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

ry, I abc 2,

In the same manner you may find the area of 2ihg2, 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 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 is 22C. 12L. or 11C. 121. of four-pole ghains. See the figure.

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Ch. L. Id=2.25*32L, half the first off-set= .7200 ed=1.65X1C. 26L.: the sum ofthe 1st and 2d 2.0790 ef=2.60X1C. 32L. the sum of 2d 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 1.04031

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Excess of left

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

SECTION V.

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 out side of the ground.

PL. 12. fig. 1.

LET 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 represented by the lines HA, HB, HC, HD, &c. Again fix your instrument at I, and take bearings to the saine 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. .
A

2614 331)
B2654 317.
С

3074
D 2384 289
E 215: 262
F 208 2863
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 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 perperdiculars 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 bBCD Eeb, there be taken bBAG Feb, 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 cČDd; and again, half the sum of Dd, and Ee multiplied by de, gives the area of the trapezium dD Ee; and the sum of these three trapezia will be the area of the figure bBC Deb.

Again, in the same manner, half the sum of Bb and Aa multiplied by ah, will give the area of the trapezium Bb Aa ; and half the sum of aA, and gG, by ag, gives the trapezium a AGg ;. to these add the trapezia gG Ff, and fFEe, which are found in the like manner, and you will have the figure bBAG FEeb, and this taken from bBCDeb, will leave the map ABCDEFGA. Q. E. F.

It will be sufficient to protract this kind of work, and from the map to determine the area as well as in plate 10. 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 part of a castle, house, or cock of hay, &c. or if all These be wanting, to a long staff with a white sheet or napkin set thereon, to render the object more conspicuous, and let this be placed on the summit of the land, and let the respective intersections so

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