*4 /tfV '!« (fj '1 »1 '1 ' 1 < . , I ,' ■ , ,' A SYSTEM OF GEOMETRY AND TRIGONOMETRY, WITH A TREATISE O^ SURVEYING; COMPRISING VARIOUS METHODS OF TAKING THE SURVEY OF A FIELD, WITH DIRECTIONS FOR PROTRACTING AND FINDING AREAS ; IN WHICH, ALSO, THE PRINCIPLES OF RECTANGULAR SURVEYING, BY WHICH AREAS MAY BE ACCURATELY CALCULATED WITHOUT PLOTTING, ARE FULLY EXPLAINED : WITH A COMPLETE SERIES OF MATHEMATICAL TABLES, AND THE NECESSARY EXPLANATIONS. BY ABEL FLINT, A. M. A im'W EDITIOir, ENZiARGISD, WITH PRACTICAL ILLUSTRATIONS ON SURVEYING, INCLUDING OTHER BRANCHES CONNECTED THEREWITH ; REMARKS ON THE MODE OF DISTRIBUTING ESTATES ; LOCATING AND SURVEYING ROADS, &c. FOR THE USE OF THE STUDENT AND PRACTICAL SURVEYOR. BY GEORGE GILLET, SURVEYOR GENERAL OF CONNECTICUT. ALSO, A CONCISE TREATISE ON LOGARITHMS, BY F. A. P. BARNARD, LATE TUTOR IN YALE COLLEGE. SEVENTH EDITION. HARTFORD, PUBLISHED BY COOKE AND GO. FOR SALE BY THE PRINCIPAL BOOKSELLERS IN THE U. STATES. 1833. DISTRICT OF CONNECTICUT, ss. Be it remembered, That on the second day of August, in the L. S» fifty-fifth year of the Independence of the United States of America, Oliver D. Cooke, of the said District, 'hath deposited in this office the title of a book, the right whereof he claims as proprietor, in the words fol- lowing, to wit : " A system of Geometry and Trigonometry, with a Treatise on Surveying ; comprising various methods of taking the survey of a field, with directions lor protracting and finding areas ; in which, also, the principles of Rectangular Surveying, by which Areas may be accurately calculated without plotting, are fully explained : with a complete series of Mathematical Tables, and the necessary Explanations. By Abel Flint, A. M. With improvements by George Gillet, Surveyor General of Connecticut. Sixth edition, revised and enlarged, by the addition of copious notes and illustrations, and a concise Trea- tise on Logarithms, by Frederic A. P. Barnard, A. B., Instructor in Mathe- matics in the Hartford Grammar School, and author of a Treatise on Arithme- tic." In conformity to the act of Congress of the United States, entitled, " An act for the encouragement of learning, by securing the copies of Maps, Charts, and Books, to the authors and proprietors of such copies, during the times therein mentioned." And also to the act entitled, " An act supplementary to an act, entitled, ' An act for the encouragement of learning, by securing the copies of Maps, Charts, and Books, to the authors and proprietors of such co- pies, during the times therein mentioned,' " and extending the benefits thereof ;o the arts of designing, engraving, and etching historical and other prints. CHARLES A. INGERSOLL, Clerk of the District of Connecticut. A true copy of record, examined and sealed by me, CHARLES A. INGERSOLL, Clerk of the District of Connecticut. /J'/f EntereJ according to act of Congress, in the year 1833, by Edward P Cooke, in the Clerk'^s Office of the District Court of Connecticut. ^- ^Pt?-c:p ADVERTISEMENT TO THE SEVENTH EDITION. The original compiler of the following work designed, m preparing it, to furnish a plain and concise system of Practi- cal SURVEYIIVG. That he did not fail of success, has been proved, by the high estimation in which this treatise has been, and is, at the present time, held by Surveyors, and by the continued and increasing demand for it. The publishers have used exer- tions to render the present edition more worthy of patronage. About thirty pages of practical matter have been added by George Gillet, Esq., Surveyor General of the state of Connecticut. A distinguishing feature of the present as well as of the previous edition, is the introduction of a new and elegant set of tables, more extensive than those which have before been attached to this work, and on a superior type. — The numbers in the traverse table are carried to 100. In the table of logarithms, and of logarithmic sines, &c., the decimals are carried to six figures, and a column of differ- ences is added for the purpose of finding intermediate num- bers. When logarithms are carried to six places, they may be taken to as few, or as many within that limit, as is desired. Prefixed to the tables will be found a brief explanation of the nature and use of logarithms by Mr. Barnard, which will be found a valuable addition. The articles on distributing estates, locating and surveying roads, and on levelling, cannot fail of being highly useful. Hartfcn-d, January ^ 1833, The subscribers have examined in manuscript, the additions to tht seventh edition of Flint's Surveying, by George Gillet, Esq., Survey- or General of Connecticut, and find them to embrace a system of cor- rect, useful, and practical matter, judiciously arranged, and clearly ex- plained to the understanding of the learner. Having long acted as Sur- veyors under public authority, we recommend this vi'ork as containing all the elementary science, and requisite tables, necessary or convenient for the learner, and the practitioner. The present is a more full and complete system than any former edition. MOSES WARREN, Dep. Sur. JV. London Co. LEMUEL INGALLS, late Dep. Svr. Windham Co DANIEL St. JOHN, Dep. Sur. Hartford Co. ASAHEL DEWEY, County Sur. J\r. London Co. JONATHAN NICHOLS, Dep. Sur. Windham Co. Connecticut, August, 1832. PREFACE TO THE FIRST EDITION. The following work is chiefly a compilation from other Books; and but very little new is added, except a more full explanation, than has yet been published, of Rectangular Surveying, or the method of calcu- latingthe area of fields arithmetically, without drawing a plot of them and measuring with a scale and dividers, as has been the common prac- tice ; and also a more particular explanation of the use of natural sines, than is contained in most mathematical hooks. The compiler has endeavoured to render thi? work so easy and intel- ligible, that a learner will require but little assistance from an Instnic- ter, except with regard to the construction and use of mathematical and surveying instruments. Before, however, he enters on the study of this book, he must be well acquainted with common Arithmetic, withdeci. mal fractions, and the square root ; and he must also know the various characters or marks used in Arithmetic. A surveyor will doubtless find many questions arise in the course of his practice, for the solution of which, no particular directions are here given ; nor is it possible to give directions for every case that may oc- cur. In all practical sciences, much mustb^'Jeft to the judgment of the practitioner, who, if he is well acquainted with the general principles of his art, will readily learn to apply those principles to particular cases. The primary design of this treatise is to teach common Field Survey, ing ; at the same time it contains the elements of Surveying upon a larger scale ; and the system of Geometry and Trigonometry with which it is introduced, with the problems for the mensuration of superficies, as also the mathematical tables at the end, will be found useful for many other purposes. It would be well, therefore, for those who do not intend to become practical surveyors, to acquaint themselves with what is here taught ; and with this view the following work is very proper to be in- troduced into academies and those higher schools which are designed to fit young men for active business in life. Indeed every person who frequently buys and sells land, should learn to calculate the contents of a field arithmetically ; a knowledge which may be acquired in a very little time, from the particular explanation here given of that method. Notwithstanding the many books already published on the subjects here treated upon, it was thought a work of this kind was really want- ed, and that if judiciously executed it would be useful. It is more par- ticularly necessary at the present time in Connecticut, as the legislature of the State have lately enacted a law on the subject of surveying, in consequence of which more attention must be paid to the theory of that art than has been common. These considerations induced the compiler to select from various publications what appeared to him important ; and to arrange the whole in a method ^est adapted, in his view, for teaching that useful art. How far he has succeeded in his endeavours to simplify the subject, and render it easy to the learner, must be submitted to the test of expe- rience. B A GENERAL VIEW OF THE CONTENTS OF THIS WORK. The system of Geometry is divided into two parts. The first con- tains geometrical definitions respecting lines, angles, superficies, &c. The second part contains a number of geometrical problems necessary for Trigonometry and Surveying. The system of Trigonometry is also divided into two parts : and teach- es the solution of questions in right and oblique angled trigonometry, by logarithms, and also by natural sines. The treatise on Surveying is divided into three parts. Part first treats of measuring land, and is divided into three sections. The first con- tains several problems respecting mensuration, and for finding the area of various right-lined figures and circles. The second b^ction teaches different methods of taking the Survey of fields ; also to protract them, and find their area in the manner common- ly practised, and likewise by arithmetical and trigonometrical calcula- tions, without measuring diagonals and perpendiculars with a scale and dividers ; interspersed with sundry useful rules and directions. The third section is a particular explanation and demonstration of Rec- tangular Surveying, or the, method of computing the area of fields from the field notes, by mathematical tables, without the necessity of plotting the field. To this section is added a useful problem for ascer- taining the true area of a field which has been measured by a chain too long or too short. Part second treats of laying out land in various shapes. Part third contains sundry problems and rules for dividing land, and determining the true course and distance of dividing lines, or from one part of a field to another. To this is added an appendix concerning the variation of the compass and attraction of the needle ; also, a rule to find the diff'erence between the present variation, and that at a time when a tract was formerly surveyed, in order to trace or run out the original lines. The mathematical tables, are a traverse table, or table of difference of latitude and departure, calculated for every degree and quarter of a degree, and for any distance up to 50 ; a table of natural sines calculated for every minute ; a table of logarithms comprised in four pages, yet sufficiently extensive for common use ; and a table of loorarithmic or ar- tificial sines, tangents, and secants, calculated for every 5 minutes of a degree. To these tables are prefixed particular explanations of the manner of using them.* * This view of the contents was drawn up by Mr. Flint, and refers of course to the original edition. GEOMETRY. GEOMETRY is a Science which treats of the properties of magnitude. PART I. Geometrical Definitions, 1. A point is a small dot ; or, mathematically considered, is that which has no parts, being of itself indivisible. 2. A line has length but no breadth. 3. A superficies or surface, called also area, has length and breadth, but no thickness. 4. A solid has length, breadth, and thickness. 5. A right line is the shortest that can be drawn between two points. Fig. 1. 6. The incHnation of two lines meeting one another, or the opening between them, is called an angle. Thus at B. Fig. 1, is an angle, formed by the meeting of the lines AB and BC. 7. If a right line CD. Fig. 2, fall upon another right line AB, so as to incline to neither side, but make the angles on each side equal, then those angles are called right angles ; and the line CD is said to be per- pendicular to the other line. Fig. 2. D 10 GEOMETRY. Fig. 3. 8. An obtuse angle is greater than a right angle ; as ADE. Fig. 3. 9. An acute angle is less than a right angle ; as EDB. Fig. 3. Note. When three letters are used to express an angle, the the middle letter denotes the angular point. Fig. 5. 10. A circle is a round figure bounded by a single line, in every part equally distant from some point, which is called the cen- tre. Fig. 4. 11. The circumference or periphery of a circle is the bounding line; as ADEB. Fig. 4:. 12. The radius of a circle is a line drawn from the centre to the circumference ; as CB. Fig. 4. Therefore all radii of the same circle are equal. 13. The diameter of a circle is a right line drawn from one side of the circumfer- ence to the other, passing through the centre ; and it divides the circle into two equal parts, called semicircles ; as AB or DE. Fig. 5. 14. The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c. Note. Since all circles are divided into the same number of degrees, a degree is not to be accounted a quantity of any determinate length, as so many inches or feet, &c. but is always to be reckoned as being the 360th part of the circumference of any circle, without regarding the size of the circle. 15. An arc of a circle is any part of the circumference ; as BF or FD. Fig. 5 ; and is said to be an arc of as many de- grees as it contains 360th parts of the whole circle. GEOMETRY, 11 Fis:. 6. 16. A chord is a right line drawn from one end of an arc to an other, and is the measure of the arc ; as HG is the chord of the arc HIG. Fig, 6. Note. The chord of an arc of 60 degrees is equal in length to the radius of the circle of which the arc is a part. 17. The segment of a circle is a part of a circle cut off by a chord ; thus the space comprehended between the arc HIG and the chord HG is called a segment. Fig, 6. 19. A sector of a circle is a space contained between two radii and an arc less than a semicircle ; as BCD or ACD. Fig, 6. 20. The sine of an arc is a line drawn Fig, 7. from one end of the arc, perpendicular to the radius or diameter drawn through the other end : or, it is half the chord of double the arc ; thus HL is the sine of the arc HB. Fig. 7. 21. The sines on the same diameter in- crease in length till they come to the cen- A| tre, and so become the radius, after wb/ch they diminish. — Hence, it is plain that the sine of 90 degrees is the greatest possible sine, and is equal to the radius- 22. The versed sine of an arc is that part of the diameter or radius which is between the sine and the circumference ; thus LB is the versed sine of the arc HB. Fig, 7. 23. The tano-cnt of an arc is a right line touching the circumference p^^d drawn perpendicular to the diameter; and is terminated J)y a hne drawn from the centre through the other end oC the arc; thus BK is the tangent of the arc BH. Fig, 7. Note. The tangent of an arc of 45 degrees is equal in Zength to the radius of the circle of which the arc is a part. 24. The secant of an arc is a line drawn from the centre B2 12 GEOMETRY. through one end of the arc till it meets the tangent ; thus CK is the secant of the arc BH. Fig. 7. 25. The complement of an arc is what the arc wants of 90 degrees, or a quadrant : thus HD is the complement of the arc BH. Fig. 7. 26. The supplement of an arc is what the arc wants of 180 degrees, or a semicircle ; thus ADH is the supplement of the arc BH. Fig. 7. [Note. It will be seen, by reference to Fig. 7, that the sine of any arc is the same as that of its supplement. So like- wise, the tangent and secant of any arc are used also for its supplement.] 27. The sine, tangent or secant of the complement of any arc is called the co-sine, co-tangent, or co-secant of the arc ; thus, FH is the sine, DI the tangent, and CI the secant of the arc DH ; or they are the co-sine, co-tangent, and co-secant of the arc BH. Fig. 7. [The terms sine, tangent and secant, are abreviated thus : sin. tan. and sec. So likewise, co-sine, co-tangent, and co- secant, are written co-sin. co-tan. and co-sec] 28. The measure of an angle is the arc of a circle contain- ed between the two lines which form the angle, the angular point being the centre ; thus, the angle HCB. Fig. 7. is mea- sured by the arc BH : and is said to contain as many degrees as the arc does. Note. An angle is esteetned greater or less according to the opening of the lines wWch form it, or as the arc in- tercepted by those lines contains more or fewer degrees. Hence it may be observed, that the size of an angle does not depend at all upon the length of +he including hnes ; for all arcs described on the same poiat, and intercepted by the same right lines, contain exactly the same number of degrees, whether the radius be longer ox shorter. 29. The sine, tangent, or secant of an arc is also the sine, tangent, or secant of the angle whose measure the arc is. Fig^S. 30. Parallel lines are such as are equal- ^ ' ^ ly distant from each other; as AB and CD. Fig. 8. C- "^ « ' D GEOMETRY. 13 31. A triangle is a figure bounded by three lines ; as ABC. Fig. 9. 32. An equilateral triangle has its three sides equal in length to each other. Fig. 9. 33. An isoceles triangle has two of its sides equal. Fig. 10. Fig. 11. 34. A scalene triangle has three unequal sides. Fig. 11. Fig. 12. 35. A right angled triangle has one right angle. Fig. 12. Fig. 13. 36. An obtuse angled triangle has one obtuse angle. Fig. 13. 37. An acute angled triangle has all its angles acute. Fig. 9, or 10. 38. Acute and obtuse angled triangles, are called oblique angled triangles, or simply oblique triangles ; in which the lower side is generally called the base, and the other two, legs. 39. In a right angled triangle the longest side is called the hypothenuse, and the other two, legs, or base and perpendieu- lar. 14 GEOMETRY. Note. The three angles of every triangle being added to- gether will amount to ISO degrees ; consequently the two acute angles of a right angled triangle amount to 90 de- grees, the right angle being also 90. 40. The perpendicular height of a trian- gle is a line drawn from one of the angles perpendicular to its opposite side ; thus, the dotted line AD. Fig. 14. is the perpendicu- lar height of the triangle ABC. Note. This perpendicular may be drawn from either of the angles; and whether it falls within the triangle, or on one of the lines continued beyond the triangle, is im- material. Fig, 15. 41. A square is a figure bounded by four equal sides, and containing four right angles. Fig. 15. 42. A parallelogram, or oblong square, is a figure bounded by four sides, the opposite ones bemg equal and the angles right.* Fig. 16. Fig. 16. Fis. 17, 43. A rhombus is a figure bounded by four equal sides, but has its angles oblique. Fis. 17. B Fig. 18. 44. A rhomboid is a figure bounded by four sides, the opposite ones being equal, but the angles obhque. Fig. 18. * Any fonr-sided figure, bavins its opposite sides parallel, is a parallelogram ; bot, in this book, the Urm is understood as it is here explained. Sometimes this figure is called a rectangle. GEOMETRY. 15 45. The perpendicular height of a rhombus or rhomboides, is a line drawn from one of the angles to its opposite side ; thus, the dotted lines AB. figs. 17 and 18, represent the per- pendicular heights of those figures. Fig. 19. 46. A trapezoid is a figure bounded by- four sides, two of which are parallel though of unequal lengths. Fig. 19 and Fig. 20 Fig. 20. Note. Fig. 19. is sometimes called a right angled trape- zium. Fig. 21. 47. A trapezium is a figure bounded by- four unequal sides. Fig. 21. 48. A diagonal is a line drawn between two opposite angles ; as the line AB. Fig. 21. 49. Figures which consist of more than four sides are called polygons ; if the sides are equal to each other they are call- ed regular polygons, and are sometimes named from the num- ber of their sides, as pentagon, or hexagon, a figure of five or six sides, &c. ; if the sides are unequal, they are called ir- regular polygons. 16 GEOMETRY. PART II, Geometrical Problems. PROBLEM I. To draw a line parallel to another line at any given distance ; as at the point D, to make a line, parallel to the line AB. Fig. 22. Fig. 22. ^ With the dividers take the nearest distance between the point D and the given hne AB ; with that distance set one foot of the dividers any where on the Hne AB, as at E, and draw the arc ; through the point D draw a Hne so as just to touch the top of the arc C. A more convenient way to draw paraHel Hnes is with a paraUel rule. [The paraUel rules, however, found in cases of mathematical instruments, are often inaccurate.] Fig, 23. PROBLEM IL To bisect a given line , to find the middle of it. Fig. 23. Open the dividers to any convenient distance, more than half the given Hne AB, and with one foot in A, describe an arc above and below the line, as at C and D ; with the same distance, and one foot in B, describe arcs to cross the former ; lav a rule from C to D, and where the rule crosses the Hne, as at E, will be the middle. Fig. 24. PROBLEM in. To erect a perpendicu- lar from the end, or any part of a given line. JS,.— UC Fig. 24. .^^ ^ GEOMETRY 17 Open the dividers to any convenient distance, as from D to A, and with one foot on the point D, from which the perpen- dicular is to be erected, describe an arc, as AEG ; set off the same distance AD, from A to E, and from E to G ; upon E and G describe two arcs to intersect each other at H ; draw a Une from H to D, and one Une will be perpendicular to the other. Note. There are other methods of erecting a perpendicu- lar, but this is the most simple. Fig. 25. PROBLEM rV. From a given point as at C, to drop a perpendicular on a , given line AB. Fig. 25. ^ ' O With one foot of the dividers in C describe an arc to cut the given line in two places, as at F and G ; upon F and G describe two arcs to intersect each other below the line as at D ; lay a rule from C to D and draw a line from C to the giv- en line. Perpendiculars may be more readily raised and let fall, by a small square made of brass, ivorv, or wood. Fig. 26 PROBLEM V. To make an angle at E, equal to a given angle ABC. Fig. 26. A Open the dividers to any convenient distance, and with one foot in B describe the arc FG ; with the same distance and one foot in E, describe an arc from H ; measure the arc FG, and lay off the same distance on the arc from H to I ; draw a line through 1 to E, and the angles will be equal. Fig. 27. PROBLEM VI. To make an acute an- gle equal to a given number of degrees, sup- pose 26. Fig. 21. 18 GEOMETRY. Draw the line AB of any convenient length ; from a scale of chords take 60 degrees with the dividers, and with one foot in B describe an arc trom the line AB : from the same scale tEike the given number of degrees, 36, and lay it on the arc from C to D : draw a line from B through D, and the angle at B will be an angle of 36 degrees. PROBLEM VII. TomaI:eanobtusean. gle, suppose of 110 degrees. Fig. 2S. B Take a chord of 60 degrees as before, and describe an arc greater than a quadrant ; set off 90 degrees from B to C, and from C to E set off the excess above 90, which is 20 ; draw a line from G through E, and the angle will contain 110 de- grees. [It is best, however, in making obtuse angles, to take from the scale the chord of half the angle, and set it off twice. This will save taking two separate chords.] Note. In a similar manner angles maybe measured; that is, with a chord of 60 decrees describe an arc on the an- gular point, and on a scale of chords measure the arc in- tercepted by the lines forming the angle. A more convenient method of making and measuring angles is to use a protractor instead of a scale and dividers. B PROBLEM VIII. To make a triangle ^ of three given lines, as BO, BL, LO. Fig. 29, anytiL'oof which are greater than the third. Draw the line BL from B to L ; from B, with the length of the line BO, describe an arc as at ; from L, with the length of the line LO, describe another arc to intersect the former ; from O draw the lines OB and OL, and BOL will be the triangle required. GEOMETRY. 19 Fig, 30. PROBLEM IX. To make a right an. gled triangle, the hypothenuse and angles be- ing given. Fig. 30. Suppose the hypothenuse CA 25 rods or chains, the angle at C 35^ 30' and consequently the angle at A 54° 30'. See note after the 39^/i Geometrical Definition, Note. When degrees and minutes are expressed, they are distinguished from each other by a small cipher at the right hand of the degrees, and a dash at the right hand of the minutes ; thus 35° 30' is 35 degrees and 30 min- utes. Draw the line CB an indefinite length ; at C make an an- gle 35° 30' ; through where that number of Degrees cuts the arc draw the line CA 25 rods, which must betaken from some scale of equal parts ; drop a perpendicular from A to B, and the triangle will be completed. [A scale of equal parts may be found on one side of Gun- ter's scale, occupying half its length. It will be known by slanting lines which cross it at each end. The length of the scale, not occupied by these oblique lines, is equally divided into several larger divisions, numbered on one side, and like- wise twice as many smaller, numbered on the other. In tak- ing distances from the scale, each of these divisions, (either the larger or the smaller, as is most convenient,) must bo considered 1, 10, 100, &c. rods, chains, or other dimensions of length. If each division be called 1, it will be easy to tako off the required number. But the scale is not usually Ion ; enough for this. When each division is called 10, as many divisions must be taken, as there are tens in the given number. For the excess of tens, in this case, the little scale, with the oblique lines, is used. Each side of this httle scale is divi- ded into 10 equal parts, and each of these parts is, of course, 1. Then, to take off the hypothenuse, 25, above, we should take in the dividers 2 of the divisions of the large scale, and 5 of those of this small one. There is one of these little scales for the greater, and one for the smaller divisions of the large scale. When each division of the large scale is called 100, each of those of the small one becomes 10, and the units are found by means of the oblique lines. These are drawn across parallel lines, running the whole length of the scale, from each division on one side, to the next higher on C 20 GEOMETRY. the other. The parallel lines divide the width of the scale into 10 equal parts. Since each oblique line, then, in cross- ing the scale passes over one division of length, it is evi- dent, that, in passing one tenth across, (that is, to the first parallel line,) it will pass over one tenth of a division of length ; in passing two tenths across (that is, to the second parallel line.) it will pass over two tenths of a division of length, and so on. The parallel lines are numbered at the end of the scale. To take off a distance, containing hundreds, then, as 234, we must place one foot of the dividers on the second division of the larger scale, and on the parallel line marked 4, and extend the other foot to the third oblique line. Decimals may evidently be taken off in a similar manner ; the divisions of the larger scale being made units, and those of the smaller, tenths and hundredths.] XoTE. The length of the two legs may be found by mea- suring them upon the same scale of equal parts from which the hypothenuse was taken. Fis. 31. PROBLEM X. To ina'ke a right an. gJed triangle, the angles and one leg be- ing given. Fig. 31. Suppose the angle at C 33' 15', and the leg AC 2S5. Draw the leg AC making it in length 2S5 ; at A erect a perpendicular an indefinite length : at C make an angfle of 33^ 15' : through where that number of degrees cuts the arc, draw a line till it meets the perpendicular at B. Note. If the given line CA should not be so long as the chord of 60", it may be continued beyond A, for the pur- pose of making the angle. Fig. 32. C PROBLEM XL To make a rigJii angled triangle, the liypothenuse and one leg being given. Fig. 32. Suppose the hypothenuse AC 40, and the leg AB 28. Draw the leg AB in length 2S ; from B erect a perpendic GEOMETRY. 21 iilar an indefinite length ; take 40 in the dividers, and setting one foot in A, wherever the other foot strikes the perpendic ular will be the point C. Note. When the triangle is constructed, the angles may be measured by a protractor, or by a scale of chords. Fig. 33. C PROBLEM XII. To make a right angled triangle, the two legs being given. Fig, 33. Suppose the leg AB 38, and the leg BC 46. Draw the leg AB in length 38 ; from B erect a perpendic ular to C in leno-th 46 ; and draw a line from A to C. Fig. 34. PROBLEM Xlll. To make an oblique angled tri- angle, the. angles and one side being given. Fig. 34. Suppose the side BC 98 ; the angle at B 45° 15', the angle at D 108° 30', consequently the other angle 26° 15'. Draw the side BC in length 98 ; on the point B make an angle of 45° 15' ; on the point C make an angle of 26° 15', and draw the lines BD and CD. Fig. 35. PROBLEM XIV. To make an oblique angled triangle, two sides and an angle op- posite to one of them being given. Fig. 35. ^ i"GLED TRIANGLE, THE SUM OF THE SQUARES OF THE TWO LEGS IS EQU.VL TO THE SQUARE OF THE HYPOTHENUSE. In the above ex.a^iple, the square ot' AB 7S.7 is 6193.69, the square of BC S9 is 7921 ; these added make 14114.69 the square root of which is nearest 119. By natural sines. The hypothenuse being found by the square root, the angle> may be foimd by nat. sines, according to the preceding case. Hyp. Leg BC. Nat. Sine. 119) 89.00000 (747S9 83 3 570 The nearest degrees and minutes 476 corresponding to the above nat. sine are 43° 24', for the angle BAG. The 940 difference between this and the angle 833 as found by logarithms is occasioned by dividing by 119, which is not the 1070 exact length of the hypothenuse, it 952 being a fraction too much. 1180 1071 109 PART II. Oblique Trigonometry. The solution of the first two cases of Oblique Trigotwnieiry depends on the following proposition. In all PLANE TRIANGLES, THE SIDES ARE IN PROPORTION TO EACH OTHER AS THE SINES OF THEIR OPPOSITE ANGLES. TRIGONOMETRY. 33 That is, as the sine of one angle is to its opposite side, so IS the sine of another angle to its opposite side. Or, as one side is to the sine of its opposite angle, so is another side to the sine of its opposite angle. [Note. When an angle exceeds 90° make use of its supple- ment, which is what it wants of 180^. Note Def. 26. Geo3i.] CASE 1. The angles and one side given to find the other sides. Fig. 46. 200 In the triangle ABC, given the angle at B 48°, the angle at C 72^, consequently the angle at A 60°, and the side AB 200, to find the sides AC and BC. To find the side AC. As sine ACB, 72° - 9.978206 : side AB, 200 - 2.301030 :: sine ABC, 480 . 9.871073 : side AC, 156 12.172103 9.978206 2.193897 To find the side BC. As sine ACB, 72° - 9.978206 : side AB, 200 - 2.301030 : : sine BAC, 60° - 9,937531 : side BC, 182 12.328561 9.978206 2.260355 By natural sines. As THE NAT. SINE OF THE ANGLE OPPOSITE THE GIVEN SIDE IS TO THE GIVEN SIDE, SO IS THE NAT. SINE OF THE ANGLE OP- POSITE EITHER OF THE REtlUIRED SIDES TO THAT REQUIRED SIDE. Given side 200 ; nat. sine of 72°, its opposite angle, 0.95115 ; nat. sine of ABC 48°, 0.743.34 ; nat. sine of BAC 60° 0.86617. Thus, 34 TRIGONOMETRY. 0.95115 ; 200 0.95115 : 200 0.74334 : 0.86617 : 156 182.. CASE 11. Two sides, and an angle opposite to one of them given, to find the other angles and side. Fig. 47. 240 In the triangle ABC, given the side AB 240, the side BC 200, and the angle at A 46^ 30' ; to find the other angles and the side AC. Tofind the angle ACB. As side BC, 200 2.301030 Angle at A : sine BAG, 46° 30' 9.860562 C : : side AB, 240 2.38u211 46° 30' 60 30 ; sine ACB, 60° 30' nearly 9.939743 12.240773 2.301030 I Sum of the three angles I Sum of two 107.00 180° 107 Angle at B The side AC will be found by case I. to be nearest 253. Note. If the given angle be obtuse, the angle sought will be acute ; but if the given angle be acute, and opposite a given lesser side, then the angle found by the operation maybe either obtuse or acute. It ought therefore to be mentioned which it is, by the conditions of the question. By natural sines. As THE SIDE OPPOSITE THE GIVEN ANGLE IS TO THE NAT. .SINE OF THAT ANGLE, SO IS THE OTHER GIVEN SIDE TO THE NAT. SINE OF ITS OPPOSITE ANGLE. One given side 200, nat. sine of 46^ 30', its opposite angle, 0.72537, the other given side 240. As 200 : 0.72537 : : 240 : 8.87044=60" 30'. TRIGONOMETRY. 35 CASE III. Two sides and their contained angle giv- «n, to find the other angles and side. Fig. 48. 2fO The solution of this case depends on the following proposi- TION. In every plane triangle, as the sum of any two sides is to THEIR difference, SO IS THE TANGENT OF HALF THE SUM OF THE two opposite angles to the tangent of half the difference between them. Add this half difference to half the sum of the angles AND you will have THE GREATER ANGLE, AND SUBTRACT THE HALF difference from THE HALF SUM AND YOU WILL HAVE THE LESSER ANGLE. In the triangle ABC, given the side AB 240, the side AC 180, and the angle at A 36° 40' to find the other angles and side. Side AB - 240 AC - 180 Sum of the two sides 420 AB AC Difference 240 180 60 The given angle BAC 36° 40', subtracted from 180°, leaves 143^ 20' the sum of the other two angles, the half of which is 71° 40'. As the sum of two sides, 420 - - 2.623249 : their difference 60 . . . 1.778151 : : tangent half unknown ang. 71° 40' 10.479695 : tangent half difference, 23° 20' nearly The half sum of the two unknown angles, The half difference between them, Add, gives the greater angle ACB Subtract, gives the lesser angle ABC The side BC may be found by case I or II. D2 12.257846 2.623249 9.634597 71° 40' 23 20 95 00 48 23 30 TRIGONOMETRY. CASE IV. The three sides given to find the an. gles. Fig. 49. The solution of this case depends on the following pRoro- SITION. In every plane triangle, as the longest side is to THE sum of the OTHER TWO SIDES, SO IS THE DIFFERENCE between THOSE TWO SIDES TO THE DIFFERENCE BETWEEN THE SEGMENTS OF THE LONGEST SIDE, MADE BY A PERPENDIC- ITLAR LET FALL FROM THE ANGLE OPPOSITE THAT SIDE. Half the difference between these segments, added to half the sum of the segments, that is, to half the length of the longest side, will give the greatest segment ; and this half dif- ference subtracted from the half sum will be the lesser seg- ment. The triangle being thus divided, becomes two right angled triangles, in which the hypothenuse and one leg are given to find the angles. In the triangle ABC, given the side AB 105, the side AC 85, and the side BC 50, to find the angles. Side AC - 85 AC . . 85 BC - 50 BC - . 50 Sum of the two sides 135 Difference As the longest side AB, 105 : sum of the other two sides, 135 : difference between those sides, 35 : difference between the segments, 45 Half the side AB . - - Half the difference of the segments Add, gives the greater segment AD Subtract, gives the lesser segment BD 35 2.021189 2.130334 1.544068 3.674402 2.021189 1.653213 52.5 . 22.5 75.0 30.0 TRIGONOMETRY. 37 Thus the triangle is divided into two right angled trian- gles, ADC and BDC ; in cacii of which the hypothenuse and one Ico: are ijjiven to fuid the angles. To find the angle DC A. Ashyp. AC, 85 - 1.929419 : radius - - lO.OOUOOO : : seer. AD, 75 - 1.875061 11.875061 1.929419 sine DCA, 61° 56' 9.945642 To iind the angle DOB. As hyp. BC, 50 - 1.698970 : radius - - 10.000000 : : seg. BD, 30 - 1.477121 sine DCB, 36° 52' 11.477121 1.698970 9.778151 The angle DCA 61° 56' subtracted from 90° leaves the angle CAD 28° 4'. The ani^ie DCB 36° 52' subtracted from 90° leaves the an- gle CBD 53° 16'. The angle DCA 61° 56' added to the angle DCB 86° 52' gives the angle ACB 98° 48'. This CASE may also be solved according to the following PROPOSITION. Ix EVERY PLANE TRIANGLE, AS THE PRODUCT OF ANY TWO SIDES CONTAINING A REQUIRED ANGLE IS TO THE PRODUCT OF HALF THE SUX OF THE THREE SIDES, AND THE DIFFERENCE BETWEEN THAT HALF SUM AND THE SIDE OPPOSITE THE ANGLE REQUIRED, SO IS THE SQUARE OF RADIUS TO THE SQUARE OF THE CO-SINE OF HALF THE ANGLE REQUIRED. Those who make themselves well acquainted with trigo- nometry will fmd its application easy to many useful purpo- ses, particularly to the mensuration of heights and distances. These are here omitted, because, as this work is designed prin- cipally to teach the art of common field-surveying, it was thought improper to swell its size, and consequently increase its price, by inserting any thing not particularly connected with that art. It is recommended tc those who design to be surveyors to study trigonometry thoroughly ; for though a common field may be measured without an acquaintance with that science, yet many cases will occur in practice where a knowledge of it will be found very beneficial ; particularly in dividing land, and ascertaining the boundaries of old surveys. Indeed no one who is ignorant of trigonometry, can be an accomplish- ed surveyor. SURVEYING. SURVEYING is the art of measuring, laying out, and di- Tiding land. PART I. Measuring Land. The most common measure for land is the acre ; which con- tains 160 square rods, poles or perches ; or 4 square roods, each containing 40 square rods. The instrument most in use, for measuring the sides of fields, is Gunter's chain, which is in length 4 rods or 66 feet ; and is divided into 100 equal parts, called links, each containing 7 inches and 92 hundredths. Consequently, 1 square chain con- tains 16 square rods, and 10 square chains make 1 acre. In small fields, or where the land is uneven, as is the case with a great part of the land in New England, it is better to use a chain of only two rods in length ; as the survey can be more accurately taken. SECTION I. Preliminary problems. PROBLEM I. To reduce two rod chains to four rod chains. Rule. If the number of two rod chains be even, take half the number for four rod chains, and annex the hnks if any ; thus, 16 two rod chains, and 37 links make 8 four rod chains and 37 links. But if the number of chains be odd, take half the greatest even number for chains, and for the reaanaining number add 50 to the links : Thus, 17 two rod chains and 42 links make 8 four rod chains and 92 links. PROBLEM II. To reduce two rod chains to rods and deci- mcd parts. SURVEYING. 39 Rule. Multiply the chains by 2, and the links by 4, which will give hundrc('ths of a rod ; thus, 17 two rod chains and 21 links make 34 rods and 84 hundredths ; expressed thus, 34.84 rods. If the links exceed 25, add 1 to the number of rods and multiply the excess by 4 : thus, 15 two rod chains and 38 links make 31.52 rods. PROBLElM III. To reduce four rod chains to rods and decimal parts. Rule. Multiply the chains, or chains and links, by 4 ; the product will be rods and hundredths : thus, 8 chains and 64 links make 34.56 rods. Note. The reverse of this rule, that is, dividing by 4, will reduce rods and decimals to chains and links : thus, 105.12 rods make 26 chains and 28 links. PROBLEM IV. To reduce square rods to acres. Rule. Divide the rods by 160, and the remainder by 40, if it exceeds that number, for roods or quarters of an acre : thus 746 square rods make 4 acres, 2 roods, and 26 rods. PROBLEM V. To reduce square chains to acres. Rule. Divide by 10 ; or, which is the same thing, cut off the right hand figure : thus, 1460 square chains make 146 acres ; and 846 square chains make 84 acres and 6 tenths. PROBLEM VI. To reduce square links to acres. Rule. Divide by 100000 ; or, which is the same thing, cut off the 5 right hand figures : thus, 3845120 square links make 38 acres and 45120 decimals. Note. When the area of a field, by which is meant its superficial contents, is expressed in square chains and links, the whole may be considered as square links, and the number of acres contained in the field, found as above. Then multiply the figures cut off by 4, and again cut off 5 figures, and you have the roods ; multiply the figures last cut off by 40, and again cut off 5 figures, and you have the rods. Example, How many acres, roods, and rods, are there in 156 square chains and 3274 square links ? 15)63274 square Unks. 4 2)53096 40 21)23840 Answer. 15 acres 2 roo(i.^ and 21 rods. Problems for finding the area of right lined figures, and also of circles. 40 SURVEYING. PROBLEM VII. To find the area of a square or rectan- gle. Rule. Multiply the length into the breadth; the product will be the area. PROBLEM Vin. To find the area of a rhombus or rhoiiihoid. Rule. Drop a perpendicular from one of the angles to its opposite side, and multiply that side into the perpendicular ; the product will be the area. PROBLEM IX. To find the area of a triangle. Rule 1. Drop a perpendicular from one of the angles to its opposite side, which may be called the base ; then multiply the base by half the perpendicular; or the perpendicular by half the base ; the product will be the area. Or, multiply the whole base by the whole perpendicular, and half the product will be the area. Rule 2. U it be a right angled triangle, multiply one of the legs into half the other ; the product will be the area. Or, multiply the two legs into each other, and half the pro- duct will be the area. Rule 3. When the three sides of a triangle are known, the area may be found arithmetically, as follows : Add together the three sides ; from half their sum subtract each side, noting down the remainders ; multiply the half sum by one of those remainders, and that product by another remainder, and that product by the other remainder ; the square root of the last product will be the area.* Example. Suppose a triangle whose three sides are 2\ 20, and 18 chains. Demanded the area. 24x20x18^62, the sum of the three sides, the half of which is 31. From 31 subtract 24, 20, and 18 ; the three remainders will be 7, 11, and 13. 31x7=217 ; 217x11=2387 ; 2387x13=31031, the square root of which is 176.1, or 17 acres 2 roods and 17 rods. By logarithms. As the addition of logarithms is the same as the muhiplica- tion of their corresponding numbers; and as the number an- swering to one half of a logarithm will be the square root of the number corresponding to that logarithm : it follows, that if the logarithm of the half sum of the three sides and the log- arithms of the three remainders be added together, the num. * Better expressed thus. From half the sum of the sides subtract each side sepa- rately. Multiply the half sum and the several remainders togetlier, and the square root of the product will be the area.— Ed. SURVEYING. 41 ber corresponding to one half the sum of those logarithms will be the area of the triangle. The half sum, 31 - - - 1.491362 The first remainder, 7 - - - 0.845098 The second remainder, 11 - - 1.041393 The third remainder, 13 - - 1.113943 The square of the area, 31030.083 nearly 4-491796 Area 176 square chains 150 square links 2.245893 Rule 4. When two sides of a triangle and their contained angle, that is, the angle made by those sides, are given, the area may be found as follows : Add together the logarithms of the two sides and the loga- rithmic sine of the angle ; from their sum subtract the loga- rithm of radius, the remainder will be the logarithm of double, the area. Example. Suppose a triangle one of whose sides is 105 rods and another 85, and the angle contained between them 28° 5'. Demanded the area. One side, 105 .... 2.021189 The other side, 85 - - - - 1.929419 Sine angle, 28° 5' - - - - . 9.672795 13.623403 Subtract radius ... - lO.uOOOOO Double area, 4200 rods nearly 3.623403 Answer, 2100 rods. Note. Radius may be subtracted by cancelling the left hand figure of the index, or subtracting 10, without the trouble of sitting down the ciphers. By natural sines. Muhiply the two given sides into each other, and that pro- duct by the natural sine of the given angle ; the last product will be double the area of the triangle. Nat. sine of the angle 28° 5' 0.47076. 105x85=8925, and 8925x0.47076=4201 the double area of the triangle. PROBLEM X. To find the area of a trapezoid. 42 SURVEYING. Rule. Multiply half the sum of the two parallel sides by the perpendicular distance between them, or the sum of the two parallel sides by half the perpendicular distance, the pro- duct will be the area. PROBLEIM XI, To find the area of a trapezium, or ir- regular four sided figure. Rule. Draw a diagonal between two opposite angles, which will divide the trapezium into two triangles. Find the area of each triangle and add them together. Or, multiply the diagonal by half the sum of the two perpendiculars let fall upon it, or the sum of the two perpendiculars by half the di- agonal, the product will be the area. Note. Where the length of the four sides and of the diagonal is known, the area of the two triangles, into which the trapezium is divided, may be calculated arithmetically, according to Prod. IX. Rule 3. PROBLEM XII. To find the area of a figure containing more than four sides. Rule. Divide the figure into triangles, and trapezia, by drawing as many diagonals as are necessary, w^hich diagonals must be so drawn as not to intersect each other; then find the area of each of the several triangles or trapezia, and add them together; the sum will be the area of the whole figure. Note. A little practice will suggest the most convenient way of drawing the diagonals ; but whichever way they are dra^^Ti, pro- vided they do not intersect each other, the whole area will be found the same. PROBLEM XIII. Respecting circles. Rule I. If the diameter be given the circumference may be found by one of the following proportions : as 7 is to 22, or more exactly, as 113 is to 355, or in decimals, as 1 is to 3.14159, so is the diameter to the circumference. Rule 2. If the circumference be given the diameter may be found by one of the following proportions : as 22 is to 7, oi cis 355 is to 113. or as 1 is to 0.31831, so is the circumfer* ence to the diameter. Rule 3. The diameter and circumference being known, multiply half the one into half the other, and the product will be the area. Rule 4. From the diameter only, to find the area : multi- ply the square of the diameter by 0.7854, and the product will be the area. Rule 5. From the circumference only to find the area : SURVEYING. » 43 multiply the square of the circumference by 0.07958, and the product will be the area. Rule 6. The area being given to find the diameter : divide the area by 0.7854, and the quotient will be the square of the diameter ; from this extract the square root, and you will have the diameter. Rule 7. The area being given to find the circumference : divide the area by 0.07958, and the quotient will be the square of the circumference ; from this extract the square root, and vou will have the circumference. SECTION II. • The following cases teach the most usual methods of taking the survey of fields ; also, how to protract or draw a plot of them, and to calculate their area. Note. The field-book is a register containing the length of the sides of a field, as found by measuring them with a chain; also the bearings or courses of the sides, or the quantity of the several an- gles, as found by a compass or other instrument for that purpose ; together with such remarks as the surveyor thinks proper to make in the field. CASE I. To SURVEY A TRIANGULAR FIELD. Measure the sides of the field with a chain, and enter their several lengths in a field book, protract the field on paper, and then find the area by prob. IX. Rule 1. Or, without plot- ting the field, calculate the area by prob. IX. rule 3. Fig, 50. FIELD BOOK. See Fig. 50. Chains. AB - - 20 BC - - 24 CA - - 18 To find the area. Ch. L. Base BC . - 24.00 Halfperp.AD. 7.34 „ ^^ 9600 44 • SURVEYING. 9600 7200 16S00 Acres 17)61600 4 Roods 2)46400 40 Rods 1S)56000 Acres Roods Rods Area 17 — 2 — 1S.56 ICoTE. When there are ciphers at the right hand of the links, they may be rejected ; remembering to cut od" a proper number of fig- ures according to decimal rules. Ohserve^ That in measuring with a chain, slant or inclined surfaces, as the sides of hills, should he measured horizontally, and not on the plane or surface of the hill : otherwise, a sur- vey cannot be accurately taken. To effect this, the lower end of' the chain must be raised from the ground, so as to have the whole in a horizontal hne ; and the end thus raised must be directly over the point where the chain begins or ends, ac- cording as you are ascending or descending a hill: which point may be ascertained by a plummet and line. CASE IL To smVEY A FIELD !>• THE FORM OF A TRAPEZIUJI. Measure the several sides, and a diagonal between two op- posite angles ; protract the field, and find the area by prob- lem XI. Or, without protracting the field, calculate the area according to the note at the end of that problem. Fis. 51. FIELD BOOK. See Fis. 51. Ji ^. "^ Ch.L. ^ T'^'^^ ^^ AB - 27.50 BC CD DA Diag. AC 11.70 21.50 14.70 72. SURVEYING. 45 To PROTRACT THIS TRAPEZIUM. Draw the side AB the given length ; with the diagonal AC 28 and the side BC 11.70 describe cross arcs as at C, from A and B as centres ; and the point of intersection will repre- .-sent that corner of the field : then, with the side CD 21.50 and the side AD 14.70, describe cross arcs as at D, from A and C as centies; and the point of intersection will represent that corner of the field. To FIND THE AREA. sndicular B a — Dm 11.34 11.10 diagonal AC 22.44 14.00 897600 2244 Acres 31)416 4 Rood 1)664 40 Acres Rood Area 31 — 1 — Rods 26)560 Rods 26.56 "Note. The perpendiculars need not be actually drawn ; their length may be obtained as follows : From the angle opposite the diagonal open the dividers so as when one foot is in the angular point, as at B, the other, being moved backwards and forwards, may just touch the diagonal at A, and neither go the least above or below it ; that distance in the dividers being measured on the scale will give the length of the perpendicular. CASE III, To SURVEY A FIELD WHICH HAS MORE THAN FOUR SIDES, BY THE CH\IN ONLY. Measure the several sides, and from some one of the angles from which the others may be seen, measure diagonals to 46 SURVEYING. them ; draw a plot of the field, and find the area by prob- LEM XII. FIELD BOOK. See Fig. 52. Fig, 52. Ch. L. AB 30.60 BC 20.40 CD 22.40 AC DE 16.20 AC EF 13.50 AE FA 28. Diagonals. Ch.L. 45. 35. 24.20 To PROTRACT THIS FIELD. Draw the side^lB, making it the given length 30.60; with the diagonal AC 45 and the side BC 20.40, describe cross arcs as at C, from the points A and B as centres, and the point of intersection will represent that corner of the field ; draw the side BC and the clotted diagonal AC ; with the dia- gonal AD 35 and the side CD 22.40, describe cross arcs as at D, from the Points A and C, and draw the Side CD and the dotted diagonal AD. Proceed in this manner till all the sides and diaofonals are drawn. To FIND THE AREA. The field being plotted, may be divided into one trapezium and two triangles ; the area of which is calculated as fot> lows : — SURVEYING. 47 The trapezium ABCD. Perpend. B a — Do Half diag. AC 11.68 17.10 27.78 22.50 143900 5756 5756 Square chains - 647.5500 The triangle AFE. Perpend. En - . 11.65 Hair side AF - . 14 4660 1165 Square chains - - 163.10 Acres Roods Rods Area 98 — — 34.4 The triangle ADE. Half perp. E m Diag. AD Square chains Trap. ABCD Triangle ADE Triangle AFE 4.90 35 2450 1470 171.50 - 647.55 - 171.50 - 163.10 Acres 98)215 4 Roods .860 40 Rods 34)400 REMARKS. As each of the sides of the several triangles, into which the precedmg plot of a field is divided, is known from the field book, the area of the field may be calculated arithmetically, by finding the area of each triangle, according to trob. IX. Rule 3 ; and then adding the whole together. This method, though it may require more time, is preferable to the other, because more accurate. Indeed it is always better to calcu- late the area of a field arithmetically than geometrically ; for in the former no two persons can differ in their calcula- tions ; whereas, according to the latter, which is the com- mon method of casting the contents of a field, it is hardly to be expected that any two persons will perfectly agree. The inaccuracy of scales, and the difficulty of determining with precision the length of sides and perpendiculars with a scale and dividers, render it almost if not quite impossible to obtain the exact area of a field, in the method commonly practised, even if the surveyor has measured it accurately in the first place. Other methods of taking the survey of a field by the chain only are mentioned in some treatises on this subject, but they E2 48 SURVEYING. are rather curious than useful ; and it is much better to ascer- tain the angles by an accurate compass, or some instrument designed purposely for taking angles. CASE IV. To SURVEY A FIELD WITH A CHAIN A>-D COMPASS. Measure the length of the sides with a chain, and take their beariug or coui-se with a compass ;* enter these in a field book; plot the field on paper, and calculate the area by the directions already given. To PROTRACT OR DRAW A MAP OF A FIELD. Draw a line to represent a meridian, or north and south line, from which lay off a bearing or course of the first side of the field, with a protractor or trom a line of chords ; and from a scale of equal parts, measure the length of the side and draw aline to represent it. At the end of this line draw a line parallel to the meridian line, and then lay off the second side of the field as before taught ; proceed in the same manner to draw parallel lines, and lay off the several sides till the whole is protracted. In protracting a field, let the top of the paper be considered as north, the bottom south, the right hand east, and the left hand west ; lay the course to the right or left of the meridian line, according as it is east or west, and upwards or down- wards, according as it is north or south. In all protractions, if the end of the last distance falls ex- actly on the point from which you began, tlie course also be- ing right, the field work and protractions are truly taken and pertormed ; if no% an error must have been committed in one of them : in such cases, make a second protraction ; if this agrees with the former, it is to be presumed the fault is in the field work ; a re-survey must then be taken. * A compass may be so constructed with two indexes, one moveable and the other fixed, as to ascertain the angle made by two sides, without reference to the bearing of those sides. Such a compass would be par- ticularly useful in surveying land where there are mineral substances which have an influence upon the compass needle, attracting it one way or the other, and thus rendering it impossible to take a course by ii with precision. SURVEYING FIELD BOOK. AB. N. T 0' BC. N. 74 CD. S. 9 DE. N. 63 20 EA. S. 74 Set Fig. 53. , Ch.L. \i W. 28.20 PBL- E. 39.50 fx. E. 38. \ \^ W. 14.55 I j W. 28.60 111 49 Fig. 53. Acres Rood Rods Area 117 — 1 — 6 REMARKS. The sides of the several triangles into which the plot of a field is divided may be found by trigonometry ; anch then the area of each triangle may be calculated according to prob. IX. Rule 3. The sum of the areas of the several triangles will be the area of the whole field. This method may re- quire more time, but it is perfectly accurate, since no de- pendence is placed on the uncertain measurement of scale and dividers. in the preceding example, suppose the field divided into three triangles. See Fig. 53. In the triangle EAB, the sides EA and AB are known from the field book, and their contained angle is known from the bearing of the sides. The other angles and the side EB may be found by oblique trig- onometry, CASE 111. ; and then there will be the three side* to find the area. In the triangle EBC, the side BC is known from the field book, and the side EB is found as above men- tioned ; the angle EBA is also found as above: this subtract- ed from the angle ABC, which may be found from the bear- ing of the sides AB and BC, will leave the angle EBC : there will then be the two sides and their contained angle to find the third side; and this being found, there will be the three sides to find the area. Jn the triangle EDC, the sides DE and DC are known from the field book, and their contained angle is known from the bearing of the sides. The side EC and the area may be found as above. It is recommended to the learner to make these calcula- 50 SURVEYING. tions, as it will improve him in the knowledge of trigonome- try. Note. Two sides and their contained angle being given, the area may be found by pros. IX. Rule 4. Another method of protracting fields. Without drawing "parallel lines at the end of each side, a field may be protracted by the angles made by the several sides ; and the angle made between any two sides may be found by the following rules. Rule 1. If the course or bearing of one of the sides is north and the other south, one east and the other west, sub- tract the less course from the greater. Rule 2. If one is north and the other south, and both east or west, add both courses together. Rule 3. If both are north or south, and one east and the other west, subtract their sum from 180^. Rule 4' If both are north or south, and both east or west, add the less course, and the supplement of the greater. In each case, the result will give the angle contained by the two sides. To protract a field according to the preceding rules is pre- ferable to the method of doing it by parallel lines, though it may not be so easy to the learner at first. It is difficult to draw parallel lines w'th perfect accuracy, and a small devia- tion from a true line may make considerable difference in the plot of a field.* Example II. FIELD BOOK. See Fig. 54. AB. N. 16° 30' E, Ch. L. 22. BC. N. 82 E. 19.60 CD. S. 17 E. 24. DE. S. 37 W. 22. EA. N. 49 W. 25.20 Area 85 acres. * See Appendix No. 7. SURVEYING. 51 Tb DRAW A PLOT OF THIS FIELD, ACCORDING TO THE PRECE- DIXG RULES. Having drawn the side AB, according to the directions be- fore given for laying off the first course and distance, com- pare the first and second courses together, and they will be found to be both north and both east ; consequently, the angle between them is found by rule 4, as follows : 16^ 30' the less course, and 98^ the supplement of the greater being added, the sum is 114° 30', for the angle at B. Compare the second and third courses, and they will be found to be one north and one south, and both east ; consequently, according to rule 2, 82^ the second course added to 17° the third course, the sum 99° is the angle at C. The third and fourth courses are both south, and one east and the other west. The angle between them at D is 126° ; for 17° the third course added to 37° the fourth course is 54°, which subtracted from 180° leaves 126°, according to rule 3. The fourth and fifth courses are one south and the other north, and both west. According to rule 2, 37° the fourth course added to 49° the fifth course, the sum 86° is the angle at E. A little practice will render this mode of protracting a field familiar and easy, and an attention to the courses will show in what direction the angle is to be made. Example III. FIELD BOOK. See Fig, 55. 52 SURVEYING. Ch. L. AB. N. 56° 15' E. 21.60 BC. N. 26 30 E. 13.44 CD. S. 71 30 E. 18.96 DE. S. 26 30 E. 13.44 EF. s. 71 30 W. 18.96 FG. s. 45 E. 8.47 GH. s. 63 30 E. 13.44 HI. N. 45 E. 8.47 IK. s. 26 30 E. 13.44 KL. s. 45 W. 8.47 LM. s. 63 30 w. 13.44 MN. N. 76 vv. 24.73 NA. N. 36 45 w. 30. Acres Rood Rods Area 167 — 1 — 30 The above field may be protracted, and its area calculated according to the directions given in the preceding examples. Several field books to exercise the learner in plot- ting FIELDS and calculating THEIR AREA. Ch. L. No. I. Rods 8. N. 53° 15' W. 11.60 1. N. 15° 0' E. 320 9. N. 36 45 E. 19.20 2. x\T. 37 30 E. 160 10. N. 22 30 E. 14. 3. East 120 11. S. 76 45 E. 12. 4. S. 11 E. 200 12. S. 15 W. 10.85 5. South 216 13. s. 16 45 W. 10.12 6. West 160 Acres 5 Roods Rods. 7. S. 36 30 W. 160 Area 110 — 2 23 8. N. 38 15 W. Acres Roods 136 Rods No .III. Rods. Area 744 - -3 - -28 1. S. 65° 40' W. 49.7 2. 3. s 67 15 W. W. 34.5 17.9 No. 11. Ch.L. s. U 1 54 1. N. 75° 0' E. 13.70 4. s. 20 W. 5.8 2. N. 20 30 E. 10.30 5. s. 7 30 E. 29.4 3. East 16.20 6. N. 83 E. 107.4 4. S. 33 30 W. 35.30 7. N. 5 50 W. 22. 5, S. 76 w. 16. 8. N. 18 30 W. 46. 6. North 9. Acres Rood Rods. 7. S. 84 w. 11.60 Area 34 — 1 — 19 SURVEYING. 53 NO. IV. Rods. Rods. 1. N. 43^ 0' vv. 12.44 16. N. 18° 30' W. 39. 2. N. 64 i) VV. 8. 17. N. 86 5 E. 26.7 3. N. 52 ^ VV. 14.60 Acres Rood Rods 4. N. 37 5 VV. 51.36 Ai -ea 48 — 1 — 12 5. N. 15 30 VV. 21.76 6. N. 20 40 W. 44.60 No. VII. Ch. L. 7. N. 88 20 E. 167.60 1 1. N. 0° 45' \W. 9. 8. S. 34 40 E. 71.20 2. N. 19 30 W. 5.35 9. S. 75 W. 69.72 3. N. 23 VV. 4.09 10. s. 55 w. 64.60 4. N. 41 35 W. 6.15 11. s. 25 w. 18.12 5. N. 3 W. 36.75 Acres Roods Rods. 6. S. 86 50 W. 13.33 Area 97 — 2 -29 1 7. N. 2 15 W. 17.65 8. N. 85 45 E. 12.56 No. Y. Rods. 9. S. 2 10 E. 8. 1. S. IP 50^ W. 34.6 10. N. 86 45 E. 7.38 2. s. 63 20 E. 93.6 11. S. 3 15 E. 13.20 3. N. 4 W. 34.9 12. N. 87 E. 29.92 4. S. 89 55 E. 40.1 13. N. 49 20 E. 4.04 5. N. 5 20 VV. 35.5 14. North 2.23 6. N. 69 40 W. 60- 15. N. 50 35 E. 6.50 7. S. 78 w. 30.6 16. S. 22 50 E. 17.94 8. N. 67 20 w. 1.2 17. S. 34 W. 3.50 9. S. 72 30 w. 10.4 18. s. 41 W. 3. 10. s. 66 55 w. 15.2 19. s. 22 50 W. 9.25 , \cres Rood Rods. 20. s. 3 40 E. 2.64 Area 41 — 1 — 34 21. s. 86 VV. 2.50 22. s. 25 W. 14.50 No. VJ [. Rods. 23. s. 2 W. 5.38 1. S. 34^ 0' E. 42.?^ 24. s. 10 E. 11.75 2. s. 29 E. 69.4 25. s. 86 W. 34.60 3. s. 64 50 VV. 53. Acres Roods Rods 4. s. 25 E. 4. Area 268 — 3 — 7 5. s. 66 30 W. 39. 6. N. 25 W. 4. No. YIII Rods. 7. S. 64 45 W. 32 2 1. S. 6° 30' E. 19.1 8. N. 30 30 VV. 18.3 2. S. 63 30 E. 14.36 9. N. 56 30 E. 34.5 3. s. 67 E. 10.68 10. N. 64 E. 12.5 4. N. 88 E. 13.3 11. N. 49 E. 14. 5. S. 31 30 W. 32.44 12. N. 26 10 W. 19.3 6. s. 31 55 W. 96.5 13. N. 21 W. 18.3 7. s. 33 25 W. 34.9 14. N. 44 10 W. 18. 8. s. 20 45 E. 3.68 15. N. 64 40 E. 30.5 9. s. 16 15 VV. 64. 54 SURVEYING. 10. N. 52°30' W. 12.8 21. N. 36° 0' E. 41..56 11. S. 45 W. 18.24 22. S. 68 E. 80.6 12. S. 69 W. 21.4 23. N. 44 30 E. 20.4 13. s. 12 40 W. 9.4 24. N. 2 30 W. 41. 14. s. 84 20 V\^. 9.5 25. N. 14 45 W. 62..32 15. N. 22 15 W. 24. 26. N. 16 w. 14.8 16. North 9.8 27. N. 1 45 w. 14.8 17. N. 29 15 W. 30.6 28. N. 82 30 E. 99. 18. N. 44 25 W. 21.8 19. N. 61 30 W. 23.1 Acres Rood Rods 20. N. 41 W. 10.8 Area 135 — 1 — 15 CASE V. To StTRVEY A FIELD FROM ONE STATION AT AXY PLACE WITHI?f THE FIELD, FROM WHICH THE SEVERAL ANGLES MAY BE SEEN. Take the bearing of the angles, and measure their distance from the station. FIELD BOOK. See Fig. 56. Plir. 56, station to A. N. 20° W. 8.70 B. N. 60 E. 10. C. N. 87 E. 11.40 . 1). S. 15 E. 10.50 E. S. 60 W. 12. F. N. 65 W. 8.78 TO PROTRACT THIS FIELD. Draw a meridian hne as N. S. From some point in that line as a centre, lay off the bearing and distance to the sev- eral angles, and draw lines from one angle to another, as AB, BC, CD, &c. SURVEYING. 5d TO FIND THE AREA. The Area may be calculated according to prob. XII. by measuring diagonals and perpendiculars ; or more accurately according to prob. IX. Rule 4. As the bearing and distance of the lines from the station to the several angles are known, two sides and their contained angle are given in each of the triangles into which the plot is divided ; the area may, therefore, be readily calculated by the RULE above referred to. Note. As in the operation, the logarithm of radius is to be subtract- ed from the sum of the other logarithms, it may be done by rejecting the left hand figure, without the trouble of putting down the ciphers and subtracting. Triangle aAB. aA, 8.70 - aB, 10 - . - Sine AaB, 80° - Doub. area, 85.67 Triangle aBC. aB, 10 - . . aC, 11.40 - SineBaC, 27*^ . Doub. area, 51.73 Triangle aCD. aC, 11.40 aD, 10.50 - SineCaD, 78° . Doub. area, 117 - 0.939519 1.000000 9.993351 1.932870 1.000000 1.056905 9.657047 1.713952 1.056905 1.021189 9.990404 2.068498 1 Triangle aDE. aD, 10.50 . aE, 12 - . . SineDaE,75° - Doub. area, 121.7 Triangle aEF. aE, 12 . - . aF, 8.78 - Sine EaF, 55° - Doub. area, 86.31 Triangle aFA. aF, 8.78 - aA, 8.70 - Sine Fa A, 45° - Doub. area, 54.01 1.021189 1.079181 9.984944 2.085314 1.079181 0.943495 9.913365 1.936041 0.943495 0.939519 9.849485 1.732499 Triangle aAB aBC aCD ^ aDE aEF aFA Double area Area 85.67 827 51.75 4 117.1 121.7 3)308 86.31 40 .54.01 12)320 516.54 Acres Roods Rods Area 25 — 3 — 12.32 25)827 56 SURVEYING. CASE VI. To SURVEY A FIELD FKOM SOME ONE OF THE ANGLES, FBOM WHICH THE OTHERS MAY BE SEEN. From the stationary angle take the bearing and distance to each of the other angles, with a compass and chain. FIELD BOOK. See Pig. 57. Ch. L. FG. N. 70° W. 14.60 FA. N. 50 W. 18.20 FB. N. 30 W. 16.80 FC. N. 10 W. 21.20 FD. N. 7 E. 16.95 FE. N. 30 E. 8.50 TO DRAW A PLOT OF THIS FIE] Draw a meridian line to pass through the stationary angle as at F. From the point F, lay off the bearing and distance to the several angles, and connect them by lines, as FG, FA, FB, &c. The area may be calculated as taught in the preceding CASE. CASE VII. To SURVEY A FIELD FROM TWO STATIONS WITHIN THE FIELD, PROVIDED THE SEVERAL ANGLES CAN BE SEEN FROM EACH STATION. SURVEYING. 5t Find the bearing from each station to the respective an. gles ; and also the bearing and distance from one station to the other. Fig. 58. FIELD BOOK. See Fig, 58. First Station. Second Station. AC. N. 38° 30' E. BC. S. 82° 0' E. AD. S. 69 E. BD. S. 17 E. AE. S. 59 W. BE. S. 28 W. AF. N. 63 W. BF. S. 49 W. AG. N. 21 W. BG. N. 76 W. AH. North. BR. N. 24 W. Stationary line AB. N. 14° E. 20 chains. To PROTRACT THIS FIELD* At the first station A, draw a meridian line and lay off the bearings to the respective angles ; draw the stationary line AB, according to the bearing and distance ; at B, draw a meridian line parallel to the other, and lay off the bearings to the angles, as taken from this station ; from each station draw lines through the degree which shows the bearing of each angle, as marked by the protractor or line of chords, and the points where those lines intersect each other will be the angles of the field. Connect those angular points togeth- er by lines, and those lines will represent the several sides of the field. CASE VIII. To SURVEY AN INACCESSIBLE FIELD. Fix upon two stations at a convenient distance from tht 58 SURVEYING. field, from each of which the several angles may be seen ; from each station take the bearing of the angles ; and take the bearing and distance from one station to the other. Fig. 59. FIELD BOOK. See Fig. 59. " ''C A i/ --'■'' ''v 1- B s S"' First Station. Second Station. AE. N. 9° 15' E. BE. N. 50° 0' W AF. N. 16 E. BF. N. 29 15 w AG. N. 14 30 E. BD. N. 24 w AD. N. 39 E. BG. N. 21 30 w AH. N. 40 E. BH. N. 5 E AC. N. 72 E. BC. N. 20 30 Ch. L. E Stationary distance AB, S. 88° 30' E. 19.20. The directions given in the last case for plotting the field, will apply in this case also ; and the area in this and the pre- ceding CASE may be calculated in the manner pointed out in CASE IV. by dividing the plot into triangles and measuring diagonals and perpendiculars. Or the sides may be found by trigonometry, and the area calculated arithmetically, as al- ready taught. CASE IX. To SURVEY A FIELD WHERE THE BOUNDARY LINES ARR SURVEYING. 59 VERY IKREGULAR, WITHOUT EVERY SHALL BEND. NOTICING WITH THE COMPASS Begin near one corner of the field, as at A, Fig. 60, and measure to the next large corner, as B, in a straight line ; noticing also the bearinoj of this line. From the line take offsets to the several bends, at right angles from the line ; noti- cing in the field book at what part of the line they are taken, as at A 1, H 2, 1 3, B 4. Proceed in the same manner round the field. In the figure the dotted lines represent the stationary lines, and the black lines the boundaries of the field. Fig, 60. Bearing and Distance. Offsets Bearing and Distance. Offsets Ch. L. AB.N.85°0'E. 11.20 at 5.40 8.26 the end Ch. L. 0.56 1.40 0.36 0.36 Ch. L. EF.S.67°50'W. 8.20 at 1.04 2.96 5.88 the end Ch. L. 0.40 0.36 0.33 1. 0.12 BC. N. 7^ 20' E. 7."G, OR AX ACCrEATE METHOD OF CALCULATI>'G THE AREA OF A FEELD ARITH31ETICAELY, FROM THE FEELD BOOK, WITHOET THE >-ECESSITT OF PEOTEACTI^"G rr, AXD MEASURES'G WITH A SCALE AMD DEVIDEKS, AS IS C03t- MOXT-Y PRACTISED. I. Survey the field in the usual method, with an accurate compass and chain, and from the field book set down, in a traverse table, the course or bearing of the several sides, and their length in chains and links, or rods and decimal parts of a rod ; as in the 2nd and 3d columns of the following EXAMPLE. SURVEYING. 63 1 ■ST o OJ t: S •— ' ^ o to . c»o 2; CO °q aj Tt< ■^ t- o o >o o ai o o> lO 1— 1 > «o (>> ^-: S' CQ o i> t^ I- i> o , 2o o ifS VO ■<:*• ■* ■^ ^ o (M Tf o 00 00 Tf O CT3 CO CO 1— 1 r—t Tj* Ti< ^co CO(N CO M 00 00 < <— 1 -^ lOCC 8S ■* «D 1 oco V, J>? COC>1 m »o H w o o TJ< -^ o o OiO> rt^-rj* S^N NO coco OOQO 00 »o o o lO .-1 CO to O -H ^(N CO -<* (» ^^ ;S;^ ^n gg t-o coo o o OtO . (N- 1>«- • i>«o l^Tt* ^ ^^ ro r: • • • ■ s^ lO »n coco ^ o o o ! o Tf* o o ■># o TO b o o ^ B in o £; 1— 1 -g .t^J «o 00 U '"' rt 3 15* CO CO ^ :?: U w (^ \^ a? ^ No. " o CO ^ O o 1 ^ 1 °° 19143.9019 Sum of South Areas. 4245.4016 North do. 2)1489«.5u03 Double Area of the Field. Acres 744)92501 Roods 3)70004 40 Rods 28)00160 Acres. Roods. Rods. Area 744 — 3 — 28 2. Calculate by right angled trigonometry, case I, or find by the table of difference of latitude and departure,* or *For an explanation of this table, and the manner of using it, see the re- marks preceding the table. 64 SURVEYING. by the table of natural sines,* the northing or southing, east- ing or westing, made on each course, and set them down against their several courses in their proper columns, marked N. S. £. W. Note. To determine whether the latitude and departure for any par- ticuJar course and distance are accurately calculated, square each of them ; and if Ihey are right, the sum of their squares will equal the square of the distance, for the following reason : the latitude and de- parture represent the two legs of a right angled triangle, and the distance the hypothenuse ; and it is a mathematical truth, that the square of the hypothenuse of any right angled triangle is equal to the sum of the squares of the two legs. 3. If the survey has been accurately taken, the sum of the northings will equal that of the southings, and the sum of the eastings v/ill equal that of the westings. If, upon adding up the respective columns, these are found to ditfer very con- siderably, the field should be again surveyed ; as some error must have been committed, either in taking the courses or measuring the sides. If the difference is small, a judicious, experienced surveyor will judge from the nature of the ground, or shape of the field surveyed, where the mistake was most probably made, and will correct accordingly. Or, the north- ings and southings, and the eastings and westings may be equalled by balancing them, as follows: subtract one half the difference from that column which is the largest, and add it to that column which is the smallest ; and let the difference, to be added or subtracted, be divided among the several courses, according to their length. lnEXA3iPLE 1. the upper numbers are the northings, &:c. as found by a table of difference of latitude and departure. The several columns being added, the northings are found to ex- ceed the southings 47 links, and the westings to exceed the eastings 24 links. [47 being uneven, drop a link from the northings, and it becomes 46. Let half of this (23) be taken from the northings, and added to the southings ;] likewise, take 12 links from the westings, and add it to the eastings. Take from the first course of the northings 12 links, from the second 7, and from the third 5 ; to the first southing add 7 links, to the second 10, and to the third 6 ; add to the first easting 3 links, to the second 3, to the third 4, and to the fourth 2 ; take from the first westing 5 links, from the second 4, and from the third 3. [These are the proportional correc- * See the remarks preceding the table of natural sines. SURVEYING. 65 tions belonging to each, as found by calculation.*] The lower numbers will then represent the northings, &c. as balanced. 4. These columns being balanced, proceed to form a depar- ture column, or a column of meridian distances ; which shows how far the end of each side of the field is east or west of the station where the calculation begins. This column is formed by a continual addition of the eastings, and subtraction of the westings ; or by adding the westings and subtracting the east- ings : see example 1. The first easting, 20.74, is set for the first number in the de- parture column; to this add 24.38, the second easting, and it makes 45.12, for the second number; to this add 30.04, the third easting, and it makes 75.16, for the third number ; to this add 9.56, the fourth easting, and it makes 84.72, for the fourth number ; the fifth course being south, it is evident the meridian distance will remain the same, therefore, place against it the same easting as for the preceding course ; from this subtract 39.95, the first westing, and it leaves 44.77, for the sixth course ; from this subtract 23.75, the second westing, and it leaves 21.02, for the seventh course ; from this subtract 21.02, the last westing, and it leaves 0.0 to be set against the last course, which shows that the additions and subtractions have been accurately made. For as the eastings and westings equal each other, it is evident that one being added and the other subtracted, there will be in the end no remainder. 5. The next step in the process is to form a second depar- ture column, the numbers in which show the sum of the meri- dian distances at the end of the first and second, second and third, third and fourth courses, &:c. The first number in this column will be the first in the other departure column ; to which add the second number in that column for the second in this ; for the third add the second and third ; and for the fourth, the third and fourth ; and so on until the column be completed. See exa3iple 1. The first number to be placed in the second departure col- umn is 20.74 ; to this add 45.12, and it makes 65.86 for the second number ; to 45.12 add 75.16, andit makes 120.28, for the third number; to 75.16 add 84.72, and it makes 159.88 for the fourth number ; to 84.72 add 84.72, and it makes 169.44 for the fifth number ; to 84.72 add 44.77, and it makes 1159.49 for the sixth number ; to 44.77 add 21.02, and it makes 65.79 for the seventh number ; to 21.02 add 0.0, and it makes 21.02 for the eighth number. 6. When the w^ork is thus far prepared, multiply the seve- ral numbers in the second departure column by the northings or southings standing against them respectively ; place the See Appendix, No. 5. C6 SURVEYING. products of those multiplied by the northings in the column of north areas, and of those multiplied by the southings in the column of south areas ; add up these two columns and sub. tract the less from the greater ; the remainder will be double the area of the field in square rods or square chains and links, whichever measure was used in the survey. [In the preceding explanations, the meridian is supposed to pass through the extreme west angle of the field. It is best always to take the extreme east or west angle.] Fig. 62. a. 9 ^ Demonstration of the preceding rules. ■ See Fig. 62, and example I. i The dotted line A 2 represents the northing, and the line 2 B the easting made by the first course ; these multiplied together, that is, 77.15x20.74=1600.0910, which is double the area of the triangle A2B, as is evident from the rule to find the area of a triangle, prob. IX. Rule I. This number is to be placed for the first number in the column of north areas. The line 3C represents the sum of the eastings made by the first and second courses, which is 45.12, the second number in the first departure column ; if to this you add 20.74 the length of the fine 2B you have 65.66, which is the second number in the second departure column, and which represents the sum of the two lines 3C and 2B. These two lines, with the line 2 3 which represents the northing made by the second course, and the line BC, one of the sides of the field, form a right angled trapezoid. Now, by the rule to find the area of such a trape- zoid, See PROB. X. 65.66x31.66=2085.1276, double the area of the trapezoid 2BC3. Place this product for the second number in the column of north areas. To the line 3C add CD 30.04, the easting made by the third course, and you have 75.16, which is the sum of the SURVEYING. 67 castings made by the three first courses, and the third number in the first departure column. To this add 9.56, the easting of the fourtli course, and you have 84.7*2, the length of the line IE, which represents the sum of the eastings made by the four first courses, and is the fourth number in the first de- parture column. These two, viz. the Hnes 3D 75.16 and IE 84.72, added together make 159.88, the fourth number in tlie second departure column; which, being multiplied by 49.15, the length of the line 3 1 which represents the southing made by the fourth course, will give double the area of the trapezoid 1ED3. The number thus produced is 7858.1020, which is to be placed for the first number in the column of south areas. The fiflh course being due south, it is evident the sum of the eastings will remain the same as at the end of the fourth course ; that is, the line 4F equals the line IE, which is 84.- 72. These added make 169.44, the fifth number in the sec- ond departure column. This, being multiplied by 54.10, the length of the line EF, which is the southing of the fifth course as corrected in balancing, and the same as the line 1 4 will give double the area of the parallelogram 1EF4, which is 9166.7040, the second number in tlie column of south areas. From the line AF 84.72 subtract 39.95, which is a we^t course, and it leaves 4G 44.77, the sum of the eastings, or the meridian distance, at the end of the sixth course, and the sixth number in the first departure column. From this subtract 23.75 the westing made by tl\e seventh course, and you have 21.02, the length of the line 5H, which is the meridian dis- tance at the end of the seventh course, and the seventh num- ber in the first departure column. The line 4G 44.77 added to the line 5H 21.02 make 65.79, the seventh number in the second departure column. This being multiplied by 32.21, the length of the line 4 5 which is the southing of the seventh course, will give double the area of the trapezoid 4GH5, which is 21 19.0959, the third number in the column of south areas. The line FI5, 21.02, is the westing of the last course, and the last number in the second departure column. This being multiplied by 26.65, length of the line 5A, and the northing of the last course, produces 560.1830, which is double the area of the triangle A5H, and the last number in the column of north areas. Note. It will be observed that against the third and sixth courses there are no areas ; the reason is that these coursos being one east and the other west, there is no northing or southing to be multi- G 68 SURVEYING. plied into them : regard can therefore be had to ihem only in form- ing the departure columns. By inspecticg the figure, and attending to the preceding illustrations, it will be seen that the three north areas repre- sent double the area of the triangle A2B, the trapezoid 2BC3, and the triangle A5H, all of which are without the boundary lines of the field: also, that the three south areas represent double the area of the trapezoid 3DEI, the parallelogram tEF4, and the trapezoid 4GH5 ; and that these include not only the field but also what was included in the north areas. Therefore the north areas subtracted from the south, the re- mainder will be double the area of the field, contained \\ithin the black Hnes. Additional dlrectio>'s and explanations. The northings and southings may be added and subtracted instead of the eastings and westings ; then there will be two latitude columns instead of departure columns, and the num- bers in the second latitude column must be multiplied into the eastings and westings, and you will have east and west areas. When the course is directly north or south, the distance must be set in the north or south column ; when east or west, in the east or west column. There w\\\ therefore sometimes be no number to be added to or subtracted from the number last set in the latitude or departure column ; then the number Icist placed in the column must be brought down and set against such course ; as in exa^iple I. at the 5th course. It may also sometimes be the case, that there will be no number to multiply into the number in the second latitude or departure column ; then that number must be omitted, and ao-ainst such course there will be no area, as in example 1. at the 3d and 6th courses. When the northings or southings, eastings or westings, be- ginning at the top, will not admit of a continual addition of the one and subtraction of the other, without running out before you get through the several courses, you may begin at such a course in the field book as will admit of a continual addition and subtraction ; and when you get to the bottom go to the top, and you will end in cipher at the course next above that SURVEYING. 69 where you began : as in example II. which begins at the 9th course to add the eastings and subtract the westings.* Example II. No. c— . ^ N. 14.2 S. E. W. I'dep. Gol. 2dep. Col. North areas. South areas. 1. N.75oO'E.I 54.8 52.9 144.1 235.3 3341.26 2 N.'20 30E.! 41.2 38.6 14.4 . . 158.5 302.6 11680.36 . . • 3. East 64.8 . . 64.8 . . 223.3 381.8 . . . 4. S. 33 30W., 141.2 117.7 77.9 145.4 368.7 . . . 43395.99 5. S. 76 W. 64 15.5 62.1 83.3 228.7 3544.85 6. North 36 36 83.3 166.6 5997.60 7. S. 84 W.: 46.4 49 46.1 37.2 120.5 590.45 8. N. 53 15 W 46.4 27.8 37.2 O.Q 37.2 1034.16 9. N.36 45E. 76.8 61.5 46 46 46 2829 10. N. 22 30E.1 56 51.7 21.4 . . 67.4 113.4 5862.78 11. S.76 45E.i 48 11 46.7 . . 114.1 181.5 . . . 1996.50 12. S. 15 W.| 43.4 41.9 11.2 102.9 217 9092.30 13. S. 16 45W.i 40.5 38.8 11.7 91.2 191.1 7531.08 Area 110 Acres, 2 Roods, 23 Rods. Note. In the above example you might begin at the 4th course to add the westings and subtract the eastings ; or at the 6th course to add the northings and subtract the southings : or at the 1 1th course to add the southings and subtract the northings. So in every survey, some place may be found where you may begin to add and subtract, without running out before you get through all the courses.* When a field is very irregularly shaped, it will often happen that parts of the same area will be contained in several different products in the columns of areas i but in the final result, one column being subtracted from the other will leave what is in- cluded within the boundary lines of the field. * Much the most convenient method is to place the course at which you be- gin to add and subtract as the head of the field book. The surveyor may easily do this, by observing in the field, which angle of the field is furthest east or west, north or south. — Ed. 70 SURVEYING. Demonstration. See Fig. 63, and Ex- AMPLB II. The area standing against the 9th course,* which is where the calculation begins, is the triangle 12K, all without the field. The area against the 10th course is the trapezoid 2KL3, also without the field. The area against the 11th course is the trapezoid 4ML3. This is a south area, and contains a part of the field and also part of the preceding north area. The area against the 12th course is the trapezoid 5NM4, part within and part without the field. The area against the 13th course is the trapezoid 6AN5, part within and part without the field. The area against the 1st course is the trapezoid 6AB7, part within and part without the field. This is a north area, and to be ultimately subtracted from the souih areas ; but this includes a part of the preceding south areas, viz. the space nAso ; it will, however, be seen hereafter that this same space is included in another south area. This north area contains also a part of the first north area, viz. the space 6no7 ; but the same space is also included in another south area. The area against the 2d course is also a north area, and is the trapezoid 7BC8. This trapezoid contains the space sBCx, without the field ; the space osxw, within the field ; and the space 7ow8, without the field. But the space osxw will be contained in the next south area ; and the space 7ow8 which was contained in the two first north areas, will be con- tained in the next south area. By examining the whole fgure in this manner, it will be * The numbering here, corresponds not to the diagram, but to the FIELD-BOOK on the last page. After the 8th course, hkewise, in the en- graving, the numbers are incorrect, the true 9th course, DE, being left without a number, and the others numbering 9, 10, &;c. when they should be 10, 11, fee.— Ed. SURVEYING. 71 seen that the north areas contain all without the field that is taken into the calculation, and some of it twice over ; they also contain part of the area within the field. The south areas contain all within the field, and all without the field that is contained in tiie north areas. They also contain, twice over, so much of the field as is included in any of the north areas ; and likewise, twice over, that part without the field which is contained twice in the north areas. So that subtracting the north from the south areas leaves double the area of the field. This method of calculating the area of a field by the north- ings, southings, eastings, and westings, divides the field, with a certain quantity of the adjoining ground, into right angled triangles, right angled trapezoids, parallelograms or squares, as may be seen by the figures. It may therefore with pro. priety be called kectangular surveying.* A useful Problem. To FIND THE true AREA OF A FIELD WHICH HAS BEEN MEASURED BY A CHAIN TOO LONG OR TOO SHORT. Calculate the area as if the chain was of a true length, then institute the following proportion : As the square of the length of the true chain ; Is to the area, as found by the chain made use of; So is the square of the length of that chain ; To the true area of the field. Example. Suppose a field, measured by a two rod chain 3 inches too long, is found to contain 41 acres 1 rood and 33 rods, what is the true area. As the square of 33 feet, the true length of a two rod chain ; is to 41 acres 1 rood and 33 rods ; so is the square of 33 feet 3 inches, the length of the chain used in the survey ; to 42 acres and 13 rods. 33 feet=396 inches. 396 X 396=156816 square inches. 41 acres 1 rood 33 rods=6633 rods. 33 feet 3 inches=399 inches. 399 X 399=159201 square inches. 159201x6633-^156316=6733 rods. 6733—160=42 acres 13 rods, the true area. • See Appendix, No. 6. G2 72 SURVEYING. PART II. Laying out land. PROBLEM L To lay out any number of acres in THE FORM OF A SQUARE. Annex 5 ciphers to the number of acres, which will turn them into square links, the square root of which will be the side of the square in links. Example L It is required to lay out 810 acres in the form of a square. Answer. Each side of the square must be 9000 links, or 90 chains. PROBLEM II. To lay out any number of acres in the FORM of a parallelogram WHEREOF ONE SIDE IS GIVEN. Divide the number of acres when turned into square links, by the given side ; the quotient will be the side required. Example. What must be the longest side of a rectangle, which is to contain 25 acres, when the shortest side is 5 chains and 50 links ? Answer. 2500000-5-550=4545 links for the longest side. PROBLEM III. To lay out any number of acres in a FIELD, 3, 4, 5, 6, &C. TIMES AS LONG AS IT IS BROAD. Divide the acres, when turned into square links, by the ratio between the length and breadth ; the square root of the quo- tient will be the shortest side. Example. It is required to lay out 100 acres 5 times as long as it is broad. Answer. 10000000-^-5=2000000 the square root of which is 1414 links for the shortest side, and the longest will be 7070 links. PROBLEM IV. To make a triangle which shall con- tain A given number of acres, being confined to a cer- tain base. Double the given number of acres, to which, annex 5 ci- phers, and divide by the base ; the quotient will be the per- pendicular height in links. Example. Upon a base of 40 chains to lay out 100 acres in a triangular form. Answer. 5000 links or 50 chains will be the length of the perpendicular. SURVEYING. 73 The perpendicular may be erected from any part of the base : thus, the triangle ABC, see Fig. 64, is the same as ABE, each containing 100 acres. Fig. 64. Fig. 65. When the given base is so situ- ated that a perpendicular of suffi- cient length cannot be erected therefrom, continue the base as from B to D,/^. 65, from which erect the perpendicular DC, and complete the triangle ABC, which will contain 100 acres. PART III. Dividing land. As different fields are so variously, and many of them ir- regularly shaped, and as they are required to be divided in many different proportions, it is difficult to give rules which will apply to particular cases. The business of dividing land must therefore be left, in a great measure, to the skill and judgment of the surveyor ; who, if he is well acquainted with trigonometry, and with measuring land, will not find it diffi- cult, after a little practice, to divide a field in such a manner as shall be desired. If he has before him a plot of the field, and knows the number of parts into which it is to be divided, and the proportion which each part is to bear to the others, he will readily find out where the dividing lines are to be drawn. A few RULES and examples will be given for the general instruction of the learner. PROBLEM I. To out off any number of acres from A gQUARK OR RECTANGLE. T4 SURVEYING. Say. as the whole number of acres in the field ; is to the length ot'the square or length or breadth of the rectangle: so is the number of acres proposed to be cut off; to their propor- tion of the len2:h or breadth. PROBLEM II. To cut off any ^-o^BER of acres by a LIXE FROH AXY AXGLE OF A TRIA>'GLE. Measure the base, or side opposite the angle from which the dividing line is to be drawn : then sav, as the number of acres in the whole triangle : is to the whole base : so is the given number of acres : to their part of the base. Fig, 66. Example. See Fis. 66. In the triangle ABC. which contains 4S acres., it is required to cut off 15 acres, by a Hne proceeding from C to the base AB. which is 40 chains. As 4S ; 40 : : 19 : 15 Lay 15 chains on the base from B to D, and draw the line CD. The triangle will then be divided as was proposed : BCD containing IS acres, PRC>BLEM III. To take off any given nxmber of acres from a MnrLTAN-GULAR FIELD. Fis. 67. D Example i. See Fis- Let ABCD, Ace. be the plot of a field containing 11 acres, from which it is required to cut off 5 acres. Join two opposite corners of the field as D and G, with the hne DG ^ which you may judge to be near the partition line) and lind the area of the part DEFG, which, suppose, may want 140 rods of the quantity proposed to be cut off. Mea- SURVEYING. 75 sure the line DG, which, suppose to be 70 rods ; divide 140 by 35 the half of DG, and the quotient 4 will be the length of the perpendicular of a triangle, whose base is 70 and the area 140. Lay off 4 rods from G to I, and draw the line DI, which will be the dividing line.* Example 11. See Fig. 68. Let ABCD, &c. be a tract of land to be divided into two equal parts, by a line from I to the oppo- site side CD ; to find arithmeticaliyA on what part of the line CD the di- viding Hne IN will fall ; or to find the distance CN. Fig. 68. FIELD BOOK. Rods. Rods AB. N. 19° 0' E. 108 FG. West 70.9 BC. S. 77 E. 91 GH. N. 360 0' W. 47 CD. S. 27 E. 115 HI. North 64.3 DE. S. 52 W. 58 lA. N. 62 15 W. 59 EF. s. 15 30 E. 76 Acres Rood Rods Whole area 152—1 — 25 Find the area of the part lABC, according to section 111. Jjage 57, as follows : set the latitude and departure of the three first sides, lA, AB, and BC, in their proper columns, in a tra- verse table ; and place as much southing, viz. 109.1, equal to the line CK, and as much westing, viz. 71.7, equal to the line Kl, as will balance the columns. This southing and westing will be the latitude and departure made by the line CI. The area of lABC will be found to be 8722 rods, which is less than half the area of the whole field by 3470 rods, the quantity to be contained in the triangle ICN. * This explanation supposes DG and AG at right angles to each other. When they are not at right angles, the height GI must not be meastired on the side ofthe field. —Ed. 76 SURVEYING. Find the bearing and distance of CI by bight angled tbi- Gor^oMETRY, CA5E IV. as tollows :* JLi CK, tiie somhins of CI, 109, nearlv • 2.037426 : radios - t . . . ' . . 10,000000 : : KI, die westing of CI, 71.7 - 1.855519 11.855519 2.037426 S. 33020 W. . . . 9.818093 A^ sine coarse 33o 20^ 9.739975 : depamire KI 71.7 1.855519 : : radios 10.000000 11.855519 9.739975 Distance IC 130.5 2 115544 Note. In this way the coojse and distance may be found firom any angle of a field to another. Haying found the line Cl, divide 3470, the number of rods to be contained in the triangle ICN, by one half the line CI, viz. 65.25 : the quotient will be the leneth of the perpendicu- lar PX, viz. 53.18. Now, by the bearings of CI and CD, it appears that they form an angle of 60- 20' : wherefore, in the triangle CPN are given the side FN 53.15. and the angle at C 60^ 20', to find the hypothenuse CN. As sine PCX 60^ 20' . - 9.938980 : side PX 53.18 - - . 1.725748 : : radius - - - - 10.000000 11.725748 9.938980 : hyp. CX 61. -2 . - - 1.786768 Thus the dividing line must go from I to a point on the line CD, which is 61.2 rods fromC. The bearing and distance of this line may be found by the directions given above for find- ing the bearing and distance of the line CI. Or, they may be found by oblique trigonometry, case III. * The mode gnren above is BDodonbtedly the moat correct, bat the use of the tntcne taUe will aTe many fisures. From tfaattabte the course and distance of CI, May be fimad al agh't — Ec. SURVEYING. 77 Another method of finding the distance CN. Having ascertained the latitude and departure of the line CI, set them down in a traverse table ; find the latitude and departure of the line CD, and place them in the table ; the difference between the northing of the line IC, and the southing of the line CD will be the southing of the line DI, viz. 6.6 ; and the sum of the eastings of those lines, as they are both easterly, will be the westing of the line DI, viz. 123.9. Pro- ceed to calculate the area of the triangle ICD, which will be found to be 6522 rods, nearly. Note . As in this triangle two sides and their contained angle are given, the area may be found by prob. IX. rule A, page 40. Having found the area of this triangle, proceed to find CN according to prob. II. page 74, as follows : A.S the area of the triangle ; is to CD the base ; so is the quantity to be contained in the triangle ICN ; to CN its pro- portion of the base. As 6522 : 115 ; : 3470 : 61.19. A THIRD METHOD OF FINDING THE DISTANCE CN. To the logarithm of double the area to be contained within the triangle ICN add radius ; from this sum subtract the loga- rithmic sine of the angle at C ; and from the remainder sub- tract the logarithm of the side IC ; the last remainder will be the logarithm of the side CN. The double area of the triangle ICN is 6940 ; the angle at Cis60^ 20'; the side IC is 130. Double area 6940 - - - 3.841359 Pwadius 10.000000 13.841359 Sine ICN 60° 20' . - . 9.938980 3.902379 Side IC 130.5 - - - 2.115544* Side CN 61.21 . - - 1.786835 * The log. of IC, as found by calculation on page 76, is employed, instead of taking from the tables, that of 130.5, which is not the exact length of IC— Ed. 78 SURVEYING. Note. Radius mav be added by placing a unit before the index of ihe logarithm for the double area, without the trouble of setting down the ciphers. By natural sixes. Divide the double area by the natural sine of the given an. gle. and that quotient by the given side ; the last quotient will be the side CX. Nat. sine of the ande at C 60= 20' 0.86S92 6940-^0.S6S9-2=:7936.92 79S6.92-:-130.5=61.2 From the above the following general rule may be drawn. To FI>D THE SIDE OF A TEIAN'GLE WHEX THE AREA IS GIVEN. AND ALSO ONE OF THE SIDES, AND THE ANGLE CONTAINED BE- TWEEN THE GIVEN SIDE AND THE SIDE REQUIRED. To THE LOGARITHM OF DOUBLE THE AREA ADD RADIUS : from: this SU3I subtract the LOGARITHMIC SIN~E OF THE, given angle, and fr03i the remainder subtract the log- akithm of the given side ; the last remainder will be the logarithm of the side required. Or. By natural sines : divide the double area by the NAT. sine of the given ANGLE, AND THAT QUOTIENT BY THE GIVEN SIDE : THE LAST QUOTIENT WILL BE THE SIDE REQUIRED. COXCLUDING REMARKS. Other methods of surveying fields are taught by some authors on this subject. The preceding, however, will be found most useful in actual practice. Other instruments be- sides those mentioned in this book are sometimes used ; such as the plain table, semicircle, peramubulator, theodo- lite,