The sun's right ascension at noon December 1st (Naut. Alm.) being 16h.28'.16", and on December 2d, 16h.32.35".

1. To find the amplitude as, in the triangle ABS.

The star's declination being north, the triangle ABS must be used, wherein BAS➡the co-lat. and BS the declination are given, to find As.

Rad x sine BS=sine AS X sine BAS.

(10+log sine BS)--log sine BAS=log sine as. or, Sine BAS=38°.31.20": rad::sine BS-20°.6'.50" sine As 33°.30'.50".

(Y) The amplitude is always of the same name as the declination; and since the variation of a star's declination is extremely smallt, the same star may be considered as having constantly the same amplitude, or to rise and set in the same point of the horizon, during the whole year, in the same latitude. 2. To find the ascensional difference AB, in the triangle ABS.

*

Rad x sine AB=cot BAS X tang. BS.

(log cot BAS+log tang BS)-10=log sine AB. or, Rad cot BAS 38°.31.20":: tang BS-20°.6.50" sine AB =27°.23′.21′′.

=

Then 27°.23'.21" x 4 1.49′.33", and 6h. + 1h.49′.33"=7h. 49′.33" the arc BADÆ, or half the time of the star's continuance above the horizon‡.

3. Find the time of the 's culminating, Prob. VII. (D. 273.) Then, From the time of the 's culm. (D. 273.)=21.35′.23". Take the semi-diurnal arc 7.49'.33".

Time of the 's rising, December 1st 13h.45.50".

*

*

And, To the time of the 's culminating (D. 273.)=21.35′.23′′. Add the semi-diurnal arc

=7b.49'.33".

Sum 29h.24.56".

Then 29h.24.56". - 24h5h.24.56". time of setting, December 2d.

+ The amplitude of this star on the same day, in the year 1796, was 33o.45′.46′′, making a variation of only 14′.56" in its amplitude during 26 years. In the first edition of this work (which was published in the year 1801) the places of all the stars were corrected to the year 1796.

When the declination of the star and the latitude of the place are of the same name, the ascensional difference added to 6 hours gives the semidiurnal arc; but when of contrary names, the ascensional difference subtracted from 6 hours, leaves the semidiurnal are.

(Z) On account of the small change in the declinations of the stars, the same star, in any latitude, may be considered as having the same ascensional difference during the year; consequently the diurnal difference of the same star's rising, culminating, and setting, in the same latitude, is nearly equal to the diurnal difference of the sun's right ascension.

The sun's mean apparent daily motion is 59'.8" nearly, which is equal to 3'.56".32", the daily difference between the rising, southing, and setting of any fixed star, in the same latitude.

PRACTICAL EXAMPLES.

1. Given the sun's amplitude=39°.48′ N. his declination= 23°.28' N.; to find the latitude of the place, time of sun rising and setting, and the length of the day and night.

Length of day=16h.24.56′′ length of night=7".35′.4".

2. Required at what time Sirius, the Dog star, will rise, culminate, and set at Greenwich lat. 51°.28'.40" N. on the 18th of December 1822.

The right ascension of Sirius being 6h.37.18", and declination 16°.28'.39" S; the sun's right ascension at noon (Naut. Alm.) being 17.42.55", and on December 19th at noon 17h.47'.21". Ascensional difference 21°.48′.43′′-1.27.13".

Semi-diurnal arc-4b.32'.47′′.

Answer. Sirius will be on the meridian at 12h.52'.

rise at 8.19.13". in the evening.

set at 5 .24.47" next morning.

Star's declination being south, the triangle AGC is used.

3. Required the amplitude, time of rising, setting, and culminating of Aldebaran at Greenwich on the 20th of November

1822.

The right ascension of Aldebaran being-4h.25'.43", declination=16°.8′.36′′ N., the sun's right ascension at noon (Naut. Alm.) being 15.41'.29", and on the 21st November, at noon, 15h.45'.41".

Ascensional difference-21°.19′.19′′-1b.25'.17".
Semi-diurnal arc 7.25'.17".

Aldebaran culminates at 12h.42'. 1".

rises at
sets at

Answer.

The amplitude

5h.16'.44". evening.

gh. 7.18". next morning. =26°.30′.53". towards the N.

4. Given the latitude 51°.32' N., the sun's amplitude=

59.48' N. of the east; required the sun's declination, ascensional difference, time of rising, setting, &c.

PROBLEM III. (Plate III. Fig. 1.)

(A) The latitude of the place, and the sun's (or a star's) declination being given, to find the altitude and azimuth, &c. at six o'clock.

In the right-angled spherical triangle ADS.

1. AD the complement of Do the sun's or star's azimuth from the North, or elevated pole.

2. DS the altitude at six o'clock.

3. As the declination.

4. SAD NO the elevation of the pole, or latitude of the place. Any two of the above quantities being given, the rest may be found.

The triangle s Zenith N is complemental to the triangle ADS; it therefore follows, that the several things concerned in the problem may be determined in this triangle, right-angled at N. (B) As the meridian of a place, with respect to the sun, is called the 12 o'clock hour circle, so that circle NSAS which cuts it at right angles, being 6 hours from the meridian, iş called the 6 o'clock hour circle.

The sun and stars are on the eastern half of this circle 6 hours before they come to the meridian, and on the western half 6 hours after they have passed the meridian.

EXAMPLE I.

In latitude 51°.32′ North, when the sun has 19°.39' North declination; required his altitude and azimuth at 6 o'clock.

1. To find the altitude ps, in the triangle ADS.

*

Rad x sine DS=sine A × sine as.

. (log sine a+log sine AS)-10=log sine Ds. or, Rad: sine as 19°.39':: sine SAD 51°.32': sine DS-15°.16',

2. To find AD the co-azimuth, in the triangle ADS.

Rad x cos Acot AC × tang ad. (10+log cos A)-log cot as

or, Cot as 19°.39': rad::cos SAD

U

log tang AD.

519.32′ tang AD=

12°.31'.23". Hence 90°-12°.31.23" 77°.28'.37" OD the azimuth from the north.

EXAMPLE II.

At what time will Arcturus appear upon the six o'clock hour line at Greenwich, latitude 51°.28'.40" N., and what will its altitude and azimuth be on the first of April 1822?

Its right ascension being 14.7.32" and declination 20°.6'.50" N. (Table VIII.), and the sun's right ascension on the first of April (Naut. Alm.) being 0.41'.11", and on the second of April Oh.44'.49".

1.

1. To find the time of the star's culminating.

By Problem VII. (D. 273.) Arcturus will culminate at 13h.24.19".

Consequently (B. 289.) it will be upon the six o'clock hour circle at (13.24.19"-6h) 7h.24.19" in the evening, in the eastern hemisphere; and at (13h.24.19"+6h) 19h.24.19", or 7h.24.19" the next morning, in the western hemisphere.

2. To find the altitude DS, in the triangle ADS.

(log sine SAD+log sine as)-10=log sine DS. Rad: sine As=20°.6′.50′′:: sine SAD=51°.28'.40′′: sine DS= 15°.36'.25".

or,

3. To find the azimuth AD from the east, in the triangle ADS.

*

Rad x cos SAD-cot AS X tang AD.

(10+log cos SAD)-log cot AS log tang AD. or, Cot as=20°.6'.50": rad:: cos SAD=51°.28.40": tang AD= 12°.50.56". Hence oD=77°.9′.4" the azimuth from the north. (C) On account of the small change in the right ascension and declination of a star, it may, without material error, be said to have the same altitude and azimuth every time it arrives at the six o'clock hour circle; and the difference of the times it arrives there, may be considered as equal to the diurnal difference of the sun's right ascension, viz. about four minutes. (Z. 288.)

PRACTICAL EXAMPLES.

1. At what time will Aldebaran appear upon the six o'clock hour circle at Greenwich, latitude 51°.2s'.40" N., and what will its altitude and azimuth be on the 20th November 1822?

11

The right ascension being 4.25'.43", and declination 16°.8'.36" N. (Table VIII.); also, the sun's right ascension on the 20th November (Naut. Alm.) = 15h.41′.29" and on the 21st 15.45′.40",

Answer. Aldebaran culminates at 12h.42' (D. 273.), therefore it is on the 6 o'clock hour circle in the eastern hemisphere at 6h.42′ in the evening; and in the western hemisphere at 66.42′ the next morning.

Its altitude at each time being 12°.32'.3", and the azimuth 79°.46′.50′′ from the north.

2. Required the sun's altitude and azimuth at 6 o'clock, in latitude 51°.32′ N. when his declination is 15°.23′ North? Answer. Altitude 11°.59.17", azimuth 80°.17.14" from the North.

3. Given the sun's declination 19°.39' North, his altitude at 6 equal 15°.16'; required the azimuth and the latitude?:

Answer. Azimuth=77°.28′.37′′ from the elevated pole, lati→ tude 51°.32′ North.

4. Given the sun's declination=19°.39′ North, his azimuth at 6 in the morning N. 77°.28′.37′′ E.; required his altitude and the latitude?

Answer. Altitude-15°.16', latitude 51°.32′ North.

5. At what time will Regulus appear upon the six o'clock hour circle NSAS at Greenwich, latitude 51°.28'.40" North, and what will its altitude and azimuth be on the 6th February 1822?

The right ascension being 9.58.53" and declination 12°. 50'.1" N. (Table VIII.), also the sun's right ascension at noon on the same day (Naut. Alm.)=21.18'.41", and on the 7th February 21.22'.41′′?

Answer. Regulus culminates at 12.38' (D. 273.), is on the 6 o'clock hour circle, at 6.38' in the evening in the eastern hemisphere; and at 6h.38′ the next morning in the western hemisphere. Its altitude 10°.1'.11", and its azimuth=81°. 55'.28" from the north.

6. At what time will Castor appear upon the 6 o'clock hour circle at Greenwich, latitude 51°.28'.40" N., and what will its altitude and azimuth be on the 1st of December 1822?

The right ascension being 7h.23.13" and declination 32°.16' 10" N. (Table VIII.), also the sun's right ascension on the first of December at Noon (Naut. Alm.)=161.28′.16′′, and on the second 16h.32′.35′′.

: Answer. Castor culminates at 14.52'.16", (D. 273.) will be on the six o'clock hour circle at 8.52.16" in the evening, in the eastern hemisphere, and at 8h.52'.16" next morning in the western hemisphere; altitude=24°.41′.21′′, azimuth=68°.31′. 55" from the north.