A SYSTEM OF GEOMETRY AND TRIGONOMETRY WITH A TREATISE ON SURVEYING, IN WHICH THE PRINCIPLES OP RECTANGULAR SURVEYING, WITHOUT PLOTTING, ARE EXPLAINED. BY ABEL FLINT, A.M. STEREOTYPE EDITION, ENLARGED, WITH ADDITIONAL TABLES. BY GEORGE GILLET, SURVEYOR GENERAL OP CONNECTICUT. HARTFORD : BELKNAP AND HAMERSLEY. 1838. ENTERED, according to the Act of Congress, in the year 1835, by EDWARD P. COOKE, in the Clerk's Office of the District Court of CONNECTICUT. v ; .-.i ~TM THE original compiler of the following work design- ed, in preparing it, to furnish a plain and concise system of PRACTICAL SURVEYING. 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. In the present edition, practical matter, has been add- ed by GEORGE GILLET, Esq., Surveyor General of the state of Connecticut ; and in addition to all the tables of the best treatises extant, it contains the only table of Natural Tangents ever published in this country. In the table of logarithms, and of logarithmic sines, &c., the decimals are carried to six figures, and a co- lumn of differences is added, for the purpose of finding intermediate numbers. The use of these tables is so fa- miliarly explained and illustrated by examples, that no other instruction upon this subject is necessary. The articles on distributing estates, locating and sur- veying roads, and on levelling, cannot fail of being highly useful to the practical surveyor. The work being now used extensively in schools and academies, it has been the chief object of the publishers to render it acceptable as a text book. M306235 The subscribers have examined, in manuscript, the additions to the seventh edition of FLINT'S SURVEYING, by GEORGE GILLET, Esq , Sur- veyor General of Connecticut, and find them to embrace a system of correct, useful, and practical matter, judiciously arranged, and clearly explained to the understanding of the learner. Having long acted as Surveyors under public authority, we recommend this work, as con- taining 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. N. London Co. LEMUEL INGALLS, late Dep. Sur. Windham Co. DANIEL ST. JOHN, Dep. Sur. Hartford Co. ASAHEL DEWEY, County Sur., N. London Co. JONATHAN NICHOLS, Dep. Sur. Windham Co. Connecticut, August, 1832. FLINT'S SURVEYING has now been before the public upwards of thirty years. During this period, it has passed through numerous editions, and been enriched, from time to time, by important contributions from the present Surveyor General, GEORGE GILLET, Esq. The distinguish- ing feature of the work, as now published, is its excellent adaptation to the every day wants of the practical surveyor, while it supplies to aca- demies and private students an eminently useful, clear, and well-digest- ed system of elementary instruction, both in the theory and practice of Surveying. I know of no work, in this respect, which equals it. E. H. BURRITT, Civil Engineer. New Britain, Con., Nov., 1835. 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 nnd 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. 6. The inclination of two lines meeting one another, or the opening between them, is called an angle. Thus, atB, Fig. 1, is an angle, formed by the meeting of the lines A B and B C. 7. If a right line C D, Fig. 2, fall upon another right line A B, 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 C D is said to be perpendicular to the other line. Fig. 1. 10 GEOMETRY. Fig. 3. 8. An obtuse angle is greater than a right angle ; as A D E, Fig. 3. 9. An acute angle is less than a right angle; asEDB, Fie. 3. NOTE. When three letter? arc used to express an angle, the middle letter denotes the angular point. 10. A circle is a round figure bounded Fig. 4. by a single line, in every part equally dis- ^ tant from some point, which is called the ^ \ centre. Fig. 4. / \ 11. The circumference or periphery of a -J \ circle is the bounding line ; as A D E B, * Fig. 4. 12. The radius of a circle is a line drawn from the centre to the circumfe- rence ; as C B, 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 circumfe- rence to the other, passing through the cen- tre ; and it divides the circle into two equal parts, called semicircles; as ABor 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 num- ber of degrees, a degree is not to be accounted a quan- tity 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 B F or F D, Fig. 5 ; and is said to be an arc of as many degrees as it contains 360th parts of the whole circle. Fig. 5. GEOMETRY. 11 16. A chord is a right line drawn from one end of an arc to another, and is the measure of the arc ; as H G is the chord of A the arc H I G, 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 H I G and the chord H G is called a segment. Fig. 6. 18. A sector of a circle is a space contained between two radii, and an arc less than a semicircle ; as B C D, or A C D, Fig. 6. 19. 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, H L is the sine of the arc H B, Fig. 7. 20. The sines on the same diameter in- crease in length till they come to the cen- A tre, and so become the radius, after whjch they diminish. Hence it is plain that the sine of 90 degrees is the greatest possible sine, and is equal to the radius. 21. The versed sine of an arc is that part of the diameter or radius which is between the sine and the circumference; thus L B is the versed sine of the arc H B, Fig. 7. 22. The tangent of an arc is a right line touching the circumference, and drawn perpendicular to the diameter ; and is terminated by a line drawn from the centre through the other end of the arc ; thus, B K is the tangent of the arc B H, Fig. 7. NOTE. , The tangent of an arc of 45 degrees is equal in length to the radius of the circle of which the arc is a part. 12 GEOMETRY. 23. The secant of an arc is a line drawn from the centre through one end of the arc till it meets the tangent; thus, C K is the secant of the arc B H, Fig. 7. 24. The complement of an arc is what the arc wants of 90 degrees, or a quadrant: thus, H D is the complement of the arc B H, Fig. 7. 25. The supplement of an arc is what the arc wants of 180 degrees, or a semicircle ; thus A D H is the supplement of the arc B H, 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, likewise, the tangent and secant of any arc are used also for its supplement. 26. 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, F H is the sine, D I the tangent, and C I the secant of the arc D H ; or they are the co-sine, co-tangent, and co-secant of the arc B H, Fig. 7. [The terms sine, tangent, and secant, are abbreviated thus : sin., tan., and sec. So, likewise, co-sine, co-tangent, and co- secant, are written co-sin., co-tan., and co-sec.] 27. 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 H C B, Fig. 7, is mea- sured by the arc B H ; and is said to contain as many degrees as the arc does. NOTE. An angle is esteemed greater or less, according to the opening of the lines which 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 the including lines; for all arcs described on the same point, and intercepted by the same right lines, contain exactly the same num- ber of degrees, whether the radius be longer or shorter. 28. The sine, tangent, or secant of an arc, is also the sine, tangent, or secant of the angle whose measure the arc is. Fig. 8. 29. Parallel lines are such as are equal- ^ -g ly distant from each other ; as A B and C D, * 8 - GEOMETRY. 13 30. A triangle is a figure bounded by three lines ; as A B C, Fig. 9. 31. An equilateral triangle has its three sides equal in length to each other. Fig. 9. 32. An isosceles triangle has two of its sides equal. Fig. 10. fig. 10. Fig. 11. 33. A scalene triangle has three unequal / sides. Fig. 11. / Fig. 12. 34. A right angled triangle has one right angle. Fig. 12. 35. An obtuse angled triangle has one obtuse angle. Fig. 13. 36. An acute angled triangle has all its angles acute. Fig. 9, or 10. 37. 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. 38. In a right angled triangle, the longest side is called the hypothenuge, and the other two, legs, or base, and perpen- dicular. 2* 14 GEOMETRY. NOTE. The three angles of every triangle being added together will, amount to 180 degrees; consequently the two acute angles of a right angled triangle amount to 90 degrees, the right angle being also 90. 39. The perpendicular height of a tri- angle is a line drawn from one of the angles perpendicular to its opposite side ; thus, the dotted line A D, Fig. 14, is the perpen- dicular 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 immaterial. Fig. 15. 40. A square is a figure bounded by four equal sides, and containing four right angles. Fig. 15. 41. A parallelogram, or oblong square, is a figure bounded by four sides, the oppo- site ones being equal, and the angles right.* Fig. 16. 42. A rhombus is a figure bounded by four equal sides, but has its angles oblique. Fig. 17. Fig. 16. Fig. 17. * Any four sided figure, having its opposite sides parallel, is a pa- rallelogram ; but, in this book, the term is understood as it is here ex- plained. Sometimes this figure is called a rectangle. GEOMETRY. 15 Fig. 18. A 43. A rhomboid is a figure bounded by four sides, the opposite ones being equal, but the angles oblique. Fig. 18. 44. The perpendicular height of a rhombus or rhomboi- des, is a line drawn from one of the angles to its opposite side; thus, the dotted lines A B, Figs. 17 and 18, represent the perpendicular heights of those figures. Fig. 19. r 45. 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. 46. A trapezium is a figure bounded by four unequal sides. Fig. 21. 47. A diagonal is a line drawn between two opposite angles ; as the line A B, Fig 21. 48. Figures which consist of more than- four sides are called polygons ; if the sides are equal to each other, they are called regular polygons, and are sometimes named from the number of their sides, as pentagon, or hexagon, a figure of five or six sides, &c. ; if the sides are unequal, they are called irregular polygons. 16 GEOMETRY. PART II. GEOMETRICAL PROBLEMS. Fig. 22, PROBLEM I. To draw a line paral- C D lei to another line at any given distance ; as at the point D, to make a line parallel _ > ____ .,..-' to the line A B. Fig. 22. A E With the dividers take the nearest distance between the point D and the given line A B ; with that distance set one foot of the dividers any where on the line A B, as at E, and draw the arc C ; through the point D draw a line so as just to touch the top of the arc C. A more convenient way to draw parallel lines is with a parallel rule. [The parallel, rules, however, found in cases of mathematical instruments, are often inaccurate.] Fig. 23. PROBLEM II. To bisect a given line ; or, to find the middle of it. Fig. 23. D Open the dividers to any convenient distance, more than half the given line A B, 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 ; lay a rule from C to D, and where the rule crosses the line, as at E, will be the middle. Fig. 24. H PROBLEM III. To erect a perpen- ^ dicular from the end, or any part of a ..-"" given line. 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 per- pendicular is to be erected, describe an arc, as A E G ; set off the same distance A D, from A to E, and from E to G ; upon E and G describe two arcs to intersect each other at H ; draw a Hne from H to D, and one line will be perpendicular to the other. NOTE. There are other methods of erecting 1 a perpendicu- lar, but this is the most simple. Fig. 25. PROBLEM IV. From a given point, as at C, to drop a perpendicular on a given line A B, Fig. 25. * Gr 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 given line. Perpendiculars may be more readily raised and let fall, by a small square made of brass, ivory, or wood. Fig. 26. PROBLEM V. To make an angle af, equal to a given angle ABC, Fig. 26. " Open the dividers to any convenient distance, and with one foot in B, describe the arc F G ; with the same distance, and one foot in E, describe an arc from H ; measure the arc F G, and lay off the same distance on the arc from H to I ; draw a line through I 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 36. Fig. 27. 18 GEOMETRY. Draw the line A B of any convenient length ; from a scale of chords take 60 degrees with the dividers, and with one foot in B, describe an arc from the line A B 5 from the same scale take 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. 28. PROBLEM VII. x To make an obtuse angle, suppose of 110 degrees. Fig. 28. B Take a chord of 60 degrees as before, and describe an arc greater than a quadrant ; set oft' 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 may be measured ; that is, with a chord of 60 degrees describe an arc on the angular point, and on a scale of chords measure the arc intercepted by the lines forming the angle. A more convenient method of making and measuring an- gles, is to use a protractor instead of a scale and dividers. Fig. 29. O B PROBLEM VIII. To make a trianglej, of three given lines, as B O, B L, L O, Fig. 29, any two of which are, greater than the, third. Draw the line B L from B to L ; from B, with the length of the line B O, describe an arc as at O ; from L, with the length of the line L O, describe another arc, to intersect the former ; from O draw the lines O B and O L, and B L will be the triangle required. GEOMETRY. 19 Fig. 30. PROBLEM IX. To make a right an- gled triangle, the hypothenuse and angles being given. Fig. 30. Suppose the hypothenuse C A 25 rods or chains, the an- gle at C 35 30', and consequently the angle at A 54 30'. See note after the 38th 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 mi- nutes. Draw the line C B an indefinite length ; at C make an angle 35 30'; through where that number of degrees cuts the arc, draw the line C A 25 rods, which must be taken 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 ta- king distances from the scale, each of these divisions, (either the larger or the smaller, as is most convenient,) must be considered 1, 10, 100, &c. rods, chains, or other dimensions of length. If each division be called 1, it will be easy to take off the required number. But the scale is not usually long 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 little scale is divided 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 the 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 20 GEOMETRY. are drawn across parallel lines, running the whole length of the scale, from each division on one side, to the next higher on the other. The parallel lines divide the width of the scale into 10 equal parts. Since each oblique line, then, in crossing the scale, passes over one division of length, it is evident 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 hun- dreds, then, as 234, we must place one foot of the dividers on the second division of the larger scale, and on the paral- lel line marked 4, and extend the other foot to the third ob- lique line. Decimals may evidently be taken off in a simi- lar manner ; the divisions of the larger scale being made units, and those of the smaller, tenths and hundredths.} NOTE. 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. Kg. 31. TJ PROBLEM X To make a right angled triangle, the angles and one being given. Fig. 31. 285 Suppose the angle at C 33 15', and the leg A C 285. Draw the leg A C, making it in length 285 ; at A erect a perpendicular an indefinite length; at C make an angle 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 C A 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. PROBLEM XL To make a right an- gled triangle, the hypothenuse and one leg being given. Fig. 32. GEOMETRY. Suppose the hypothenuse A C 40, and the leg- A B 28. Draw the leg A B in length 28; from B erect a perpen- dicular an indefinite length; take 40 in the dividers, and set- ting one foot in A, wherever the other foot strikes the perpen- dicular 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. PROBLEM XII. To make a right angled triangle, the two legs being given. Fig. 33. B Suppose the leg A B 38, and the leg B C 46. Draw the leg A B in length 38 ; from B erect a perpen- dicular to C, in length 46 ; and draw a line from A to C. Fig. 34. PROBLEM XIII. To make an oblique angled tri- angle, the . angles and one side being given. Fig. 34. Suppose the side B C 98; the angle at B 45 15', the an- gle at D 108 30', consequently the other angle 26 15'. Draw the side B C in length 98 ; on the point B make an angle of 45 15' ; on the point C make an angle of 26 1 5', and draw the lines B D and C D. Fig. 35. PROBLEM XIV. To make an ob- lique angled triangle, two sides and an angle opposite to one of them being given. Fig. 35. B Suppose the side B C 160, the side B D 79, and the angle at C 29 9'. 3 GEOMETRY. Draw the side B C in length 160 ; at C make an angle of 29 9', and draw an indefinite line through where the degrees cut the arc ; take 79 in the dividers, and with one foot in B lay the other on the line CD; the point D will be the other angle of the triangle. Fig. 36. PROBLEM XV. To make an oblique angled triangle, two sides and their contained angle being given. Fig. 36. Suppose the side B C 109, the side B D 76, and the angle atB 101 30'. Draw the side B C in length 109 ; at B make an angle of 101 30', and draw the side B D in length 76; draw a line from D to C, and it is done. Fig. 37. PROBLEM XVI. Fig. 37. To make a square. Draw the line A B the length of the proposed square ; from B erect a perpendicular to C, and make it of the same length as A B ; from A and C, with the same distance in the dividers, describe arcs intersecting each other at D, and draw the lines A D and D C. Fig. 38. D PROBLEM XVII. tangle. Fig. 38. To make a rec- Draw the line A B equal to the longest side of the rectan- gle ; on B erect a perpendicular the length of the shortest side to C ; from C, with the longest side, and from A, with the shortest side, describe arcs intersecting each other at D, and draw the lines A D and C D. GEOMETRY. PROBLEM XVIII. To de- scribe a circle which shall pass through any three given points, not lying in a right line, as A, B, D. Fig. 39: Draw lines from A to B, an.d from B to D ; bisect those lines by PROBLEM II, and the point where the bisecting lines intersect each other, as at C, will be the centre of the circle. PROBLEM XIX. To find the centre of a circle. By the last PROBLEM it is plain, that if three points be any where taken in the given circle's periphery, the centre of the circle may be found as there taught. Directions for constructing irregular figures of four or more sides, may be found in the following treatise on SUR- VEYING, TRIGONOMETRY. TRIGONOMETRY is that part of practical GEOMETRY, by which the sides and angles of triangles . are measured ; whereby three things being given, either all sides, or sides and angles, a fourth may be found ; either by measuring with a scale and dividers, according to the PROBLEMS IN GEOME- TRY, or more accurately by calculation with logarithms, or with natural sines. TRIGONOMETRY is divided into two parts, rectangular and oblique-angular. PART I. RECTANGULAR TRIGONOMETRY. This is founded on the following methods of applying a circle to a triangle. Fig. 40. PROPOSITION I. In every right angled triangle, as A B C, Fig. 40, it is plain from Fig. 7, compared with the Geometrical definitions to which that Figure refers, that if the hypothe- A, nuse A C be made radius, and with it \ an arc of a circle be described from each end, B C will be the sine of the angle at A, and A B the sine of the angle at C; that is, the legs will be sines of their opposite angles. i. PROPOSITION II. If one leg, A B, Fig. 41, be made radius, and with it on the point A an arc be described, then B C, the other leg, will be the tangent, and A C the secant of the angle at A ; and if B C be made radius, and an arc be described with it on the point C, then A B will be the tangent, and A C the se- cant of the angle at C ; that is, if one Fig. 41. GEOMETRY. 25 leg be made radius, the other leg will be a tangent of its op- posite angle, and the hypothenuse a secant of the same an- gle. Thus, as different sides are made radius, the other sides acquire different names, \Vhich are either sines, tangents, or secants. As the sides and angles of triangles bear a certain propor- tion to each other, two sides and one angle, or one side and two angles being given, the other sides or angles may be found by instituting proportions, according to the following rules. RULE I. To find a side, either of the sides may be made radius, then institute the following proportion: As THE NAME OF THE SIDE GIVEN, (which will be either radius, sine, tangent, or secant;) Is TO THE LENGTH OF THE SIDE GIVEN ; So IS THE NAME OF THE SIDE REQUIRED, (which also will be either radius, sine, tangent, or secant ;) To THE LENGTH OF THE SIDE REQUIRED. RULE II. To find an angle, one of the given sides must be made radius, then institute the following proportion : AS THE LENGTH OF THE GIVEN SIDE MADE RADIUS j Is TO ITS NAME, (that is, radius ;) So IS THE LENGTH OF THE OTHER GIVEN SIDE } To ITS NAME, (which will be either sine, tangent, or se- cant.) Having instituted the proportion, look for the correspond- ing logarithms, in the logarithms of numbers for the length of the sides ; and in the table of artificial sines and tangents, for the logarithmic sine, tangent, or secant. ADD TOGETHER THE LOGARITHMS OF THE SECOND AND THIRD TERMS, AND FROM THEIR SUM SUBSTRACT THE LOGA- RITHM OF THE FIRST TERM .* THE REMAINDER WILL BE THE LOGARITHM OF THE FOURTH TERM, WHICH SEEK IN THE TA- BLES, AND FIND ITS CORRESPONDING NUMBER, OR DEGREES AND MINUTES. See the introduction to the table of logarithms ; which should be attentively studied by the learner, before he pro- ceeds any further. NOTE. The logarithm for radius is always 10, which is the logarithmic sine of 90, and the logarithmic tangent of 45. 26 GEOMETRT, The preceding PROPOSITIONS and RULES being duly at- tended to, .the solution of the following CASES of Rectangu- lar Trigonometry will be easy.* CASE I. The angles and hypothenuse given, to find the, legs. Fig. 42. A- In the triangle AB C, given the hypothenuse A C 25 rods or chains; the angle at A 35 30': and consequently the angle at C 54 30': (See Note, GEOM. Def. 38.) to find the legs. Making the hypothenuse radius, the proportions will be : To find the leg A B. As radius 10.000000 : hyp. AC, 25 - 1.397940 : : sine ACB, 54 30' 9.9 10686 11.308626 10.000000 To find the leg E C. As radius - - - 10.000000 : hyp. AC, 25 - 1.397940 : : sineCAB, 3530' 9.763954 11.161894 10.000000 : leg AB, 20. 35 nearly 1.308626 :legBC, 14.52 nearly 1. 161894 NOTE. When the first term is radius, it may be substract- ed by cancelling the first figure of the sum of the other two terms. As secants are excluded from the tables, such cases in Trigonometry as are solved by secants are also excluded. [The learner should exercise himself, in this and the fol- lowing rules in TRIGONOMETRY, in stating all the propor- tions which can be made, until he is able to do it with fa- cility.] - * See Practical Trigonometry in, the Appendix. TRIGONOMETRY. 27 BY NATURAL SINES. This CASE may be solved by natural sines,* according to the following proportions : As UNITY, OR 1, IS TO THE LENGTH OF THE HYPOTHENUSE, SO IS THE NATURAL SINE OF THE SMALLEST ANGLE, TO THE LENGTH OF THE SHORTEST LEG. OR, SO IS THE NATURAL SINE OF THE LARGEST ANGLE, TO THE LENGTH OF THE LONG- EST LEG. Or, which is the same thing, MULTIPLY THE NATURAL SINES OF THE TWO ANGLES BY THE HYPOTHENUSE J THE PRODUCTS WILL BE THE LENGTH OF THE TWO LEGS. EXAMPLE. Nat. sine of 35 30' Nat. sine of 54 30' 0.58070 0.81412 Hyp. 25 Hyp. 25 290350 407060 116140 162824 14.51750 20.35300 LegBC 14.52 Leg AB 20.35 NOTE. The third decimal figure in the first product be- ing 7, the preceding figure may be called one more than it is, viz. 2. And whenever in any product, &c. there are more places of decimals than you wish to work with, if the one at the right hand of the last which you wish to retain is more than 5, add a unit to the last, be- cause a greater decimal number than 5 is more than half. CASE II. The angles and one leg given, to find the hypothenuse and the other leg. Fig. 43. B * See the Introduction to the Table of Natural Sines. 28 TRIGONOMETRY. In the triangle ABC, given the leg A B 325, the angle at A 33 15', and the angle at C 56 45'; to find the hypothe- nuseand the leg BC. Making the hypothenuse radius, the proportions will be ; To find the. hypothenuse. As sine of C, 56 45' 9.922355 :leg AB, 325. 2.511883 To find the leg BC. As sine of C, 56 45' 9.922355 : leg- AB, 325 2.511883 Radius 10.000000 : : Sine of A, 33 15' 9.739013 12.511883 9.922355 12.250896 9.922355 Hyp. 388.6 2.589528 : leg. B, C. 2131 2.328541 NOTE. If the leg B C had been given, instead of the leg A B, the proportions would have been the same, the obvious changes being made. BY NATURAL SINES. To solve this CASE by natural sines, institute the follow- ing proportions : To find the hypothenuse. As THE NATURAL SINE OF THE ANGLE OPPOSITE THE GIVEN LEG, IS TO THE LENGTH OF THE LEG, SO IS UNITY, OR 1, TO THE LENGTH OF THE HYPOTHE- NUSE. Or, which is the same thing, DIVIDE THE GIVEN LEG BY THE NATURAL SINE OF ITS OPPOSITE ANGLE, AND THE QUO- TIENT WILL BE THE HYPOTHENUSE. To find the other leg. As THE NATURAL SINE OF THE AN- GLE OPPOSITE THE GIVEN LEG, IS TO THE LENGTH OF THE GJVEN LEG, SO IS THE NATURAL SINE OF THE ANGLE OPPO- SITE THE OTHER LEG, TO THE LENGTH OF THE OTHER LEG. EXAMPLE. Given leg 325. Nat. sine of 56 45', the angle opposite the given leg, 0.83629. Nat. sine of 33 15', the angle op- posite the other leg, 0.54829. As 0.83629 : 325 : : 1 : 388.6. As 0.83629 : 325 :: 0.54829 : 213.07. TRIGONOMETRY. 29 CASE III. The hypothenuse and one leg given, to find the angles and the other leg. Fig. 44. In the triangle A C B, given the hypothenuse A C 50, and the leg A B 40, to find the angles and leg B C. Making the hypothenuse radius, the proportion to find the angle A C B will be : As hyp. 50 : radius : : leg A B 40 - 1.698970 10.000000 1.602060 11.602060 1.698970 : sine A C B, 53 8' 9.903090 The angle A C B being 53 8', the other is consequently 36 52'. The angles being found, the leg B C may be found by either of the preceding CASES. It is 30. BY NATURAL SINES. The angle opposite the given leg, may be found by the fol- lowing proportion : As THE HYPOTHENUSE IS TO UNITY, OR 1, SO IS THE GIVEN LEO TO THE NAT. SINE OF ITS OPPOSITE ANGLE. Or, which is the same thing, DIVIDE THE GIVEN LEG BY THE HYPOTHENUSE, AND THE QUOTIENT WILL BE THE NAT. SINE. EXAMPLE. The leg A B 40, divided by the hypothenuse 50, gives a 30 TRIGONOMETRY. quotient 0.80000 ; which, looking in the table of nat. sines, the nearest corresponding number of degrees and minutes will be found to be 53 8', the angle A C B. BY THE SQUARE ROOT. In this CASE the required leg may be found by the square root, without finding the angles ; according to the following PROPOSITION : IN EVERY RIGHT ANGLED TRIANGLE, THE SQUARE OF THE HYPOTHENUSE IS EQUAL TO THE SUM OF THE SQUARES OF THE TWO LEGS. Hence, THE SQUARE OF THE GIVEN LEG BEING SUBSTRACTED FROM THE SQUARE OF THE HYPOTHENUSE, THE REMAINDER WILL BE THE SQUARE OF THE REQUIRED LEG. As in the preceding EXAMPLE; the square of the leg A B 40 is 1600 ; this substracted from the square of the hypothe- nuse 50, which is 2500, leaves 900, the square of the legBC, the square root of which is 30, the length of leg B C, as found by logarithms. CASE IV. The legs given, to find the angles and hypothenuse. Fig. 45. 73.7 In the triangle A B C, given the leg A B 78.7, and the leg B C 89 ; to find the angles and hypothenuse. Making the leg A B radius, the proportion to find the angle BAG will be, TRIGONOMETRY. 31 As leg A B, 78.7 - 1.895975 : radius - 10.000000 : : leg B C, 89 - 1.949390 11.949390 1.895975 : tariff. BAG, 48 31' 10.053415 The angle A C B is consequently 41 29'. Making the leg B C radius, the proportion to find the angle B C A will be similar, with the obvious differences. The angles being found, the hypothenuse may be found by CASE II. It is nearest 1 19. BY THE SQUARE ROOT In this case, the hypothenuse may be found by the square root, without finding the angles ; according to the following PROPOSITION. IN EVERY RIGHT ANGLED TRIANGLE, THE SUM OF THE SQUARES OF THE TWO LEGS IS EQUAL TO THE SQUARE OF THE HYPOTHENUSE. In the above EXAMPLE, the square of A B 78.7 is 6193.69, the square of B C 89 is 7921 ; these added make 14114.69, the square root of which is nearest 1 19. BY NATURAL SINES. The hypothenuse being found by the square root, the angles may be found by nat. sines, according to the preceding CASE. 32 TRIGONOMETRY. Hyp. Leg B C. Nat. Sine. 119) 89.00000 (74789 83.3 570 The nearest degrees and minutes 476 corresponding to the above nat. sine, are 48 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 hypothenuso it 952 being a fraction too much. 1180 1071 109 BY NATURAL TANGENTS. Leg A B. Rad. B. C. 78,7 : 1 : : 89.0 89.0 Nat. Tang. 78.7) 89.0 (: 1.13087=48 31' 787 1030 787 2430 2361 6900 6296 6040 5509 TRIGONOMETRY. 33 PART II. OBLIQUE TRIGONOMETRY. The solution of the first two CASES of Oblique Trigonom- etry depends on the following- PROPOSITION. IN ALL PLANE TRIANGLES, THE SIDES ARE IN PROPORTION TO EACH OTHER, AS THE SINES OF THEIR OPPOSITE ANGLES. 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 sup- plement, which is what it wants of 180. Note, Def. 25. GEOM. CASE I. The angles and one side given, to find the other sides. Fig- 46. Fig. 46. C B In the triangle ABC, given the angle at B 48, the angle at C 72, consequently the angle at A 60, and the side A B 200, to find the sides A C and B C. To find the side AC. As sine A C B, 72 9.978206 : side AB, 200 - 2.301030 : : sine ABC, 48 -9.871073 12.172103 9.978206 side AC, 156 - 2.193897 To find the side B C. As sine A C B, 72 - 9.978206 :side AB, 200 - 2.301030 : : sineB AC, 60 -9.937531 12.328561 9.978206 : sideB C, 182 - 2.260355 34 TRIGONOMETRY. 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 AN- GLE OPPOSITE EITHER OF THE REQUIRED SIDES TO THAT REQUIRED SIDE. Given side 200 ; nat. sine of 72, its opposite angle, 0.951 15 ; nat. sine of ABC 48, 0.74314; nat. sine of B A C 601 0.86617. Thus, 0.95115 : 200 : : 0.74314 : 156. 0.95115 : 200 : : 0.86617 : 182, CASE II. Two sides, and an angle opposite to one of them given, to find the other angles and side. Fig. 47. In the triangle ABC, given the side A B 240, the side B *G 200, and the angle at A 46 30' ; to find the other angles and the side A C. To find the angle A C B. As sideB C, 200 - 2.301030 : sine B AC, 46 30' 9.860562 : : side AB, 240 - 2.380211 12.240773 2.301030 :sineACB,6030' nearly9.939743 Angle at A C 46 30' 60 30 107.00 Sum of the three angles 180 Sum of two Angle at B 107 73 The side A C 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 op- posite a given lesser side, then the angle found by the operation may be either obtuse or acute. It ought therefore to be mentioned which it is, by the conditions of the question. TRIGONOMETRY. 35 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 an- gle, 0.72537, the other given side 240. As 200 : 0.72537 : : 240 : 0.87044=60 30'. CASE III. Two sides and their contained angle given, to find the other angles and side. Fig. 48. The solution of this CASE depends on the following PROP- OSITION. IN EVERY PLANE TRIANGLE, AS THE SUM OF ANY TWO SIDES ISTO THEIR DIFFERENCE, SOIS 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 SUBSTRACT THE HALF DIFFERENCE FROM THE HALF SUM, AND YOU WILL HAVE THE LESSER ANGLE. In the triangle ABC, given the side A B 240, the side A C 180, and the angle at A 36 40', to find the other angles and side. Side A B - 240 A B - 240 A C - 180 AC - 180 Sum of the two sides 420 Difference 60 The given angle BAG 36 40', substracted from 180, leaves 143 20', the sum of the other two angles, the half of which is 71 40'. TRIGONOMETRY. As the sum of two sides, 420 : their difference 60 - : : tangent half unknown ang. 71 40' : tangent half difference, 23 20' nearly The half sum of the two unknown angles, The half difference between them, Add, gives the greater angle A C B Substract, gives the lesser angle ABC The side B C may be found by CASE I. or II. 2.623249 1.778151 10.479695 12.257846 2.623249 9.634597 71 40' 23 20 95 00 48 23 CASE IV. The three sides given, to find the Fig. 49. A 105 B B The solution of this CASE depends on the following PROP- OSITION. 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 PERPEN- DICULAR 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 difference substracted from the half sum will be the lesser segment. The triangle being thus divided, becomes two right angled triangles, in which the hypothenuse and one leg are given to find the angles. TRIGONOMETRY. 37 In the triangle ABC, given the side A B 105, the side A G 85, and the side B C 50, to find the angles. Side A C - 85 AC 85 B C - 50 B C 50 Sum of the two sides 135 As the longest side A B, 105 : sum of the other two sides, 135 : : difference between those sides, 35 : difference between the segments, 45 Half the side A B Half the difference of the segments Add, gives the greater segment A D Difference 35 2.021189 2.130334 1.544068 3.674402 2.021189 1.653213 - 52.5 - 22.5 - 75.0 Substract, gives the lesser segment B D 30.0 Thus the triangle is divided into two right angled trian- gles, ADC and B D C ; in each of which the hypothenuse and one leg are given to find the angles. To find the angle D C A. As hyp. A C, 85 : radius : : seg. A D, 75 1.929419 10.000000 1.875061 11.875061 1.929419 : sine DC A, 61 56' 9.945642 To find the angle D C B. As hyp. BC, 50 - 1.698970 : radius - - 10.000000 ::seg. B D, 30 - 1.477121 11.477121 1.698970 : sine DCB, 36 52' 9.778151 The angle D C A 61 56', substracted from 90, leaves the angle C A D 28 4'. The angle D C B 36 52', substracted from 90, leaves the angle CBD53 16'. The angle D C A 6 1 56', added to the angle D C B 36 52', gives the angle A C B 98 48'. This CASE may also be solved according to the following PROPOSITION. IN EVERY PLANE TRIANGLE, AS THE PRODUCT OF ANY TWO SIDES CONTAINING A REQUIRED ANGLE IS TO THE PRODUCT 4* 38 TRIGONOMETRY. OF HALF THE SUM OF THE THREE SIDES, AND THE DIFFER- ENCE 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 find 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 in- crease its price, by inserting any thing not particularly con- nected with that art. It is recommended to 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. In- deed no one who is ignorant of TRIGONOMETRY, can be an accomplished surveyor. SURVEYING. SURVEYING is the art of measuring, laying out, and di- viding land. PART I. MEASURING LAND. The most common measure for land is the acre ; which contains 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 contains 16 square rods, and ten 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 links, 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 remaining number add 50 to the link : Thus, 17 two rod chains and 42 links makes 8 four rod chains and 92 links. PROBLEM II. To reduce two rod chains to rods and de- cimal parts. * As there is no standard of long measure established by the General Government, the standard yard, which is kept in the State Treasurer's office, is adopted as a standard of long measure for Connecticut. Each chain 6f 2 rods, should be 11, and each of 4 rods, 22 yards, in length. 40 SURVEYING. RULE. Multiply the chains by 2, and the links by 4, which will give hundredths 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 one to the number of rods, and multiply the excess by 4: thus, 15 two rod chains and 38 links make 31.52 rods. PROBLEM 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, cutoff 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 1 15)63274 square links. 4 2)53096 40 21)23840 Answer. 15 acres 2 roods and 21 rods. PROBLEMS for finding the area of right lined figures, and also of circles. SURVEYING. 41 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 VIII. To find the area of a rhombus or rhomboid. RULE. Drop a perpendicular from one of the angles to its opposite side, and multiply that side into the perpendicu- lar; 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 maybe called the base; then mul- tiply the base by half the perpendicular, or the perpendicu- lar by half the base ; the product will be the area. Or, mul- tiply the whole base by the whole perpendicular, and half the product will be the area. RULE 2. If 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 product will be the area. RULE 3, When the three sides of a triangle are known r the area may be found arithmetically, as follows : Add together the three sides ; from half their sum substract 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 24, 20, and 18 chains. Demanded the area. 24+204-18=62, the sum of the three sides, the half of which is 31. From 31 substract 24, 20, and 18; the three remainders will be 7, 11, and 13. 31X7=217; 217x11=2387; 2387X1 3= 3 1031, 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 multipli- cation of their corresponding numbers ; and as the number answering 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 * Better expressed thus. From half the sum of the sides substract each side separately. Multiply the half sum and the several re- mainders together, and the square root of the product will be the area. ED. 42 SURVEYING. the logarithms of the three remainders, be added together, the number corresponding to one half the sum of those loga- rithms will be the area of the triangle. The half sum, 31 1.491362 The first remainder, 7 - 0.845098 The second remainder, 1 1 1.041393 The third remainder, 13 1.11 3943 The square of the area, 31030.083 nearly 4.491796 Area, 176 square chains 150 square links 2.245898 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 substract the lo- garithm 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 Substract radius - - 10.000000 Double area, 4200 rods nearly 3.623403 Answer, 2 100 rods. NOTE. Radius may be subtracted by cancelling the left hand figure of the index, or substracting 10, without the trouble of setting down the ciphers. BY NATURAL SINES. Multiply 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. SURVEYING. 43 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 product will be the area. PROBLEM 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, multi- ply the diagonal by half the sum of the two perpendiculars let fall upon it, or the sum of the two perpendiculars by half the diagonal, 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 PROS. 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, which diago- nals 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 drawn, pro- vided they do not intersect each other, the whole area will be found the same. PROBLEM XIII. Respecting circles. RULE 1. If the diameter be given, the circumference may be found ,. 7 96 at 2.36 4.28 1 e end 0.20 0.36 0.96 0.30 F G. S. 27 40' E. 7 06 at 2. the end 1.20 0.24 016 CD. N. 62 0' W 4.6b at 4 34 0.30 030 GA. S.^WW. 6.48 at 3.80 the end 0.80 0.40 DE. N.I PUT W. 4.20 TO PROTRACT THIS FIELD. Draw the stationary lines according to the directions in CASE IV. From A make an offset of 56 links to I ; measure from A to H 540 links, and make the offset, H 2, 140 links; measure from A to I 826 links, and make the offset I 3, 36 links ; at B make the offset B 4, 36 links. Proceed in the same manner round the field, and connect the ends of the off- sets by links, which will represent the boundaries of the field. 62 SURVEYING. TO FIND THE AREA. Find the area within the stationary lines as before taught j then of the several small trapezoids, rectangles, and trian- gles, made by the stationary lines, offsets, and boundary lines, and add the whole together : thus, add 56 links, the offset A 1, to 140 links, the offset H 2, and multiply their sum, 196, by half 540, the length of the line A H, and the product 52920 square links, will be the area of thetrapezoid A H 21 ; again, add 140, the offset H 2, to 36, the offset I 3, and multiply their sum, 176, by half 286, the length of the line H I, and the pro- duct, 25168 square links, will be the area of the trapezoid H I 3 2. Proceed in the same manner to calculate the area of all the trapezoids, triangles, &c. CASE X. TO SURVEY A FIELD BY TAKING OFFSETS BOTH TO THE RIGHT AND LEFT; THAT is, WITHIN AND WITHOUT THE FIELD, AS OCCASION SHALL REQUIRE, IN CONSEQUENCE OF THE STATIONARY LINES CROSSING THE BOUNDARY LINES J ALSO, BY INTERSECTIONS, THAT IS, TAKING THE BEARING OF AN INACCESSIBLE CORNER FROM TWO STATIONS." - The directions given in the preceding CASE, together with the following FIELD BOOK, will show the learner how to sur- vey a field like the following, and also to protract it when surveyed. SURVEYING. FIELD BOOK. See Fig. 61. Fig&l. Offsets, to the Left. Bearing and distance. Offsets, to the Right. Remarks. Ch. L. 1.12 3.40 1.25 Ch. L. AB. N. 880'W. 22.12 at 4.25 7.40 13. Ch. L. A tower bears from A. N. 48 W. 1.45 B C. N. 27 45' W. 21.12 at 4 10 10.25 15. 1.20 1.15 From B the tower bears N. 38 30' E. 01. S. 8215 / E. 5.45 1 2 N. 70 E. 13.25 2 D. N. 20 E. 3.36 From C go into the field to 1, on account of some impediment on or near the boundary line. AtD, you get into another cor- ner of the field. D P. S. 35 0' E. 15.15 E. an inaccessible cor- ner, bears from. D. S. 65 30' E. 2.20 2.32 FA. S. 15 15' E. 15.10 at 1.20 7.45 12.25 1.36 E. the inaccessible corner, bears from F. N. 4 W. NOTE. To draw a tree, house, tower, or any other remarkable ob- ject, in its proper place, in the plot of a field from any two stations, while surveying the field, take the bearing of an object, and the intersection of the lines, which represent the bearings, will deter- mine the place of the object, in the same manner that the tower is drawn in ihe figure. 64 SURVEYING. TO FIND THE AREA OF THE ABOVE FIELD. Find the area within the stationary lines, and then of the several small trapezoids, &c. remembering 1 to distinguish those without the stationary lines from those which are with- in. Substract the area of those within the stationary lines from the area of those without, and add the remainder to the area contained within the stationary lines ; the sum will be the whole area of the field. [Or, add the areas of those without the stationary lines, to the area contained within those lines, and substract from the sum the areas of the several triangles, trapezoids, &c. with- in the stationary lines.] SECTION III. RECTANGULAR SURVEYING, OR AN ACCURATE METHOD OF CALCULATING THE AREA OF A FIELD ARITHMETICALLY, FROM THE FIELD BOOJC, WITHOUT THE NECESSITY OF PROTRACTING IT, AND MEASURING WITH A SCALE AND DIVIDERS, AS IS COMMONLY 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 EX- AMPLE. SURVEYING. 65 EXAMPLE I. II i i i i 1 1 1 1 91fi6.7040 ci 1 1 1 i I co' w s 11 1600.0910 1 1 I l I 1 1 1 (L,^ c* 1 S 1 1 ffi g P 5* l i J_ i- TT i'- oS i {: 5 "" oci 55 o o H i 11 II 11 cioi | y o ^ 02 I i i 11 00 ^ s CO t5 * ~H r-J \ , 1 ; 1 o QO s s o i js 1 O N. 15 0' E. H w w o oi 1 s c^ No. ' H * rj to ^ 00 19143.9019 sum of South Areas. 4245.4016 North do. 2)14898.5003 Acres 744)92501 4 Roods 3)70004 40 Rods 28)00160 Acres Roods Rods Area 7 44 3 28 66 SURVEYING. 2. Calculate by RIGHT ANGLED TRIGONOMETRY, CASK I, or find by the table of difference of latitude and departure,* or by the table of natural sines,f the northing or southing, easting or westing, made on each course, and set them down against their several courses in their proper columns, marked N. S. E. W. NOTE. To determine whether the latitude and departure for any particular course and distance are accurately calculated, square each of them ; and if they are right, the sum of their squares will equal the square of the distance, for the following reason: the la- titude and departure represent the two legs of a right angled tri- angle, and the distance the hypothenuse; and it is a mathematical truih, that the square of the hypothenuse of any right angled tri- angle, is equal io 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 will equal that of the westings. If, upon adding up the respective columns, these are found to differ 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 smafl, 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 northings and southings, and the eastings and westings, may be equalled by balancing them, as follows : substract 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 substracted, be divided among the several courses, according to their length. In EXAMPLE I., 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 exceed 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 ;] like- wise, take 12 links from the westings, and add it to the east- ings. 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 * For an explanation of this table, and the manner of using it, see the remarks preceding the table. t See the remarks preceding the table of natural sines. SURVEYING. 67 fourth 2 ; take fronvthe first westing 5 links, from the second 4, and from the third 3. [These are the proportional correc- 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 de- parture 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 col- umn is formed by a continual addition of the eastings, and substraction of the westings ; or by adding the westings and substracting the eastings: see EXAMPLE 1. The first easting, 20.74, is set for the first number in the departure 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 prece- ding course; from this substract 39.95, the first westing, and it leaves 44.77, for the sixth course ; from this substract 23.75, the second westing, and it leaver 21.02, for the sev- enth course; from this substract 21.02, the last westing, and it leaves 0.0 to be set against the last course, which shows that the additions and substractions have been accurately made. For as the eastings and westings equal each other, it is evident that the one being added and the other substract- ed, there will be in the end no remainder. 5. The next step in the process is to form a second de- parture column, the numbers in which show the sum of the meridian 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 se- cond and third; and for the fourth the third and fourth ; and so on, until the column be completed. See EXAMPLE I. 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, and it 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 * See Appendix, No. 5. 63 SURVEYING. 129.49 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 work is thus far prepared, multiply the several numbers in the second departure column by the northings or southings standing against them respectively; place the 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 substract the less from the greater ; the remain- der will be double the area of the field in square rods or square chains and links, which ever 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. Demonstration of the preceding rules. See Fig. 62. and EXAMPLE 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 A 2 B, as is evident from the rule to find the area of the triangle, PROB IX. Rule I. This num- ber is to be placed for the first number in the column of north areas. The line 3 C represents the sum of the eastings made by the first and second courses, which is 45.12, the se- cond number in the first departure column; if to this you add 20.74, the length of the line 2 B, you have 65.86, which is the second number in the second departure column, and which represents the sum of the two lines 3 C and 2 B. These two lines, with the line 2 3, which represents the SURVEYING. 69 f northing made by the second course, and the line B C, one of the sides of the field, form a right angled trapezoid. Now, by the rule to find the area of such a trapezoid, see PROB. X., 65.86X31.66=2085.1276, double the area of the trapezoid 2 B C 3. Place this product for the second number in the column of north areas. To the line 3 C add C D 30.04, that easting'made by the third course, and you have 75.16, which is the sum of the eastings made by the three first courses, and the third num- ber in the first departure column. To this add 9.56, the easting of the fourth course, and you have 84.72, the length of the line 1 E, which represents the sum of the eastings made by the four first courses, and is the fourth number in the first departure column. These two, viz., the lines 3D 75.16, and i E 84.72, added together, make 159. 88, the fourth number Jn the second departure column ; which being multi- plied 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 1 E D3. The number thus pro- duced is 7858.1020, which is to be placed for the first num- ber in the column of south areas. Tha fifth 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 se- cond departure column. This, being multiplied by 54.10, the length of the line E F, 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 1 E F4, which is 9166.7040, the second number in the column of south areas. From the line 4 F 84.72, substract 39.95, which is a west course and it leaves 4 G 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 substract 23.75, the westing made by the 7th course, and you have 21.02, the length of the line 5 H, which is the meridian distance at the end of the seventh course, and the seventh number in the first departure column. The line 4 G 44.77, added to the line 5 H 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 4 G H 5, which is 21 19.0959, the third number in the column of south areas. The line H 5, 21.02, is the westing of the last course, and 7 70 SURVEYING. the last number in'the second departure column. This being multiplied by 26.65, length of the line 5 A, and the northing of the last course, produces 560.1830, which is double the area of the triangle A 5 H, and the last number in the col- umn of north areas. NOTE. It will be observed that against the third and sixth courses there are no areas ; the reason is, that these courses being one east and the other west, there is no northing or southing to be multiplied into them ; regard can therefore be had to them only in forming the departure columns. By inspecting 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 A 2 B, the trapezoid 2 B C 3, and the triangle A 5 H, all of which are without the boun- dary lines of the field : also, that the three south areas rep- resent double the area of the trapezoid 3 DE 1, the parallelo- gram 1 E F 4, and the trapezoid 4 G H 5 ; and that these in- clude not only the field, but also what was included in the north areas. Therefore the north areas substracted from the south, the remainder will be double the area of the field, con- tained within the black lines. ADDITIONAL DIRECTIONS AND EXPLANATIONS. The northings and southings may be added and substract- ed instead of the eastings and westings ; then there will be two latitude columns instead of departure columns, and the numbers 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 and south, the distance must be set in the north or south column ; when east or west, in the east or wesfcolumn. There will therefore sometimes be no number to be added to or substracted from the num- ber last set in the latitude or departure column ; then the number last placed in the column must be brought down and set against such course ; as in EXAMPLE 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 against such course there will be no area, as in EXAMPLE I. at the 3d and 6th courses. When the northings or southings, eastings or westings, beginning at the top, will not admit of a continual addition of the one and the substraction of the other, without run- SURVEYING!. 71 ning 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 substraction ; and when you get to the bottom, go to the top, and you will end in cipher at the course next above that where you began. EXAMPLE II. No. 1. 2. 3. 4. 5. 6. 7. a 9. 10. IL 12. 13. Courses. Diet, rods. N. S. E. W. 112 11.7 77.9 62.1 1 Dep. Col. 46.0 67.4 114.1 102.9 91.2 144.1 158.5 223.3 145.4 83.3 83.3 37.2 00.0 2 Dep. Col. North areas. South areas. N. 36 45' E. N. 22 30 E. & 7645 E. S. 15 00 W. S. 16 45 W. N. 75 00 E. N. 20 30 E. East S. 33 30 W. 8. 76 00 W. North S. 84 00 W. N. 53 15 W. 76.8 56.0 48.0 43.4 40.5 54* 41.2 64.8 141.2 64.0 36.0 46.4 46.4 61.5 51.7 4.2 38.6 36.0 27.8 ii.o 41.9 38.8 117.7 15.5 46.0 21.4 46.7 529 14.4 64.8 46.0 113.4 181.5 217.0 194.1 235.3 302.6 381.8 368.7 228.7 166.6 120.5 37.2 2829.00 5862.78 3341.26 11680.36 1996.50 9092.30 7531.08 5997.60 1034.16 43395.99 3544.85 590.45 4.9 46.1 37.2 110 Acres, 2 Roods, and 23 Rods. DEMONSTRATION. Fig. 63. The area standing against the 1st course, is the triangle 12t, lying without the field. 72 That against the 2d course is the trapezoid 23