Ex. A line was measured 17.55 chains on ground having a continuous rise of 9o; required the horizontal length of the line.

The whole reduction

= 17.55 × 1·23 links,

=21.5865 links;

and this subtracted from 17.55 chains leaves the horizontal length

= 17.334135 chains;

or, at once, the horizontal length

=17.55 × ·9877 chains,

= 17.334135 chains.

It is usual to neglect all quantities below links, unless amounting to half a link.

263. NOTE.-The student who understands a little Trigonometry will know that Ab can be obtained from AB, by measuring the BAb, and looking in a Table of logarithms for what is termed the cosine of BAb. This cosine is the value of the fraction

Ab

AB

For example, if BAU were 18°, its cosine

We have observed that it is often needful to set off a straight line at right angles to another straight line upon the surface of the ground, as in measuring the perpendicular altitudes of triangles, or the lengths of offsets.

For measuring such perpendiculars, when very short, we may with a tape or offset-staff guess, with sufficient accuracy by the eye, the required perpendicular position of the lines; but for long lines, where any error would be more serious, we employ an instrument called

THE CROSS-STAFF.

264. The CROSS-STAFF, in its simplest form, consists

of a circular board, as ABCD, about two inches thick, mounted on a pole about 5 feet high, and indented on its upper surface by two grooves, AC, BD, passing through the centre, 0, at right angles to each other.

The use of it is to set off straight lines at right angles to other

straight lines, as mentioned above.

Thus, for example, (see fig. p. 276) the surveyor, in measuring BC, wishes to determine the point D, where the perpendicular from A meets BC. When he comes to the point, which he supposes to be nearly right, he there fixes his Cross-Staff in the ground, and turns it round, until one of the grooves is exactly over the line BC. He then looks along the other groove for A; and, without turning the board, or deviating from the line BC, he moves the whole instrument to the right, or to the left, until he sees A through the second groove. The foot of the staff is then on the required point D.

Similarly, in (255), the points a, b, c,... from which the offsets had to be measured, would be determined as above, unless the offsets from those points were so short, that it might be considered sufficiently correct to fix their position solely by the eye.

The accuracy of the instrument may thus be tested. Let it be so placed, that through CA, BD, two objects a, b, may be seen respectively: turn the cross half-round, and if a, b, are now seen through BD, AC, respectively, the instrument is correct.

265. The Chain itself also can be used sometimes conveniently for constructing a right angle. For from (43, Part 1.) we know, that the square of the hypothenuse is equal to the sum of the squares of the other sides in a right-angled triangle.

Take, then, 12 links of the chain, and having laid down 4 of them in the direction AB of the line to which you wish to draw a perpendicular, so that the ends cannot move, divide the remaining 8 links into two lengths of 5 and 3 respectively, and pull them tight: the three lengths will form a right angled triangle ABC, where CB will be at right angles to AB, because 532+42, or 25=9+16.

PB

Since 3, 4, and 5, links are very short lines on the ground, the above method of setting off a right angle, although theoretically true, requires some modification in practice, because a short line of 3, or 4, links could not be continued without risk of serious error. But, remem

bering the numbers 3, 4, 5, we may adopt any multiple of those numbers at pleasure, as 30, 40, 50. Thus, if AB be measured 40 links, and 80 links of the chain be made to measure AC and BC, viz. AC=50, and BC=30, then ABC is a right angle, since

302+402=2500=50°.

We now proceed to consider the difficulties arising from the intervention of obstacles, or any other cause of inaccessibility.

266. To continue the measurement of a straight line with the chain, when some obstacle, as a pond, or building, or river, intervenes.

Let AB be the straight line which the Surveyor is measuring, and B the point where the obstacle O ́interrupts the continuance of the measurement.

A

B

C

b

с

1st. At B set off

by the Cross-Staff, or with the chain alone according to

(266), the line Bb at right angles to AB, and of such a length that it will not only clear the obstacle O, but terminate in a point b, where a parallel to AB can be readily set off and measured. Measure Bb; at b set off, at right angles to Bb, the line bc, of such a length that it will clear O, and allow a line to be set off and measured at right angles to it. Measure bc; and at c set off cC at right angles to bc, and equal to Bb. Lastly, at C set off a line CD at right angles to cC. Then it is plain, that, if BC be supposed joined, BhcC is a parallelogram, and BC required is equal to bc, which was measured. Also CD is in the same straight line with AB, and therefore the measurement can be proceeded with as if the obstacle had not intervened.

If the obstacle does not impede the surveyor's view beyond it, an assistant may set up two poles at C and D, in the straight line continuous with AB, and then it will not be necessary for him either to measure cC, or set off a right angle at C.

A

2ndly. The measurement of BC, and the line of

B

E

direction CD, may D both be found by the Chain alone, without the Cross-Staff, as follows:

:

In the direction of the line measure the portion AB equal to any convenient length, say 50 links; and then set off the equilateral triangle ABb, by stretching a chain double the length

of AB, in this case 100 links. Continue the measurement in the direction bcE, where E is a point, from which, as nearly as can be judged by the eye, a line to O would be perpendicular to the direction of AB produced. Measure bE, and take Ec=AB= 50 links; and, as before, form the sides Ed, dc, each 50 links, then Ecd is equilateral. Continue the measurement in the

direction Ed, making ED= AE. In DE measure De=50 links, and form the sides DC, Ce, each = 50 links. Then CD is in the same direction as AB, and A, D, E, are the angular points of an equilateral triangle; also, since AB, CD are equal to Ab, cE, BC= bc, which has been already measured.

Srdly. Let the obstacle be a river, or steep ravine, which prevents measurements being taken as in either of the former cases.

1st. An assistant on the other side of Ꭰ the river, or ravine, fixes a staff at C to range with A and B, and another at D in the same line. At B set off a line Bb at right angles to AB,

and equal to any convenient even number of links of the chain; bisect it in E, and there fix a staff. At b set off bc at right angles to Bb; measure bc of such a convenient length that the staff C on the other side of the river, and in a line with AB, is seen in the same line with E and c. Then bEc, BEC, are similar triangles, and

BC= bc, as before.

N.B. BEb may be drawn at any angle to AB, provided there be means at hand for setting off bc at the same angle with BEb produced, so that the triangles may still be similar.

267. To measure a wood, or a lake.

As an example of difficulties met with in obtaining the area of a piece of land from the intervention of obstacles, we may take the case where it is required to measure an area of very irregular form, but bounded by lines nearly straight, and containing wood, or water, so that it cannot easily be traversed in every direction. It will then be best measured by observations taken from without.