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The foregoing field-book may be otherwise kept,




Remarks and intersection.

Deg. Offset


Ch.L. 318 Int. to a tower

1 358 1.12 4.25

3.40 7.40
1.25 | 13.00


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1, 2-1f=2f-le=fe, lemldeed.

Then ldx ida=lda, by prob. 6, page 183, and 1 ed da+fc=befc, and 2f * } fc=cf9; the sum

of all which will be labc21; the area contained between the stationary line 1, 2, and the bound

ary, I abc 2.

In the same manner you may find the area of Lihg2 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 retrogade order.

2. Multiply the last of these remainders by I 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 TIC. 12. of four pole-chains. See the figure.

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Ch. L.
id=2.95 x 32L. half the first off-set=

7200 ed=1.65X1C. 26L.; the sum of the stand 2d 2.0790 ef=2.60 x IC. 32L. I the sum of 21 and 3d=3.4320 2f=4.62x37L. half the last off set= 1.7094 Content of left off-sets on the first dist. in square four-pole chains

7.9404 In like manner the rest are performed. 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 1.02031

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Excess of left hand off-sets above the right hand ones, 1A. OR. 6P. to be added to the area within the stationary lines.


To find the area of a piece of Ground by intersections only, when all the angles of the field can be seen from any two stations on the outside of the ground.

PL. 12. fig. 1.

LET ABCDEFG be a field, H and I two

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 Hand 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 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 ABCDEFGHA required.

Bear. 180 Dis. 28C. of the Sta. H and I.

| No. Bear. Bear.

А 261 3311
B 265 3171
С 248 3071
D 238' 289
E 215| 262
F 208: 2861
G 220 300

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, AIH, and the base XI 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, Gp, 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 BCD Eeb, there be taken bB.AG Feb, the remainder will be the map ABCDEFGA.

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