How to cast up off-sets by the pen. Plate XI. fig. 2. 1, 2—1f,=2f, 1f,-le-fe, 1e-1d-ed. = Then 1dx da 1da, by prob. 6. page 183, anded × da +ebadeb by the doctrine of trapezia; also fexeb+fc-befc, and 2fxfc=cf2; the sum of all which will be 1abc.21; the area contained between the stationary line 1, 2, and the boundary, 1 abc 2. In the same manner you may find the area of 2ibg2 of ik3i as well as what is without and withinside of the stationary line 7, 1. If therefore the left hand off-sets exceed the right hand ones, it is plain, the excess must be added to the area within the stationary lines, but if the right hand off-sets exceed the left hand ones, the difference must be deducted from the said area; if the ground be kept on the right hand as we have all along supposed; or in words, thus; To find the contents of off-sets. 1- From the distance line, take the distance to the preceding off-set, and from that the distance of the one preceding it, &c. in four-pole chains; so will you have the respective distances from off-set to off-set, but in a retrograde order. 2. Multiply the last off these remainders by the first off-set, the next by the sum of the first and second, the next by half the sum of the second and third, the next by half the sum of the third and fourth, &c. The sum of these will be the area produced by the off-sets. Thus, in the foregoing field book, the first stationary line is 22C. 12L. or 11C, 12L. of fourpole chains. See the figure. Ch. L. From 11.12=1,2 Ch. L. 4.62=2f 1d=2.25×32L. half the first off-set, = 1.7094 ed=1.65×1C. 26L. the sum of the 1st & 2d 2.0790 ef=2.60x 1C. 32L. the sum of 2d & 3d 3.4320 2f=4.62×37L. half the last off-set Content of left off-sets on the first dist. in square four-pole chains In like manner the rest are performed. 7.9404 The sum of the left hand off-sets will be 14.0856 And the sum of the right hand ones 3.6825 Excess of left hand off-sets in squ. 4 pole C. 10.4031 Acres 1.04031 .16124 4 Perches 6.4496 Excess of left hand off-sets above the right hand ones, 1A. OR. 6P. to be added to the area within the stationary lines. OF INTERSECTION S. How to find the Area of a Piece of Ground by Intersections only, when all the Angles of the Field can be scen from any two Stations on the Outside of the Ground. Plate XII. fig. 1. L' ET ABCDEFGA be a field, H and I two places on the outside of it, from whence an object at every angle of the field may be seen. Take the bearing and distance between H and I, and set that at the head of your field-book, as in the annexed one. Fix your instrument at H, from whence take the bearings of the several angular points, A, B, C, D, &c. as they are here represented by the lines HA, HB, HC, HD, &c. Again fix your instrument at I, and take bearings to the same angular points, represented by the lines IA, IB, IC, ID, &c. and let the first bearings be entered in the second column, and the second bearings in the third column of your field-book; then it is plain that the points of intersection, made from the bearings in the second and third columns of every line, will be the angular points of the field or the points A, B, C, D, &c. which points being joined by right lines, will give the plan ABCDEFGA required. Bea. 180 Dis. 28C. of the Sta. H and I. The same may be done from any two stations within-side of the land, from whence all the angles of the field can be seen. This method will be found useful in case the stationary distances, from any cause, prove inaccessible, or should it be required to be done by one party when the other in whose possession it is, refuses to admit you to go on the land. To find the content of a field by calculation, which was taken by intersection. In the triangle AIH, the angles AHI, AII, and the base HI being known, the perpendicular Aa, and the segments of the base Ha, AI may be obtained by trigonometry: and in the same manner all the other perpendiculars Bb, Cc, Dd, Ee, Ff, Gg, and the several segments at b, c, d, e, f, and gif therefore the several perpendiculars be supposed to be drawn in the scheme (which are here omitted to prevent confusion arising from a multiplicity of lines) it is plain that if from bBCDEeb, bBCDEeb, there be taken bBAGFeb, the remain der will be the map ABCDEFGA. As before, half the sum of Bb, and Cc, multiplied by bc, will be the area of the trapezium 6BCc; after the same manner, half the sum of Cc, and Dd, multiplied by cd, will give the area of the trapezium cCDd; and again, half the sum of Dd, and Ee multiplied by de, gives the area of the trapezium dDEe; and the sum of these three trapezia will be the area of the figure bBCDeb. Again, in the same manner, half the sum of Bb, and a multiplied by ab, will give the area of the trapezium bВAa; and half the sum of aA, and gG, by ag, gives the trapezium aAGg; to these add the trapezia gGFf, and ƒFEe, which are found in the like manner, and you will have the figure bВAGFЕeb, and this taken from 6BCDeb, wil leave the map ABCDEFGA. Q. E. D. It will be sufficient to protract this kind of work, and from the map to determine the area, as well as in plate X. fig. 3. to find the areas of the pieces 3, 4, 5, 6, 3, and 6, 7, 7, 6, from geometrical constructions. How to determine the station where a fault has been committed in a field-book, without the trouble of going round the whole ground a second time. From every fourth or fifth station, if they be not very long ones, or oftener if they are, let an intersection be taken to any object, as to any particular |
