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standing on the ground at any considerable distance, seldom appear, even through good telescopes, sufficiently defined to permit the angles to be taken to that precision which is evidenly necessary when a satisfactory result is required; for a small error in either angle will produce a very considerable one in the distance. Thus, in the foregoing example, suppose an error or variation of 3" in the angle OCB, or let it be 2′ 12′′ instead of 2' 15",

Then, as 12" (the difference of 2′ 12′′ and 2'), is to 2', so is 400 to 4000 yards the distance CB, instead of 3200.

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Again, Let the base AC 300 yards, and suppose the an gles at A and C are 3' 20" and 4', respectively;

Then, as 40" (their difference), is to 3' 20", so is 300 yards, to 1500 yards = CB.

Now admit an uncertainty of 3" in each angle, and take that at A 8' 23", and the other at C 3' 57', and we have

=

As 34" (the difference) : 3' 23" :: 300 yards: 1794 yards; the difference is 294 yards in about a mile; an uncertainty per haps as great as that in an estimate by the eye at the same distance.

11. If CS be a line of cavalry; to determine the wheeling intervals between half squadrons marching en echellon from the right, when halted and formed on the line AB which is inclined to CS in a given angle.

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NE; and from ʼn, e,

and make ne NE, eh EH; hQ, QT each = Q, T, draw lines parallel to CS, and on those lines make parallelograms each equal to GN for the half squadrons: then if the half squadrons wheel on the pivots n, e, Q, T, till their fronts are in the line AB, the extent TB will be equal to CS, with the proper interval between the squadrons, or he HE. =

Calculation. We want the perpendicular distances OI, IP, and VR, DR. 24 yards = 72 feet. 40 yards = 120 feet.

en EN
eQ= NH

In the right angled triangles eIn, QRe the angles at Q and e are 35o.

As rad. :72 (en) :: sin. 35° : 41·3 feet = nl, whence OI=19 feet, nearly. rad. :72: cosin. 35° : 59 feet = el, whence IP = 13 feet, nearly. rad. : 120 (=eQ) :: sin. 35°: 68'8 feet eR, whence VR-46 feet, nearly. rad.: 120 :: cosin. 35° : 98 feet, nearly, whence DR = 26 feet.

But the measurement of those lines from construction, will be sufficiently correct for practical purposes.

234. In the preceding examples, the angles subtended by distant objects are supposed to be in an horizontal, or in a vertical plane: We shall now give the method of computation when they are measured in planes oblique to the horizon.

Angles oblique to the horizon are usually taken with a sextant or Hadley's quadrant, which is held in a position so that its plane passes through both objects and the eye of the observer. And elevations are found by reflecting the object from an artificial horizon. But whoever intends to observe with a sextant must acquire the method of using it from practice under the direction of a person who is master of the several adjustments, &c.; for which reason we shall not attempt a description of the instrument.

EXAMPLES.

1. Suppose ON is an object standing on the horizontal plane HNP; HA and PC two stayes or rods equal in height to that

of the eye; and let the plane ABC be parallel to the horizontal plane HNP; also suppose HP or AC is a base of 250 yards; and that the angles taken in the plane OCA, are OAC = 56° 46′, and OCA = 62° 54′; the angles of elevation OAB, OCB being 6° 40′, and 7° 6', respectively. Hence the height, and horizontal distances AB, CB, are required?

When one of the sides (AC) including an angle (OAC) oblique to the plane of the horizon, is horizontal, the angle is reduced to the corresponding horizontal angle by the following proportion,

H

As the cosine of the angle of elevation (OAB),
Is to the cosine of the given ang. (OAC),

So is the radius or sine of 90°,

To the cosine of the reduced angle (BAC).

For let DBO be a vertical plane and the angle ADO a right one; then the triangles ABO, DBO being also right angled at B, we shall have, (Case I. 221)

Sine ABO, 90° : AO :: sine AQB: AB;

Sine ADO, 90° AO: sine AOD: AD;

Therefore by equality,

sine AOB: sine AOD :: AB: AD :: sine ADB, 90°: sine ABD; or, sine AOB: sine AOD :: sine 90° : sine ABD:

But AOB is the complement of the elevation; AOD the complement of the observed angle OAC; and ABD that of the reduced angle BAC; therefore, &c.

As cosine 6° 40′

To cosine 56° 46′

So sine 90°

log. 9:997053

0 002947

log. 9.738820

log. 10000000

To cosine 56° 31′ the reduced angle BAC log. 9741767

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Therefore the angles of the triangle AOC BAC reduced to the horizontal plane are

BAC

56° 31′

ACB

62 40

ABC

60 49

And the side AC being 250 yards, we shall have (by Case I. 221) AB= 254·4, and CB = 238'8 yards; whence BO = 297 yards: to this add NB the height of the observer's eye above the horizontal plane HNP, and the sum will be the whole height NO.

But the distances AB, CB, and height BO may be calculated without any reduction of angles; for AC and all the angles of the triangle AOC being given, the sides AO, CO are found by Case I. and then the right angled triangles ABO, CBO, will give AB, CB, and BO at three proportions.

And should it be necessary, the reduced angles may be found from the sides of the triangle ABC, by Case IV. (228).

2. If A and C are two stations on sloping ground; O an ob ject on the top of a hill: and the angles OCA, OAC (measured with a sextant) equal to 79° 29', and 63° 11', respectively; also suppose the angle of elevation at A is 6° 36', at C5° 22': What are the horizontal distances and height of the object; AC being 410 yards?

Let GG be perpendicular, and AG, CB, parallel to the horizon: then AG, CB are

the horizontal distances.

B

G

In the triangle AOC )

the angles are

Whence (221) AO

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LAOC

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37 20

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6647, CO = 603-4, these hypotenuses, with the

angles of elevation OAG, OCB, in the right angled triangles AGO, CBO, give

AG = 660-3, OG 764, CB 600-7, OB 564 yards.

And the difference of OG and OB is 20 yards = BGCP the differ ence in the heights of the stations, AP being supposed horizontal.

The sides AC, CP, will give AP. And the angles of the triangle AOC when reduced to the horizon, may be found from the horizontal distances AP, AG, CB, taken as the sides of a triangle (228).

3. At a mile-stone N on the ascending road NS we observed the angle SNW between the next mile-stone S and the windmill W on the top of a hill, and found it to be 46° 37'; the elevation of W, or angle WNP was 3° 49'; next, at the mile-stone S, the angle NSW measured 91° 4'. Hence the horizontal distance NP, and height PW are required?

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In this example, no reduction is necessary on account of the inclination of the base NS to the horizon.

4. Let BC be a measured base of 370 yards on the plane ABC; and suppose marks are set up at the stations A, B, C, and the following angles taken with a sextant to the elevated object 0:

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