:5 
 
 OC-H««rF,, 
 
 
 Vol. 93. 
 
 (VIATHEiyiATIG AL POR TION OF THE SERIES. 
 
 A KUDIMENTAKY TREATISE 
 
 ON 
 
 MENSUIIATION, 
 
 FOR 
 
 7n.y .xj^-R ov SCHOOLS 
 
 AND 
 
 FOR PRACTICAL MEN. 
 
 BY T. BAKET, C.E. 
 
 Price Is. 
 
 LONDON : JOHN WEALE. 
 
LIBRARY 
 
 OF THE 
 
 University of California 
 
 ...^.CXjGaJ^ Jkju^...^^ 
 
 Class 
 
RUDIMENTARY TREATISE 
 
 MENSUEATION, 
 
 FOR THE USE OF SCHOOLS AND PRACTICAL MEN: 
 
 COMPEEHENDINQ THE ELEMENTS OF 
 
 MODERN ENGINEERING. 
 
 ®^e JFiftfj ©lition:, tottfj ftWittons anJ Corrtctions. 
 
 By T. baker, C.E., 
 
 Author of "Theodolite Surveying, Levelling-, &c.," in the Ninth Edition of Nesbit'a 
 
 SiUTeying ; ' * Railway Engineering ; " " Laud and Engineering Surveying ; " 
 
 •'Statics and Dynamics, Elements of Mechanism ; " " Integration 
 
 of Differentials ; " &c. &c. 
 
 LONDON : 
 JOHN WEALE, 59, HIGH HOLBOEN. 
 
 1859. 
 

 LONDON : 
 BRADBUEY AND EVANS, PRINTERS, WHITEFRIARS. 
 
of. 
 
 NOTICE BY THE PUBLISHER. 
 
 In issuing a Fifth Edition of the present work, I feel it incumbent 
 upon me, as the Publisher, to offer a few remarks in reference to an 
 article in the Athenmtm of March 13th, 1858, which article, under 
 the assumed character of a Review of Baker's Mensuration, contains 
 charges injurious alike to the Author and to myself, and to give 
 publicity to which seems to have been the sole object of the writer. 
 
 The Reviewer has coupled together, in his production, two books 
 on Mensuration — one by Mr. Elliot, and the other by Mr. Baker ; 
 and the single point he aims at establishing is, that the latter Author 
 has plagiarised from the former. 
 
 Upon his own showing, the Reviewer has been in previous cor- 
 respondence with Mr. Elliot, and has been supplied with letters, 
 passing between Mr. Elliot, Mr. Baker, and myself ; by aid of these 
 he is enabled, he thinks, to substantiate a charge of plagiarism 
 against Mr. Baker, and consequently to implicate me in dealing in 
 stolen goods. 
 
 Now, to those who have not read the Reviewer's remarks, it is 
 necessary to state that all the alleged plagiarisms consist of certain 
 schoolboy questionSy with which books on Mensuration usually 
 abound, and which are given as exercises for schoolboys to work 
 out. As everybody knows, these questions involve in general but 
 arbitrary suppositions, and seldom have any reference to actually 
 existing circumstances or matters of fact. I here quote two, which — 
 although the nunibers are altered in Mr. Baker's statements of them 
 — are adduced as plagiarisms from Mr. Elliot : 
 
 1. ^* A ladder is to be placed so as to reach a window, the sill of 
 which is 67J feet from the ground : the foot of the ladder cannot be 
 brought nearer than 36 feet from the wall : what length of ladder 
 will be sufficient? Answer, 76J feet." 
 
 In Mr. Baker's version, the given numbers being different, the 
 answer is 38^ feet. 
 
 2. *' A ladder standing upright beside a wall 100 feet high, just 
 reaches the top ; how far may the foot of the ladder be removed 
 from the wall, and still reach within 6 inches of the top ? Answer, 
 10 feet." 
 
 This question too, though substantially the same in Baker (see 
 question 11, p. 21), is nevertheless considerably modified, and is 
 certainly not copied. 
 
 Now, I would ask any candid person, familiar with schoolbooks of 
 this kind, whether questions such as the above are not common to 
 them all ; and whether there is a particle of merit in framing them, 
 or a particle of demerit in copying them ? 
 
 1118?6 
 
If the Reviewer wish to further employ his talents in this kind 
 of criticism, I can furnish him with an ample crop of materials. Or, 
 if he have ever written a hook on Mensuration or on Algebra himself, 
 I will undertake to supply him with the like materials from his own 
 work, so that he may profitably review himself. 
 
 Did the Reviewer ever see the question about the post being so 
 much in mud, so much in water, and so much above the water ? 
 
 I dare say he can tell me in what book on Algebra that occurs : 
 can he tell in what book on Algebra it does not occur ? 
 
 I might make the same inquiry about the Jlshy which every book 
 on Algebra serves up — head, body, and tail. And he ought to know 
 that questions about ladders strutting across streets, and reaching 
 windows, are quite as hackneyed as these. 
 
 Such is the nature of the literary delinquencies brought against 
 the author of this book: but the Reviewer gathers — not from the 
 book — but from the private letters before alluded to, that Mr. Baker 
 has affirmed what is not true. Mr. B. denies having seen Mr. Elliot's 
 work, and the Reviewer dwells a good deal upon his moral delin- 
 quency. Is it the oflB.ce of the Reviewer to assume the function of a 
 public corrector of morals, even though the offence be confined to a 
 private letter ? 
 
 But, even on this point, the Reviewer's evident desire to damage 
 the Author and his work, has hurried him to hasty conclusions. 
 Mr. Elliot has written two works on Mensuration, in size and 
 appearance totally unlike, and put forth by different publishers. 
 
 Mr. Baker, as was his duty, collected all the books on the subject 
 he could during the compilation of his own ; and he did so expressly 
 with the view of selecting questions : I presume the general practice 
 of all compilers of such treatises. He has no doubt about having 
 seen one of Mr. Elliot's two works, but a very different book from 
 which he is charged with copying, I am persuaded he never did see. 
 As the Reviewer is so rigid a judge in matters of plagiarism, I 
 think he ought, in consistency, to rebuke Mr. Elliot for copying even 
 from himself. 
 
 I have only to say, in conclusion, that it has always been my most 
 anxious endeavour to secure, for my series of Rudimentary Treatises, 
 writers of ability and reputation ; and I respectfully, but confidently, 
 leave the public to judge whether this endeavour has not been on 
 the whole successful. And I cannot but complain that any writer 
 in a literary journal should attempt to frustrate my honourable 
 efforts, and injure property on which I have expended a very Mi-ge 
 outlay, by insinuations, as unjust as they are injurious, that I 
 connive at an infringement of the rights of others. 
 
 JOHN WEALE. 
 
INTRODUCTION. 
 
 It will at once be seen that condensation of the materials pro- 
 duced by previous authors, and the introduction of a judicious 
 selection of matter, adapted to the expanded intellect of the 
 present age, are the proper requisites for a work on Mensuration. . 
 To this plan, the author trusts, from his long experience in en- 
 gineering pursuits, that he has strictly adhered. In the first 
 part, on Practical Geometry, numerous examples are intro- 
 duced, wherein the dimensions of certain parts are given to find 
 the dimension of their corresponding parts, which has been rarely 
 or never done by previous authors. This part is succeeded by a 
 second part, on the Mensuration of Lines; which is not 
 added for the sake of novelty only, but because it seemed to be 
 the natural order of a work of this kind. The third and fourth 
 Parts treat of the Mensuration of Superficies and of 
 Solids ; while in all the three last-named Parts the rules are 
 not only given in words at length, in the usual way, but the same 
 rules are expressed by Formulas, together with other formulae 
 depending thereon, by which the rules receive considerable exten- 
 sion. Some of the rules and examples are taken verbatim from 
 Dr. Huttons Mensuration; for the author conceives that it would 
 be disreputable to attempt, by verbal alterations in such rules, to 
 give an air of originality to his work, as all other authors have 
 done since Dr. H.'s time : the originality of this work consists in 
 the new matter, everywhere added, to adapt it to the wants of 
 modern times. Timber measuring and Artificers' work, the latter 
 with considerable modern improvements, are next introduced, 
 with concise and practical methods of finding the surfaces and 
 solidities of vaulted roofs, arches, domes, &c. 
 
 Concise, and the author trusts, clear systems of land surveying, 
 levelling, laying out railway curves and finding the contents of 
 raiivvay cuttings, complete the work, and serve as an introduction 
 to the author's Land and Engineering Surveying, which con- 
 tains everything adapted to modern practice that can be de- 
 sired, an extension to this subject having been first given by the 
 author, not found in any work previous to those written by him, 
 see pages 179 and 203 of that work. 
 
 The demonstrations of all the rules and formulae, in the four 
 
VI INTRODUCTION. 
 
 leading parts of the work, will be found in Dr. Huttotis Large 
 Mensuration and in the Budimentary Geometry ; the remainder 
 of the demonstrations will be fomid in the author's Railway En- 
 gineering or in his Land and Engineering Surveying, 
 
 Conic Sections and their solids are very briefly treated of in 
 the four preceding parts of the work, and chiefly in as far as 
 they may be useful to those who may intend to become excise 
 officers, whose actual practice is best learnt from an experienced 
 officer. Thus an extended article, such as is usually given by 
 other authors, is avoided, as not being generally useful to prac- 
 tical men. 
 
 The weights and dimensions of balls and shells may be found 
 by Prob. VIII., Part IV., in conjunction with the Table and 
 Rules for finding the specific gravities of bodies, if required. 
 
 The method of piling balls and shells, finding their number 
 in a given pile, and the quantity of powder contained in a given 
 shell or box, form no essential part of a work on mensuration, 
 being only useful in an arsenal, and are, therefore, also omitted. 
 The author has thus secured ample space for the discussion of 
 subjects really useful to the great majority of students and prac- 
 tical men, in the compass of a volume less than half the size and 
 one-fifth of the price of the works of his predecessors ; besides 
 adding matter, adapted to the wants of modern times, not found 
 in any existing work on mensuration. 
 
 The plan being thus briefly detailed, it will now be proper, 
 previous to studying the following work, to give the following 
 
 DIRECTIONS FOR BEGINNERS. 
 
 The beginner, for a first course, may omit the Problems be- 
 yond the thirty-second in Practical Geometry, and Problems 
 III., VIII., IX., XI., and XII., in the Mensuration of Lines, 
 with the formulse and examples depending on them. He may 
 also omit all the formulae in the Mensuration of Superficies and 
 Solids, with the examples depending on them, as well as the 
 Problems beyond the tenth in the Mensuration of Solids, except 
 it is required he should learn the method of gauging casks, in 
 which case omit only the two last problems. But if he require 
 an extensive knowledge of some or all the subjects here treated 
 of, he will do well to learn the use of such of the formulae and 
 the other parts, omitted in the first course, according to what he 
 may require as a practical man. 
 
CONTENTS. 
 
 PART I. 
 
 PAQB. 
 
 Practical Geometry — Definitions . . , . . . .1 
 
 Problems in Practical Geometry (40 Problems) 4 
 
 Geometrical Theorems 16 
 
 Explanation of the Principal Mathematical Characters used ia this Work 1 8 
 
 PART II. 
 Mensuration OF Lines— Table of Lineal Measure . . , .19 
 Twelve Problems on the Mensuration of Lines, -with Formulae . .ibid, 
 
 PART IIL 
 Mensuration op Surfaces — Table of Square Measure . , .34 
 Sixteen Problems on the Mensuration of Surfaces, with Formulce . . ibid. 
 Promiscuous Exercises 52 
 
 PART IV. 
 Mensuration of Solids — Definitions ..*... 52 
 
 Table of Solid Measure 54 
 
 Twenty-one Problems on the Mensuration of Solids, Gauging, &c., with 
 
 Formulte 55 
 
 The Sliding, or Carpenters' Rule— Seven Problems . . .73 
 
 Timber Measuring — Three Problems 75 
 
 Artificers' WORK—Bricklayers' Work 80 
 
 Masons' Work 82 
 
 Carpenters' and Joiners* Work 83 
 
 Slaters' and Tilers' Work 86 
 
 Plasterers' Work 87 
 
 Painters' Work — Glaziers' Work 88 
 
 Paviors' Work — Plumbers' Work 89 
 
 Arched and Vaulted Roofs 90 
 
 Specific Gravity • , 91 
 
 Land Surveying— The Chain 92 
 
 Offset Staff-- Cross 93 
 
 Directions for Measuring Lines on the Ground . . . . . ibid. 
 
 The common Drawing Compasses— Plotting Scales . . ■ . 95 
 
 Planning Surveys— The Field Book 96 
 
 To Survey with the Chain and Cross— Tables and five Problems, with 
 
 numerous Examples . . . . . . . . 97 
 
IV CONTENTS. 
 
 PAGE. 
 
 To Survey with the Chain only — Four Problems and numerous Ex- 
 amples, with the method of Surveying a Large Estate . .103 
 
 Levelling — Definition — Levelling Instruments . . . . . 115 
 
 The Y Level ihid. 
 
 Levelling Staves (Gravatt's) 118 
 
 Correction for Curvature ihid. 
 
 for Refraction, and Examples 119 
 
 To find the differences of Levels of several points on the Earth's surface 120 
 To draw a Sectional Line of several points in the Earth's surface, the 
 
 Levels of which have been taken 121 
 
 Level Book 122 
 
 Datum Line Hid. 
 
 Practical Level Book 123 
 
 Levels for the formation of a Section 124 
 
 Level Book for plottiug the Section 125 
 
 The Method of Laying out Railway Curves on the Ground . .127 
 pROB. I. — To lay out a Railway Curve by the common Method — Cases 
 
 I. and II ihid. 
 
 Prob. II. — To lay out a Railway Curve by Ofisets from its Tangent. — 
 
 Cases Land II 129 
 
 Contents of Railway Cuttings— Tables 130 
 
 Prob. I. — To find the Contents of Railway Cuttings from their Depths 132 
 Prob. II. — To find the Contents of Railway Cuttings from Sectional Areas 133 
 
 General Rule for finding the Contents of Solids, with Examples . .135 
 
 Tables, No. 1.— Areas of Segments of Circles 137 
 
 No. 2.— Ofi'sets to Railway Curves 138 
 
 No. 3.— Correction for Curvature and Refraction . . .ihid. 
 
OF THE '^ 
 
 UNIVERSITY 
 
 OF 
 
 MENSURATION 
 
 Mensuration treats of the various methods of measuring and 
 estimating the dinlensions and magnitudes of figures and bodies. 
 It is divided into four parts, viz.. Practical Geometry, and Men- 
 suration of Lines, of Superfices, and of Sohds, with their several 
 applications to practical purposes. 
 
 PART I. 
 
 PRACTICAL GEOMETRY. 
 
 DEFINITIONS. 
 
 L A point has no dimensions, neither length, breadth nor 
 thickness. 
 
 2. A line has length only, as A. 
 
 3. A surface or plane has length and 
 breadth, as B. 
 
 B 
 
 4. A right or straight line lies wholly in the same direction, 
 as AB. 
 
 5. Parallel lines are always at the same ^ ^ 
 
 distance, and never meet when prolonged, q _t) 
 
 as A B and C D. 
 
 6. An angle is formed by the meeting of 
 two lines, as AC, C B. It is called the 
 angle A C B, the letter at the angular point C -^^ 
 C being read in the middle. 
 
 7. A right angle is formed by one right j^ 
 line standing erect or perpendicular to an- , j) 
 other ; thus, ABC is a right angle, as is / 
 also ABE. / ^ 
 
 8. An acute angle is less than a right e B 
 angle, as D B C. 
 
 9. An obtuse angle is greater than a right angle, as D B E. 
 
 I 
 
2 PRACTICAL GEOMETRY. 
 
 10, A plane tria7igle is a space included by three right lineSj 
 and has three angles. 
 
 11. A right angled triangle has one right 
 angle, as A B C. The side A C, opposite the 
 right angle, is called the hypothenuse; the 
 sides A B and B C are respectively called the 
 base and perpendicular. 
 
 12. An obtuse angled triangle has one obtuse 
 angle, as the angle at B. 
 
 13. An acute angled triangle has all its 
 three angles acute, as D. 
 
 14. An equilateral triangle has three equal 
 sides, and three equal angles, as E. 
 
 15. An isosceles triangle has two equal sides, 
 and the third side greater or less than each of 
 the equal sides as F. 
 
 \6, A quadrilateral figure is a space bounded by four right 
 lines, and has four angles ; when its opposite 
 sides are equal, it is called a parallelogram, 
 
 17,-4 square has all its sides equal, and all 
 its angles right angles, as G. 
 
 \^, A rectangle is a right angled parallelogram, whose length 
 exceeds its breadth, as B, (see figure to definition 2). 
 
 19, A rhombus is a parallelogram having 
 all its sides and each pair of its opposite angles 
 equal, as I. 
 
 20. A rhomboid is a parallelogram having 
 its opposite sides and angles equal, as K. 
 
 21. A trapezium is bounded by four 
 straight lines, no two of which are parallel 
 to each other, as L. A line connecting 
 any two of its angles are called the diagonal^ 
 asAB. 
 
PRACTICAL GEOMETRY. 
 
 22. A trapezoid is a quadrilateral, having 
 two of its opposite sides parallel, and the re- 
 maining two not, as M. 
 
 23. Polygons have more than four sides, and receive par- 
 ticular names, according to the number of their sides. Thus, 
 2k pentagon has five sides; a hexagon^ six; Vi heptagon, seven; 
 an octagon, eight ; &c. They are called regular polygons, 
 when all their sides and angles are equal, otherwise irregular 
 polygons. 
 
 24. A circle is a plain figure, bounded by a j^^ — . ^ 
 curve line, called the circumference, which is j \ 
 everywhere equidistant from a point C within, I c j 
 called the centre. \ — y 
 
 25. An arc of a circle is a part of the circumference, as 
 AB. 
 
 26. The diameter of a circle is a 
 straight line A B, passing through the 
 centre C, and dividing the circle into two 
 equal parts, each of which is called a 
 semicircle. Half the diameter A C or 
 CB is called the radius. If a radius CD 
 be drawn at right angles to A B, it will 
 divide the semicircle into two equal parts, each of which is 
 called a quadrant, or one fourth of a circle. A chord is a right 
 line joining the extremities of an arc, as F E. It divides the 
 circle into two unequal parts called segments. If the radii C F, 
 C E be drawn, the space, bounded by these radii and the arc 
 F E, will be the sector of a circle. 
 
 27. The circumference of every circle is supposed to be 
 divided into 360 equal parts, called degrees, and each degree 
 into 60 minutes, each minute into 60 seconds, &c. Hence a 
 semicircle contains 180 degrees, and a quadrant 90 degrees. 
 
 28. The measure of an angle is an arc of 
 
 any circle, contained between the two lines .-'-—.. A 
 
 which form the angle, the angular point 
 
 being the centre ; and it is estimated by the 
 
 number of degrees contained in that arc :-— 
 
 thus the arc A B, the centre of which is C, 
 
 is the measure of the angle A C B. If the 
 
 angle A C B contain 42 degrees, 29 minutes, ?ind 48 seconds, 
 
 it is thus written 42° 29' 48". 
 
PRACTICAL GEOMETRY. 
 
 I 
 
 A 
 
 PROBLEMS IN PRACTICAL GEOMETRY. 
 
 (In solving the five following problems only a pair of common compasses and 
 a straight edge are required ; the problems beyond the fifth require a scale of 
 equal parts ; and the two last a line of chords ; all of which will be found in a 
 common case of instruments.) 
 
 Problem I. 
 .0. ' To divide a given straight line A B into iioo 
 
 equal parts. 
 
 From the centres A and B, with any radius, 
 B or opening of the compasses, greater than half 
 A B, describe two arcs, cutting each other in C 
 .. , and D; draw CD, and it will cut A B in the 
 
 ^ middle point E. 
 
 Problem IL 
 At a given distance E, to draw a straight line C D, parallel 
 to a given straight line A B. 
 
 E From any two points m and r, in 
 
 ^ g the line A B, with a distance equal to 
 
 C T^-— ^r — - D E, describe the arcs n and s : — draw 
 
 C D to touch these arcs, without cut- 
 
 ^ — m r -^ *^"S them, and it will be the parallel 
 
 required. 
 Note. This problem, as well as the following one, is usually performed by an 
 instrument called the parallel ruler. 
 
 Problem III. 
 Through a given point r, to draw a straight line CD "parallel 
 to a given straight line A B. 
 
 From any point n in the line A B, with 
 
 ^7, s^ the distance wr, describe the arc rm\ — from 
 
 /'^•-.. i centre r, with the same radius, describe the 
 
 \ ''••.. / arc n s: — take the arc mr in the compasses, 
 
 ^^ and apply it from n to s : — through r and 5 
 
 draw C D, which is the parallel required. 
 
 Problem IV. 
 JFV'om a given point P in a straight line A'B to erect a per^ 
 pendicidar, 
 
 1 When the point is in or near the middle of the line. 
 
 On each side of the point P take any two 
 
 C equal distances, P m, Vn\ from the points 
 
 m and w, as centres, with any radius greater 
 
 than P m^ describe two arcs cutting each 
 
 other in C; through C, draw C P, and it will 
 
 ^""-g be the perpendicular required. 
 
PRACTICAL GEOMETRY. 
 
 2. IFhen the point P is at the end of the line, 
 
 With the centre P, and any radius, de- 
 scribe the arc nrs; — from the point w, 
 with the same radius, turn the compasses 
 twice on the arc, as at r and s : — again, 
 with centres r and s, describe arcs inter- 
 secting in C : — draw C P, and it will be 
 the perpendicular required. -^ j 
 
 Note. This problem and the following one are usually done with an instru- 
 ment called the square. 
 
 Problem V. 
 From a giv€?i point C to let fall a perpendicular to a given line, 
 1 When the point is 72early opposite the middle of the line. 
 From C, as a centre, describe an arc to cut 
 
 A B in 7w and n ; — with centres m and w, 
 
 and the same or any other radius, describe 
 
 arcs intersecting in o\ through C and o draw 
 
 C o, the perpendicular required. 
 
 2. When the point is nearly opposite the end of the line. 
 
 From C draw any line C m to meet B A, 
 in any point m\ — bisect C m in w, and with 
 the centre ^^, and radius Cw, or mn^ describe 
 an arc cutting B A in P. Draw C P for the 
 perpendicular required. 
 
 Problem VI. 
 
 To construct a triangle with three given right lines^ any two 
 of which must he greater than the third, (Euc. I. 22.) 
 
 Let the three given lines be 5, 4 and 3 
 yards. From any scale of equal parts lay 
 off the base A B = 5 yards ; with the 
 centre A and radius A C = 4 yards, de- 
 scribe an arc ; with centre B and radius A ' 'g 
 
 C B = 3 yards; describe another arc cutting the former arc in- 
 C ; — draw A C and C B ; then A B C is the triangle required. 
 Problem VII. 
 
 Given the base and perpendicular^ with 
 the place of the latter on the base^ to con- C 
 
 struct the triangle. 
 
 Let the base AB = 7, the per- 
 pendicular C D = 3, and the dis- 
 tance A D = 2 chains. Make AB 
 r= 7 and AD = 2; — at D erect the 
 
 - p 
 
6 
 
 PRACTICAL GEOMETRY. 
 
 perpendicular D C, whicli make = 3 : — draw A C and C B ; 
 then A B C is the triangle required. 
 
 Problem VIII. 
 
 To describe a square, whose side shall he of a given length. 
 
 Let the given line A B he three feet. 
 At the end B of the given line erect the 
 perpendicular B C, (by Prob. IV. 2.) which 
 make = A B : — with A and C as centres, 
 and radius A B, describe arcs cutting each 
 other in D : draw A D, D C, and the 
 square will be completed. 
 
 Problem IX. 
 
 To describe a rectangled j^arallelogram having a given length 
 and breadth. 
 
 Let the length A B = 5 feet, 
 and the breadth B C = 2. At B 
 erect the perpendicular B C, and 
 make it = 2 : — with the centre A 
 and radius B C describe an arc ; and 
 with centre C and radius AB describe 
 another arc, cutting the former in 
 
 3>i 
 
 B 
 
 D : join AD, D G to complete the rectangle. 
 
 Problem X. 
 
 The base and two perjpendiculars being given to construct a 
 trapezoid. 
 
 Let the base A B = 6, and the 
 perpendiculars A D and B C, 2 and 
 3 feet respectively. Draw the per- 
 pendiculars A D, D C, as given above, 
 and join D C, thus completing the 
 trapezoid. 
 
 Problem XI. 
 
 To construct a right angled triangle having a given base and 
 perpendicular i and to find the hypothenuse. 
 
 Let the base A B = 6 feet, and the perpen- 
 dicular B C = 8. Draw B C perpendicular 
 to A B, and join A C ; then ABC will be the 
 triangle required, and A C being measured will 
 be found = 10 feet. 
 
PRACTICAL GEOMETRY. 
 
 Problem XII. 
 Having given the base and hypothenuse to construct the right 
 angled triangle, and find the perpendicular, 
 
 {See figure to last Problem.) 
 Let A B = 6 feet and A C = 10. — Draw the perpendicular 
 B C indefinitely ; take A C = 10 feet in the compasses, and with 
 one foot on A apply the other to C ; join A C, which completes 
 the triangle, and B C will be found = 8 feet. 
 
 EXAMPLE. 
 
 A ladder 50 feet in length is placed with its foot 14 feet from 
 a wall, the top of the ladder just reaching to the top of the wall ; 
 required the height of the wall. 
 
 Here 14 feet is the base of the right angled triangle, and 50 
 feet, = length of the ladder is the hypothenuse, with which the 
 triangle being constructed, as in the last Problem, the perpen- 
 dicular will be found = 48 feet. 
 
 Problem XIII. 
 
 To divide a given angle ABC into two equal parts. 
 
 From the centre B, with any E 
 
 distance, describe the arc A C. 
 From A and C, with one and the 
 same radius, describe arcs inter- 
 secting in m. Draw the line B »^, 
 and it will bisect 
 quired. 
 
 Problem XIV, 
 
 To set off an angle to contain a given number of degrees. 
 
 Let the angle be required to \^' 
 contain 41 degrees. Open the 
 compasses to the extent of 60° 
 upon the line of chords, and 
 setting one foot upon A, with 
 this extent, describe an arc cut- 
 ting x\ B in B ; then taking the A^ -^ L^ 
 
 the angle as re- 
 
 . 3w iw 
 
 extent of 41° from the same line of chords, set it off from B to 
 C ; join A C ; then B A C is the angle required. 
 Problem XV. 
 To ineasure an angle contained by two straight lines » 
 
 (See last figure.) 
 Let A B, A C contain the angle to be measured. Open the 
 compasses to the extent of 60°, as before on the line of chords, 
 and with this radius describe the arc B C, cutting A B, A C 
 
8 
 
 PRACTICAL GEOMETRY. 
 
 produced, if necessary, in B and C ; then extend the compasses 
 from B to C, and this extent, appUed to the hne of chords, will 
 reach to 41°, the required measure of the angle BAG. 
 
 A right angle, or perpendicular, may be laid ofi by extending 
 the arc B C, and setting off the extent of 90° thereon. Also an 
 angle greater than 90° may be laid off, by still further extending 
 the arc, and laying the excess of the arc above 90°, from the end 
 of the 90th degree. 
 
 Note. Angles are more correctly and expeditiously laid off and measured 
 Ijy an instrument called the protractor, to be hereafter described. 
 
 Problem XVI. 
 To find the centre of a circle.. 
 Draw any chord K B, and by Prob. I. 
 bisect it perpendicularly with C D, which 
 will be a diameter. Bisect C D in the point 
 O, and that will be the centre. 
 
 Problem XVII. 
 
 To describe the circumference of a circle through three given 
 points ABC. 
 
 From the middle point B draw chords 
 to the two other points A, C, bisect these 
 chords perpendicularly by lines meeting 
 in O, which will be the centre ; then from 
 the centre O, at the distance O A, or O B, 
 or O C, describe the circle. 
 
 Note. In the same manner may the centre of an arc of a circle be found. 
 
 Problem XVIII. 
 Through a given point A to draw a tangent to a given circle. 
 Case I. When A is in the circumference of the circle. 
 
 B . -— . A C From the given point A, draw A O to 
 
 the centre of the circle ; then through A 
 draw B C perpendicular to A O, and it will 
 be the tangent as required. 
 Case II. When the given point is B not in 
 the circumference. 
 From B draw B O to the centre of the circle ; and on B O 
 describe the semicircle B A O, cutting the circle in A : then 
 through B and A draw BAG, and it will be the tangent re- 
 quired. 
 
PRACTICAL GEOMETRY, 
 
 ^ o 
 
 Problem XIX. 
 
 To make a regular pentagon on a given line A B. 
 
 Make B m perpendicular and equal 
 to half A B ; draw A m, and produce 
 it till m 71 be equal to B m ; with centres 
 A and B, and distance B n describe 
 arcs intersecting in o, which will be 
 the centre of the circumscribing circle ; 
 then with the centre o, and the same 
 
 radius, describe the circle ; and about A ^ B 
 
 the circumference of it apply A B the proper number of times. 
 
 Problem XX. 
 
 To make a hexagon on a given line A B. 
 
 With the distance A B, and the 
 centres A and B, describe arcs inter- 
 secting in ; with the same radius and 
 centre o describe a circle, which will 
 circumscribe the hexagon ; then apply 
 the line A B six times round the cir- 
 cumference, marking out the angular 
 points, and connect them wath right lines. 
 
 Problem XXI. 
 
 To make an octagon on a given line A B. 
 
 Erect A F and B E perpendicular 
 to A B ; produce A B both ways, and 
 bisect the angles wz A F and /z B E 
 with the lines A H and B C, each 
 equal to A B ; draw C D and H G 
 parallel to A F or B E, and each equal 
 to A B ; with the distance A B, and 
 centres G and D, cross A F and B E 
 in F and E : then join G F, F E, E D, 
 and it is done. 
 
 Problem XXII. 
 
 To make any regular polygon on a 
 given line A B. 
 
 Draw A o and B o making the 
 angles A and B each equal to half 
 the angle of the polygon, by Prob. 
 XIV., with the centre o and distance 
 o A describe a circle : then apply the 
 line A B continually round the cir- 
 cumference the proper number of times, and it is 
 
 ■ E , 
 
10 
 
 PRACTICAL GEOMETRY. 
 
 Note. The angle of any polygon, of which the angles o A B and o B A are 
 each one half, is found thus : divide the whole 360 degrees by the number of 
 sides, and the quotient wull be the angle at the centre o ; then subtract that 
 from 180 degrees, and the remainder will be the angle of the polygon, and is 
 double of A B or of B A. And thus you will find the numbers of the fol- 
 lowing table, containing the degrees in the angle o, at the centre, and the 
 angle of the polygon, for all the regular figures from 3 to 12 sides. 
 
 
 
 Angle 
 
 Angle 
 
 Angle 
 
 No of sides. 
 
 Name of the Polygon. 
 
 at the 
 
 of the 
 
 A B or 
 
 
 . 
 
 centre. 
 
 polygon. 
 
 oB A. 
 
 3 
 
 Trigon . 
 
 120° 
 
 60° 
 
 30° 
 
 4 
 
 Tetragon 
 
 
 
 90 
 
 90 
 
 45 
 
 5 
 
 Pentagon 
 
 
 
 72 
 
 108 
 
 54 
 
 6 
 
 Hexagon 
 
 
 
 60 
 
 120 
 
 60 
 
 7 
 
 Heptagon 
 
 
 
 51f 
 
 128f 
 
 64|- 
 
 8 
 
 Octagon 
 
 
 
 45 
 
 135 
 
 67 
 
 9 
 
 Nonagon 
 
 
 
 40 
 
 140 
 
 70 
 
 10 
 
 Decagon 
 
 
 
 36 
 
 144 
 
 72 
 
 11 
 
 Undecagon 
 
 
 
 32tV 
 
 147t\ 
 
 73tV 
 
 12 
 
 Dodecagon 
 
 
 
 30 
 
 150 
 
 75 
 
 Problem XXIII. 
 
 In a given circle to inscribe any regular 'polygon ; or to divide 
 the circumference into any number of equal ]parts» 
 
 (See the last figure,) 
 
 At the centre o make an angle equal to the angle at the centre 
 of the polygon, as contained in the third column of the above 
 table of polygons : then the distance A B will be one side of the 
 polygon, which being carried round the circumference the proper 
 number of times, will complete the figure. Or, the arc A B will 
 be one of the equal parts of the circumference. 
 
 Problem XXII. 
 
 About a given circle to circumscribe any regular polygon. 
 
 Find the points m, n, p, &c., as in the 
 last problem ; to which draw radii m o, 
 n o, &c., to the centre of the circle ; then 
 through these points m, w, &;c., and per- 
 pendicular to these radii, draw the sides 
 of the polygon. 
 
 EXAMPLE. 
 
 Let the radius of the given circle be five feet ; then, having 
 described a regular pentagon round it, the side of the figure 
 
PRACTICAL GEOMETRY. 
 
 11 
 
 will be found = 7 feet 3 J inches. If the figure to be described 
 round the same circle be a regular hexagon, its side will be 
 found = 5 feet 9} inches : and so on for any other regular 
 polygons. 
 
 Problem XXV. 
 
 To find the centre of a given polygon^ or the centre of its in- 
 scribed or circumscribed circle. 
 
 Bisect any two sides with the per- 
 pendiculars m 0, no, and their intersec- 
 tion will be the centre ; then with the 
 centre o, and the distance o m, describe 
 the inscribed circle ; or with the distance 
 to one of the angles as A, describe the 
 circumscribing circle. 
 
 Problem XXVI. 
 
 In any given triangle to inscribe a circle. 
 
 Bisect any of two of the angles with 
 the lines A <?, B o ; and o will be the 
 centre of the circle ; then with the 
 centre o, and radius the nearest dis- 
 tance to any one of the sides, describe 
 the circle. 
 
 example. 
 
 Let the sides of the given triangle be 5, 4, and 3 feet ; then, 
 having inscribed a circle therein, its radius will be found = 
 1 foot. 
 
 Problem XXVII. 
 
 About a given triangle to circumscribe a circle. 
 
 Bisect any of the two sides A B, B C, C 
 
 with the perpendiculars m o, n o,-, with 
 the centre o, and distance to any one of 
 the angles, describe the circle. 
 
 EXAMPLE. 
 
 Let the sides of the given triangle be 
 15, 14, and 13 feet ; then having de- 
 scribed a circle about it, the radius will 
 be found = 8 feet 1| inches. 
 
 Problem XXVIII. 
 
 Jw, or about, a given square to describe a circle. 
 
 Draw the two diagonals of the square, and their inter- 
 
12 
 
 PRACTICAL GEOMETRY. 
 
 section o will be the centre of both the 
 
 circles : then with that centre, and the 
 
 nearest distance to one side, describe the 
 
 inner circle, and with the distance to one 
 angle, describe the outer circle. 
 
 EXAMPLE. 
 
 Let the side of the given square be 3 feet : then, having de- 
 scribed circles in and about it, the radius of the former will be 
 found — l\ feet, and that of the latter = 2 feet 1|- inches 
 nearly. 
 
 Problem XXIX. 
 
 In J or about, a given circle, to describe a square or an octago7i. 
 
 Draw two diameters A B, CD, perpen- 
 dicular to each other; then connect their 
 extremities, and that will give the inscribed 
 square A C D B. Also through their ex- 
 tremities draw tangents parallel to them, and 
 they will form the outer square m nop. 
 
 Note. If any quadrant, as A C, be bisected in ^, it will give one-eighth of 
 the circumference, or the side of the octagon. 
 
 Problem XXX. 
 
 In a given circle to inscribe a trigon, a hexagony or a dodecagon* 
 
 A The radius of the circle is the sid-e 
 
 of the hexagon ; therefore from any^ 
 point A in the circumference, with the 
 distance of the radius, describe the arc 
 B O F : then is A B the side of the 
 hexagon ; and therefore carrying it 
 six times round will form the hexagon, 
 or will divide the circumference into 
 six equal parts, each containing 60 
 degrees. — The second of these, C, will 
 give A C the side of the trigon, or equilateral triangle ACE, and 
 the arc A C one-third of the circumference, or 120 degrees. — 
 Also the half of A B, or A n, is one-twelfth of the circumference, 
 or 30 degrees, which gives the side of the dodecagon. 
 
 Note. If tangents to the circle be drawn through all the angular points of 
 any inscribed figure, they will form the sides of a like circumscribing figure. 
 
 EXAMPLE. 
 
 In a circle, the radius of which is 10 feet, inscribe a trigon, a 
 hexagon, and a dodecagon. — Having measured a side of the 
 
PRACTICAL GEOMETRY. 
 
 13 
 
 several figures, that of the trigon will be found =1/ feet 4 
 inches, that of the hexagon 10 feet, and that of the dodecagon 
 5 feet 2 inches. 
 
 Problem XXXI. 
 
 In a given circle to inscribe a pentagon or a decagon, 
 
 Dri% the two diameters A P, m n A 
 
 perpendicular to each other, and bisect 
 the radius o w at </ ; with the centre q 
 and the distance q A, describe the arc 
 A r ; and with the centre A, and 
 radius A r, describe the arc r B : then 
 is A B one-fifth of the circumference ; 
 and A B carried five times over will 
 form the pentagon. Also the arc A B 
 bisected in *, will give A s the tenth 
 part of the circumference, or the side of the decagon. 
 
 Note. Tangents being drawn through the angular points will form the 
 circumscribing pentagon or decagon. 
 
 EXAMPLE. 
 
 In a circle, the radius of which is 10 feet, inscribe a pentagon 
 and decagon. — Having measured a side of each of the figures, 
 that of the pentagon will be found =11 feet 9 inches, and that 
 of the decagon = 6 feet 2 inches. 
 
 Problem XXXII. 
 
 To make a triangle similar to a given triangle ABC. 
 
 Make a B equal to the base of the required 
 triangle ; through a draw a c parallel to A C : 
 then a B c is the triangle required. 
 
 example. 
 
 Let A B = 12, A C = 15 and B C = 9 
 feet ; and the side « B of the required tri- 
 angle = 8 feet. — Then having drawn a c 
 parallel to A C, the side ac will be found =10, and B c = 6 feet. 
 
 Problem XXXIII. 
 
 To make a figure similar to any 
 other given figure ABODE. 
 
 From any angle A draw diagonals 
 to the other angles ; take A 6 a side 
 of the figure required ; then draw h c 
 parallel to B C, and c c? to CD, and 
 c? e to D E, &c. ^ b B b 
 
14 
 
 PRACTICAL GEOMETRY. 
 
 Problem XXXIV. 
 
 To make a square equal to two given 
 squares P and Q. 
 
 Set two sides A B, B C, of the given 
 squares, perpendicular to each other ; 
 join their extremities A, C ; so skril the 
 square B, constructed on A C, be equal 
 to the two P and Q taken together. 
 (Euc. I., 47.) 
 
 Problem XXXV. 
 
 To make a square equal to the difference between two given 
 squares, P, R. 
 
 {See the last figure,) 
 
 On the side A C of the greater square, as a diameter, describe 
 a semicircle ; in which apply A B the side of the less square ; 
 join B C, and it will be the side of a square equal to the differ- 
 ence between the two P and B as required. 
 
 Problem XXXVI. 
 
 To make a square equal to the sum of any numher of squares 
 taken together. 
 
 Draw two indefinite lines K m, K n, perpendicular to each 
 other at the point A. On the one of these set off A B the side 
 of one of the given squares, and on the other 
 A C the side of another of them ; join B C, 
 and it will be the side of a square equal to the 
 two together. Then take A D equal to B C, 
 and A E equal to the side of the third given 
 square. So shall D E be the side of a square 
 equal to the sum of the three given squares. — 
 And so on continually, always setting more sides 
 of the given squares on the line A n, and the sides of the suc- 
 cessive sums on the other line A m. 
 
 Problem XXXVII. 
 
 To construct a figure resembling an ellipse, by circular arcs 
 from four centres. 
 
 On a line F f of convenient length, describe two equilateral 
 triangles F P /, Y p f; prolong the sides of the triangles ; 
 join P p, as shown in the figure. With centres P, p and 
 radius P H = ^ A describe the arcs H I, A i, meeting the 
 prolonged sides of the triangles, and such that the diameter C D 
 
 jD B 
 
PRACTICAL GEOMETRY. 
 
 15 
 
 may be equal to the required 
 width of the figure ; with 
 centres, F, / and radius II F 
 = I /= &c., describe the 
 arcs H A /i, I i, and the 
 figure will be completed. j 
 
 Note. If the longer diameter 
 A B be not obtained of the re- 
 quired length by the above opera- 
 tion, the triangles F P /, F jt? / 
 
 may be enlarged or diminished, ^ M't> 
 
 or made isosceles, till by trials the " ^' ^ 
 
 proper dimensions are obtained. — This method of drawing the ellipse is prac- 
 tised by the picture-frame makers. 
 
 Problem XXXVIII. 
 
 To describe a true ellipse. 
 
 Let T R be the trans- 
 verse, C O the conjugate, 
 and c the centre. With 
 the radius T c and centre 
 C, describe an arc cutting 
 T R in the points F /; 
 which are called the two 
 foci of the ellipse. 
 
 Assume any point P in 
 the transverse ; then with 
 
 the radii P T, P E, and centres F, f, describe two arcs inter- 
 secting in I ; which will be a point in the curve of the ellipse. 
 
 And thus, by assuming a number of points P in the trans- 
 verse, there will be found as many points in the curve as you 
 please. Then, with a steady hand, draw the curve through all 
 these points. 
 
 otherwise, WITH A THREAD. 
 
 Take a thread of the length of the transverse, T R, and fasten 
 its ends with two pins in the foci F, f. Then stretch the thread, 
 and it will reach to I in the curve ; and by moving a pencil 
 round, within the thread, keeping it always stretched, it will 
 trace out the ellipse. 
 
 Problem XXXIX. 
 
 To describe or construct a parabola, 
 
 V P being an absciss, and P Q its given ordinate ; bisect 
 P Q in A, join A V, and draw A B perpendicular to it ; 
 
16 
 
 PRACTICAL GEOMETRY. 
 
 irv 
 
 then transfer P B to V F and 
 V C in the axis produced. So 
 shall F be what is called the 
 focus. 
 
 Draw several double ordinates 
 S R, S &c., perpendicular to V P. 
 Then with the radii C R, &c. and 
 the centre F, describe arcs cut- 
 ting the corresponding ordinates 
 in the points S &c. Then draw 
 the curve through all the points 
 S&c. 
 
 Problem XL. 
 
 To construct or describe cm hyperbola. 
 
 Let D be the centre of the 
 
 ^ >Jif^ hyperbola, or the middle of 
 
 the transverse A B ; and B C 
 
 perpendicular to A B, and 
 
 equal to half the conjugate. 
 
 With centre D, and radius 
 D C, describe an arc, meet- 
 ing A B produced in F and /, 
 which are the two focus points 
 of the hyperbola. 
 Then assuming several points E in the transverse produced, 
 with the radii A E, B E, and the centres /, F, describe arcs 
 intersecting in the several points G ; through all which points 
 draw the hyperbolic curve. 
 
 :x 
 
 GEOMETRICAL THEOREMS. 
 
 {Necessary to be known by Beginners.) 
 
 Theorem I. 
 
 Angles vertically opposite are equal : — 
 tlius the angle A G E = angle H G B, 
 and E G B = A G H. (Euc. 1. 15.) 
 
 • B 
 ■D 
 
 F Theorem IL 
 
 (See last figure,) 
 A right line E F cutting two parallel right lines A B^ C D. 
 
PRACTICAL GEOMETRY. 1> 
 
 makes the alternate angle equal, ^c. : thus the angles A G H, 
 G H D are equal ; also the exterior angle E G B is equal to the 
 interior and opposite G H D. (Euc. I. 29.) 
 
 Theorem III. 
 
 The greatest side of every triangle is opposite the greatest 
 angle. (Euc. I. 18.) 
 
 Theorem IV. 
 
 Let the side of AB the triangle ABC be 
 produced to D, the exterior angle C B D 
 is equal to the interior angles at A and C ; 
 -?L also the three interior angles of the triangle 
 are equal to two right angles. (Euc. I. 32.) 
 Whence any two angles of a triangle heing given the third be- 
 comes known. 
 
 Theorem V. 
 
 (^See figure to Definition 11.) 
 
 Let ABC be a right angled triangle, having a right angle at 
 B ; then, the square on the side A C is equal to the sum of the 
 square on the sides A B, B C. (Euc. I. 47.) Whence any two 
 sides of a right angled triangle being given the third becomes 
 known. 
 
 Theorem VI. 
 
 In any triangle A B C, let D E be 
 drawn parallel to one of its sides, C B ; 
 then, AB is to AYi asBCis to D E ; 
 and the triangles are said to be similar. 
 (Euc. VI. 2.) 
 
 Theorem VII. 
 
 {See last figure.) 
 
 Let A B C, A E D be similar triangles ; then, the triangle 
 A B C f 5 to the triangle A E D «5 the square A B is to the square 
 q/* A E : that is, similar triangles are to one another in the 
 duplicate ratio of their homologous sides. (Euc. VI. 19.) 
 
 Theorem VIII. 
 
 All similar figures are to one another as the squares of their 
 homologous, or like, sides. (Euc. III. 20.) 
 
 Theorem IX. 
 
 All similar solids are to one another as the cubes of their like 
 linear dimensions. (Euc. VL 24.) 
 
18 PRACTICAL GEOMETRY. 
 
 EXPLANATION OF THE PRINCIPAL MATHEMATICAL 
 CHARACTERS USED IN THIS WORK. 
 
 The sign for equality = is read "equal ;" tlius 12 inclies = 
 1 foot. 
 
 The sign for addition + is read ''plus or more ;" thus 2 + 3 
 = 5. « + 5, &c. 
 
 The sign for subtraction — is read ''minus or less;" thus 
 5 — 2 = 3, « •- 6, &c. 
 
 The sign for multiplication X is read ''into;" thus 5x3 
 = 15, « X b, or a b, &c. 
 
 The sign for division -i- is read "by;" thus 15 h- 3 = 5, or 
 
 15 a 
 
 y = 5, or -, &c. 
 
 The signs for proportion, as : : : : " as, is to, so is, to ;" 
 thus as 2 ; 5 : : 8 : 20, or as a : b : : c : d, the fourth 
 number being found by multiplying the second by the third, 
 
 and dividing the first, or — - — =20, and — = d. 
 
 The signs ( ) or < > ^ is called vinculam or brace : 
 
 thus (5 + 4) X 2 = 9 X 2 = 18, or 5 + 4l 2 = 18, {a + 5, 
 X c, or a -\- b~\ c, &c. 
 
 The signs 2, ^, &c., placed above a quantity, represent re- 
 spectively the square, cube, &c., of that quantity ; thus 5^ = 
 5 X 5 = 25, 53 = 125, ^~+~V[ 3 = 72 = 49, 4 (5 + 3)2 = 
 4 X 8'^ = 256 ; and or' and a^ represent the square and cube 
 of «, also {a + 6)2 c^ signifies that the square of the sum of a 
 and b is to be multiplied by the cube of c, &c. 
 
 The sign V ov ^J placed before a quantity, or \ placed above 
 represents the square root of that* quantity ; thus \/3^ = 6, 
 \/9 X 16= 12, and \/a X b ov \/ ab signifies the square 
 root of the product of a and b, &c. 
 3 
 
 The sign V placed before a quantity, or i placed above it, 
 denotes the cube root of that quantity ; thus 
 
 V 12 X 2 X 3 - 8, or ^(12 X 2 X 3 - 8) = a/72 — 8 = 
 
 v' 64 = 4, V 6 { (« + b)^ — e d} denotes the cube root of the 
 
MENSURATION OF LINES. 
 
 19 
 
 (liiference of the square of the sum of a and h and the product 
 of c and d multiplied into e. Also, the value of 
 
 8 
 
 >/ e { (a + by — c d^y when « = 2, 5 = 7, c = 5, c? = 9, and 
 
 e = G is V 6 ( (2 + 7)^ — 5 X 9 } = >7 6 (81 - 45) 
 ^G X 3G = ^/216 = 6. 
 
 PART II. 
 MENSURATION OF LINES. 
 
 The Mensuration of Lines is applied to find the lengths of 
 straight or curved lines, from the given lengths of other lines, 
 on which these straight or curved lines depend. 
 
 
 table 
 
 of line 
 
 AL MEASURE. 
 
 Inches. 
 
 Feet. 
 
 
 
 12 
 
 1 
 
 Yards. 
 
 
 36 
 
 3 
 
 1 
 
 
 
 
 
 Poles. 
 
 
 198 
 
 16i 
 
 5i 
 
 1 
 
 Furlongs. 
 
 7920 
 
 660 
 
 220 
 
 40 
 
 1 
 
 63360 
 
 5280 
 
 1760 
 
 320 
 
 8 
 
 Mile. 
 1 
 
 711- = 7-92 inches = 1 link. 
 22 yards = 4 poles = 1 chain of J 00 links. 
 69i English miles = 60 geographical miles = 1 degree. 
 
 Problem I. 
 
 To find one side of a right angled triangle, having the othei* 
 two sides given. 
 
 The square of the hypothenuse is equal to both the squares of 
 the two legs. (Euc. I. 47.) Therefore, 
 
 Rule I. — To find the hypothenuse ; add the squares of the 
 two legs together, and extract the square root of the sum. 
 
 Rule II. — To find one leg ; subtract the square of the other 
 leg from the square of the hypothenuse, and extract the square 
 root of the difference. 
 
20 MENSURATION OF LINES. 
 
 i C Let A B C be a triangle, right angled at B ; 
 
 then by Theorem V. page 17, we shall have 
 the following 
 
 Formulae. 
 Put the base A B = 5, the perpendicular B C 
 B = i^5 and the hypothenuse A C = A ; then 
 
 A = V 6^ + jo", 6 = // A^ — 'P^i and p z=z ^ h? — p^, 
 
 EXAMPLE. 
 
 1 . Required the hypothenuse of a right angled triangle, the 
 base of which is 40 and the perpendicular (iO feet. 
 
 By Bule I. By first Formula. 
 
 40 30 V402 + 302 = 50 = A C. 
 
 40 30 
 
 1600 900 
 900 
 
 2500 (50 = hypothenuse A C. 
 25 
 
 00 
 
 Note. The student ought to solve this and all the following examples by 
 geometrical construction, as in Problems XI. and XII., Part I. — Thus : make 
 A B = 40 feet ; draw B C = 30 feet perpendicular to A B, and join A C ; 
 then A C, being measured, will be found to be 50 feet. 
 
 The construction of example 2, will be as follows. Make A B = 56 feet, 
 and perpendicular thereto draw B C indefinitely ; take AC = 65 in the com- 
 passes, and with one foot on A apply the other foot to C ; then B C, being 
 measured will be found to be 33 feet. 
 
 Remark. The triangle ABC, being for the purpose of illustrating the Pro- 
 blem generally, is not drawn to correspond to any of the dimensions given in 
 the examples. 
 
 2. What is the perpendicular of a right angled triangle, whose 
 base A B is 56, and the hypothenuse A C 65 feet? 
 56 65 ' 
 
 56 65 
 
 336 325 
 
 280 390 
 
 3136 4225 
 3136 
 
 1089 (33 feet = perpendicular BC. 
 9 
 
 63) 189 
 189 
 
MENSURATION OF LINES. 21 
 
 3. A ladder is to be placed so as to reach the top of a wall 
 33f feet high, and the foot of the ladder cannot be placed nearer 
 the wall than 1^ feet ; what must be the length of the ladder? 
 
 Jns, SSi/eet. 
 
 4. The side of a square is 100 yards ; what is the length of 
 its diagonal? A7is. 141*4 yards. 
 
 5. A line of 320 feet will reach from the top of a precipice, 
 standing close by the side of a river, to the opposite bank : re- 
 quired the breadth of the river ; the height of the precipice being 
 103 feet. Jns. S0297feet. 
 
 6. A ladder of 50 feet long being placed in a street, reached 
 a window 28 feet from the ground on one side ; and by turning 
 the ladder over, without removing the foot out of its place, it 
 touched a moulding 36 feet high on the other side : required the 
 breadth of the street. Ans. 7Q'l2Sf€ef. 
 
 7. The width of a house is 48 feet, and the height of ridge 
 above the side walls 10 feet ; required the length of one of the 
 rafters. 
 
 In the annexed figure A B is the 
 width of the house, or length of 
 the tie-beam of the rafters A C, 
 B C ; and C T) the height of the 
 ridge or length of the king-post; 
 and since D is the middle point of 
 A B, we shall have A D = ^ AB 
 
 = 24 feet. Whence by the first formula ; 
 A C = ^/TD^qfTc J)^= ^ 242+ iQ2^ ^^Q ^ 26 feet, 
 
 the required length of one of the rafters. 
 
 8. Required the height of an equilateral triangle, the side of 
 which is 10 feet. Ans. % feet 8 in. nearly. 
 
 9. The base of an isosceles triangle is 2.5 feet, and its two sides 
 are each 32^ feet; required the perpendicular. Ans. SO feet. 
 
 10. The diagonal of a square is 10 yards, required the length 
 of one of its sides., A7is. 7 yds. 0/eet 2\ in. 
 
 11. A ladder, standing upright against a wall 100 feet high, 
 was pulled out at the foot 10 feet from the wall ; how far did the 
 top of the ladder fall ? Ans 6 inches nearly, 
 
 12. The upright axle of the horse-wheel of a thrashing 
 machine is placed with its centre Z\ yards from a wall ; but the 
 shafts of the axle are 5 yards in length, measured from the cen- 
 tre : how much of the wall must be removed to admit it to revolve ? 
 
 Ans, 7 yds. ft. 5 in. 
 
22 
 
 MENSURATION OF LINES. 
 
 Problem II. 
 
 Having given any two of the dimensions of the figure A B C, 
 and one of the corresponding dimensions of a similar figure ahc,to 
 find the other corresponding dimension of the last figure. 
 
 Rule.— Let ABC, 
 
 « 5 c be two similar 
 triangles, then by The- 
 orem VI., page 17. 
 
 AB:BC::a6:«c, ora5:«c::AB :BC. 
 
 The same proportion holds with respect to the similar lineal 
 parts of any other similar figures, whether plane or solid. 
 
 EXAMPLES. 
 
 1 . The shadow of a cane 4 feet long, set perpendicularly, is 
 5 feet, at the same time that the shadow of a lofty tree was 
 found to be 83 feet ; required the height of the tree, both 
 shadows being on level ground. 
 
 Let 6 c be the cane, and B C the tree, their shadows being 
 respectively represented by a 5 and A B : the upper extremities 
 of the cane and tree being joined with the extremities of their 
 shadows, giving the parallel lines a c, A C for the directions of 
 the sun's rays, and thus constituting similar triangles ah c^ 
 ABC: whence «6 : 6 c : : AB : BC, 
 that is 5 : 4 : : 83 : 66f 
 4 
 
 5)332 
 
 of the tree. 
 
 66f feet = 
 
 BC, the height 
 
 2. The side of a square is 5 feet, and its diagonal 7*071 feet, 
 what will be the side of a square, the diagonal oiwhich is 4 feet. ? 
 
 Ans. 2 ft. 10 in. nearly, 
 
 3. In the ground plan of a building 120 feet long and 50 
 broad, the length, as laid down, is 1 inches ; what must be its 
 breadth ? Ans. 4-J- inches. 
 
 4. The scale of the Ordnance survey of Ireland is 6 inches to 
 1 mile, what length of paper will be sufficient for the map of that 
 country, its length being 300 miles ? Ans, 50 yards. 
 
MENSURATION OF LINES. 23 
 
 5. The length of the shadow of the Monument (London) is 
 151|- feet, while the shadow of a post 4 feet high, is 3 feet; 
 required the height of the Monument. Jns, 202 feet. 
 
 Problem III. 
 
 The two sides and the base of a triangle (A B C) are given to 
 ^nd the perpendicular (C D) . 
 
 Rule. — The segments of the base A D, D C must be first 
 found. Let B C be the greater of the two sides, then B. D will 
 C be the greater of the two seg- 
 
 ments. Then, as the base AB 
 ^v is to the sum of the sides B C + 
 
 X^^ CA, so is the difference of the 
 
 \v sides B C — C A to the diiference 
 
 -^ of the segments of the base B D 
 
 — D iV. Half this difference, 
 being added to and subtracted from half the base A B, will give 
 respectively the segments B D, DA; though only one of the seg- 
 ments is required to be found. Now, either of the sides, and its 
 adjacent segments constitute a right angled triangle, whence the 
 perpendicular C D may be found by Rule IL, Prob. I. 
 
 Formulae. 
 Put AB=flr, BC = 5 and C A = c ; then from the propor- 
 tion in the Rule 
 
 B D — D A = 'tjZLf ; whence 
 
 BT)=z\fa -f ^^ "-'^^ \ , and 
 
 examples. 
 
 1. The three sides of a triangle are 42, 40, and 26 feet; re- 
 quired the perpendicular on the longest side. 
 By the Rule 
 
 AB:BC + CA::BC — CA:BD — DA, that is, 
 42 : ^Q : : 14 : 22, and 
 
 i (42 — 22) = 10 feet = A D 
 Or by the last Formula 
 
 ■HA 1 / 402 — 263\* \^n. 
 D A = i /42- — -^ j = 10 feet, 
 
 and C D = ./AC^ — DA^ = ^262—102 = 24 feet. 
 
24 
 
 MENSURATION OF LINES. 
 
 2. The base of a triangle is 30, and the two sides 25 and 35 ; 
 required the perpendicular. Ans, 2Afeet 6 in. nearly, 
 
 3. A house 21 feet in width, has a roof with unequal slopes, 
 the lengths of which, from the eaves to the ridge, are 20 and 13 
 feet ; required the height of the ridge above the eaves. 
 
 Ans. \2feet. 
 
 Note. All the preceding examples may be readily solved by construction, 
 by first laying down the triangles, as in Prob. VI., Practical Geometry, and 
 then letting fall the perpendicular, as in Prob. V. 
 
 Problem IV. 
 
 The side A^qfa regular 'polygon being given to find the radii 
 O C and O A of its inscribed and circumscribed circles. 
 
 Rule. — Multiply the side of the polygon by the number op- 
 posite its name in the following Table, in the column headed 
 "Rad. Inscribed Circle," or in that headed "Rad. Circumsc. 
 Circle," accordingly as the one or the other radius may be re- 
 quired. • 
 
 Formula. 
 
 Let r and R be the radii of the in- 
 scribed and circumscribed circles respec- 
 tively, q and p their respective tabular 
 radii, and I = side of the polygon ; 
 
 then r =^lq, and R = /^ ; also 
 
 - IT "^ R ' 
 
 TABLE OF POLYGONS. 
 
 No. 
 
 
 Rad. 
 
 Rad. 
 
 
 of 
 
 Name. 
 
 Inscribed 
 
 Circums. 
 
 Area. 
 
 Sides. 
 
 
 Circle. 
 
 Circle. 
 
 
 3 
 
 Trigon or equi-triangle . 
 
 •2887 
 
 •5773 
 
 •4330 
 
 4 
 
 Tetragon or square . 
 
 
 •5000 
 
 •7071 
 
 1-0000 
 
 5 
 
 Pentagon . . . 
 
 
 
 •6882 
 
 •8506 
 
 1-7205 
 
 6 
 
 Hexagon . 
 
 
 
 
 
 •8660 
 
 1-0000 
 
 2-5981 
 
 7 
 
 Heptagon . 
 
 
 
 
 
 1-0383 
 
 1-1524 
 
 3-6339 
 
 8 
 
 Octagon 
 
 
 
 
 
 1-2071 
 
 1-3066 
 
 4^8284 
 
 9 
 
 Nonangon . 
 
 
 
 
 
 1-3737 
 
 1^4619 
 
 6-1818 
 
 10 
 
 Decagon 
 
 
 
 
 
 1-5388 
 
 1-6180 
 
 7-6942 
 
 11 
 
 IJndecagon 
 
 
 
 
 
 1-7028 
 
 1-7747 
 
 9'3^bQ 
 
 12 
 
 Dodecagon 
 
 
 
 
 
 1-8660 
 
 1-9319 
 
 1M962 
 
MENSURATION OF LINES. 25 
 
 EXAMPLES. 
 
 1. The side of a regular pentagon is 5 feet 1 inch, what are 
 the radii of its circumscribed and inscribed circles ? 
 
 Ans. Aft, 3*7 in.y and nearly 3 ft. 6 in, 
 
 2. The side of an octagonal grass plot is 203^ yards, and four 
 walks are made therein, joining the middle of each of the op- 
 posite sides ; required the united length of the four walks. 
 
 A71S. 1998 yards. 
 
 3. A circular grass plot of 50 yards diameter is to be com- 
 passed by a regular octagonal iron pailing, and the eight spaces, 
 between the grass plot and the pailing, to be planted with 
 shrubs ; required the whole length of the pailing, and the greatest 
 width of each of the eight spaces. 
 
 Problem V. 
 
 To find the diameter and circumference of a circle, the one 
 from the other. 
 
 Rule I. — As 7 is to 22, so is the diameter to the circum- 
 ference. 
 
 As 22 is to 7, so is the circumference to the diameter. 
 
 Rule II. — As 1 is to 3 "14 16, so is the diameter to the cir- 
 cumference. 
 
 As 3*1416 is to 1, so is the circumference to the diameter. 
 
 formula. 
 
 Let d = diameter, c = circumference, and 7r= 3*1416;* then 
 
 c 
 c^= d IT, and d = — , 
 
 TT 
 
 examples. 
 
 1. To find the circumference of a circle, whose diameter is 10. 
 By Rule 1. 
 7 : 22 : : 10 : 31f 
 
 10 Bf Ha 
 
 7 ) 220 
 31f 
 or 31*42857^^5 
 
 * The true circumference of a circle, the diameter of which is unity, is 
 3-14159265358979, &c. This number has been determined by Machin to'lOO 
 places of decimals, by others to still more ; but these results are more curious 
 than useful, since the first four or five decimals are quite suflacient for all 
 practical purposes. 
 
 2 
 
26 MENSURATION OF LINES. 
 
 By Rule II. or first Formula c = dir = SV416, which is 
 nearer the truth. 
 
 2. To find the diameter when the circumference is 50 feet. 
 
 By Rule I. 
 
 7 X 25 175 
 22 : 7 : : 50 : — j — = -jj = ^m =15-9090 Ans. 
 
 c 
 Bv Rule II. or second Formula df= - = 15-9156 feet. 
 
 3. If the diameter of the earth he 7958 miles, as it is very 
 nearly, what is the circumference, supposing it to be exactly 
 round? Ans. 25000-8528 miles, 
 
 4. To find the diameter of the globe of the earth, supposing 
 its circumference to be 25000 miles. Ans, 725 7 J nearly, 
 
 5. Required the diameter of a coach wheel, that turns round 
 500 times in travelling a mile. Ans. 3 ft, 5*05 in, 
 
 6. The driving wheel of a locomotive engine is 6 feet in 
 diameter, how often does it turn in a second, when travelhng at 
 the rate of 60 miles in an hour ? Ans, 4f times nearly. 
 
 Problem VI. 
 
 The chord (B E) and the height or versed sine (C D) of an 
 arc (B C E) o/* « circle being given to find the diameter (A C) 
 and the chord of half the arc (B C). 
 
 Rule. — Divide the square of half the chord B E, by the 
 height C D ; to the quotient add C D, and the sum will be the 
 diameter A C ; half of which is the radius B O or C O. 
 
 The chord B C of half the arc is found by Prob. I. 
 
 FORMULiE. 
 
 G Put C = |- chord of the whole arc = 
 
 i B E = B D, c = chord of i the arc = 
 ^ Ny B C, w = height or versed sine = C D 
 
 and d = diameter A C • then 
 
 B^ 
 
 ; C3 
 
 d=~+v (1), c= VC3 + z;3(2) 
 
 V 
 
 Also d=:^{3),v~{A),c:=^ s/dv(5). 
 
 EXAMPLE. 
 
 . The chord of an arc is 48 feet, and its height 18 ; required the 
 diameter of the circle of which the arc is a part. 
 
MENSURATION OF LINES. 27 
 
 By the Rule 
 
 24 = i B E = 
 24 
 
 BD 
 
 CD = 
 
 18)576 
 
 32 
 
 18 = CD 
 
 
 50 feet = A C. 
 whence the radius B = 25 feet. 
 
 
 2. The span (chord) of the iron arch of Sunderland bridge 
 is 240 feet, and the rise (height) of the crowns of the arch 
 34 feet : with what radius w^as the arch drawn ? 
 
 By Formula (1). 
 
 1203 
 The diameter = -34"+ 34 = 440*41 feet. 
 
 whence the required radius = 440*41 ~- 2 = 220*205 = 220 
 feet 2^ inches. 
 
 3. On a parliamentary map of 4 chains to an inch, the chord of 
 a railway curve measured 40 inches, and its height 5 inches, re. 
 quired the radius of the curve both on the map and on the ground- 
 
 20^ 
 The diameter = -r- + 5 = 85 inches, whence 
 
 the radius = 85 -r- 2 = 42|- inches on the map. 
 
 And, since the scale of the map is 4 chains to an inch, we 
 shall have 42^ X 4 = 170 chains = 2| miles, the radius of the 
 curve on the ground. 
 
 4. The chord of the whole arc is 48 feet, and its height 7 ; 
 required the chord of half the arc. • 
 
 By Formulae (2). 
 
 c = \/C^ + v^ = \/24^ + 73 = 25 feet, the required chord. 
 
 5. The. chord of half the a^c of a bridge is 24 feet, and the 
 rise of the cro^n of the arc 16 feet; required the radius of tiie 
 circle of which the arch is a part. 
 
 Ans. By formula (3) the diameter is found = 36 feet, whence 
 the required radius is 18 feet. 
 
 6. A circular grass plot of 100 yards diameter is cut by a walk 
 through the centre, this walk is cut at right angles by another 
 walk through the middle of the radius ; required the length of 
 the last named walk. 
 
 By transposing formula (1) C = >/ v (d—v) = V25 (100—25) 
 = 43*3 yards the double of which is the length of the walk. — 
 The same result may be obtained from the right angled tri- 
 angle B D O. 
 
28 MENSURATION OF LINES. 
 
 7. The rise of the circular arch of a bridge is 12 feet and the 
 radius of the whole circle is 100 feet; required the distance 
 from the spring of the arch to the crown, viz., the chord of 
 half the arch. Ans, A9 feet nearly. 
 
 Problem VII. 
 
 To find the length of any arc of a circle. 
 
 Case I. — When the degrees in the arc and the radius are given. 
 
 Rule I. — As 180° is to the number of degrees in the arc, so 
 is 3*1416 times the radius to its length. 
 
 Case II. — When the chord of half and the whole arc are given. 
 
 Rule II. — From 8 times the chord of half the arc subtract 
 the chord of the whole arc, and take \ of the remainder for the 
 length of the arc nearly. 
 
 FORMULAE. {See last figure^ 
 
 Put r = radius B O, A = 180°, S = degrees in the arc B E, 
 and TT = 3* 1416, and I = length of the arc ; then 
 
 rhiT ^ Z A 
 
 I -^>andr = ;^ 
 
 EXAMPLES. 
 
 1 . To find the length of an arc of 30 degrees, the radius being 
 9 feet. 
 
 By Rule I. 3-1416 
 
 9 
 
 180 : 30 : : 28*2744 : 47124 feet. 
 9x30x3-1416 3x3-1416 
 By first Formula I = jg^ == 2 =4-7124 
 
 2. The length of the arc of a circle of 30 degrees is 9 feet 
 5 inches, required its radius. 
 
 Ans. By the second formula, IS feet nearly. 
 
 3. The chord B E of the whole arc being 4*65374 feet, and 
 the chord B C of the half arc 2*34947 ; required the length of 
 the arc. 
 
 By Rule II. 2*34947 
 
 ^ 8 
 
 18-79576 
 4-65874 
 
 3)14-13702 
 
 Ans. 4- 71 234 /cc^. 
 
MENSURATION OF LINES. 29 
 
 3. Required the length of an arc of 12 degrees 10 minutes, or 
 12^ degrees, the radius being 10 feet. 
 
 By Rule I., 2' \2Z4 feet, Ans. 
 
 4. Required the length of the iron arch, in example 2, 
 Prob. VI. 
 
 First, the chord of J the arch, or distance from spring to 
 crown, by Formula 2, Prob. VI., will be found 124'724 feet. 
 Whence, by Rule II. of this Problem, we shall have the re- 
 quired length of the arch = 252 feet 7 inches. 
 
 5. Find the length of one of the arcs of the six equal seg- 
 ments of an iron girder, the whole span of the arch being 120 
 feet, and the radius 180. Ans. 20 feet 4*67 inches. 
 
 Rule III. is not sufficiently accurate for finding the length of 
 the arc, when it is greater than ^ of the circumference of the 
 circle : in such cases, (see figure to Prob. VI.) the chord of ^ 
 of the arc B C E = chord of |- the arc B C (not shown in the 
 figure) must be found by the formula. 
 
 Chord of i of arc B C E = V i ^ (^ — \7 ^^ — c^). 
 
 in which d and c are the same as in Prob. VI. ; after which Rule 
 II. may be applied with sufficient accuracy to find the length 
 of the ^ arc B C, which, being doubled, will give the whole 
 length B C E. 
 
 6. Required the length of a circular iron girder, the span 
 (B E) of which is 48 feet, and the rise (C D) at the crown 
 18 feet. 
 
 By Formulae 1 and 2, Problem VI., d = AC is found = 50 
 feet, and c = B C = 30 ; whence, by the formula just given, 
 
 the chord of i of arc B C E = V 25 (50— V50^— 30'0 ^= 
 15-8113, and by Rule II., (15*8113 X 8 — 30) -r- 3 = 32-1635 
 feet = arc B C, the double of which is 64-3270 feet = the re- 
 quired length of the arch B C E. But by using Rule II., 
 without the above formula, the length of the arch is found to 
 be 64 feet, or nearly 4 inches short of its more accurate length, 
 as previously found. 
 
 Note. The true method of finding the length of an arc of a circle is to find 
 the natural sine of the angle BOD (figure to Prob. VI.) and its corresponding 
 number of degrees, minutes, &c., which, being doubled, give the angular mea- 
 sure of the whole arc B C E ; whence the length of the arc may be accurately 
 found by Rule I. But the first part of this operation is the province of Tri- 
 gonometry ; moreover, sufficient accuracy for all practical purposes may be 
 obtained by Rule II. for arcs less than a quadrant ; and like accuracy may be 
 
30 MENSURATION OF LINES. 
 
 secured by means of the formula used in Example 6, in cases where the arc 
 approaches near to a semicircle. 
 
 Remark. — When the arc is greater than a semicircle, the remaining part of 
 the circumference must be found by Rule II., with the help of the formula 
 used in Example 6, if necessary. This remark does not apply to cases where 
 the degrees of the arc are given, which are solved by Rule I. 
 
 Problem VIII. 
 
 To find the diameter of a circular zoney its two parallel chords 
 A B, CD;, and its breadth E F, being given. 
 
 I^ This and the following Problems 
 
 may be omitted by the student, as not 
 ^ being much required in practice. 
 
 E, 
 
 ,_^ \\h formulae. 
 
 — 7D 
 
 Let C and c be the half chords 
 C F and A E respectively, b the 
 breadth E F, and d the diameter 
 K L = twice radius O B ; then 
 
 GH 
 
 Also AC = BD= v(j2 . c— c)' ^°^ 
 
 10 ft. 
 
 EXAMPLES. 
 
 1. The parallel sides of a circular zone are 6 and 8 feet, and 
 its breadth 7 feet ; required the diameter of the circle. 
 
 By the first Formulae the diameter 
 
 r? = V I 72 + 2 (42 + 32) + ( ^ T^T } = ^/ 49 H- 50 + 1 = 
 
 2. Find the chord BD and the height G H of the zone in 
 the preceding Example. 
 
 Here, the diameter d is first found, as above ; then by the 
 second and third Formulae, 
 
 B D = V (72 H- 4 — 32) = V49 + 1 = 7-07 feet, and 
 GH = I 10 — i ^/|(4 + 3)2 + ^l!z-_i!j' j = 5_ X V4£rrr= 1-465 ft. 
 
 3. The parallel chords of a zone are the same as in the Ex- 
 ample 1, and its breadth 1 foot ; required the diameter. 
 
 Ans, 14 feet. 
 
MENSURATION OF LINES. 
 
 31 
 
 Note. In this example the two chords are l)oth on the same side of the 
 centre of the circle. 
 
 4. The two parallel chords of a circular zone are 16 and 12 
 feet, and the diameter of the circle 20 feet ; required the hreadth 
 of the zone. Ans. 14 feet. 
 
 Note. 1 . The breadth of the zone, in this example, is found by squaring and 
 transposing the first formula, whence there results a quadractic equation, from 
 which the value of b is found. 
 
 Note 2. When the chord B D = A C, and the height G H have been found, 
 the lengths of the equal arcs A C, B D are found by the Prob. VII. 
 
 Problem IX. 
 
 In an ellipse are given any three of the four following parts to 
 find the fourth, viz. the transverse axis T R, the conjugate axis 
 C O, the abscissa H Q, and the ordinate P Q. 
 
 semiconjugate = C H» 
 = P Q ; then 
 
 FORMULA. 
 
 Put a = semitransverse = H R, h ■ 
 X = abscissa = H Q, and y = ordinate 
 
 a b 
 
 0? = - />/ b'^ — y^, y = - Va^ — x^, 
 b a 
 
 a = _^ and b =-^g=== 
 
 ^b^ — y^ V«^— ^2 
 
 Also the focial distance from 
 the centre. 
 
 HF==H/= V«^^^^ 
 
 EXAMPLES. 
 
 1. The transverse axis is 30, the conjugate 20, and the 
 abscissa 3 feet. 
 
 By the second formula, 
 
 P Q = y = If V 1^^—32 = 9-798 feet. 
 
 2. The transverse T R = 70 feet, the conjugate C O = 50, 
 and the ordinate P Q == 20 ; required the abscissa H Q. 
 
 Ans. By the first formula, H Q = 21 feet. 
 
 3. The transverse is 180 inches, the ordinate 16, and the 
 abscissa 54 ; required the conjugate. 
 
 Ans. By the fourth formula, the conjugate = 40 inches, 
 
 4. If the conjugate be 50 feet, the ordinate 20, and the 
 abscissa 2 1 ; v^hat is the transverse ? 
 
 Ans, By the third formula, the transverse = 1^ feet. 
 
32 MENSURATION OF LINES. 
 
 5. The transverse TR= 100 yards, and the conjugate C O 
 = 60 ; required the distance of the foci F/from the centre H. 
 
 Ans, By the last formula, H F = Hy = 40 yards. 
 
 6. The ratio of the major and minor axes of the earth's 
 orbit is as 1 to w, the former being about 190,000,000 miles 
 = 2 05, How much is the earth nearer to the sun in winter than 
 in summer? 
 
 Ans. The distance here required is twice the focial distance 
 from the centre of the earth's elliptical orbit, which, by the last 
 formula is found to be 2 « >/Y3^. 
 
 7. Required the distance of the foci of an elHptical section, 
 passing through the poles of the earth, the earth's axes being 
 7926 and 7899 miles. 
 
 Ans, 654 milesy or 327 miles each from the earth's centre. 
 
 Problem X. 
 
 The axes of an ellipse are given to find its circumference. 
 
 Rule I. — Multiply half the sum of the two axes by 3* 14 16, 
 and the product will give an approximate length of the circum- 
 ference, which ivill be found near enough for most practical 
 •purposes. 
 
 Rule II. — To half the sum of the two axes add the square 
 root of half the sum of their squares, and multiply half the sum 
 by 3*1416 for the circumference very nearly. 
 
 FORMULAE {see last figure). 
 
 Let 2 a and 2 5 represent the axes, as in the last problem, 
 and TT = 3*1416 ; then, 
 Circumf. = tt (« + 5), or == i tt (« + 5 + ^Y'(^ + b^)). 
 
 EXAMPLES. 
 
 1. The axes of an ellipse are 15 and 10 feet; required the cir- 
 cumference by Rule I. 
 
 Ans. 39 feet 3 inches. 
 
 2. The axes being the same as in the last example ; required 
 the circumference by Rule II. 
 
 Ans. 39 feet 8 inches nearly. 
 
 3. Find the meridional circumference of the earth, the axes 
 being as given in the last example of Prob. IX. 
 
 Ans. 24,858 miles nearly. 
 
MENSURATION OF LINES. 
 
 Problem XI. 
 
 33 
 
 In a parabola I V II, the focus of which is F, any two of the 
 three following parts, viz., the parameter P Q, the abscissa V G, 
 and the ordinate G II being given, to find the third part. 
 
 FORMULyE. 
 
 Put P Q = parameter = jo, V G == 
 abscissa = x, and C H = ordinate = y ; 
 then 
 
 X = 
 
 =y\ ,, = 
 
 y == ^ p X, SLlidp =^. 
 
 p X 
 
 EXAMPLES. 
 
 1. The parameter P Q of a parabola is 50, and its ordinate 
 G H == 60 feet, required the abscissa V G. 
 
 J?is. By the first formulae; = G H = Vtt = ^"2. 
 
 2. The parameter of a parabola is 10, and its ordinate 4; 
 required the abscissa. Ans. 1*6. 
 
 3. The abscissa of a parabola is 4, and its corresponding 
 ordinate 10 ; required the parameter. Ans. 25. 
 
 Problem XII. 
 
 To find the length of the arc of a parabola y its ordinate and 
 abscissa being given, (See last figure.) 
 
 FORMULA. 
 
 Let X and y represent the same parts, as in the last ProbiCm ; 
 then The i arc V II = VF^ +1/^ i^early 
 
 EXAMPLES. 
 
 1 . Required the half arc V Q of a parabola, V F being = 3 
 feet, and F Q = 6. 
 
 Ans. V Q = v^A 33 ^ Q2 ^ Qfget 11| inches. 
 
 2. The abscissa is 2, and the ordinate 6 ; required the length 
 of the half arc of the parabola. Ans. 6.4291. 
 
 Note. 1. The parabola is the path of projectiles in vacuo ; it is also used 
 in the astronomical theory of comets. 
 
 Note 2. The student who wishes for further information concerning this 
 curve, as well as concerning the ellipse and hyperbola, may consult the various 
 works on conic sections. 
 
34 
 
 MENSURATION OF SUPERFICIES. 
 
 PART III. 
 
 MENSURATION OF SUPERFICIES OR SURFACES. 
 
 The area of any surface is estimated by the number of squares 
 in that surface, without regard to its thickness ; the side of those 
 squares being one inch, one foot, one yard, &c. Hence the area is 
 said to be so many square inches, or square feet, or square yards, &c. 
 
 A TABLE OF SQUARE MEASURE. 
 
 Sq. Inches. 
 
 Sq. Feet. 
 
 
 
 
 
 144 
 
 1 
 
 Sq. Yards. 
 
 
 1,296 
 
 9 
 
 1 
 
 
 
 
 
 
 Sq. Poles. 
 
 
 
 39,204 
 
 272:^ 
 
 30^ 
 
 1 
 
 Sq. Rods. 
 
 
 
 1,089 
 
 1,210 
 
 40 
 
 1 
 
 Acres. 
 
 
 43,560 
 
 4,480 
 
 160 
 
 4 
 
 1 
 
 
 
 3,097,600 
 
 102,400 
 
 2,560 
 
 640 
 
 Sq. Mile. 
 1 
 
 Problem I. 
 
 To find the area of a 'parallelogram ; whether it he a square^ 
 a rectangle^ a rhomhus, or a rhomboid. 
 
 Rule. — Multiply the length by the breadth or perpendicular 
 height, and the product will be the area. 
 
 rORMULJE. 
 
 Let I = length of the figure, b = 
 its breadth, and A = its area {which 
 also represents the areas in all the 
 following Problems) ; then 
 
 A A 
 
 K =^ Ibi also / = — , and b = — . 
 b I 
 
 When the figure is a square, then 
 the length is equal to the breadth, 
 which put = 5 = side of the square ; 
 then 
 
 A = 5^ and s =^ s/ A., 
 
 EXAMPLES. 
 
 1. The length of a rectangular board is 7 feet, and its breadth 
 4 feet; required its area in square feet. (See first figure.) 
 
 By the Rule. 7 X 4 = 28 square feet, the area required. 
 
 2. The side of a square is 18 inches; required its area in 
 square feet. (See last figure.) 
 
 4 ;--^---J— -j- p^ 
 
MENSURATION OF SUPERFICIES. W 
 
 18 
 18 
 
 ri2 
 144 J. 
 112 
 
 324 
 
 27 
 
 2 J square feet, the area required. 
 
 3. Find the area of a rhombus, the length of which is 6*2 feet, 
 and its perpendicular breadth 5 '45. (See second figure.) 
 
 Ans. 33*79 = 33f square feet nearly, 
 
 feet, in. 
 
 4. The length of a table is 7 feet 8 inches, 7 8 
 and its breadth 3 feet 10 inches; required its 3 10 
 area. 
 
 23 
 Here the operation is performed by duo- 6 4 8 
 
 decimals, and the area is found to be 29 square 
 
 feet, 4 inches or 12ths, and 8 parts or yf^ths. 29 4 8 
 
 5. "What length must be cut off a rectangular board, the 
 breadth of which is 9 inches, to make a square yard? 
 
 A square yard contains 1296 square inches, whence by the 
 second formula. 
 
 9)1296 
 
 144 inches = 12 feet, the length required. 
 
 6. How many square feet of deal will make a box 6 feet long, 
 5 broad, and 2 feet 8 inches deep ? 
 
 Ans. 116 square feet 2' 8". 
 
 7. How many square yards are contained in a floor 23 feet 
 long 14^- feet wide? Ans. ^1^-^ square yards. 
 
 8. The base of the largest Egyptian pyramid is a square, the 
 side of which is 693 feet; required the number of acres it 
 occupies. Ans. \\a. Or. 4p, 
 
 9. A square court yard is 42 feet long, and 23 feet lOJ inches 
 broad ; what did it cost paving at 4*. \Qd. per square yard. 
 
 Ans. £26 \Ss. 6id. 
 
 10. Required the side of a square, the area of which is 500 
 square feet. 
 
 By the fourth formula s = side of the square = V A, that is 
 s = V500 = 22-3607 feet = 22 feet 4^ inches nearly. 
 
 11. "What is the side of a square the area of which is an 
 acre? Ans. 69' 6 yards nearly, 
 
 12. A square in a city contains 6^ acres of ground, required 
 the side of the square. Ans. 173-92 yards. 
 
36 
 
 MENSURATION OF SUPERFICIES. 
 
 Problem IL 
 
 To find the area of a triangle. 
 
 EuLE, I. — Multiply tlie base by the perpendicular heigh t, 
 and take half the product for the area. 
 
 Rule II. — When the three sides only are given : add the 
 three sides altogether, and take half the sum ; from the half 
 sum subtract each side separately ; multiply the half sum and 
 the three remainders continually together ; and take the square 
 root of the last product for the area of the triangle. 
 
 FORMULA. 
 
 Let the base A B = 6, and the 
 C perpendicular CD =j9 ; then 
 
 == — , and 6= — 
 h p 
 
 When all the three sides of the 
 triangle are given, let them be re- 
 presented by a, b and c, and their 
 half sum by s ; then 
 
 A = Vs{s — a) (s — b) (s — c). 
 
 EXAMPLES. 
 
 1. Let the base AB = 42 feet, and the perpendicular CD=33 
 feet ; required the area in square yards. 
 
 By Rule I. 42 X 33-4-2=693, and 693-^9=77 square yards. 
 
 2. To find the number of square yards in a triangle, the sides 
 Df which are 13, 14, and 15 feet. 
 
 By Rule II. 
 
 13 21 
 
 14 6 
 
 ID 
 
 2)42 
 
 
 
 126 
 
 7 
 
 \ sum 21 
 13 
 
 21 
 14 
 
 7 
 
 21 
 15 
 
 6 
 
 882 
 8 9 
 
 remainders 8 
 
 7056(84 square feet, 
 64 — 9-^ sq, yds. Ans 
 
 
 
 
 164)656 
 6b6 
 
MENSURATION OF SUPERFICIES. S/ 
 
 3. The base of a triangle is 40, and its perpendicular 30 feet ; 
 required the area in square yards. A71S. 66^ square yards, 
 
 4. Find the area of a triangle, the three sides of which are 
 20, 30 and 40 feet. Ans, 32-27 square yards. 
 
 5. The base of a triangle is 49 and its height 25^ feet, how 
 many square yards does it contain ? 
 
 Ans. 68*736 square yards, 
 
 6. The base of a triangle is 18 feet 4 inches, and its height 11 
 feet 10 inches; required the area. Ans. 108 feet 5' 8". 
 
 7. The hypothenuse of a right angled triangle is 102^^ feet, 
 and its base 100 ; required the area in square yards. 
 
 Ans, 125 square yards, 
 
 8. The side of an equilateral triangle is 5*1 feet, required the 
 area. Ans, ll'2626 square feet. 
 
 9. The base of a triangle is 121 yards ; required its perpendi- 
 cular, when it contains an acre of land. Ans, 80 ya7'ds, 
 
 10. The equal sides of an isosceles triangle are each 50 feet, 
 and its base 28 ; how many square yards does it contain ? 
 
 Ans. 74|- square yards. 
 
 Problem III. 
 To find the area of a trapezoid. 
 
 Add together the two parallel sides ; multiply that sum by the 
 perpendicular distance between them, and take half the product 
 for the area. 
 
 EXAMPLES. 
 
 1. In a trapezoid the parallel lines are A B 7'^i and D C 12-25, 
 also the perpendicular distance A P is 15.4 feet; required the 
 area. 
 
 12-25 
 7-5 
 
 19-75 A B 
 
 15-4 
 
 7900 
 9875 
 1975 
 
 2)304-150 
 152075 square feet^ Ans, 
 
38 MENSURATION OF SUPERFICIES. 
 
 2. How many square feet contains the plank, whose length is 
 12 feet 6 mches, the breadth at the greater end 1 foot 8 inches, 
 and at the less end 11 inches? A^is \Z\\feet. 
 
 3. Required the area or a trapezoid, the parallel sides being 21 
 feet 3 inches and 1 8 feet 6 inches, and the distance between them 
 8 feet 5 inches. Ans. 167 square feet, 3' 4" &" , 
 
 Problem IV. 
 
 To find the area of a trapezium. 
 Case I. — For any trapezium. 
 
 Divide it into two triangles by a diagonal ; then find the areas 
 of these triangles, and add them together. 
 
 Or, if two perpendiculars be let fall on the diagonal, from the 
 other two opposite angles, the sum of these perpendiculars being 
 multiplied by the diagonal, half the product will be the area of 
 the trapezium. ^ 
 
 Case II. — When two opposite angles are supplements of each 
 other. 
 
 Add all the four sides together, and take half the sum ; next 
 subtract each side separately from the half sum ; then multiply 
 the four remainders continually together, and take the square root 
 of the last product for the area of the trapezium. 
 
 EXAMPLES. 
 
 1 . To find the area of the trapezium A B C D, the diagonal 
 A C being 42, the perpendicular B E 18, and the perpendicular 
 DF16. 
 
 18 
 B 16 
 
 ^y'^K 34 Sum 
 
 \.i/ 136 
 
 2)1423 
 714 Ans, 
 
 2. In the trapezium A B C D, the side A B is 15, DC 13, 
 CD 14, AD 12, and the diagonal A C is 16 : required the 
 area. 
 
MENSURATION OF SUPERFICIES. 39 
 
 AC 16 
 AB 15 
 BC 13 
 
 
 
 AC 16 
 CD 14 
 AD 12 
 
 
 
 2)44 
 22 
 16 
 
 22 
 15 
 
 22 half sum 
 13 
 
 2)42 
 21 
 16 
 
 21 
 14 
 
 21 half sum 
 12 
 
 6 
 
 7 
 
 7 
 
 9 
 
 5 
 
 7 
 
 7 
 
 9 
 
 42 
 9 
 
 
 
 35 
 9 
 
 
 
 378 
 22 
 
 
 
 315 
 21 
 
 
 
 756 
 756 
 
 
 
 315 
 630 
 
 
 
 V'8316 = 9M921 
 
 The triangle ABC .. 
 The triangle ABC 
 
 V(3615 = 
 
 .. 9M921 
 
 , .. 81-3326 
 
 =81.3326 
 
 The trapezium ABC D 172-5247, Ans. 
 
 3. If a trapezium have its opposite angles supplements to each 
 other, and have four sides 24, 26, 28, 30 ; required its area. 
 
 By Rule II. the area is 723-989. 
 
 4. How many square yards of paving are in the trapezium, 
 the diagonal of which is 65 feet, and the two perpendiculars let 
 fall on it 28 and 38-5 feet? Ans. 222^-^ yards. 
 
 5. What is the area of a trapezium, the south side being 
 27-40 chains, east side 35-75 chains, north side 37-55 chains, 
 west side 41-05 chains, and the diagonal from south-west to 
 north-east 48*35 chains? Ans. I23a. Or. 11.8656^. 
 
 6. What is the area of a trapezium, the diagonal of which is 
 108|- feet, and the perpendiculars 65^ and 60f feet. 
 
 A71S. 705^ square yards. 
 
 7. What is the area of a trapezium, the four sides being 
 12, 13, 14, 15? having its opposite angles supplemental. 
 
 Ans. 180-997. 
 
 8. In the four sided field A B C D, on account of obstructions 
 in the two sides A B C D, and in the perpendiculars B F, D E, 
 the following measures only could be taken: namely, the two 
 sides B C 265 and A D 220 yards, the diagonal A C 378 yards. 
 
40 
 
 MENSURATION OF SUPERFICIES. 
 
 and the two distances of the perpendiculars from the ends of the 
 diagonal, namely A E 100, and C F 70 yards: required the 
 area in acres, -when 4840 square yards make an acre. 
 
 Ans. 17a. 2r. 2lp, 
 9. When A B = 314. B C = 232, C D = 228^, D A = 266| 
 
 and the diagonal A C 
 yards. 
 
 : 4 1 7^ feet ; required the area in square 
 Jns. 864 ^ square yards. 
 
 Problem V. 
 
 To find the area of an irregular polygon. 
 
 Rule. — Draw diagonals dividing the figure into trapeziums 
 and triangles ; then find the areas of all these separately, and 
 add them together for the content of the whole figure. 
 
 examples. 
 1 . To find the content of the irregular figure ABCDEFGA, 
 "" in which are given the follow- 
 
 ing diagonals and perpendicu- 
 lars : namely, 
 
 AC 5-5 
 5-2 
 4-4 
 1-3 
 1-8 
 1.2 
 0-8 
 2-3 
 3rd 
 For triangle GCD. 
 4-4 
 2-3 
 
 GDEF. 
 
 31 
 5-5 
 
 1-55 
 15-5 
 
 2-0 
 5-2 
 
 10-4 
 
 132 
 
 88 
 
 10-12 
 
 1705 double ABCG 
 10-40 double GDEF 
 10-12 double GCD 
 
 2)37*57 double the whole. 
 
 18-785 Ans, 
 
MENSURATION OF SUPERFICIES. 
 
 41 
 
 2. Required the area of the figure A B C D E F G, when 
 
 AC= 12, FD = 11, GC = 9i, Gm = 3}, Bn- 
 Ep = li, and I>q=^4^ feet. 
 
 : 4, G = 2i 
 
 Problem VI. 
 
 To find the area of a regular polygon » 
 
 Rule I. — Multiply the sum of the sides or perimeter of the 
 polygon by half the perpendicular from its centre to one of its 
 sides, and the product will be the area. 
 
 Rule II. — Multiply the square of the side of the polygon by 
 the number opposite its name, in the column headed ** Areas," in 
 the Table to Prob. IV., Part II. and the product will be the area. 
 
 FORMULAE. 
 
 Let 5 = A B = side of the polygon, 
 jp := CP perpendicular from the centre 
 on A B, w = number of sides of the po- 
 
 lygon, and a = its tabular area; 
 A = ^np s, and A, = a s^. 
 
 then 
 Also 
 
 ^ = v/-= 
 
 A 2 A 
 
 71 p 
 
 A 2A 
 
 and /3 = 
 
 n s 
 
 examples. 
 
 1 . Required the area of a regular pentagon, the side A B of 
 which is 25 feet, and the perpendicular C P == 17*205. 
 
 By Rule I. 
 
 17-205 
 25 X 5 = 125 
 
 = perim. 
 
 By Rule II. 
 1-7205 table area, 
 625 = 252 
 
 86025 
 34410 
 
 17205 
 
 86025 
 34410 
 103230 
 
 2)2150-625 
 
 1075-3125 sq.feet. 
 
 1075'3125 sq.feet. Ans. 
 
 2. To find the area of the hexagon, the side of which is 20 
 feet. Ans. 1039*23 square feet. 
 
 3. To find the area of the trigon, or equilateral triangle the 
 side of which is 20 feet. Ans. 173-205 square feet. 
 
 4. Required the area of an octagon, the side of which is 20 
 feet. Ans. 1931-37 square feet. 
 
42 MENSURATION OF SUPERFICIES. 
 
 5. What is the area of a decagon, the side of which is 20 
 feet. Ans. 3077*68 square feet, 
 
 6. Required the side of a decagon, the area of which is 16 
 square feet. 
 
 A 
 
 By the third formula, the side5=/v/ — , that is, 
 
 a 
 
 ^'T-'KoJo "^ 1*442 /ee^ = I foot 5*3 inches. Ans, 
 
 7. The fence of an octagonal inclosure, within a square in a 
 city, cost ^840 at 45. Sd, per foot ; what will be the cost of the 
 gravelling the surface at lO^d. per square yard? 
 
 Ans. £IS2 Os. G^d. 
 
 8. The corners of a square are cut off so as to form an octagon; 
 required the area of the octagon, the side of the square being 
 200 feet. Ans. 3681*8 square yards. 
 
 Problem VII. 
 
 To find the area of a circle ivhen the radius, or half diameter 
 is given. 
 
 Rule I. — Multiply the square of the radius by 3*1416 for the 
 area. 
 
 To find the area of a circle when the circumference is given. 
 
 Rule II. — Multiply the square of the circumference by 
 •07958. 
 
 Put the radius A C = r, the circum- 
 ference = c, and 3*1416 = tt ; then 
 
 ^ ' A = TT r^i and r =\/— ; also 
 
 TT 
 
 A= - = \ r c, and c == //4A tt, 
 47r 
 
 examples. 
 
 1. Required the area of a circle, the "radius of which is 
 5 feet. 
 
 By Rule I., or the first formula. 
 
 3*1416 X 52=3*1416 X 25 = 78*54 square feet. 
 
 2. The circumference of a circle is 18*4 feet, what is its area? 
 
 Ans. 26*92 square feet. 
 
 3. A circular pleasure ground is to be laid out to contain 
 
MENSURATION OF SUPERFICIES. 43 
 
 exactly an acre, required the length of the chord with which the 
 circle must be traced. 
 
 By the second formula, the length of the chord, or 
 r = VoTTTTg = ^^i y^^^^ ^^^y nearly. 
 
 4. How many square yards are in a circle whose diameter is 
 3} feet? Ans. 1-069. 
 
 5. How many square feet does a circle contain, the circumfe- 
 rence being 10* 995 6 yards. Ans. 86*19543. 
 
 6. The area of the piston of a steam engine is required to be 
 1192 square inches to give it the requisite power; required the 
 interior diameter of the cylinder, and its exterior circumference 
 the thickness of the metal being one inch. 
 
 J J Interior diameter 39 inches nearly, 
 
 ' \ Exterior circumference \0 feet 8f inches, 
 
 7. The circumference of the circular paling of a plantation was 
 found to be 235^ yards, what is its area. 
 
 Ans, 4400 square yards, 
 
 8. What is the circumference of a circle, the area of which is 
 an acre? A7is, 246 yards \ foot \0^ inches. 
 
 Problem VIII, 
 
 To find the area of a sector of a circle. 
 
 Rule I. — Multiply the radius, or half the diameter, by half 
 the arc of the sector, for the area. Or, multiply the diameter 
 by the arc of the sector, and take \ of the product. 
 
 Note. The arc may be found By Prob. III. 
 
 Rule II. — As 360 is to the degrees in the arc of a sector, so 
 is the whole area of the circle, to the area of the sector. 
 
 Note. For a semicircle take one half, for a quadrant, one quarter, &c., of 
 the whole circle. 
 
 FORMULiE. 
 
 1 2 A 
 
 A = — r X arc, and r = — . 
 
 2 arc 
 
 example. 
 
 1 . What is the area of the sector C A D B, the radius being 
 10, and the chord A B 16? 
 
44 
 
 MENSURATION OF SUPERFICIES. 
 
 By Rule 1 100 =:AC^ 
 
 64 = AE2 
 
 36(6 = CE 
 10 = CD 
 
 
 4==DE 
 
 
 16 = DE3 
 
 G4 = AE3 
 
 
 80 (8-9442719 = 
 8 
 
 = AD 
 
 71-5541752 
 16 
 
 
 3)55-5541752 
 
 2)18-5180584 
 
 9-2590297 
 
 10 
 
 arc ADB 
 
 = half arc 
 = radius 
 
 92-590297 Ans. 
 2. Required the area of a sector, the arc of which contains 96 
 degrees, the diameter being 3 feet. 
 
 •7854 = i7 
 9 = 32 
 
 area of the whole circle. 
 
 7-0686 
 Then by Rule II., 
 
 as 360° : 96° : : 7*0686 
 or, as 30° : 8° : : 7*0686 : 1-88496 square feet. Ans, 
 
 3. What is the area of a sector, the radius of which is 10 feet. 
 
 and the arc 20 ? 
 
 Ans, \\\ square yards. 
 
 4. Required the area of a sector, the radius of which is 18 
 feet, and the chord of its arc 12? Ans. 110^ square feet, 
 
 5. How many square yards are in a sector of 187° 37', the 
 radius of the circle being 289 ? Ans. 15194 square yards. 
 
 6. Required the area of a sector, the radius of which is 25 
 feet, and its arc contains 147° 29'. 
 
 Ans. 804*4 square feet nearly. 
 
 7. What is the area of a sector, the chord of the arc of which 
 is 24 feet, and its height 6 ? Ans. 208.572 square feet. 
 
MENSURATION OF SUPERFICIES. 45 
 
 8. Required the area of a sector greater than a semicircle, the 
 chord of its arc being 12, and its diameter 15 feet. 
 
 Arts. 124-2- square feet. 
 Problem IX. 
 To find tjie area of a segment of a circle. 
 
 Rule I. — Find the area of a sector having the same arc as the 
 segment, by the last problem ; find also the area of the triangle, 
 formed by the chord of the segment and the two radii of the 
 sector : then the difference of these two areas is the area of the 
 segment. See Note 1. 
 
 Rule II. Divide the height or versed sine of the segment by 
 the diameter, and find the quotient in the column of versed sines, 
 in Table I., at the end of the book. Take out the corresponding 
 area, in the next column on the right hand, and multiply by the 
 square of the diameter for the area. 
 
 FORMULA. 
 
 Put r = A E, C = A B, i> = C D, 
 
 p = E D, t = tabular area, and « = 
 
 arc A C B ; then /' \ [ -^ /\ 
 
 A = i («r ~ Cp) = I «; VC' + I ^ ^. / \ j /' \ 
 
 Note 1. "When the segment is greater than \ ; / 
 
 a semicircle, find the area of the remaining \^ ; 
 
 segment, and subtract from the whole area of '"* — Ju..— ''' 
 
 the circle for the required area. F 
 
 Note 2. The first rule or formula gives an 
 approximate value of the area, not very far from 
 the truth ; the second and third are still nearer 
 the truth ; and the last rule or formula may be 
 considered as exactly true. 
 
 EXAMPLES. 
 
 1 . Required the area of the segment A C B D A, its chord 
 A B being 12, and the radius A E or C E 10 feet. 
 
 First find C D and A C from the properties of the figure, and 
 the length of the arc A C B by Prob . YII., P art II. ; then fi nd 
 the area by Rule I. ; thus D E = VAE^- AD^ =.j^/W—^= 
 8, CD = CE — DE= 10-8=2, and A C = VAD^ -f CD-= 
 a/62 + 2^ = 6-324555 ; whence — — 
 
 38-59644 = arc AC B, and by Rule I., \ (38-59644 X 10) - 
 i (12 X 8) = 16-3274 square feet, Ans, 
 
 * This formula is due to B. Gorapertz, Esq., F.R S. 
 
46 MENSURATION OF SUPERFICIES. 
 
 By Rule II. The example being the same as before, we have 
 C D equal to 2 ; and the diameter 20. 
 
 Then 20)2 (-1 
 
 And to -1 answers ...» -040875 per Table 1. 
 
 Square of diameter. . . . 400 
 
 Ans. 1 6' 3 500 square feet. 
 By the second formula, the same example being still used, 
 A == I V VC^ + t ^'^ = 4 V122 + T 22 = 16-3511 square feet, 
 which is very near truth. 
 
 2. What is the area of the segment, the height of which is 2, 
 and the chord 20 feet. Jns, 26*36046. 
 
 3. What is the area of the segment, the height of which is 
 18, and the diameter of the circle 50 feet? Ans, 636*375. 
 
 4. Required the area of the segment, the chord of which is 16, 
 the diameter being 20 feet, A7is. 44*7292. 
 
 5. What is the area of a segment, the arc of which is a sextant, 
 the whole circumference of the circle being 25 feet ? 
 
 Ans. I '431 2 square feet, 
 
 6. The chord of a segment is 40, and its height 8 feet? what 
 is its area by the third formula? Ans. 219*73 square feet. 
 
 Problem IX. 
 
 To find the area of a circular zone. 
 
 {See figure to Prob. VIII., Part II.) 
 
 Rule. — The zone being first divided into a trapezoid (ABCD) 
 and two equal segments (B H D and A C), find tlie area of the 
 trapezoid by Prob. III., and the areas of the two segments by 
 Prob. IX. ; which areas, being added together, will give the area 
 of the zone. 
 
 examples. 
 
 1. The breadth of a zone is 42 feet, and its parallel chords are 
 48 and 36 feet, required the area. 
 
 Ans. 253*08 square yards. 
 
 2. The two parallel chords of a circular zone are each 100 
 yards, and the radius of the circle 72 yards ; required the area 
 of the zone. Ans. 13508|^ square yards. 
 
 3. The parallel chords of a circular zone are each 2|- hety and 
 the radius of the circle If ; required the area. 
 
 Ans, 6|- square feet nearly. 
 
MENSURATION OF SUPERFICIES. 47 
 
 Problem X. 
 
 To find the area of a circular ring, or space included between 
 two concentric circles. 
 
 Take the difference between the two circles, for the ring ; or 
 multiply the sum of the radii by their difference, and multiply 
 the product by 3*1416 for the answer. 
 
 FORMULAE. 
 
 A = TT (R2 - r3) ^ J_ (C2 — c2) ; in which R and r are 
 Air 
 the greater and lesser radii, and C and c the greater and lesser 
 circumferences. 
 
 EXAMPLES. 
 
 1 . The diameters of the two concentric circles being A B 20 
 and DG 12 feet, required the area of the ring contained be- 
 tween their circumferences A E B A, and D F G D. 
 
 E 
 
 AC = 10 
 DC = 6 
 
 3-1416 
 64 
 
 sum 16 
 dif. 4. 
 
 12-5664 
 188-496 
 
 64 
 
 201-0624 
 
 2. The diameters of two concentric circles being 20 and 10 
 feet ; required the area of the ring between their circumferences. 
 
 Ans, 235*72 square feet, 
 
 3. What is the area of a ring, the diameters of its bounding 
 circles being 6 and 4 feet? Ans. 15-708. 
 
 4. The circular fences on each side a gravel walk, surrounding 
 a shrubbery, are 800 and 714 feet in length ; what is the area of 
 the walk, and what did it cost in laying with gravel at \s. 4^d, 
 per square yard. j J Area 1151 square yards. 
 
 ••{; 
 
 Cost £79 2s, 7 id. 
 Problem XI. 
 
 To find the area of a lune A C B D A. 
 
 Rule. — Find by Prob. VIII. the areas of 
 the segments A C B and A D B, formed by 
 the chord A B of the two arcs of the lune, 
 and the difference of these areas will be the 
 area required. 
 
48 MENSURATION OF SUPERFICIES. 
 
 EXAMPLES. 
 
 1. What is tlie area of lune, the chord A B of which is 24 ft. 
 and the heights of its two arcs 5 and 3^ ft. ? Ans. 25f sq.ft. 
 
 2. The chord of a lune is 40 feet, and the heights of its arcs 4 
 and 20 feet ; required the area. Ans, 57'S67 square yards. 
 
 Problem XII. 
 
 To find the area of an ellipse. 
 
 Rule. — Multiply the product of the semiaxes TP, CP by 
 3*1416 for the area. 
 
 C m 
 
 FORMULA. 
 
 iV. = abir, in which a and h /^ 
 are the semiaxes. r^ ( 
 
 EXAMPLES. \^ 
 
 1. The axes of an elliptical — 
 shrubbery in a park are 300 and O n 
 200 feet ; required the area. 
 
 Ans. 5236 square yards, = 1 acre 396 square yards, 
 
 2. Required the area of an ellipse, the axes of which are 70 
 and 50 yards. Ans, 274S square yards 8 feet. 
 
 Problem XIII. 
 
 To find the area of an elliptical segment, the chord of which is 
 parallel to one of the axes. (See last figure.) 
 
 Rule. — Divide the height of the segment by that axis of the 
 elHpse of which it is a part ; and find in the table of circular 
 segments at the end of the book, a circular segment having the 
 same versed sine as this quotient. Then multiply continually 
 together, this segment, and the two axes, for the area required. 
 
 EXAMPLES. 
 
 1 . What is the area of an elliptic segment m'Rn, whose height 
 R r is 20 ; the tranverse T R being 70, and the conjugate C O 
 50 feet? 
 
 70) 20 ( •285-f- the tabular versed sine. 
 The corresponding segment 
 is -185166 
 70 
 
 12-961620 
 50 
 
 648*081000 square feet, the area required. 
 
MENSURATION OF SUPERFICIES. 
 
 4i) 
 
 2. What is the area of an elHptIc segment, cut off parallel to 
 the shorter axis, the height being 10, and the axes 2,0 and 
 3o feet ? Ans. \Q2'02\ square feet, 
 
 3. "What is the area of the elliptic segment, cut oif parallel to 
 tlie longer axis, the height being 5, and the axes 25 and 
 35 feet? Ans. 97*8458 square feet. 
 
 Problem XIV. 
 
 To find the area of a parabola. 
 
 Rule. — Multiply the axis or height VE 
 by the base or double ordinate D F, and -| 
 of the product will be the area. 
 
 FORMULA. 
 
 A = -| a d, in which a is the axis, and 
 d the double ordinate. 
 
 EXAMPLES. 
 
 1. Required the area of the parabola 
 A V C, the axis V B being 2, and the D 
 double ordinate A C 1 2 feet. 
 
 -| X 12 X 2 = 16 square feet, the area required. 
 
 2. The double ordinate of a parabola is 20 i^tt, and its axis or 
 height 18 ; required the area of the parabola. 
 
 Ans. 240 square feet. 
 
 Problem XV. 
 
 ' To find the area of a parabolic frustrum A C F D. 
 
 Cube each end of the frustrum, and subtract the one cube from 
 the other ; then multiply that difference by double the altitude, 
 and divide the product by triple the difference of their squares, 
 for the area. 
 
 FORMULA. 
 
 Q3 ^3 
 
 A=|«.— ~, in which a is the altitude, and C and c the 
 
 iu^ & 
 
 parallel chords. 
 
 EXAMPLES. 
 
 1. Required the area of the parabolic frustrum A C F D, A C 
 being 6, D F 10, and the altitude B E 4 feet. 
 
50 MENSURATION OF SUPERFICIES. 
 
 Ends. Squares. Cubes. 
 
 DF=10 100 1000 
 
 AC= 6 36 216 
 
 64 dif, 784 
 
 3 8=2BE 
 
 192 ) 6272 ( 32|f| = 32| Ans, 
 512 
 
 384 
 
 128 
 
 2. "What is the area of the parabolic frustrum, the two ends of 
 which are 6 and 10, and its altitude 3 feet. Ans. 24^ square feet. 
 
 Note. Those who wish for further information on the areas of the conic 
 sections, are referred to the works of Emerson, Hamilton, &c., it being foreign 
 to the object of this work to give more on this subject. 
 
 Problem XVI. 
 
 To find the areas of irregular figures whether hounded hy straight 
 lilies or curves. 
 
 Case I. — When the figure is long and narrow. 
 
 Rule. — Take the perpendicular breadth at several places, at 
 equal distances ; to half the sum of the first and last two breadths, 
 add the sum of all the intermediate breadths, and multiply the 
 result by the common distance between the breadths for the 
 area. 
 
 Case II. — When the breadths or perpendicidars are taken at 
 unequal distances, the figure being long and narrow. 
 
 Rule I. — Find the areas of all the trapezoids and triangles 
 separately, and add them together for the area. 
 
 Rule II. — Add all the breadths together, and divide the 
 sum by the whole number of them for the mean breadth, which 
 multiply by the length for the area. — This method is not very 
 correct, but may do where great accuracy is not required. 
 
 EXAMPLES. 
 
 1 . The perpendicular breadths, or offsets of an irregular figure 
 at five equidistant places are A D = 8*2, m p := 1'4:, n q = 9*2, 
 o r = 10*2, B C = 8*6 feet ; and the common distances A m = 
 m w = &c. = 50 feet ; required the area. 
 
MENSURATION OF SUPERFICIES. 
 
 51 
 
 By Rule L, Case \. 
 
 8-2 
 8-6 
 
 2)16-8 = sum 
 
 8-4 = ^ sum 
 7-4 
 9-2 
 10-2 
 
 35-2 
 50 
 
 y/w.9. 1 760*0 square feet. 
 
 2. The length of an irregular plank is 25 feet, and its perpen- 
 dicular breadth at six equidistant places are 17*4, 20*6, 14*2, 
 lG-5, 20*1, and 24*4 inches; required the area. 
 
 Ans, 30| square feet. 
 
 3. Take the dimensions and 
 find the area of the annexed 
 irregular figure, by Rule I. and 
 II., Case II. 
 
 Case III. When the breadth of the figure is large and its 
 boundary curved or crooked. 
 
 Rule. — Divide the figure into trapeziums and triangles, in the 
 most convenient manner, taking offsets to the curved or crooked 
 portion of the boundary. Find the areas of the trapeziums, 
 triangles, and the offset pieces separately, which, being added 
 together, will give the required area of the whole figure. 
 
 EXAMPLE. 
 
 The annexed figure is 
 divided into two trape- 
 ziums A B F G, B C E F, 
 and one triangle C D E, 
 with offsets on A B, A G, 
 CD, andDE. It is re- 
 quired to measure the se- 
 veral parts of the figure, 
 and to find its area. 
 
 The areas of the trape- 
 ziums are found by letting fall perpendiculars on the diagonals 
 A F, B E by Prob. IV., and the area of the triangle by Prob. II., 
 the areas of the several offset pieces being found by one or other 
 of the cases of this Problem . 
 
D'2 MENSURATION OF SOLIDS, 
 
 PROMISCUOUS EXERCISES. 
 
 1 . The sides of three squares are 6, 8, and 24 feet ; required 
 the side of a square that shall have an area equal to all the 
 three. Ans. 26 feet. 
 
 2. In cutting a circle, the largest possible, out of a card-hoard 
 5 feet square, how much will he wasted. 
 
 Ans. 5*365 square feet. 
 
 3. The area of a square is 72 square feet ; required the length 
 of its diagonal. Ans. \2 feet. 
 
 4. A ditch 13 yards wide surrounds a circular fortress, the 
 circumference of the fortress being 704 yards ; required the area 
 of the ditch. Ans. 2 acres nearly. 
 
 5. What is the area of a circular table the diameter of which 
 is 59 inches. Ans. 19 square feet nearly . 
 
 6. What is the area of an isosceles triangle, the base of which 
 is 5 feet 10 inches, and each side 8|- feet ? 
 
 Ans. 23 square feet 41^ inches. 
 
 7. Required the side of a decagon the area of which is 9 square 
 feet. Ans. 1 foot 1 inch nearly. 
 
 8. The side of a square is 50 yards, and its corners are cut off 
 so as to form an octagon ; required the area of the octagon. 
 
 Ans. 2071 square yards. 
 
 PART IV. 
 MENSURATION OF SOLIDS. 
 
 DEFINITIONS. 
 
 1. A Solid has three dimensions, length, breadth, and 
 thickness. 
 
 2. A prism is a soUd, or body, whose ends are any plane 
 figures, which are parallel, equal, and similar ; and its sides are 
 parallelograms. 
 
 A prism is called a triangular one when 
 
 ^^^^^^^A its ends are triangles; a square prism, 
 
 <~r "T^^^X^ when its ends are squares; a pentagonal 
 
 ^ ^^^^^ ~^~ ^ 1 prism, when its ends are pentagons; and 
 
 so on. 
 
MENSURATION OF SOLIDS. 
 
 3. A. cube is a square prism, having six 
 sides, which are all squares. It is like a 
 die, having its sides perpendicular to one 
 another. 
 
 4. A parallelopipedon is a solid having 
 six rectangular sides, every opposite pair of 
 which are equal and parallel. 
 
 5. A cyUnder is a round prism, having 
 circles for its ends. 
 
 Note. A prism is called a right one, when its sides 
 are perpendicular to its ends ; and an oblique prism 
 when its sides are inclined to its ends. 
 
 6. A pyramid is a solid having any 
 plane figure for a base, and its sides are 
 triangles, the vertices of which meet in a 
 point at the top, called the vertex of the 
 pyramid. 
 
 The pyramid takes names according to the figure of its base, 
 like the prism; being triangular, or square, or hexagonal, &c. 
 
 7. A cone is a round pyramid, having 
 a circular base. 
 
 8. A sphere is a solid bounded by one 
 continued convex surface, every point of 
 which is equally distant from a point within, 
 called the centre. — The sphere ma}'^ be con- 
 ceived to be formed by the revolution of a 
 semicircle about its diameter, which remains 
 fixed. 
 
 9. The axis of a solid, is a line drawn from the middle of one 
 end, to the middle of the opposite end ; as between the opposite 
 ends of a prism. Hence the axis of a pyramid, is the line from 
 the vertex to the middle of the base, or the end on which it is 
 supposed to stand, as O Y. And the axis of a sphere, is the 
 same as a diameter, or a line passing through the centre, and ter- 
 minated by the surface on both sides. 
 
 Note. It is called a right pyramid when the axis is perpendicular to the 
 base, but ^Yhen inclined to the base, it is called an oblique pyramid. 
 
54 MENSURATION OF SOLIDS. 
 
 10. The height or altitude of a solid, is a line drawn from its 
 vertex or top, perpendicular to its base. — This is equal to the 
 axis in a right prism or pyramid ; but in an oblique one, the 
 height is the perpendicular side of a right-angled triangle, whose 
 hypothenuse is the axis. 
 
 1 1 . Also a prism or pyramid is regular or irregular, as its base 
 is a regular or an irregular plane figure. 
 
 12. The segment of a pyramid, sphere, or any other solid, is 
 a part cut off the top by a plane parallel to the base of that 
 figure. 
 
 13. A frustrum or trunk, is the part that remains at the 
 bottom, after the segment is cut off. 
 
 14. A zone of a sphere, is a part intercepted between two 
 parallel planes. When the ends, or planes, are equally distant 
 from the centre, on both sides, the figure is called the middle 
 zone. 
 
 15. The sector of a sphere, is composed of a segment less 
 than a hemisphere or half sphere, and of a cone having the 
 same base with the segment, and its vertex in the centre of the 
 sphere. 
 
 16. A circular spindle, is a solid generated 
 ^ ^ by the revolution of a segment of a circle about 
 
 its chord, which remains fixed. 
 
 17. A regular body, is a solid contained under a certain 
 number of equal and regular plane figures of the same sort. 
 
 18. The faces of the soUd are the plane figures under which it 
 is contained ; and the linear sides, or edges of the soUd, are the 
 sides of the plane faces. 
 
 19. There are only five regular bodies : namely, 1st, the tetra- 
 hedron, which is a regular pyramid, having four triangular faces : 
 2nd, the hexahedron, or cube, which has 6 equal square faces : 
 3rd, the octahedron, which has 8 triangular faces : 4th, the do- 
 decahedron, which has 12 pentagonal faces: 5th, the icosahe- 
 dron, which has 20 triangular faces. 
 
 TABLE OF SOLID MEASURE. 
 
 1728 cubic inches =1 cubic foot. 
 
 27 cubic feet =1 cubic yard. 
 
 277-274, or 
 
 27 7^ nearly 
 
 ^^ I cubic inches ^, ; =1 gallon. 
 
MENSURATION OF SOLIDS. b5 
 
 Problem I. 
 To find the solidity of a cube. 
 
 Rule. — Cube one of its sides for the content ; that is, mul- 
 tiply the side by itself, and that product by the side again. 
 
 FORMULA. 
 
 Let I = length of the side of the cube, S its solidity, and s its 
 surface ; {which two last are also used to repi'esent the solidities 
 and siafaces of all the solids in the following i^roblems) then, 
 S = Z3, and Z = 3 ^ S. Also s=^P, 
 
 examples. 
 1 . If the side A B, or A C, or B D, of a cube be 24 inches, 
 what is its solidity or content ? 
 By the Rule or the first Formula. 
 
 24 
 
 24 A B 
 
 96 
 
 48 
 
 b7(S 
 24 
 
 2304 
 1152 
 
 13824 Ans. 
 
 2. How many solid yards are in a cube the side of which is 
 22 feet? Ans. 394 solid yards lOfeet. 
 
 3. Required how many solid i^Qi are in the cube the side of 
 which is 18 inches? Ans. 3|-. 
 
 4. What is the content of a cube, measuring G feet 8 inches 
 every way? Ans, 296 cubic feet 3'. 6". 8"'. 
 
 5. A cubical box contains 343 cubic feet; required the length 
 of its side. 
 
 By the second formula Z = ^ ^ S = ^ ^ 343 = 7 feet. 
 
 6. How many square feet of deal will make a cubical box, lid 
 included, each side of the box being 3 feet ? 
 
 By the last formula, « = 6 Z^ = 6 X 3^ = 54 square feet. 
 
 Problem II. 
 To find the solidity of a parallelopipedon. 
 Rule. — Multiply the length, breadth, and depth, or altitude. 
 
56 MENSURAIION OF SOLIDS. 
 
 all continually together, for the solid content : that is, multiply 
 the length by the breadth, and that product by the depth. 
 
 FORMDLiE. 
 
 Put I = length, h = breadth, and d = depth of the solid ; 
 
 s s s 
 
 then S = I b d, I = — -, b = ■— -, d = --^, Also, s = 
 
 b d Id Lb 
 
 2 [l{b + d) -Vbd)\ 
 
 EXAMPLES. 
 
 1. Required the content of the parallelopipedon, whose 
 length A B is 6 feet, its breadth A C 2|^ i^^iy and altitude B D 
 
 If feet? 
 
 1-75 = BD 
 6 = AB 
 
 10-50 
 2-5 = AC 
 
 — i 5250 
 
 ^ ^ 2100 
 
 26-250 Ans. 
 
 2. Required the content of a parallelopipedon, the length of 
 which is 10*5, breadth 4-2, and height 3-4. Ans, 149-94. 
 
 3. How many cubic feet are in a block of marble, the length 
 of which is 3 feet 2 inches, breadth 2 feet 8 inches, and depth 
 2 feet 6 inches? Ans, 21|. 
 
 4. A stone in the ruins of the walls of Balbec is 36 feet in 
 length, 14 in breadth, and 12 in thickness; required its content, 
 and its weight at the rate of 180 lbs. per cubic foot. 
 
 Ans. 11088 cubic feety and iveiyht 891 tons. 
 
 5. A rectangular cistern is to be made 32 feet in length and 
 
 12 in breadth, and to hold 1920 cubic feet of water ; what must 
 
 be its depth ? 
 
 S 19'^0 
 
 By the third formula the depth <f = _- = " = 5 feet. 
 
 ^ ^ I b 32 X 12 
 
 6. What quantity of deal is there in a box 3|- feet long, 2 
 wide, and H deep ? 
 
 By the last formula 5 = 2 | 3^ (2 + 1 1) + 2 x 1^ j = 
 
 square feet. 
 
 30i 
 
MENSURATION OF SOLIDS, 57 
 
 Problem III. 
 
 To find the solidity of any prism or cylinder. 
 
 Rule. — Find the area of the base or end ; which multiply by 
 the height or length ; and the product will be the content. 
 
 To find the area of the surface of a prism or cylinder. 
 
 Rule. — Multiply the circumference of the base or end by the 
 length or height, and the product will be the area required. 
 
 Note. If the whole surface be required, the area of the two ends must be 
 added to the area found by the rule. 
 
 FORMULAE. 
 
 Put I = length or height, as before ; a the area, and c the cir- 
 cumference of the base ; then 
 
 S = « / = — , l=— . Also s == c I -j- 2a = surface of the 
 
 4 TT a 
 
 prism, including the two ends, and s = tt r I = convex surface 
 of the cylinder, exclusive of the ends, r being the radius of 
 the base. 
 
 EXAMPLES. 
 
 1. Required the content of a triangular prism, the length 
 A C of which is 1 2 feet, and each side of its equilateral base 
 2\ feet. 
 
 By the Rule or first Formula. 
 
 •433013 tabular No. 
 
 6i = (2i)3 ^^^^^ B 
 
 2-598078 
 108253 
 
 a= 2706331 area of end 
 /= 12 length 
 
 Ans, 32-475972 solid feet, 
 
 2. Required the solidity of a triangular prism, the length of 
 which is 10 feet, and the three sides of its triangular end or base, 
 are 5, 4, 3 feet? Ans. 60 cubic feet. 
 
 3. What is the content of a hexagonal prism, the length being 
 8 feet, and each side of its end 1 foot 6 inches. 
 
 Ans. 4iQ'7Q^ cubic feet. 
 
 4. Required the content of a cylinder, the length of which is 
 20 feet, and circumference 5|^ feet. 
 
 By the second formula, 
 
 ) X -07958 = 48-146 cu 
 
 3* 
 
 S ==— = (5^)2 X 20 X -07958 = 48-146 cubic feet. 
 
 4 IT 
 
58 
 
 MENSURATION OF SOLIDS. 
 
 5. What is the convex surface of a cylinder, the 
 length of which is 16 feet, and its diameter 2 feet 3 
 inches ? 
 
 By the last formula, 
 5 =V r I = 3-1416 X 2|^ X 16 = 113-0976 sq. feet. 
 
 6. Required the whole superficial area of a cylinder, 
 the length of which is 15 feet, and diameter 5^ feet? 
 
 A71S. 32|^ square yards. 
 
 7. The whole superficial area of a triangular 
 prism is 143 square feet, and each side of its equi- 
 lateral ends 5 feet ; required its length ? 
 
 By transposing the third formula. 
 
 z = : 
 
 2a 
 
 == 6 feet 8 inches nearly 
 
 8. The diameter of a cylinder is 12 feet, and its 
 length 20 ; required the content ? 
 
 Ans. 2262 cubic feet nearly, 
 
 9. How many cuhic i^ei of stone is there in a 
 round pillar, the height of which is 16 feet, and 
 diameter 2 feet 3 in? Ans. 63*62 cuhic ft. 
 
 10. How many square yards of painting are there 
 in the convex surface of a column, the length of 
 which is 20 feet, and its diameter 2 feet ? 
 
 Ans. 13 square yards S^feet nearly. 
 
 Problem IV. 
 
 To find the solidity of any cone or any pyi^amid. 
 
 Rule. — Compute the area of the base, then multiply that area 
 by the height, and take -J of the product for the content. 
 
 To find the convex surface of a right cone, or the slant sur- 
 facQ of a right pyramid. 
 
 Rule. — Multiply the circumference of the base by the slant 
 height, or length of the side, and take half the product for the 
 surface. 
 
 formulae. 
 
 3S 
 
 3 S 
 
 % = \al,a = -^,l ^— . Also 5 = i c Z', V being the 
 I a 
 
 slant height. When the whole surface is required, the area of 
 the base must be added. 
 
 EXAMPLES. 
 
 1 . What is the sohdity of a cone, the height C D of which is 
 12^^ feet, and the diameter AB of the base 2J? 
 
MENSURATION OF SOLIDS. 
 
 59 
 
 Here 2^ x 2^ x -^ x 
 
 25 
 4 
 
 = 6i = AB2. 
 
 Then '7854 
 H 
 
 4-7124 
 19635 
 
 490875 area of base 
 12i height CD 
 
 58*90500 
 2-454375 
 
 3)61-359375 
 20 453125 A71S. 
 2. What is the solid content of a pentagonal pyramid, its height 
 being 12 feet, and each side of its base 2 feet? C 
 
 1-720477 tab. area 
 4 square side 
 
 6-881908 area base 
 
 4 1 of height C O 
 
 Ans. 27-527632 cubic feet, 
 
 3 What is the content of a cone, its height being lOJ i'QQiy 
 and the circumference of its base 9 feet ? 
 
 Ans, 22*561 cithicfeet. 
 
 4. Required the content of a triangular pyramid, its height 
 being 14 feet 6 inches, and three sides of its base, 5, 6, 7. 
 
 Ans. 71-0351 cubic feet. 
 
 5. What is the content of a hexagonal pyramid, the height of 
 which is 6-4, and each side of its base 6 inches. 
 
 Ans. 1-38 cubic feet. 
 
 6. If the diameter of the base AB be 5 feet, and the side of 
 the cone A C 1 8, required the convex surface. 
 
 3-1416 
 
 5 diameter 
 
 15-7080 circumference 
 18 
 
 125664 
 15708 
 
 2)282-744 
 Ans. lil.37 2 square feeL 
 
60 MENSURATION OF SOLIDS. 
 
 7. What is the convex surface of a cone, the slant side of 
 which is 20, and the circumference of its base 9 feet ? 
 
 Ans* 90 square feet. 
 
 8. Required the convex surface of a cone, the slant height of 
 which is 50 feet, and the diameter of its base 8 feet 6 inches? 
 
 A71S. 667 '^9 square feet. 
 
 9. The side of the equilateral base of a triangular pyramid is 
 5 feet, and its solid content 62|- cubic feet ; required its per- 
 pendicular height. 
 
 Ans. By the third formula, the height if found 1/ feet 4 
 inches nearly. 
 
 10. Required the weight of a hexagonal pyramid of marble, 
 each side of the base of which is 1 foot 3 inches, and the vertical 
 height 10 feet, the weight of the marble being 1/0 lbs. per 
 cubic foot. Ans. 1 ton. civt. 18|^ lbs. 
 
 1 1 . A cone contains 8 solid feet, and its height is 2 feet ; 
 what is the circumference of its base? Ans. 12'28 ft nearly. 
 
 12. The circumference of the base of a cone is 33 feet, and 
 the slant height 8 feet 9 inches ; required the content ? 
 
 A71S. 202*65 cubic feet. 
 
 Problem V. 
 
 To find the solidity of the frustrum of a cone, or any pyramid. 
 
 General Rule. — To the area of the two ends add the 
 square root of their product, and multiply the sum by \ of the 
 height for the solidity. 
 
 FORMULAE. 
 
 If A and a be the areas of the greater and lesser ends ; then, 
 
 S = i (A -f « + \A«) I' 
 IFhen the solid is the frustrum of a cone, or of a pyramid, 
 having its ends regular polygons. 
 
 Rule. — To the sum of the squares of the radii of the ends, 
 if a cone, or of the sides of the ends, if a pyramid, add their 
 product; and multiply the sum by 3* 14 16, if a cone, or by the 
 tabular number of the polygon, if a pyramid, and again by ^ of 
 the height for the content. 
 
 FORMULA. 
 
 S = i(R'-^ + r2 + Rr) In, 
 in which R and r are the radii of the ends, if a cone, or the 
 sides of the ends, if a pyramid. In the latter case tt represents 
 
MENSURATIO>i OF SOLIDS. 
 
 61 
 
 the tabular number of the polygon. If R and r be taken as the 
 circumferences of the ends of a cone, then tt must be taken 
 = -07958. 
 
 To find the convex surface or frustruni of a cone, or the slant 
 surface of a pyramid. 
 
 Rule. — ^lultiply the sum of the circumferences of the two 
 ends by ^ the slant height of the frustrum for the required 
 surface. 
 
 Note. AVheu the whole surface is required the areas of the two ends must 
 be added to the result of the Rule. 
 
 EXAMPLES. 
 
 1 . What is the content of a frustrum of a cone, the height of 
 which is 20 inches, and the diameters of its two ends 28 and 20 
 inches ? 
 
 28 
 28 
 
 28 
 20 
 
 20 
 20 
 
 224 
 56 
 
 560 
 
 784 
 400 
 
 400 
 
 784 
 
 
 1744 
 •2618 
 
 
 
 13952 
 1744 
 10464 
 
 3488 
 
 
 
 456-5792 
 
 20 = 
 
 PQ 
 
 Ans. 9131 -5840 solid inches, 
 
 2. Required the content of a pentagonal frustrum, the height 
 of which is 5 feet, each side of the base 1 foot 6 inches, and 
 each side of the less end 6 inches. 
 
62 
 
 MENSURATION OF SOLIDS. 
 
 18 
 18 
 
 18 - 6 
 6 6 
 
 144 
 
 18 
 
 .S-24 
 
 108 36 
 324 — 
 36 
 
 
 3)468 
 
 
 
 156 i of sum. 
 1-720477 tab. area. 
 
 10322862 
 8602385 
 1720477 
 
 144 
 
 }i 
 
 268-394412 mean area 
 5 height P Q 
 
 1341-972060 
 111-831005 
 
 9-319250 Ans. in cubic feet. 
 
 3. What is the soHdity of the frustrum of a cone, the altitude 
 being 25, the circumference at the greater end 20, and at the 
 less end 10 feet? ' Ans. 464*205 cubic feet, 
 
 4. How many solid feet are in a piece of timber, whose bases 
 are squares, each side of the greater end being 15 inches, and 
 each side of the less end 6 inches ; also the length, or perpendi- 
 cular altitude is 24 feet? Aiis. \9\ cubic feet. 
 
 5. To find the content of the frustrum of a cone, the altitude 
 being 18, the greatest diameter 8, and the least 4 feet? 
 
 Alls. 527*7888 cubic feet. 
 
 6. What is the solidity of a hexagonal frustrum, the height 
 being 6 feet, the side of the greater end 18 inches, and of the 
 less 12 inches? Ans 24*68 cubic feet. 
 
 7. The girts of the trunk of a tree at its two ends are 15 and 
 10 feet, and its length 48 feet ; how many solid feet does it 
 contain? (R and r being taken for the girts in the second 
 formula, &c.) Ans. 604f nearly. 
 
 8. The height of the frustrum of an octagonal pyramid is 48 
 feet and the sides of its ends 26 and 19; required the solid 
 content. Ans. 118279 cubic feet. 
 
MENSURATION OF SOLIDS. 
 
 63 
 
 9. The sides of the ends of the frustrum of a 
 square pyramid are 6 and 4 feet, and its slant 
 length 20 feet, required its slant surface, 
 
 ^ ^ 1i^^^i^ circumf. ofends. 
 4 X 4 = 16 J 
 
 40 sum 
 10=^ 
 
 length 
 
 9)400 
 
 44f square yards. 
 Note. The slant length is measured from the middle of 
 one side to that of its corresponding side. 
 
 10. The slant height of tower, in the form of 
 a hexagonal pyramid, is 74 feet, each side of the 
 base 7ii each side of the top 2|- feet ; required the area of the 
 sides, and the expense of painting it at 1^. 3d. per square yard. 
 
 Ans. 2220 square feet, and £15 8s. 4d. 
 
 1 1 . What is the convex surface of the frustrum of a cone, the 
 slant height of the frustrum being 12*5, and the circumferences 
 of the two ends 6 and 8*4 feet? Ans. 90 square feet. 
 
 12. Required the convex surface of the frustrum of a cone, 
 the side of the frustrum being 10 feet 6 inches, and the circum- 
 ferences of the two ends 2 feet 3 inches, and 5 feet 4 inches ? 
 
 Ans. 39^f square feet. 
 
 13. The perpendicular height of the frustrum of a cone is 3 
 feet, and the circumferences of the base and top 9 and 6 feet ; 
 required the whole surface? Ans. 68*35 square feet. 
 
 Problem VI. 
 
 To find the solidity of a wedge. 
 
 Rule. — To the length of the edge add twice the length of 
 the back or base, and reserve the sum ; multiply the height of 
 the wedge by the breadth of the base ; then multiply this pro- 
 duct by the reserved sum, and take -^ of the last product for the 
 content. 
 
 on 
 
 FORMULA. 
 
 S = -J- (2 Z + Z') hhy the symbols denoting the parts shown 
 the following figure. 
 
 EXAMPLES. 
 
 1. What is the content in feet of a wedge, the altitude AP 
 of which is 14 inches, its edge AB 21 inches, and the length of 
 its base D E 32 inches, and its breadth CJ)A\ inches? 
 
64 
 
 21 
 32 
 32 
 
 85 
 
 IflENSURATION OF SOLIDS. 
 
 14 A 
 
 56 
 
 7 
 
 63 
 
 85 
 
 a 
 
 315 
 504 
 
 6 5355 
 
 ri2 892*5 Aiis. in cubic inches. 
 1728.^ 12 74-375 
 
 Ll2 6-197916 
 
 -516493 Ans. in cubic feety or little i)iore than 
 half a cubic foot. 
 
 2. The edge and base of a wedge are respectively 9 feet, and 
 5 feet 4 inches in length, the base is 2 feet 8 inches in breadth, 
 and the height 3 feet 6 in. ; required the content of the wedge. 
 
 Ans. 30 cubic feet, T. 1". 4"^ 
 
 3. The height and length of edge, the length and breadth of 
 base of a wedge are each 2 feet ; what is its soUdity ? 
 
 Ans. 4 cubic feet. 
 
 Problem YII. 
 
 To find the solidity of a pi'ismoid. 
 
 Definition. — The ends of a prismoid are parallel and dis- 
 similar rectangles or trapezoids ; the solid is, therefore, the 
 frustrum of a wedge, the part of the wedge next the edge being 
 cut off. 
 
 Rule. — Add into one sum, the areas of the two ends and 4 
 times the middle section parallel to them, and ~ of that sum 
 will be a mean area ; which being multiplied by the height, 
 will give the content. 
 
 Note. For the length of the middle section, take half the sum of the lengths 
 of the two ends ; and for its breadth, take half the sum of the breadths of 
 the two ends. 
 
 FORMULA. 
 
 S = -J- (L B + ^ ^ 4- 4 M m)y the symbols representing the 
 parts shown in the following figure. 
 
 EXAMPLE. 
 
 1. How many cubic feet are there in a stone, the ends of 
 
MENSURATION OF SOLIDS. 
 
 65 
 
 which are rectangles, the length and breadth of the one being 
 14 and 12 inches; and the corresponding sides of the other G 
 and 4 inches : the perpendicular height being 30^ feet? 
 
 14 10 6 
 
 12 8 4 ^ 
 
 168 
 
 80 
 4 
 
 320 
 
 168 
 
 24 
 
 5) 512 
 
 24 
 
 M 
 
 85^ mean area in inches 
 30i height 
 
 2560 
 424 
 
 144 
 
 
 2 I 2602-6 
 2 I 2^6-*8 
 
 18-074* A71S. 
 
 2. What is the content of a railway coal waggon, of which 
 the length and breadth at top are 81|- and 55 inches at bottom, 
 the length and breadth are 41 and 2 9|^ inches, and the height 
 47^- inches? A7is, 73^ cubic feet. 
 
 Note. Several railway cuttings are in the form of a prismoid, with dissimi- 
 lar trapezoidal ends. The following is an example of this kind, the bottom 
 width being the same throughout. 
 
 3. The top widths of a railway cutting are 120 and 90 feet, 
 their respective depths 30 and 20 feet, the bottom width 30 feet, 
 and the length of the cutting 3 chains or 66 yards ; required the 
 content in cubic yards. Ans. 12466f cubic yards. 
 
 Problem VIII. 
 To find the solidity of a sphere or globe. 
 Rule. — Multiply the cube of the diameter by '5236. 
 
 FORMULA. 
 
 S = ^ D3 TT, and D = 3 V 
 
 6S 
 
66 
 
 MENSURATION OF SOLIDS. 
 
 EXAMPLES. 
 
 1 . The diameter of a sphere is 1 2 feet re- 
 quired its soHdity ? 
 
 123 X •5236 = 904-7808 cubic feet, 
 
 2. Find the content and weight of an ivory 
 ball 3|- inches in diameter, the weight of ivory 
 being 1 820 omices (Av.) per cubic foot. 
 
 Ans, Content 22*448 cubic in., and weight 24 ounces nearly. 
 
 3. A 2|^ inch cube of ivory is turned into a sphere of the 
 same diameter ; what weight of ivory will be lost ? 
 
 Ans. 7*68 ounces. 
 
 4. Required the solid content of the earth, supposing its cir- 
 cumference to be 25000 miles ? 
 
 Ans. 263858149120 cubic miles. 
 
 Problem IX. 
 
 To find the solidity of a spherical segment. 
 
 Rule. — To three times the square of the radius of its base, 
 add the square of its height ; then multiply the sum by the 
 height, and the product again by *5236. 
 
 FORMULA. 
 
 S = ^ (3 ?- + h"^) h TT, in which r = AB, and A = C D. 
 
 examples. 
 
 1. Required the content of a spherical segment, its height 
 being 4 inches, and the radius of its base 8? 
 
 A 
 
 8 
 
 4 
 
 •5236 
 
 ^^^^^ 
 
 8 
 
 4 
 
 832 
 
 l^^^^^v 
 
 
 
 
 :^^B 
 
 64 
 
 16 
 
 10472 
 
 3 
 
 192 
 
 15708 
 
 
 
 
 41888 
 
 1 / 
 
 192 
 
 208 
 4 
 
 
 ' ,*""' 
 
 435-6352 ^?Z5. 
 
 ' 832 
 
 2. What is the solidity of the segment of a sphere, the height 
 of which is 9, and the diameter of its base 20 feet ? 
 
 Ans. 1795-4244 cubic feet. 
 
 3. Required the content of the spherical segment, the height 
 of which is 2J, and the diameter of its base 8*61684 feet? 
 
 Ans. 71*5695 cubic feet. 
 
MENSURATION OF SOLIDS. 67 
 
 Problem X. 
 
 To find the solidity of a spherical zone or f rust mm. 
 
 Rule. — Add together the square of the radius of each end, 
 and ^ of the square of their distance, or of the height ; then 
 multiply the sum by the said height, and the product again by 
 1-5708.' 
 
 FORMULA. 
 
 S = -J{3 (R2 + 7-2) + A-} 7^77. 
 
 EXAMPLES. 
 
 1. What is the solid content of a zone, its greater diameter 
 being 12 inches, the lesser 8, and the height 10 inches? 
 
 G3 = 36 ^^.---i ^^ 
 
 jL X 10^ == 33J- /^^Z~ — '. !I^k 
 
 ^ mXm 
 
 85i X 10 X 1-5708 = 1340-416 cubic in., \":::::±:::^^ 
 
 the content required. 'n^ : ^.''' 
 
 2. Required the content of a zone, the greater diameter is 12, 
 less diameter 10, and height 2 feet. Ans. 195*8264 cubic ft. 
 
 3. What is the content of a middle zone, the height being 8 
 feet, and the diameter of each end 6 feet ? 
 
 Ans. 494-2784 cubic ft. 
 
 4. A cask is in the form of the middle zone of a sphere, its 
 top and bottom diameters being 5 feet 8 inches, and its height 
 5 feet, inside measure ; how many gallons will it contain ? 
 
 Ans. 11 93 J gallons. 
 
 Problem XI. 
 
 To find the convex surface of a sphere, also of a segment and 
 zone thereof 
 
 For the sphere. 
 Rule. — Multiply the square of the diameter by 3-1416. 
 
 For the segment or zone. 
 
 Rule.—- Multiply the circumference of the whole sphere bj 
 the height of the segment or zone. 
 
68 
 
 MENSURATION OF SOLIDS, 
 
 FORMULA. 
 
 A = a^TT = c c?= — for the sphere. 
 
 TT 
 
 A = c? A TT = c A for the segment or zone. 
 
 EXAMPLES. 
 
 1 . Required the convex surface of a sphere, the diameter of 
 which is 2 feet. 
 
 22 X 3-1416 = 12-4264 square it^i, 
 
 2. The circumference of a spherical stone is 4 feet, required 
 its convex surface. 
 
 Here 1. = 42 
 
 TT 
 
 3-1416 = 5-0928 square feet. 
 
 3. Required the area of the surface of the earth, its diameter 
 or axis being 795 7f miles, and its circumference 25000 miles. 
 
 Arts, 198943750 square miles, 
 
 A. The axis of a sphere being 42 inches, what is the convex 
 superficies of the segment, whose height is 9 inches? 
 
 Ans, 1187*5248 square inches. 
 
 5. Required the convex surface of a spherical zone, the breadth 
 or height of which is 2 feet, and cut from a sphere of 1 2^ feet of 
 diameter. Ans, 78-54 square feet. 
 
 Problem XII. 
 
 To find the solidity of a circular spindle. 
 
 Rule. — From -g- of the cube of the length AB subtract twice 
 the product of the area of the circular segment A C B and the 
 central distance EO, and multiply the difference by 3* 141 6 for 
 the solidity. 
 
 FORMULA. 
 
 S = (^ AB2 -2E0area of seg. ACB) tt. 
 
 C Note. The central distance E 0, and 
 
 the area of the segment ACB may be 
 found by the Problems for the purpose in 
 Parts II. and III.; however, the whole 
 operation shall be given in the following 
 example. 
 
 EXAMPLES. 
 
 1. The length A B of a cir- 
 cular spindle is 24 feet, and its 
 middle diameter C D = 18; what is its content ? 
 
 Diameter II I = ^ +CE=:_-f9=25 feet. 
 CE 9 
 
OF THE 
 
 UNIVERSITV . 
 
 SNg^URAOntON OF SjBLIDS. 69 
 
 Then, by Rule II., ProC. vTlir, page 45, 
 
 9 _^ 25 = -36 = tabular lieight or versed sine. 
 
 The area of corresponding seg. Tab. 1, is -25455, ^Yhich mnl- 
 plied by II V = 25- gives = 159-09375 = area of seg. A C B. 
 
 Now EO = CO — CE=V— ^ = 3-5, or 2 E O = 7. 
 
 Whence S = (^AB3--2EO x area of seg. A C B) tt = 
 (.1-24^ — 7 X 159-09375) x 3-1416 = 3739 ?r cubic feet. ^ 
 
 2. The length of the circular spindle is 6 feet, and its middle 
 diameter 2 J feet ; required the content? Ans. 16^ eft. nearly. 
 
 Problem XIII. 
 
 To find the content of the middle zone or fnistrim of a circular 
 spindle. 
 
 Rule. From thrice the square of the length of the whole 
 spindle subtract the square of the length of the middle frustrum, 
 and multiply the difference by 
 \ of the said length of the 
 frustrum. Again, multiply 
 four times the central distance 
 by the area of the middle sec- 
 tion A D E C B F.— Subtract 
 the latter product from the 
 former, and multiply the re- 
 mainder by -7854 for the con- 
 tent. 
 
 Note. The middle frustrum of a circular spindle is one of the varieties of 
 casks in guaging, the practical rule for finding the content of which, though 
 not so strictly accurate, is much more concise than the one just given. 
 (See Guaging). We shall, therefore, not give an example worked out at 
 length, as in the preceding Problems, but shall give, for the exercise of the 
 student, the following 
 
 EXAMPLE. 
 
 How many gallons (Imperial) are contained in a cask, in the 
 form of the middle frustrum of a circular spindle, its head dia- 
 meter (A D) being 12, its bung diameter (E F) 16, and its 
 length (D C) 20 inches? Ans. 12*6 gallons. 
 
 Problem XIV. 
 To find the coyivex surface of a circular spindle A C B D, 
 (fig. to Prob. XII.) or of any segment A P D, or zone A D C B, 
 (fig. to Prob. XIII.). 
 
 FORMULiE. 
 
 Let I = length of the spindle, segment, or zone, a = its re- 
 volving arc, r = the radius of the generating circle, and c = 
 the central distance : then A = 2 (l r — ac.) it. 
 
 V 
 
 
70 MENSURATION OF SOLIDS. 
 
 EXAMPLE. 
 
 What is the surface of a circular spindle, the length of which 
 is 24, and its middle breadth 18 feet ? Aiis, 1 181} 55'. ft. 
 
 Problem XV. 
 
 To find the solidity of a cylindrical ring. 
 
 To the thickness of the ringj add the inner diameter ; then 
 multiply that sum by the square of the thickness, and the pro- 
 duct again by 2*4674, or \ of the square of 3*1416, for the 
 solidity. 
 
 FORMULA. 
 
 Let R and r be the inner and outer radii of the ring ; then 
 S = i (R - r)3 (R + r)7T\ 
 
 EXAMPLES. 
 
 1. Required the solidity of a 
 ring, the inner diameter being 
 12 and the thickness 2 inches. 
 
 (12 + 2) X 22 X 2-4674. 
 D 138-1744 square inches. 
 
 2. What is the content of a 
 ring, the thickness of which is 
 4, and inner diameter 1 6 feet 1 
 
 Ans. 789*568 square feet. 
 
 Problem XVI. 
 
 To find the surface of a cylindrical ring. 
 
 Rule. — Multiply the sum of the outer and inner radii of the 
 ring by their difference, or the thickness of the ring, and the 
 product by 9-8696, or the square of 3*1416 for the surface. 
 
 formula. 
 
 A = (R + r) (R — r) tt^. 
 
 EXAMPLES. 
 
 1. The inner and outer radii of a ring are 6 and 8 inches ; 
 required the area of its surface. 
 
 (8 + 6) (8 — 6) X 9-8696 = 276*3188 sq. inches. Ans. 
 
 2. What is the surface of a ring, the inner diameter of which 
 is 16, and the thickness 4 inches 1 Ans, 789*568 sq. in. 
 
 Problem XVII. 
 
 To find the solidity of a spheroid. 
 
 Square the revolving axis, multiply that square by the fixed 
 axis, and multiply the product by '5236 for the content. 
 
MENSURATION OP SOLIDS. 
 
 71 
 
 FORMULA. 
 
 S 3= i^^/r, r being the revolving axis, and /the fixed axis. 
 
 Note. In the prolate spheroid, the revolving axis is the conjugate ; but, in 
 the oblate, the revolving axis is the transverse. 
 
 EXAMPLES. 
 
 1 . Required the solidity of the prolate spheroid A C B D, the 
 axes being A B 50 and C D 30. 
 
 30 
 30 
 
 •5236 
 45000 
 
 900 
 50 
 
 26180000 
 20944 
 
 45000 
 
 23562-0000 Ans 
 
 2. Required the content of an oblate spheroid, the axes of 
 which are 50 and 30 inches. Ans, 22*7257 cubic feet. 
 
 3. What is the solidity of a prolate spheroid, the axes of 
 which are 9 and 7 feet ? Ans. 230*9076 cubic feet. 
 
 4. The earth is an oblate spheroid, the equatorial and polar 
 axes being 7925| and 7899 miles respectively ; what is its con- 
 tent in cubic miles ? 
 
 Problem XVIII. 
 
 To find the content of the middle zone of a spheroid^ the ends 
 being circular arid perpendicular to the fixed axis. 
 
 Rule. — To double the square of the middle diameter, add the 
 square of the diameter of one end ; multiply this sum by the 
 length of the frustrum, and the product again by '2618 for the 
 content. 
 
 Note. The middle zone or frustrum of a spheroid is one of the varieties 
 of casks in gauging. 
 
 EXAMPLES. 
 
 1 . Required the content of the middle zone of a spheroid, the 
 middle diameter being 30, the diameter of each end 18, and the 
 length 40 inches. Ans. 22242*528 cubic inches, 
 
 2. How many gallons are contained in a cask in the form of 
 the middle zone of a spheroid, the bung and head diameters 
 being 20 and 16 inches, and the length 18 inches, all internal 
 measures? Ans. \1\^ imperial gallons. 
 
72 
 
 MENSURATION OF SOLIDS. 
 
 Problem XIX. 
 
 To find the solidity of a paraboloid. 
 
 Note. A paraboloid is generated by the revolution of a parabola on its axis. 
 
 Rule. — Multiply the square of the diameter of the base by 
 the height or axis, and the product again by '3927 for the, 
 content. 
 
 EXAMPLE. 
 
 What is the content of the parabolic conoid, the height of 
 which is 42, and the diameter of its base 24 .'' 
 
 Ans, 9500-1984. 
 Problem XX. 
 To find the solidity of the frustrum of a paraboloid. 
 Rule. — Square the diameter of the two ends, add those two 
 squares together, multiply that sum by the height, and the pro- 
 duct again by '3927, for the content. 
 
 EXAMPLES. 
 
 1 . Required the content of a paraboloidal frustrum, the dia- 
 meters being 20 and 40, and the height 22 j inches? 
 
 Jns. 17671-1 cubic inches. 
 
 2. What is the content of the frustrum of a paraboloid, the 
 greatest diameter being 30, the least 24, and the altitude 9 ? 
 
 Ans. 5216-6268. 
 
 Problem XXI. 
 
 To find the superficies or solidity of any regular body. 
 
 Rule I. — Multiply the proper tabular area (taken from the 
 following table) by the square of the linear edge of the solid, 
 for the superficies. 
 
 Rule II. — Multiply the tabular solidity by the cube of the 
 linear edge, for the solid content ? 
 
 SURFACES AND SOLIDITIES OF REGULAR BODIES. 
 
 No. of 
 faces. 
 
 Names. 
 
 Surfaces. 
 
 Solidities. 
 
 4 
 
 6 
 
 8 
 
 12 
 
 20 
 
 Tetrahedron 
 
 1-73205 
 600000 
 3-46410 
 20-64573 
 8-66025 
 
 0-11785 
 1-00000 
 0-47140 
 7-66312 
 2-18169 
 
 Hexahedron or 
 Octahedron . . . 
 Dodecahedron 
 Icosahedron . . 
 
 Cube. . 
 
 
 Note. These are frequently called the Platonic bodies. 
 
carpenters' rule. 73 
 
 examples. 
 
 1. If the linear edge or side of a tetrahedron be 3 feet, re- 
 quired its surface and solidity ? 
 
 1-73205 X 3- = \b'5S84o square feet = surface, 
 . 0-11785 X 33= 3-18195 cubic feet = solidity. 
 
 2. What arc the superficies and solidity of the octahedron, 
 the linear side of which is 2 feet ? 
 
 J J Superficies 13-8564 square feet. 
 ^^^- t Solidity S' 7 7 12 cubic feet, 
 
 3. What are the surfaces and solidities of a dodecahedron 
 and icosahedron, each of the linear sides of which are 2 feet ? 
 
 r Dodecahedron / Surface 82-583 square feet.. 
 
 Ans i ^ • • I Solidity. . . .61-305 cubic feet. 
 
 It 11 f Surface . . . .34-641 square feet. 
 
 I Icosahedron . . . . < c? tj-a 1^7^-0 z- /^ 
 
 *- (^ bolidity. . . .17*4y3 cubic feet. 
 
 THE SLIDING OR CARPENTERS' RULE. 
 
 This instrument is otherwise called the sliding rule, and it is 
 made use of in measuring of timber and artificers' works, both for 
 taking the dimensions, and computing the contents. 
 
 The instrument consists of two equal pieces, each a foot in 
 length, which are connected together by a folding joint. 
 
 One side or face of the rule, is divided into inches, and halt 
 quarters or eights. On the same face also are several plane 
 scales divided into 12th parts by diagonal lines ; which are used 
 in planning dimensions that are taken in feet and inches. The 
 edge of the rule is commonly divided decimally, or into tenths ; 
 namely, each foot into 10 equal parts, and each of these into 10 
 parts again : so that, by means of this latter scale, dimensions are 
 taken in feet, in tenths and hundredths, and then multiplied as 
 common decimal numbers, which is the best way. 
 
 On the one part of the other face are four lines, marked A, B, 
 C, D ; the two middle ones, B and C, being on a slider, which 
 runs in a groove made in the stock. The same divisions serve 
 for both these two middle lines, the one being above the numbers, 
 and the other below. 
 
 These four lines are logarithmic ones, and the three A, B, C, 
 which are all equal to one another, are double lines, as they pro- 
 ceed twice over from 1 to 10. The other or lowest line D, is a 
 single one, proceeding from 4 to 40. It is also called the 
 
 4 
 
74 carpenters' rule. 
 
 girt line, from its use in computing the contents of trees and 
 timber. 
 
 On the other part of this face there is a table of the value of 
 a load, or 50 cubic feet of timber, at all prices, from 6 pence to 
 2 shillings a foot. 
 
 When I at the beginning of any line is accounted 1, or unit, 
 then the 1 in the middle will be 10, and the 1 at the end 100 ; 
 and when 1 at the beginning is accounted 10, then the 1 in the 
 middle is 100, and the 1 at the end 1000: and so on. All the 
 smaller divisions being altered proportionally. 
 
 Problem I. 
 
 To multiply numbers together. 
 
 Suppose the two numbers 13 and 24. — Set 1 on B to 13 on 
 A; then against 24 on B stands 312 on A, which is the re- 
 quired product of the two given numbers 13 and 24. 
 
 Note. In any operations, when a number runs beyond the end of the line, 
 seek it on the other radius, or other part of the line ; that is, take the 10th 
 part of it, or the 100th part of it, &c., and increase the result proportionally 
 10 fold, or 100 fold, &c. 
 
 In like manner the product of 35 and 19 is 665. 
 And the product of 270 and 54 is 14580. 
 
 Problem II. 
 
 To divide by the sliding rule. 
 
 As suppose to divide 312 by 24. — Set the divisor 24 on B to 
 the dividend 312 on A; then against 1 on B stands 13, the 
 quotient, on A. 
 
 Also 396 divided by 27 gives 14-6. 
 And 741 divided by 42 gives 17* 6. 
 
 Problem III. 
 
 To square any numher. 
 
 Suppose to square 23. — Set 1 on B to 23 on A; then against 
 23 on B, stands >529 on A, which is the square of 23. 
 
 Or, by the other two lines, set 1 or 100 on C to the 10 on 
 D, then against every number on D, stands its square in the line 
 C. So, against 23 stands 529 
 
 against 20 stands 400 
 
 against 30 stands 900 
 
 and so on. 
 
 If the given number be hundreds, &c., reckon the 1 on D for 
 100, or 1000, &c., then the corresponding 1 on C is 10,000, or 
 100,000, &c. So the square of 230 is found to be 52,900 
 
TIMBER MEASURING. 75 
 
 Problem IV. 
 
 To extract the square root. 
 
 Set 1 or 100, &c., on C to 1 or 10, &c., on D; then against 
 every number found on C, stands its square root on D. 
 Thus, against 529 stands its root 23 
 against 400 stands its root 20 
 against 900 stands its root 30 
 against 300 stands its root 17.3 
 and so on. 
 
 Problem V. 
 
 To find a mean 'proportional between two numbers. 
 
 As suppose between 29 and 430. — Set the one number 29 on 
 C to the same on D ; then against the other number 430 on C, 
 stands their mean proportional 111 on D. 
 
 Also the mean between 29 and 320 is 96*3. 
 
 And the mean between 71 and 274 is 139. 
 
 Problem VI. 
 
 To find a third proportional to two numbers. 
 
 Suppose to 21 and 32. — Set the first 21 on B to the second 
 32 on A; then against the second 32 on B stands 48*8 on A; 
 which is the third proportional sought. 
 
 Also the third proportional to 17 and 29 is 49"4. 
 
 And the third proportional to 73 and 14 is 2'5. 
 
 Problem VII. 
 
 To find a fourth proportional to three numbers : or^ to perfoimi 
 the Rule-of-Three. 
 
 Suppose to find a fourth proportional to 12, 28, and 114. — 
 Set the first term 12 on B to the second term 28 on A; then 
 against the third term 114 on B, stands 266 on A, which is the 
 fourth proportional sought. 
 
 Also the fourth proportional to 6, 14, 29, is (d7'Q* 
 And the fourth proportional to 27, 20, 73, is 54*0. 
 
 TIMBER MEASURING. 
 Problem I. 
 To find the area, or superficial content , of a board or plank. 
 Multiply the length by the mean breadth. 
 
 Note. When the board is tapering, add the breadth at the two ends to- 
 gether, and take half the sum for the mean breadth. 
 
 4 2 
 
TIMBER MEASURING. 
 
 BY THE SLIDING RULE. 
 
 Set 1 2 on B to the breadth m inches on A ; then against the 
 length in feet on B, is the content on A, in feet and fractional 
 parts. 
 
 EXAMPLES. 
 
 L What is the value of a plank, at Hd, per foot, whose 
 length is 12 feet 6 inches, and mean breadth 11 inches? 
 
 By decimals. By duodecimals. 
 
 12-5 12 6 
 
 11 11 
 
 12 
 IM. is-i- 
 
 137-5. 
 11-46 
 l5. 5d. 
 
 Vns. 
 
 Ud. is i 
 
 5 in. is -r^ 
 
 11 
 
 5 6 
 
 Is. 
 
 
 4id. 
 
 Is. rtd. Ans, 
 
 BY THE SLIDING RULE. 
 
 As 12 B : 11 A : : 12i B : Hi A. 
 *fhat is, as 12 on B is to 11 on A, so is 12^ on B to 1 1|- on A. 
 
 EXAMPLES. 
 
 2. Required the content of a board, whose length is 11 feet 2 
 inches, and breadth 1 foot 10 inches. 
 
 Arts. 20 feet 5 inches 8". 
 
 3. What is the value of a plank, which is 12 feet 9 inches 
 long, and 1 foot 3 inches broad, at 2\d. a foot ? 
 
 Ans. 35. 3|J. 
 
 4. Required the value of five oaken planks at Zd. per foot, 
 each of them being 1 7|- feet long ; and their several breadths 
 as follow, namely, two of 13|- inches in the middle, one of 14| 
 inches in the middle, and the two remaining ones, each 18 inches 
 at the broader end, and 11:J- at the narrower, 
 
 Ans. £1 5s. M. 
 
 Problem II. 
 
 To find the solid content of squared or four-sided timber. 
 
 Multiply the mean breadth by the mean thickness, and the 
 product again by the length, and the last product will give the 
 content. 
 
 BY THE SLIDING RULE. 
 C D D C 
 
 As length : 12 or 10 : quarter girt : content. 
 
 That is as the length in feet on C, is to 12 on D when the 
 quarter girt is in inches, or to 10 on D when it is in tenths of 
 feet ; so is the quarter girt on D, to the content on C. 
 
TIMBER MEASURING. 
 
 77 
 
 Note 1. If the tree taper regularly from the one end to the other, either 
 take the mean breadth, and thickness in the middle, or take the dimensions at 
 the two ends, and half their sum for the mean dimension. 
 
 Note 2. If the piece do not taper regularly, but is unequally thick in some 
 parts and small in others ; take several different dimensions, add them all toge- 
 ther ; and divide their sum by the number of them, for the mean dimension. 
 
 Note 3. The quarter girt is a geometrical mean proportional between the 
 mean breadth and thickness, that is the square root of their product. Some- 
 times unskilful measurers use the arithmetical mean instead of it, that*is half 
 their sum ; but this is always attended with error, and the more so as the 
 breadth and depth differ the more from each other. 
 
 1. The length of a piece of timber is 18 feet 6 inches, the 
 breadths at the greater and less end 1 foot 6 inches and 1 foot 
 3 inches, and the thickness at the greater and less end 1 foot 
 3 inches and 1 foot ; required the solid content. 
 
 Decimals. Duodecimals. 
 
 I'D 1 6 
 
 1-25 13 
 
 2) 2-75 
 1-375 
 
 mean breadth 
 
 1-25 
 1- 
 
 2) 2-25 
 
 M25 
 1-375 
 
 5625 
 7875 
 3375 
 1125 
 
 1-5468875 
 18-5 
 
 mean depth 
 mean breadth 
 
 2) 2 
 
 1 
 
 9 
 
 4 
 
 6 
 
 1 
 1 
 
 3 
 
 
 
 2) 2 
 
 3 
 
 
 1 
 1 
 
 1 
 
 4 
 
 6 
 6 
 
 1 G 
 
 4 G 
 
 6 
 
 length 
 
 1 
 
 18 
 
 7734375 
 
 
 
 
 
 
 
 12375000 
 
 
 
 27 
 
 10 
 
 1 
 
 6 
 
 1546875 
 
 
 content 
 
 
 9 
 
 3 
 
 4 
 
 28-6171875 
 
 28 
 
 7 
 
 4 
 
 10 
 
 
 
 BY THE SLIDING 
 
 RULE. 
 
 
 
 
 B 
 
 A 
 
 B A 
 
 
 
 
 
 As 1 : 
 
 m 
 
 : : IGi : 22-3, 
 
 the mean squaie. 
 
 
 c 
 
 D 
 
 C D 
 
 
 
 
 
 As 1 : 
 
 1 
 
 : : 223 : 14-9, 
 
 quarter 
 
 girt. 
 
 
 
 c 
 
 D 
 
 D C 
 
 
 
 
 
 As Ifl : 
 
 12 
 
 : : 14-9 : 28*6, 
 
 the content. 
 
 
 
78 TIMBER MEASURING. 
 
 2. What is the content of the piece of timber, whose length i^ 
 24^ feet, and the mean breadth and thickness each 1*04 feet? 
 
 Ans, 26^ feet. 
 
 3. Required the content of a piece of timber, whose length is 
 20 '38 feet, and its ends unequal squares, the side of the greater 
 being 19|, and the side of the less 9|- inches. 
 
 Ans. 297-56 feet. 
 
 4. Required the content of the piece of timber, whose length 
 is 27'36 feet; at the greater end the breadth is 1*78, and the 
 thickness 1*23; and at the less end the breadth is 1*04, and 
 thickness 0.91. Ans. 4l'27S feet. 
 
 Problem III. 
 
 To find the solidity of round or unsquared timber. 
 
 Rule I., or common rule. — Multiply the square of the 
 quarter girt, or of \ of the mean circumference, by the length, 
 for the content. 
 
 BY THE SLIDING RULE. 
 
 As the length upon C : 12 or 10 upon D : : quarter girt, in 
 12ths or lOths, on D : content on C. 
 
 Note. 1. When the tree is tapering, take the mean dimensions as in the 
 former Problems, either by girting it in the middle, for the mean girt, or at 
 the two ends, and take half the sum of the two. But when the tree is very 
 irregular, divide into several lengths and find the content of each part sepa- 
 rately : or else, add all the girts together, and divide the sum by the number 
 of them, for the mean girt. 
 
 Note 2. This rule, which is commonly used, gives the answer about \ less 
 than the true quantity in the tree, or nearly what the quantity would be after 
 the tree is hewed square in the usual way ; so that it seems intended to make 
 an allowance for squaring the tree. When the true quantity is desired, use the 
 2nd Rule given below. 
 
 EXAMPLES. 
 
 1 . A piece of round timber being 9 feet 6 inches long, and its 
 mean quarter girt 42 inches ; what is the .content ? 
 
 Decimals. Duodecimals. 
 
 3-5 quarter girt 3 6 
 
 3.5 3 6 
 
 175 
 105 
 
 12-25 
 9-5 length 
 
 6125 
 11025 
 
 10 
 
 6 
 
 
 1 
 
 9 
 
 
 12 
 
 3 
 
 
 9 
 
 6 
 
 
 110 
 
 3 
 
 
 6 
 
 1 
 
 6 
 
 116-375 content 116 
 
TIMBER MEASURING. 79 
 
 BY THE SLIDING RULE. 
 C D D C 
 
 As 9-5 : 10 : : 35 : 116^ 
 Or 9-5 : 12 : : 42 : 116^ 
 
 2. The length of a tree is 24 feet, its girt at the thicker end 
 14 feet, and at the smaller end 2 feet; required the content. 
 
 Ans, 96 feet. 
 
 3. What is the content of a tree, whose mean girt is 3*15 
 feet, and length 14 feet 6 inches ? Am. S'9d29feet. 
 
 4. Required the content of a tree, whose length is 17i feet, 
 which girts in five different places as follows, namely, in the 
 first place 9*43 feet, in the second 7*92, in the third 6*15, in the 
 fourth 4-74, and in the fifth 3-16. Ans. 42-5195. 
 
 KuLE II. — Multiply the square of ^ of the mean girt by 
 double the length, and the product will be the content, very 
 near the truth. 
 
 BY THE SLIDING RULE. 
 
 As the double length on C : 12 or 10 on D :: -J- of the girt,. 
 in 12ths or lOths, on D : content on C. 
 
 EXAMPLES. 
 
 1. What is the content of a tree, its length being 9 feet 6 
 inches, and its mean girt 14 feet? 
 
 Decimals. Duodecimals. 
 
 2-8 i of girt 2 9 7 
 
 2-8 2 9 7 
 
 224 5 7 2 
 
 56 2 13 
 
 1 8 
 
 7-84 
 19 
 
 7056 
 
 784 
 
 148-96 content 
 
 BY THE SLIDING RULE. 
 
 ODD 
 
 As 19 : 10 : : 28 
 Or 19 : 12 : : 33y«o 
 2. Required the content of a tree, which is 24 feet long, and 
 mean girt 8 feet. Ans. 122'SS feet. 
 
 7 
 
 10 
 
 1 
 
 19 
 
 
 
 48 
 
 11 
 
 7 
 
 -" 
 
 c 
 
 
 : 
 
 149 
 
 
 
 149 
 
 
80 artificers' work. 
 
 3. The length of a tree is 14^ feet, and mean girt 3*15 feet; 
 what is the content ? Ans. II' 51 feet. 
 
 4, The length of a tree is 17i feet, and its mean girt 6*28; 
 what is the content? Am. 54'425 feet. 
 
 ARTIFICERS' WORK. 
 
 Artificers compute the contents of their works hy several 
 different measures. 
 
 As glazing by the square foot. 
 
 Painting, masonry, plastering, paving, &c., by the yard of 9 
 square feet. 
 
 Flooring, partitioning, roofing, tiling, &c., by the square of 
 100 square feet. 
 
 And brick-work, either by the yard of 9 square feet, or by the 
 square rod of 27 2^ square feet, or 30:^ square yards, being the 
 square of the rod of 1 6^ feet or 5^ yards long. 
 
 As this number 272:| is a troublesome number to divide by, 
 the i is often omitted in practice, and the content in feet divided 
 only by the 272. But when the exact divisor 2 72 J is to be used, 
 it will be easier to multiply the feet by 4, and then divide suc- 
 cessively by 9, 11, and 11. Also to divide square yards by 30^, 
 first multiply them by 4, and then divide twice by 1 1 . 
 
 All works, whether superficial or solid, are computed by the 
 rules proper to the figure of them, whether it be a triangle, or 
 rectangle, a parallelopiped, or any other figure. 
 
 BRICKLAYERS' WORK. 
 
 Brick- work is usually estimated by the rod at the rate of a 
 brick and a half thick ; so that if a wall be more or less than this 
 standard thickness, it must be reduced to it, as follows : Multiply 
 the superficial content of the wall by the number of half bricks 
 in thickness, and divide the product by 3. And to find the su- 
 perficial content of a wall, multiply the length by the height in 
 feet, and divide the product by 272, for the content in rods. 
 
 Chimneys are by some measured as if they were solid, deducting 
 only the vacuity from the hearth to the mantle, on account of the 
 trouble of them. 
 
 All windows, doors, &c., are to be deducted out of the contents 
 
BRICKLAYERS WORK. 
 
 81 
 
 of the wall in which they are placed, and a distinct charge must 
 be made for window heads, sills, quoins, &c. 
 
 A rod of brickwork of standard thickness contains about 305 
 cubic feet, and 4,500 bricks, making due allowance for mortar 
 joints, &c. Assuming the cost of bricks at the kiln to be 3(}5. 
 per thousand, the cost of bricks per rod will be 36s, X 4|- = 
 ^8 2^. The cost for building a rod of brickwork may be taken 
 at SGs. The cost of cartage and mortar varies with the locality 
 of the building, this may be taken to average about 205. per rod. 
 These items being collected, there results - 
 
 Cost of bricks 
 Cost of labour 
 Cost of carta^ie and mortar 
 
 Profit at 10 per cent 
 
 Cost per rod . . . . = 11 19 10,or^l2 
 per rod. 
 
 Note. — The student must, of course, make his estimation according to tne 
 locaUty of the building, this being only a specimen of the method adopted by 
 inteUigent architects and builders. 
 
 EXAMPLES. 
 
 1 . How many rods of standard brickwork are in a wall whose 
 length or compass is 57 feet 3 inches, and height 24 feet 
 G inches ; the walls being 2|- bricks, or 5 half bricks thick ? 
 Decimals. Duodecimals. 
 
 57*25 57 3 
 
 24-5 24 6 
 
 £ s. 
 = 82 
 = 1 16 
 = 1 
 
 
 
 
 
 10 18 
 = 1 1 
 
 
 10 
 
 28625 
 22900 
 11450 
 
 1402-625 
 5 
 
 half bricks 
 
 3 
 272 
 
 7013-125 
 
 2337*7081 square feet. 
 
 234 
 
 
 
 
 114 
 
 
 
 28 
 
 7 
 
 6 
 
 1402 
 
 7 
 
 6 
 5 
 
 3 7013 
 
 1 
 
 6 
 
 72 2337 
 
 S 
 
 6 
 
 8-594 rods. 8r, 16\/L 8' 6" 
 
 2. A triangular gable is raised 17^ feet high, on an end wall 
 whose length is 24 feet 9 inches, the thickness being 2 bricks ; 
 required the reduced content in square yards. Ans. 320-8} ijds. 
 
82 masons' work. 
 
 3 The end wall of a house is 28 feet 10 mches long, and 55 
 feet 8 inches high to the eaves, 20 feet high is 2^ bricks thick, 
 other 20 feet high is 2 bricks thick, and the remaining 15 feet 
 8 inches is 1|- brick thick, above which is a triangular gable of 
 1 brick thick, which rises 42 courses of bricks, of which every 
 four courses make a foot. What is the whole content, and cost 
 
 at *i- per roa . A7is, | ^^^^ ^^^^ ^2^^ 2^/. 
 
 MASONS' WORK. 
 
 To masonry belongs all sorts of stone work ; and the measure 
 made use of is a square foot or square yard. 
 
 Walls, columns, blocks of stone or marble, &c., are measured 
 by the cubic foot, and pavements, slabs, chimney-pieces, &c., by 
 the superficial or square foot. 
 
 Cubic or solid measure is used for the materials, and square 
 measure for the workmanship. 
 
 In the solid measure, the true length, breadth, and thickness, 
 are taken, and multiplied continually together. In the super- 
 ficial, there must be taken the length and breadth of every part 
 of the projection, which is seen without the general upright face 
 of the building. 
 
 The cost of stones for walling varies with the locality, as 
 already noticed in bricklayers' work. 
 
 A square yard of rubble walling 2 feet thick weighs 1| tons, 
 and assuming the cost of rubble stones at the quarry to be 6d. 
 per ton, the cost of cartage Is. per ton, the cost of mortar 4c?., 
 and that of labour 1*. 8d, per square yard, we shall have the fol- 
 lowing 
 
 Estimate /or a square yard of ruhhle walling 2 feet in thickness, 
 
 s, d. 
 Cost of materials , , == 6d, x 
 Cost of cartage . . = I5. x 
 Cost of labour .... 
 Cost of mortar .... 
 
 = 
 
 7i 
 
 = 1 
 
 3 
 
 = 1 
 
 8 
 
 = 
 
 4 
 
 3 
 
 m 
 
 = 
 
 H 
 
 Profit at 10 per cent . 
 
 Total cost per square yard . . . r= 4 3 
 An additional charge must be made for quoins, window heads 
 and sills, when used in houses constructed of rubble work : also. 
 
carpenters' and joiners' work. 83 
 
 when the fronts of houses are constructed entirely with Ashler 
 work, a separate estimate must be made for it. Ashlers usually 
 average 9 inches in the bed, this width must, therefore, be de- 
 ducted from whole width of the wall, and the remainder estimated 
 as rubble work, to which the additional cost at the quarry, and 
 of hewing the ashler, must be added. 
 
 EXAMPLES. 
 
 1 . Required the square yards and cost of a rubble wall of the 
 specified thickness, the length of which is 53 feet 6 inches, and 
 the height 12 feet 3 inches, at 4*. Sd. per square yard. 
 
 53-5 X 12-25 -^ 9 = 72*82 square yards, 
 
 and 72-82 X 4^. Sd. = 2^15 9^. 5J^. cost. 
 
 BY THE SLIDING RULE. 
 
 B A B A 
 
 9 : 53^ : : 12J : 72f square yards. 
 
 2. Required the value of a marble slab, at 8s. per foot ; the 
 length being 5 feet 7 inches, and breadth 1 foot 10 inches. 
 
 Ans. £4 Is. lOid. 
 
 3. In a chimney piece, suppose the length of the mantle and 
 
 slab, each . . . , . . 4 ft. 6 in. 
 
 breadth of both together . . .32 
 
 length of each jamb . . . .44 
 
 breadth of both together . . .19 
 
 required the superficial content. Ans. 2\ft. 10 in. 
 
 CARPENTERS' AND JOINERS' WORK. 
 
 To this branch belongs all the wood-work of a house, such as 
 flooring, partitioning, roofing, &c. "" 
 
 The large and plain parts are usually measured by the square 
 of 100 feet ; but enriched mouldings, and some other articles, 
 are often estimated by running or lineal measure, and some things 
 are rated by the piece. 
 
 Joists are measured by, multiply the depth, breadth, and 
 length all together, for the content of one joist ; multiply that 
 by the number of the joists. Note, that the length of the joists 
 will exceed the breadth of the room by the thickness of the wall 
 and ^rd of the same, because each end is let into the wall aboi:t 
 ■|rd of its thickness. 
 
 Partitions are measured from wall to wall for one dimension. 
 
84 carpenters' and joiners' work. 
 
 and from floor to floor, as far as they extend, for the other ; then 
 multiply the length by the height. 
 
 In measuring joiners' work, the string is made to ply close to 
 every part of the work over which it passes. 
 
 In roofing, the length of the rafters is equal to the length of 
 a string stretched from the ridge down the rafter, and along the 
 eaves-board, till it meets with the top of the wall. This length 
 multiphed by the common depth and breadth of the rafters, gives 
 the content of one, and that multiplied by the number of them, 
 gives the content of all the rafters. 
 
 King post roofs, ^c, all the timbers in a roof are measured 
 
 in the same manner as the 
 joists, &c., in flooring. In the 
 annexed figure, representing 
 a truss for a roof, all the 
 beams, as the tie-beam, king- 
 post, braces, &c., are mea- 
 sured to their full lengths, 
 breadths, and thicknesses, in- 
 cluding the lengths of tenons ; also the parts cut out on each 
 side of the king-post, to form abutments for the braces, are in- 
 cluded; unless their lengths exceed 2 feet each by 3 inches 
 breadth, w^ien their solidities must be deducted, pieces of smaller 
 size, being considered of little or no value, are, therefore, included 
 in the measurement. 
 
 For stair -cases, take the breadth of all the steps, by making 
 a line ply close over them, from the top to the bottom ; and 
 multiply the length of this line by the length of a step for the 
 whole area. — By the length of a step, is meant the length of 
 the front and the returns at the two ends ; and by the breadth, 
 is to be understood the girt of its two outer surfaces, or the 
 tread and rise. 
 
 For the hali^trade, take the whole length of the upper part 
 of the hand-rail, and girt over its end till it meet the top of the 
 newel post, for one dimension ; and twice the length of the 
 baluster upon the landing, with the girt of the hand-rail, for the 
 other dimension. 
 
 For loainscoting , take compass of the room for one dimen- 
 sion ; and the height from the floor to the ceiling, making the 
 string ply close into all the mouldings, for the other dimension. 
 — Out of this must be made deductions, for windows, doors, and 
 chimneys, <S:c. 
 
 For doors, it is usual to allow for their thickness, by adding 
 it into both the dimensions of length, and breadth, and then 
 
CARPENTERS* AND JOINERS* ^VORK. 85 
 
 multiply them together for the area. — If the door be pannelled 
 on both sides, take double its measure for the workmanship ; 
 but if one side only be pannelled, take the area and its half for 
 the workmanship. — For the surrounding arcJdtravei girt it 
 about the outermost part for one dimension, and measure over 
 it as far as it can be seen when the door is open, for the other. 
 
 Window-shutters, bases, tjr., are measured in the same 
 manner. 
 
 EXAMPLES. 
 
 1. Required the content of a floor 48 feet 6 inches long, and 
 21 feet 3 inches broad. 
 
 Decimals. Duodecimals. 
 
 48-5 48 6 
 
 24i 24 3 
 
 1940 204 
 
 970 dQ 
 
 12-125 12 1 
 
 1176-125 feet 1176 1 6 
 
 11-76125 squares Ans. 11*76 1 6 
 
 2. A floor being 36 feet 3 inches long, and 16 feet 6 inches 
 broad, how many squares are in it ? Ans, 5 squares 9S\feet, 
 
 3. How many squares are there in 173 feet 10 inches in 
 length, and 10 feet 7 inches height, of partitioning ? 
 
 Ans. IS' ^97^ squares, 
 
 4. What cost the roofing of a house at 105. 6d, sl square; 
 the length, within the walls, being 52 feet 8 inches, and the 
 breadth 30 feet 6 inches : reckoning the roof f of the flat ? 
 
 Ans, £\2 I2s. Uid, 
 
 5. To how much, at 6s. per square yard, amounts the wain- 
 scoting, of a room ; the height, taking in the cornice and 
 mouldings being 12 feet 6 inches, and the whole compass 83 
 feet 8 inches ; also the* three window shutters are each 7 feet 8 
 inches by 3 feet 6 inches, and the door 7 feet by 3 feet 6 inches ; 
 the door and shutters, being worked on both sides, are reckoned 
 work and half work? Jns. £36 I2s, 2\d, 
 
 6. In a naked floor there are 2 girders, each 20 feet long, 
 and 1 foot 2 inches by 1 foot ; there are 1 6 bridging joists, each 
 20 feet long, and 6\ inches by 3 ; 16 binding joists, each 9 feet 
 in length, and %\ inches by 4 : 48 ceiling joists, each 6 feet 
 long, and 4 inches by 2\ : required the content in cubic feet. 
 
 Ans. 144 cubic feet. 
 
86 slaters' and tilers' work. 
 
 7. What will the wainscoting of a room cost at 4*. per square 
 yard ; the height of the room, including cornice and the mould- 
 ings, is 1 2|- feet and the compass 1 25^ feet ; there are three 
 window shutters, each 7 feet 8 inches by 3^ feet, and the door 
 7 feet by 3|- feet ; the door and shutters, being worked on both 
 sides, are reckoned half work additional? Ans. £36 \2s. 2\d. 
 
 SLATERS' AND TILERS' WORK. 
 
 In these articles, the content of a roof is found by multiply- 
 ing the length of the ridge by the girt over from eaves to eaves ; 
 making allowance in this girt for the double row of slates at 
 the bottom, or for how much one row of slates or tiles is laid 
 over another. 
 
 When the roof is of a true pitch, that is, forming a right 
 angle at top ; then the breadth of the building with its half 
 added, is the girt over both sides. 
 
 In angles formed in a roof, running from, the ridge to the 
 eaves, when the angle bends inwards, it is called a valley ; but 
 when outwards, it is called a hip. 
 
 Deductions are made for chimney shafts or window holes. 
 
 EXAMPLES. 
 
 1 . Required the content of a slated roof, the length being 45 
 feet 9 inches, and the whole girt 34 feet 3 inches. 
 
 Decimals. 
 45-75 
 34i 
 
 Duodecimal 
 45 9 
 34 3 
 
 s. 
 
 18300 
 13725 
 114375 
 
 205 6 
 135 
 11 5 
 
 3 
 
 9)1566-9375 feet 
 ds 174104 
 
 9)1566 11 
 1 74 yds. 
 
 3 
 11' 3' 
 
 2. To how much amounts the tiling of a house, at 25^. 6d. 
 per square ; the length being 43 feet 10 inches, and the breadth 
 on the flat 27 feet 5 inches, also the eaves projecting 16 inches 
 on each side, and the roof of a true pitch ? Ans. £24 9s. o^d. 
 
plasterers' work. 87 
 
 PLASTERERS' WORK. 
 
 Plasterers' work is of two kinds, namely, ceiling, which is 
 plastering upon laths ; and rendering, which is plastering upon 
 walls : which are measured separately. 
 
 The contents are estimated either by the foot or yard, or square 
 of 100 feet. Enriched mouldings, &;c., are rated by running or 
 lineal measure. 
 
 Deductions are to be made for chimneys, doors, windows, &c. 
 
 EXAMPLES. 
 
 1. How many yards contain the ceiling, which is 43 feet 3 
 inches long, and 25 feet 6 inches broad ? 
 
 Decimals. 
 43-25 
 25i 
 
 21625 
 8650 
 21625 
 
 Duodecimals. 
 
 
 43 
 
 3 
 
 
 25 
 
 6 
 
 
 221 
 
 3 
 
 
 86 
 
 
 
 21 
 
 7 
 
 6 
 
 9) 1102-875 9) 1102 10 6 
 
 yards 122-541 Ans. 122 yds. 4/t. 10' 6". 
 
 2. To how much amounts the ceiling of a room, at 10c?. per 
 yard ; the length being 21 feet 8 inches, and the breadth 14 feet 
 10 inches? A7is. £1 9s. S^d. 
 
 3. The length of a room is 18 feet 6 inches, the breadth 12 
 feet 3 inches, and height 10 feet 6 inches ; to how much amounts 
 the ceiling and rendering, the former at 8c?. and the latter at Sd. 
 per yard ; allowing for the door of 7 feet by 3 feet 8, and a fire- 
 place of 5 feet square? Ans. £\ \Zs. 3d. 
 
 4. Required the quantity of plastering in a room the length 
 being 14 feet 5 inches, breadth 13 feet 2 inches, and height 9 feet 
 3 inches to the under side of the cornice, which girts 8|- inches, 
 and projects 5 inches from the wall on the upper part next the 
 ceiling ; deducting only for a door 7 feet by 4. 
 
 'bZyds. bft. 3 in. of rendering. 
 Ans. <( 18 5 6 of ceiling. 
 
 39 0^ of cornice. 
 
 {■ 
 
88 painters' and glaziers' work. 
 
 PAINTERS' WORK. 
 
 Painters' work is computed in square yards. Every part is 
 measured where the colour Hes ; and the measuring Une is forced 
 into all the mouldings and corners. 
 
 Windows are done at so much a piece ; and it is usual to allow 
 double measure for carved mouldings, &c. 
 
 EXAMPLES. 
 
 1 . How many yards of painting contains the room which is 
 65 feet 6 inches in compass, and 12 feet 4 inches high? 
 
 Ans. 89 yards 6 feet 10'. 
 
 2. The length of a room being 20 feet, its breadth 14 feet 6 
 inches, and height 10 feet 4 inches ; how many yards of painting 
 are in it, deducting a fire-place of 4 feet by 4 feet 4 inches, and 
 two windows each 6 feet by 3 feet 2 inches? Jns, IZ-^^yds. 
 
 3. What cost the painting of a room at 6d. per yard; its 
 length being 24 feet 6 inches, its breadth 1 6 feet 3 inches, and 
 height 12 feet 9 inches; also the door is 7 feet by 3 feet 6, and 
 the window shutters to two windows each 7 feet 9 by 3 feet 6, 
 but the breaks of the windows themselves are 8 feet 6 inches 
 high, and 1 foot 3 inches deep : deducting the fire-place of 5 feet 
 by 5 feet 6? Ans, £Z Zs. \^\d. 
 
 GLAZIERS' WORK. 
 
 Glaziers take their dimensions either in feet, inches, and parts, 
 or feet, tenths and hundredths. And they compute their work in 
 square feet. 
 
 In taking the length and breadth of a window, the cross bars 
 between the squares are included. Also windows of round or 
 oval forms are measured as square, measuring them to their 
 greatest length and breadth, on account of the waste in cutting 
 the glass. 
 
 EXAMPLES. 
 
 1. How many square feet contains the window which is 4*25 
 feet long, and 2*75 feet broad? 
 
 Decimals. Duodecimals. 
 
 2-75 2 9 
 
 4i 4 3 
 
 11-00 11 
 
 •6875 8 
 
 11-6875 Ans. 11 8 
 
PAVIORS* AND PLUMBERS* WORK. SS* 
 
 2; What will the glazing a triangular sky-light come to at 
 10c?. per foot; the base being 12 feet 6 inches, and the perpen- 
 dicular height 6 feet 9 inches? Ans. £\ 155. \^d, 
 
 3. There is a house with three tier of windows, three windows 
 iii each tier, their common breadth 3 feet 1 1 inches ; 
 
 now the height of the first tier is 7 feet 1 inches, 
 of the second 6 8 
 
 of the third 5 4 
 
 Required the expense of glazing at lAd. per foot. 
 
 Ans, £13 Us. \^d. 
 
 4. Required the expense of glazing the windows of a house at 
 \3d, a foot ; there being three stories, and three windows in each 
 story : the height of the lower tier is 7 feet 9 inches, 
 
 of the middle G C 
 
 of the upper 5 3 
 
 and of an oval window over the door 1 \0\ 
 
 The common breadth of all the windows being 3 feet 9 inches. 
 
 Ans, £\2 bs. Gd. 
 
 PAVIORS' WORK. 
 
 Paviors' work is done by the square yard, and the content is 
 found by multiplying the length by the breadth. 
 
 EXAMPLES. 
 
 1. What cost the paving a foot-path at 3^. 4d. per yard ; the 
 length being 35 feet 4 inches, and breadth 3 feet 3 inches ? 
 
 Ans. Content 32 yards 3 feet 6'. Cost So 7s. ll^d. 
 
 2. What will be the expense of paving a rectangular court yard, 
 whose length is 63 feet, and breadth 45 feet ; in which there is 
 laid a foot path of 5 feet 3 inches broad, running the whole 
 length, with broad stones, at 3*. a yard ; the rest being paved 
 with pebbles at 2^. 6d. a yard? Ans. ^40 5s. lO^d, 
 
 PLUMBERS* WORK. 
 
 Plumbers' work is rated at so much a pound, or else by the 
 hundred weight of 112 pounds. 
 
 Sheet lead used in roofing, guttering, &c., is from 7 to 12fts. 
 to the square foot. And a pipe of an inch bore is commonly 
 13 or 14 ibs. to the yard in length. 
 
90 ARCHED AND VAULTED ROOFS. 
 
 EXAMPLES. 
 
 1. How much weighs the lead which is 39 feet 6 inches long, 
 and 3 feet 3 inches broad, at 8^ lbs. to the square foot ? 
 Decimals. Duodecimals. 
 
 39-5 39 6 
 
 3i 3 3 
 
 118-5 
 
 9-875 
 
 128-375 
 
 8i 
 
 1027000 
 64-1875 
 
 118 
 
 6 
 
 
 9 
 
 10 
 
 6 
 
 128 
 
 4 
 
 6 
 
 Si 
 
 
 
 1024 
 
 
 64 
 
 
 
 2f 
 
 
 
 OH 
 
 
 
 1091-1872 Ans. 1091^-^ lbs. 
 
 2. What cost the covering and guttering a roof with lead, at 
 185. the cwt. ; the length of the roof being 43 feet, and breadth 
 or girt over it 32 feet; the guttering 57 feet long, and 2 feet 
 wide: the former 9-831 lbs., and the latter 7-373 lbs. to the 
 square foot? Ans. £115 9s. l^d. 
 
 ARCHED AND VAULTED ROOFS. 
 
 To find the concave surface of a circular^ gothic, or elliptical 
 vaulted roof 
 
 Rule. — Multiply the length of the arch by the length of the 
 vault for the required surface. 
 
 Note, — The length of the arch, for most practical purposes, is found by 
 applying a small chord along its concavity, and then measuring its length. 
 
 EXAMPLE. 
 
 What is the concave surface of Caerfylly bridge, over the Taafe ; 
 which is a segment of a circle, span 140, height 35, and width 12 
 feet? Ans. \9^A'4: squai^efeet. 
 
 To find the content of a circular ^ gothic, or elliptical roof. 
 Rule. — Multiply tlie area of the eud by the length of the 
 roof for the content of the vacuity. 
 
 To find the solid content of the materials. 
 From the solid content of the whole arch take that of the 
 vacuity for the solid content of the materials. 
 
 EXAMPLE. 
 
 Required the solidity of the vacuity and of the materials of a 
 
SPECIFIC GRAVITY. 
 
 91 
 
 circular vault ; span 36 feet, height 18 feet, thickness of walls 
 at the spring G feet, thickness of crown 4 feet, and length of the 
 vault 100 feet? 
 
 J r 1884-96 cubic yards solidity of vacuity. 
 '^** \ 2026*15 cubic yards solidity of materials. 
 
 To find the surface and solidity of a dome, the height and 
 dimensions of the base being given. 
 
 Rule. — Take twice the area for the surface, and multiply the 
 area of the base by -|rd of the height. 
 
 Note. Although these rules are only true when the domes are hemi- 
 spherical, yet they are sufficiently near the truth for all practical purposes. 
 
 EXAMPLE. 
 
 Required the surface and soUdity of a hemispherical dome, the 
 diameter of its base being 60 feet. 
 
 J f Surface 314*16 square yards, 
 ^'^'^- I Solidity 2094*4 cubic yards. 
 
 Note 1. The surface of a saloon is found in the same manner as a vaulted 
 roof. 
 
 Note 2. Rules might have been here given for the measurement of hay- 
 stacks, coal-heaps, &c ; but these may be readily resolved into two or more 
 of those solids, the methods of finding the content of which are given in the 
 Mensuration of Solids, Part IV. Moreover, haystacks are usually sold by 
 weight, and seldom or never by measurement. 
 
 SPECIFIC GRAVITY. 
 
 The specific gravity of bodies are their weights when com- 
 pared with an equal bulk of pure water, which, at the tempera- 
 ture of 40°, weighs 1000 ounces avoirdupois per cubic foot. The 
 following table, therefore, contains the weights of a cubic foot of 
 several bodies in ounces. 
 
 A table of tke specific gravity of bodies. 
 
 Platinum . 
 
 21470 
 
 Light earth 
 
 1984 
 
 Gold 
 
 19260 
 
 Solid Gunpowder 
 
 1745 
 
 Mercury . 
 
 13600 
 
 Sand 
 
 1520 
 
 Lead 
 
 11325 
 
 Coal . from 1030 
 
 to 1300 
 
 Silver 
 
 10470 
 
 Pitch 
 
 1150 
 
 Copper 
 
 9000 
 
 Box-wood . 
 
 1030 
 
 Cast brass 
 
 8400 
 
 Sea-water . 
 
 1030 
 
 Steel 
 
 7850 
 
 Common water . 
 
 1000 
 
 Iron 
 
 7704 
 
 Mahogany . 
 
 1065 
 
 Cast Iron 
 
 7065 
 
 Oak . 
 
 925 
 
 Tin 
 
 7320 
 
 Ash . 
 
 755 
 
 Granite 
 
 3950 
 
 Beech 
 
 700 
 
 Flint Glass 
 
 3000 
 
 Elm . 
 
 600 
 
 Marble . 
 
 2700 
 
 Fir . . . 
 
 540 
 
 Freestone 
 
 2500 
 
 Cork . 
 
 240 
 
 Clay 
 
 2160 
 
 Air . . , 
 
 1. 
 
 Brick 
 
 2000 
 
 
 
92 LAND SURVEYING. 
 
 To find the weiyht of a hody from its bulk, 
 
 Rule. — Multiply the content of the hody, in cubic feet, by 
 its tabular specific gravity for its weight in avoirdupois ounces. 
 
 EXAMPLES. 
 
 1 . Required the weight of a block of marble, whose length is 
 63 feet, and breadth and thickness each 12 feet, being the dimen- 
 sions of one of the stones in the walls of Balbec. 
 
 63 X 122 X 2700 ^^^ ^ , . , . 
 
 — 1 fi y n 2 X 20 — ~ ^^^16" ^^^s, which is equal to the 
 
 burden of a second rate East India ship. 
 
 2. What is the weight of a block of dry oak, which measures 
 10 feet long, and 3 feet by 2\ ? Ans. 4336 lb. nearly. 
 
 To find the maynitude of a body from its weiyht. 
 
 Rule. — Divide its weight in avoirdupois ounces by its tabular 
 specific gravity for its content in cubic i^^i, 
 
 EXAMPLES. 
 
 1. Required the content of an irregular block of freestone, 
 which weighs 1 cwt. 
 
 112 X 16 ,. ^ 
 2520 ~ ^^^^^ ^^^^' 
 
 , 112 X 16 X 1728 ,. . , 
 
 and 9~"~9n ~ 1228*8 cubic inches. 
 
 2. How many cubic feet are there in a ton of dry oak ? 
 
 Ans. 38^ff cubic feet, 
 
 3. A cast iron pipe is 6 inches diameter in the bore and 1 inch 
 in thickness ; required the weight of a running foot. 
 
 Ans. 67*45 lbs. 
 
 LAND SURVEYING. 
 
 Description of Instruments used for Measuring and 
 Planning Surveys. 
 
 the chain. 
 The chain, usually called Gunter's chain, is almost generally 
 used in the British dominions, for measuring the distances re- 
 quired in a survey. It is (!)(S feet, or 4 poles, in length, and is 
 divided into 100 links, which are joined by rings. The length 
 of each link, together with half the rings connecting it with the 
 
SURVEYING INSTRUMENTS, &C. 93 
 
 adjoining hnivs, is consequently — - ot a toot, or ____=:/ -92 
 
 inches. At every tenth hnk from eacli end is attached a piece 
 of hrass ^vith notches ; that at the tenth Hnk has one notch, 
 that at the 20th two notches, that at the 30th three, that at the 
 40th four, the middle of the chain, or the 50th link heing 
 marked with a large round piece of hrass ; hence, any distance 
 on the chain may be readily counted. Part of the first link, at 
 each end, is formed into a large ring for the purpose of hokling 
 it with the hand. 
 
 The chain acquires extension by much use, it should there- 
 fore, be frequently examined, and adjusted to the proper length 
 by taking out some of the rings between the links : for this pur- 
 pose, chains having three rings between each link are to be pre- 
 ferred to those having only two. 
 
 THE OFFSET STAFF. 
 
 The offset staff is used to measure short distances, called offsets; 
 hence its name. It is usually ten links in length, the links 
 being numbered thereon with the figs. 1, 2, 3, &c. It is 
 usually pointed with iron at one end, for the purpose of fixing 
 it in the ground, as an object for ranging lines, for marking 
 stations, »S:c. 
 
 THE CROSS. 
 
 The cross is an instrument used by surveyors to erect per- 
 pendiculars. It is usually a round piece of sycamore, box, or 
 mahogany, about four inches in diameter, with two folding 
 sights at right angles to each other, or more commonly with 
 two fine grooves sawed at right angles to each other, which 
 answer the purpose of sights. It is sometimes fixed on a staff 
 of convenient length for use, pointed with iron at the bottom, that 
 it may be fixed firmly in the ground : but it is found more com- 
 modious in practice to have a small pocket cross, which may be 
 readily fitted to the offset-staff, either by an iron spike on the cross 
 being inserted in a hole made in the offset-staff, or the offset-staff 
 being passed through a hole made in the cross, to about the 
 eighth link from the piked end, at which place the staff must be 
 shouldered, that the cross may rest firmly. 
 
 DIRECTIONS FOR MEASURING LINES ON THE GROUND. 
 
 Besides the instruments already described, ten arrows must be 
 provided, about 12 inches long, pointed at the end, so as to be 
 readily pressed into the ground, and turned at the other end, so 
 as to form a ring to serve for a handle. 
 
94 MEASURING LINES ON THE GROUND. 
 
 In using the chain, marks are to be set up at the extremities 
 of the hne to be measured, as well as its intermediate points, if 
 its extremities cannot be seen from one anollier, on account 
 of hills, woods, hedges, or other obstructions. Two persons 
 are then required by the surveyor to perform the measurement. 
 The chain leader starts with the ten arrows in his left hand, 
 and one end of the chain in his right ; while the follower re- 
 mains at the starting point, who, looking at the staff or staves, 
 that mark the line to be measured, directs the leader to extend 
 the chain in the direction of the staif or staves. The leader 
 then puts down one of his arrows, and proceeds a second chain's 
 length in the same direction, while the follower comes up to the 
 arrow first put down. A second arrow being now put down 
 by the leader, the first is taken up by the follower ; and the 
 same operation is repeated till the leader has expended all his 
 arrows. Ten chains, or 1000 links, having now^ been measured 
 and noted in the field book, the follower returns the ten arrows 
 to the leader, and the same operation is repeated as often as 
 necessary. When the leader arrives at the end of the line, 
 the number o£ arrows in the follower's hand shows the number 
 of chains measured since the last exchange of ar- 
 rows noted in the field book, and the number of 7000 
 links extending from the last arrow to the mark or 600 
 
 staff at the extremity of the line, being also added, 83 
 
 gives the entire measurement of the line. Thus, if 
 
 the arrows have been exchanged seven times, and 7683 
 if the follower have six arrows, and from the arrow last put 
 down to the end of the line be 83 links, the whole measurement 
 will be 7683 links, or 76 chains 83 links, which is usually 
 written thus — 76*83 chains, the two last figures being decimals of 
 a chain. 
 
 In using the chain, care must be taken to stretch it always with 
 the same tension, as it will extend by much use, and will there- 
 fore require to be examined occasionally, and shortened, if neces- 
 sary. But a good chain may be used sev^eral days on tolerably 
 smooth ground, without any material extension. 
 
 The surveyor must mark, or caused to be marked, every station 
 on the line, while it is being measured, with a staff or cross on the 
 ground, entering its distance in the field book. 
 
 When a survey is made for a finished plan, all remarkable 
 objects should be noted down ; as buildings, roads, rivers, ponds, 
 footpaths, gates, &c. 
 
 The boundary of the estate measured ought to be carefully 
 observed. If the ditch be outside the boundary fence, it usually 
 
PLOTTING INSTRUMENTS. 95 
 
 belongs to the estate, and vice versd ; although this is not uni- 
 formly the case ; therefore, inquiry ought to be made with re- 
 spect to the real boundary. 
 
 In some places five links from the hedge-post or roots of the 
 quickwood are allowed for the breadth of the ditch, but this 
 breadth varies to as far as even ten links, especially in swampy 
 countries. 
 
 All ditches and fences must be measured with the fields to 
 which they belong, when the full quantity on the plan is re- 
 quired : but when the growing crops only are to be measured, 
 only so much as is occupied by the crops. 
 
 Instruments for Laying Down or Plotting Surveys. 
 
 THE common drawing COMPASSES. 
 
 This instrument consists of two legs moveable about a joint, 
 so that the points at the extremities of the legs may be set at 
 any required distance from one another : it is used to transfer 
 and measure distances, and to describe arcs and circles. 
 
 Note. There are several other kinds of compasses, used for planning ; as 
 those with moveable points, for the introduction of black lead pencils, or ink 
 points, beam compasses for taking large distances, proportional compasses, 
 &c., the uses of which are easily learned. (See Heather s Treatise on Mathe- 
 matical Instruments.) 
 
 PLOTTING SCALES. 
 
 Plotting scales, also called featlier-edged scales, are straight 
 rulers, usually about 12 inches long. Each ruler has scales of 
 equal parts, decimally divided, placed on its edges, which are 
 made sloping, so that the extremities of the strokes marking 
 the divisions He close to the paper. The primary divisions re- 
 present chains, and the subdivisions ten links each, the inter- 
 mediate links being determined by the eye. Plotting scales may 
 be procured in sets, each with a different number of chains to 
 the inch. They are usually made of ivory or box, and each pro- 
 vided with a small scale called an offset scale for laying down the 
 offsets. In using these scales, the first division or zero, on the 
 plotting scale, is placed coincident with the beginning of the line 
 to be plotted, and so as just to touch that line with the feather- 
 edge : the end of the offset scale is then placed in contact with 
 the edge of the plotting scale; and thus the offsets may be ex- 
 peditiously pricked off: for which purpose an instrument called 
 
96 FIELD BOOK. 
 
 a pricker is used, but a hard black lead pencil with a fine point, 
 is greatly to be preferred, as it does not injure the paper. 
 
 The vernier scale and protractor, the uses of which will be 
 readily understood, are best adapted to laying down extensive 
 surveys, where great accuracy is required. (See Baker's Land 
 and Engineering Surveying, Weale's Series.) 
 
 PLANNING SURVEYS. 
 
 In planning or plotting surveys, the upper part of the paper 
 or book, on which the plan is made, should always, if possible, 
 be the north. The chain lines, buildings, fences, &c., ought 
 first to be drawn with a fine black lead pencil : the first should 
 then be dotted with ink, and the latter neatly drawn. Great 
 care is required in the construction of the plan, when the dimen- 
 sions are to be measured therefrom with the scale. The scale 
 should never be more than three chains to an inch, for when the 
 parts of a plan are large, the dimensions may be taken with 
 greater accuracy. After having found the content of the field or 
 fields, &c., of which any plan consists, it may be laid down by 
 any scale to give it a more convenient size. 
 
 THE FIELD BOOK. 
 
 The method now generally adopted in setting down field notes, 
 and which has long been found to be the best in practice, is to 
 begin at the bottom of the page and write upwards. 
 
 Each page of the book is usually divided into three columns. 
 The middle column is for distances measured on the chain line, 
 at which hedges are crossed, or offsets, stations, or other marks 
 are made ; and in the righl and left columns, those offsets, 
 marks, and any other necessary observations thereon, must be 
 entered, accordingly as they are situated on the right or left of 
 the chain-line. 
 
 The crossing of roads, rivers, hedges, &c., are, by some sur- 
 veyors, shewn in the field book, by lines drawn across the mid- 
 dle column at the distances where they are crossed, and by 
 others these crossings are shewn by lines drawn on part of the 
 right and left hand columns, opposite the distances where they 
 are crossed by the chain line ; and buildings, turns of fences, 
 corners of fields, to which offsets are taken, are usually shewn 
 by lines sketched in a similar situation to the middle column, 
 as the fences, buildings, &c., have to the chain line. Thus a re- 
 presentation of the chief objects in the survey may be sketched 
 in the field, which will give essential assistance in laying down 
 the plan. The stations are usually numbered, for the sake 
 
LAND SURVEYING. 
 
 97 
 
 of reference, and marked thus © . The hearing of the first main 
 hne is usually taken hy surveyors, from which the position of 
 the plan with respect to the north is determined. This may be 
 done by a common pocket compass, where great accuracy is not 
 required. 
 
 R. of O 2, and L. of O 5, &c., denote that the following 
 lines are measured to the right of station 2, and to the left of 
 station 5, respectively. 
 
 To Survey with the Chain and Cross. 
 
 An acre of land is equal to 10 square chains, that is 10 chains 
 in length and one in breadth, or 1000 links in length and 
 100 in breadth; an acre, therefore, contains 100,000 square 
 links, as per table of square measure below. Hence the con- 
 tents in square links are, in the following examples, divided by 
 100,000, or what is the same thing 5 figures to the right are 
 cut off for decimals, the figures remaining on the left being 
 acres. The decimals are then multiplied by 4 for rods, and 
 again by 40 for poles. 
 
 The following tables exhibit the number of chains and links 
 in the different units of lineal measure, and the number of 
 square chains and links in the different units of square 
 measure. 
 
 A TABLE OF LINEAR MEASURE. 
 
 Links. 
 25 
 
 100 
 
 1,000 
 
 8,000 
 
 Feet. 
 
 Yards. 
 
 Poles. 
 
 
 
 16^ 
 
 5^ 
 
 1 
 
 Chains. 
 
 
 66 
 
 22 
 
 4 
 
 1 
 
 Furlongs. 
 
 660 
 
 220 
 
 40 
 
 10 
 
 1 
 
 5,280 
 
 1,760 
 
 320 
 
 80 
 
 8 
 
 Mile. 
 1 
 
 A TABLE OF SQUARE MEASURE. 
 
 Sq. Links, j Sq. Feet. 
 
 625 
 
 10,000 
 
 25,000 
 
 100,000 
 
 64,000,000 
 
 272^ 
 4,356 
 10,890 
 43,560 
 27,878,400 
 
 Sq. Yards. 
 
 Sq. Poles 
 or Perches. 
 
 
 
 
 m 
 
 1 
 
 
 
 
 Sq.Chs. 
 
 
 484 
 
 16 
 
 1 
 
 
 
 
 
 Rods. 
 
 
 1,210 
 
 40 
 
 H 
 
 1 
 
 Acres. 
 
 4,840 
 
 160 
 
 
 
 4 
 
 1 
 
 3,097,600 
 
 102,400 
 
 6,400 
 
 2,560 
 
 640 
 
 5 
 
 Sq.Mile. 
 
 1 
 
98 LAND SURVEYING 
 
 Problem I. 
 
 TRIANGULAR FIELDS. 
 
 1 . Let A B C be a triangle, of which the survey, plan, and 
 content are required. 
 
 Set up poles or marks at the angles A, B, and C, and mea- 
 sure from A towards B, and when at or near D, trj with the 
 cross for the place of the perpendicular C D ; plant the cross 
 
 and turn it till the marks A 
 ^^ and B can he seen through 
 
 one of the grooves ; then look 
 through the other groove, 
 and if the mark at C can 
 be seen through it, the cross 
 is in the right place for the 
 "g perpendicular ; if not, move 
 the cross backward or for- 
 ward till the three marks can be seen as before directed. Sup- 
 pose the distance A D to be 625 hnks, and the whole AB, to 
 be 1257 links; return to D, and measure the perpendicular 
 D C, which suppose to be 628 hnks, thus completing the 
 survey of the triangle. 
 
 CONSTRUCTION. 
 
 From a scale of equal parts, or plotting scale, lay off the 
 base AB = 1257 links; on which take AD = 625 links; at 
 D erect the perpendicular D C, which make = 628 links ; join 
 AC, C B, then A B C is the plan of the triangle. 
 
 TO FIND THE CONTENT. 
 
 Rule. — Multiply the base by the perpendicular, and half the 
 product will be the area. 
 
 examples. 
 
 1. The dimensions being the same as found above, required 
 the content. 
 
 Ans. 1257 X 628 -^ 2 = 3*94698 acres = Sa. 3r. 322), 
 
 2. The distance from the beginning of the base to the place 
 of the perpendicular is 375 links, the whole base 954, and the 
 perpendicular 246 ; what is the area of the triangle. 
 
 954 X 246 -^ 2 = 1*17342 = la. Or. 27^;?. the content. 
 
 3. Measuring the base of a triangle the place of the perpen- 
 dicular was found at 863 links, and its length 645 ; the whole 
 base was 1434 links ; required the plan and area. 
 
 Area. 4a, 2r, 20p. 
 
WITH THE CHAIN AND CROSS. 
 
 99 
 
 Problem II. 
 
 FIELDS IN THE FORM OF TRAPEZIUMS. 
 
 Fields in this form are usually divided into two triangles by 
 a diagonal, which is a base to both the triangles. 
 
 Let A B C D be a field in the form of a trapezium, the plan 
 and area of which is required. 
 
 Measure from A towards C ; and let the place of the perpen- 
 dicular m B be at 5-52, and its length 3*76, also let the place of 
 the perpendicular n D be at 11-82, and its length 3*44, and the 
 length of the whole diagonal A C be 13*91 chains, which com- 
 pletes the survey : but it is usual also to measure the other dia- 
 gonal B D for a proof line, which is found to be 9*56 chains. 
 
 Note 1. The construction of each of the two triangles, forming the trape- 
 zium, is the same as the construction given to the first example in Prob. I. 
 
 Note 2. The longer of the two diagonals should always be selected for the 
 base of the two triangles forming the trapezium, for sometimes the perpen- 
 dicular will not fall on the shorter diagonal, without its being prolonged ; and 
 when this is the case with both diagonals, one of the sides may be taken for 
 a base, or two of the sides, if necessary. 
 
 TO FIND THE CONTENT. 
 
 Rule. — Multiply the sum of the two perpendiculars by the 
 diagonal, and half the product will be the content. 
 
 EXAMPLES. 
 
 1 . Let the measurement of a trapezium be as above found , 
 required the content. 
 344 
 376 
 
 
 720 
 1391 
 
 
 27820 
 9737 
 
 2) 
 
 10-01520 
 
 
 5-00760 
 4 
 
 
 0-03040 
 40 
 
 J -21600 
 
 Ans. 5a. Or. 1/?, 
 
100 
 
 LAND SURVEYING 
 
 2. From the following notes, plan and find the content of a 
 field. 
 
 Perpendiculars 
 on left. 
 
 Base or 
 Station Line. 
 
 Perpendiculars 
 on right.' 
 
 
 to© C 
 
 
 
 3250 
 2504 
 
 1046 D 
 
 B 1278 
 
 1272 
 
 
 Begin 
 
 at O A 
 
 and range West. 
 
 Content. 37«. 3?-. 2p. 
 3. Give the plan and area of a field from the following notes. 
 
 B652 
 Beffin at 
 
 545 C 
 
 and range East. 
 
 Area, ba. Or. 3op. 
 
 ANOTHER METHOD. 
 
 A four- sided field may frequently be surveyed by dividing it 
 into two triangles and a trapezoid, by perpendiculars on its 
 longest side. 
 
 TO FIND THE CONTENT. 
 
 EuLE. — Multiply the sum of the two perpendiculars by their 
 distance on the base line for the double area. The double areas 
 of the two triangles must be found as in Prob. I., and both be 
 added to the double area of the trapezoid ; the sum being divided 
 by 2, will give the content required. 
 
 EXAMPLES. 
 
 1. Required the survey and area of the following field. 
 
 Measure the base A B, and put down in the field book the 
 distances of P and Q, where the perpendiculars rise, &c., as 
 follows: — 
 
WITH THE CHAIN AND CROSS. 
 
 101 
 
 
 to0D. 
 
 
 AB = 1097 
 
 QD = 595 
 
 AQ= 745 
 
 PC = 352 
 
 AP = 110 
 
 Perpen. 
 
 From O A go E 
 
 Trapezoid P C D Q 
 rPerp. 
 
 595 P 
 
 745 sum. 
 635 = PQ 
 
 715 
 110 
 
 635 = PQ 
 
 Triangle ACP 
 352 
 110 
 
 38720 
 
 Triangle QDB 
 595 
 352 = QB 
 
 9975 
 3570 
 
 1785 
 
 4735 
 2841 
 5682 
 
 601345 
 38720 
 
 217175 ^ P 
 
 Q B 
 
 1097 
 755 
 
 352 = QB 
 
 217175 
 
 2)8-57240 
 4-2862 
 4 
 
 1-1448 
 40 
 
 Area, 4a. Ir. b^p. 5-7920 
 
 2. Required the plans and areas of two fields from the foh 
 lowinor notes. 
 
 
 AB 
 
 
 A B 
 
 
 
 1169 
 
 
 1448 
 
 
 E 
 
 G15 
 
 339 D D 513 
 
 1102 
 
 E 
 
 G 
 
 234 
 
 461 E C 683 
 
 436 
 
 G 
 
 From 
 
 Ago 
 
 W. From 
 Problem III. 
 
 ©A 
 
 goE 
 
 
 
 
 
 TO SURVEY FIELDS CONTAINED BY MORE THAN FOUR 
 SIDES. 
 
 Field or plots of gromid hounded hy more than four sides, 
 may be surveyed by dividing them into trapeziums and tri- 
 angles. -^Thus, a field of five sides may be divided into a tra- 
 pezium and a triangle ; of six sides, into two trapeziums ; of 
 seven sides into two trapeziums and a triangle ; &c. 
 
 TO FIND THE CONTENT. 
 
 Rule. — By the two last Problems, find the double areas of 
 each trapezium and triangle in the field ; add all the double 
 areas together, and half their sum will be the content. 
 
 EXAMPLES. 
 
 1. Lay down a field and find its area from the following 
 dimensions. 
 
102 
 
 LAND SURVEYING 
 
 
 toQD 
 
 
 520 
 
 V 
 
 288 
 
 G 120 
 
 206 
 
 
 Go to O F 
 
 
 to0 G 
 
 
 440 
 
 D230 
 
 152 
 
 
 L. of O C 
 
 
 to C 
 
 
 550 
 
 B 180 
 
 410 
 
 m 
 
 135 
 
 Begin 
 
 at A 
 
 80 E 
 
 
 
 n 
 
 130 G 
 range E. 
 
 CONSTRUCTION. 
 
 The above field is divided into two trapeziums A B C G, G D E F 
 and the triangle G C D. Dravr the diagonal A C, which make 
 = 550 links; at 135 and 410 set off the perpendiculars m C = 
 130, and n B = 180 links respectively ; join AB, BC, CG, and GA, 
 and the first trapezium will be completed. Then on CG, lay off 
 C^ = 152, and draw the perpendicular ^ D = 230; join CD, 
 D G, and the triangle is completed. Lastly, with centre G and 
 radius o G = 120 describe an arc; and with centre D and radius 
 o D = 314 (=: 520 — 206) describe another arc, intersecting the 
 former in o : through o draw the diagonal DF = 520 links, upon 
 which, at 288 links, draw the perpendicular^ E ; join DE, EF, 
 FG, and the figure will be completed. 
 
 130 
 180 
 
 440 
 230 
 
 120 
 80 
 
 Double areas. 
 170500 trap, 
 tri. 
 
 310 
 550 
 
 13200 
 
 880 
 
 200 
 520 
 
 104000 
 
 2)3-75700 
 1-87850 
 
 15500 
 1550 
 
 101200 
 
 104000 
 
 ABCG 
 
 CDG 
 
 DEFG 
 
 ■ 1«. 3r. 20tp. 
 
 170500 
 
"WITH THE CHAIN AND CROSS. 
 
 103 
 
 2. Required the plan and areas of two fields from the folio w- 
 ino; dimensions. 
 
 First Field. 
 
 Second Field. 
 
 
 to O A 
 
 
 504 
 
 E98 
 
 233 
 
 leturn 
 
 to© B 
 
 
 to D 
 
 
 673 
 
 C 207 
 
 472 
 
 
 427 
 
 Begin 
 
 at O A 
 
 Base 
 
 Diaff. 
 
 268 B 
 range W. 
 
 Area, la, Sr. 15/?. 
 
 
 to F 
 
 
 970 
 
 
 520 
 
 E 290 
 
 413 
 
 
 R.of0D 
 
 
 to D 
 
 
 744 
 
 C 161 
 
 386 
 
 
 303 
 
 Begin 
 
 at A 
 
 Diag. 
 181 B 
 
 Diaff. 
 
 333 B 
 range W. 
 
 Area. 4a. Or. 9lp. 
 
 ANOTHER METHOD. 
 
 A small piece of land, having several sides, may sometimes 
 be most conveniently measured by taking one diagonal, and 
 upon it erecting perpendiculars to all the angles on each side of 
 it. The piece will thus be divided into right angled triangles 
 and trapezoids, the areas of which must be calculated as in the 
 two last Problems. 
 
 EXAMPLES. 
 
 1 . Required the plan and area of a field from the following 
 notes. 
 
 
 to G 
 
 
 1020 
 
 F470 
 
 890 
 
 e 
 
 610 
 
 D320 
 
 585 
 
 C70 
 
 440 
 
 b 
 
 315 
 
 Begin 
 
 at A 
 
 / 
 
 50 E 
 d 
 c 
 
 350 B 
 goE. 
 
 Note. The method of planning the above field is sufficiently clear, from the 
 preceding field-notes, and from what has been already done. 
 
104 
 
 LAND SURVEYING. 
 
 Triangle A C c. 
 A c = 440 
 Cc= 70 
 
 Trape. CcdD. Trape. D df F.I Tri. F/ G. 
 D d = 320D d = 320l/*G= 130 
 
 C c = 70 F/ = 470 
 
 30800 
 
 F/ = 
 
 sum = 790sum = 790 
 cd = 145 ef = 395 
 
 13050 
 435 
 
 56550 
 
 3950 
 2370 
 
 240950 
 
 Trapezoid B b eB. 
 B 6 = 350 
 E e = 50 
 
 sum = 400 
 b e =295 
 
 Triangle E e G. 
 Ge = 410 
 
 Ee = 50 
 
 20500 
 
 470 
 
 9100 
 52 
 
 Tri. A J B. 
 A 5 = 315 
 B 5 = 350 
 
 15750 
 945 
 
 61100 
 
 110250 
 
 Double areas, 
 
 30800 
 
 56550 
 
 240950 
 
 61100 
 
 110250 
 
 118000 
 
 20500 
 
 2)638150 
 
 3-19075 
 
 3a. 1/-. 38i^. Area. 
 
 Problem IV. 
 
 FIELDS INCLUDED BY ANY NUMBER OF CROOKED OR 
 CURVED SIDES. 
 
 "When a field or estate is bounded by crooked fences^ a line 
 must be measured as near to each of them, as the angles or 
 bends will permit ; in doing which an oifset must be taken to 
 each corner or bend in the fence. The offsets or perpendiculars 
 thus erected, will divide the whole offset space into right angled 
 triangles and trapez^oids, the areas of which may be found as 
 already shewn. 
 
 Note 1. When the offsets are short, that is, not greatly exceeding a chain 
 in length, their places on the line may be found by laying the offset-staff at 
 right angles to the chain, as nearly as can be judged by the eye ; but when the 
 offsets are large, and correctness is required, their places must be found by 
 the cross, and measured by the chain. 
 
 Note 2. The quickest method of laying down offsets, is, by laying the 
 feather-edge of the plotting scale against the base or chain line, and shding 
 the offset scale along the feather edge to the several distances of the offsets, 
 and pricking off their lengths, corresponding to their several distances. 
 
WITH THE CHAIN AND CROSS. 
 
 105 
 
 Note 3. Unskilful surveyors usually add all the offsets taken on one line 
 together and divide the sum by their number for a mean breadth ; but this 
 method is very erroneous, especially where the offsets vary greatly in length, 
 and should therefore be avoided where great accuracy is required. 
 
 EXAMPLES. 
 
 1. Required the plan and content of a riglit-lincd piece of 
 ground by offsets, from the following notes. 
 
 
 to B 
 
 
 
 955 
 
 n 91 
 
 785 
 
 mb7 
 
 634 
 
 I 88 
 
 510 
 
 lc70 
 
 340 
 
 i84 
 
 220 
 
 h62 
 
 45 
 
 
 
 00 
 
 Begin 
 
 atO A 
 
 Ac d e f g 
 
 range E. 
 
 A c = 45 
 cA = 62 
 
 90 
 270 
 
 2790 
 
 cA = 62 
 di =84 
 
 146 
 cc? = 175 
 
 730 
 1022 
 146 
 
 25550 
 
 di =. 8\ 
 e /j = 70 
 
 154 
 c?e = 120 
 
 18400 
 
 eh 
 fl 
 
 of 
 
 = 70 
 
 = 88 
 
 158 
 = 170 
 
 
 11060 
 158 
 
 
 26860 
 
 / I = 88 
 
 145 
 fg = 124 
 
 580 
 
 290 
 145 
 
 17980 
 
 
 
 
 
 Double areas. 
 
 (J m =57 
 
 h! 
 
 B 
 
 = 170 
 
 2790 
 
 h' n =91 
 
 h! 
 
 n 
 
 = 91 
 
 25550 
 18480 
 
 148 
 
 
 
 170 
 
 26860 
 
 (/A' = 151 
 
 
 
 1530 
 
 17980 
 22348 
 15470 
 
 148 
 
 
 
 15470 
 
 740 
 148 
 
 
 
 
 
 
 
 
 2)1-09478 
 
 22348 
 
 0-64739 = 0«. 2r. 23». 
 
 
 
 
 
 5 * 
 
106 
 
 LAND SURVEYING 
 
 Calculation by the erroneous method (See Note 3). 
 
 00 
 62 
 84 
 70 
 88 
 ^7 
 91 
 00 
 
 955 
 
 AB. 
 
 56|- = mean breadth. 
 
 5730 
 4775 
 477 
 
 8)452 
 
 56* 
 
 0*53957 = 0«. 2r. 6p. Content by this method, 
 tvhich is 17 perches too little. For 
 this method is always erroneous except 
 when the offsets stand at equal distances 
 from one another, and when the first 
 and last offsets are both 0. 
 
 Some omit all the offsets that are 0, dividing the sum of the 
 offsets by the number of real offsets ; by this method we shall 
 have 
 
 955 
 7H 
 
 6)452 
 
 75i 
 
 4775 
 6685 
 318 
 
 0*71947 = 0«. 2r. 34^., which is 11 poles 
 too much. 
 
 2. To lay down a crooked piece of land, adjoining a river from 
 the following notes. 
 
 
 to O F 
 
 197 
 
 720 
 
 163 
 
 600 
 
 226 
 
 500 
 
 139 
 
 350 
 
 80 
 
 200 
 
 74 
 
 100 
 
 
 
 000 
 
 Begin 
 
 at O A 
 
 [E. 
 range 
 
 The content is found by the same method as in the preceding 
 example. 
 
WITH THE CHAIN AND CROSS. 107 
 
 3. Plan and find the area of a field from the subjoined notes. 
 
 C 
 
 
 to O A 
 
 
 
 
 480 
 
 
 37 
 
 350 
 
 
 28 
 
 160 
 
 
 
 
 000 
 
 L. of O C 
 
 
 
 to o C 
 
 
 
 585 
 
 
 
 
 450 
 
 57 
 
 
 320 
 
 40 
 
 
 200 
 
 72 
 
 
 100 
 
 47 
 
 
 000 
 
 
 
 
 L.of0B 
 
 
 
 to O B 
 
 
 
 743 
 
 
 C382 
 
 290 
 
 P 
 
 From 
 
 A 
 
 goE. 
 
 Havmg found the area of the triangle ABC, the areas of the 
 offsets on the line B C must be added thereto, and the sum of the 
 areas of the insets on the line C A must be subtracted from the 
 sum, and the remainder will be the content of the field. 
 
 Note. The area of the triangle ABC may be found, when the measurement 
 of all three sides are given, (which is the case in the present example,) either 
 Hy calculation, as shall hereafter be shewn, or by measuring the perpendicular 
 from the plan, which, as already shewn, maybe laid down from the three sides 
 of the main triangle. — The areas of fields, having a great number of crooked 
 and curved fences, are found by the method given in the last Example, page 51. 
 
 Problem V. 
 
 TO MEASURE A LINE ACROSS A WIDE RIVER. 
 
 Let the annexed figure be a river, which 
 is required to be crossed by the chain line 
 P B. Fix, or cause to be fixed, a pole or 
 mark at B, at or near the margin of the 
 river, in the hne to be measured; erect 
 the perpendicular A D, measuring A C 
 and C D of any equal lengths ; at D erect 
 the perpendicular I) E ; on arriving at E^ 
 in the direction B C, the distance D E will 
 be equal to A B, the required breadth of 
 the river. 
 
108 LAND SURVEYING BY THE CHAIN ONLY. 
 
 From the arrangement of the lines in the figure, it is evident 
 that the triangles CAB, C D E are equiangular, and since A C 
 was made = C D, the triangles are equal in all respects, and con- 
 sequently A B = D E. 
 
 Note 1. For various other methods of measuring obstructed lines, under 
 different circumstances, see Baker^s Land and Engineering Surveying, Chap= III. 
 
 Note 2. A sufficient detail of methods of surveying by the help of the cross, 
 which, though not much used by experienced surveyors, is a simple instrument, 
 and its use readily understood by students. This method is, therefore, a 
 proper introduction to the higher branches of surveying ; besides, in rural 
 districts, villages, &c., few sm*veyors use the more expensive instrument, the 
 chain and cross being found quite sufficient to measure the quantities of grow- 
 ing crops, and other such small surveys as may be there required. 
 
 LAND SURVEYING BY THE CHAIN ONLY. 
 
 This method of surveying has long been adopted by experi- 
 enced surveyors, who have found it, in general, more accurate 
 and expeditious, as well as better adapted to laying down ex- 
 tensive surveys especially where no serious obstructions from 
 woodlands, water, buildings, &c., exist ; the use of the cross, in 
 this method, being entirely excluded by some surveyors, and 
 by others only used for secondary purposes, as for taking oc- 
 casionally long offsets, or for squaring of lines obstructed by 
 buildings, water, &c. Instead of the cross some use the optical 
 square for these purposes ; while some erect perpendiculars with 
 the chain only. See Chap. III., Baker's Land and Engineering 
 Surveying, 
 
 Problem I. 
 
 TRIANGULAR FIELDS. 
 
 When a triangular field, or piece of ground in that shape, is 
 to be surveyed, set up poles or marks at each corner, and mea- 
 sure each side, leaving marks in at least two of the lines, and 
 entering their distances in the field book; then measure the 
 distance between the two marks for a proof line : — or, one 
 mark only may be left in one of the lines, which may be con- 
 nected with its opposite angle for a proof line. 
 
 EXAMPLES. 
 
 1 . Required the construction and area of a field from the fol- 
 lowing dimensions. 
 
LAND SURVEYING BY THE CHAIN ONLY. 
 
 109 
 
 
 to O n 
 
 Proof 
 
 384 
 
 From 
 
 © m 
 
 
 to O A 
 
 
 1244 
 
 Qn 
 
 700 
 
 
 L© C 
 
 
 to © C 
 
 
 852 
 
 
 L© B 
 
 
 to © B 
 
 
 1338 
 
 
 1000 
 
 m 
 
 600 
 
 From 
 
 ©A 
 
 line. 
 
 A m D B 
 
 "When the triangle A B C is 
 constructed, the proof line m n 
 will be found to measure 384 
 links, shewing that there has 
 been no error in the work : 
 also the perpendicular C D will 
 be found to be 770 links ; 
 whence the area of the triangle 
 Range E. = 1338 x 770 -f- 2 =5*15130 
 = 5a. Or. 24^. the area. 
 
 Note. If the proof line measured from the plan does not exactly, or very 
 nearly, agree with that measured in the field, some error has been made, and 
 the work must be repeated. 
 
 TO FIND THE AREA OF A TRIANGLE FROM THE THREE 
 SIDES. 
 
 Rule. — From half the sum of the three sides subtract each 
 side severally and reserve the three remainders ; multiply the 
 half sum continually by the three remainders, and the square root 
 of the product will be the area. 
 
 Note. By this rule the area of a triangle may be found without laying it 
 down, or finding the perpendicular. 
 
 Adopting the preceding example, we have by the rule, 
 
 1338 + 852 + 1244 , ,, n ^ ^ 
 
 ' = 1717 = half sum of the three sides. 
 
 Then 1717 -- 1338 = 379 = 1st remainder; 1717— 852 = 
 865 = 2nd remainder ; 1717 — 1244 = 473 = 3rd remainder ; 
 whence ^{1117 X 379 X 865 X 473) = 5-15022 = ba. Or. 24p. 
 the sum as the area already found by measuring the perpendicular 
 from the plan. 
 
 Note. This method of finding the areas of triangles is very little used in 
 practice, on account of its requiring a tedious calculation, which may, however, 
 be more readily performed by logarithms. 
 
 2. It is required to lay down a survey and find its content from 
 the following field notes. 
 
LAND SURVEYING BY THE CHAIN ONLY. 
 
 
 to O A 
 
 
 
 2504 
 
 
 
 
 2000 
 
 74 
 
 O 7n 
 
 1860 
 
 351 to © 
 
 
 1-650 
 
 137 
 
 
 1430 
 
 90 
 
 
 1220 
 
 144 
 
 
 850 
 
 30 
 
 
 425 
 
 110 
 
 
 000 
 
 
 
 
 L. ©C 
 
 
 
 to O C 
 
 
 
 
 1346 
 
 
 80 
 
 1072 
 
 
 128 
 
 708 
 
 
 98 
 
 458 
 
 
 
 
 000 
 L. OB 
 
 
 
 to O B 
 
 
 
 1946 
 
 
 
 
 1490 
 
 96 
 
 
 1200 
 
 152 
 
 
 1000 
 
 112 
 
 n 
 
 600 
 
 
 
 520 
 
 50 
 
 
 000 
 
 
 From 
 
 ©A 
 
 go N. E. 
 
 Having drawn the figure, the proof line m n will be found 
 to measure 351 links, as in the field notes ; and the perpen- 
 dicular B 5 to be 1056 links. 
 
 Double areas. 
 
 2644224 Triangle ABC 
 655676 Offsets on A B and A C 
 
 3299900 Sum 
 200616 Insets on B C 
 
 2)3099284 Difference 
 15-49642 = \5a. 2r. Op. nearly, the area required. 
 
LAND SURVEYING BY THE CHAIN ONLY. 
 
 Ill 
 
 Problem II. 
 
 FOUR SIDED FIELDS. 
 
 When a field has four sides, straight or crooked, measure the 
 four sides, or lines near them, if crooked, taking the oiFsets : 
 also measure one or both the diagonals, one of which will serve 
 as a base in plotting the work, and the other for a proof-line ; 
 or the proof-line may be measured in any other direction that 
 may be most convenient. 
 
 Sometimes the measurement of both the diagonals is pre- 
 vented by obstructions, in such cases it will be sufficient to 
 measure tie-lines across two of the angles of the trapezium, at 
 the distance of from two to five chains from each angle, ac- 
 cording to the size of the field. These tie lines with their dis- 
 tances from the angles on the main lines will be found sufficient 
 for planning the lines and proving them. 
 
 EXAMPLE. 
 
 In the annexed figure the 
 lines A B, B C, CD, DA are 
 measured, marks being left at p, 
 q, and r, and their respective 
 distances on the lines noted in 
 the field book, thus furnishing 
 the following method of laying 
 down the plan. 
 
 On A B, as a base, take A p 
 = given distance, and with the 
 distances A r, 2^ '% and centres 
 A and p describe arcs cutting in 
 r ; then prolong A r, and lay off 
 thereon the given length A D. 
 In the same manner construct 
 the triangle pl^q, and make B C = its givei^ length. Lastly, 
 join D C, which must be of the length shewn in the field book, 
 otherwise there has been some mistake either in the measure- 
 ment, or in laying it down. Should this be the case, the whole 
 of the work, firstly on the plan, and secondly in the field, must 
 be gone over again till the error be discovered. 
 
 Problem III. 
 
 FIELDS HAVING MORE THAN FOUR SIDES. 
 
 Various methods will suggest themselves to the surveyor for 
 taking lines to lay down a field that requires more than four 
 main lines to take its boundary. The method of dividing such 
 
112 
 
 I.AND SURVEYING BY THE CHAIN ONLY. 
 
 fields into trapeziums and triangles is, in most cases, circuitous, 
 and displays little skill on the part of the surveyor, especially 
 where all the sides are crooked, and where a plan is required. 
 A few methods of surveying fields of this kind will, therefore, 
 be presented to direct the student ; although their variety of 
 shape is so endless, that no general rule can be given for laying 
 out lines on the ground, that shall give an incontestably accurate 
 plan. To tie every angle in succession, though true in principle, 
 is by no means a safe method, especially where there are a great 
 number of angles to be tied, as an error in one of the tie-hnes 
 will derange the whole of the work, without affording the means 
 of detecting where the error lies. 
 
 Note. The following examples of surveys of this kind occurred in part of 
 the author's extensive practice, as a surveyor of parishes under the Tithe- 
 Commissions. The student is recommended to sketch the following speci- 
 mens on a large scale, and find their contents by the usual methods. 
 
 EXAMPLE. 
 
 Here a field of five 
 sides is surveyed by 
 the same number of 
 lines, viz. A B, B C, 
 C D, D my and A n, the 
 last two intersecting 
 in E. These lines 
 evidently constitute a 
 decisive proof among 
 themselves, and all of 
 them are available in 
 takhig the boundary. 
 In surveying this field (poles or natural marks being supposed 
 to be fixed at A, B, C, D, and E) commence close to the river's 
 edge, in the line A B prolonged backwards, enter the offsets and 
 the station A in fhe field-book. On arriving at© m, in the 
 direction E D, enter its distance, and so on to © B, measuring 
 the line to the fence ; from B proceed to C, in like manner, 
 measuring beyond the station to the fence. The place of the © 
 71 is to be noted, on arriving in the direction E A, while mea- 
 suring CD. D m is next measured, the place of the © E being 
 noted. Lastly, go from m to A, and measure A w, entering the 
 place of the © E a second time, all the offsets being supposed 
 to be taken during the operation. 
 
 Construction of the 'plan. Select the distances Aw?, A E, and 
 E m from the field-book, and with them construct the triangle 
 A m E; prolong the sides to their entire lengths, up to the 
 
LAND SURVEYING BY THE CHAIN ONLY. 113 
 
 boundaries, and fix the places of the stations B, n, and D. 
 Now, if the measured length of D 7i just fit between D and n, 
 the work is right with respect to the triangles A Em, E D n. 
 Lastly, prolong D n to the O C, and, if the distance from thence 
 to the O B be the same as shewn by the field book, the whole 
 of the work is right. But, if the distance D ?i do not agree, the 
 work must be examined from the beginning ; if only the distance 
 B C fail, then only that distance and the portions m B, C n need 
 be examined. 
 
 Note. For a variety of other methods of surveying irregular fields, small 
 estates, &c., &c., see Baker^s Land and Engineering Surveying, Chap. III. 
 
 Problem IV. 
 
 SURVEYING LARGE ESTATES OR PARISHES BY THE CHAIN ONLY. 
 
 Having perambulated the boundary of the estate, parish, or 
 lordship to be surveyed, if you find that its boundary approaches 
 somewhat near to that of a four- sided figure, or trapezium, the 
 system of fundamental lines, adopted by order of the Tithe 
 Commissioners of England and Wales, is to be preferred. 
 These fundamental lines are six in number, of which four must 
 run close by, or as nearly as possible to, the boundary in question, 
 thus forming a trapezium, four lofty station poles being placed at 
 each angle, as objects for running the lines ; the other two lines 
 must form the diagonals of this trapezium, and therefore pass 
 through the central parts of the survey, intersecting each other, 
 the points of intersection being noted on measuring each line, so 
 that when the system of lines are laid down on the plan, the 
 proof of the accuracy of the work may be fully established, before 
 the minor operations, or filling up, as it is called, is commenced. 
 It will be necessary, moreover, in almost every case, to range the 
 lines between every two of the main stations with long slender 
 ranging poles, as the intervention of hills, fences, trees, buildings, 
 &c., will frequently interrupt the view of even the loftiest station 
 poles that can be obtained; and more especially so, when the 
 main stations are at a great distance, which depends on the mag- 
 nitude of the survey, and is sometimes as much as ten miles. 
 In measuring these main lines, every fence, road, stream, build- 
 ing, &c., which is passed or crossed must be noted in the field 
 book, the several crossings, and bends being sketched therein, 
 to the latter of which offsetts must be taken. Stations must 
 also be left on these main lines, at convenient situations for taking 
 the interior fences, &c., of the survey, and their distances care- 
 fully noted in the field book. From and to the stations, thus 
 
114 
 
 LAND SURVEYING BY THE CHAIN ONLY. 
 
 left, or from and to points near them, secondary lines must be 
 run, as near the interior parts of the survey as possible, the cross- 
 ings, offsets, and other remarks being made in the field book, as 
 already directed for the measurement of the main lines. Tl ese 
 secondary lines will accurately fit between the points from and to 
 which they have been measured, when laid down on the plan ; 
 thus forming a net work of small triangles within the four large 
 triangles, into which the survey is divided by the six funda- 
 mental lines. This principle of proof is founded on the obvious 
 property of triangles having a common angle always fitting one 
 vrithin another, the common angle of both being coincident. 
 The lines marked with the figures 1 to 6, represent the system 
 in question, those without figures are the secondary lines. 
 
 The main lines are numbered with the figures 1, 2, 3, &c., in 
 small circles, as the most convenient method of reference to the 
 field book : the secondary lines must have these numbers con- 
 tinued on them, for the same purpose, but this is not done in the 
 diagram, to avoid confusing it. 
 
 It will be seen that the secondary lines mn, rs are ' prolonged 
 beyond the system of main lines, to give stability to the parts of 
 the survey that protrude beyond line 2. 
 
 Note. For the method of keeping the field book in exten?ke surveys, the 
 description and use of the theodolite and other surveying instruments, see 
 Bakers Land and Engineering Surveying. 
 
LEVELLING. 
 
 115 
 
 ENGINEERING SURVEYING. 
 
 LEVELLING. — DEFINITION OF LEVELLING. 
 
 By the art of levelling the inequalities of the upper boundary 
 of any section of the earth's surface may be shewn, and thence 
 may be determined the several heights of any number of points 
 in tliat boundary, above or below an assumed hne, called a level 
 line ; though, in reality, this line is a great circle of the earth, 
 and is such as would be derived from a section of the surface of 
 still water. 
 
 LEVELLING INSTRUMENTS. 
 
 (1.) Levelling instruments all depend on the action of gra- 
 vity ; of these the plumb-line, on which the mason's level 
 depends, Is the most simple ; but it cannot be used in extensive 
 operations, on account of its practical inconvenience. The fluid, 
 or water level, in all its modifications, is slso found inconvenient 
 for extensive practice. 
 
 (2.) Spirit levels are now commonly used, as the most ac- 
 curate instruments for finding the differences of level, or vertical 
 distances between two stations : of these there are several, we 
 shall only here describe the Y level. 
 
 THE Y LEVEL. 
 
 The foregoing figure represents this instrument. A is an 
 achromatic telescope, resting on two supporters, which in shape 
 resemble the letter Y ; hence the name of the instrument. The 
 lower ends of these supporters are let perpendicularly in a strong 
 
116 LEVELLING INSTRUMENTS. 
 
 brass bar, which carries a compass box C. This compass is con- 
 venient for taking bearings, and has a contrivance for throwing 
 the needle off its centre, when not in use. One of the Y sup- 
 porters is fitted into a socket, and can be raised or lowered by the 
 screw B. 
 
 Beneath the compass box, which is generally of one piece with 
 the bar, is a conical axis passing through the upper of two 
 parallel plates, and terminating in a ball supported by a socket. 
 Immediately above the upper parallel plate is a collar, which can 
 be made to embrace the conical axis tightly by turning the 
 clamping screw E ; and a slow horizontal motion can be given to 
 the instrument by means of the tangent screw D. The two 
 parallel plates are connected together by the ball and socket 
 already mentioned, and are set firm by four mill-headed screws, 
 which turn in sockets fixed to the lower plate, while their heads 
 press against the under side of the upper plate, and thus serve 
 the purpose of setting the instrument truly level. 
 
 Beneath the lower parallel plate is a female screw, adapted to 
 the staff head, which is connected with brass joints to three 
 mahogany legs, which support the instrument. 
 
 The spirit level 1 1 is fixed to the telescope by a joint at one 
 end, and a capstain headed screw at the other, to raise or de- 
 press it for adjustment. 
 
 (3.) Previous to using this instrument the following adjust- 
 ments must be attended to. 
 
 1. The adjustments of the telescope for parallax and collima- 
 tion, 
 
 2. The adjustment of the huhhle tube, 
 
 3. The adjustment of the axis of the telescope perpendicularly 
 to the vertical axis. 
 
 1. The adjustment for parallax and collimation. Move the 
 object-glass b}^ the screw, and the eye-glass with the hand, till 
 distant objects and the cross wires within the telescope, appear 
 clearly defined; and the adjustment for parallax will be com- 
 pleted. Next, direct the telescope to some well-defined object 
 at a great distance ; and see that the intersection of the cross 
 wires cut it accurately ; then loose the clips that confine the 
 telescope in the Ys, and turn it round on its axis, observing 
 whether the centre of the wires still continue to cut the object, 
 during a whole revolution. If it does, it is in adjustment ; if 
 not, the line of collimation, or optical axis of the instrument, is 
 not in the line joining the centres of the eye and object-glasses. 
 To correct this error, turn the telescope on its axis, and by 
 
LEVELLING INSTRUMENTS. 117 
 
 means of the four conjugate screw «, «, &c., that move the cross 
 wires, correct for half the error, alternately loosing one screw and 
 tightening its opposite one, till the cross wires cut the same point 
 of the distant object, during an entire revolution of the telescope 
 round its axis. 
 
 2. The adjustment of the bubble tube. — Move the telescope 
 till it lies in the direction of two of the parallel plate screws, and 
 by giving motion to these screws bring the air-bubble to the 
 centre of its run. Now reverse the telescope carefully in the 
 Ys, that is, change the places of its ends ; and should the bubble 
 not settle in the same point of the tube as before, it shows 
 that the bubble tube is out of adjustment, and requires cor- 
 recting. The end to which the bubble retires must then be 
 noticed, and the bubble made to return one- half the distance by 
 turning the parallel plate screws, and the other half by turning 
 the capstain headed screws at the end of the bubble tube. The 
 telescope must now again be reversed, and the operation repeated, 
 until the bubble settles at the same point of the tube, in the 
 centre of its run, in both positions of the instrument. The ad- 
 justment is then perfect, and the clips, that confine the telescope 
 in the Ys should be made fast. 
 
 3. The adjustment of the axis of the telescope perpendicularly 
 to the vertical axis. — Place the telescope over two of the pa- 
 rallel plate screws, and move them, unscrewing one while 
 screwing up the other, until the bubble of the level settles in 
 the centre of its run ; then turn the instrument half round on 
 its vertical axis, so that the contrary ends of the telescope may 
 be over the same two screws, and, if the bubble does not again 
 settle in the same point as before, half the error must be cor- 
 rected by turning the screw B, and the other half by turning the 
 two parallel plate screws, over which the telescope is placed. 
 Next turn the telescope a quarter round, that it may be over 
 the other two screws, and repeat the same process with these 
 two screws ; and when, after a few trials, the bubble maintains 
 the same position in the centre of its run, while the telescope is 
 turned round on the vertical axis, this axis will be truly ver- 
 tical ; and the axis of the telescope being horizontal, by reason 
 of the previous adjustment of the bubble tube, will be perpen- 
 dicular to the vertical axis, and remain truly horizontal, while 
 the telescope is turned completely round. The adjustment is 
 therefore perfect. 
 
 There are several other highly approved levelling instruments, 
 as Troughton's and Gravatt's levels, &c., for the descriptions of 
 which, see Baker* s Land and Engineeinng SurveTjing, 
 
118 CORRECTION FOR CURVATURE, &C 
 
 LEVELLING STAVES. 
 
 (6.) The best constructed levelling staff (Gravatt's) consists of 
 three parts sliding one within another, and, when opened out for 
 use, form a staff 17 feet long, jointed together something after 
 the manner of a fishing rod. The whole length is divided into 
 hundredths of a foot, alternately coloured black and white, and 
 occupying half the breadth of the staff ; but for distinctness the 
 lines denoting tenths of feet are continued the whole breadth, 
 every half foot, or five- tenths, being distinguished by a conspi- 
 cuous black dot on each side, the whole feet being numbered with 
 the figures 1, 2, 3, &c. 
 
 CORRECTION FOR CURVATURE. 
 
 (7.) Let B D E be a horizontal line, 
 that is, such as would be given by the 
 line of sight of a level, properly ad- 
 justed ; B C F an arc of a great circle 
 of the earth, and A its centre. It will 
 at once appear from the figure, that 
 the heights D C, E F, of the apparent 
 level B E, above the true level increase 
 successively from the point B. The 
 height E F of the apparent level above 
 the true, is equal to the square of the 
 distance B E divided by twice the 
 earth's radius A B, that is E F = 
 
 B E2 ^ . . B D2 
 
 cTir^ and similarly D C = ^~\~Wi &c., therefore the correc- 
 tions for carvature, D C, E F, &c., vary as the squares of the 
 distances B D, BE, &c., since 2 A B is a constant quantity. 
 
 Taking the earth's radius to be 3979 miles, and assuming the 
 distance B D to be 1 mile, then the correction for curvature 
 DC = BD3--2AB = 12-- 7958 = y^V-g of a mile = 7*962 
 inches = nearly 8 inches. If the distance B E = 3 miles, then 
 the correction EF = BE3-r-2AB = ^-^\^ = 71*658 inches, 
 or nearly 6 feet. 
 
 Let any distance B D = ^ in miles, and the correction for cur- 
 vature for 1 mile be taken = 8 inches = f of a foot, which it is 
 very nearly ; then 
 
 2 d'- 
 correction = ~o~ feet, 
 
 for any distance d in miles : 
 
 * The demonstration of this formula is given in my edition of NeshWs Sur- 
 veying ^ p. 348. 
 
IN LEVELLING. 119 
 
 C 
 
 
 
 and let -r^ = d, c being chains ; then 
 
 80 
 
 2 ^2 2x12 c3 c2 . ^ 
 correction = -3-=-^^^^^= 800" ^^"^"'- 
 for any distance c in chains. 
 
 CORRECTION FOR REFRACTION. 
 
 (8.) The effect of the earth's curvature is modified by another 
 cause, arising from optical deception, namely, refraction ; the 
 correction for which varies with the state of the atmosphere, but 
 it may generally be taken at i of the correction for curvature, 
 as an average ; and since refraction makes objects appear higher 
 than they really are, the correction for it must be deducted from 
 that for curvature. 
 
 EXAMPLES. 
 
 1 . Required the correction for curvature and refraction, when 
 the distance of the object is 2^ miles. 
 
 2 2x6-25 
 
 V X (2*52 o — = 4* 166 cor. for curvature, 
 
 •f of which is '693 cor. for refraction. 
 
 Difference 3*473 feet, cor. required. 
 
 2. Required the correction, as in the last example, when the 
 distance is 60 chains. 
 
 60^ -T- 800 = 4*5 cor. for curvature, 
 -f of wdiich is . . '643 cor. for refraction. 
 
 Difference .... 3'857 inches, cor. required. 
 
 3. From a point in the Folkestone road, the top of the keep 
 of Dover Castle was observed to coincide with the horizontal wire 
 of a levelling telescope, when adjusted for observation, and there- 
 fore was apparently on the same level ; the distance of the instru- 
 ment from the castle was 4|- miles, required the correction for 
 curvature and refraction, that is, the true height of the keep of 
 the castle above the point of observation. 
 2 ^ 40-5 
 
 o" X (4*5)^ = -^ — = 13*5 feet, cor. for curvature, 
 
 ■f of which = 1*93 feet, cor. for refraction. 
 
 Difference = 11 '57 feet, cor. required. 
 
 See aUo the tables for these corrections at the end of the book. 
 
120 LEVELLING. 
 
 PRINCIPLES AND PRACTICE OF LEVELLING. 
 
 To find the differences of the levels of several points on the 
 surface of the earth, 
 
 (7.) Before entering on this subject, it will be proper to 
 state that the corrections for curvature and refraction, already 
 explained, are seldom applied in the practice of levelling, the 
 spirit level being usually placed midway between the stations, 
 the levels of which are to be observed, hence the resulting cor- 
 rection for each station are equal, and therefore the difference 
 of the levels at the two stations is as truly shown by the dif- 
 ference of the readings of the two staves, fixed thereon, as if the 
 corrections had been made. Thus the trouble of making these 
 corrections is avoided by simply placing the instrument midway 
 between the two staves, 
 
 (8.) Let it be required to find the difference of level between 
 the points A and G. A levelling staff is erected at A, the in- 
 strument is set up and adjusted at B, another staff is also erected 
 at C, at the same distance from B that B is from A, as nearly 
 as can be judged by the eye ; the reading of the two staves are 
 
 then noted ; the horizontal lines, connecting the staves with the 
 instrument, represent the visual ray or level line of sight. The 
 instrument is then conveyed to D, and the staff that stood at 
 A is now removed to E, the staff C retaining its former position, 
 only its graduated side turned to the instrument, and from 
 being the fore staff at the last observation, it is now the back 
 staff; the reading of the two staves are again noted, and the 
 instrument removed to F, and the staff C to the point G, the 
 staff at E retaining its position, now in its turn becomes the 
 back staff, and so on to the end of the work, which may thus 
 be continued to any extent. The difference of the readings of 
 the staves at A and C will show the difference of level between 
 the points or stations A and C, because the visual line of the 
 instrument is virtually level, and the same is true with respect 
 to every two consecutive stations. 
 
 EXAMPLE. 
 
 Back sight on staff A 10*66 feet 
 
 Fore sight on staff C 1 178 „ 
 
 The fall from A to C 1 ' 12 difference. 
 
LEVELLING. 121 
 
 Because when the front reading is the greater the ground falls, 
 and vice versd. 
 
 Back sight on staff C 13*36 
 
 Fore sight on staff E D* 16 
 
 The rise from C to E 4*20 difference. 
 
 Subtract the fall from x\ to C . . M 2 
 
 The rise from A to E 3-08 difference. 
 
 Because the rise from C to E is greater than the fall from A to 
 C, their difference shows the total rise. 
 
 Back sight on staff E 7*62 
 
 Fore sight on staff G 8*16 
 
 The fall from E to G 0-54 difference. 
 
 This fall taken from the rise from A to E, that is, 
 
 3-08 
 0-54 
 
 gives the total rise from A to G 2*54, or nearly 2 
 
 feet 6^ inches. 
 
 The difference of the sums of the back and fore readings of 
 the staves, will more readily give the difference of level between 
 A and G : thus. 
 
 Back sights. Fore sights, 
 
 feet. feet. 
 
 10-66 at A 1178 at C 
 
 13-36 at G 9-16 at E 
 
 7-62 at E 8-16 at G 
 
 sums 31-64 29*10 
 
 29*10 
 
 2*54 difference of level, the same as before. 
 
 TO DRAW A SECTIONAL LINE OF SEVERAL POINTS IN THE 
 earth's SURFACE, THE LEVELS OF WHICH HAVE BEEN 
 TAKEN. 
 
 Let «, 5, c, di e,f, and g be the several points ; then, in order 
 to draw the section to show the undulations of the ground be- 
 tween a and g, the distances of the several points from «, in 
 addition to their levels, must be taken ; this is usually done 
 during the operation of levelling. These distances, with the 
 back and fore sights, may be arranged in a level book of the fol- 
 lowing form, which, though not the form practically used, will 
 probably be more clearly understood. {See Fig, page 124.) 
 
 6 
 
122 
 
 LEVELLING. 
 
 LEVEL BOOK. 
 
 Back 
 
 Fore 
 
 Fall. 
 
 Hise. 
 
 Reduced 
 
 Distances in Chains, 
 
 Sights. 
 
 Sights. 
 
 Levels. 
 
 and Remarks. 
 
 3-50 
 
 5-65 
 
 2-15 
 
 
 2-15 
 
 4-60 at b on road. 
 
 4-10 
 
 10-85 
 
 6-75 
 
 
 8-90 
 
 7-80 at c. 
 
 5-04 
 
 9-25 
 
 4-21 
 
 
 13-11 
 
 11-60 at 6?. 
 
 3-84 
 
 12-91 
 
 9-07 
 
 
 22-18 
 
 15-20 at e. 
 
 4-12 
 
 7-65 
 
 3-53 
 
 
 25-71 
 
 bottom of canal, distance 2*16 
 
 10-49 
 
 3-92 
 
 . 
 
 6-57 
 
 19-14 
 
 21-00 at/. 
 
 12-96 
 
 3-03 
 
 
 9-93 
 
 9-21 
 
 27-00 at ff. 
 
 44-05 
 
 53-26 
 
 dilf. 
 
 44-05 
 
 the I 
 
 >ame 
 
 as the last of the reduced levels. 
 
 9-21 
 
 In this level book it will be seen that the differences 2*15 
 and 6-75, in the column marked Fall, are added together, 
 making 8*90, thus giving the fall at c, in the cohimn marked 
 Reduced Levels : to this sum the succeeding falls are added, 
 one by one, till we get the fall 25-71 at the bottom of the canal, 
 which is the lowest point. Then the differences in the column 
 marked Rise, are subtracted successively from 25-71 for the 
 falls at / and ^ ; the latter of which is 9-21, the total fall from 
 a to c/, which, agreeing with the difference of the sums of the 
 back and fore sights, shows the truth of the castings. The 
 last column shows the distances of the several points, b, c, &c., 
 from a, in chains, with other remarks. 
 
 DATUM LINE. 
 
 The section might be plotted by laying off the distances 
 in the last column in the preceding level book on a horizontal 
 line, and setting off their corresponding numbers of feet, in the 
 column marked Reduced Levels, perpendicularly below the line ; 
 but it is found inconvenient in practice to plot a section in all 
 cases after this method, as in extensive operations the reduced 
 levels would repeatedly fall above and below the line in question, 
 and thus confuse the operation ; therefore a line A G, called "the 
 datum line/' is assumed at 100, 200 feet, &c., below the first 
 station a ; thus making that line always below the sectional line 
 af, of which a clearer view may be obtained. 
 
 In the following practical level book the rise or fall is re- 
 
LEVELLING. 
 
 123 
 
 spectively added to, or subtracted from, the assumed distance 
 of the datum line, and the next rise or Ml again added to, or 
 subtracted from, the sum or diiference ; — thus 2" 15, being a fall, 
 is subtracted from 100 (the assumed distance of the datum line) 
 leaving 97-85 feet, the height of the ground at b ; the next fall 
 G-75 is then subtracted from 97-85, leaving 91-10 feet for the 
 height at c ; and so on to 3*53, which is tlie last fall : — the next 
 6-57, being a rise, is added, as well as 9*93; — thus the last re- 
 duced level is 90*79 feet, which taken from the datum 100 
 leave 9*21 feet, agreeing with the diiferences of the sums of the 
 back and fore sights, and of the sums of the rises and falls, and 
 showing the work of casting to be correct. Thus are obtained a 
 series of vertical heights to be set off perpendicularly to the datum 
 Ihie, through the upper extremities of which the sectional line 
 must be drawn. 
 
 PRACTICAL LEVEL BOOK. 
 
 (Datum line 100 feet below the bench mark at A.) 
 
 Back 
 
 Sights. 
 
 Fore 
 
 Sights. 
 
 Rise. 
 
 Fall. 
 
 Reduced 
 Levels. 
 
 a^j:t ^--'^^- 
 
 feet. 
 
 feet. 
 
 feet. 
 
 feet. 
 
 feet. 
 
 chains. 
 
 3-50 
 4-10 
 
 5-65 
 10-85 
 
 
 2-15 
 6-75 
 
 100-OOd 
 97-85 
 91-10 
 
 .^r. fBMonroad 
 7. go 1 to lime kilns. 
 
 5-04 
 
 9-25 
 
 
 4-21 
 
 86-89 
 
 11-60 
 
 3-84 
 
 12-91 
 
 
 9-07 
 
 77-82 
 
 15-20 r Bottom of 
 
 4-12 
 
 7'65 
 
 
 3-53 
 
 74-29 
 
 .... < canal, distant 
 
 10-49 
 
 3-92 
 
 6-57 
 
 
 80-86 
 
 21-00 [2-80 chains. 
 
 12-96 
 
 3-03 
 
 9-93 
 
 
 90-79 
 100-00 
 
 27-00 to B M at ^. 
 
 44-05 
 
 53-26 
 
 16-50 
 
 25-71 
 
 
 44-05 
 
 
 16-50 
 
 
 r/liA? l^M-turao.n Incf 
 
 
 9-21 
 
 diff. = 
 
 9-21 = 
 
 = 9'21 < reduced level and j 
 [ datum. 1 
 
 In laying down the sectional line from the above columns of 
 reduced levels and distances, the former are always taken from 
 a much larger scale than the latter, otherwise the undulations 
 on the surface of the ground would in many cases be hardly 
 perceptible. 
 
 Draw the horizontal line A G, setting off the distances A B, 
 AC, &c., as in the column of distances, that is A B = 4*60 
 
124 
 
 LEVELLING. 
 
 chains, AC = 7'80, &c., then draw A « = 100 feet, perpen- 
 dicular to A G and parallel to A a draw B 5, C c, &c., setting 
 
 
 
 
 
 
 
 
 
 
 ^1 
 
 
 
 
 
 
 
 
 
 
 
 
 ll 
 
 
 
 1 
 
 
 ~ T 
 
 -~~~^~~~^ 
 
 e 
 
 «^^ 
 
 ./* 
 
 y 
 
 1 
 
 ^1 
 
 ^1 
 
 
 
 
 1 
 
 1 
 a; 
 
 ^1 
 
 A\ 
 
 460 
 
 780 
 
 1160 
 
 1520 
 
 
 T\ 
 
 G\ 
 
 
 2100 
 
 ~ 2500 
 
 off their heights 97*85, 91*10, &c., respectively, from the column 
 of reduced levels, and through the points «, 6, c, &C.3 draw the 
 required sectional hne a g. 
 
 Note. The above operations, though extremely simple, require great care, 
 otherwise, in extensive works of this kind, errors creep in imperceptibly, to 
 check which the agreement of the differences in the level book is essential. 
 
 LEVELS FOR THE FORMATION OF A SECTION. 
 
 In this case it is required to take the levels of a line of 
 country, whei-e the ground plan is already made, and the line of 
 section determined upon, and marked out on the plan. Here, in 
 addition to what is required in running or check levels, the distances 
 to the several stations of the levelling staves from the starting 
 point must be measured. 
 
 Two additional assistants are required in this case to measure 
 the distances of the stave stations along the lines while the 
 operation of levelling goes on, which is the same in every respect 
 as that already described, excepting that, in this case, the opera- 
 tion is conducted upon a line, on the surface plan, a copy of which 
 must be in the surveyor's possession to direct him, and the dis- 
 tances of the several stave stations must be noted in the level 
 book, in the column marked " Distances." 
 
 The following is the level book of an example, showing the 
 manner of keeping it, and also the method of reducing the levels, 
 to obtain the actual heights of each station above the datum 
 line, which is placed 100 feet below the starting point, for con- 
 venience of drawing the section. The whole operation being 
 similar to that already given at page 123, excepting that here 
 we give the particular manner of performing the several parts 
 of the field work, in order that it may be clearly understood by 
 those who are unacquainted with the subject, as it is presumed 
 that, in a short time, railways will become the common means 
 
LEVELLING. 
 
 125 
 
 of transit, both for passengers and goods, throughout every country 
 of the civihzed world. 
 
 THE LEVEL BOOK FOR PLOTTING THE SECTION. 
 
 (Datum 100 feet below the station A.) 
 
 Back 
 Sights. 
 
 Fore 
 Sights. 
 
 Rise. 
 
 Fall. 
 
 Reduced 
 Levels. 
 
 Distances 
 
 Remarks. 
 
 feet. 
 
 feet. 
 
 feet. 
 
 feet. 
 
 feet. 
 100-00 D 
 
 links. 
 
 
 13-71 
 
 7-88 
 
 5-83 
 
 
 105-83 
 
 519 
 
 B.;M. side of road. 
 
 9-40 
 
 16-30 
 
 
 6-90 
 
 98-93 
 
 1315 
 
 
 3-87 
 
 11-71 
 
 
 7-84 
 
 91-09 
 
 1542 
 
 
 2-63 
 
 12-41 
 
 
 9-78 
 
 81-31 
 
 1850 
 
 
 14-62 
 
 0-95 
 
 13-67 
 
 
 94-98 
 
 2358 
 
 
 17-00 
 
 1-45 
 
 15-55 
 
 
 110-53 
 
 2698 
 
 
 10-66 
 
 15-40 
 
 
 4-74 
 
 105-79 
 
 3357 
 
 
 2-87 
 
 17-00 
 
 
 14-13 
 
 91-66 
 
 3758 
 
 
 3-40 
 
 10-32 
 
 
 6-92 
 
 84-74 
 
 3976 
 
 
 5-73 
 
 2-24 
 
 3-49 
 
 
 88-23 
 
 5077 
 
 
 16-54 
 
 0-85 
 
 15-99 
 
 
 103-92 
 
 5904 
 
 
 16-08 
 
 0-89 
 
 1519 
 
 
 119-11 
 
 6124 
 
 
 14-56 
 
 0-73 
 
 13-83 
 
 
 132-94 
 
 6437 
 
 
 10-36 
 
 14-06 
 
 
 3-70 
 
 129-26 
 
 7467 
 
 
 9-84 
 
 1-36 
 
 8-48 
 
 
 137-72 
 
 8369 
 
 
 9-80 
 
 7-00 
 
 2-80 
 
 
 140-52 
 
 9303 
 
 
 2-30 
 
 10-96 
 
 
 -8-66 
 
 131-86 
 
 
 Centre of road at 21 5 
 
 10-96 
 
 14-46 
 
 
 3-50 
 
 128-36 
 
 9679 
 
 [links. 
 
 2-08 
 
 15-05 
 
 
 12-97 
 
 115-39 
 
 9936 
 
 
 1-75 
 
 16-58 
 
 
 14-83 
 
 100-56 
 
 10164 
 
 
 1-84 
 
 17-10 
 
 
 15-26 
 
 85-30 
 
 10576 
 
 
 0-00 
 
 7-43 
 
 
 7.43 
 
 77-87 
 
 11423 
 
 Forward at corner 
 
 5-38 
 
 3-50 
 
 1-88 
 
 
 79-75 
 
 13066 
 
 [of wood. 
 
 8-50 
 
 4-50 
 
 4-00 
 
 
 83-75 
 
 14954 
 
 
 5-30 
 
 1-36 
 
 3-94 
 
 
 87-69 
 
 15650 
 
 
 10-20 
 
 9-40 
 
 0-80 
 
 
 88-49 
 
 17345 
 
 
 6-86 
 
 0-40 
 
 6-46 
 
 
 94-95 
 
 19135 
 
 
 11-00 
 
 3-96 
 
 704 
 
 
 101-99 
 
 19359 
 
 
 11-80 
 
 3-53 
 
 8-27 
 
 
 110-26 
 
 19631 
 
 
 10-53 
 
 2-68 
 
 7-85 
 
 
 118-11 
 
 19841 
 
 Forward ©at end of 
 
 8-82 
 
 1-38 
 
 6-84 
 
 
 124-95 
 
 20561 
 
 [wood. 
 
 8-76 
 
 2-20 
 
 6-56 
 
 
 131-51 
 
 21671 
 
 
 14-00 
 
 14-50 
 
 
 0-50 
 
 131-01 
 
 
 Road at 450 links. 
 
 14-50 
 
 4-32 
 
 10-18 
 
 
 141-19 
 
 22710 
 
 
 9-14 
 
 1-00 
 
 8-14 
 
 
 149-33 
 100-00 
 
 23221 
 
 
 304-19 
 
 254-86 
 
 166-49 
 
 117-16 
 
 254-86 
 
 
 117-16 
 
 
 
 fDifferei 
 i andl 
 
 ice between Datum 
 ast Reduced level, or 
 
 49-33 
 
 — 
 
 49-33 
 
 — 
 
 49-33 
 
 
 
 
 
 
 L heigl 
 
 it of B above A. 
 
126 LEVELLING. 
 
 The several differences of the sums of the back and fore sights, 
 of the sums of the rises and falls, and of the last reduced level 
 and the datum, exactly agreeing, proves the accuracy of the 
 arithmetical operation in the preceding level book, all these dif- 
 ferences being 49*33 feet, which is the height of the last station 
 above the first. 
 
 It is advisable for the surveyor to reduce the levels in the field 
 as he proceeds, as it will occupy very little time, and can be easily 
 done while the staffman is taking a new position. The surveyor 
 will thus be enabled to detect with the eye if he is committing 
 any glaring error ; for instance, inserting a number in the column 
 of rises, when it ought to be in that of falls, the surface of the 
 ground at once reminding him that he is going doAvnward instead 
 of ascending. 
 
 It is seldom the case in practice that the instrument can be 
 placed precisely equi- distant from the back and fore staves, on 
 account of the inequalities of the ground, ponds, &c. ; it would 
 appear, therefore, to be necessary, to make our results perfectly 
 correct, to apply to each observation the correction for curvature 
 and refraction as explained at page 118: this, we believe, is 
 seldom done unless in particular cases, where the utmost possible 
 accuracy is required, on account of the smallness of such cor- 
 rection, as may be seen by referring to the table at the end of 
 the book, where this correction for 1 1 chains is shown to be no 
 more than -^^-^ part of a foot ; and as the difference in the dis- 
 tances between the instrument and the fore and back staves can 
 in no case equal that sum, it is evident that such correction may 
 be safely disregarded in practice. Besides, it is not necessary to 
 have the level placed directly between the staves while making 
 observations, as it is frequently inconvenient to do so, for reasons 
 just given, nor does a deviation from a line of the staves, in this 
 respect, in the, least affect the accuracy of the result. 
 
 The distances in the sixth column of the level book are 
 assumed to be horizontal distances, and in measuring them, care 
 should be taken that they are as nearly such as possible, or they 
 must be afterwards reduced thereto, otherwise the section will 
 be longer than it ought to be. For the purpose of assisting the 
 surveyor in making the necessary reduction from the hypothe- 
 nusal to the horizontal measure, when laying down the section, a 
 table is given in Baher^s La7id and Engineering Surveying, page 
 146, shewing the reduction to be made on each chain's length 
 for the several quantities of rise, as shewn by the reading of 
 the staves. 
 
 Note. For extensive information on this subject see Baker's Land and 
 Engineering Surveying, where an engraved plan and section, adapted to this 
 example; are given at the end of the work. 
 
RAILWAY CURVES. 
 
 127 
 
 THE METHOD OF LAYING OUT RAILWAY CURVES 
 ON THE GROUND. 
 
 In railway practice, the curve adopted is always an arc of a 
 circle, to which the straight portions of the railway are tangents 
 at each extremity of the arc. Sometimes the curve consists of 
 two, three, or more circular arcs with their concavities turned 
 in the same or different directions, as in the compound and ser- 
 pentine curves. 
 
 Problem I. 
 
 To lay out a railway curve on the ground by the common 
 method. 
 
 Case I. — Let H A, q^ q^ he the tangental portions of a rail- 
 way, the extremities A and q^ oi which are required to he 
 united by the circular curve iV q^, to which II A, q^ q^ shall be 
 tangents ; the radius of the curve being supposed to be previ- 
 ously determined. 
 
 Let the radius 
 in this case be 
 80 chains or one 
 mile; prolong the 
 tangent 13 A a 
 distance A p = 
 1 chain; then op- 
 posite 80 in table 
 No. 2, at the end 
 of the book, is 
 found4" 95 inches 
 ^=pqi which set off at right angles to A^, thus giving the first 
 point in the curve. In the direction A q, measure qp^^^ ^ 
 chain, and set off ^2 ^2 — twice ^ q = 4-95 X 2 = 9*9 inches, 
 at right angles to q P2 '> then q^ is the second point in the 
 curve. This last operation must be repeated till the curve 
 shall have been set out to the point 5^^. Lastly q4^2h being 
 measured = 1 chain, in the direction ^3 q^, the offset ^5 q^ will 
 be found = 4*95 inches = the first offset p q, thus proving the 
 accuracy of the work. In this manner the operation is con- 
 ducted, whatever be the length of the curve. 
 
 Case II. — Let A = r, and 5 = A. p = q p^ &c., which 
 may be either less or greater than one chain ; then the general 
 
 5 2 
 length of the first and last offsets p q, 2^5 Ps is — , and the 
 
 length of each of the other offsets is — , or twice the first or 
 
 r 
 
128 RAILWAY CURVES. 
 
 last offset ; but the length of the offsets given n tne table is 
 
 represented by — ; therefore, if A p, qp^i ^^^^y ^^ taken as 2, 
 2 r ' 
 
 3, 4, &c., chains, the value of — must be multiplied by 2^ = 
 
 4, 3^ =: 9, 42 = 16, &c., respectively to find p q, and the re- 
 sult, in each case, multiplied by 2 for each of the offsets ^2 Q^y 
 P3 ^3, &c. In this manner the curve may be set out more 
 speedily, and with less liability to error, on account of the less 
 number and greater length of the lines required in the opera- 
 tion. 
 
 EXAMPLE. 
 
 Let A O = ?' = 120 chains, and S = 4 chains ; then - = — 
 
 2r 2r 
 
 X 16 = 3*3 X 16 = 52*8 inches 4 feet 4*8 inches =p q; whence 
 
 4 feet 8 inches X 2 = 4*8 feet 9*6 inches •=P2 92y Ps Qs = &c. 
 
 Note 1. When the curve has been correctly set out, as m Case II., the inter- 
 mediate stumps may be put in at the end of every chain, if required, by the 
 method given in Case I. The distances of the intermediate stumps, thus put 
 in, will not, in most cases, exceed a fraction of an inch ; because the lengths of 
 the offsets q,P2 5'2» <^c., is so small, that the curvilinear lengths kq, qq q^i 
 &.C., can never greatly exceed those A p, qp^y &c. 
 
 Note 2. The method given in Case II., is sufficiently accurate when 8 does 
 not exceed -^-^ of the radius of the curve. Besides, at the closing point of the 
 curve, as at q^^ the distance q^p^ is most commonly less or greater than 5. Let 
 
 q^p. — d\ then the offset jy^ q^^ at the end of the curve is = i li.; and, 
 
 2r 
 
 when 5 = 1 chain, p'^ q'^ =^^ L ; or the tabular number for the given 
 
 2 r 
 radius must be multiplied by (5 + d)d, or by (1 + d)d, according as A j», qp^^ 
 Slc, is taken = chains or 1 chain, to give the last offset p^, q^; ^ of which 
 is =jt?g 5-5, the offset to the tangent q^ q^. 
 
 EXAMPLE. 
 
 Let r = 1 20, and d = 4 chains, as in the last example, and 
 let q^ p^ =z d =^ 2*68 chains ; then j9^ ?4 = — ^ (d -]- d) cl = 
 
 L V 
 
 3-3 X (4 + 2-68) X 2*68 = 59-07 inches; \ of which, viz., 
 29*535 inches is == j^s ^5. 
 
 Note 3. ^Vhen exceeds -^^ of the radius r of the curve, the following for- 
 mula ought to be used for finding the offsets. 
 
 p q = r V r^ — 5^, 
 52 
 andjo^ q- — Sec. = 
 
 V r^ — :| §2 
 See BaJcer^s Land and Engineering Surveying , page 164. 
 
 Note 4. By this method the greater part of both British and foreign rail- 
 
RAILWAY CURVES. 129 
 
 way curves have been laid out. It was invented by the autlior about 30 years 
 ago, when the Stockton and Darlington ' Railway was laid out, and eagerly 
 adopted by engineers as it involves veiy little calculation, and does not require 
 the use of a theodolite. It is, however, defective in practice, on account of 
 its requiring so very many short lines connected together, as errors vill un- 
 avoidably creep in and multiply, and more especially so where the ground is 
 rough ; thus tlie curve has frequently to bo retraced several times before it 
 can be got right ; hence the author prepared the methods in the following 
 Problem. 
 
 Problem II. 
 
 To Jay out a railway curve on the ground, by offsets from its 
 tanyentSy no obstructions being supposed to prevent the use of the 
 chain on the convex side of the curve. 
 
 Case I. — When the length of the curve does not exceed | of 
 its radius. 
 
 Let A B, D C be 
 straight portions of a 
 railway, the points C and 
 B being required to be 
 joined by a circular curve 
 B C, to which A B, D C 
 shall be tangents, the 
 radius B O of the curve 
 being supposed to be pre- 
 viously determined from 
 an accurate plan of the 
 intended railway. 
 
 Range the tangents A B, D C till they meet at T ; and let 
 the radius B O = 80 chains = 1 mile ; measure on B T the dis- 
 tance B ^ = 1 chain ; and, at right angles to B T, lay oif the 
 offset qp — 4*95 inches, by Table No. 2, as in Problem I. ; then 
 p is the first point in the curve. Next measure qq^ =i \ chain, 
 and lay off the offset ^g q^ = 4-95 X 4 for the second point in 
 the curve. The successive offsets, at the end of every chain, 
 being 4, 9, 16, &c., or 2'\ 3^, 4^, &c., times the first offset pq, 
 which may also be found opposite the given radius in the Table 
 No. 2., as in Prob. I. 
 
 When the offsets have been thus laid out, till the last one 
 ^5^5 falls little short of T ; lay off the same offsets on T C as 
 were laid off in B T, but in au inverted order, making the first 
 distance on T C = T^'^ ; thus completing the curve to C. 
 
 Note, It can rarely happen in practice that the last offsets, from both tan- 
 gents, will meet at the middle point p^ of the short curve, as shewn in the 
 figure ; but will either intersect one another or fall short of the middle point ; 
 but this is a matter of no consequence. 
 
1 30 RAILWAY CURVES. 
 
 EXAMPLE. 
 
 Let the radius of the curve be 1 60 chains, required the offsets 
 at the end of every chain, from the tangent to the curve. 
 p q per Table (No. 1.) = 2*475 inches. 
 
 P2 ^2 = 2-475 X 4 = 9-9 
 
 Ps ^3 = 2-475 X 9 = 22'275 
 
 p^q^ = 2'475 X 16 = 39-6 
 
 &c. = &c. = &c. 
 
 Case II. — To lay out the curve lohen it is any required length. 
 
 In a long curve (of which there are some more than two miles 
 in length) the tangents, if prolonged to their point of meeting, 
 would necessarily fall at a great distance from the curve, thus 
 giving an inconvenient length to the offsets, which in practice 
 should never exceed two chains. To remedy this inconvenience 
 the curve must be divided into two or more parts, by introducing 
 one or more additional tangents, thus the offsets may be confined 
 within their proper limits. Thus the tangent T C may, ill this 
 case, be extended, another tangent applied, and the offsets laid 
 off, thus repeating the operation of Case I. a second time : if the 
 curve be not yet completed, the operation may be repeated a 
 third, fourth, &c., time, till it be completed. 
 
 Note. For a complete development of this important subject, see Baker* s 
 Land and Engineeriny Surveing, Part II., Chap. II., where two other methods 
 of laying out railway curves are given ; also methods of laying out compound, 
 serpentine and deviation curves, with original formulae ; all of which methods, 
 as well as the tvv^o already given, were first drawn up by the author. See 
 page 179 of the work above referred to, where a short history of the inven- 
 tion is given. See, also, Tate^s Geometry, page 247. 
 
 CONTENTS OF RAILWAY CUTTINGS, &c. 
 
 TABLES. 
 
 The General Earthwork Tables, in conjunction with Two 
 Auxiliary Tables, on the same sheet, in Baker'' s Raikvay En- 
 gineering, or the numbers for the slopes in Bidder's Table, are 
 applicable to all varieties of ratio of slopes and widths of forma- 
 tion level in common use ; and with the help of Barlow's table of 
 square roots, these tables will apply to sectional areas, with all 
 the mathematical accuracy that can be attained, with very little 
 more calculation than adding the contents between every two 
 cross-sections, as given by the General Table, — The contents in 
 the General Table are calculated to the nearest unit, as are also 
 
RAILWAY CUTTINGS. 
 
 131 
 
 those in the AiLxiHary Table, No. 2, which is for the decimals of 
 feet in the depths. Tlie AuxiUary Table, No. 1, shews tlie 
 depths of the meeting of the side-slopes below the formation- 
 level, with the number of cubic yards to be subtracted from tbc 
 contents of the General Table for each chain in length, for eight 
 of the most common varieties of ratio of slope. 
 
 The following diagrams and explanations will further illustrate 
 the method of taking the dimensions of railway cuttings, pre- 
 paratory to using the above named tables. 
 
 Let ABD C c abd, he a railway 
 cutting, of which ABDC, ahdc 
 are the cross sections, A B = « i = 
 width of formation level, MM', mwb 
 the middle depths of the two cross- 
 sections ; the side-slopes AC, D B, 
 a c b d, w^hen prolonged two and 
 two, will intersect at N and n, at 
 which points the prolongations of 
 M M', m m will also meet, thus constituting a prism A B N 
 nabi the content of which is to be deducted from the whole 
 content, given by the General Table, by means of the Table No. 
 1 . ; in wdiich the depth M' N = ?;« ?i is also given, as already 
 stated, to several varieties of slope and bottom width. 
 
 To place this subject in a more practical point of view, let the 
 annexed figure represent a longitudinal and vertical section of a 
 cutting, passing through the middle A E of the formation level. 
 H I, the line of the rails, and a hj the line on which the slopes, 
 if prolonged, would meet. 
 
 It will 
 cutting 
 
 be seen tbat the 
 
 K b c d Yi com- 
 mences and runs out on the 
 formation level A E, and 
 that the depth A « = B e 
 = Cf = &c., is to be added 
 to the several depths B b, 
 
 C c, D c? of the cutting, the . 
 
 first and last depth at A ^ -^ 9 7i 
 
 and E being each = ; or, what amounts to the same thing, the 
 several depths must be measured from the line a h : thus A «, be, 
 eg, &c., are the depths to be used. And since the depth A « is 
 given in Table No. 1, for all the most common cases, or it may be 
 readily found by calculation for all cases, (see Railway Engineer- 
 ing), the line corresponding to a h must, therefore, be ruled on 
 the railway section, at the proper distance below A E, from 
 which the several depths must be measured; or the vertical 
 
132 
 
 RAILWAY CUTTINGS. 
 
 scale may be marked with Indian ink, (which may be readily 
 rubbed off) at the same distance, and this mark may then be 
 apphed to the formation level A E, for the purpose of measuring 
 the several depths. — In the case of an embankment, the line for 
 the several depths must be placed at a like distance above the 
 formation level. 
 
 Problem I. 
 
 The several depths of a railway cutting to the meeting of the side 
 slopes, the width of formation level, and the ratio of the slopes heing 
 given, to find the content of the cutting in cubic yards, from the 
 Tables referred to, the distances of the depths being one chain each. 
 
 Rule. — Take the several quantities, corresponding to every 
 two succeeding depths of a cutting, or embankment, measured to 
 the meeting of the side slopes, at the distance of 1 chain each, 
 from the General Table in Baker's Railway E7igineering , and 
 multiply their sum by the ratio of the slopes ; from the product 
 subtract the cubic yards, corresponding to the given bottom 
 width and ratio of slopes from Table No. 1., multiplied by the 
 whole length of the cutting, and the remainder will be the con- 
 tent of the cutting in cubic yards. 
 
 But, when the distances of the depths are greater or less than 
 1 chain, the quantities of the General Table must be multiplied by 
 their respective distances. — And, when the distances are given in 
 feet, the quantities must be multiplied by those distances, and the 
 final result divided by 66 for the content in cubic yards, as in the 
 following 
 
 EXAMPLES. 
 
 1. Let the depth of the railway cutting 
 or embankment to the meeting of the 
 side-slopes, at the end of every chain, be 
 as in the following table, the bottom-width 
 30 feet, and the ratio of the slopes as 2 to 
 1 ; required the content in cubic yards. 
 
 Note. In the annexed table the quantity 1238, 
 corresponds to the depths 10 and 33 feet, in the 
 General Table ; the quantity 3175 to the depths 
 33 and 39, and so on for the succeeding depths. 
 By the Auxiliary Table No. 1, it will be seen, that 
 the depth to be added below the formation level, 
 for the given width and ratio of slopes, is 7*50 = 
 7i feet, therefore, the cutting begins and ends with 
 a depth of 10 — 7^ = 2^ feet. The corresponding 
 numberof cubicyards,tobe deducted for each chain 
 
 Dist. in 
 chains. 
 
 Depth 
 in feet. 
 
 Qnts.per 
 G. Table. 
 
 
 
 10 
 
 
 1-00 
 
 33 
 
 1238 
 
 2-00 
 
 39 
 
 3175 
 
 3-00 
 
 35 
 
 3350 
 
 4-00 
 
 10 
 
 1355 
 
 For slope 1 to 1 
 
 .9128 
 
 For slope 2 to 1 . 
 
 2 
 
 18256 
 
 Subtract l 
 
 275 X 4 j ~ 
 Content in l _ 
 
 cubic yds. J 
 
 = 1100 
 17156 
 
RAILWAY CUTTINGS. 
 
 133 
 
 in length, is multiplied by 4 chains the whole length of the cutting, thus 
 giving the whole quantity to be deducted, the remainder being the true 
 content in cubic yards of the cutting. 
 
 2. The several depths 
 of a railway cutting to the 
 meeting of the side slopes 
 are as in the annexed 
 table, the bottom width 
 being 30 feet, and the 
 ratio of the slopes 1^ to 
 1 ; required the content 
 of the cutting. 
 
 Note. When any of the dis- 
 tances between two succeed- 
 ing depths is greater or less 
 than 1 chain, the correspond- 
 ing quantity from the General 
 Table must be multiplied by 
 that particular distance ; as the 
 distances between the depths 
 20 and 25, and between 32 and 
 39, &c., the distance being 2 
 chains. The last distance, viz., 
 that bet ween 30 and 10, is 1-4 6, 
 in this case 2 figures must be cut off for decimals, after multiplying, 
 
 3. Let the depths of a 
 railway cutting to the 
 meeting of the side slopes, 
 and their distances in feet 
 he as in the annexed table, 
 the bottom wddth 30 feet, 
 and the ratio of the slopes 
 1 1- to 1 ; required the con- 
 tent in cubic yards. 
 
 Note. When the distances 
 are in feet the quantities from 
 General Table must be re- 
 spectively multiplied by their 
 distances, the quantity from 
 Table No. 1, by the whole dis- 
 tance, and the result divided 
 by 66, the feet in 1 chain, for the content in cubic yards. 
 
 Dist. in Depths 
 chains, in feet. 
 
 Products for Dist. 
 
 Total 
 
 greatcrthanl chain. 
 
 quantities. 
 
 
 
 10 
 
 
 
 1-00 
 
 16 
 
 • • • • 
 
 420 
 
 2-00 
 
 20 
 
 • • . • 
 
 705 
 
 4-00 
 
 25 
 
 1234 X 2 
 
 2486 
 
 5-00 
 
 32 
 
 , 
 
 1996 
 
 7-00 
 
 39 
 
 3091 X 2 
 
 6182 
 
 8-00 
 
 45 
 
 • • . « 
 
 4319 
 
 10-00 
 
 50 
 
 5520 X 2 
 
 11040 
 
 12-00 
 
 40 
 
 4971 X 2 
 
 9942 
 
 13-00 
 
 30 
 
 • . . . 
 
 3015 
 
 14-46 
 
 10 
 
 1059 X 1-46 
 
 1546 
 
 For side slones 1 to 1 .... 
 
 41741 
 
 For side slopes ^ to 1 
 
 For side slopes 1 4 to 1 
 
 20870 
 
 62611 
 
 366-67 X 14-46 = 
 
 5302 
 
 
 Conte 
 
 nt in cubic yards 
 
 67309 
 
 Dist. i] 
 feet. 
 
 
 
 90 
 
 178 
 
 278 
 
 Depths 
 in feet. 
 
 37 
 50 
 61 
 39 
 
 Quantities multi- 
 plied by length. 
 
 4660 
 7554 
 6210 
 
 90 
 
 100 
 
 Total 
 quantities. 
 
 419100 
 664752 
 621000 
 
 For slopes 
 For slopes 
 
 1 tol 1705152 
 
 itol 852576 
 
 For slopes 14 to 1 2557728 
 
 366-67 X 278 = 101933 
 
 66)2455795 
 
 Content in cubic yards 37209 
 
 Problem II. 
 
 Case I. — The areas of two cross sections of a railway cutting 
 to the intersection of the side slopes, its length in chains, bottom 
 width, and the ratio of the slopes are given; required the content 
 of the cutting in cubic yards. 
 
 RxTLE. — With the square roots of the given areas as depths, 
 
 7 
 
134 
 
 RAILWAY CUTTINGS. 
 
 find tlie content from the General Table, as in tlie last Problem, 
 from which subtract the quantity answering to the given width, 
 and the ratio of side slopes from Table No. 1, and the remainder, 
 being multiplied by the length, will be the content required. 
 
 i^GTK If the length be given in feet, proceed as in Example 3, last Problem. 
 
 EXAMPLE. 
 
 1. Let the two sectional areas of a cutting be 4761 and 1296 
 square feet, the bottom width 36 feet, the length 3*25 chains, 
 and the ratio of the side slopes 2 to 1 ; required the content in 
 cubic yards. 
 
 s/4m = 69 
 
 v/1296 == 36 
 
 For bottom v>^idth 36 and slopes 2 to 1, per 
 Table No. 1 
 
 > content per General Table 6959 
 
 } 
 
 Content for 1 chaui in length 
 
 396 
 
 . 6563 
 3i 
 
 19689 
 1641 
 
 Content for 3-15 chains 21330 cubic yds. 
 
 Case II. — In measuring co7itract ivorJc, where great accuracy 
 is required, the i^Qths of afjot, or second decimals, must he used 
 in the calculation, ty taking for them ^^th of their respective 
 quantities in Table No. 2. 
 
 EXAMPLE. 
 
 The areas of seven cross sections of a 
 railway cutting to the meeting of the side 
 slopes and their distances are as in the 
 annexed table ; the bottom width is 30 feet, 
 and the ratio of the slopes l^ to 1 ; required 
 the cubic yards in the cutting. 
 
 Ans. The content, per General Table, 
 and Table No. 2, is 172318 cubic yards, 
 from which the quantity corresponding to 
 the given bottom width and ratio of slopes 
 X by the whole length, viz. 275 x 18 = 4950 cubic yards 
 
 Dist. in 
 
 Areas in. 
 
 chains. 
 
 sq. feet. 
 
 
 
 2727 
 
 2-00 
 
 3136 
 
 600 
 
 4221 
 
 9-00 
 
 4100 
 
 1400 
 
 5141 
 
 16-00 
 
 3759 
 
 18-00 
 
 2161 
 
CONTENTS OF SOLIDS. 135 
 
 must be deducted, wliicH leaves 167568 cubic yards, the content 
 required. 
 
 Note. For further explanations aud numerous examples of the methods 
 of finding the contents of earthwork, sco Baker's Land and Engineering 
 Surveying. See, also, Tate's Geometri/, page 252. 
 
 GENERAL RULE FOR FINDING THE CONTENTS 
 OF SOLIDS. 
 
 The wedge, the prismoid, the pyramids, and their frustums ; 
 the whole or a segment, or any portion of the whole, contained 
 between two parallel planes perpendicular to the axis of a sphere, 
 of an ellipsoid, of a paraboloid, of an hyperboloid, may be found 
 by the following general formula. 
 
 Let A and B be the areas of the ends of the solid, C the 
 area of a section parallel to and equidistant from the ends, and 
 L the distance between the ends ; then 
 
 ^, ..^. A + B + 4C -. 
 The sohdity = x L : 
 
 The following investigation of this very general Rule was given 
 by B. Gomperiz, Esq., F.R.S., &c., in the Gentlemen's Mathe- 
 matical Companion for 1822. 
 
 Let X be the variable distance from a given point of another 
 section parallel to the two ends, and a, 6, c be given quantities, 
 the area of the said section will be a + 6 ^ + c a;-, as this will 
 contain the cases of the sections of the solids, enumerated in 
 the Rule ; for instance, in the pyramid or cone, the area of the 
 section may be expressed by c aP, in the wedge the areas may be 
 expressed by « + 6 a; 4- c ^-, a, h, c being constant for the 
 same point and w^edge for any parallel section to a plane given 
 in position, as long as the section has the same number of 
 sides. In the paraboloid all planes perpendicular to the axis, 
 if the given point be the vertex, will have the areas of their 
 sections expressed by h x. In the ellipsoid or hyperboloid, the 
 point being in that vertex of the axis about which it is re- 
 volved, the area of the sections may be expressed hjhx-\-c x^, 
 c being negative in the ellipsoid and positive in the hyper- 
 boloid. And I observe from the method of equidistant ordi- 
 nates in curves, or of sections in solids, see the method of 
 differences, if A, B represent the areas of the two ends, C the 
 
136 CONTENTS OF SOLIDS. 
 
 area of the section in the middle between them, and L the 
 length; then 
 
 The solidity = ^ + ^ + ^ ^ ^ L Q. E. T. 
 
 EXAMPLES ON THE FOREGOING RULE. 
 
 1 . The length of a railway cutting is 5 chains or 110 yards, 
 the top width and depth at one end are respectively 120 and 
 30 feet, the top width and depth at the other end are respec- 
 tively 90 and 20 feet, and the bottom width 30 feet ; required 
 the content of the cutting in cubic yards. 
 
 Ans, 20777|- cubic yards, 
 
 2. Required the content of a sphere, the diameter of which is 
 12 feet. Ans, 33*5104 cuhic yards, 
 
 Note. Here the areas of the extreme sections are each = 0. 
 
 3. A cask is in the form of the middle zone of a sphere, its 
 top and bottom diameters being 34 inches, and its height 30 
 inches, inside measure ; how many gallons will it hold ? 
 
 Ans. 149-22. 
 
 4. How many gallons are contained in a cask, in the form of 
 the middle zone of a spheroid, the bung and head diameter being 
 40 and 32 inches, and the length 36 inches, all internal measures ? 
 
 Ans, 143 J imperial gallons, 
 
 5. How many cubic feet are there in a parabolic conoid, the 
 height of which is 42, and the diameter of its base 24 inches ? 
 
 Ans, 5-4978 cubic feet. 
 
 6. A 5 inch cube of ivory is turned into a sphere of the same 
 diameter; what weight of ivory will be lost, its weight being 
 1820 ounces (Av,) per cubic foot? Ans, 61-44 ounces. 
 
137 
 
 TABLE No. 1.— THE AREAS OF SEGMENTS OF CIRCLES. DIAMETER UNITT. 
 
 Height. 
 
 Area Segment. H 
 
 sight. 
 
 Area Segment H 
 
 eight. 
 
 Area Segment. 
 
 •001 
 
 •000042 
 
 050 
 
 .014681 
 
 099 
 
 •040276 
 
 •002 
 
 •000119 
 
 051 
 
 •015119 
 
 100 
 
 •040875 
 
 •003 
 
 •000219 
 
 052 
 
 •015561 
 
 101 
 
 •041476 
 
 •004 
 
 •000337 
 
 053 
 
 •016007 
 
 102 
 
 •042080 
 
 •005 
 
 •000470 
 
 054 
 
 •016457 
 
 103 
 
 •042687 
 
 •006 
 
 •000618 
 
 055 
 
 •016911 
 
 104 
 
 •043296 
 
 •007 
 
 •000779 
 
 056 
 
 •017369 
 
 105 
 
 •0i3908 
 
 •008 
 
 •000951 
 
 057 
 
 •017831 
 
 106 
 
 •044522 
 
 •009 
 
 •001135 
 
 058 
 
 •018296 
 
 107 
 
 •045139 
 
 •010 
 
 •001329 
 
 059 
 
 •018766 
 
 108 
 
 •045759 
 
 •Oil 
 
 •001533 
 
 060 
 
 •019239 
 
 109 
 
 •046381 
 
 •012 
 
 •001746 
 
 061 
 
 •019716 
 
 110 
 
 •047005 
 
 •013 
 
 •001968 
 
 062 
 
 •020196 
 
 111 
 
 •047632 
 
 •014 
 
 •002199 
 
 063 
 
 •020680 
 
 112 
 
 •048262 
 
 •015 
 
 •002438 
 
 064 
 
 •021168 
 
 113 
 
 •048894 
 
 •016 
 
 •002685 
 
 065 
 
 •021659 
 
 114 
 
 •049528 
 
 •017 
 
 •002940 
 
 066 
 
 •022154 
 
 115 
 
 •049165 
 
 •018 
 
 •003202 
 
 •067 
 
 •022652 
 
 116 
 
 .050804 
 
 •019 
 
 •003471 
 
 068 
 
 •023154 
 
 117 
 
 •050446 
 
 •020 
 
 •003748 
 
 069 
 
 •023659 
 
 118 
 
 •051090 
 
 •021 
 
 •004031 
 
 070 
 
 •024168 
 
 119 
 
 •052736 
 
 •022 
 
 •004322 
 
 071 
 
 •024680 
 
 120 
 
 •052385 
 
 •023 
 
 •004618 
 
 072 
 
 •025195 
 
 121 
 
 •053036 
 
 •024 
 
 •004921 
 
 073 
 
 •025714 
 
 122 
 
 •054689 
 
 •025 
 
 •005230 
 
 074 
 
 •026236 
 
 123 
 
 •054345 
 
 •026 
 
 •005546 
 
 075 
 
 •026761 
 
 124 
 
 •055003 
 
 •027 
 
 •005867 
 
 076 
 
 •027289 
 
 125 
 
 •056663 
 
 •028 
 
 •006194 
 
 077 
 
 •027821 
 
 126 
 
 •056326 
 
 •029 
 
 •006527 
 
 078 
 
 •028356 
 
 127 
 
 •057991 
 
 •030 
 
 •006865 
 
 •079 
 
 •028894 
 
 128 
 
 •057658 
 
 •031 
 
 •007209 
 
 080 
 
 •029435 
 
 129 
 
 •058327 
 
 •032 
 
 •007558 
 
 081 
 
 •029972 
 
 130 
 
 •059999 
 
 •033 
 
 •007913 
 
 082 
 
 •030526 
 
 131 
 
 •060672 
 
 •034 
 
 •008273 
 
 083 
 
 •031076 
 
 132 
 
 •061348 
 
 •035 
 
 •008698 
 
 084 
 
 •031629 
 
 133 
 
 •062026 
 
 •036 
 
 •009008 
 
 085 
 
 •032186 
 
 134 
 
 •062707 
 
 •037 
 
 •009383 
 
 086 
 
 •032745 
 
 135 
 
 •063389 
 
 •038 
 
 •009763 
 
 087 
 
 •033307 
 
 136 
 
 •064074 
 
 •039 
 
 •010148 
 
 088 
 
 •033872 
 
 137 
 
 •064760 
 
 •040 
 
 •010537 
 
 089 
 
 •034441 
 
 138 
 
 •065449 
 
 •041 
 
 •010931 
 
 090 
 
 •035011 
 
 139 
 
 •066140 
 
 •042 
 
 •011330 
 
 091 
 
 •035585 
 
 140 
 
 •066833 
 
 •043 
 
 •011734 
 
 092 
 
 •036162 
 
 141 
 
 •067528 
 
 •044 
 
 •012142 
 
 093 
 
 •036741 
 
 142 
 
 •068225 
 
 •045 
 
 •012554 
 
 094 
 
 •037323 
 
 143 
 
 •068924 
 
 •046 
 
 •012971 
 
 095 
 
 •037909 
 
 144 
 
 •069625 
 
 •047 
 
 •013392 
 
 096 
 
 •038496 
 
 145 
 
 •070328 
 
 •048 
 
 •013818 
 
 097 
 
 •039087 
 
 146 
 
 •071033 
 
 •049 
 
 •014247 
 
 098 
 
 •039680 
 
 147 
 
 •071741 
 
 Note. When the tabular height exceeds the heights given in this Tablo 
 the segment must be divided into two equal parts and their common versed 
 hfne found by Prob. VL, page 26. The tabular heights will then fall within 
 this Table, whence the area of the whole segment may be found. 
 
138 TABLES OF OFFSETS FOR RAILWAY CURVES. AND CORRECTION I 
 
 
 
 OF LEVELS FOR CURVATURE, ETC. 
 
 
 
 No. S.- 
 
 -Offsets at the end of the first chain from tangent point 
 
 of railway curves, j 
 
 Radius 
 
 Ofifsets in 
 
 Radius 
 
 Offsets in! 
 
 Radius 
 
 Offsets in 
 
 Radius 
 
 Offsets in 
 
 of curve 
 
 nches and 
 
 of curve 
 
 nches and 
 
 of curve 
 
 inches and 
 
 of curve 
 
 inches and 
 
 ill chns. 
 
 decimals. 
 
 ill chns. 
 
 decimals. 
 
 in chns. 
 
 decimals. 
 
 in chns. 
 
 decimals. 
 
 40 
 
 9-9000 
 
 64 
 
 6-1875 
 
 88 
 
 4-5000 
 
 120 
 
 3-3000 
 
 i 41 
 
 9-6588 
 
 65 
 
 9-0923 
 
 89 
 
 4-4496 
 
 122 
 
 3-2459 
 
 42 
 
 9-4285 
 
 66 
 
 60000 
 
 90 
 
 4-4000 
 
 124 
 
 3-1935 
 
 43 
 
 9-2093 
 
 67 
 
 5-9104 
 
 91 
 
 4-3516 
 
 125 
 
 3-1680 
 
 44 
 
 9-0000 
 
 68 
 
 5-8235 
 
 92 
 
 4-3043 
 
 1-26 
 
 3-1428 
 
 45 
 
 8-8000 
 
 69 
 
 5-7391 
 
 93 
 
 4-2581 
 
 128 
 
 3-0937 
 
 46 
 
 8-6087 
 
 70 
 
 5-6571 
 
 94 
 
 4-2128 
 
 130 
 
 3-0461 
 
 47 
 
 8-4255 
 
 71 
 
 5-5774 
 
 95 
 
 4-1684 
 
 132 
 
 3-0000 
 
 48 
 
 8-2500 
 
 72 
 
 5-5000 
 
 96 
 
 4-1250 
 
 134 
 
 2-9552 
 
 49 
 
 8-0816 
 
 73 
 
 5-4246 
 
 97 
 
 4-0825 
 
 135 
 
 2-9333 
 
 50 
 
 7-9200 
 
 74 
 
 5'3513 
 
 98 
 
 4-0408 
 
 136 
 
 2-9117 
 
 51 
 
 7-7647 
 
 75 
 
 5-2800 
 
 99 
 
 4-0000 
 
 138 
 
 2-8645 
 
 52 
 
 7-6154 
 
 76 
 
 5-2105 
 
 100 
 
 3-9600 
 
 140 
 
 2-8285 
 
 53 
 
 7-4717 
 
 77 
 
 5-1428 
 
 102 
 
 3-8824 
 
 142 
 
 2-7887 
 
 54 
 
 7-3333 
 
 78 
 
 5-0769 
 
 104 
 
 3-8077 
 
 144 
 
 2-7500 
 
 55 
 
 7-2000 
 
 79 
 
 5-0126 
 
 105 
 
 3-7714 
 
 145 
 
 2-7310 
 
 56 
 
 7-0714 
 
 80 
 
 4-9500 
 
 106 
 
 3-7358 
 
 146 
 
 2-7123 
 
 57 
 
 6-9473 
 
 81 
 
 4-8889 
 
 108 
 
 3-6667 
 
 148 
 
 2-6756 
 
 58 
 
 6-8276 
 
 82 
 
 4-8292 
 
 110 
 
 3-6000 
 
 150 
 
 2-6400 
 
 59 
 
 6-7118 
 
 83 
 
 4-7711 
 
 112 
 
 3-5352 
 
 152 
 
 2-6052 
 
 60 
 
 6-6000 
 
 84 
 
 4-7143 
 
 114 
 
 3-4736 
 
 154 
 
 2-5714 
 
 61 
 
 6-4918 
 
 85 
 
 4-6588 
 
 115 
 
 3-4435 
 
 155 
 
 2-5548 
 
 62 
 
 6-3871 
 
 86 
 
 4-6046 
 
 116 
 
 3-4138 
 
 156 
 
 2-5384 
 
 63 
 
 6-2857 
 
 87 
 
 4-5517 
 
 118 
 
 3-3599 
 
 158 
 
 2-5063 
 
 TABLE OF cor 
 
 IRECTION 
 
 3 FOR CI 
 
 JRVATURU 
 
 , ETC. 
 
 
 No. 3.—] 
 
 Difference be 
 
 tween api 
 
 :)arent and 
 
 No. 4.— 
 
 Diflercuce b 
 
 stweeu apparent and 
 
 true 
 
 level for dis 
 
 tances in 
 
 chains. 
 
 true 
 
 level for di 
 
 stances in 
 
 miles. 
 
 Co] 
 
 -rection in d 
 
 ecimals of 
 
 feet. 
 
 Corr 
 
 ection in fe 
 
 et and de 
 
 cimals. 
 
 Distance 
 
 For curva- 
 
 Distance 
 
 For curva- 
 
 Distance 
 
 For curva- 
 
 Distance 
 
 For curva- 
 
 ill 
 
 ture and 
 
 in 
 
 ture and 
 
 in 
 
 ture and 
 
 in 
 
 ture and 
 
 cliaius. 
 
 refraction. 
 
 chains. 
 
 refraction. 
 
 miles. 
 
 refraction. 
 
 miles. 
 
 refraction 
 
 H 
 
 •001 
 
 14 
 
 •017 
 
 i 
 
 4 
 
 •03 
 
 10 
 
 57-17 
 
 4 
 
 •002 
 
 144 
 
 -019 
 
 i 
 
 •15 
 
 lOi 
 
 61-30 
 
 H 
 
 •002 
 
 15 
 
 -020 
 
 1 
 
 •32 
 
 11 
 
 69-16 
 
 5 
 
 •003 
 
 154 
 
 •021 
 
 1 
 
 •58 
 
 m 
 
 75-59 
 
 54 
 
 •003 
 
 16 
 
 •023 
 
 H 
 
 1-29 
 
 12 
 
 82-29 
 
 6 
 
 -003 
 
 16i 
 
 •024 
 
 2 
 
 2-29 
 
 121 
 
 89-29 
 
 64 
 
 •004 
 
 17 
 
 •026 
 
 2i 
 
 3-57 
 
 13 
 
 96-58 
 
 7 
 
 •004 
 
 17i 
 
 -027 
 
 3 
 
 5-14 
 
 134 
 
 104-14 
 
 T4 
 
 •005 
 
 18 
 
 •029 
 
 34 
 
 7-00 
 
 14 
 
 112-00 
 
 8 
 
 •006 
 
 18i 
 
 -031 
 
 4 
 
 9-15 
 
 IH 
 
 120-15 
 
 84 
 
 •006 
 
 19 
 
 •033 
 
 H 
 
 11-62 
 
 15 
 
 128-57 
 
 9 
 
 •007 
 
 194 
 
 •034 
 
 5 
 
 14-29 
 
 151 
 
 137-29 
 
 H 
 
 •008 
 
 20 
 
 •036 
 
 ^ 
 
 17-30 
 
 16 
 
 146-29 
 
 10~ 
 
 •009 
 
 20J 
 
 •038 
 
 6 
 
 20-58 
 
 161 
 
 155-57 
 
 104 
 
 •009 
 
 2l' 
 
 •039 
 
 61 
 
 24-15 
 
 17 
 
 165-15 
 
 11 
 
 •Oil 
 
 214 
 
 •041 
 
 7 
 
 28-01 
 
 174 
 
 175-00 
 
 Hi 
 
 •012 
 
 22 
 
 •043 
 
 74 
 
 32-16 
 
 18 
 
 185-14 
 
 12 
 
 •013 
 
 224 
 
 .'.0f6" 
 
 8 
 
 36-59 
 
 181 
 
 195-59 
 
 12i 
 
 •014 
 
 23 , 
 
 ■ -HX'- 
 
 8h 
 
 41-31 
 
 19 
 
 206-29 
 
 13 
 
 •016 
 
 234 /(' 
 
 .:m:" 
 
 1 ^ 
 
 46-30 
 
 191 
 
 217-29 
 
 134 
 
 •016 
 
 24^ 
 
 U-O^iv . 
 
 l.:-2^ . 
 
 5Ui60 
 
 20 
 
 228-60 
 
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 Orders, plan-drawing 
 
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 General Clauses 
 
 Foundiitions 
 
 Well 
 
 Artificial Foundations 
 
 Brickwork 
 
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 mingled 
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 TREDGOLD'S CARPENTRY. FOURTH EDITION. 
 
 In One large Vol. Ato., 21. 2.<f., in crtra cloih. 
 
 THE ELEMENTARY PRINCIPLES OP 
 CARPENTRY ; 
 
 A Treatise on the pressure and equilibrium of timber framing, the resistance of 
 timber, and the construction of floors, arches, bridges, roofs, uniting iron and stone 
 ■with timber, «S:c., with practical rules and exami)lcs ; to which is added, an essay on 
 the nature and properties of timber, including the method of seasoning, and the 
 causes and prevention of decay, ■with descriptions of the kinds of wood used in 
 building ; al^o numerous tables of the scantlings of timber for different purposes, 
 the specific gravities of materials, &:c. 
 
 By THOMAS TREDGOLD, Civil Engineer. 
 
 Illustrated by fifty-three Engravings, a portrait of the author, and several Woodcuts. 
 Fourth Edition, corrected and considerably enlarged. With an Appendix, containing 
 specimens of various ancient and modern roofs. 
 
 Edited by Peter Barlow, F.R.S. 
 
 1 Equilibrium and pressure of 
 
 beams. 
 
 2 Pressure- of beams and centre 
 
 of gravity 
 
 3 Equilibrium and pressure of 
 
 beams and framing 
 
 4 Nailed flooring 
 
 5 to 9 Roofs 
 
 10 Roofs that bave been erected 
 
 11 Roof of tbe riding-bouse at 
 
 12 Domes [Moscow 
 
 13 Partitions and centre 
 
 14 Centres for stone bridfjes; 
 
 centre used for the brid-^-e nt 
 Neuilly; for the "VVaterloo 
 Bridge, and Conon Bridge 
 
 15 Centres for stone bridges 
 
 16 Bridges 
 
 17 Ditto, double plate 
 
 18 Construction of bridges 
 
 19 Bridges, double plate 
 
 20 Bridges and joints 
 
 21 Joints 
 
 22 Joints and straps 
 
 23 Roof and construction of the 
 
 Pantheon, Oxford Steet 
 
 24 Ditto 
 
 25 Section of Roof of Hall, Park- 
 
 hurst Prison 
 
 26 Section of roof of New Saloon, 
 
 Academy of Arts, Florence 
 
 CONTENTS OF PLATES. 
 
 27 Longitudinal section of ditto 
 
 28 Truss of the roof of the Ducal 
 
 riding house, Modena, double 
 plate 
 
 29 Truncated roof of ditto 
 
 30 Truss of roof of ditto. 
 
 31 Section of the roof over the 
 
 Exchange, Geneva, double 
 
 32 Foot of truss of roof over the 
 
 new theatre at Ancona , 
 ditto, Palazzo Vecchio, 
 Florence ; ditto. Cathedral, 
 Florence, double. 
 
 33 Roofs of the Cathedral at 
 
 Leghorn, double 
 
 34 Details of roof of Christ's 
 
 Hospital, ditto 
 
 35 Ditto 
 
 36 Longitudinal section of St. 
 
 Dunstan's Church, Fleet 
 Street, double 
 
 37 Roof and plan of ditto, ditto 
 
 38 l»etails of ditto, ditto 
 
 39 Truss at the Thames Plate 
 
 Glass AVorks; truss at tlie 
 Princess's Theatre, Oxford 
 Street ; truss at a house in 
 Berkeley Square 
 
 40 Roof of iron and timber at 
 
 Nottingham Water Works, 
 treble plate 
 
 41 Caf^t-ironroof over the model- 
 
 room of the Butterley Com- 
 pan^, treble 
 
 42 Cast-iron roof over the smith- 
 
 ery of the ditto, ditto 
 
 43 Iron and timber roof over 
 
 the engine manufactory of 
 the ditto, ditto 
 
 44 Roof to King's College Cha- 
 
 pel, Cambridge, ditto 
 
 45 Ditto, transverse section 
 
 46 Details, ditto 
 
 47 Details, ditto 
 
 43 Sectionof roof Great Northern 
 Railway, London Terminus, 
 Passenger Sfetion 
 
 49 Sections and details of ditto 
 
 50 Great Northern Railway, Lon- 
 
 don Terminus, Goods' Sta- 
 tion, transverse section 
 
 51 Iron Roof made for the Clyde 
 
 Trustees for the Quay" at 
 Glasgow, with details 
 
 52 Details of iron roof erected for 
 
 Messrs. Joseph Whitworth & 
 Co., Manchester 
 
 53 Details, &c., of an iron roof, 
 
 erected at the Galway Teruii- 
 nu8 
 
 HANDY BOOK FOR ACTTJAHIES, BANKERS, INSURANCE OFFICES, 
 AND COMMERCIAL MEN IN GENERAL. 
 
 In 12wio, cloih, price 5s. 
 
 THEORY OF COMPOUND INTEREST AND 
 
 ANNUITIES, 
 
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 By FEDOR THOMAN, of the Societe Credit Mobilier, Paris. 
 
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 With 8 copper plates, and numerous woodciUs, hi roan tuck, 63. 
 
 THE ENGINEER'S, ARCHITECT'S, AND 
 CONTRACTOR'S POCKET BOOK. 
 
 Published annually. With Diary of Events and Data connected with Engineering, 
 Architecture, and the kindred Sciences, professionally and otherwise revised. 
 
 Alloys. Almarack 
 
 Ballasting. Barluw 
 
 Barrel Drains 
 
 Bessemer ou the Manufacture of 
 Iron and Steel 
 
 Boilers and Entwines (Proportions 
 of) 
 
 Boilers, Furnaces, and Cliiinneys 
 
 Calendar 
 
 Carpentry and Joinery 
 
 Cask and Malt Gauging: 
 
 Castings, sundry for Sewers, Gas- 
 works, &c. 
 
 Cast-iron Columns and Girders 
 
 Chairs for Railways 
 
 Chimneys, dimensions of 
 
 Circumference of Circles 
 
 Circular Arcs (Tables of) 
 
 Circle, Cylinder, Sphere, &c. 
 
 Coal Expeiiments; Economic 
 \alues of Coals 
 
 Coking (evaporative Powers of 
 Coal, and Results of) 
 
 Columns, Posts, &c. 
 
 Copper Mines (Synopsis of) in 
 Devon and Cornwall 
 
 Cornwall Pumping-Eiigiaes 
 
 Current Coins 
 
 Du Buat • 
 
 Earthwork 
 
 East London Waterworks 
 
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 Elastic Properties of Steam 
 
 Ellipses, Cones, Frustrums, «&c. 
 
 Ephemerides of the Planets 
 
 Fairbairn on the Met hanical Pro- 
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 various ten>peratures; Tubular 
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 Resistance of Tubes to Collapse 
 
 Flooring 
 
 French and English Scales 
 
 Friction 
 
 Fuel on the American Railways 
 and on English llailwaya 
 
 Gas Engineers' Calendar 
 
 CONTENTS FOE. 1860. 
 
 Gauges (List of) and Weights of 
 Galvanized Tinned Iron sheets 
 
 Girders (Cast-iron) 
 
 Hawksley 
 
 Heat (Effects of) 
 
 High Water at London Bridge 
 
 Howard 
 
 Hydraulics 
 
 Hydrodynamics 
 
 Institution of Civil Engineers 
 (List of Members of) 
 
 Institute of British Architects 
 (List of Members of) 
 
 Iron Bar 
 
 Iron 
 
 Roofs 
 
 Knot Tables 
 
 Latitudes and Longitudes [&c. 
 
 Log. of Sines, Cosines, Tangents, 
 
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 Marine Screw Propulsion 
 
 Maristier 
 
 Masonry 
 
 Mensuration (Epitome of) 
 
 Morin's Experiments on Friction ; 
 on Ropes 
 
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 Nfcville, on Retaining Walls 
 
 Notes to accompany the Abbre- 
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 Peninsular and Oriental Steam 
 Fleet 
 
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 I^roportions of Marine Engines 
 
 Proportional Sizes and Weights 
 of Hexagon-heads and Nails for 
 Bolts 
 
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 Rennie (G.); Messrs. Reunie 
 
 Roofs 
 
 Ropes, Experiments of 
 
 Sewers 
 
 Sleepers for Railways 
 
 Smith's Sewer. Sound 
 
 Specific Gravity of Gases 
 
 Square, Rectangle, Cube, &c. 
 
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 Strength of Materials of Con- 
 struction 
 
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 Stones 
 
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 Castings for Timber Ptoofs 
 
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 ent Kinds of Timber 
 
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 Cylinder of Cast-iron, &c. 
 
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 of Strength of Cast-Iron 
 
 Bars 
 
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 of Weights and Measures 
 
 of Natural Sines 
 
 Teeth of Wheels 
 
 Telford's Memorandum Book 
 
 Thermometers 
 
 Timber for Carpentry and Joinery 
 
 Tredgold's Rule 
 
 Waterworks 
 
 Weights of Copper, Brass, Steel, 
 Hoop-Iron, &c. 
 
 Weights and Measures 
 
 Weishts of Rails 
 
 Wickstead 
 
 Woods 
 
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