Εικόνες σελίδας
PDF

of the field-book, annexing thereto the words Int, for boundary : the crossing or inter section of these two bearings will determine the point p, and of course the boundary 6p7 is also determined.

If your view will then reach in the first station, take its bearing, stationary line, and offsets as before, and you have the field-book completed. Thus,

[merged small][merged small][ocr errors][ocr errors][merged small][ocr errors][merged small][merged small]

If you would lay down a tower, house, or any other remarkable object in its proper place, from any two stations take bearings to the object, and their intersection will determine the place where you are to insert it, in the manner that the tower is set out in the figure, from the intersection taken at the first and second stations of the above field-book.

A protraction of this will render all plain, on which lay off all your offsets and intersections, and proceed to find the contents by any of the methods in section the fourth.

*

The foregoing Field-book may be otherwise kept, thus:

* . No L. hand Dist R. handl Remarks and Intersection. s. Deg. Offset. h | Offset. - ch. l. * * | ch. l. 318 Int. to a tower - 1 || 358 1.12 || 4.25 3.40 7.40 1.25 | 13.00 22. 12 232; Int, for ditto - - || 2 || 297; 4.10 | 1.20 10.25 | 1.15 * 13.10 0.45 | 15.00 21.21 3 || 1724 5.45 4 200 13.25 5 250 3.36 155] Int, for boundary | 6 || 125 15.15 274 Int. for boundary 7 105 2.20 | 1.20 - 2,32 7.45 11.25 12.25 || 0.36 15.10

[merged small][ocr errors]

and the boundary labcz.

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 offsets exceed the right-hand ones, it is plain the excess must be added to the area within the sta

tionary lines; but if the right-hand offsets 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 offsets.

1. From the distance line take the distance to the preceding offset, and from that the distance of the one preceding it, &c. in four-pole chains; so will you have the respective distances from offset to offset, but in a retrograde order. 2. Multiply the last of these remainders by half the first offset, the next by half 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 offsets. Thus, in the foregoing field-book the first stationary line is 22ch. 12l., or 11ch. 12l. of four-pole chains. See the figure. ch. l. ch. l. ch. l. From 11.12=1, 6.50=1s 3.90=1e Take 6.50=1s 3.90=16. 2.25=10

[ocr errors][ocr errors]

Contents of left offsets on the first distance in * square four-pole chains, } 7.9404

In like manner the rest are performed.

The sum of the left-hand offsets will be 14.0856 And the sum of the right-hand ones 3.6825

Excess of left-hand offets in sq. four-pole chains, 10.4031 w Acres 1,04031

[ocr errors][merged small]

Excess of left-hand offsets above the right-hand ones, 1A. 0R. 6P., to be added to the area within the stationary lines.

SECTION W.

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 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 ABCD, &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.

[merged small][ocr errors][ocr errors][merged small]

The same may be done from any two stations withinside 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 lands."

[ocr errors]

To find the contents 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 perpendiculars, Bb, Cc, Dal, Ee, Ff, Gg, and the several segments at b, c, d, e,f,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 b BCDEeb there bes 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 b BCc ; after the same manner, half the sum of Cc and Dd multiplied by cal will give the area of the trapezium coBd; and again, half the sum of Dd and Ee multiplied by de gives the area of the trapezium dI)Ee; and the sum of these three trapezia will be the area of the figure bBCDEeb.

Again, in the same manner, half the sum of Bb and Aa multiplied by ab will give the area of the trapezium Bb4a, and half the sum of a A and g(; by ag gives the trapezium a AG g; to these add the trapezia gCo Ff and f FEe, which are found in the like manner, and you will have the figure bPAGFEeb, and this taken from b|3CDEeb will leave the map ABCDEFGA. Q. E. I.

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 fieldbook, 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 taken be inserted on the left-hand side of the field-book opposite to the stations from whence they were respectively taken.

In your protraction as you proceed let every intersection be laid off from the respective stations from whence they were taken, and let these lines be continued; if they all converge or

[ocr errors]
« ΠροηγούμενηΣυνέχεια »