=Log. 14.1. 
 
 And 4395.6 -f 13.7 = 4409.3 for the difference of levels, 
 to which, adding 1 metre, we have 4410.3 metres. And 
 again, adding 35.55, we have 4445.83, the elevation of the 
 summit of Mont Blanc above the Lake of Geneva. 
 
 Francoeur makes it 4436.2 above the level of the Lake 
 of Geneva. 
 
 Now the difference of levels is determined in the for- 
 mulse by the multiplication of five factors, three of which 
 are nearly equal to 1, and their product is nearly equal 
 to 1 ; the other two factors are 18336 and the barometer 
 factor, i. e. the 1st factor; the height of Mont Blanc is 
 
 K 
 
362 TRIGOXOMETKY. P.IRT X, 
 
 tliree miles nearly, or about TaVo ^^ ^^^^ earth's radius ; 
 the barometer factor being increased by this quantity, and 
 multiplied into 18336, gives 14 metres for the correction, 
 which Francoeur makes only 3.55 metres. 
 
 M. Coraboeuf has, by Geodesic operations, made by 
 means of triangles of the first order, and the accuracy of 
 which may be depended on, found for the height of Mont 
 Blanc, 4435.92 metres. 
 
 The height of the level of the Lake of Geneva above the 
 level of the sea is 376.165 metres, whence the summit of 
 Mont Blanc is elevated 4812.085 metres, or 15788 feet 
 above the level of the sea ; the number of feet in the 
 Cabinet Atlas is stated at 15735 feet, which is not very 
 different in such a great height. 
 
 2. At Leith quay, latitude 56°, the barometer stood at 
 29.567 inches, attached thermometer 55-}, detached ther- 
 mometer 54°. At Arthur's seat, barometer 28.704 inches, 
 attached thermometer 51^, detached thermometer 50|. 
 Required the elevation of Ai'thur's seat above Leith quay. 
 
 Barometers. Attached therm. Detached therm. Lat. 
 
 Lower 29.567 55.25 54.0 56° 
 
 Upper 28.704 51.75 50.5 x2-90 
 
 gives 22° 
 Diff. 3.5 Sum -64=40.5 
 
 x4 -h900 
 
 90000 )14- .045 
 
 Subtract .00016 2d factors 1.045 
 
 Log. lower = .47080 
 Log. upper = .45794 
 Difference = .01286 
 Subtract = 16 
 
 1st factor = .01270 Log. . . =8.10380-10 
 Which is less than .07 Log. 1.045 =0.01912 
 Log. 60345.6 = 4.78065 
 
 {2. 90357 = Log. 800.9 
 Log. .002837 =7.45287 - 10 
 Log. sm. 22^ = 9.57358-10 
 
 9. 93002 -10= Log. -85 = cor- 
 rection for latitude which is subtractive. 
 
 Ans. Height of Arthur's seat above Leith quay = 800.05 feet. 
 The height by geometrical measurement is S02. 66 feet. 
 
 3. Find the height of the summit of Mont Blanc above 
 the level of the sea in English feet from the following 
 data. Height of barometer at an elevation of 116.6 feet 
 
PART X. TRIGONOMETRY. 363 
 
 above tlie level of the Lake of Geneva 29.086 inches, 
 thermometers 82°. 85 Faln-enheit : at 3.3 feet below the 
 summit, barometer 17.101 inches, thermometer 26°. 834 
 Fahrenheit ; the altitude of the Lake of G eneva above the 
 level of the sea being 1234.2 feet. Ans. 15S19. 7 feet. 
 
 4. Find the height of Guanaxuato in Mexico, from the 
 following observations of H. Von Humboldt : — 
 
 Lower station — Height of barometer 763.15, attached 
 thermometer 25.3 centigrade, detached thermometer the 
 same. Upper station — height of barometer 600.95, ther- 
 mometers 21.3. Ans. 2083.6 metres. 
 
 5. Find the elevation of Puy-de-Dome on the Alps 
 above Clermont, latitude 45° 46', from the following ob- 
 servations of M. Ramond : — 
 
 Lower station — Barometer 728.52, attached thermo- 
 meter 24°. 7 centigrade, detached thermometer 28°.3. 
 Upper station — barometer 705.65, attached thermometer 
 27.8, detached thermometer 25.5. A72S. 287.2 metres. 
 
 6. Suppose the barometer at the lower station to be 
 29.4 inches, attached thermometer 50° of Fahrenheit, de- 
 tached thermometer 45°; and at the upper station baro- 
 meter 25.19 inches, attached thermometer 46°, detached 
 thermometer 39° ; the latitude 45° : required the elevation 
 of the upper station above the lower in feet. Ans. 4129 feet. 
 
 ON THE FIGURE AND IIAGNITUDE OF THE EARTH. 
 
 In the foregoing examples of heights and distances, we 
 supposed the measurements of the base lines to be taken 
 on a plane surface ; but the curvature of the earth's sur- 
 face, has a considerable influence, when the distances are 
 considerable. This influence is so much the greater, as the 
 magnitude of the earth is smaller, other things remaining 
 the same. We shall take for granted that the earth is 
 somewhat of a spherical form, and revolves uniformly on 
 its axis in about 24 hours, and proceed to investigate its 
 magnitude. 
 
 Supposing the earth, therefore, to be a sphere, if any 
 arc of the meridian be measured which bears a known 
 proportion to the whole meridian circle, we shall obtain 
 the magnitude of the meridian, and therefore the length 
 of the radius of the sphere. Norwood, by measuring from 
 R 2 
 
364 TRIGONOMETRY. PART X. 
 
 York to London, found a degree of the meridian to answer 
 to 69-^ geographical miles : the circle of the meridian is 
 therefore 360 x 69^, or 25020 miles : dividing this by 
 3.1416, we have 7964 miles for the diameter of the 
 sphere.* 
 
 If the degrees of the meridian at different latitudes are 
 unequal, it is manifest that the figure of the earth is not 
 spherical ; and, by trigonometrical measure taken in India 
 and Sweden, such is found to be the case. 
 
 The figure of the earth has been investigated on the 
 principles of hydrostatics, considering the earth to have 
 been originally a fluid mass. Assuming this hypothesis, 
 and considering the only force concerned to be gravity, 
 the figure takes a spherical form to insure equilibrium : 
 but, in taking into account the diurnal rotation, a part of 
 the gravitating force is destroyed in consequence. If in 
 addition to the assumption of fluidity, the specific gravity 
 of it is considered to be the same in every part, Newton 
 has demonstrated, that the figure of equilibrium, is that of 
 an ellipse revolving on its shorter diameter, and that the 
 ratio of the diameters, is that of 230 to 231. 
 
 "V\T:iatever be the figm^e of equilibrium, and in whatever 
 manner the specific gravity of the fluid mass may vary, 
 all the forces concerned at any point may be reduced to 
 one, which one must be perpendicular to the surface at 
 that point. 
 
 The force produced by the rotation is called the centri- 
 fugal force ; and, when a body revolves in a circle, this 
 force is proportional to the square of the linear velocity 
 divided by the revolving radius. This force acts in the 
 direction of the prolongation of the radius, but the square 
 of the linear velocity, is as the square of the revolving 
 radius multiplied into the square of its angular velocity. 
 The centrifugal force is therefore as the revolving radius 
 multiplied by the square of the angular velocity, or as the 
 revolving radius divided by the square of the time of 
 rotation : this is the centrifugal force at any latitude in 
 tbe direction of the prolongation of the radius of the 
 
 * He took the sun's meridian altitude at London near the Tower, 11 June, 1633, 
 and deduced the latitude 51° 30' ; and also at York, near the middle of the city, and 
 deduced the latitude ef York 53° 58', having 2° 28' for the difference of latitudes ; 
 and he measured from York to London 9149 chains, each chain being 6 poles, each 
 pole ISJ feet, that is, every chain 99 feet ; then as 148', the difference of latitudes : 
 9149 chains : 60' : : 3709 chains, or 3G7196 feet, or 367200 feet in a degree ; whence 
 he deduces 25036 miles for the circumference of the meridian, and 7966 miles for 
 the diameter — See Norwood's Seaman's Practice, pp. 6, 7, 8. London, 1637. 
 
I'AKT X. TRIGONOMETliY. 365 
 
 parallel of that latitude. To find its efficacy in the direction 
 of the perpendicular to the surface of equilibrium, it must 
 be diminished in the ratio of the cosine of the latitude to 
 radius. This is nearly the ratio of the revolving radius to 
 the distance from the surface to the centre. The centri 
 fu"-al force opposed to gravity is therefore very nearly as 
 the square of the revolving radius, divided by the product 
 of the square of the period of rotation by the distance 
 from the surface to the centre. 
 
 Newton has shown that the effective force of gravity is 
 at any point on the surface inversely as the distance of 
 that point from the centre ; and the distance of this point 
 irom the plane of the equator reckoned by the perpen- 
 dicular to the surface, is inversely as the distance of this 
 point from the centre. It follows tliat the effective power 
 of gravity on any point on the surface is proportional to 
 the distance of that point from the plane of the equator, 
 reckoned by the perpendicular to the surface. This agrees 
 with Stirling's construction {Phil. Trans. No. 438. for 
 part of the year 1735). If, therefore, we can obtain the 
 ratio of the effective gravities at two points of the earth's 
 surface, we can also obtain the ratio of the distances of 
 these jDoints from the centre, and, consequently, the ratio 
 of the diameters of the elliptic meridian ; this if the fluid 
 mass be homogeneous. 
 
 Now we may estimate gravity by the effect it produces ; 
 as, for example, by the space through which a heavy 
 body falls in a small portion of time ; and this space by 
 writers on mechanics is demonstrated to be proportional 
 to the length of a pendulum performing its vibrations in 
 tlie same time : the force of gravity is then proportional to 
 the length of the pendulum. 
 
 The length of a pendulum vibrating seconds : space 
 tlirough which a heavy body falls in a second : : twice 
 the square of the radius of a circle : square of the semi- 
 circumference; or, as 2 : 9.916, or 1 : 4.953. The length 
 of the pendulum vibrating seconds in the latitude of Lon- 
 don is 39.1393 inches, whence in one second a heavy 
 body falls 16.155 feet. The length of a pendulum vi- 
 brating seconds in Melville Island, in lat. 74° 47', i^ 
 39.21 inches, and at the North Pole it may be reckoned 
 at 39.22 ; and the descent of a heavy body at 16.6 feeu 
 and 16 feet at the equator. 
 
 In comparing the force of gravity with other forces, 
 R 3 
 
366 trigoxo:metry. paet x. 
 
 the double of the fall is taken as its measure ; if, there- 
 fore, we compare it with the centrifugal force at the 
 equator, we must take it in the ratio of 32 feet to the 
 square of the velocity divided by the revolving radius. 
 
 Then, as 24 hours : 25020 miles :*. 1 second * 
 
 25020X52^ feet=M£l^ feet=1529 feet. 
 24x60x60 9 
 
 T>^-' 32 feet :||^feet:: 1:283. 
 
 Thus at the equator the centrifugal is -zr^^ of gravity. 
 
 "Whence the gravity at the pole : gravity at the equa- 
 tor : : 283 : 282j if the earth were a solid sphere of rota- 
 tion. 
 
 And Newton's investigation leads to the conclusion, that 
 for the fluid spheroid, equatorial diameter .* polar diameter 
 : : 4 times the above polar gravity .* o times the above 
 equatorial gravity — polar gravity :: 1132 : 1127 :: 
 226.4 ; 225.4 
 
 In the fluid spheroid of equilibrium he deduces the 
 conclusion that the effective force of gravity on any point 
 of the surface is inversely as the distance of that point 
 from the centre, whence the polar gravity : equatorial 
 gravity : *. 226.4 : 225.4. 
 
 He has also shown that the effective force of gravity is 
 proportional to the length of a pendulum vibrating in 
 small circular arcs in a very small portion of time ; thus 
 the length of a pendulum vibrating seconds at the pole : 
 lens^th of a seconds pendulum at the equator :: 226 
 : 2^25.* 
 
 Li an ellipse, when the difference of the diameters is 
 very small in comparison to the greater diameter, the excess 
 of the equatorial radius above the distance of any point 
 on the surface to the centre : equatorial radius : *. differ- 
 ence of the diameters of the spheroid divided by the 
 equatorial diameter and multiplied by the square of the 
 sine of the latitude of the given point : square of radius. 
 
 EXAJMPLE. 
 
 Let a point on the surface be in latitude 30° ; then sine 
 30° = i radius and (sine 30°)'-^ =i (radius) -\ Hence the 
 
 * The above is only to be considered an approximation. Newton, by using raore 
 accurate numbers, finds the abr,ve ratios to be 289 : 288, and 231 *. 230. 
 
PATvT X. TRIGONOMETRY. 367 
 
 above ratio becomes t-I-^x^ (radius)- : (radius)^ :: ^4^ 
 
 : 1. 
 
 Hence equatorial radius : distance from the centre : * 1 
 
 • : ^4^ : : 924 : 923. 
 
 Similarly for a point at latitude 45° equatorial radius ; 
 distance :: 1 : 1-t:^tX-1 :: 462 : 461 :: 924 : 922. 
 
 Hence distance of the former point : distance of the 
 latter : : 923 .* 922 : : gravity of the latter point : gravity 
 of the former : I length of the seconds pendulum in latitude 
 45° : length of the seconds pendulum in latitude 30°. 
 Now in fact the lengths of the pendulums areas 39.12650 
 inches : 39.06393 inches, or as 923 : 921.5, which does 
 not agree with the former ratio ; neither do the trigonome- 
 trical measurements of arcs of the meridian : the hypothesis 
 of a uniform specific gravity is therefore untenable. 
 
 Abandoning the assumption of homogenity, but retain- 
 ing the hypothesis of fluidity, the conditions of equili- 
 brium are still fulfilled if we suppose a spheroid consisting 
 of spheroidal strata, in each of which every particle is of 
 the same specific gravity, the density varying from one 
 spheroidal surface to another according to any assigned 
 law. AndClairaut has deduced this remarkable conclusion : 
 Whatever be the law of the earth's specific gravity, if its 
 ellipticity be added to the ratio which the excess of the 
 polar above the equatorial gravity has to the equatorial 
 gravity, the sum will be equal 2^ times the ratio which 
 the centrifugal force at the equator bears to the equatorial 
 gravity. 
 
 Or, as he has stated it in other terms : If from twice the 
 ratio which the difierence of the polar and equatorial 
 diameters bears to the equatorial diameter, on the suppo- 
 sition of homogenity, there be taken the ratio which in the 
 actual case the difference of the polar and equatorial 
 gravities bears to the equatorial gravity, the difference 
 will be equal to the ratio, which in the actual case the 
 difference of the polar and equatorial diameters bears to 
 the equatorial diameter. 
 
 Now Airy has shown (Mathematical Tracts, p. 106. 
 No. 63. edit, of 1826) that the increase of gravity above 
 the equatorial gravity is as the square of the sine of the 
 latitude ; and the increase of the pendulum's length is in 
 the same proportion. 
 
 We may therefore, from known observations of the 
 
368 TRlGONO:irETRY. PAET X. 
 
 seconds pendulums at any two latitudes, find the ratio of 
 the polar and equatorial gravities thus. 
 
 RULE. 
 
 oSIultiply the length of each of the pendulums, by the 
 square of the sine of the latitude, corresponding to the 
 other pendulum ; take the difference of these products, 
 divided by the square of radius for a dividend, which di- 
 vide by the difference of the lengths of the pendulums ; 
 the quotient gives the ratio of the equatorial gravity, to 
 the increase of the polar gravity, above the equatorial. 
 
 EXAMPLE. 
 
 At Madras, lat. 18° 4', the length of the seconds pen- 
 dulum was observed to be 39*0237 inches. At MelviUe 
 Island, lat. 74° 47', length of seconds pendulmn was 
 observed 39*207 inches. 
 
 Sine 18° 4' . . . Log.= 9.491535 
 
 Log. sine square . . =18.983070 
 Log. 39.207 . . . = 1.593364 
 
 Sine Log =20.576434 
 
 10^0x3.37708. 
 
 Sine 74° 47' . . Log. = 9.984500 
 Log. sine square . . =19.969000 
 Log. 39.0237 . . . = 1.591328 
 
 Sine =21.560328 
 
 10-0 X 36.33530. 
 
 Diff.= 1020x32.95822 = Divisor 1020=(radius)-\ 
 39.2070-39.0237. . . . = .1833=dividend 
 Log. 32.95822 . . . . =1.517964 
 
 Lo^. .1833 . ar. co. =0.736838 
 
 Diff. Log. 180 ... =2.264802 
 
 Hence, difference of polar and equatorial gravities 
 equat. grav. :: 1 : 180. 
 
 By Clairaut's rule, we have for the ellipticity 2 x ^}i 
 
 1 
 
 362 
 
 -231 
 
 1 .-? 1 _ 
 
 - 1 
 
 ■ISO 
 
 ~231 
 
 Xl80~ 
 
 4 1 5 t; ~ 
 
 -^1 
 
 Therefore, polar diameter : equatorial :: 316 .* 317. 
 This is conformable to experience. 
 
 2. The length of a pendulum vibrating seconds at Lon- 
 don, lat. 51° 3r, is 39.13929 inches ; and atUnst, lat. 60° 
 454-', it is 39.17146 inches. Required the ratio of the 
 
PAKT X. TKIGONOMETllY. 369 
 
 polar, and equatorial gravities, and the ratio of the polar 
 and equatorial diameters. Ans, 179 io 180, 321 to 322. 
 
 PROBLEM I. 
 
 To find the length of a 'pendulum vibrating seconds in 
 any latitude. 
 
 RULE. 
 
 Multiply the square of the sine of the latitude by the 
 Humber .20862 ; divide the product by the square of the 
 radius ; add the result to 39.01146 ; the sum is the length 
 of the seconds pendulum in inches. 
 
 EXAMPLES. 
 
 1. Required the length of the pendulum vibrating se- 
 conds in lat 30°. 
 
 Here (sine 30°)2 -- (radius)^ = i, and \ of .20862 = 
 .05215. 
 
 Whence 39.01146 + .05215 = 39.06361 inches, the 
 length required. 
 
 2. Required the length of the seconds pendulum in the 
 following latitudes, viz. 0°, or at the equator, at lat. 45°, 
 at lat. of London 51° ZV, and at the north and south 
 poles. 
 
 A71S. 39.01146, 39.11577, 39.13929, and 39.22008 
 inches. 
 
 Note. It has been found by measurement of meridian arcs, that the ratio of the 
 earth's diameter is more nearly as 299 to 300 ; from this the polar is to the equa- 
 torial gravity as 188 to 187: thi"s ratio, combined with the measure of the pendulum 
 by Captain Kater, at London, has furnished the above rule. 
 
 To determine the magnitude as well as the ratio of the 
 diameters, recourse must be had to trigonometrical 
 surveys. 
 
 PROBLEM n. 
 
 To determine the figure and magnitude of the eartJCs 
 diameters^ having given the lengths of the arcs of meridian 
 corresponding to two given latitudes. 
 
 RULE. 
 
 Divide each measured arc by the number of degrees 
 and parts contained in it ; call the greatest quotient the 
 R 5 
 
370 TRIGONOMETRY. PART X. 
 
 first term, the other the second term, and the difference 
 is the third term. To double the third term, add three 
 times the second term multiplied bj the square of the sine 
 of the latitude nearest the equator, corresponding to the 
 first term, and divided by the square of the radius, and 
 from the sum subtract three times the first term multi- 
 plied by the square of the sine of the less latitude corre- 
 sponding to the second term divided by the square of the 
 radius ; call this result the fourth term ; then, as fourth 
 term : third term : *. equatorial diameter .* difference of the 
 equatorial and polar diameters. 
 
 Divide the third term by the fourth, ^nd call the quo- 
 tient the fifth term. Multiply three times the fifth term 
 by the square of the least latitude, corresponding to the 
 first or second term, and add 1 to it, and from the sum 
 subtract twice the fifth term ; call this last result the sixth 
 term; then as sixth term : 2 :: 1st or 2d term (accord- 
 ing to which the latitude has been used) .* 7th term*, 
 which, multiplied by 57-29578, gives the equatorial 
 diameter. 
 
 The equatorial diameter diminished by the product of 
 itself and the fifth term gives the polar diameter. 
 
 EXA3IPLES. 
 
 1. Let the arc of the meridian measured from lat. 30° to 
 lat. 32° be 137.756 miles, and the arc of the meridian from 
 lat. 45° to lat 46° be 69.047 miles. Required the equa- 
 torial and polar diameters. 
 
 Here are 2 degrees of arc in the first, and 1 degree in 
 the latter arc. 
 
 ^ = 68.878; hence this is the second term, and 
 
 69.047 the first, and .169 the third. Lat. for 1st term 
 45° ; its square divided by radius squared = \ ; lat. for 
 second term 30°, its square divided by radius squared 
 
 2 X .169-1-3 X 68.878 x ^=103.655 
 3x69.047x1-= 51.785 
 
 Diff. = 5 1.870= 4th term. 
 A.S 51.870 : .169 : ! 306 : 1 : equatorial diameter: 
 difference of equatorial and polar diameters. 
 Hence -j^g- is the fifth term. 
 
 * This seventh term is the length of two degrees on the equator. 
 
PART X. TRIGONOMETRY. 371 
 
 Taking lat. 45" square of its sine, divided by square of 
 radius =-^5 which belongs to the 1st term. 
 612-4-3 — 4 
 .M +13^^ x^-^e =— 612— = '-^^^^^ *^^- 
 
 Then, as ^ : 2 :: 69.047 : 138.320; and 138.320 x 
 
 57.29578 = 7925.153 miles, the equatorial diameter ; and 
 7925.153 -4- 306 = 25.899 ; whence 7925.153-25.899 = 
 7799.254 miles, the polar diameter. 
 
 »T . ,.., «n^-n diameter X 180 
 
 Note. 57.29578=-^ j . 
 
 circumference 
 
 2. Colonel Lambton, in India, measured an arc of the 
 meridian extending from latitude 8° 9' 38.''4 ; to latitude 
 10° 59' 48''.9 its length =1029100.5 feet =194.905 miles. 
 Svanberg, in Sweden, measured an arc of the meridian 
 extending from latitude 65° 3V Z2".2 to latitude 67° 8^ 
 49^8; its length was found =593277.5 feet = 112.363 
 miles. Find from these data the ratio and magnitudes of 
 the equatorial and polar diameters. 
 
 Ans. Ratio 302 to 303; equatorial diameter 7924 
 miles ; polar diameter 7897.85 miles. 
 
 The lines drawn perpendicular to the surface of a sphe- 
 roid do not, as in the sphere, meet at the centre ; the 
 latitudes assumed above are the angles which their per- 
 pendiculars make with the plane of the equator, and are 
 immediately determinable from astronomical observations. 
 If we call the true latitudes those angles which lines drawn 
 from the surface to the centre make with the plane of the 
 equator, the above may be called the elliptic latitudes, 
 which are always greater than the true latitudes. The 
 difference of these two latitudes is equal to the angle 
 which the two straight lines, one perpendicular to the sur- 
 face the other drawn from the same point of the surface 
 to the centre, make with each other at the surface — this 
 is called the angle of the vertical. This difference is 
 greatest at latitude 45°, and bears the same proportion to 
 
 ■J ono 
 
 ^ /.^ ' ^^ = 57° 17' 44''.8, that the difference of the 
 
 equatorial and polar diameters bears to the equatorial 
 diameter. If we take this ratio as 1 to 300, this greatest 
 difference is 11'' 27''.5: at equal angular distance from the 
 equator and pole the difference is equal : thus, in latitude 
 15° the difference is 5' 43''.75, the same at latitude 75°. ^ 
 R 6 
 
372 TRIGONOMETRY. PART X. 
 
 PROBLEM ni. 
 
 Tojiiid the true latitude, having the elli^ytic latitude 
 given, and conversely. 
 
 RULE. 
 
 Take tlie difference between 45° and the given latitude ; 
 then, • 
 
 As radius '. cosine of twice this difference ! ! 11' 21" 5 
 : angle of the vertical. 
 Then, true lat.=: elliptic lat. - angle of the vertical ; 
 elliptic lat. —true lat. + angle of the vertical. 
 
 EXAMPLE. 
 
 Find the true latitude answering to 65° elliptic latitude^ 
 and the elliptic latitude answering to 32° true latitude. 
 A?is. First angle of the vertical 8' 43'^6 ; triie lati- 
 tude 64° 51' 16" A ; second angle of the vertical 
 10' 18''; elliptic latitude 32° 10' 18". 
 
 Note. A more recent discussion of the figure and magnitude of the earth gives 
 the e]lipticity =55! and 69.156 miles, the length of an equatorial degree ; whence 
 2489G. 16 miles, the circumference of the equator, and 7924.7 miles, the equatorial 
 
 7924.7 
 diameter ——^=26.4 miles difference of diameters, and 7898.3 miles, the polar 
 
 oOO 
 diameter, which values we shall follow. 
 
 PROBLEM IV. 
 
 To find the length of an arc of the meridian from the 
 equator to any given elliptic latitude. 
 
 RULE. 
 
 Multiply 69.156 by the degrees and parts of the latitude; 
 from the product subtract its 600th part ; subtract besides 
 9.9 multiplied by the sine of twice the latitude, divided by 
 radius ; the result is the answer in miles.* 
 
 EXAMPLES. 
 
 1. Required the arc of the meridian from the equator 
 to 15° of latitude. 
 
 * The whole subtractive quantity shows by how much the proposed arc of the 
 meridian is less than a similar arc of the equator, the quantity G9. 156 — its GOOth 
 part = 69.041. See Prob. IX. p. 374. 
 
PART X. TRIGONOMKTIIY. 373 
 
 Here, sine 2x 15°-T-radius = J,. 
 
 .•.69.156x15=1037.340 / . the 600tli = 1.729 
 Subtract 6.679. .9.9x4 =4.95 
 Sum =6.679 
 A?is. 1030.661 miles. 
 Or tlius : — 
 69.041 X 15 = 1035.615 and 1035.61 5 -4.95 = 1030.665, as 
 before, nearly. 
 
 2. Required the arc of the meridian from the equator 
 to the parallel of 54° 18^=54.3°. Ans. 3739.5 miles. 
 
 3. Required the arc of the meridian from the equator 
 to the pole. Ans. 6213.67 miles. 
 
 PROBLEM y. 
 
 To find the length of mi arc of the meridian from the 
 equator to any given true latitude. 
 
 RULE. 
 
 Multiply 69.041 by the degrees and parts in the lati- 
 tude, and add 3.302, multiplied by the sine of twice the 
 latitude divided by the radius. 
 
 EXA3IPLES. 
 
 1. Required the arc of the meridian from the equator 
 to 15° oflat. 
 
 Here, sine 2 x 1 5° -r- radius =i. 
 
 69.041 X 15 = 1035.615 3.302 x 1 = 1.651 
 Add . . . 1.651 
 
 Ans. 1037.266 miles. 
 
 2. Required the arc of the meridian from the equator 
 to the parallel of London 51° 32'. Ans. 3561.13 miles. 
 
 3. Find the length of the elliptic quadrant, and the 
 whole meridian, and show the difference between the 
 meridional circumference and the equatorial. 
 
 Ans. Quadrant =^62\2>. 61 miles ; whole meridian 
 = 24854.68, ivhich is 41.493 miles shorter than 
 the equator. 
 
 Xote. According to the estimate of the French mathematicians, the length of 
 the quadrant of the meridian, reduced to English measure, is 6213.822 Sec. miles, 
 or 32808992 feet, the ten-millionth part of which is 3.2808992 feet, the length of 
 the French metre, which is tlie basis of their measures. 
 
374 TRIGONOMETRY. PART X. 
 
 PROBLEM. VI. 
 
 To find the arc of the meridian between two given 
 latitudes. 
 
 RULE. 
 
 Find the length of distance of each lat. from the 
 equator by 4 or 5, and take their difference. 
 
 PROBLEM TH. 
 
 To find the length of a degree of the meridian to any 
 elliptic latitude, iti the middle of the degree. 
 
 RULE. 
 
 Subtract from 69.041 the quantity .346 multiplied by 
 the cosine of twice the latitude divided by radius, if the 
 latitude be less than 45° ; otherwise, add. The differ- 
 ence or sum is the length of the degree in miles requii'ed. 
 
 exa:\lples. 
 
 1. "Wliat is the length of the degree from lat. 29° 30' 
 to lat. 30° 30'. And from lat. 59° 30' to lat. 60° 30'. 
 
 Ans. \st. 68.868; 2d. 69.214 miles. 
 
 2. What is the length of the mean degree bisected by 
 the parallel of London, whose true lat. is 51° 30', elliptic 
 lat. 51° 41'. Ans. 69.121 
 
 PROBLEM Vni. 
 
 To find the length of a meridional degree to any true 
 latitude, in the middle of the degree. 
 
 RULE. 
 
 From 69.156 subtract .231 multiplied by the square of 
 the sine of the latitude divided by the square of the ra- 
 dius ; the difference is the answer in miles. 
 
 EXAT^rPLES. 
 
 1. Required the length of the degree of the meridian 
 from lat. 29° 30' to lat. 30° 30'. Ans. 69.098 miles. 
 
 2. What is the length of the mean degree bisected by 
 the parallel of London ? Ans. 69.014 
 
PART X. TRIGONOMETRY. ti / O 
 
 PROBLEM IX. 
 
 To find the length of the mean degree by elliptic latitude, 
 and what latitude it refers to. 
 
 69.156 
 600 
 tends from lat. 44° 30' to 45° 30^ 
 
 The same results from using the true latitude 
 
 This de.gree is 69.156 ^'^^-=69-041 miles, and ex- 
 
 Note- This mean degree is the sum of all the degrees divided by their number. 
 Similarly, the mean length of the seconds pendulum is the sum of all the se- 
 conds pendulums divided by their number, and is equal to the mean of the equa- 
 torial and polar pendulums, it is 39.11577 inches, and answers to lat. 45°. 
 
 PROBLEM X. 
 
 To find the distance of any point on the earth's surface 
 to the centre, corresponding to any given latitude. 
 
 RULE. 
 
 From 3962.35 subtract 13.21 multiplied by the square 
 of the sine of the latitude divided by the square of radius. 
 The result is the answer in miles. 
 
 EXASIPLE. 
 
 In lat. 30° 3962.35— 3.30=3959.05 m7e5. Ans. 
 
 PROBLEM XL 
 
 To find the length of a degree of longitude on any 
 parallel of latitude. 
 
 RULE. 
 
 1st. For elliptic latitude- 
 Multiply .115 by the square of the sine of the latitude 
 divided by the square of radius. Add the result to 
 69.156, and multiply the sum by the cosine of the latitude 
 divided by radius. The result is the answer in miles. 
 2d. For true latitude. 
 
 Multiply .115 by the square of the sine of the latitude 
 divided by the square of radius. Subtract the result from 
 69.156, and multiply the difference by the cosine of the 
 latitude divided by radius, the result is the answer in 
 miles. 
 
376 TIUGONOMETKY. PART X. 
 
 EXAMPLE. 
 
 Let tlie elliptic latitude be 60°, then square of sine of 
 60° -^- square of radius =4, and cosine of 60° -H- radius =?,. 
 And (69.156 + f of .llo) x^= 39.621 miles. Ans. 
 If the true latitude be 60°, then 
 
 (69.156—1: of .115) X i= 39.535 miles. Aiis. 
 
 Note. If the number of miles in the degree of the parallel be multiplied by 
 67.29578, the result is the radius of the parallel in miles ; and multiplying by 
 360 gives the circumference of the parallel in miles. Log. 59.29578=1.758123. 
 The ratio of the miles of any parallels of latitude is the ratio of their velocities in 
 the diurnal rotation. 
 
 In the same manner as the mean meridional degree of 
 the true latitude was found, may be found the mean dis- 
 tance from the surface to the centre, this is the mean of 
 the equatorial and polar semidiameters, it is 3955.75 miles 
 and answers to the latitude of 45°. 
 
 The mean parallel of latitude is the sum of all the 
 parallels of latitude divided by their number, and similarly 
 of the mean degree of longitude. Its length is 44.075 
 miles, and answers to the elliptic latitude 50° 25/ 
 
 By using the true latitude, it comes out 43.977 miles, 
 and answers to 50° 34' true latitude, the mean is 44.026 
 miles. 
 
 By finding the surface of a sphere, having its diameter 
 equal to the equatorial diameter, and subtracting the 
 450th part, we obtain the surface of the earth as a spheroid, 
 this is equal to the surface of a sphere having a diameter 
 equal to the equatorial diameter diminished by a 900th 
 part. This answers to 196856266.488 miles. 
 
 By finding the solidity of a sphere having its diameter 
 equal to the equatorial diameter, and diminishing it by 
 its 300th part we obtain the solidity of the earth, this is 
 equal to the solidity of a sphere having a diameter equal 
 to §{;§ of the equatorial diameter, the same will result by 
 multiplying the surface by one-third of the equatorial 
 diameter, and subtracting a three hundredth part in this 
 manner, we find the solid content 518282128509.519 
 cubic miles. 
 
 That sphere which has its surface and solidity equal to 
 the terrestrial spheroid, has its radius equal to the distance 
 from the surface to the centre of the terrestrial spheroid, 
 at a latitude the square of whose sine is I the square of 
 radius, this latitude is 35° 16^ true latitude this distance 
 is 3957.95 miles. 
 
 This is the radius used in the discussion of the tides. 
 
PAKT X. 
 
 TlilGONO-AIETRY. 
 
 377 
 
 ON TUE MEASUREMENT OF AN ARC OF THE JMERIDIAN. 
 
 The folloAving extract from Keith's Trigonometry 
 will fully explain the method of conducting a trigono- 
 metrical survey for the purpose of being furnished with 
 such data as those given in Ex. 2. p. 371. 
 
 A country may be surveyed, and the distances of re- 
 mote objects may be obtained by means of a series of 
 triangles. Thus, measure a base AB, and let C and D be 
 two objects which can be seen from A and B. Measure 
 the horizontal angles BAG, -^ 
 
 BAD, ABC, ABD, and cal- f 
 
 culate the distance DC, as in 
 Example 12, and likewise 
 the distances DB, DA, BC, 
 AC. In like manner, if E 
 and K be two objects visible 
 from C and D, the distance 
 EK may be found, and its 
 position with respect to CD ; 
 and thus the measurement 
 may be continued to any dis- 
 tance. But, in order that 
 the conclusion may be more 
 accurate, the mensuration 
 from one base to another ^ 
 
 may be carried on by dif- ~ 
 
 ferent sets of triangles ; for instance, two objects V and W 
 might have been chosen instead of A and B, by means of 
 which the distance EK might have been ascertained ; and, 
 it is plain, that by taking several series of triangles which 
 lead to the same two objects, the mean of the results will 
 be more accurate than the measurement obtained from 
 one triangle. When a series of triangles has been carried 
 on for a considerable distance, tlie interval of two objects, 
 whose distance has been determined by calculation, should 
 be actually measured, in order to detect any error which 
 may have been made in the calculated distance. This is 
 called a Base of verificatiori. The trigonometrical sur- 
 vey of England and AYales had its first commencement in 
 1784; and from a base of 27404.2 feet measured on 
 Hounslow Heath, and continued by a series of triangles 
 to Salisbury Plain, the medium distance between Beacon 
 
 ..'-:A 
 
 O 
 
378 TUIGONOMETRY. PAKT X. 
 
 Hill and Old Sarum was 36574.3 feet, and from actual 
 admeasurement the distance was 36574.4 feet, a difference 
 of little more than one inch.* 
 
 By the assistance of a survey, executed ^Yith such ac- 
 curacy, the length of an arc of a meridian may be mea- 
 sured. Let NS be the meridian of any place A, and let 
 © represent the point of the horizon where the sun sets ; 
 measure the angle B A ; then, by having the latitude of 
 the place A given, the amplitude NA © is easily ascer- 
 tained, and consequently their difference BAN is given ; 
 also if from BAG there be taken BAN, the remainder 
 CAN is given. Let BF, CI, be drawn perpendicular to 
 the meridian, then because AB and the angle BAN are 
 given, AF and FB are easily found : in the manner, be- 
 cause AC and the angle CAN are given, the lines AI and 
 IC may be found. 
 
 If from the angle ABC there be taken the angle ABF, 
 the remainder FBC, added to DBC, gives DBF; by means 
 of which, and the given side DB, DG, or FL will be found, 
 and likewise BG. By adding AF to FL we get AL, and 
 by deducting BG from BF we shall have GF=DL. 
 
 In a similar manner the several distances from the point 
 A to the perpendiculars, let faU from each of the angular 
 points of the figure upon the meridian NS, may be de- 
 termined, as also the perpendiculars themselves. By this 
 method the extensive survey of the kingdom of France 
 was carried on from the measure of nineteen bases. 
 
 The measures being all reduced to the meridian NS, 
 the difference of latitude between A and D may be de- 
 termined. 
 
 The latitudes of A and D determined direct from astro- 
 nomical observations are elliptic latitudes. 
 
 OBSEHYATIONS ON THE ADMEASUREMENT OF A BASE LINE. 
 
 Where the ground is perfectly level, the manner of 
 measuring a straight line from one object to another ap- 
 pears to be simple and easy ; yet, on account of the curvature 
 of the earth, no two points on its surface can be exactly 
 situated in the same horizontal line ; the chord of the arc, 
 and not the arc itself, being the horizontal distance. Now 
 
 * Trigonometrical Survey of England and Wales, vol. i. pp. 279, 280., published 
 by Mr. Faden, Charing Cross, London. 
 
FART X. 
 
 TlliGONOMETKY. 
 
 G79 
 
 the radius of one circle is to the radius of an}' other 
 circle, as any arc of the former is to a similar arc of the 
 latter. If we take, for instance, the base line mea- 
 sured on Ilounslow Heath 27404.2 feet, the radius of 
 the earth 3958 miles (the radius of the equal sphere at 
 lat. 35° 16^ of the spheroid) or 20898240 i^Qi, we shall 
 
 27404 '> 
 have 20898240 feet : 27404.2 feet : 1 : ^^^^ feet, 
 
 20898240 ' 
 
 the length of the measured arc in parts of the radius. 
 
 But the difference between any arc and its chord, the 
 
 radius being 1, is 4\ of the cube of the length of the arc ; 
 
 (27404 9 \"3 
 20898^40 / ^ -T =-000000000094 will express the 
 
 difference in parts of the radius, which multiplied by 
 20898240 feet, the radius of the earth, produces .001965 
 feet, the extent by which a terrestrial arc of upAvards of 
 5 miles, exceeds the chord of the same arc, a difference 
 scarcety worth notice, even where the greatest accuracy 
 is required. 
 
 When the ground on which a base line is to be mea- 
 sured is sloping, it will be necessary, in some cases, to 
 reduce it to a horizontal level. Thus, after having deter- 
 mined the direction of the base AF*, by poles LA, Im, Hn, 
 
 Go, pointed at one end, and fixed perpendicularly in the 
 ground by means of a plumb-line ; the sum of the liori- 
 zontal distances Lm, Jn, Ho, GF v,^ill evidently be equal 
 to the whole horizontal distance AE ; and if t}ie heights 
 AL, ??il, nil, Go, be successively measured, their sum 
 will give the whole height EF. 
 
 If the ground be irregular, or if it ascend and descend 
 alternately, it is evident that tlie difference between the 
 heights of the poles must be added when ascending, and 
 subtracted when descending, in order to determine the 
 different elevations and depressions of the gi'ound. 
 
 Surveyors generally ascertain the altitudes of irregular 
 
 * The point A, and the summits of the hills, ni, n, o, F, should be connected, so 
 as to form a regular slope AF. 
 
380 
 
 TRIGONOMETRY. 
 
 hills bj tlie assistance of a spirit-level, and perpendicular 
 poles placed at convenient distances from each other. 
 This practice is called levelling. 
 
 A base line, on a sloping ground, may likewise be mea- 
 sured by taking angles at its extremities with a theodolite. 
 Thus, let l?n represent a theodolite, AL a pole hxed per- 
 pendicular to the horizon and equal in length to the 
 height of the instrument ; also, let KI be a horizontal line 
 (which may be ascertained by the bubble of air in the 
 spirit-level of the telescope resting in the middle) and 
 IvIL the angle of depression between the top of the pole 
 AL and the horizontal line KI. 
 
 Then, because KI is parallel to AB, the angle KIL is 
 
 equal to the angle niAB ; then, rad. : Atn : : cos. Z. KIL 
 
 .^ . T> A?72 COS. Z. KIL 
 : AB, or AB= ^ 
 
 rad. 
 
 If Awi=400 yards, and Z KIL =4°, AB will be 399.026 
 yards, hence the difference between Am and AB is less 
 then 1 yard. It appears from this example, that when 
 the measured base is inclined to the horizon in a small 
 angle, a reduction of this kind will be unnecessary, except 
 in cases where great accuracy is required. 
 
 OF THE ERRORS WHICH OCCUR IX TAKING ANGLES OF ELE- 
 VATION AND DEPRESSION WITH A THEODOLITE. 
 
 When the observer 
 is at a considerable 
 distance from the ob- 
 ject, the altitude taken 
 with a theodolite will 
 require correction. In 
 ■ the first place, the ho- 
 rizon of the observer 
 and that of the object 
 observed are not the 
 same. Let C be the 
 centre of the earth, D 
 the summit of a moun- 
 tain, and HPOB the 
 horizon of the ob- 
 server. Through D 
 draw EDF perpendi- 
 cular to DC, and it 
 
PAUT X. TRIGONOMETRY. 381 
 
 will be the horizon of the point D. Now SD -will be the 
 true height of the mountain above the horizontal line PS ; 
 DPS the true angle of elevation, and DPB the observed 
 angle, from which the height of the mountain is deter- 
 mined to be BD, instead of SD. 
 
 Again, the L PBD is supposed to be a right angle, in 
 oi-dinary calculations, but in reality it is equal to the sum 
 of the two interior angles BPC and PCB (32 Euclid, I.) 
 = 90° + ZC. 
 
 The French academicians measured an inclined base at 
 
 Peru, the length PD of which was found to be 6274.057 
 
 toises* the angle of elevation DPB was 1° 5^ 43'^ (the 
 
 effect of refraction being deducted). Now rad. : PD : : 
 
 ■n^i^T-^ TiT-w 1 -r»Tx PD sin. DPB ^,^^^ 
 
 sm. DPD : BD, hence BD = . = 119.93 toises. 
 
 rad. 
 
 In order to calculate BS, PD may be used instead of PB, 
 and CS and CB may be considered as equal to each other 
 without sensible error. And as (2CS-fBS) x BS=PB^ 
 (36 Euclid, in.), it follows that 2CS.BS = PD2, and 
 
 PD- 
 hence BS=2cs. 
 
 If the diameter of the earth be taken = 6543373 toises, 
 as deduced from the admeasurements in Lapland, Paris, 
 
 and Peru, BS will be found =^^^ = 6.0158 toises. 
 
 By adding BS to BD the true height SD of the mountain 
 = 125.9458 toises. Had the height DS been determined 
 from the triangle DPS, by the most exact calculation it 
 would have been = 125.97 toises. Hence it appears that 
 in most cases the angle at C may be rejected, that PB 
 may be taken =PD, the /_TBT>= Jf right angle, and CS 
 = CB without material error. 
 
 Angles of elevation or depression taken with a theodo- 
 lite may be corrected thus : Let D be the place of the 
 telescope when the theodolite stands on the vertical line 
 CD ; P the situation of the telescope on the vertical line 
 CP. Then if the telescope at P be directed to an object 
 at D, the elevation of that object above the horizontal line 
 HPB is the /_ DPB ; and when the telescope is at D, and 
 directed to an object at P, the angle of depression, below 
 the horizontal line EDF, is the /_ FDP. 
 
 • Bouguer, Figure de la T«rre. A toise=6 French feet, and 107 French feet=» 
 114 English feet. 
 
382 TRIGONOMETRY. PAP.T X. 
 
 Now, because PB touclies the circle, and PS cuts it, the 
 L BPS (measured by half the arc PS*) = -i^l C ; there- 
 fore ZDPS=:izC + / of elevation DPB. " 
 
 Through D draw DG- parallel to SP, then the L G-DP 
 = /_ DPS (29 Euclid, L), hence/ GDP= ^ Z C + Z of de- 
 viation DPB ; and because the tiiangles CPS and CDG 
 are isosceles, and the Z C is common to both of them, the 
 ZCPS = ZCDG; but the angles CPB and CDF are 
 right andes, therefore the ZBPS (=iZC) ^ZFDG. 
 
 Again^ L FDP^ L GDP + /. FDG == i Z C + Z of ele- 
 vation DPB -|-^ZC; therefore ZFDP=:ZC+Z of ele- 
 vation DPB ; to each of these equals add the Z DPB, then 
 Z of depression FDP + Z of elevation DPB= Z C + twice 
 Z of elevation DPB ; hence 
 
 (Zofdepr. FDB-^Zofelev. DPB)-ZC ,, , , . 
 
 ^^ ■ '- = the true Z of 
 
 £i 
 
 elevation. 
 
 '^Yhen the angles are both elevations or both depres- 
 sions, their .dilFerence must be diminished bj the Z C, and 
 half the remainder will be the true Z of elevation of the 
 higher of the two objects. 
 
 The Z C is generally very small, and where the mea- 
 sured base does not exceed six or seven hundred yards, it 
 may be rejected. 
 
 EXAMPLE. 
 
 Suppose D and P to be two objects fixed exactly at the 
 same height above the ground as the height of the tele- 
 scope of the theodoHte ; now if the Z FDP of depression 
 be 26', and the ZDPB of elevation 14', what will be the 
 error in observation ? the arc PS, or distance of the 
 stations, being 8000 feet. 
 
 The length of a degree in latitude 51° 9' is 364950 
 feetf ; 364950 feet : 60' : : 8000 feet : V 19'' nearly= Z 
 
 C. Then (26' + 1^0-1' ^^^' ^ly 20.5'^ the true Z of 
 
 elevation DPB; hence 19' 20.5''-14'=5' 20.5^' is the 
 error of the instrument, or the quantity by which the Z 
 of elevation was too small, or the Z of depression too 
 large. 
 
 * The angle formed between the tangent of any arc and its chord, is measured by 
 half that arc. 
 
 For it is equal to the angle in the alternate segment (33 Euclid, III.), which is 
 an angle at the circumference, which is half the angle at thecentre (20 Euclid, III.) 
 
 t Trigonometrical Survey of England and Wales, vol.ii. part 2. p. 113. 
 
PAET 
 
 TRIGONOMETRY. 3S3 
 
 THE NATURE OF TERRESTRIAL REFRACTION, AND ITS 
 EFFECTS ON ANGLES OF ELEVATION.* 
 
 As terrestrial refraction arises from the gross vapours, 
 and exhalations of various kinds, which are suspended in 
 the air near the surface of the earth, and which are per- 
 petually changing, it is very difficult to ascertain the 
 exact quantity of it at any particular time. 
 
 The course of a ray of light in its passage through the 
 atmosphere is, in general, that of a curve which is con- 
 cave towards the earth, and the observer views the object 
 in the direction of a tangent to this curve ; hence the 
 apparent, or observed angle of elevation is always greater 
 than the true angle. 
 
 The altitudes of the heavenly bodies when within o*^ 
 or 6° of the horizon, should never be used wliere a very 
 accurate result is required. The figures of the sun and 
 moon, when near the horizon, are sometimes elliptical, 
 having the minor axis perpendicular, and the major axis 
 parallel to the horizon. This change of figure arises from 
 the refraction of the under limb being greater than that 
 of the upper. But a perpendicular object, situated on the 
 surface of the earth, will not have its length altered by 
 refraction, the refraction of the bottom being the same as 
 that of the top. 
 
 The allowances usually made for refraction are too 
 uncertain for any reliance to be placed on them, as scarcely 
 two writers agree on tliis subject. Dr. Maskelyne makes 
 it Jq of the intermediate arc PS between the observer 
 and the object ; Bouguer -J- ; Legendre -^l. ; General Roy 
 from -1- to t/^ ; and in the second vohime of the Trigo- 
 nometrical Survey, the variation is found to be from -} to 
 J^ of the intermediate arc.f This difference does not 
 arise from inaccuracy of observation, but from circum- 
 stances which cannot be avoided, as the evaporation of 
 rains, dews, &c. which produce variable and partial re- 
 fractions. 
 
 The following method is used in the Trigonometrical 
 Survey J for ascertaining the quantity of refraction : 
 
 * See a paper by Mr. Huddart, in the Philosophical Transactions for 1797, 
 p. 2D. ; and another by the Rev. S. Vince, 170'J, p. 13. Also the Trigonometrical 
 Survey of England and Wales, vol. i. p. 175. 
 
 t Pages 177, 178. part I. t Vol. i. p. 175. 
 
S84 thigono3ii:tky taut x. 
 
 Let C be the centre of the earth, P and S two stations 
 on its surface ; PB, S A the horizontal lines at right an- 
 gles to CA and CB ; also, suppose A and B to be the true 
 places of the objects observed, and a and h their apparent 
 places. Then the /. ^>PS will be the refraction at P, and 
 the Z «SP that at S. * 
 
 In the quadrilateral figure CSOP, the angles at P and 
 S are right angles, therefore the l_ SOP4- Z. C= two right 
 angles, but the three angles of the triangle SOP = two 
 right angles, hence Z. OSP-r Z. SOP+ Z OPS=i Z SOP+ 
 Z C, consequently Z OSP+ Z OPS= Z C, which is mea- 
 sured by the intermediate arc PS. 
 
 Now the sum of both refractions Z«SP+ Z ^I'S=(Z 
 ASP+ ZBPS)-(Z AS« + ZBP^») =(z OSP + Z OPS) 
 - ( Z ASa + Z ~SPb)= Z C - ( Z ASa + Z BP^). Hence 
 the following 
 
 RULE. 
 
 Subtract the sum of the two depressions from the con- 
 tained arc, and half the remainder is'the mean refraction. 
 
 If one of the objects (A), instead of being depressed, be 
 elevated, suppose to the point e, the Z of elevation being 
 eSA ; then the sum of the angles /wSP, twPS will be 
 greater than ZOPS+ZOSP (the Z C, or contained arc 
 PS) by the Z of elevation cSA. Hence /_ 7«SP+ / mPS 
 = Z C + Z cSA, from each of these equals take the Z BP^, 
 then the sum of the two refractions Zw2SP+Z^PS=: 
 Z C + Z ^SA — Z BPZ» ; that is, subtract the depression 
 from the sum of the contained arc and elevation, and half 
 the remainder is the mean refraction. Perhaps it may be 
 necessary to remark, that previous to the observations the 
 error of the instrument must be accounted for. (See p. 38 1.^ 
 
 EX.OIPLE. 
 
 The refraction between Dover Castle and Calais church 
 was thus determined, j" 
 
 Let C be the centre of the earth, PS the surface ; D the 
 station on Dover Castle ; A the top of the great balustrade 
 of Calais steeple : EDF the horizontal line ; also let 
 PG=SD ; then theZFDG=4ZC, or half the arc PS. 
 
 * In observing these angles two instruments are used, one at P and another at 
 S.; and the reciprocal observations are made at the same instant of time by means 
 of signals, or by watches previously regulated for that purpose. The observer at 
 P takes the depression of S, at the same moment which the observer at S takes 
 the depression of P. 
 
 t Trigonometrical Survey, vol. i. p. 173. 
 
PART X. TRIGONOMETRY. 385 
 
 The distance from Dover to Calais is 137455 feet, hence 
 364950 feet : 60' : : 137455 feet : 22' 35'' the L C ; hence 
 =FDG=ir 171". 
 
 The height of D above low-water spring tides =479 feet. 
 The height of A (communicated from France) = 140 1- feet. 
 
 AG=328|~ 
 
 The triangle DGA may be considered as isosceles, and 
 DG or DA= 137455 feet, the distance between Dover 
 and Calais. Hence 4 AG : rad. : : DA : sec. Z. DAG=: 
 89° 55' 53", the double of which deducted from 180°, leaves 
 8' 14" for the Z GDA, to which add the Z FDG= 1 1' 171", 
 and the whole angle FDA=19' 31-^" supposing there was 
 no refraction ; but the /. FDA was determined from ob- 
 servation to be 17' 59", hence the refraction was (19' 311^' 
 -17' 59"=) l'32i", being about -^ of the contained 
 arc. 
 
 Mr. Huddart is of opinion, that a true correction for 
 the effect of terrestrial refraction cannot be obtained by 
 taking any part of the contained arc * ; for different points, 
 though nearly at the same distance from the observer, will 
 have various refractions. 
 
 OF THE REDUCTION OF ANGLES TO THE CENTRE OF THE 
 STATION. 
 
 In surveys of kingdoms and counties, where signals on 
 the steeples of churches, vanes of spires, &c. are used for 
 points of observation, the instrument cannot be placed 
 exactly at the centre of the signal, and consequently the 
 angle observed will be different from that which would 
 have been found at the centre. The correction is generally 
 very small, and is only necessary where great accuracy is 
 required. 
 
 The observer may be considered in three different po- 
 sitions with respect to the centre, viz. he is either in a 
 line with the centre and one of the objects ; or a line 
 drawn from the centre through his situation would, if 
 produced, pass between the objects ; or a line drawn from 
 the centre to the place of the observer, when produced, 
 would pass without the objects. 
 
 * Philosophical Transactions for 1797, p. 29; etscq. 
 
 s 
 
386 TRIGONOMETHY. PAET X. 
 
 First. Let the ob- 
 server be at T>, in a line 
 between the obj ects B and 
 C, viz. on one side of the .-./ 
 
 triangle ABC ; B being y./ 
 
 the proper centre of the .'' / 
 
 station and the ZABC ,-' / 
 
 that required. It is plain £ •;■ / 
 
 that the L CDA, being \ ^::s:'" •■-'—-''■ 
 
 the exterior L of the ^ B 
 
 triangle ADB, is too 
 large by the interior L DAB. 
 
 Therefore /. ABC == z CD A - L DAB. 
 Secondly. Let the observer be at O, within the triangle 
 ABC, and let B be the centre of the station, and the 
 L ABC that required. Now L AOC + L OAC + L OCA 
 = L ABC + L OAB + L OAC -f L OCA + L OCB, each of 
 the sums being equal to two right angles ; therefore the 
 L AOC= L ABC + L OAB + L OCB, that is, the L AOC 
 is greater than the L ABC by the sum of the angles OAB 
 and OCB. Therefore Z ABC* = Z AOC-(Z OAB-|- 
 ZOCB). 
 
 Thirdly. Let the observer be at E, without the tri- 
 angle ABC, and let A be the centre of the station, and 
 the/. CAB that required. 
 
 Now L CAB + L ACB + L ABE + L EBC= 
 L CEB + / ECA -f L ACB + L EBC, 
 each of the sums being equal to two right angles ; therefore 
 /_ CAB + L ABE= L CEB + L ECA, and consequently 
 L CAB=: i_ CEB + L ECA— Z ABE. 
 Assume L CAB= /. CEB, from which it does not much 
 differ, and Z ABC=Z: AOC, to which it is nearly equal, 
 and find the angle ACB then in the triangle ABC, a side 
 being given and all the angles, the remaining sides may 
 be found by Prob. I., Oblique Angled Triangles. 
 
 In the triangle ACE there are known AC, AE and the 
 angle CEA, whence the angle ECA may be found. 
 
 Similarly in the triangle BEA, the sides AB, AE and 
 the angle BEA are known, whence the angle ABE may 
 be found. 
 
 [ If E is upon the side AC, angle ECA=0, this becomes 
 the .first case, and angle CAB = angle CEB — angle ABE.] 
 
 * If the proper centre of the station were at O, and the observer at B, It is plain 
 that the angles OAB and OCB must be added to the/ ABC to obtain the /.AOC. 
 
PART X. TRIGONOMETRY. 387 
 
 Li a similar manner in the triangles BCO, ABO the 
 angles OCB, OAB may be found. 
 
 Thus the angles CAB, ABC may be corrected with 
 these corrected angles, and a side given, the other sides 
 of the triangle ABC may be corrected. 
 
 EXAMPLE. 
 
 Let A and B represent the vanes on two steeples, E the 
 situation of the theodolite upon the steeple A, and its 
 situation upon the steeple B. Then, suppose 
 
 AE=12feet 
 ZCEB=74°32^ 
 ZCEA=:139° 39' 
 
 B0= 10.5 feet 
 ZAOC=49°27'' 
 ZCOB=137°o5' 
 
 It is required to find the angles CAB and ABC, the 
 distance AB being 5000 feet. 
 
 Solution. 
 
 The L CEB = ZCAB nearly, and Z. AOC=: Z.ABC 
 nearly, with these angles and AB, find AC, BC=4581.7 
 and 5811.5. Then, 
 
 AC : sin.zCEA :: AE : sin.zECA=5' 50'' 
 AB : sin.zBEA :: AE : sin.ZABE=7' 29'' 
 Hence Z. CAB =:: 74° 30' 21''. 
 
 BC : sin. Z COB :: BO : sin.zOCB=4' 10" 
 
 AB : sin.Z.AOB :: BO : sin. Z. OAB =0' 6&' 
 
 HenceZ ABC=:49° 21' 54". 
 
 With the corrected angles CAB, ABC, and the distance 
 
 AB, the sides AC and BC may be determined, viz. 4570 
 
 and 5803. 
 
 OF THE REDUCTION OF ANGLES FROM ONE PLANE TO 
 ANOTHER. 
 
 Angles which are inclined to the horizon, may be re- 
 duced to the corresponding horizontal angles, in cases 
 where very great accuracy is required. Let the lines PS, 
 PB, BS, be three chords of terrestrial arcs, that is, let the 
 points P, B, and S, be all equally distinct from the centre 
 of the earth, and let the point I) be elevated so as to be 
 farther from the centre of the earth than any of the points 
 P, B, S, it is required to reduce the triangle BDP to the 
 triangle BSD. 
 
 The line SB may be supposed to be perpendicular to 
 8 2 
 
888 
 
 TRIGONOMETRY. 
 
 PART X. 
 
 each of the chords SB and SP without sensible error, 
 though strictly speaking the angles DSB and DSP arc 
 each equal to 90° + i the arc which the chords SB and SP 
 subtend, by art. 
 
 Likewise the chords SB 
 and SP may be used instead 
 of their corresponding arcs. 
 By inspection of the figure 
 it is plain that BD is 
 greater than BS, and PD 
 greater than PS; but the 
 base PB is common to the 
 two triangles BDP and 
 BSP, therefore the ZBSP 
 is greater than the Z. BDP. 
 (21 Euclid, I.) 
 
 The horizontal angle SBP is obtained from the angle 
 of elevation DBS, and the oblique angle DBP by the fol- 
 lowing proportion : 
 
 As cosine of the angle of elevation DBS : cosine of the 
 oblique angle DBP : *. radius : cosine of the horizontal 
 angle SBP. 
 
 Exactly in the same manner the angle SPB may be 
 found. 
 
 Then angle BSP=180°—(ang. SBP+ang. SPB). 
 
 But the angle BDB taken on the elevated station may 
 be reduced to the corresponding angle BSP by using only 
 the observed angles BDP, and the angles of depression 
 mJ)B and wDP. The angle mJ)B = ang. DBS, and the 
 ang. nDP=ang. DPS. Then by De Lambre's formula 
 
 Sine half ang. BSP=rad. x V- 
 
 "sine 1 (BDP 
 sinei (BDP 
 
 DBS-DPS) X 
 DPS-DBS) 
 
 COS. DBS X COS. DPS 
 
 } 
 
 which is in a form adapted for logarithmic computation. 
 
 This formula of De Lambre's may be applied to any 
 two triangles, whether their bases be in the same hori- 
 zontal plane, and their vertices elevated, or their vertices 
 be in the same horizontal plane, and the extremities of 
 the base of the one triangle be elevated above the ex- 
 tremities of the base of the other. 
 
 Li the annexed figure, if the angle PDB be measured 
 with a sextant, and the vertical angles BDO, PDS be 
 measured with a theodolite. 
 
PAET X. 
 
 TRIGOIJOMETRY. 
 
 389 
 
 by M. de Lambre's formula, viz, 
 
 V 
 
 Jn.^Z.SDO=rad. 
 
 sin. i (PDB + PDS-BDO) 
 
 COS. PDS . COS. BDO 
 
 ^ (PDB+BDO— PDS) 
 
 EXAMPLE. 
 
 From a station at D in the horizontal plane DSO, I 
 took the angle PDB, subtended by the tops of two towers 
 = 37° 53' 20'', and also the angles of elevation BDO = 
 4° 23' 55", and PDS = 4° 17' 21". The height of the 
 tower BO is known to be 40 yards, and that of PS 30 
 yards ; from which it is required to find the horizontal 
 distance of my station from each of the towers, and their 
 horizontal distance from each other. 
 
 A71S. Z. SDO = 38° 0' ; DS = 400 7/ards, DO = 520, 
 a7id SO =320 7/ards. 
 
 SOilE TRIGONOMETRICAL QUESTIONS ^VfilCH REGARD THE. 
 
 earth's curvature. 
 
 At the distance of 25 miles from a tower its top just 
 appeared in the horizon ; required its height. The dia- 
 meter of the earth being 7916 
 miles, and its circumference 24869 
 miles. 
 
 Solution. 
 
 360° :: 25 : 21' 
 
 24869 miles : 
 43" = Z. DAB. 
 
 In the right angled triangle 
 ABD, AB and the angle DAB 
 are given, by which DB may be 
 found = 25 miles. 
 
 Then CE = 7916 miles : DB : : DB : DC very nearly 
 419 feet. 
 
 s 3 
 
390 
 
 TRIGONOMETRr. 
 
 PART X. 
 
 EXAMPLE. 
 
 Supposing it Vv^ere possible to see a light-house or other 
 object D, in the horizon, at the distance of 200 miles, it is 
 required to find its height CD, the diameter EC of the earth 
 being 7916 miles, and its circumference 25019.7024 miles. 
 "^Ans. The L DAB = 2° 52^ 40'^ AD = 3987.028 miles, 
 andT>C=5.02Smiles:=26547.S4feet,bei7igD9U.S4: 
 feet higher than Chimborago, the highest of the 
 Andes; and 285.S4: feet higher than Dhawaligiri 
 ill Bootan, the highest known mountain in the 
 xvorld. 
 Similarly, when DC and the diam eter CE are given 
 DB=^/CExDC: 
 
 EXA3IPLE. 
 
 Let DC = 26.262 feet, the height of Dhawaligiri, and 
 CE = 7916 miles to find the extent of the visible distance 
 DB. Ans. 197.6 miles. 
 
 OF THE DIP, OR DEPRESSION OP THE HORIZON AT SEA. 
 
 The dip or depression of the 
 horizon at sea is the angle con- 
 tained between the horizon of the 
 observer, and the farthest visible 
 point on the surface of the sea. 
 
 For, if an observer whose eye 
 is situated at D, takes the alti- 
 tude of a celestial object by a 
 sextant, or Hadley's quadrant, and 
 brings that object to the surface 
 of the water at B, instead of the ^ _ 
 
 horizon DF, he evidently makes the altitude too 
 theZFDB=ZBAC. 
 
 Log. tangent dip FDB=6.49047-f i log- of DC m feet. 
 
 Let the bright DC be 40 ; required the dip. 
 
 Ans. 6^ 44''. 
 
 re at by 
 
METHOD OF LEVELLING, 
 
 AS GEISTERALLY PKACTISED BY SURVEYORS, 
 EXGINEEES, ETC., FOR THE RAILWAYS. 
 
 Leyelli^^g is the art of determining the heights or de- 
 pressions of points on the earth's surface, or, when the 
 extent of ground to be levelled is inconsiderable, with 
 respect to an horizontal plane passing through some given 
 point on the ground. 
 
 In order to obtain the difference of level between any 
 number of places, reference must be had to some fixed 
 mark, from which an imaginary line, called the datum 
 line, is dravv'-n, upon which line all the levels are based ; 
 and all variations on the surface of the ground are reck- 
 oned from this assumed line, some fixed mark being 
 generally chosen at either end of the line of levels, to 
 which they are all referred. It is sometimes deemed 
 advisable to fix the datum line lower than either end, 
 which may be done by assuming a line of 100 or any 
 number of feet lower than the fixed mark, on which 
 assumed datum line the levels will be based. 
 
 In setting out lines of railways it is very obvious that 
 the centre line of the road should be very accurately de- 
 termined, and for this purpose a stake should be firmly 
 driven into the ground at every chain's length, or at every 
 100 feet, as may be deemed most advisable : this being done, 
 the centre line should be carefully levelled and sights 
 taken by the spirit-level at each stake respectively, which 
 are termed bench marks ; these stakes should be numbered 
 in regular succession ; each section must be plotted and 
 the gradient marked thereon, showing the declivity or 
 acclivity of the line, as also the height of the embank- 
 ment or depth of the cutting at each stake, the respective 
 widths should also be noted, and the names of the pro- 
 prietors of the grounds thi'ough which the line passes, the 
 s 4 
 
392 LEVELLING. 
 
 quality and quantity of land that may be required, and 
 the estimated damage they may probably sustain either 
 by severance or otherwise. 
 
 It is not a difficult matter to set out a straight line of 
 railway, but to do it correctly a good theodolite is 
 indispensable : should any impediments, such as woods, 
 buildings, &c., obstruct your line, recourse must be had 
 to tie lines, see Prob. 8, Land Surveying ; or two sides 
 of an equilateral triangle may be measured, by first 
 setting off the supplementary angle of 120° between the 
 first side of the triangle and the line run ; let the station 
 forming the triangle be placed in such a position as to 
 form an angle of 60° with another station forming the 
 second side of the triangle, which two sides being mea- 
 sured equal, the third side and angles will also be 
 determined and equal. 
 
 When a line is curved the straight line becomes a 
 tangent to it, which admits of a variety of trigonometri- 
 cal calculations; but this is disregarded by surveyors 
 generally, it being evident that the circle cannot be struck 
 from a centre, but that the curve may be easily formed 
 by a combination of straight lines. 
 
 When the gradient of a line is known, you may easily 
 determine the respective widths that will be required on 
 every part of the line : knowing the embankment or 
 cutting where stakes are driven, multiply the height or 
 depth of the section by 4, to which add the width of 
 the top of the railway and the ditches, if any, and you 
 will have the width of the bottom ; this rule is founded 
 (though subject to variations ) on the principle that every 
 foot of perpendicular height requires two feet of base. 
 The gradients or rates of clivity may be ascertained by 
 dividing the difference of level by the distance in chains; 
 the quotient will be the rise or fall at each chain's length. 
 
 The method of calculating the contents of embank- 
 ments or earth-work must be performed by the rules for 
 the prismoid, Prob. 10, Part 4, as also Probs. 3 and 4, 
 Part 6. When the sections are numerous the calculations 
 by these rules become very tedious, whence the reader is 
 referred to Bidder's tables, which will be found very useful 
 to railway surveyors in performing their various calcula- 
 tions. 
 
 The figure of the earth is said to be known by a sur- 
 
LEVELLING. 
 
 393 
 
 face perpendicular to the direction of gravity, or the 
 direction which heavy bodies fall or gravitate towards the 
 centre, the visible horizon being a tangent, or at right 
 angles to the line of direction. 
 
 Two or more places are on a true level when they are 
 equally distant from the centre of the earth; also, one 
 place is higher than another, or out of level with it, when 
 it is farther from the centre of the earth, and a line 
 equally distant from that centre in all its points is called 
 the line of true level. But the earth being round, that 
 line must be a curve, which forms a part of the earth's 
 circumference, or at least parallel to it, or concentrical 
 with it, as the line BCFG, which has all its points equally 
 distant from A, the centre of the earth, considering it as a 
 perfect globe. But 
 the line of sight 
 BDEH, given by 
 the operations of 
 levels, is a tangent, 
 or a right line per- 
 pendicular to the 
 semi-diameter of 
 the earth at the 
 point of contact B, 
 rising always high- 
 er above the true 
 line of level, which 
 is also called the 
 apparent line of level. Thus CD is the height of the 
 apparent level above the true level, at the distance BC or 
 BD ; also EF is the excess of height at F, and GH at 
 G, &c. The difference, it is evident, is always equal to 
 the excess of the secant of the arc of distance above the 
 radius of the earth. 
 
 Now the difference CD, between the true and apparent 
 level, at any distance BC or BD may be found thus. By 
 a well-known property of the circle, 2AC -+• CD : BD 
 : ; BD : CD ; or, because the diameter of the earth is 
 great with respect to the line CD, at all distances to 
 which an operation of levelling commonly extends ; sup- 
 pose 2AC be taken for 2 AC + CD in that proportion, 
 without any sensible error, it will be 2AC : BD : : BD 
 : CD, which is = BD2 -- 2AC, or BC2 -- 2AC, that is, 
 s 5 
 
394 LEVELLING. 
 
 the difference between the true and apparent level is equal 
 to the square of the distance between the places, divided 
 by the diameter of the earth; consequently it is always 
 proportional to the square of the distance. 
 
 Suppose the diameter of the earth to be 7958 miles ; 
 if we first take BC = 1 mile, then the excess BC^ ~ 
 2AC becomes ^V^ of a mile, which is 7.962 inches, or 
 nearly 8 inches, for the height of the apparent above the 
 true level at the distance of one mile. In order to make 
 proper correction for the curvature of the earth, observe 
 the following rules. 
 
 When the distance is in yards, 
 
 KULE I. 
 
 Multiply the square of the distance in yards by 2.57, 
 and divide by 1000000, or cut off six places on the right- 
 hand for decimals ; the rest is the curvature in inches. 
 
 When the distance is in chains, 
 
 RULE U. 
 
 Divide the square of the distance in chains by 800, the 
 quotient will be the depression in inches. 
 
 When the distance is in miles. 
 
 RULE m. 
 Multiply the square of the distance in miles by Q6 feet 
 4 inches, and divide by 100, or prick off two decimals, 
 and you will have the depression in feet. Or, two-thirds 
 of the square of the distance in miles will be the amount 
 of curvature in feet, nearly. 
 
 Note. It must be observed that an allowance of one-seventh is deducted 
 for refraction, it being opposed to the curvature. The terrestrial refraction is 
 found to vary with the state of the atmosphere in regard to heat, cold, and 
 humidity, so that what may be deemed correct in one state of the atmosphere will 
 not answer correctly for another. 
 
 EXAMPLE I. 
 
 Suppose the distance of the line of level to be 900 
 yards, what is the curvature ? 
 
LEVELLING. 395 
 
 Here 900^ x 2.57 = 810000 x 2.57 = 2081700, and six 
 decimals being cut off leaves 2.0817 inches ; then 2.0817 
 — 1^=1.7833 inches, the true curvature. 
 
 EXAMPLE n. 
 
 Suppose the distance of the line of level is 40 chains. 
 Here 402^800— |=1600--800—f =2— |=1.7l in. 
 
 EXAMPLE in. 
 
 Suppose the distance of the line of level to be 2^ 
 miles, required the curvature. 
 
 Here 2.52 x 66.33 -- 100 = 6.25 x 66.33 -^ 100 = 
 414.5625 -=- 100= 4.145625; then deduct ]■ leaves S.55 
 feet for curvature. Or, 2.52 x f- 1- = 6.25 xf — \ = 
 4.23 1=3.6 feet. 
 
 Note. The correction for refraction of one-seventh will be found sufiBciently 
 correct for every state of the atmosphere, except in very extensive trigonometrical 
 operations. It will be necessary to observe that in the common practice of levelling, 
 the surveyors and engineers for the railways seldom make any allowance for either 
 curvature or refraction. 
 
 A DESCRIPTION 
 
 OF THE 
 
 INSTRUMENTS 
 
 USED IN SURVEYING AND LEVELLING. 
 
 Spibit-Level is a tube of glass nearly filled with spirits 
 of wine or distilled water, and hermetically sealed at both 
 ends, so that when held with its axes in an horizontal posi- 
 tion, the air which occupies the part not filled with the 
 spirit or water places itself contiguously to the upper 
 surface. The tube being supposed to be perfectly cylin- 
 drical, the exact horizontality of its axes is ascertained by 
 the extremities of the air-bubble being at equal distances 
 from the middle point in the length of the glass. 
 
 The spirit-tube is used in determining the relative 
 
 heights of ground at two or more stations, in order to 
 
 render it available : for this purpose it is placed within a 
 
 brass case having a Ion q; opening on the side which is 
 
 's 6 
 
396 
 
 LEVELLING. 
 
 the uppermost, and is attached to a telescope ; the teles- 
 cope and tube are then fitted to a frame or cradle of 
 brass, which is supported on three legs ; in the interior 
 of the telescope, at the common focus of the object-glass 
 and eye-glasses, are fixed two wires, at right angles 
 to each other, their intersection being in the line of colli- 
 mation, or that which joins the centres of all the lenses. 
 
 fffriF-TTTTiTnTni^ ^ 
 
 The case containing the spirit-tube is made to turn on 
 a joint at one extremity, as «, by the revolutions of a 
 screw b, at the opposite extremity ; and the telescope 
 rests, near each end, within two arms at the top of a 
 small pillar, A or B, the pillar and its arms resembling 
 the letter Y, and the interior sides of the arms being 
 tangents to the tube of the telescope. One of these 
 pillars is made capable of a small movement in a vertical 
 direction by turning a screw, C, at its base, for the pur- 
 pose of elevating or depressing one end of the telescope 
 and spirit-tube, and, in the more perfectly constructed 
 instruments, both the pillars may be so moved. The 
 pillars are at the extremities of a strong brass plate, EF, 
 the under side of which is connected with the tripod- 
 stand, w^hicli supports the wdiole instrument; and a 
 compass-box, G, is attached immediately to the plate, as 
 in the cut, or is raised above the telescope by means of 
 four small pillars. A hollow conical socket, H, of brass, 
 is screwed to the under side of the plate, and is intended 
 
LE^-ELLING. 397 
 
 to receive a piece of bell-metal of a corresponding form, 
 which constitutes the -upper part of the stand. This 
 piece serves as a vertical axis, upon which the telescope, 
 the spirit-level, and the compass are turned round hori- 
 zontally; sometimes, however, the conical pivot, projects 
 from the under part of the plate, EF, and the socket is 
 on the stand. 
 
 The three legs which are to support the instrument are 
 firmly fixed to a circular plate, K, perforated at its centre, 
 and having about the perforation a hollow spherical zone, 
 resembling a small inverted cup. In the simpler kinds of 
 sj^irit-levels a circular plate L, of the same dimensions as 
 the last, carries above it the pivot before mentioned ; and 
 from below it projects a stem terminating in a ball, 
 which fits the inverted cup or socket. By means of four 
 screws which pass through one of those two plates (the 
 Tipper j)late in the cut) nearly at the extremities of two 
 diameters at right angles to one another, the upper plate 
 is made parallel to the horizon, and consequently the 
 conical pivot which it carries is brought to a vertical 
 position. 
 
 The telescope should, by a proper opening of the legs 
 of the stand, be first reduced as nearly level as can be 
 estimated by the eye: then, being turned so as to lie ver- 
 tically above the line joining two opposite screws in what 
 are called parallel plates, K and L, the spirit tube is 
 brought to an horizontal position by relaxing the screw 
 nearest to its higher end, and tightening that which is 
 opposite to it. The like operation is to be performed 
 with the other pair of screws, after placing the telescope 
 vertically above them. In order to render the spirit-tube 
 parallel to the axes of the telescope, after the bubble of 
 air has been made to occupy the middle place by the pro- 
 cess just mentioned, let the telescope be reversed in the 
 arms, the Y's as they are called ; then if the bubble does 
 not still occupy the middle, it must be made to do so by 
 successive trials, endeavouring to correct half the error 
 by means of the screw b, and the other half by the 
 screw G. 
 
 The eye-piece of the telescope must be moved inwards 
 or outwards till the wires in the field of view are dis- 
 tinctly seen; and the object-glass must also be moved by 
 means of the pinion M, till the station-staff, placed at any 
 
398 LEVELLING. 
 
 convenient distance, suppose 100 yards, is also distinctly 
 seen. By a few trials, the distance between the eye and 
 the object-glass may be made such that the intersection 
 of the wires will appear to remain constantly at one point 
 on the staff, while the observer on looking through the 
 telescope varies the position of his eye. It is necessary 
 besides that the intersection of the wires should be pre- 
 cisely in the line of collimation, or the optical axes of the 
 telescope ; for this purpose the point of intersection 
 should be directed to some weU-defined mark at a con- 
 siderable distance. The telescope must then be turned on 
 its axis ; and if the intersection remains constantly on 
 the mark, the adjustment is complete, otherwise it must 
 be rendered so by means of the screws, c, d, &c., on the 
 telescope ; those screws being placed at the extremities 
 of two diameters at right angles to one another, on being 
 turned they move the plate carrying the wires in the 
 direction of those diameters. In order that the correction 
 may be made, the apparent displacement of one of the 
 wires, in consequence of the telescope being turned half 
 round on its axis, should be observed, and the screws 
 turned tiU half the displacement is corrected ; the like 
 observation and correction may then be made for the 
 other wire; a few repetitions of each adjustment will 
 probably be necessary before the error is wholly removed. 
 
 The level constructed by the late Mr. Troughton differs 
 from that which has been above described in having the 
 spirit-tube sunk partly in the telescope, and the latter, 
 being incmpable of a movement about its axis, does not 
 admit of a separate adjustment for the intersection of the 
 wires. 
 
 Mr. Gravatt, who has within a few years made con- 
 siderable improvements in the mechanism of these instru- 
 ments, recommends the follovv^ing method by which the 
 error in the positions of the cross-wires and spirit-tube 
 may be ascertained and corrected. 
 
 Let three pickets be driven into the ground in a line 
 and at equal distances from one another, and let the spirit- 
 level be set up successively in the middle between the 
 first and second, and between the second and third 
 pickets ; then having by the screws of the instrument 
 adjusted the spirit-tube so that the bubble of air may 
 retain the same place while the telescope is turned round 
 
LEVELLING. 399 
 
 on the vertical axis, direct the object-end of the telescope 
 successively to the station-staves held up on the dijQferent 
 pickets, read the several heights, and take the differences 
 between those on the first and second, and on the second 
 and third staves. Now the staves being at equal dis- 
 tances from the instrument, it is obvious that any error 
 which may have existed in the line of coUimation, or from 
 the spirit-tube not being parallel to that line, will be 
 destroyed, and the differences between the readings on the 
 staves are the differences in the levels of the heads of the 
 pickets; but unless the adjustments are perfect, this 
 will not be the case if the instrument be set up at 
 any point which is unequally distant from all the 
 pickets ; therefore from such point direct the telescope to 
 the staves, and take the differences of the readings as 
 before ; on comparing these differences with the former, a 
 want of agreement will prove that the intersection of the 
 wires is not in the optical axis, and the error may be 
 corrected by means of screws belonging to the wire plate- 
 After the agreement has been obtained, should the bubble 
 of air not stand in the middle of the tube, it may be 
 brought to that position by the screw 6, at one extremity 
 of the case, and the instrument is then completely 
 adjusted. 
 
 The spirit-level is usually provided with a clamp, N, 
 and a screw, P, by which, when the axis of the telescope 
 has by hand been brought near the object, the coincidence 
 may be accurately made by a slow and steady motion 
 about the vertical axis. 
 
 The spirit-tube or level which is employed for the 
 adjustment of transit telescopes or astronomical circles is 
 contained in a case with feet or with loops at its ex- 
 tremities, in order that it may rest above or be suspended 
 below the horizontal axis of the instrument to be levelled : 
 also the upper part of the case is furnished with a gradu- 
 ated 'scale, the divisions of which are numbered on each 
 side of a zero point, this point being usually placed near 
 each of the two extremities of the air-bubble when the 
 tube is in an horizontal position. Having set up or 
 suspended the spirit-tube, the two particular graduations 
 at which the extremities of the air-bubble rest are 
 marked ; and half the sum, or half the difference of these 
 numbers, according as the extremities of the bubble are 
 
400 LEVELLING. 
 
 in the same or in opposite directions from the two zero- 
 points, being taken, gives the distance of the centre of 
 the bubble from the middle between those points. The 
 level being then reversed, the graduations at which the 
 air-bubble rests are again marked, and half the sum or 
 half the difference is taken as before. 
 
 A mean of the two distances thus found is the true 
 distance of the centre of the bubble ^frorn the middle 
 point on the scale ; and the screw which elevates or 
 depresses one end of the axis of the telescope being then 
 turned, till either extremity of the bubble has moved, in 
 a direction contrary to that in which the centre of the 
 bubble had moved from the middle of the scale, through 
 a number of divisions equal to that mean distance, that 
 axis will be brought to an horizontal position. This 
 method is used in preference to that of successive trials, 
 in order to avoid the trouble of making several reversions 
 of the whole instrument. 
 
 The levelling staff is a rod consisting of two parts, 
 each six feet long, which, by being made to slide on 
 one another, will indicate differences of levels nearly as 
 great as twelve feet. The face of the rod is divided into 
 feet, inches and tenths, and a vane or cross-piece of 
 wood, perforated through the middle, is moved up or 
 down upon the rod by an assistant till a chamfered edge 
 at the perforation is seen by the observer at the spirit- 
 level to coincide with the horizontal wire in the telescope. 
 The height from the ground to the vane must be read by 
 the assistant. 
 
 The staves now in general use, invented by ]Mi\ Gravatt, 
 are without any vane, and are divided into feet, tenths, 
 and hundredth parts of a foot, the lines and numbers being 
 sufficiently distinct to enable the observer to note the 
 reading of the instrument, with the telescope now applied 
 to spirit-levels, at the distance of ten chains : the arrange- 
 ment of the staves is simple ; they are in three pieces, 
 with joints similar to a fishing-rod, and when put 
 together for use are 17 feet in length. Mr. Sopwith and 
 IVIr. W. P. Barlow have improved on Mr. Gravatt's 
 invention ; the divisions are nearly the same, but the 
 subdivisions are more minute ; when closed it is 5 feet in 
 length, but draws out to 15 feet, and a strong catch 
 retains each joint in its place. "" 
 
LEVELLING. 401 
 
 Anotlier staff has been invented by Mr. Peter Bruff, 
 author of a treatise on Surveying and Levelling ; this staff 
 is also from 5 to 15 feet in length, but the figures are 
 inverted, whereby, when viewed through an inverting 
 telescope, they appear in their natural order ; this staff 
 has a shoe attached to it, which the author considers of 
 considerable importance in practice. 
 
 THE THEODOLITE. 
 
 The Theodolite is an instrument used for measuring 
 horizontal angles ; it consists of a circular plate, which is 
 to be set parallel with the horizon, which is divided quite 
 round the circle into 360°. A semicircular arc for taking 
 vertical angles is also fixed to the Y's, the whole mo- 
 ving in a vertical plane ; from the centre of the circle 
 describing this arc are projecting arms resting on stan- 
 dards, which are fixed to the upper plate of the instrument 
 qh which the verniers are marked, which plate is called 
 the limb. 
 
 Bruff, in his treatise on Surveying and Levelling, says 
 he used a five-inch theodolite, which he recommends as the 
 most useful size for general purposes, made by Troughton 
 and Simms. 
 
 Captain Everett's improved theodolite is said to have 
 some advantages over those in common use ; it has three 
 verniers for taking horizontal angles, the mean angle 
 being taken, as also the two vertical angles. 
 
 Kamsden's great theodolite, the circle of which is three 
 feet in diameter, was used for a triangulation to connect 
 the observatories of Greenwich and Paris ; the principal 
 triangles of the English, Irish and Indian surveys have 
 been observed with this instrument, or with those nearly 
 identical in size and construction. 
 
 The course of observation, after the instrument is pro- 
 perly adjusted, is very simple. The question is to measure 
 tlie horizontal angles between two objects. Turn the 
 telescope two or three times round in the direction in 
 which you intend to observe, bisect one of the objects, 
 read off the verniers, take a mean, and the difference of 
 the two means is the angle required. By means of a 
 repeating table the operation is continued thus : bring the 
 telescope back on the first object by the motion of the 
 repeating table, using its clamp and tangent screw, and by 
 
402 LEVELLING. 
 
 the motion of the instrument, bring the telescope on the 
 second object ; it is evident the motion of the repeating- 
 table has nearly restored the telescope to its original 
 direction, without altering the readings of the circle ; and 
 that if the telescope be turned on the second object by its 
 motion alone, without distribViting the circle, the difference 
 between the mean of these new readings and the preced- 
 ing mean will also be the angle required : and by continu- 
 ing the process, the angle may be measured as often as 
 the observer pleases, or, in order to bisect objects correctly, 
 a slow-motion screw is attached to the upper plate, the 
 clamp-screw securing it when the object is nearly bisected, 
 and the slow-motion screw, moving the vernier through 
 the least possible space, will complete the bisection. A 
 vernier is fixed to the plate through which the arc for 
 taking vertical angles passes; two spirit-bubbles are 
 attached to this plate at right angles to each other, for the 
 purpose of setting the instrument in an horizontal posi- 
 tion. A small compass is also attached to this plate, the 
 line NS ranging with the line of sight ; the bearings at 
 the different stations being noted will serve as a check 
 on the angles, the parallel plates being mounted similar 
 to those of the spirit-level. 
 
 The method of performing adjustments in the theodo- 
 lite is similar to that of the level ; the axis of the hori- 
 zontal plate must be vertical, that is, set up the instrument 
 as level as you can, the telescope lying over the plate- 
 screws ; then, by means of the clamp and slow -motion 
 screws attached to the vertical arc, bring the bubble into 
 the centre of the tube, move the instrument half round, 
 and the telescope will be over the other plate-screws ; 
 and if the bubble remain in the centre of the tube it is 
 right, if not, it may be corrected thus, one-half by the 
 clamp and slow-motion screw attached to the vertical arc, 
 and the other half by the parallel plate -screws ; this pro- 
 cess may be repeated, if necessary, until the result is 
 satisfactory, and the bubble stands in the centre of the 
 tube. 
 
 The adjustment of the vertical arc : — when the vernier 
 stands at zero, and the former adjustments are perfect, it 
 is correct ; if not, alter the vernier by means of the screws, 
 attaching it to the plate until it does, and note the error, 
 if any ; then allow for it in each vertical angle ; and for 
 
LEVELLING. 403 
 
 the purpose of determining this error accurately, it will 
 be necessary to take the vertical angle of some conspicuous 
 object with the telescope, reversed in the Y's ; this must 
 be done by turning the instrument half round ; the tele- 
 scope will then occupy the same position as at first, but 
 the vertical arc will be reversed ; and the mean of these 
 two readings will be the error required. 
 
 To TAKE Angles with the Theodolite. 
 
 Fix the theodolite exactly over some station, which 
 can be easily done with the assistance of a plummet and 
 line being suspended under the centre from a hook at- 
 tached to the stand ; fix it level with the parallel plate- 
 screws, clamp the plates together, turn the instrument 
 towards some station or B mark ; clamp the lower plate, 
 and the bisection must be made with the slow-motion 
 screw, always observing to bisect the object as near the 
 ground as possible; observe the number of degrees, 
 minutes, and seconds, taking the mean of two or three 
 verniers, (two, however, should always be used); the upper 
 plate may now be loosed, but the lower one must remain 
 clamped and kept steady ; turn towards another station, 
 which clamp as before, then bisect with its own slow- 
 motion screw, observe the number of degrees, minutes, 
 and seconds, as before, and their difference will be the 
 angle required. But in general practice the vernier is 
 set at 360° and clamped ; the instrument is then to be 
 turned quite round, in the direction of the first object ; 
 this being done, clamp the lower plate, and bisect with its 
 slow-motion screw, release the upper plate, the lower 
 remaining clamped; then turn the instrument in the 
 direction of the second station, which clamp and bisect, 
 and you will have the angle as before. The latter method 
 of taking angles is generally adopted by surveyors, 
 although not so correct as the former, it being a difficult 
 point to set the instrument exactly to 360°, and in exten- 
 sive operations, where many angles are required to be 
 taken from one station, the former method is decidedly 
 preferable. At each principal station the bearing is 
 the angle pointed out by the compass needle, which 
 will serve as a check on the accuracy of the angle when 
 
404 
 
 LEVELLING. 
 
 the bearing can be read correctly; set the plates and 
 needle at zero, and the angle on the limb, when the object 
 is bisected, will be the bearing. 
 
 Vertical angles are taken for the purpose of reducing 
 hypothenusal measure to horizontal ; on one side of the 
 vertical arc are the degrees and minutes for determining 
 the angle, and on the other the number of links to be 
 deducted from each chain's length, to reduce it to hori- 
 zontal measure : to do this, set up a mark at the exact 
 height of the optical axis of the telescope, also at the 
 extreme point to be measured to the angle on which this 
 deduction is to be made, which can be allowed for in the 
 field ; thus suppose the angle of elevation to be 18° 15', 
 on the other side of the arc is five links to be deducted 
 from the hj-pothenusal line ; or the chain may be drawn 
 forward on the ground five links to bring it to horizontal 
 measurement. Surveyors generally find it the most ad- 
 vantageous to make the allowances in the field, especially 
 where many cross fences occur and you take a good 
 quantity of offsets ; it would be found very troublesome in 
 plotting, and to make the necessary allowance for each 
 distance ; but if great accuracy is required, the angle 
 should be noted, and the reduction made in plotting the 
 work, as it will be seen you can very well allow for even 
 links on each chain in the field, but not for the decimal 
 parts, which might be considerable on a line of great 
 extent ; hence it depends on the description of the work 
 whether the allowance is made in the field or office. 
 
 Tt is required to find the difference of level between the 
 two points A and L. Set up the level as at B, also the 
 staff" A, the spirit-level must be brought to an horizontal 
 position, and on looking through the telescope attached to 
 it, you have the dotted line of sight towards the staff" at A, 
 
LEVELLING. 405 
 
 which is graduated or divided into feet, inches, &c., which 
 can be read by the observer with the aid of the telescope. 
 Suppose the line of sight to cut the staff A at 3 feet 
 6 inches, the spirit-level being turned horizontally in the 
 direction of the staff at C, which is bisected at 7 feet ; then 
 the difference between the two points A and C will be 
 3 feet 6 inches. The spirit-level is then moved to D, the 
 staff at C still remaining; the spirit-level being in an 
 horizontal position towards the staff at C, which it bisects 
 at 1 foot 6 inches, it is then turned towards the staff at E, 
 which it bisects at 7 feet 9 inches ; the difference of level 
 between the points C and E will be 6 feet 3 inches, and 
 from A to E 9 feet 9 inches. Remove the spirit-level to 
 
 F, the staff being placed alternately at E and G ; at E it is 
 3 feet, and at G 7 feet, which is a fall of 4 feet from E to 
 
 G, making altogether the fall from A to G 13 feet 9 inches. 
 Remove the level to H, and on the staff at G is 5 feet 
 3 inches, then on the staff at I is 2 feet 3 inches, which is 
 a rise of 3 feet from G to I, making a fall from A to I 
 10 feet 9 inches; place the level at K, and on the 
 staff at I is 4 feet, and L is 2 feet 9 inches, making a rise 
 of 1 foot 3 inches ; the two rises together make 4 feet 
 6 inches, which being deducted from 13 feet 9 inches will 
 be equal to 9 feet 6 inches, the difference of the level from 
 A to L ; and in this manner the process may be continued 
 to any length, adding or subtracting as the ground rises 
 or falls. The difference of the back and fore sights, 
 between the two extreme points A and L, is the difference 
 of level. 
 
 Ft. 
 
 In. 
 
 Ft. 
 
 In. 
 
 A . . 3 
 
 6 
 
 . 7 
 
 
 
 C . . 1 
 
 6 
 
 . 7 
 
 9 
 
 E . . 3 
 
 
 
 . 7 
 
 
 
 G . . 5 
 
 3 
 
 . 2 
 
 3 
 
 I . . 4 
 
 
 
 . 2 
 
 9 
 
 ± sights 17 
 
 3 
 
 26 
 
 9 fore sights. 
 
 
 
 17 
 
 3 
 
 Diff. of level between A and L 9 6 
 
 But in order to draw the various sections so as to form 
 a plan and show the undulation of the ground between the 
 
406 
 
 LEVELLING. 
 
 points A and L, you must measure the respective distances 
 between A and C, C and E, &c. The staff being set up at 
 each change of inclination of the ground, and knowing the 
 distance from one staff to another, you may proceed to 
 plot the section, and the form of the field-book would be 
 similar to the following : — 
 
 FIELD-BOOK. 
 
 B. Sights. 
 
 F. Sights. 
 
 Red. Levels. 
 
 Distances. 
 
 Remarks. 
 
 3 6 
 1 6 
 
 3 
 5 3 
 
 4 
 
 7 
 7 9 
 7 
 2 3 
 2 9 
 
 0.00 0.00 Datum. 
 
 3.60 2.00 
 
 9.9 4.00 
 13.9 6.00 
 10.9 8.00 
 
 9.6 10.00 
 
 Difference, the same as the last. 
 
 17 3 
 
 26 9 
 17 3 
 
 9 6 
 
 In order to prove the correctness of the castings, add 
 up the back and fore-sight columns, and if their difference 
 be equal to the last reduced level, it is correct ; you may 
 then form the column of reduced levels, and proceed to 
 draw the section, taking the level of the ground at A as 
 the datum line. In plotting, the reduced level numbers 
 are always taken from a larger scale, or the vertical scale 
 to which the sections are plotted is always different to the 
 horizontal, otherwise the variations of the ground would 
 scarcely be perceptible ; the horizontal scale used here is 
 2 chains to an inch, and the vertical scale is 10 feet to an 
 inch. Having drawn the horizontal section or datum-line 
 at 2, 4, 6, 8, and 10 chains, set off the vertical sections 
 3-6, 9-9, &c., which draw perpendicular to the datum-line 
 diagram, and through the points of intersection draw the 
 ground line. 
 
LEVELLING 
 
 407 
 
 Suppose the following field-book to form part of a con- 
 tract section of a railway : — 
 
 FIELD-BOOK. 
 
 Elevation. 
 
 Back Set. 
 
 Fore Set. 
 
 Depression. 
 
 Total Rise. 
 
 Distance. 
 
 Remarks. 
 
 
 
 
 
 50.00 
 
 B. M. 
 
 
 .90 
 
 6.10 
 
 5.20 
 
 - 
 
 50.90 
 
 0.00 
 
 
 
 5.80 
 
 6.84 
 
 1.04 
 
 49.86 
 
 .50 
 
 
 
 7.45 
 
 8.80 
 
 1.35 
 
 48.51 
 
 1.40 
 
 
 1.20 
 
 8.64 
 
 7.44 
 
 - 
 
 49.71 
 
 3.00 
 
 
 
 6.40 
 
 8.10 
 
 1.70 
 
 48.01 
 
 4.00 
 
 
 
 5.82 
 
 8.22 
 
 2.40 
 
 45.61 
 
 6.00 
 
 
 
 5.24 
 
 9.00 
 
 3.76 
 
 41.85 
 
 8.00 
 
 
 
 4.44 
 
 8.10 
 
 3.66 
 
 38.19 
 
 9.00 
 
 
 
 5.12 
 
 6.82 
 
 1.70 
 
 36.49 
 
 10.00 
 
 
 
 6.18 
 
 6.84 
 
 .66 
 
 35.83 
 
 11.00 
 
 
 
 6.42 
 
 6.84 
 
 .42 
 
 35.41 
 
 12.50 
 
 
 
 4.83 
 
 6.92 
 
 2.09 
 
 33.32 
 
 14.00 
 
 
 .14 
 
 6.26 
 
 6.12 
 
 . 
 
 33.46 
 
 16.00 
 
 
 1.42 
 
 7.24 
 
 5.82 
 
 _ 
 
 34.88 
 
 18.00 
 
 
 2.07 
 
 7.11 
 
 5.04 
 
 . 
 
 36.95 
 
 19.00 
 
 
 1.44 
 
 8.16 
 
 6.72 
 
 - 
 
 38.39 
 
 20.00 
 
 
 
 6.46 
 
 7.84 
 
 1.38 
 
 37.01 
 
 22.00 
 
 
 
 4.78 
 
 5.98 
 
 1.20 
 
 35.81 
 
 23.00 
 
 
 
 2.88 
 
 5.64 
 
 2.76 
 
 35.05 
 
 25.00 
 
 
 
 2.48 
 
 5.82 
 
 3.34 
 
 29.71 
 
 26.00 
 
 
 
 4.74 
 
 5.28 
 
 .54 
 
 29.17 
 
 28.00 
 
 
 
 2.16 
 
 4.24 
 
 2.68 
 
 27.09 
 
 B. M. 
 
 
 .10 
 
 4.00 
 
 3.90 
 
 _ 
 
 27.19 
 
 31.00 
 
 
 1.43 
 
 5.87 
 
 4.44 
 
 - 
 
 28.62 
 
 32.00 
 
 
 1.04 
 
 4.08 
 
 3.04 
 
 . 
 
 29.66 
 
 34.00 
 
 
 
 1.98 
 
 4.66 
 
 2.68 
 
 26.98 
 
 36.00 
 
 
 
 1.46 
 
 4.66 
 
 3.20 
 
 23.78 
 
 38.00 
 
 
 
 2.82 
 
 4.72 
 
 1.90 
 
 21.88 
 
 40.00 
 
 
 9.74 
 
 144.92 
 
 173.04 
 
 144.92 
 
 28.12 
 
 37.86 
 
 9.74 
 
 28.12 
 
 28.12 
 
 
 
 The method o^ reducing the columns in the above field- 
 book v/ill be easily understood. In the second and third 
 columns are entered the back and fore-sights opposite 
 each other ; in the first column is the difference of the two 
 when the ground rises ; and when it falls, the difference 
 is entered in the fourth column ; the first is called eleva- 
 tion, the fourth depression, the differences being cast out 
 and entered in the respective columns, as the work pro- 
 
408 LEVELLING. 
 
 ceeds. The fiftli column contains the reduced levels, which 
 are found by adding or subtracting the diiFerences in the 
 first and fourth columns to the last reduced level, begin- 
 ning at the datum number. In the sixth column are en- 
 tered the distances between each observation, which are 
 continued from the commencement to the end of the 
 section; and the last is for general remarks, such as 
 crossing of roads, rivers, brooks, &c., or the bearings of 
 the line of levels or stations. 
 
 REMARK. 
 
 The first four columns in the field-book should always 
 be added up at each page, and if the work is correct, the 
 difference between the first and fourth columns will be 
 equal to the difference between the second and third ; and 
 in order to prove the fifth column, subtract the last re- 
 duced level from the datum number, and if correct, it will 
 be equal to the two last differences. 
 
 The method of surveying for railways differs somewhat 
 from that given in the Land Surveying, part 3 ; it being 
 obvious, that companies seldom require more than one 
 field in breadth, through which their lines pass. The 
 most prompt method of proceeding is to measure base 
 lines, beginning at station A, then B, C, D, &c., follow 
 in succession. Place them as far distant from one another 
 as can be conveniently seen, and all within the precincts 
 of the line, leaving false stations at or near every fence 
 you cross, which may be placed at either side, and in such 
 a position that the line, or side of the field, when measured, 
 will pass through it. This must be observed at the cross- 
 ing of every fence, noting the distance or chains of th« 
 base line in the field-book, at each station and fence, until 
 you arrive at station B. Return to station A ; measure 
 the sides and angles of each field, as also the necessary 
 offsets. When the angles are taken correctly by the theo- 
 dolite, or other angular instrument, the sides will be placed 
 in their true position, and you may depend upon the ac- 
 curacy of your work being correct. It will not be un- 
 necessary to observe, that the chain should be frequently 
 tested. Having completed your survey to station B, pro- 
 ceed in a similar manner with the base line to station C. 
 
THE 
 
 BUILDER'S DICTIONARY, 
 
 CONTAINING 
 
 AN EXPLANATION OF THE 3I0ST GENERAL TERMS MADE 
 USE OF IN ARCHITECTURE. 
 
 A. 
 
 Abacus. A square table, list, or plinth, in the upper part 
 of the chapiters of columns, especially those of the Co- 
 rinthian order. 
 
 Abrea^oirs. The seams or joints between stones and 
 bricks, in laying them. 
 
 Acanthus. The herb whose leaves are represented in the 
 capital of the Corinthian column. 
 
 Acroters. Sharp and spirj battlements or pinnacles, 
 that stand in ranges, with rails and balusters, upon flat 
 buildings. Also pedestals upon the corners and middle 
 of pediments, to support statues. 
 
 Alcove. A recess in a room for a bed of state ; also ai: 
 arched building in a garden or pleasure-ground. 
 
 A31PHITHEATRE. An edifice of an oval or circular form, 
 with rows of seats rising one above another, in order to 
 accommodate the people who attend in it, to see public 
 diversions. 
 
 Antichamber. The room that leads to the chief apart- 
 ment ; also a room in a nobleman's house, in which 
 strangers stay till the party to be spoken with is at 
 leisure. 
 
 Aqut:duct. a construction of timber or stone, made on 
 uneven ground to preserve the level of the water, and 
 convey it, as by a canal, from one place to another. 
 
 The Romans were extremely sumptuous and magnifi- 
 cent in their Aqueducts, some of which extended 100 
 miles. Frontinus and Blasius inform us that the city 
 of Rome was supplied with 500,000 hogsheads of water 
 every twenty-four hours, from 9 Aqueducts, which 
 t 
 
410 AKC AST 
 
 emptied themselves tlirougli 13,514 pipes of an iiicli in 
 diameter. 
 Arch. A curved line; a vault. See page 195. 
 Architect. A person wlio professes the science of Archi- 
 tecture ; and whose business it is, if required, to give 
 draughts or designs of intended buildings, with estimates 
 of the expenses necessary to complete them ; to super- 
 intend the workmen, and to give directions concerning 
 the manner and m.ethod of executing the whole. 
 Architecture. The science which teaches the method 
 . of building, being a skill obtained by the art of design- 
 ing, aided by the precepts of Geometry. By it. Rules 
 are given for planning and raising all sorts of structures, 
 according to Geometry and proportion. 
 
 The scheme or projection of a building, is easily laid 
 down in three several draughts or designs. The first 
 is a plan, which exhibits the extent, division, and distri- 
 bution of the ground into apartments, and other conve- 
 niences. The second shews the heights of the diiferent 
 stories, and the outward appearance of the whole build- 
 ing ; and is called the design or elevation. The third, 
 called the section or profile, shews the inside of each 
 apartment. 
 
 From these three draughts or designs, the Architect 
 or Undertaker makes a computation of the probable ex- 
 pense of the building, and the time required to com- 
 plete it. 
 Architrave. The moulding or ornament immediately 
 above the capital of a column ; it being always the next 
 member below the frieze. Also the chief or principal 
 beam of a building. 
 
 Doors and windows which have architraves on the 
 jambs, and over the cap-pieces, are also denominated 
 architrave-doors and architrave-windows. 
 Ashlar. Freestone, as it comes rough out of the quarry, 
 before it receives form and shape from the tool of the 
 stone-cutter. 
 AsHLERiNG. Quartering in garrets, from the floor to the 
 under side of the rafters. It is generally perpendicular 
 to the floor ; and 2\ or 3 feet in height. 
 Astragal. A little round moulding in the form of a ring, 
 serving as an ornament at the top or bottom of a column. 
 The shaft of a column always terminates at the top with 
 an astragal, and at the bottom with a fillet. 
 
ATI BE A 411 
 
 Attic. A little order, placed above another much greater ; 
 for instead of piHars, this order has nothing but pilas- 
 ters, with a cornice architraved, for an entablement. 
 This order is sometimes used in conjunction with the 
 Ionic and Doric orders. 
 
 The word is also used to denote the uppermost part 
 of a building, or the highest story. 
 
 B. 
 
 Balcony. A projection beyond the naked front of a v>'all 
 or building, supported by pillars, and encompassed with 
 a balustrade. Also a kind of open gallery. 
 
 Balk. A large piece of timber. 
 
 Balustrade. A term used to signify a row of turned 
 pillars, called balusters, made of marble, iron, wood, or 
 stone, with a hand-rail upon the top ; and fixed upon a 
 terrace, the top of a building, a balcony, a stair-case, 
 &c. &c. 
 
 Band. Any flat member that is broad, but not very deep. 
 A list, a fillet. 
 
 Barge-course. A term used for that part of the tiling 
 which projects over the gable-end of a building. 
 
 Base. A rest, a support, a foundation ; the bottom of any 
 thing. It is used to signify any body which bears up 
 another ; but it is particularly applied to the bottoms 
 of columns and pedestals. 
 
 Batten. A name given by workmen, to a piece of wooden 
 stuff, from two to four inches broad, and about one inch 
 in thickness. 
 
 Batten-door. A door having the panels nailed upon the 
 boards of which it is composed ; whereas, in a wainscot- 
 door, the panels are grooved into the framing. 
 
 Batter. A term used by workmen, to signify that a wall, 
 a piece of timber, &c. does not stand upright ; but in- 
 clines from you. If it inclines towards you, they say it 
 over-hangs or hangs-over. 
 
 Battlements. Indented notches in the top of a wall or 
 other building, made for the purpose of looking through, 
 &c. 
 
 Bead. A little round moulding, commonly made upon the 
 edge of a member, in the Corinthian and Composite or- 
 ders ; and other works of ornament. It is cut or curved 
 in short embossments, resembling beads in a necklace. 
 t2 
 
412 BEA BOU 
 
 Bea3I. a large piece of timber lying across a building, 
 into which the feet of the principal rafters are framed. 
 It is generally called the tie-beam. Also, any large 
 piece of timber. 
 
 Beaivi-fillestg. The method of filling up the vacant space 
 between the raising-plate and the roof of a building. 
 The materials with which the space is filled up. 
 
 Bearing. The distance that the ends of tie-beams, gir- 
 ders, joists, &c. are let into the walls of buildings. Also, 
 the distance between the fixed extremities of beams, 
 girders, joists, &c. is sometimes called their bearings. 
 
 Bed-moulding. All the parts of a cornice that are placed 
 under the corona. It commonly consists of these four 
 members; viz. (1.) an OG ; (2.) a list; (3.) a large 
 boultine ; and (4.) another list under the coronet. 
 
 Bevel. A kind of square used by Masons, Joiners, &c. 
 Its two legs are moveable on a point or centre ; conse- 
 quently they may be set to any angle. 
 
 Bevil-angle. An angle greater or less than a right-angle. 
 
 Binding-joists. Those joists in any floor, into which the 
 trimmers of well-holes and chimney-ways are framed. 
 Also, those joists which are framed into the girders, to 
 support the bridging-joists. 
 
 Block. A large piece of marble or stone, as it comes out 
 of the quarry, before it assumes any form from the hand 
 of the workman. 
 
 Blocking-course, or sim^lj Blocking, in Masonry, a course 
 of stones laid on the top of the cornice, crowning the 
 walls. 
 
 Blocking-courses were used by the ancients to terminate 
 the walls of a building as well as attics ; and are made 
 sometimes equal to the breadth of the pilasters. They 
 serve as an ornament, and prevent the water from fall- 
 ing to the front of the building, turning it either way 
 into lead gutters or spouts made for the purpose. 
 
 Boast. A term used among workmen, to signify the taking 
 off the superfluous parts of door-posts, window-jambs, 
 mouldings, &c. &c. 
 
 Border. Those pieces of deal wliich are put round the 
 slab of a chimney -hearth. Also, the outer part or edge 
 of any thing. 
 
 Boultine. A moulding whose convexity is just one-fourth 
 of a circle ; being the member next below the plinth, in 
 the Tuscan and Doric capitals. 
 
BRA CAV 413 
 
 Brace. A piece of timber framed in with bevel-joints, in 
 order to keep the roof of a building, &c. from swerving. 
 When braces are framed into king-posts and principal 
 rafters, they are, by some workmen, called struts. 
 
 Brackets. Pieces of wood that support shelves, the steps 
 of stairs, &c. 
 
 Brest-su:\evier. A large piece of timber, in the internal 
 part of a building, into which the girders of the naked 
 floor are framed. In the ground-floor, it is called a 
 sill ; and in a garret-floor, a beam. See summer-tree. 
 
 Brick. A square mass of clay burnt hard, for the use of 
 building. 
 
 Bridging- JOISTS. Those joists which lie upon, and are 
 supported by, the binding-joists ; and upon which the 
 boarded-floor is laid. 
 
 When those joists are supported by the walls of the 
 building, or framed into the girders, without any bind- 
 ing-joists, they are generally called common-joists. 
 
 BuTiiiENT. That which supports the foot of an arch. A 
 shoulder. 
 
 Buttress. A pier made against the outside of a wall or 
 building, to support and strengthen it. 
 
 Camber-beam. A piece of timber cut in an arching direc- 
 tion, or with an obtuse angle in the middle. 
 
 Capital. The upper part of a column. Such as have no 
 ornaments, are the Tuscan and Doric, which are caUed 
 capitals with mouldings, and the rest wliich have leaves 
 with other ornaments, are denominated capitals with 
 sculptures. Also, the head or top of any pillar or 
 pilaster. 
 
 Cap or cap-piece. The highest or uppermost part of any 
 thing. 
 
 Cartouses. a kind of modillions. 
 
 Caseiment. a hollow moulding, which some Ai'chitects 
 make one-sixth of a circle, and others one-fourth. 
 Also, that part of a window which opens. 
 
 Casings. The boards put round door-ways, window-open- 
 ings, &c. 
 
 Cavetto. a concave moulding, which has quite a con- 
 trary effect to the quarter-round ; the workmen call it 
 T 3 
 
414 
 
 CAU CEM 
 
 mouth when in its natural position, and throat when 
 turned upside down. 
 Caultcoll. The carved scrolls, under the abacus, in the 
 
 Corinthian capital. 
 Ceiling. The lathing and plastering at the top of a room, 
 
 upon the under-side of the joists of the next room. 
 Celling- JOISTS. The joists to which the ceiling-laths are 
 
 nailed, when there is no chamber above. 
 Cement. A strong, glutinous, binding composition. 
 
 A good cement for uniting bricks together, for carv- 
 ing, scrolls, capitals, &c. &c. may be made of resin, bees- 
 wax, brick-dust, and chalk boiled together. The bricks 
 must be rubbed even ; then heated, and rubbed together, 
 with the cement between them. 
 
 A strong and useful cement for joining marble, and 
 all other hard stones, may be made as foUows : Melt two 
 pounds of bees-wax, and one pound of resin together ; 
 then add one pound and a half of the same kind of 
 matter, burnt and well pulverized, as the body to be 
 cemented is composed of ; let the whole mass be kneaded 
 with water, and applied warm, to the parts of the body 
 to be cemented, which must also be heated. 
 
 An exceedingly strong cement that will become as 
 hard as stone, and last for ages, may be made in the fol- 
 lowing manner : take lime, well slacked, and sand in 
 equal proportions ; and temper them with linseed oil to 
 the consistency of mortar : then beat it well in a trough 
 or upon a floor, and it will be fit for use. 
 
 When an old stone or brick wall is to be covered with 
 this cement or plaster, let the face be chipped a little 
 with a bricklayer's hammer, or a mallet and chisel; 
 then drench it with linseed oil and white lead, until it 
 will drink no more. 
 
 Dr. Higgins had a patent in 1779, now expired, for 
 his invention of a water cement, or stucco, as follows : 
 fifty-six pounds of pure coarse sand, forty-two pounds 
 of pure fine sand, mix them together, and moisten them 
 thoroughly with lime-water; to the wetted sand add 
 fourteen pounds of pure fresh burnt lime, and whiLe 
 beating them up together, fourteen pounds of bone-ash ; 
 the quicker and more perfectly these materials are beaten 
 together, and the sooner used, the better will the ce- 
 ment be. Fine sand alone, or coarse sand alone, will 
 
ciiA cu 415 
 
 do for some works ; but the finer the sand, the more 
 lime must be used. 
 
 Cankers of a forge well beaten with lime and sand, 
 make a good cement for furnaces. 
 
 Chambers. The rooms between the ground story and the 
 garrets. 
 
 Channel. That part of the Ionic capital which is under 
 the abacus, and lies open upon the echinus or eggs ; and 
 has its centres or turnings on every side, to form the 
 volutes. Also, the ornamental part of the Doric tri- 
 glyphs ; a hollow bed of running waters. 
 
 Chapiter. The top or head of a pillar or column 
 
 Chainlet. A small ornament carved into round pearls or 
 beads. 
 
 Cheek. Any piece of metal or timber which answers to 
 another of the same sort, opposite to it, in any utensil, 
 machine, or building. 
 
 Cheek-boards. The boards which form the sides of door- 
 cases, window-cases, &c. 
 
 Chemney. The passage through which the smoke ascends 
 from the fire. 
 
 Chemney -JAMBS. The sides of a chimney, on the extremi- 
 ties of which the mantle-tree rests. 
 
 Chemney-piece. An ornamental piece of wood or stone, 
 standing on the foreside of the jambs, and projecting 
 over the mantle-tree. 
 
 Cima-recta. a wave, called by the English workmen 
 ogee or OG, which is of two kinds ; viz. cima-recta or 
 OG, and cima reversa, or the back OG, whose beauty 
 consists in having its height and projection equal to 
 each other. 
 
 CmcLE. A round figure comprehended under one curved 
 line, called the circumference. The diameter is a right 
 line passing through the centre, and terminating in the 
 circumference on each side. 
 
 Cistern. A receptacle for water. 
 
 Cla3ip. a kiln built above the ground, for the purpose of 
 burning bricks. Also, pieces on the ends of tables, 
 window-shutters, &c. 
 
 Cleansed-work. Any stone-work that is polished or 
 made smooth, by scouring or rubbing it. It is fre- 
 quently called polished-work. 
 
 Clinkers. Those bricks which, by the violence of the fire, 
 are run and glazed over. 
 
 T 4 
 
4W GDI BEl. 
 
 Coins or quoins. The corners of brick or stone walk 
 Also, the stones in the angles of buildings, whether 
 plain, rustic, or otherwise. 
 
 CoLLAR-BEAJM. A beam framed crosswise between tv/o 
 principal rafters. 
 
 Column. A round pillar, composed of a base, a shaft, and a 
 capital ; and serves to support the entablement. Also, 
 any round pillar. 
 
 Conduit. A canal of pipes for the conveyance of water. 
 
 Console. A projecting ornament cut upon the key of an 
 arch, or other member ; and which occasionally serves 
 to support little cornices, busts, vases, &c. 
 
 Coping. The top or covering of a wall made sloping to 
 carry off the water. 
 
 Corbel. An ornament representing a basket ; a short 
 piece of timber or stone jutting out of a wall for a sup- 
 porter. 
 
 Corbel-holes. Arches in the walls to receive images or 
 statues. 
 
 Cornice. The third and uppermost member of the en- 
 tablement, which is different in the several orders. 
 The word cornice is likewise applied to every promi- 
 nent or jutting member that crowns any body. Cor- 
 nices are also placed on the top of the wainscot, under 
 the eaves of houses, &c. 
 
 Corona or crown. Any thing that finishes an ornament. 
 A large flat member which crowns, not only the cor- 
 nice, but the entablement. The larmier. 
 
 Coves or covings. Thin stones placed against the jambs 
 and back of a chimney. They are of two kinds ; viz. 
 side-coves, and back-coves. 
 
 Coussinet. The stone which crowns a pier, or that lies 
 immediately over the capital of the impost and under 
 the sweep. The bed of it is level below, and curved 
 above, receiving the first rise or spring of the arch or 
 vault. 
 
 Cupola. See dome. 
 
 D. 
 
 Dado or die. The plain parts of the pedestal of a column, 
 between the base and the cornice. 
 
 Denticles or dentils. Ornaments in a cornice, resem- 
 bling teeth. They are frequently made in the Ionic and 
 Corinthian orders. 
 
DOn ENT 417 
 
 Bo^iE or Cupola. A spherical roof, resembling tlie bell 
 of a clock, raised over the middle of a building, as a 
 church, hall, pavilion, vestibule, stair-case, &c. by way 
 of crowning. Domes are generally supported by columns, 
 both for the sake of ornament and strength. 
 Sometimes domes are of a polygonal form. 
 
 Door-case. The frame in which a door is inclosed. 
 
 DoRitfAN-TREE. See summer-tree. 
 
 DoRiMANT-AViNDOW. A window made in the roof of a 
 house ; being raised upon the rafters. 
 
 Dove-tail. A kind of joint, made by joiners, resembling 
 the tail of a dove or pigeon. 
 
 Doucine. See cima-recta. 
 
 Drag. A door is said to drag, when in opening or shut- 
 ting, it stops or grates upon the floor. 
 
 Draught or Design. The figure of an intended building 
 described upon paper ; in which are laid down, by a 
 scale, in due proportion to the whole building, the several 
 devised apartments, with doors, windows, passages, and 
 other conveniences. 
 
 Dress. To prepare stones, timber, &c. for any purpose. 
 
 Drops. Ornaments in the Doric entablature, representing 
 drops or little bells, immediately under the triglyphs. 
 
 E. 
 
 Eaves. The margin of the roof of a building, which pro- 
 jects beyond the wall, to throw off the water. 
 
 Echinus. The member or ornament near the bottom of 
 the Ionic, Corinthian, and Composite capitals. It is, 
 by some workmen, called eggs and anchors; and by 
 others, quarter-round. 
 
 Embossing. A kind of sculpture or engraving, in which 
 the figures project from the plane upon which they are 
 engraved. It is called bass-relief, mean-relief, or high- 
 relief according as the figures are more or less pro- 
 tuberant. 
 
 Enriched-mouldings. Cornices or mouldings ornamented 
 with modillions, beads, denticles, leaves, and other de- 
 sirable additions. 
 
 Entablement or Entablature. The architrave, the 
 frieze, and the cornice taken together. 
 
 Entry. A passage leading from the outer door into the 
 house. Any door, gate, or passage. 
 
418 EYE FLU 
 
 Eye. The middle of an Ionic volute or scroll, cut in the 
 form of a rose. A round window made in a pediment, 
 or at the top of a dome, &c. 
 
 Fabric, A church, a house, or any other building. 
 
 Face. Any member that has a considerable breadth, and 
 but a small projection ; as the architrave in the front 
 of a building, &c. The front of any thing. 
 
 Facia. Any flat member that is not very broad. It is, 
 by some workmen, called a string, and by others a 
 riband. 
 
 Feather-edged. Thinner on one edge than the other. 
 
 Fell. To hew down ; to cut down. 
 
 Festoon. An ornament of carved work, in the form of 
 a wreath, or garland of flowers, fruits, and leaves in- 
 termixed or twisted together ; being thickest in the 
 middle, and suspended by the two extremities, whence 
 it hangs down with a graceful sweep. 
 
 These ornaments were formerly much used on tri- 
 umphal arches, &c. 
 
 Fillet. A little square ornament or moulding, which 
 accompanies or crowns a larger. A list, a band. 
 
 Fire-place. That part of a room in which the fire is 
 made. 
 
 Fire-stone. Stone that will bear the fire well, without 
 being soon consumed. 
 
 There are also fire-bricks, which have the same pix)- 
 perty, and are much used about ovens, stoves, &c. 
 
 Floating. A term used by Plasterers, for their best 
 ceilings. 
 
 Floor. The bottom part of a room on which we walk. 
 Floors are of different kinds ; some are of earth, some 
 of brick, some of stone, and some of wood. 
 
 The framed timbers of a floor, viz. the summers, 
 girders, and joists, are, by Carpenters and Joiners, de- 
 nominated naked flooring, in order to distinguish it 
 from the boarded-flooring. 
 
 Flush. A term used by Carpenters and Joiners, when, 
 the work is even or smooth. It is also used by Masons, 
 to signify the breaking off" of any part of a stone. 
 
 Flutes or Flutings. Hollows or channels made in the 
 shaft of a column, from the base to the capital. The 
 
FLY GIR 419 
 
 Doric, Ionic, Corinthian, and Composite columns va'q 
 
 commonly fluted ; but the Tuscan column scarcely ever. 
 Flyeks. Stairs which do not wind, but go straight ; and 
 
 whose steps are all in the form of rectangles. 
 Foliage. Ornaments wrought in the form of branches 
 
 and leaves. 
 FooT-rACE or h.nxf-pace. A part of a pair of stairs, 
 
 whereon you arrive, after having ascended four, six, or 
 
 more steps. Here two or three paces may generally be 
 
 taken before you ascend another step, by which means 
 
 your legs are eased. A kind of landing. 
 Foundation. The lowest part of a building, upon which 
 
 the walls of the superstructure are raised. It is gene- 
 rally laid under ground. 
 Fra3ie. Any thing constructed *o as to inclose or admit 
 
 something else. 
 Frieze or Frize. A large flat member, which separates 
 
 the architrave from the cornice. 
 Fret-work. Plain bordering around ceilings, &c. It is 
 
 made with divers fillets or bands, which form a great 
 
 variety of figures by their turnings. 
 Front. The principal face or side of a building ; or that 
 
 which presents the chief aspect or view. The face or 
 
 forepart of any thing. 
 Frontispiece. The decorations on the forepart of any 
 
 thing. 
 Funnel. The narrow part of a chimney. 
 Furring. A piece of timber put upon a joist or rafter, 
 
 to strengthen or repair it. 
 Fust. The shaft, trunk, or body of a column, extending 
 
 from the base to the capital. 
 
 G. 
 
 Gable-end. The upright triangular end of a house, from 
 
 the eaves to the top of the roof. 
 Gallery. A kind of immovable scaffold, in the inside 
 
 of a room, accommodated with seats ; as in a place of 
 
 worship, a theatre, &c. 
 Garland. An ornament in imitation of branches, leaves, 
 
 fruits, and flowers, twisted and involved into each other. 
 
 It is sometimes called a wreath. 
 Girders. Large pieces of. timber into which the ends of 
 
 the joists of a naked floor are framed. The ends of the 
 T 6 
 
420 GOT IIAU 
 
 girders are sometimes framed into the summers and 
 brest-summers : and sometimes they are supported by 
 the walls of the building. 
 
 GoTHic-ARCHiTECTUKE. That species of architecture 
 which is far removed from the manner and proportions 
 of the antique; having its ornaments ^vild and chi- 
 merical, and its profiles generally very incorrect. It is 
 frequently very heavy, solid, and massive ; but some- 
 times, on the contrary, exceedingly light, delicate, an^ 
 rich. Almost all the ancient cathedrals are built after 
 the gothic manner. 
 
 Gothic-roof. A roof formed by two circular arches, 
 struck from different centres, and meeting in a point 
 over the middle of the span. 
 
 Granaby. a store-house for corn. 
 
 Grange. A farm, with all the appendages ; as a dwell- 
 ing-house, barns, stables, cow-houses, granaries, and 
 other necessary places for husbandry, at a distance from 
 neighbours. 
 
 Grate. The range of bars within which fires are made. 
 
 Groin. A kind of arch or roof formed by the intersection 
 of vaults. 
 
 Groove. A term used by Joiners, to signify the channel 
 made by their plough, in the edge of a moulding, or 
 stile, to receive the panels, in wainscotting. 
 
 Grotesque. Distorted, unnatural; formed without due 
 proportion. 
 
 Grotto. A cave, a 'cavern, made for coolness and plea- 
 sure. 
 
 Ground-floor. The lower story of the house. 
 
 Ground-plot. The ground upon which a building is to 
 be erected. 
 
 Gutter. A channel for the conveyance of water. 
 
 Guttering. The boarding and bearers upon which the 
 lead is laid that forms the gutters, in roofing. 
 
 H. 
 
 Hall. A manor-house. Also, the first large room at the 
 
 entrance of a house. 
 Hand-rail. See balustrade. 
 Haunches or Hances. The ends or bases of elliptical 
 
 arches. The word is frequently used to denote the 
 
 ends of any arch. 
 
HEA KIN 4"^- 
 
 He^vd. The top or upper part of any thing. 
 
 He.aj)ERS. a term used among Bricklayers, wlien bricks 
 are laid endwise in a wall; but when they are laid 
 lengthwise, they call them stretchers. 
 
 Hearth. The floor or pavement of a chimney, or lire- 
 place. 
 
 Heel. An inverted Ogee. A talon. 
 
 Hew. To cut with an edged instrument. 
 
 Hinges. Those appendages, by which doors, the lids of 
 boxes, &c. make their motion. 
 
 Hips. In Carpentry, are those pieces of timbers which are 
 placed at the corners of a roof. They are much longer 
 than the rafters, by reason of their oblique position. 
 
 Hollow. A concave moulding, by some called a case- 
 ment. Any excavation. 
 
 House. A habitation ; a place of human abode. 
 
 Housing. A term used by Bricklayers, when a tile or 
 brick is cast crooked or hollow, or has become so by 
 burning ; they then say, such a tile or brick is " hous- 
 ing." 
 
 J. 
 
 Jambs. The posts of a door. Also the posts or piers on 
 the sides of window^s, fire-places, &c. 
 
 larposTS. The capitals of pilasters, that support arches. 
 That part of a pier that supports the arch. 
 
 Intercoluiviniation. The space between two columns. 
 
 Joint. The place where any two pieces of matter form a 
 junction. 
 
 Joists. Those pieces of timber in a naked-floor, that 
 are framed into the girders and summers. They re- 
 ceive various names, according to the positions in which 
 they are laid ; as trimming-joists, binding-joists, bridg- 
 ing joists, and ceiling-joists. 
 
 K 
 
 Kerf. The aperture or slit made in a piece of timber or 
 
 board, by a saw. 
 Key-stone. The principal stone in the crown or middle 
 
 of an arch. 
 King-post or king-piece. A piece of timber placed in 
 
 a perpendicular direction, upon the middle of the tie 
 
 beam, and extending to the top of the principal rafters. 
 
122 KIT MAE 
 
 in order to support them. The rafters, in general, are 
 
 also supported by braces or struts, proceeding from the 
 
 king-post. 
 Kitchen. A room in a house where the provisions are 
 
 cooked, &c. 
 Knee-piece. A piece of timber cut crooked, or with an 
 
 angle. 
 
 Landing. The top of a stair-case. 
 
 Larmier. The flat square member that forms the utmost 
 projection of a cornice. The corona. 
 
 Laths. Long, thin, narrow slips of wood used in tiling 
 and ceiling. 
 
 Lead. A soft heavy metal, used for roofs, gutters, pipes, 
 cisterns, sinks, &c. 
 
 Leaves. Well-known ornaments with which ceilings, 
 cornices, &c. are enriched. 
 
 Ledgers. Pieces of timber used by Masons and Brick- 
 layers, in making scaffolds to work upon. The ledgers 
 are laid parallel to the wall of the building ; and made 
 fast, by ropes, to the scaffold-poles. 
 
 Lime. Chalk or stone burnt, of which mortar, cement, 
 &c. is made. 
 
 Lining-boards. See casings. 
 
 Lintels. Pieces of timber that lie horizontally over the 
 top of the door-posts, window -jambs, &c. to bear the 
 weight of the walls. The upper part or head of a door- 
 frame is sometimes called the lintel. 
 
 List. A little square moulding, serving to crown or ac- 
 company a larger one. The space which separates the 
 flutings of a column. It is sometimes called a fillet, 
 and sometimes a band. 
 
 Lobby. An antichamber. 
 
 M. 
 
 Mallet. A wooden hammer used by Joiners, Masons, 
 
 Stonecutters, &c. 
 Mantle or blantle-tree. The head-piece which rests 
 
 upon the tops of chimney -jambs. 
 ]\L^BLE. A fine, hard stone, of various colours, used in 
 
 statues and elegant buildings. 8ec page 181. 
 
MEM NAV 423 
 
 Member. Any part of an integral. 
 
 ]Metope or 3IETOPA. The space between every triglypli 
 of the Doric frieze. 
 
 Model. An artificial pattern made of Avood, stone, clay, 
 or other matter, with all its parts and proportions, in 
 order to give an idea of the etfect that will be produced, 
 when the work is executed on a larger scale. 
 
 MoDiLLioxs. Small consoles or brackets, set as orna- 
 ments under Ionic, Corinthian, and Composite cornices. 
 
 Module. A kind of a measure used to regulate the pro- 
 portions of the several members of columns. 
 
 Monument. A building erected to preserve the memory 
 of some illustrious personage, remarkable event, or great 
 achievement. 
 
 MoKTAR. A cement or composition formed of lime, sand, 
 and water ; and used by Masons and Bricklayers, in 
 building. 
 
 Mr. John Phillips says, that equal parts of Ifme and 
 sand, make exceedingly good mortar ; especially if sharp 
 river sand be used ; as fine sand makes much weaker 
 mortar than coarse. (See Crosby's Builder s Price 
 Book, page 45.) 
 
 Mortise -HOLE. A hole cut into a piece of timber, that 
 another piece, called the tenon, may be put into it, and 
 form a joint. 
 
 Mosaic -work. Curious kind of work, consisting of small 
 inlaid pebbles, cockles, and shells of various colours. 
 
 Mouldings. Under this name are comprehended all those 
 juttings or projections beyond the naked of walls, co- 
 lumns, &c. which serve only for ornaments ; whether 
 they be square, round, straight, crooked, concave, or 
 convex. There are seven kinds of mouldings more 
 considerable than the rest ; viz. the doucine or cima- 
 recta ; the talon or heel; the ovolo or quarter-round; 
 the plinth ; the astragal ; the denticle ; and the cavetto. 
 
 MuNNiONS. The upright posts that divide the lights in a 
 window-frame. 
 
 N. 
 
 Naked of a wall. The surface or plain from which 
 the mouldings and other projections arise. 
 
 Nave. The middle part or body of a church, distinct from 
 the aisles or wings. 
 
424 NEW CRD 
 
 Newel-post. The upright post in a winding stair-case, 
 
 round which the steps are set. 
 Niches. Hollows or cavities in walls, in which statues 
 
 are sometimes placed. 
 Nosings. The edges or front parts of steps ; part of the 
 
 sides of marble chimney-pieces ; also, pieces put on at 
 
 the ends or front of any thing. 
 
 0. 
 
 Obelisk. A quadrangular pyramid, generally erected 
 in some public place, to serve as a monument of some 
 memorable action, or event, &c. 
 
 Ogee or o g. A sort of moulding, consisting of a round 
 and a hollow, something resembling an S. 
 
 Openings. The breaks or apertures that are left in the 
 walls of buildings, for the doors, windows, &c. 
 
 Orders. The different forms and proportions of columns, 
 &c. There are commonly reckoned five orders ; viz. 
 the Tuscan, the Doric, the Ionic, the Corinthian, and 
 the Composite. 
 
 The Tuscan is the most simple of the five orders ; 
 having neither triglyphs, dentils, nor modillions. It is 
 sometimes called the Rustic Order ; being very strong 
 and massive. The height of the Tuscan column is seven 
 diameters ; and the height of the entablement one-fourth 
 of the column's height. 
 
 The Doric order seems the most natural, and best 
 proportioned of all tiie orders. The frieze is adorned 
 with triglyphs, drops, and metopes ; and the height of 
 the column is eight diameters ; and that of the en- 
 tablement one-fourth of the column's height. 
 
 The Ionic, or third order, is a kind of mean between 
 the strong and delicate orders. The height of the co- 
 lumn is nine diameters ; and that of the entablement 
 one-fifth of the column's height. The capital is orna- 
 mented with volutes; and the cornice with denticles. 
 The shaft of this column is generally fluted. 
 
 The Corinthian order is the noblest, richest, and 
 most delicate of all the orders. The capital of this 
 column is adorned with two rows of acanthus leaves ; 
 and the entablement with modillions, and sometimes 
 with both modillions and dentils. 
 
OKL PED 425 
 
 The height of the column is ten diameters ; and that 
 of the entablement one-fifth of the column's height. 
 
 The Composiie or fifth order, is so called, because it 
 is formed out of the other orders. It borrows a quar- 
 ter-round from the Tuscan and Doric ; a row of leaves 
 from the Corinthian ; and volutes from the Ionic. Its 
 cornice has simple modillions or dentils. The height 
 of the column and entablement is the same as that of 
 the Corinthian order. 
 
 The Doric, Ionic, and Corinthian orders were in- 
 vented bj the Greeks ; and the Tuscan and Composite 
 by the Romans. 
 
 Those who wish to make themselves acquainted with 
 the method of drawing and working the five orders, are 
 referred to Nicholson's Student^s Instructor. 
 
 Orlo. a fillet under the ovolo, or quarter-round. 
 
 Ornaments. All the sculpture or carved work with which 
 a piece of architecture is enriched. 
 
 Ovolo. A round moulding, whose profile or sweep is 
 about one-fourth of a circle. It is commonly enriched 
 with figures resembling eggs and anchors, or arrow- 
 heads placed alternately. 
 
 Panel. A square or rectangular piece of wainscotting, 
 sometimes framed or grooved in between thicker pieces ; 
 as in doors, window shutters, &c. Also the face of a 
 hewn stone. 
 
 Pantry. A room in which victuals are kept. 
 
 Parapet. A wall breast high, serving either as a rest for 
 the arm, or an inclosure about a key, bridge, terrace, &c. 
 
 Parlour. A lower room designed for the entertainment of 
 company. 
 
 Partition. That which divides one apartment from an- 
 other. 
 
 Passage. An entry or narrow place, serving for a tho- 
 roughfare into other rooms, &c. 
 
 Pavement. Stones or bricks laid upon the ground, in 
 order to make the passage easy. 
 
 Pedestal. The lower member of a column, or that upon 
 which it stands. The base of a statue. 
 
 Pedbient. An ornament or crowning, which finishes 
 the front of a building ; and serves as a decoration over 
 
426 PER POR 
 
 doors, windows, niches, &c. It is generally of a trian- 
 gular shape : but sometimes forms the arch of a circle. 
 
 The ridges of houses first gave Architects the idea of 
 
 this noble part. 
 Perquettixg. a floor composed of divers small figures, 
 
 as triangles, squares, rhombuses, &c. 
 Perron. A flight of steps leading to the front door of a 
 
 great house, when the floor of the lower story is raised 
 
 a little above the ground. 
 Persian Order. That which has figures of Persian slaves, 
 
 instead of columns, to support the entablement. 
 Pla^zza. a portico. 
 Pier. The column which supports the arch of a bridge. 
 
 The space between two windows in a room. 
 Pilaster. A square column, sometimes insulated, but 
 
 oftener set within a wall ; and only shewing a fourth or 
 
 a fifth part of its thickness. Pilasters frequently have 
 
 bases and capitals the same as columns. 
 Pile. A strong piece of wood di-iven into the ground, to 
 
 make a good foundation. 
 Pillar. A round, square, or pentagonal column. 
 Pinnacle. A turret or elevation above the rest of the 
 
 building. A high spiring top. 
 Pitch. The elevation of a roof. See page 183,, note 4. 
 Plaster. A composition made of lime, sand, hair, and 
 
 water, with which walls, ceiling, &c. are overlaid. 
 Plastic. The art of forming figures or ornaments in 
 
 plaster. 
 Plat-band. Any flat, square moulding, having less pro- 
 jection than height. 
 Platford. The ceiling of a chamber, or other room ; the 
 
 sofiit. 
 Platform. The ichnography or ground-plot of a house. 
 
 The roof or covering of a building that is flat at the 
 
 top. Any horizontal plane. 
 Plinth. A flat square member under the mouldings of 
 
 the bases of columns, pedestals, &c. A square projec- 
 tion near the bottom of a wall. 
 Pole. A long slender piece of timber. 
 Polished-work. Any stone-work that is made smooth, 
 
 by scouring or rubbing it. In some places it is called 
 
 cleansed-ivork. 
 Porch. A roof supported with pillars before a door. An 
 
 entrance ; a portico ; a covered walk. 
 
roK PUT 427 
 
 PoRTAiLS. The decorations of a front-gate ; or in the 
 front of a church, &c. 
 
 Portal. A gate ; an entrance ; the arch under which a 
 gate or door opens. 
 
 PoETico. A kind of gallery, raised upon arches, where 
 people walk under shelter. It has sometimes a soffit or 
 ceiling, but is more commonly vaulted. Also, a roof, 
 gallery, or other projection before a door, in the front 
 of a hall, church, &c. supported by columns detached 
 from the building. 
 
 Post. A piece of timber placed in a perpendicular direc- 
 tion. 
 
 Prickpost. a post placed between two principal posts. 
 
 Principal. Capital; essential; chief; of the first rate. 
 
 Profile. The figure or draught of a building, &c. in 
 which are expressed the several heights, breadths, and 
 thicknesses, such as they would appear if the building 
 were cut perpendicularly from the top of the roof to 
 the foundation ; hence it is frequently called the section. 
 This word is used to denote the outline of any mem- 
 ber of a building ; as that of the base, a cornice, &c. It 
 also denotes the prospect of a building, or any piece of 
 architecture, viewed sideways. 
 
 Project. All mouldings, &c. are said to project, when 
 they jut out, or are beyond the adjacent surface. 
 
 Projecture. The prominency or embossment which 
 mouldings and other members have beyond a column, 
 the naked of a wall, &c. 
 
 Proportion. The justness of the members in each part of 
 a building, and the relation they bear to the whole. 
 
 Pcl^tnata. a frieze swelling or bulging out like a pillow. . 
 
 PuNCHiN. A short piece of timber so placed as to bear 
 some considerable weight. 
 
 PuRLiNGS. Those pieces of timber that lie in a horizontal 
 position ; and support the inferior rafters. They are 
 sometimes called sida-pieces or side-wavers. 
 
 Putlogs. Pieces of timber used by Masons and Brick- 
 layers in making scaffolds. They are laid at right- 
 angles to the wall of the building, with their ends in 
 the scaffold-holes and upon the ledgers, by which means 
 they are supported. 
 
428 QUA EUS 
 
 0- 
 
 Quadrant. The fourth part of a circle. 
 
 Quarters. Those slight upright pieces of timber, placed 
 between the punchins and posts, against which the 
 weather-boarding and laths are nailed, in making par- 
 titions. 
 
 Quarter-round. Any moulding whose contour is one- 
 fourth of a circle, or nearly so. 
 
 Quirk. A piece taken off the corner of a ground-plot, to 
 make a court-yard. 
 
 Quor>?s. See Coins. 
 
 R. 
 
 Rafters. Those pieces of timber which extend from the 
 ridge to the eaves of a roof. The feet of the principal 
 rafters are always framed into the tie-beam. 
 
 Rails. The horizontal pieces of timber between the panels 
 of wainscotting. A number of cross beams supported 
 by upright posts. 
 
 Range. A grate or fire-place. Also, when work runs 
 straight, without breaking into angles, it is said to 
 range. ^ 
 
 Regula. See orlo. 
 
 Return. The continuation of a moulding, &c. after having 
 turned a corner. The side that reclines from the front 
 of any straight work. 
 
 Ridge. The top of a roof, where the ends of the rafters 
 meet each other. 
 
 Ridge-stones. The stones which cover the top of a roof. 
 
 Riser. A board placed edge-ways to support the upper 
 part of a step called the tread. The perpendicular face 
 of a step. 
 
 Rising-pieces. Pieces of timber laid under the ends of tie- 
 beams, girders, &c. 
 
 Roof. The covering of a house; but particularly the 
 timber-work. 
 
 Rose. An ornament cut in imitation of a rose. It is used 
 in friezes, abacuses, cornices, the ceiHng of churches, 
 &c. and sometimes it is made between the modillions, 
 in the Corinthian and Composite orders. 
 
 Rustic. A manner of building quite rude, rather in imi- 
 tation of nature, than according to the rules of art. 
 
RUS SLA 429 
 
 Rustic-work. Any stone-work, the face of which is 
 hacked or picked into holes, with the point of a hammer, 
 &c. 
 
 S. 
 
 Saloon. A lofty, spacious hall, the side-Avalls of which are 
 
 connected to a flat roof- or ceiling, in the middle, by 
 
 arches on each side. See page 195. 
 Sand. Small particles of stone not conjoined. Sand is 
 
 one of the principal ingredients in making mortar. 
 Scaffold. A temporary gallery or stage, raised by the 
 
 side of a building, for the Avorkmen to stand upon. 
 Scaffold-holes. Holes left in the sides of a building, 
 
 for the end of the putlogs. 
 Scantling. The breadth and thickness of a piece of 
 
 timber. 
 Scotia. A concave member, whose contour is the arc of 
 
 a semicircle, or nearly so. It is commonly placed in 
 
 the base of a column, between the torus and the as- 
 tragal. 
 Scribe. To fit one irregular piece of stuff to another, by 
 
 marking them with compasses, &c. Also the mark 
 
 made upon a piece of timber with any pointed instru 
 
 ment. 
 Scribe-iron. An instrument with which timber is 
 
 marked. 
 Scroll. See volute. 
 Sculpture. The art of carving wood, stone, &c. into 
 
 images and other ornaments. 
 Section. See 'profile. 
 
 Semicircle. A half-round ; half of a circle. 
 Sewer. A passage for water to run through. 
 Sill. The bottom part of a window-frame. 
 Shaft. The trunk or body of a column. See fust. 
 Shaky or sh^njken. Such stuff as is cracked by the heat of 
 
 the sun, &c. is called shaky or shaken stuff. 
 Shingles. Thin pieces of timber used to cover houses, 
 
 instead of tiles or slates. 
 Shoulder. The hutments of a tenon. A prominence ; a 
 
 support. 
 Skirting-boards. The narrow boards fitted round the 
 
 under-side of wainscotting, next the floor. 
 Slabs. The sappy planks cut off the outsides of timber. 
 
 Also the foot-pace to a fire-place is called a slab. 
 
130 SLA STU 
 
 Slates. Thin stones with which houses are covered. 
 They are chiefly of two kinds, blue and grey. 
 
 Slating. The covering of a house with slates. 
 
 Sleeper. The oblique rafter that lies in the gutter of a 
 roof. 
 
 SLL^CE. A water-gate ; a flood-gate ; a vent for water. 
 
 Soffit. In great edifices, it signifies the ceiling or wains- 
 cot of any apartment, formed by cross beams or flying 
 cornices, and having the square panels of its compart- 
 ments enriched with sculpture, painting, gilding, &c. In 
 common buildings, it denotes the board at the top of a 
 window-opening, opposite to the window-board. 
 
 Sole. The stone upon which window -jambs are placed. 
 The bottom part of any thing. 
 
 Span. The chord line of an arch, or the direct distance 
 between its extremities. It is particularly applied to 
 bridges and other buildings with arched roofs, to denote 
 the distance that the feet of the arches are from each 
 other. 
 
 Spar. A small beam of timber. The inferior rafters are 
 sometimes called spars. 
 
 Stairs or stair-case. The passage by which we ascend 
 and descend from one story of a house to another. It 
 generally consists of steps, landing-places, and a balus- 
 trade. 
 
 Statues. Embossed figures or images, either of wood, 
 stone, or metal, representing some person distinguished 
 by his birth, merits, or achievements, &c. They serve 
 to perpetuate the memory of the person they represent. 
 
 Stiles, The upright pieces that extend from the bottom 
 to the top in wainscot. A set of steps or rails to pass 
 from one inclosure to another, is also called a stile. 
 
 Stone. A hard substance generated in the earth ; and 
 much used in building. See page 181. 
 
 Stove. A hot-house, for preserving exotic plants. A 
 close place in which fire is made. 
 
 Strike. To form by a sweep, — as "to strike a circle.* 
 To put in motion ; to stamp ; to impress. 
 
 Strings. Pieces of timber which support the steps of 
 wooden stairs. 
 
 Strut. A piece of timber framed into the king-post and 
 principal rafters. See brace. 
 
 Stuff. A general term for aU kinds of timber upon which 
 Joiners work. 
 
SUM 
 
 TOO 431 
 
 SusDiER-TREE. A large piece of timber, in the internal 
 part of a building, having a row of mortise-holes on 
 each side, into which the ends of the girders are framed. 
 Brest-summers are laid adjoining the walls of a 
 building, and consequently have mortise-holes only on 
 one side. See Brest-siimmer. 
 
 Sweep. The compass of any continued motion. The 
 direction of motion in compassing any thing. 
 
 Systyle. That manner of placing columns, so that the 
 space between them consists of two diameters. 
 
 I 
 
 Tabling. Flat stones laid upon the gable-ends of a house, 
 
 from the eaves to the ridge, to keep the slates fast. 
 Talon. A moulding consisting of square fillets, which 
 
 sometimes terminates the ornaments of Joiners' Work. 
 
 A heel. 
 Tassels. Pieces of timber that lie under the ends of 
 
 mantle-trees, beams, &c. 
 Tenon. That part of a piece of timber which is put into a 
 
 mortise-hole. 
 Terrace. An open walk or gallery ; a flat roof of a.. 
 
 house ; also a kind of strong mortar, chiefly used in- 
 making basins, cisterns, wells, and other reservoirs for 
 
 water. 
 Thorough-lighted. Having windows at both sides, or ■ 
 
 both ends. 
 Threshold. The stone or step under a door, upon which' 
 
 the door-posts are placed. The bottom part of a door-- 
 
 frame. 
 TiE-BEASL The beam into which the feet of the principal 
 
 rafters are framed. 
 Tiles. Thin plates of burnt clay, used in covering houses* 
 Timber. All kinds of wood used by Carpenters, Joiners, 
 
 Turners, &c. &c. For a description of Timber Trees, 
 
 see page 166. 
 Tooled-work. Implies the manner of finishing ; hence 
 
 any stone struck with parallel strokes, by either a broad 
 
 or narrow chisel, is so called. 
 
 The door-posts, window-jambs, chimney -jambs, &c. 
 
 of common buildings, are always tooled ; but those of 
 
 elegant buildings are most commonly polished. In 
 
 buildings of mediocrity, the door-posts, window -jambs. 
 
432 TOP VOL 
 
 &c. are frequently tooled ; and the chimney-jambs, 
 coves, slabs, &c. polished. See polished-work. 
 
 Top-rail. The upper part of a window-frame. 
 
 Torus. A large round moulding in the base of a column, 
 between the plinth and the list. 
 
 Transom. A piece that is framed across a double light 
 window. 
 
 Traverse. To plane a board, &c. across the grain. 
 
 Tread. The upper or horizontal part of a step. 
 
 Triglyph. a member of the Doric frieze, placed between 
 the metopes. 
 
 Tri3I. When workmen fit a piece of timber, stoni^, &c. 
 into some other work, they say, they trim in a piece. 
 
 Trimmers. Pieces of timber framed into the binding- 
 joists, against well -holes, and chimney- ways. 
 
 Trunk. The shaft or a fust of a column. The body of 
 a tree. 
 
 Truss. That part of a roof which supports the purlins 
 and inferior rafters. It generally consists of the tie- 
 beam, principal rafters, king -posts, braces, and punchins. 
 See page 187. 
 
 Tusk. A bevil shoulder, made to strengthen the tenon of 
 a joist, where it is let into the girder. 
 
 V. 
 
 Valleys. The gutters over the sleepers, in the roof, of 
 
 a building. 
 Vault. A cellar ; a cave ; a cavern ; a continued arch. 
 
 See page 187. 
 Vestibule. A large, open space before the door, or at 
 
 the entry of a house. A porch. 
 Volute. A kind of scroll, wreath, or spiral contortion, 
 
 used in the Ionic and Composite capitals, of which it is 
 
 the chief ornament. 
 
 It is supposed, by some, to represent the bark of trees, 
 
 twisted into spiral bines ; and by others, the head- 
 dresses of virgins in their long hair. 
 
 The volute is also used in the Corinthian capital ; and 
 
 asit is an ornament that contributes very greatly to the 
 
 beauty of columns, Architects have invented various 
 
 ways of delineating it. 
 
 Consoles, modillions, and other ornaments, have like • 
 
 wise their volutes. 
 
VllN WRE 433 
 
 \jRS, or Vase. A vessel with a mouth narrower than tne 
 body ; and serving as an ornament or crowning over 
 bahistrades, chimney- piece?, columns, pyramids, funeral 
 monuments, tlie tops of buildings, 8cc. 
 
 It is also considered as an attribute to rivers, foun- 
 tains, cascades, jets, &c. 
 
 W. 
 
 Waixscott, or Wainscotting. The inner wooden cover- 
 ing of a wall. It commonly consists of panels, stiles, 
 rails, and a cornice. 
 
 Watek-board. a board nailed over the bottom of an 
 outer-door, to carry off the water. 
 
 "Wateh-spout. a trough placed immediately under the 
 eaves of a house, to receive the water that falls upon 
 the roof. It always has a communication with the 
 water-trunk, or conductor. 
 
 "Watek-tadle. a kind of ledge left in a stone or brick 
 wall, about 1 8 or 20 inches from the ground, at which 
 place the wall is decreased in thickness. 
 
 Water-truxk. a pipe by which water is conveyed from 
 the top or eaves of a building. It is sometimes called a 
 conductor. 
 
 WExVTnER-BOAEDLNG. The boards nailed against the 
 quarters, in making timber partitions. 
 
 Well-hole. The space left in a floor for a stair-case. 
 
 AYreatii. An ornament resembling branches, leaves, 
 fruits, and flowers, interwoven and entwined one into 
 anotiicr. It is frequently called a garland. 
 
 u 
 
434 qi:t:stions for exa^hnation. tart l 
 
 QUESTIONS 
 
 THE EXAMINATION OF THE PUPIL, 
 
 TO BE ANSWERED VERBALLY. 
 
 The figures refer to the pages which contain the answers ; and 
 fliagrams must be drawn by the pupil, when the questions cannot be 
 properly explained without them. 
 
 PART I. 
 
 GEO:iIETRICAL DEFIXITIONS. 
 
 Page 
 
 IYhat is Geometry ? — "^Yliat is a point, — a line ? — 
 How many kinds of lines are there ? — What is a 
 right or straight line ? — What is a curved line ? - 1 
 
 What are parallel lines? — What is a superficies? — 
 Can a superficies be contained within one line? — 
 What is a solid? — What is an angle? - - 1 
 
 How many kinds of angles are there? — What is a 
 rio-ht angle? — What is an acute angle? — What is 
 the complement of an angle? — What is an obtuse 
 angle? — "What is the supplement of an angle? — 
 What is a triangle ? - - - ," ^ 
 
 "Wliat is an equilateral triangle, —an isosceles tri- 
 angle, — a scalene triangle, — a right-angled tri- 
 angle, — an acute-angled triangle ? - - 3 
 
 What is an obtuse-angled triangle? — Which is the 
 base, and which the perpendicular, of any figure? 
 
 What is a figure of four sides denominated? — 
 
 What is a paraUelogram, — a square, — a rectangle ? 
 
 — Wliat is a rhombus ? - - ' . " ^ 
 
 What is a rhomboid or rhomboides, — a trapezium, 
 — a diagonal, — a trapezoid ? - - - 4 
 
 By what names are plane figures of more than four 
 
 sides generally called ? — Name the polygons ? - 4 
 What is a regular polygon, — an irregular polygon? 5 
 
TART I. QUESTIONS FOR EXAMINATION. 435 
 
 Pag3 
 
 What is a circle, — the diameter, — the radius, and 
 the arc of a circle ? - - - - 5 
 
 What is a chord, — a segment, — a semicircle, — a 
 quadrant, — a sector ? - -■ - - 5 
 
 How is the circumference of every circle supposed to 
 be divided ? - - - - - 5 
 
 How many degrees does the arc of a quadrant con- 
 tain? - - - - - -5 
 
 Wliat is an ellipse ? - - - - 6 
 
 Wliat are the names of the longest and shortest dia- 
 meters that can be drawn within an ellipse ? - 6 
 
 PART I. 
 
 GEOMETRICAL PROBLEMS. 
 
 Draw a right line, and shew the manner of bisecting it 7 
 Make an angle, and shew the manner of bisecting it 7 
 How do you draw one line parallel to another, so as 
 
 to pass through a given point ? - - - 7 
 
 Shew the method of drawing one line parallel to 
 
 another, at a given distance - - - 7 
 
 How do you erect a perpendicular, from a given 
 
 point, near the middle of a given line ? - - 8 
 
 How do you proceed when the point is near the end 
 
 of the given line ? - - - - 8 
 
 How do you let fall a perpendicular, from a point 
 
 nearly opposite to the middle of a given line ? - 8 
 How do you proceed when the point is nearly oppo- 
 site to the end of the line ? - - - 9 
 How do you find the centre of a given circle ? - 9 
 Shew the method of constructing a triangle, when the 
 
 three sides are given - - - - 9 
 
 How do you construct a triangle, when the base and 
 
 perpendicular are given ? - - - 10 
 
 Shew the manner of describing a square, when the 
 
 side is given - - - - - 10 
 
 How do you make a rectangle, when the length and 
 
 breadth are given ? - - - - 11 
 
 How do you make a regular rhombus, when its side 
 
 is given ? - - - - - 12 
 
 How do you construct an irregular rhombus, when 
 
 its side and perpendicular height are given ? - 12 
 
 u 2 
 
4:36 QUESTIONS FOR EXAMINATION. rART I. 
 
 Page 
 
 Shew tlie method of constructing a regular rhomboid, 
 
 when its sides are given - - - - 12 
 
 How do you construct an irregular rhomboid, when 
 
 its base and perpendicular are given ? - - 12 
 
 Shew the method of constructing an ellipse, when the 
 
 transverse and conjugate diameters are given - 12 
 How do you make a regular pentagon, upon a given 
 
 line? - - - - - - 13 
 
 Shew the manner of describing a regular hexagon, 
 
 upon a given line - - - - 1 3 
 
 How do you construct a regular octagon, upon a 
 
 given line ? - - - - - 14 
 
 Shew the method of inscribing a circle in a given 
 
 triangle - - - - - -14 
 
 How do you circumscribe a circle about a given 
 
 triangle? - - - - - 14 
 
 How do you make a triangle equal to a given 
 
 trapezium ? - - - - -15 
 
 Shew the method of making a right angle by the line 
 
 of chords on the plane scale - - - lo 
 
 Describe the method of making an acute angle - 15 
 
 How do you make an obtuse angle ? - ♦•15 
 
 How do you find the number of degrees contained in 
 
 any angle ? - - - --16 
 
 How do you make a regular polygon of any number 
 
 of sides, upon a given line ? - - - 16 
 
 Shew the method of inscribing a polygon of any 
 
 number of sides in a given circle - - - 16 
 
 How do you find a mean proportional between two 
 
 given lines ?- - - - -17 
 
 How do you raise a perpendicular by a scale of equal 
 
 parts? - - - - - - 17 
 
 If the chord line and versed sine of any arch be 
 
 given, how do you find the radius that Afili strike 
 
 the arch ? - - - - - 17 
 
 PART I. 
 
 GEOMETRICAL THEOREMS. 
 
 Draw two right lines to intersect each other ; and 
 shew which of the angles, made by the intersection, 
 are equal - - - - - 18 
 
PART II. QUESTIONS FOK EXAMINATION. 437 
 
 Page 
 Wliat side of a ti'iangle is opposite to the greatest 
 
 angle ?- - - - - -18 
 
 When a right line intersects two parallel right lines, 
 
 which of- the angles made by the intersection are 
 
 equal? - - - - - - 19 
 
 Draw a diagram, and repeat tlie fourth theorem - 19 
 
 Draw a diagram, and repeat the fifth theorem - * 19 
 
 Draw a figure, and repeat the sixth theorem - 20 
 
 Wliat is expressed in the seventh theorem ? - 20 
 
 'S^Hiat does the eighth theorem express ? - - 20 
 
 Repeat the ninth theorem - - - - 20 
 
 Draw a diagram, and repeat the tenth theorem - 21 
 
 What is expressed in the eleventh theorem ? - 21 
 
 Draw a diagram, and repeat the twelfth theorem - 21 
 
 Draw a figure, and repeat the thirteenth theorem - 22 
 "What property is contained in the fourteenth 
 
 theorem? - - - - - 22 
 
 Draw a diagram, and repeat the fifteenth theorem - 22 
 
 Repeat the sixteenth theorem - - - 22 
 What are the properties contained in the seventeenth 
 
 theorem? - - - - - 23 
 
 Repeat the eighteenth theorem - . - - 23 
 What properties are contained in the nineteenth 
 
 theorem? - - - - - 23 
 
 Repeat the twentieth theorem - - - 23 
 
 PART n. 
 
 MENSUEATION OF SUPERFICIES. 
 
 What is the area of a figure ? - - - 25 
 
 In what are the dimensions of figures taken ; and 
 
 how are their areas estimated ? - - - 25 
 
 Repeat the table of lineal measure - - - 25 
 
 Repeat the table of square measure - - - 26 
 
 How do you find the area of a square ? - -26 
 
 When the area of a square is given, how do you find 
 
 its side ?- - - - - ■■26 
 
 How do you find the area of a rectangle ? - - 27 
 
 When the area of a rectangle and one of its sides are 
 
 given, how do you find the other side ? - - 28 
 
 How do you determine the area of a rhombus or 
 
 rhomboides ? - - - - - 29 
 
 u 3 
 
438 QUESTIONS FOR EXAMINATION. PART 11. 
 
 Page 
 
 If the area of a rliombus or rliomboides, and one of 
 its dimensions be given, how do you find the other 
 dimension ? - - - - - 29 
 
 When the base and perpendicular of a triangle are 
 
 given, how do you find the area ? - - 30 
 
 If the area of a triangle and one of its dimensions be 
 
 given, how do you find the other dimension ? - 30 
 
 How do you find the area of a triangle, when the 
 
 three sides only are given ? - - - 31 
 
 Repeat the rule for performing multiplication by 
 logarithms - - - - - 32 
 
 How do you perform division by logarithms ? - 32 
 
 How is the rule of three performed by logarithms ? - 32 
 Repeat the rule for involution, or raising powers, by 
 logarithms? - - - - - 32 
 
 How is evolution, or the extraction of roots, per- 
 formed by logarithms ? - - - - 32 
 
 When the two legs of a right-angled triangle are 
 
 given, how do you find the hypothenuse ? - 33 
 
 "When the hypothenuse and one of the legs are given, 
 
 how do you find the other leg ? - - - 33 
 
 How do you find the area of a trapezium ? - - 35 
 
 How may the area of a trapezium be found, when it 
 
 can be inscribed in a circle ? - - - 35 
 
 How do you find the area of a trapezoid ? - - 36 
 
 How do you find the area of an irregular polygon ? - 37 
 
 If the side of a regular polygon be given, how do you 
 
 find the radius of its inscribed or circumscribing 
 
 circle ?- - - - - -39 
 
 How do you find the area of a regular polygon ? - 40 
 How do you find the area of a regular polygon, when 
 the side only is given ? - - - - 41 
 
 Repeat the first rule for finding the circumference of 
 a circle from the diameter, or the diameter from 
 the circumference. — Repeat the second rule. — 
 Repeat the third rule - - - - 42 
 
 What is the first rule for finding the length of any 
 arc of a circle ? - - - - - 44 
 
 Repeat the second rule - - - - 44 
 
 Repeat the third rule - - - - 46 
 
 How do you find the area of a circle ? - - 47 
 
 How do you find the area of a sector or circle ? - 48 
 
rAKT 111. QUESTIONS FOR KXAMINATION. 439 
 
 Page 
 
 Repeat the first rule for finding the area of tlie seg- 
 ment of a circle - - - - - 50 
 
 Repeat the second rule. — Repeat the third rule - 51 
 
 Explain the rule for finding the area of a circular zone 52 
 
 How do you find tlie area of a circular ring ? " - 53 
 
 How do you find the area of a lune ? - - 54 
 
 How do you find the area of an ellipse ? - - 55 
 
 How do you find the area of an elliptical ring ? - 55 
 Repeat the rule for finding the area of an elliptical 
 segment - - - - - -56 
 
 Explain the rule for finding the area of any curvi- 
 
 lineal figure, by means of equi-distant ordinates - 57 
 
 PART HI. 
 
 L AND-SURVEYrNG . 
 
 With what is land commonly measured ? - - 63 
 
 Wliat length is a chain, and a link ? - - 63 
 
 AYhat is an acre of land ? - - - - 63 
 
 In what is the measurement of land generally given? 63 
 
 "Wliat is a statute pole or perch ? - - - 63 
 
 '\\niich is the best method of keeping the field-book ? 63 
 
 How do you measure a field of three sides ? - - 65 
 
 How do you measure a field of four sides ? - - 70 
 How must you proceed, when neither of the diagonals 
 
 can be measured, in consequence of obstructions ? 70 
 How Avould you measure a trapezium, when two per- 
 pendiculars cannot be taken upon either of the dia- 
 gonals ? - - - - - - 70 
 
 How do unskilful surveyors sometimes measure tra- 
 peziums ? - - - - - 70 
 
 In what manner do you proceed in order to find the 
 area of a field whose sides are crooked, by the me- 
 thods called " castings"? - - - - 74 
 
 How do you measure a field of more than four sides ? 76 
 How do you measure an irregular narrow piece of 
 
 land? - - - - - - 80 
 
 How should paring and reaping be measured ? - 80 
 
 How do you measure a mere or wood ? - - 82 
 
 How do you measure roads, rivers, or canals ? - 83 
 Explain the method of finding tlie breadth of a river, 
 
 without crossing it - - - - 86 
 
440 QUESTIONS FOR EXAMINATION. P.\JRT lY. 
 
 How do you measure tlie distance of an inaccessible 
 object? - - - - - - 86 
 
 Explain the method of measuring a line upon which 
 there is an impediment - - - - 87 
 
 How do you erect a perpendicular in the field, by the 
 chain only ?- - - - -87 
 
 How do you part any proposed quantity of land from 
 a rectangle, by a line parallel to one of its sides ? 87 
 
 Explain the method of parting land from a trapezium, 
 or any irregular polygon - - - 87 
 
 How do you divide a trapezium, or any irregular 
 polygon, among any number of persons ? - 88 
 
 Repeat the rules for reducing statute-measure to cus- 
 tomary ; and vice versa - - - - 92 
 
 Explain the first method of surveying estates or lord- 
 ships - - - - - - 94 
 
 Explain the second method - - - 99 
 
 Explain the third method - - - - 100 
 
 Explain the fourth method - - - - 100 
 
 PAUT IV. 
 
 Section I. 
 
 SIENSUKATION OF SOLIDS. 
 DEFINITIONS. 
 
 What is a solid ? - - - - - 121 
 
 What is the measurement of a solid called ? - - 121 
 
 By what are the contents of solids estimated ? - 121 
 
 ^Yhat is a cube, — a parallelopipedon .^ - - 121 
 
 Wliat is a prism ? ----- 121 
 Wliat different names have prisms ? - - 121 
 
 What is a cylinder, — a pyramid ? - - - 122 
 
 What is a cone, — the frustum of a pyramid or cone, 
 
 — a wedge, — a prismoid, — a sphere or globe ? - 122 
 What is the segment of a sphere, — the zone of a 
 
 -sphere, — a circular spindle, — a cylindi-ical ring, 
 
 — a regular body ? - - - - 123 
 
 How many regular bodies are there ; and what are 
 
 their names ?- - - - -124 
 
 PROBLEMS. 
 
 How do you find the solidity oi a cube, — its sur- 
 fare? - - - - - - 124 
 
PART IV. QUESTIONS FOR EXA:\IIXATI0X. 441 
 
 rage 
 
 Wlien the solidity of u cube is given, liow do you find 
 its side ? - - - - - - 124 
 
 How do you find the solidiff/ of a paralielopipedon, 
 
 — iti surface? - - - - _ 126 
 How do you find the soUdift/ of a prism, — its surface? 127 
 By what rule do you determine the solidity of' a cy- 
 linder, — its surface? - _ . . i29 
 
 What is the rule for finding the solidity of a pyramid, 
 
 — lis surface? ----- 130 
 Repeat, the general rule for finding the solidity of the 
 
 frustum of a pyramid - - - - 131 
 
 Repeat the second rule - - - - 181 
 
 AVhat is the rule for finding the surface of the frus- 
 tum of a pyramid ? - - - -131 
 
 Repeat the rule for finding the solidity of a cone, — 
 its surface - - - - - 132 
 
 Repeat the first rule for finding the solidity of the 
 frustum of a cone - - - - 133 
 
 Repeat the second rule - - _ . 133 
 
 How do you find the surface of the frustum of a cone? 134 
 
 How may the solidity of a cuneus or wedge be found, 
 — its surface ? - - - - - 135 
 
 "What is the rule for finding the solidity of a prismoid, 
 — its surface? - - - _ _ 137 
 
 Repeat the rule for finding the solidity of a sphere 
 or globe - . - - . - . 133 
 
 Wliat is the rule for finding the convex surfiice of a 
 globe? - - - - - - 138 
 
 Repeat the first rule for finding the solidity of the 
 segment of a globe - - - - 13S 
 
 Repeat the second rule - - - - 139 
 
 How do you find the surface of the segment of a 
 sphere ? - - - - - - 139 
 
 What is the rule for finding the solidity of the frus- 
 tum or zone of a sphere ? - _ . 139 
 
 How do you find the convex surface of a zone of a 
 sphere? ----_. 140 
 
 Explain the rule for finding the solidity of a circular 
 spindle, — its surface? - - - - 140 
 
 How do you find the solidity of the middle frustum 
 of a circular spindle ? - - - - 141 
 
 What is the rule for finding the solidity of a cylin- 
 drical ring, — its surface? - - - 142 
 
442 QUESTIONS FOR EXAIVHNATION. PAET IV. 
 
 Page 
 What is the rule for finding the solidity or superficies 
 
 of any regular body ? - - - - 143 
 Repeat the first rule for finding the solidity of an 
 
 irregular solid - - - - 145 
 
 Kepeat the second rule _ - _ - 145 
 
 Repeat the third rule - - - - 145 
 
 Repeat the fourth rule - - - - 145 
 How do you find the solidity of a body from its 
 
 weight ? _ . _ - - 147 
 How do you find the weight of a body from its 
 
 solidity? - - - - - 148 
 
 PART TV. 
 
 Section 1L 
 
 the use of the sliding rule. 
 
 How do you perform multiplication by the sUding 
 rule, — Division? _ ^ - _ 150 
 
 Hqw do you square any number by the sliding rule, — 
 extract the square root, — find a mean proportional 
 between two numbers ? - - - - 151 
 
 How do you perform the rule of three by the sliding 
 rule? - 152 
 
 How do you find the area of a rectangle or rliom- 
 boides by the sliding rule, — of a triangle, — and of 
 a trapezium ?- - - - -153 
 
 What is the rule for finding the area of a regular 
 polygon, by the sliding rule? - - - 153 
 
 How do you find the diameter and circumference of 
 a circle, the one from the other ? - - - 153 
 
 How do you find the area of a cii'cle ? - - 153 
 
 What is the rule for finding the solidity of a cube by 
 the sliding rille, — of a parallelopipedon, — of a 
 prism or cylinder, — of a cone or pyramid? - 153 
 
 PART TV. 
 
 Section HI. 
 
 TIMBER MEASURE. . 
 
 How do you find the superficial content of a board 
 or plank? - . . - 154 
 
PART IV. QUESTIONS FOR EXAmNATION. 44o 
 
 Page 
 
 How do you find the solidity of squared or four-sided 
 
 timber? - - - - - -loo 
 
 Repeat the first rule for finding the solidity of round 
 
 or unsquared timber - - - - lo6 
 
 'Wliat is the proportion by the sliding rule ? - 157 
 
 Repeat the second rule for round timber - - 157 
 
 What is the proportion by the sliding rule ? - 157 
 
 Is the first rule correct for round timber ? - - 157 
 Does the second rule give the true content of round 
 
 timber? - - - - - - 157 
 
 Describe the method of measuring and valuing stand- 
 
 insr timber ----- 160 
 
 If more questions be thought necessary, they may be easily supplfcd 
 from the Table of Contents, by the Tutor. 
 
 J^ote. — The reader is reftvred for further ir.formatien to Ncsbit'a 
 Land Surveying, v.-here the subject is fully treated upon. 
 
# 
 
 447 
 
 APPENDIX, 
 
 £ 
 
 5. 
 
 d. 
 
 
 
 
 
 
 
 1 
 
 2 
 
 01 
 
 3 
 
 6 
 
 1^ 
 
 Having amply treated on the subject of gauging as 
 practised by the Inland Revenue olhcers, and in order to 
 render the work more complete and valuable, we have 
 given a list of the various Duties to be charged by the 
 Officers of the Excise in the United Kingdom. 
 
 EXCISE LICENSES. 
 
 All excise licenses must be taken out annually, and 
 pay for such licenses, as follows : — 
 
 Auctioneers. — Every auctioneer - 
 
 Beer. — Every person keeping a common 
 
 inn or alehouse; whose premises are rated 
 
 under 20/. - 
 
 If at 20/. or upwards - - - 
 
 2^ole Licenses to sell beer not to be drunk on the premises, £1 Is. O^d. ; to be 
 
 drunk on the premises, £3 Os. \%d. 
 
 beer for sale, not ex- 
 
 d. £ s. d, 
 
 6 - 7,000 barrels - 11 16 3 
 - 10,000 " - 15 15 
 6 - 20,000 « - 31 10 
 - 30,000 " - 47 5 
 - 40,000 " - 63 
 6 and 40,000 upwards 78 15 
 Brewers using sugar - - - 1 
 Every other brewer, or common brewer - 5 10 3 
 Sellers of beer only, not being brewers - 3 6 IJ 
 Sellers, but not to be drunk on the pre- 
 mises - - - - -12 01 
 Coffee, tea, cocoa-nuts, chocolate and pepper Oil 6^ 
 
 Hackney carriages. — Annual license in 
 
 London - - - - - 1 
 
 Every 
 
 brewer 
 
 of 
 
 ceeding 
 
 
 
 
 
 
 £ 
 
 s. 
 
 20 barrels ■ 
 
 ' 
 
 10 
 
 50 
 
 c 
 
 ■ 1 
 
 1 
 
 100 
 
 c 
 
 - 1 
 
 11 
 
 1,000 
 
 ( 
 
 - 2 
 
 2 
 
 2,000 
 
 (( 
 
 - 3 
 
 3 
 
 5,000 
 
 ( 
 
 - 7 
 
 17 
 
448 
 
 APrENDIX. 
 
 Malt — Every maltster to take a license 
 annually - _ _ . 
 
 If the quantity made exceed 50 quarters 
 and not 100 - - . . 
 
 Not exceeding 150 qrs. - . - 
 
 200 " - 
 
 £ s. d. 
 - 7 101 
 
 a 
 
 250 
 
 a 
 
 300 
 
 
 350 
 
 
 400 
 
 
 450 
 
 
 500 
 
 
 550 
 
 Exceeding 
 
 550 
 
 Malt roasters _ - . . 
 
 Dealers in roasted malt 
 
 Paper. — Every paper maker 
 
 Passage vessels. — Every master or com- 
 mander to retail wine, beer, spirituous liquors, 
 and tobacco - - - - - 
 
 Postmasters to let horses for hire - 
 
 Persons keeping 1 horse or 1 carriage 
 
 Not above 2 horses or 2 carriages 
 
 ct 4 
 
 11 
 
 3 
 
 " 8 
 
 a 
 
 6 
 
 " 12 
 
 a 
 
 9 
 
 " 16 
 
 a 
 
 12 
 
 ■" 20 
 
 u 
 
 15 
 
 Above 20 
 
 li 
 
 15 
 
 Above 20 horses, then for every additional 
 10 horses, and for any additional number less 
 than 10, over and above 20, or any other 
 multiple of 10 horses, the further sum 
 
 Race horse duty - - - - 
 
 Spirits. — Distillers . - - 
 
 Rectifiers - - - 
 
 Dealers not retailers - - - 
 
 Dealers for retailing foreign liquors 
 Makers of methylated spirits 
 Makers of stills in Scotland and Ireland - 
 Chemist or any other person requiring 
 the use of a still in England 
 
 15 
 
 9 
 
 1 3 
 
 71 
 
 1 11 
 
 6" 
 
 1 19 
 
 ^ 
 
 2 7 
 
 3 
 
 2 15 
 
 u 
 
 3 3 
 
 0" 
 
 3 10 
 
 101 
 
 3 18 
 
 9 
 
 4 6 
 
 n 
 
 4 14 
 
 6 
 
 20 
 
 
 
 10 
 
 
 
 4 4 
 
 
 
 1 1 
 
 7 
 
 10 
 
 
 
 12 
 
 10 
 
 
 
 20 
 
 
 
 
 
 30 
 
 
 
 
 
 40 
 
 
 
 
 
 50 
 
 
 
 
 
 60 
 
 
 
 
 
 70 
 
 
 
 
 
 10 
 
 
 
 3 17 
 
 
 
 10 10 
 
 
 
 10 10 
 
 
 
 10 10 
 
 
 
 2 2 
 
 
 
 10 10 
 
 
 
 10 
 
 6 
 
 10 
 
APrENDIX. 419 
 
 In Scotland and Ireland - - - 
 
 Retailers of spirits^ whose premises arc 
 rated under 10/. in England and Ireland 
 At 10/. and under 20/. 
 " 20/. " 25/. - 
 
 " 25/. " 30/. - 
 
 " 30/. " 40/. - 
 
 " 40/. " 50/. - 
 
 " 50/. or upwards _ _ . 
 
 Retailers of spirits a?idf?€er, whose premises 
 are rated under 10/. per annum in Scotland - 
 At 10/. and under 20/, 
 " 20/. " 25/. - 
 
 " 25/. " SO/. - 
 
 " 30/. « 40/. - - 
 
 " 40/. " 50/. - 
 
 " 50/. or upwards - - - 
 
 Retailers of spirits, in Ireland, being duly 
 licensed to sell coffee, tea, &c., whose premises 
 under 25/. per annum _ - _ 
 
 At 25/. and under 30/. 
 " 30/.- " 40/. - 
 
 " 40/. " 50/. - 
 
 " 50/. and upwards . - _ 
 
 Siveets. — Retailers in the United Kingdom 
 Tobacco and snuffs manufacturers of, not 
 above 20,000 lbs. - - - . 
 
 Above 20,000 lbs. and not above 40,000 lbs. 
 " 40,000 " 60,000 " 
 
 " 60,000 " 80,000 " 
 
 " 80,000 " 100,000 " 
 
 Above 100,000 " 
 Beginners, and a surcharge on the quantity 
 made - 
 
 Dealers in tobacco and snuff 
 Vinegar makers - - - - 
 
 V/ine — Every dealer in foreign wine, who 
 shall not have a license for retailing spirits 
 and beer - - - _ _ 
 
 Plaving a license for beer, and not spirits 
 Having a license for beer and spirits 
 
 A'o^d?.— Licenses taken out by brewers, distillers, publicans, and retailers of beer, 
 ■wine, spirits, or sweets, expire oa the 10th October, annually; all otiiers on the 
 5th of July. 
 
 Beginners in any part of ihe year, viz., three quarters, or less period, to Lu 
 charged a proportionate part of the annual license. 
 
 £ s. 
 
 d. 
 
 10 
 
 () 
 
 2 4 
 
 1 
 
 4 8 
 
 2i 
 
 3i 
 
 6 12 
 
 7 14 
 
 4 
 
 8 16 
 
 4f 
 
 9 18 
 
 5i 
 
 11 
 
 6 
 
 4 4 
 
 
 
 5 5 
 
 
 
 9 9 
 
 
 
 10 10 
 
 
 
 11 11 
 
 
 
 12 12 
 
 
 
 13 13 
 
 
 
 9 18 
 
 5| 
 
 11 
 
 6 
 
 12 2 
 
 6i 
 
 13 4 
 
 7 
 
 14 6 
 
 n 
 
 1 2 
 
 Oi 
 
 5 5 
 
 
 
 10 10 
 
 
 
 15 15 
 
 
 
 21 
 
 
 
 26 5 
 
 
 
 31 10 
 
 
 
 5 5 
 
 
 
 5 
 
 3 
 
 5 5 
 
 
 
 10 10 
 
 
 
 4 8 
 
 21 
 
 2 4 
 
 r 
 
450 APPENDIX. 
 
 EXCISE DUTIES. 
 
 £ s. d. 
 
 Hops. — Per lb. (and 5 per cent, addi- 
 tional) - - - - - 2 
 
 3Ialf. — Per bushel (and 5 per cent, ad- 
 ditional) - - - - - 2 7 
 
 Beer or bigg in Scotland and Ireland, and 
 consumed there (and 5 per cent, additional) - 2 
 
 Hackney carriages. — Duty per week in 
 IjOndon - - _ - _ 
 
 If worked every day . - - 
 
 Paper. — Duty on. all kinds of paper per 
 lb. (and 5 per cent, additional) 
 
 Spirits. — All spirits made in England, or 
 Scotland, per proof gallon - - - 
 
 Spirits made in Ireland - - - 
 
 Ditto, imported from the Channel Islands 
 into Great Britain, per proof gallon - 
 
 Ditto, into Ireland . - _ 
 
 
 
 6 
 
 
 
 
 
 7 
 
 
 
 
 
 
 
 li 
 
 
 
 8 
 
 
 
 
 
 8 
 
 
 
 
 
 9 
 
 
 
 
 
 9 
 
 
 
 THE END. 
 
 LONDON 
 
 PRINTED BY SPOTTISWOODE AND CO, 
 
 WliW-STKEEX SQUAEB. 
 
m 
 

 
 •X 
 
 541154 
 
 UNIVERSITY OF CALIFORNIA LIBRARY