13. Make a rough sketch of the field ABC, and calculate its area from the accompanying field-book; the chainlines are all within the field.

14. Make a rough sketch of the field ABCD, and calculate its area from the accompanying field-book. The side AD was not measured, but it was a straight line without offsets.

15. Lay down the field ABCDEFG, and find its area from the following dimensions:

16. Make a rough sketch of a field from the following field-book: find the area of the whole field, assuming that the triangle BCD contains 416732 square links, and that the piece at C between the boundary and the offsets to CD and CB respectively contains 300 square links.

XLVII. PROBLEMS.

417. In our account of Land Surveying we have confined ourselves to illustrating the use of the chain and the cross. In surveys of great extent or of extreme accuracy, instruments are employed for measuring angles, and the calculations are effected by the aid of the science of trigonometry. With these resources also problems relating to the distance of an inaccessible object are usually solved. Nevertheless some of these problems may be treated sufficiently for practical purposes in a simpler manner: we will give examples.

418. To find the breadth of a river.

Let A be an object close to the river; B an object on the other side, directly opposite to A, and also close to the river.

Draw a straight line AC at right angles to AB, of any convenient length, and fix a picket at C. Produce the straight line AC to a point D, such that CD is equal to AC.

B

From D draw a straight line at right angles to AD, and in it find the point E so that BCE may be in a straight

line.

Then the triangles CAB and CDE are equal in all respects, and DE is equal to AB. We can measure DE, and thus we find the length of AB, that is, the breadth of the river.

419. The preceding Article requires us to be able to draw a straight line at right angles to another straight line; we have shown in Chapter XLV. how this may be done. We will however now solve the problem by another method which does not require a right angle.

420. To find the distance between two points, one of which is inaccessible.

Let A and B be the two points, B being inaccessible on account of a river, or some other obstacle.

From A measure any straight line AC. Fix a picket at any point D in the direction BČ. Produce CA to F, so that AF may be equal to AC; and produce DA to E so that AE may be equal to AD. Fix pickets at

B

Fand E. Then find the point G at which the directions of BA and FE intersect; that is, find the point from which A and B appear in one straight line, and E and F appear in another straight line.

Then the triangles DAB and EAG are equal in all respects, and GA is equal to AB. We can measure GA, and thus we find the length of AB..

421. In Articles 418 and 420 we assume that we have the command of sufficient ground to enable us to trace a straight line of the same length as that which we wish to measure. These methods would be practically inapplicable if the inaccessible objects were at a considerable distance. We shall therefore give a solution which could be used in such a case.

422. To find the distance between two points, one of which is inaccessible and remote.

Let A and B be the two points, B being inaccessible and remote.

Measure any length AC in the direction of BĂ; and from C in any convenient direction measure CD equal to CA. Take two strings, each equal to CA; fix an end of one at A, and an end of the other at D. Then stretch the strings out, so that they form straight lines, and so that their other D

E F

ends shall meet at a point; let E be this point. Then ACDE is a rhombus. Place a picket at F, the point at which the directions AE and BD intersect.

The triangles FAB and FED are similar. Therefore EF is to ED as AF is to AB. Thus if we measure EF and FA, we can find AB from this proportion.

XLVIII. DUODECIMALS.

423. Examples relating to square measure and to solid measure are sometimes worked by a method which is called Cross Multiplication or Duodecimals. This method is found convenient in practice, and the theory of it is instructive; so that the explanation which we shall now give deserves attention. We shall first consider the case of square measure.

424. The student of course knows perfectly well what is meant by a square foot and what is meant by a square inch; he must now become familiar with another area, namely, a rectangle which is twelve inches long and one inch broad: we will call this a superficial prime.

Thus we have the following addition to the Table of square measure:

12 square inches make 1 superficial prime,

12 superficial primes make 1 square foot.

425. Any number of square inches greater than 12 can be separated into superficial primes and square inches. Thus, for example,

17 square inches=1 superficial prime 5 square inches, 32 square inches=2 superficial primes 8 square inches, 54 square inches=4 superficial primes 6 square inches. Any number of superficial primes greater than 12 can be separated into square feet and superficial primes. Thus, for example,

19 superficial primes 1 square foot 7 superficial primes,

45 superficial primes

primes,

=

3 square feet 9 superficial

54 superficial primes = 4 square feet 6 superficial primes.

T. M.

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