EXAMPLES FOR PRACTICE,

BY TWO STATIONS.

The following examples are given for the learner's practice, who, unless he understand trigonometry, and can from its rules, which is the proper mode, calculate the required distances, must plot them from the notes, and then measure the lengths of AD, DE, &c., by the scale, and compare them with their correct lengths as given below.

Answer to 1st Example:

AD 14.20 chs.; DE-22 chs.; EB-18.45 chs.; BF 11.70 chs.; FG 11.60 chs.; GA=14.23 chs.

Answer to 2nd Example:

AD=17.70 chs.; DE=27-75 chs.; EB-20.60 chs.; BF 14.10 chs.; FG-13.70 chs.; GA=16.80 chs.

Useful and practical Problems.

PROBLEM I.

To measure a base line across a river.

Let DA be the direction of the line which has been

measured up to the river. It is required to ascertain the distance BA at once upon the ground; so as to continue the measurement of the line.

B

D

On the line DA, take any point, B, whence a perpendicular, BC, can be taken, which will be free from obstruction, so that BC can be accurately measured; carry the range on across the river, and at A set up a flag;、on BC, take any point C, whence A can be seen; and at C erect a perpendicular to AC, intersecting the line AD in D.

Measure DB; then, because ACD is a right angle.

BC2

BD

=AB (Euclid, lib. vi. Prob. 8.)

These angles can either be taken by a cross-staff or by the chain, with the distances of 30 links, 40 links, and 50 links; 50 being the hypothenuse of a right-angled triangle when the base and perpendicular are respectively 30 and 40.

EXAMPLE 1. Was engaged in the measurement of a base line, that unfortunately crossed a river too wide for the chain measured up to the river 261 chains 45 links. Sent a man across in a boat, with a flag to carry on the

K'

range, and to plant the flag in the line on the other side. At 261 chains, at right angles with the base line on the right, measured 4 chains 50 links to the water's edge, whence only, on account of the trees near the river's side, the flag was visible.

At this point, at right angles with an imaginary line to the flag, measured to a point on the base line, which on trial I found to be 259 chains 25 links: required the width of the river?

Width of river, 11 chains 12 links.

EXAMPLE 2. Measured BC, (B being at the water's edge) 3·85 chś.; found the right angle ACD, to intersect AD, at a distance of 5'00 chs. from B;-what is the length of BA? Answer 2.96 chs.

PROBLEM II.

To measure the width of a river with a base line alongside

of it.

Take at either end of the base line, with a theodolite, the angle made between the base line and a flag placed at the edge on the other side of the river. Com

A

pute the length of the sides by
the first of the three cases of trigonometry.

Then AO nat. sine ▲ A; or OB nat. sine ≤B= the width. Or construct the triangle AOB by a scale of equal parts, and measure the altitude = the perpendicular width of the river.

EXAMPLE 1. Took a base line by the side of a river, 12

chains, and observed the angles at its ends to the flag on the other side, found them 25° and 55°: what is the perpendicular width ? Answer 4.22 chs. EXAMPLE 2. Given the base 20 chs.; and the angles 44° and 42°; what is the width?

PROBLEM III.

Answer 9.30 chs.

To find the distance of one object from another, where a river

divides them, without using the theodolite.

Let O and A be the given objects, and AO the distance required. From A draw AB at any angle to AO, and produce OA to C, measuring AC about one third length of AB; from C

A

B

D

measure CD, parallel to AB*; and such that OBD may be in one straight line.

Then, because AB and CD are parallel, the triangles are similar, and therefore CD: AB:: CO: AO;

and CD-AB: AB:: CO-AO, or CA: AO

.:. AO

AB.CA,
CD-AB

the distance required.

If the perpendicular distance be required, we can obtain ' it, either by similar triangles, CA: AO:: the perpendicular distance between the parallels: to the perpendicular distance across the river.

Or, we can so select the point A, as to have OAC at right

To make CD parallel to AB: at A and B erect equal perpendiculars which can be done either by a cross staff or by means of the unit proportion of the sides of a right angled triangle: viz. 3, 4, and 5, or any multiples of them whatever.