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How to cast up off-sets by the pen.
Plate XI. fig. 2.
1, 2-1f=2f-1c-fe, le-deed.
Then ld x įda=lda, by prob. 6, page. 183, and į ed x da + fc=befc, and 2f xifc=cf9; the sum of all which will be labc2l; 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 2ihg2 of iksi, 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 of 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 22C. 12L. or 110. 12L. of fourpole chains. See the figure. C. L. C. L.
C. L. From 11.12=1,2 6.50=lf 3.90=le Take 6.50=lf 3.90=le 2.25=1d
2.60 = ef
Ch. L. 1d-2.25 X 32 L. half the first off-set = .7200 ed-1.65 x 1C. 26 L.; the sum of the 1st & 2d 2.0790 ef-2.60 x 1C. 32L. the sum of 2d and 3d 3.4320 AM1 half the last off-set 1.7094 Content of let fl-sets on the first dist. in square four-pole chains
In like manner the rest are performed. The sum of the left hand off-sets will be And the sum of the right hand ones
Excess of lef: band off-sets in squ. 4 pole C. 10.4031
Excess of left hand off-sets above the right hand ones, IA. OR. 6P. to be added to the area within the stationary lines.
How to find the area of a piece of Ground by in
tersections only, when all the angles of the field can be seen from any two Stations on the outside of the ground.
Plate XII. fig. 1.
ET ABCDEFG 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, 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 re- , presented by the lines HA, HB, HC, HD, &. 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 beari's 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 point being joined by right lines, will give the plan ABCDEF GHA required.
Bea. 180 Dif. 28C. of the Sta. H. and I.
No, Bear. Bear.
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 partý, 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, rohich
was taken by intersection.
In the triangle AIH, the angles AHI, AIH, 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 g; if therefore the several perpendiculars be supposed to be drawn into the scheme (which are here omitted to prevent confusion arising from a multiplicity of lines) it is plain that if from BCDeb,
there be taken bBAGFeb, the remainder will be the map ABCDEFGA.
As before half the sum of Bb, and Cc multiplied by bc, will be the area of the trapezium bBCC; after the same manner, half the sum of Cc, and Dd, multiplied by cd, will give the area of the trapezium CDd; and again, half the sum of Dd, and Ee multiplied by de, gives the area of the trapezium dDĒe ; 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 Aa multiplied by ah, 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 fFEe, which are found in the like manner, and you will have the figure BAGFEeb, and this taken from 6BCDeb, will 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