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In the solution of the above triangle it will be observed that we have found each of the unknown parts in terms of the two given, and have not employed one of those first calculated to obtain another. This is agreeable to the principle laid down at Art. 41, of plane trigonometry, and the reason is the same. Such a method of proceeding is always practicable in the solution of right angled spherical triangles.

In the examples which we have taken above, we have supposed the base and perpendicular of a right angled spherical triangle given. Any other two parts being given, each of the unknown parts may be calculated by the aid of Napier's rules, in a manner entirely similar to what has been just exhibited.

EXERCISES.

1. In the spherical triangle ABC right angled at A, given the hypothenuse a 65° 5' and the angle c 48° 12' to find в, b and c

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3. Given the two oblique angles в 1110 11', c 910 11' to find the three sides.

a 89° 32' 28" Ans. b 111 48 43

с 91 16 8

N. B. If the given quantities in a right angled triangle be a side, and its opposite angle, there will be legitimate ambiguity in the solution.

In all other cases no ambiguity properly exists, but to avoid error it is necessary to

- observe the two following principles.

1. The greater side is opposite to the greater angle.

2. An angle and the opposite side are of the same affection, i. e., both greater or both less than 90°.

EXAMPLE I.

92. Given the sun's declination to find the time of his rising and setting at any place whose latitude is known.

Let n ESQ represent the meridian of the place, z being the zenith, and Ho the horizon, and let s' s" be the apparent path of the sun on the proposed day, cutting the horizon in s. Then the arc EZ will be the latitude of H the place, and consequently EH, or its equal qo, will be the colatitude, and

this measures the angle OAQ; also RS

E

S

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will be the sun's declination, and ar, expressed in time, will express the time

of sunrise from 6 o'clock, for nas is the 6 o'clock hour circle.

Hence, in the right angled triangle SAR, we have given Rs, and the opposite angle a to find AR, the time from 6 o'clock.

Required the time of sunrise at latitude 40° 43', when the sun's decli

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It should be here remarked that the time thus determined is apparent time, which is that which would be shown by a clock so adjusted as to pass over 24 hours during one apparent revolution of the sun, or from its leaving the meridian to its return to it again, the index pointing to 12, when the sun is on the meridian. But it is impossible that any clock can

* Degrees are converted into hours by multiplying by 4 and dividing by 60, which is equivalent to dividing by 15.

be so adjusted, because the interval between the successive returns of the sun to the meridian is continually varying, on account of the unequal motion of the sun in its orbit, and of the obliquity of the ecliptic; each of these varying intervals is called a true solar day, and it is the mean of these during the year which is measured by the 24 hours of a well regulated clock, this period of time being a mean solar day; hence, at certain periods of the year, the sun will arrive at the meridian before the clock points to 12, and at other periods the clock will precede the sun; the small interval between the arrival of the index of the clock at 12 and of the sun to the meridian, is called the equation of time, and it is given on pages I. and II. of each month of the Nautical Almanac for every day in the month; this correction, therefore, must always be applied to the apparent time determined by trigonometrical calculation to obtain the mean time or that shown by a well regulated clock or chronometer, or vice versâ, and the Nautical Almanac always indicates whether this correction is additive or subtractive.*

A third kind of time is called siderial time. A siderial day is the period of revolution of the earth upon its axis with reference to the fixed stars, or it is the time which elapses after a fixed star passes the meridian of any place until the same star comes to that meridian again. Owing to the apparent motion of the sun from west to east along the stars about 360°

3654

daily, occasioned by the real motion of the earth in its annular orbit,

the solar day is a little longer than the siderial, because when the meridian of a place has revolved with the earth on its axis from west to east to come under a certain star, the sun which the day before may have been on the meridian with the star having moved a little to the east, the meridian has a little farther to revolve towards the east to come under the sun again, and thus complete the solar day. The difference between the siderial and solar day is about 3" 57". The same fixed stars cross the meridian, rise and set about this much later every day. The siderial day is divided into 24 siderial hours, the hour into 60 siderial minutes, and these each into 60 siderial seconds. At pages 584, 585, 586, 587 of the

*To solve the above problem very accurately it would be necessary to compute the sun's declination at the time of sunrise as deduced approximately above, and then to go over the calculation again. The Nautical Almanac gives the declination of the sun at noon for every day in the year, and of the process for determining its declination at any other time of the day we shall have numerous examples in Part. V.

+ The Nos. of these pages change a little every year.

Greenwich Nautical Almanac are what are called tables of time equivalents. On pages 584, 585 will be found for any given interval of mean solar hours, or minutes, or seconds, the equivalent interval in siderial time. And similarly on pages 586, 587, for any given siderial interval will be found the equivalent mean solar interval.

EXAMPLE.

Required the equivalent of an interval of 7 28 30 of mean solar time in siderial time. From page 584 the equivalent

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Required the solar equivalent of 22 sid. h 30m 27′ (page 586 N. A.)

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An astronomical clock is one which keeps siderial time. A common clock may be made to do this by shortening a little the pendulum. The weight attached to the pendulum is usually furnished with a screw by which it may be lengthened and shortened at pleasure, and this should be done till the clock goes just 24 hours from the time a star makes its transit over the meridian till the same star makes its meridian transit again. The exact instant of a star's crossing the meridian is observed with a "Transit Instrument."

This instrument consists of a telescope ab supported by a horizontal axis, each half of which c and d is a hollow cone of brass, at the outer ends of which are short solid cylindrical pivots which rest upon stone pillars e and f, called piers, the latter being imbedded in a mass of masonry extending a few feet below the surface of the

e

ground, in order that the vibrations occasioned by passing vehicles or the tread of the observers may not be felt. The pivots of the axis do not rest immediately upon the piers, but upon flat pieces of brass about 4 inches square, and an inch in thickness, which are screwed to the top of the stone. These brass pieces have a notch technically called a y or v from its shape, in which the pivot of the axis rests. The part of the piece of brass having the notch or v is detached and movable, by means of a screw arranged differently at the opposite ends of the horizontal axis c d, so that one end may be moved horizontally, and the other vertically.

The exact line of vision directed to a distant object is marked by two threads of spider's web, technically called wires, crossing each other at right angles, at a point in or near the optical axis of the telescope. They are stretched across a ring or diaphragm to which they are fastened with wax, and this ring, which is smaller in diameter than the tube of the telescope, is held in its place by screws passing through the tube, having their heads outside. By loosening the screw on one side of the tube and tightening the other, the diaphragm, and consequently the point in which the wires cross, receives a lateral motion. The diaphragm is placed at the focus of the object glass near the eye end b.f

The whole instrument just described has to be so placed that as it turns on the pivots of the horizontal axis the line of vision along the optical axis of the telescope shall describe the plane of the meridian; narrow trap doors in the roof and sides of the transit room serve to expose the meridian to view. As this plane is vertical, if the line of vision above mentioned be exactly perpendicular to the axis cd, whilst at the same time the axis is exactly horizontal, and finally the telescope point due north and south, then will the required position be attained. For this,

1. The adjustment of the line

therefore, three adjustments are requisite. of collimation* in a perpendicular to the supporting axis c d. 2. The adjustment of the supporting axis to a horizontal position. 3. The adjustment of the line of collimation to the meridian.

The method of making these several adjustments we shall describe in their order.

1. To collimate the instrument.-Bring the intersection of the wires upon a well defined point of some distant terrestrial object; take the instrument out of the y and reverse the supporting axis end for end; bring the telescope upon the same distant point, and if the intersection of the wires covers it exactly, the instrument is collimated; if not, move

* i. e. the line of vision determined by the intersection of the wires.
For the illumination of the wires see p. 363.

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