EXAMPLE VII.

Wanting to know the angle ACB formed by two walls AC and BC, I measured DC=840 yards, Ec=735 yards, and DE= 741 yards: hence the angle ACB=DCE is required?

Answer. ACB 55°. 40'.

EXAMPLE VIII.

Suppose ACB to represent the flanked angle of a bastion, (which of course will be inaccessible on account of the covertway, ditch, &c.) and it be required to determine the measure of that angle.

=

Place a pole at D, in the direction of the face BC of the bastion (out of the reach of musket shot). With a cross-staff at D direct a person to set up another pole at F perpendicular to DC. Measure DF 375 paces or 9374 feet, make FK perpendicular to DF, and set up another pole at H in the direction of the face Ac of the bastion. Prolong CH, and measure the three sides of the triangle GHI; suppose GH=10 yards, GI=12 yards, and HI=10 yards, then will the angle GHI be equal to the flanked angle ACB. Required this angle?

Answer. The angle ACB=74°. 36', the proper flanked angle of a regular pentagon.

(B) In the same manner (except measuring the triangle GHI,) we may draw a line KF parallel to the inaccessible face BC of the bastion, in order to place a battery at K, to produce the greatest effect; viz. at about 375 paces distance, and that the direction of the fire may be nearly perpendicular to of the face BC.

EXAMPLE IX.

Wanting to know the distance of an object at D, from two others A and B, and also the distance between A and B, I set up a pole at c in a right line with AB, at c the angle ACD was 57.0'; I then measured CD=784-8 yards, and at D the angles CDA and ADB were 14°. 0′ and 41°.30. Required the several distances above mentioned?

Answer. AD=696'1, DB=7124, and AB=499.3 yards.

EXAMPLE

the rules of plane trigonometry already explained, together with the use of certain instruments for taking angles.

(S) Horizontal and vertical angles are usually measured with a theodolite furnished with one or two telescopes, and a vertical arc; and if the horizontal and vertical arcs of the instrument be described with a radius of not less than 34 inches, the observed angles may be measured to half a minute, or the 120th part of a degree.

(T) Angles which are oblique to the horizon are generally taken with a sextant; which must be held in such a position, that its plane may coincide with the two objects and the eye. When vertical angles are taken with this instrument, an artificial horizon must be used, and the reflected image of the object from the glasses of the sextant must be brought to coincide with the reflected image of the same object in the artificial horizon.

(V) Base lines are generally measured with rods, or the four pole Gunter's chain; but common tape of 50 or 100 feet in length is often preferred both for accuracy and expedition; especially if it be kept dry, and the ground be tolerably level.

(W) The use of instruments must be acquired under the direction of a person well skilled in their several adjustments, as but little information can be obtained from written description*, and even the most expert observer will find it necessary, in several cases, to apply corrections to the different angles according to the situations of the objects. See Chapter IV.

EXAMPLE I.

Being on one side of a river, and wanting to know the distance of a fort, or other object, on the other side, suppose I measured 500 yards along the side of the river in a straight line AB, and found the two angles between this line and the object to be CAB — 74°. 14', and CBA=49°. 23'. Required the distance between each station and the object?

AL

·H

*Copper plates, well executed, of theodolites, sextants, quadrants, and other instruments for taking angles, are given in Adams' Geometrical and Graphical Essays, edited and published by Mr. Wm. Jones, mathematical instrument maker in Holborn, London.

(X) Or, produce AB till Ba be equal to it, and CB till the angle a be equal to the angle A; then will the distances Bc and ac, be equal to the distances BC and AC 210.

(Y) Ór, without any instrument to measure an angle, the distances AC and BC may be found. Make AF of any length. in a line with AC, and BG in a line with BC; measure AF, AB, BG, FB and AG; then the three sides of the triangles BAF and ABG will be given, with which the angles BAF and ABG may be found; their supplements will give the angles CAB and CBA, and hence the distances AC and BC may be found as in the example.

(Z) The perpendicular distance DC may be readily determined, for having found BC, the angle в and BC will be given; or find AC, then AC and the angle a will be given to find pc.

The perpendicular distance Dc may be found without any instrument for taking angles. With a cross-staff fixed at D observe the object at c, and set up a staff in a line with it at E, and another at в at right angles to CE, move the cross-staff from D to E, and set up a pole or staff at H perpendicular to EC. Measure the distances DB, DE, and also EG in the direction EH till G, B, and C make one straight line. Then

EG-DB: DE :: EG: EC from which take DE.

OR, EG-DB: DE::DB: DCI

The four following examples are referred to the foregoing figure, and serve to exercise the above observations.

EXAMPLE II.

An engineer wanted to know the breadth of a river, (over which the general intended to pass the whole army,) in order to determine how many pontoons of 5 feet broad and six feet asunder, he should have occasion to make use of. Perceiving an object (c) on the opposite bank of the river close to the edge of the water, he measured 144 paces or 360 feet along the edge of the river (as AB). The angle at a between the object and the base line was 83°. 57', and at B it was 80°. 32'; required the perpendicular breadth (DC) of the river, and the number of pontoons sufficient to form a passage for the troops?

Answer. The perpendicular breadth of the river, is 528 paces or 1320 feet, a military pace being 24 feet; and 120 pontoons will answer the purpose.

Two ships of war intending to cannonade a fort at c, sepa

EXAMPLE III.

I

rated

EXAMPLE X.

Wanting to know the distance between a church at A and an obelisk at в; both objects could only be seen from a particular place as D; at D. I took the angles ADC=89°.0', adb =72°. 30', and BDE = 54°. 30'; I then measured DE = 200 yards, and at E I took the angle BED=88°. 30'; and lastly, I measured DC = 200 yards, and at c took the angle DCA= 53°. 30'. Required the distance AB?

E

Answer. AD=237.62%, DB=332.215, and AB=345 yards.

EXAMPLE XI.

I wanted to know the distance between two places A and B, but could not meet with station any from whence I could see both objects. I measured a line CD= 200 yards; from c the object A was visible, and from D the object B was visible, at each of which places I set up a pole. I also measured Fc=200 yards, and

136

B

DE 200 yards, and at F and E set up poles.-I then took the angle AFC 83°, Acf=54°. 31′, ACD=53°. 30′, BDC=156°. 25', BDE=54°. 30', and BED=88°, 30. Required the distance AB? Answer. DB = 332·215, AD = 237-627, Ac=293.93; the angle ADC 83°. 55', consequently ADB=72°. 30', hence AB=

345.5.

XEXAMPLE XII.

Wanting to know the distance between two inaccessible objects A and B, I measured a base line CD=300 yards: at c the angle BCD was 58°. 20′ and ACB 37°; at D the angle CDA was 53°. 30' and ADB 45°. 15'. Required the distance AB?

Answer. First find CA= 465.97 by Example I, and also CB=761 46; and then find AB 479-79 yards by Example vi.

B 2 2 2

EXAMPLE

EXAMPLE XIII.

Suppose the base line in the foregoing example had been 908.36 feet, the angle ACB=14°. 34', and ACD=60°.50′; also the angle ADC 96°. 44', and BDC=115°. 23'; what would have been the distance AB in this case?

Answer, 674-62 feet.

Wanting to prolong the inaccessibleline CD, in order to place a mortar battery behind the obstacle E in the direction of CD, I measured a base line AB 414-2 yards, at A, I took

EXAMPLE XIV.

E

the angles CAD 49°. 13′, CAB=123°.45', and the angle CAE 100°, having first set up a pole at E behind the obstacle. At B, I took the angles DBC 33°. 45′ and DBA=67°. 30'. Hence it is required to find the angle AEF, which will determine the position of EF with respect to CD?

Answer. Exactly in the same manner as in the two preceding examples find AC = 601·6, AD=622, and the angle ACD= 67°. 30', then ACD+CAE=AEF=167°. 30'; if therefore at E, an angle of 167°. 30′ be made with A, the point F will be The distances CD or determined, in the same line with CD.

CE are easily calculated if necessary.

EXAMPLE XV.

From a station at D I perceived three objects A, B, C, whose distances from each other I knew to be as follow, AB=12 miles, BC 7 miles, and Ac=8 miles: at D, I took the angles CDB=25°, and ADC=19°. Hence it is required to find my distance from the objects.

CONSTRUCTION.

Make a triangle ABC whose three sides are 12, 7, and 8 (O. 53), viz. make AB=12, BC=7, and Ac=8. At в make the angle EBA=ADE= 19°, and at a make the angle EAB= BDE=25°; draw AE and BE, and through the point of intersection E, and the points A, B, describe a circle. From c through E draw CED, join BD and AD, then will AD, CD, and BD be the distances required. From this construction it appears, that if

D

-B

the