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produced to the celestial sphere becomes the axis of the heavens about which all the stars appear to revolve daily.

The ecliptic is a great circle which makes an angle of about 23° 28 with the equator. It is the path which the sun appears to describe among the stars once a year.

The points in which the two great circles above defined intersect, are called equinoctial points.

The one at which the sun crosses the equator in the spring about the 21st of March, is called the vernal equinox.

The other which is where the sun crosses in the autumn, viz. about the 23d of September, is called the autumnal equinox.

The meridians are great circles, the planes of which pass through the axis and the circumferences of which all intersect in the poles or points where the axis meets the surface of the celestial sphere. They are sometimes called hour circles. The sun appears to move about the earth once in 24 hours; 360°

24

15° is the number of degrees through which the sun moves in an hour.

The angle contained between the meridian of a place and that meridian which passes through the sun at any given moment, is called the hour angle and converted into hours, 15° to the hour, will show the time of day, if we reckon from noon instead of midnight as astronomers do.

This time may be either A. M. or P. M. It is what is called apparent time, which varies a little from mean time, the time given by the clocks, in consequence of the slightly unequal motion of the sun in its annual revolution.

The hour angle of a star is similar to that of the sun.

The horizon of any place is a great circle whose plane touches the surface of the earth at that place, and extends to the celestial sphere. This is called the sensible horizon; the real horizon is a plane parallel to this through the centre of the earth. When the sun or fixed stars are in question, the distances of which from the earth are so great that its radius is as nothing comparatively, these two horizons may be

regarded as coincident. The zenith is the pole of the horizon directly over head. The nadir is the opposite pole.

The position of a heavenly body is fixed on the celestial sphere, like that of a place on the globe, by its latitude and longitude, only it must be observed that on the former these are measured from and upon the ecliptic instead of the equator.

Similar measurements from and upon the celestial equator are called the declination and the right ascension,the former corresponding to the latitude, and the latter to the longitude. Longitude upon the earth is reckoned from some fixed meridian as that of Greenwich.

Longitude upon the celestial sphere is reckoned from the vernal equinox which is called the first of Aries; right ascension also from the same point; the former upon the ecliptic the latter upon the equator.

The azimuth of a celestial object is an arc of the horizon, comprehended between the meridian of the observer and the meridian which passes through the object.

80. We are now prepared with materials for a practical application of the formulæ of spherical trigonometry, and we commence with that already demonstrated.

Let E in the annexed diagram

be the equinoxial point, EQ a

portion of the equator, Es a por- E

tion of the ecliptic, s the place of

the sun, and so a portion of a

meridian through the sun; then

S

so will be the 's declination, EQ his right ascension, and Es his longitude.

Given the 's declination* equal to 20°, required his longitude.

In the right angled triangle EQs right angled at a we know E23° 28' the opposite side sa = 20° required the hypothenuse E Hence the proportion

The declination of the sun may be found by taking its meridian altitude

14

with

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Hence Es

log. sin. of 20° = 9.534052
log. of R = 10.000000

sum rejecting 10 =

9.933934= log. sin Es 59° 11' 29" the longitude of the sun required. Let the student try the following modification of the problem as an exercise.

Given the 's longitude equal to 90° to find his declination.

81. By means of the proportion for right angled triangles, and of which an application has just been given, one may be derived for triangles in general.

Let ABC be any spherical

triangle; let fall from a the

arc AD perpendicular to the

.A.

side BC, the given triangle B

will be divided into two right

angled triangles ABD and

C

D

with the same instrument and in the same manner as was described at Art. 11. This observation should be made about noon repeatedly, and the greatest observed altitude will be the meridian altitude. A piece of colored glass will be required for the purpose. Let p be a

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N. B. The altitude of the uppermost point of the circumference of the sun should be first taken, then of the lowermost point, and half their difference added to the latter will give the altitude of the 's centre.

ACD. In the right angled triangle ABD we have the proportion (art. 78)

R:

sin B : : sin AB : sin AD

and in the right angled triangle ACD, the proportion

R sin c sin AC : sin AD

Multiplying the extremes and means of each of these proportions we have the equations

and

RX Sin AD sin B X sin AB

R × Sin AD = sin c × sin ac

The first members of these equations being the same the second numbers are equal, hence

sin B X Sin AB = sin c × sin ac

substituting for the sides AB and AC the small letters of the same name with the angles opposite to them the last equation may be written

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that is, the sines of the angles of a spherical triangle are as the sines of the opposite sides.

EXAMPLE.

Let z be the zenith, p the pole of the equator, and s the place of a star; zs will be the coaltitude of the star, Zps its hour angle, ps its codeclination or polar distance, and SZP the azimuth. Let

the azimuth, coaltitude and S

hour angle be given to find

Z

the polar distance; z, zs and P are given and Ps required.

P

* The student will recollect that a proportion is an equality of ratios, and that ratio, as commonly understood, is the quotient of two quantities.

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Let P 32° 26′ 6′′, z=49° 54′ 38′′, and zs or p=44° 13′ 45′′

=

Às sin P 32° 26' 6' ar. comp. 0.270558

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Since the sine of an arc is equal to the sine of its supplement (Art. 15.) the required side may be also the supplement of 84° 14′ 40′′ or 95° 45' 20".

To illustrate this double solution by the diagram, let the student make or conceive to be made the following construction. Draw an arc from s making with PZ an angle equal to z, meeting pz in a point which we will call z'. sz' will then be equal to sz; prolong PZ and Ps till they meet in the opposite pole which we will call p'; a triangle will be formed z'P's, in which the angles z' and P', and the side sz' will be equal to those given in the above example, but in which the side p's is the supplement of Ps.*

The polar distance of the fixed stars will be found to be always the same, hence they describe eircles about the poles. in their apparent daily motion.

82. We shall next demonstrate a formula which will express one of the angles of a spherical triangle in terms of the three sides.

* A rule for determining when there are two solutions, and when but one in such cases, is given at Art. 127.

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