(altitude zero), mounts higher in the sky until it reaches the observer's meridian, then sinks towards, and sets in, the west; it is, accordingly, continually changing its altitude and azimuth.

S. Z (Zenith)

71. The equator system: Positions described by declination and hour angle. The north celestial pole is the principal point of this system. The celestial equator is the great circle of which that point is the pole; it is evidently the projection of the earth's equator upon the celestial sphere. The celestial equator and the meridian of the observer are the reference circles in the system now being described. In P(Pole) Fig. 54, P is the north celestial pole, S the south celestial pole, EQ the celestial equator; also, HR is the horizon and Z the zenith for some particular place on the earth's surface. As said in Art. 68, the stars move in parallel circles whose axis is PS; these circles are, accordingly, parallel to the equator EQ. The

H

E

$2

Horizon

Equator S

FIG. 54

R

S.

angular distance of a star from the equator is called the declination (denoted by D or 8) of the star; north (or+) declination when the star is north of the equator, and south (or —) declination when the star is south. Thus the declination of S, is SSg. The angular distance of a star from the north pole is called its north polar distance; this is evidently the complement of the star's declination.*

In 24 (sidereal) hours a star appears to make a complete revolution (ie. to pass over 360°) about the celestial polar axis; hence, the star passes over 15° in 1 hour.† The great circles, passing

through the poles are called hour circles. Thus PSS is the hour circle of S. The hour angle (denoted by H. A.) of a star is the angle between the meridian of the observer and the hour circle of the

* The declination of the stars change by an exceedingly small amount in the course of a year.

†The interval of time between two successive passages of the observer's meridian by the sun (i.e. from noon to noon) is about 4 minutes longer than the interval of time between two successive passages of the meridian by any particular star. (This difference is due to the yearly revolution of the earth about the sun. See text-books on astronomy.) The second interval is called a sidereal day; it is divided into 24 sidereal hours.

star. This angle is measured towards the west. Thus, suppose that a star is on the meridian at S4; its hour angle is then zero. Twelve hours later the star will be at So, and will have an hour angle 180°. After a while it will be at S1, just rising above the horizon, and its hour angle will be 180° + S.PS1; later it will be at S, having the hour angle 180° + SPS,; later still it will be on the meridian at S4, and its hour angle will be zero again. The hour angle is usually reckoned in hours from 1 to 24, 1 hour being equal to 15 degrees. Thus, when the star is at So its hour angle is 12 h. The hour angle of a star is partly local; for only places on the same meridian of longitude have the same celestial meridian. Moreover, the hour angle of a star is continually changing, and its magnitude depends upon the time of observation. In Arts. 76, 77, the positions of stars are described in terms which are independent of the time and place of observation.

In Arts. 73, 74, 75, the astronomical ideas so far obtained, are used in the solution of two simple problems.

72. The altitude of the pole is equal to the latitude of the place of observation. This theorem, which is necessary in Arts. 73, 74, is the fundamental and most important theorem of spherical astronomy. In Fig. 55, C represents the

R

E

centre of the earth, P its north pole, and EQ its equator; O is the place of observation, say some place in the northern hemisphere, Z is its H zenith and HR its horizon; CPP1 is the celestial polar axis, P1 being the north celestial pole. Draw OP2 parallel to CP1, P2 being on the celestial sphere. The angle ROP is the altitude of the pole at 0, since (see Arts. 68, 70) P1 and P2 are in the same direction from 0. The latitude of a place is equal to the angle between the plumb line and the plane of the equator. Thus, the latitude of O is equal to OCE. Since OR and OP, are respectively perpendicular to CZ and CE, the angle ROP1 = OCE; that is, the altitude of the pole as observed at O is equal to the latitude of 0.

1

FIG. 55

73. The time of day can be determined at any place whose latitude is known, if the declination and the altitude of the sun at that time and place are also known.

NOTE 1. The sun, unlike the stars, changes in declination from 2310 south (about Dec. 22) to 231° north (about June 21), and then returns south. Its declination is zero, that is, it is on the celestial equator, about March 20 and Sept. 22. This change in declination is due to the revolution of the earth about the sun, and to the fact that the plane of the earth's equator is inclined about 2310 to the plane of its orbit about the sun. The latter plane is called the plane of the ecliptic. The declination of the sun is given for each day of the year in the Nautical Almanac. The altitude of the sun can be observed with a sextant.

NOTE 2. The student should consult a text-book on astronomy for an account of the special precautions and corrections necessary in connection with this and similar astronomical problems.

H

E

M

2

90-h

Z(Zenith)

902

90-D

Horizon

Equator

FIG. 56

P(Pole)

R

In Fig. 56, P is the north celestial. pole, EQ the celestial equator, S the sun, and SSS is the small circle on which the sun is moving at the given time; Z is the zenith, and HR the horizon, of the place of observation; ZSM is the sun's vertical circle, and PSN is its hour circle.

It is midnight when the sun is at So, and noon when the sun is at Sn From noon to noon is 24 hours. Hence, to find the time when the sun is at S, determine the angle ZPS in hours (15° 1 h.); subtract the number of hours from 12, if it is forenoon; and add, if it is afternoon.

=

Let 1, h, D, respectively, denote the latitude of the place, and the altitude and declination of the sun.

Then

PR=1 (Art. 72), SM=h, SN = D.

In ZPS, whose vertices are the sun, zenith, and pole,

ZP 90° - 1, ZS = 90° - h, SP=90° - D.

=

Hence, the angle ZPS can be found.

EXAMPLES.

1. In New York (lat. 40° 43′ N.) the sun's altitude is observed to be 30° 40'. What is the time of day, given that the sun's declination is 10° N., and the observation is made in the forenoon?

2. In Montreal (lat. 45° 30′ N.) at an afternoon observation the sun's altitude is 26° 30'. Find the time of day, given that the sun's declination

is 8° S.

3. In London (lat. 51° 30' 48" N.) at an afternoon observation the sun's altitude is 15° 40'. Find the time of day, given that the sun's declination is 12° S.

4. As in Ex. 2, given that the sun's declination is 18° N. 5. As in Ex. 3, given that the sun's declination is 22° N. 6. As in Ex. 1, given that the sun's declination is 10° S.

74. To find the time of sunrise at any place whose latitude is known, when the sun's declination is also known. This is a special case of the preceding problem; for at sunrise the sun is on the horizon and its altitude is zero. The problem can also be solved by means of the triangle RPS1 (instead of ZPS, which is employed in Art 73). For, in RPS1

S1P 90° - D, PR=1, PRS1 = 90°.

=

Z (Zenith)

P(Pole)

Equator

FIG. 57

The angle RPS1 (i.e. SPS1) reduced to hours, gives the time of sunrise (after midnight). If ZPS, is found, then ZPS, reduced to hours and subtracted from 12 (noon), gives the time of sunrise. The time of sunset is about as many hours after noon as the time of sunrise is before it.

=

In Fig. 57 the sun is north of the equator. When the sun is south of the equator, PS, 90° + D, and RPS, > 90° for places in the northern hemisphere. The student can make the figure and investigate this case, and also the case in which the place is in the southern hemisphere.

EXAMPLES.

Find the approximate time of sunrise at a place in latitude 1, when the sun's declination is D, in the following cases:

1. 7 = 40° 43′ N. (latitude of New York), D equal to : (a) 4° 30' N. (about April 1); (b) 15° 10′ N. (about May 1); (c) 23° N. (about June 10); (d) 5° N. (about Sept. 10); (e) 6° S. (about Oct. 8); (ƒ) 15° S. (about Nov. 3); (g) 23° S. 2. 7 = 51° 30' 48" N. (latitude of London), D as in Ex. 1.

3. 760° N. (latitude of St. Petersburg), D as in Ex. 1.

4. 770° 40′ 7′′ N. (latitude of Hammerfest, Norway, D as in Ex. 1.

5. 29° 58′ N. (latitude of New Orleans), D as in Ex. 1.

=

6. 7 = 33° 52′ S. (latitude of Sydney, N. S. W.), D as in Ex. 1.

7. Find the approximate time of sunrise for other days and places.

75. Theorem. If the latitude of the place of observation is known, then the declination and hour angle of a star can be determined from its altitude and azimuth, and vice versa. For, in the triangle ZPS (Fig. 56), ZP = 90° — 1, SP = 90° – D, SZ= 90° - h, SPZ= 360° — H. A., PZS = A 180°. Hence, if the latitude and any two of the four quantities, viz., altitude, azimuth, declination, hour angle, be known, then the remaining two can be found by solving the triangle SPZ.

E,

P (Pole)

76. The equator system: Positions described by declination and right ascension. In the system in Art. 71 the circles of reference were the equator and the meridian of the observer. In the system in this article the circles of reference are the equator and the circle passing through the celestial poles and the vernal equinox. The vernal equinox is one of the points where the ecliptic intersects the equator; namely, the point where the sun, in its (apparent) yearly path among the stars, crosses the equator in spring. (See text-book on astronomy.) This point may be called the Greenwich of the celestial sphere. (The ecliptic is the projection of the plane of the earth's orbit on the celestial plane of the ecliptic are See Art. 73, Note 1.)

E

Equator

P

Ecliptic

M

FIG. 58

C

sphere. The plane of the equator and the inclined to each other at an angle of 231°.