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SECTION II. MENSURATION OF HEIGHTS AND DISTANCES.

1st. Of Heights.
PL. 5. fig. 18.

THE instrument of least expense for taking heights, is a quadrant divided into ninety equal parts or degrees; and those may be subdivided into halves, quarters, or eighths, according to the radius, or size of the instrument: its construction will be evident by the scheme thereof. From the centre of the quadrant let a plummet be suspended by a horse-hair, or a fine silk thread, of such a length that it may vibrate freely, near the edge of its arc ; by looking along the edge AC, to the top of the object whose height is required, and holding it perpendicular, so that the plummet may neither swing from it, nor lie on it, the degree then cut by the hair or thread will be the angle or altitude required. If the quadrant be fixed upon a ball and socket on the threelegged staff, and if the stem from the ball be turned into the notch of the socket, so as to bring the instrument into a perpendicular position, the angle of altitude by this means can be acquired with much greater certainty. An angle of altitude may be also taken by any of the instruments used in surveying; as has been particularly shown in treating of their description and uses. Most quadrants have a pair of sights fixed on the edge AC, with small circular holes in them ; which are useful in taking the sun's altitude, requisite to be known in many astronomical cases; this is effected by letting the sun's ray, which passes through the upper sight, fall upon the hole in the lower one ; and the degree then cut by the thread will be the angle of the sun's altitude ; but those sights are useless for our present purpose,_for looking along the quadrant's edge to the top of the object will be sufficient as before.

To take an angle of altitude and depression with the quadrant.
PL. 14. fig. 6, 7.

Let A be any object, as the sun, moon, or a star, or the top of a tower, hill, or other eminence : and let it be required to find the angle ABC, which a line drawn from the object makes above the horizontal line B.C.

Place the centre of the quadrant in the angular point, and move it round there as a centre, till with one eye at D, the other being shut, you perceive the object A through the sights; then will the arc GH of the quadrant, cut off by the plumbline BH, be the measure of the angle ABC, as required. The angle ABC of depression of any object A, below the horizontal line BC, is taken in the same manner; except that here the eye is applied to the centre, and the measure of the angle is the arc GH, on the other side of the plumb-line." Demonstration. In taking the angle of altitude, the angle ABG is a right angle; and because the plumb-line BH is perpendicular to the horizon, the angle CBH is also a right angle; hence if the angle CBG be taken from these equals, the remaining angles will be equal, that is ABC=GBH, or equal to the arc HG. Q. E. D. * , In like manner, the angle GBH (in taking the angle of depression) is equal to the angle ABC.

PROBLEM I.

PL. 5. fig. 19.

To find the height of a perpendicular object at one station, which is on a horizontal plane.

A steeple. The angle of altitude, 53 degrees. Distance from the observer to the foot of the steeple, or the base, 85 feet. Height of the instrument, or of the observer, 5 feet. Required the height of the steeple. The figure is constructed and wrought, in all respects, as Case 2 of right-angled trigonometry; only there must be a line drawn parallel to and beneath AB, of 5 feet, for the observer's height, to represent the plane upon which the object stands; to which the perpendicular must be continued, and that will be the height of the object. Thus, AB is the base, A the angle of altitude, BC the height of the steeple from the instrument, or from the observer's eye,

Given

* In finding the height of an object, let the observed angle be as near 45° as possible, for then a small error committed in taking it makes the least error in the computed height of the object. In taking the height of a perpendicular object, if the observed angle be 45°, the height of the object above the horizontal line is equal to the base line, and if the observed angle should be 60°, three times the square of the base line is equal to the square of the perpendicular object above the horizontal line; hence b extracting the square root of three times the square of the base or hori . zontal line, will give the height of the object above that line, to which add i. lo of the observer's eye above the horizon, and you have the true

cight,

:

if he were at the foot of it; DC the height of the steeple above the horizontal surface.

Various statings for BC, as in Case 2 of right-angled plane trigonometry.

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Their sum is 117.8 or 118 feet, the height of the steeple re

quired. PROBLEM II,

PL. 5. fig. 20. To find the height of a perpendicular object, on a horizontal plane, by having the length of the shadow given. Provide a rod, or staff, whose length is given, let that be set perpendicular, by the help of a quadrant, thus; apply the side of the quadrant AC to the rod, or staff; and when the thread cuts 90°, it is then perpendicular; the same may be done by a carpenter's or mason's plumb. Having thus set the rod or staff perpendicular, measure the length of its shadow when the sun shines, as well as the length of the shadow of the object whose height is required, and you have the proper requisites given. Thus, ab, the length of the shadow of the staff, 15 feet. bc, the length of the staff, 10 feet. AB, the length of the shadow of the steeple, or object, 135 feet. Required BC, the height of the object. The triangles abc, ABC are similar, thus; the angle b=B, being both right; the lines ac, A C are parallel, being rays, or a ray of the sun; whence the angle a=A (by part 3, theo. 3,

sect. 4), and consequently c=C. The triangles, being therefore

mutually equiangular, are similar (by theo. 16, sect.4), it will be, ab : be : ; AB : BC. 15 10 135 90, the steeple's height required,

The foregoing method is most to be depended on ; however, this is mentioned for variety's sake.

PROBLEM III.

- Pl. 5. fig. 21. To take the altitude of a perpendicular object at the foot of a hill, from the hill's side. Turn the centre A of the quadrant next your eye, and look along the side AC, or 90 side, to the top and bottom of the object; and noting down the angles, measure the distance from the place of observation to the foot of the object. Thus, Angle to the foot of the object, 55#9 or 55° 15'. Given } Angle to the top of it, 314° or 31° 15' Distance to the foot of it, 250 feet. Required the height of the object.

By Construction.

Draw an indefinite blank line AD at any point in which A makes the angles EAB of 55° 15', and EAC of 31° 15'; lay 250 from A to B; from B draw the perpendicular BE (by prob. 7 of geometry) crossing AC in C: so will BC be the height of the object required.

In the triangle ABC there is given,

ABE the complement of EAB to 90°, which is 34°45'.

CAB the difference of the given angles 24°00'.

The side AB 250. Required BC. To" is performed as Case 2 of oblique angular trigonometry.

uS,

180 — the sum of ABE 34° 45' and CAB 24°00' = ACB 121° 15'. Then,

S.ACB : AB : : S.C.A.B.: BC.

121° 15' 250 24°00' 119, the height required.

PROBLEM IV.
PL. 5. fig. 22.

To take the altitude of a perpendicular object on the top of a hill at one station, when the top and bottom of it can be seen from the foot of the hill.

As in prob, 1, take an angle to the top and another to the bottom of the object, and measure from the place of observation to the foot of the object, and you have all the given requisites. Thus,

A tower on a hill.

Angle to the bottom 48° 30'.
Given (Angle to the top 67°00'.
Distance to the foot of the object 136 feet.
Required the height of the object.
Py Construction.

Make the angle DAB-48° 30', and lay 136 feet from A to B; from B let fall the perpendicular BD, and that will be the height of the hill. Produce BD upwards by a blank line. Again, at A make the angle DAC=67°00' by a blank line, and from C, where that crosses the perpendicular produced, draw the line CB, and that will be the height of the object required.

Let AC be drawn.

In the triangle ABC there is given,

The angle A CD the complement of DAC=23° 00'.

CAB the difference between the two given angles = 18° 30'.

And the side AB 136. To find BC,

S.C : AB : : S. CAB : BC.
23° 136 18° 30' 110}

If BD were wanted it is easily obtained by the first case of

right-angled plane trigonometry.

PROBLEM W.
PL. 5. fig. 23.
To take an inaccessible perpendicular altitude on a horizontal plane.

This is done at two stations. Thus, Let DC be a tower which cannot be approached, by means of a moat or ditch, nearer than B; at B take an angle of altitude to C : measure any convenient distance backward to A, which note down; at A take another angle to C; so have you the given requisites. Thus, First angle 55° 00'. go Stationary distance 87 feet. Second angle 37°00'. The height of the tower CD is required.

- By Construction.

Upon an indefinite blank line lay off the stationary distance 87 from A to B; from B set off your first, and from A your 4.

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