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PROBLEM VIII.

To furvey a field by the theodolite.

I. From a point within the fieldPlace the theodolite at 0, (Fig. laft Prob.) and turn it about till the fixed sights point to any object, as A ; screw the instrument fast, and turn about the moveable index till through the moveable lights you see B, and note the degrees cut on the limb of the instrument in a field-book, or rather on an eye-draught, then turn the index in the directions C, D, E, and record the angles as above. Then measure the lines OA, OB, OC, OD, OE, and it is done.

2d, From one of the angles. Choose any angle for a station, from whence all the other angles may be seen. Suppose angle A, as in fig. Prob. 7. Find the number of degrees in the angles BAC, CAD, DAE, and mark them down upon the eye-draught, each against the corresponding parts of the field; then measure the straight lines AB, AC; AD, AE; mark these upon the eye-draught; so the plan may be made out, and the area found, as shewn above.

PROBLEM IX

To furvey a field by two stations.

By this method, grounds may be planned and surveyed without entering úpon them. This is performed by choosing two stations, either within or without the field, from whence all the angles, ponds, houses, cross-hedges, roads, rivers, &c. &c. may be seen. Either the theodolite or plain table may be used.

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Let ABCDEF be a field whose plan is required, and let the points 1. 2. be eminences, from which a sufficient view of the field is obtained. First, Place the instrument at 1, and take the angles AIB, BIC, CID, DIE, EIF, FIA, and draw lines in the several directions of these angles, and any other remarkable object. Then measure the distance from 1 to 2, and place your instrument at 2; thence drawing lines in the directions of all the angles, and of such other objects as have been noted from the former station. Then thefe obfervations being planned, the intersection of the lines will point out the several corners and objects required. When two stations are not sufficient, three or more may

be chosen, as the surveyor hall see cause, measuring the distance between each station. And the intersections of the lines point

out the objects, with their proper places on the plan. By this . method very extensive surveys may be taken.

OF DIVIDING, OR LAYING OUT GROUND,

PROBLEM I.

To lay out rectangular ground.

RULE.

Divide the given area by the given side, and the quotient will be the fide required.

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EXAMPLE, I.

What length of a rectangular field, whose breadth is 409 links, will make 3 acres 2 roods?

400)3.50000

Anf. 875

Ex. 2. What length of a ridge, s ells broad, will make 15 falls ?

Anf. 674 ells. Ex. 3. What length of a ridge, 40 feet feet broad, will make 3 roods of land ?

Anf. 816 feet. Ex. 4. A field contains 32 acres 3 roods Scots, the breadth being 5 10 links, required the length. Anf. 6422 links nearly. Ex.

5.

The length of a rectangular field, being 6575 links, required its breadth to contain 328 acres 3 roods.

Anf. 5000 links. Ex. 6. A square field of 15 acres 2 roods 20 poles: It is required to find the length of its side. Anf. 1250 links.

Ex. 7. Required the side of a square field in yards, whose content is 30 English agres.

Anf. 831 nearly. Ex. 8. What length of a rectangular field will make 25 Englifh acres, the breadth being 193} yards ? Anf. 625 yards.

PROBLEM II.

To lay out a triangular field.

RULE I.

When the base is given, to find the perpendicular: Divide the area by the base, and twice the quotient will give the perpendicular.

RULE 2. When the perpendicular is given, to find the base : Divide the area'by twice the perpendicular, and the quotient is the base.

Rule 3. When any part of a given triangle is to be cut off by a line parallel to one of the Edes, it must be remembered that fimilar surfaces are to one another as the squares of their corresponding sides; and wie verfa.

EXAMPLE

EXAMPLE I.

Required the perpendicular breadth of a triangular field, the base being 520 links, and content 9 acres.

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Ex. 2. A triangular field of 630 acres is to be divided equally between two farmers; the base measures 6000 links, and the march is to be drawn parallel to the base ; required how much of the perpendicular will fall to each.

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Note. The greater part of the perpendicular will fall to him whose division lies towards the vertex.

Ex. 3. Suppose a field in the form of a right-angled triangle, whose base is 2500, and perpendicular 3000 links of the English chain, and that a hedge is planted parallel to the perpendicular, cutting off 20 acres, required the expence of planting the hedge, at is. 6d. per yard.

Anf. 361. zs. 0 d. Ex. 4. How long is the base of a field of 35 acres 3 roods 20 poles, the perpendicular being 3550 links ?

Anf. 2021 to nearly.

PROBLEM

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