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minating, 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 m. 56 sec. 32", the daily difference between the rising, southing, and setting of any fixed star, in the same latitude.

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(N) 3. To find the time of a star's culminating.

RULE. Subtract the right ascension of the sun for the given day from the right ascension of the star, and the remainder will be the time of the star's culminating nearly.-If the sun's right ascension exceed the star's, add 24 hours to the star's before you subtract.

Take the increase of the sun's right ascension in 24 hours, and add to it 24 hours. Then,

This sum is to 24 hours as the star's right ascension diminished by the sun's, is to the time of the star's culminating.

Continuing the present example.

*'s right ascension (1813)+24 hours O's right ascension, December 1st, 1813

Time of the 's culminating nearly

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=38". 7. 8".

=16".29'. 2".

=21.38′. 6".

O's right ascension, December 1st, 1813, at noon =16".29. 2". O's right ascension, December 2d, 1813, at noon =16".33.21".

Increase of the O's right ascension in 24 hours = 0". 4′.19′′.

24.4.19" 24" :: 21.38'.6" : 21.34.13" the true time of Arcturus's culminating, or 9.34.13", December 2d, † in the morning.

(0) 4. To find the time of a star's rising and setting.

RULE. The difference between the time of the star's culminating and the semidiurnal arc leaves the time of rising, and their sum gives the time of setting.

Continuing the present example.

=

21.34'13"-7.49.52" 13".44'.21" time of the star's rising December 1st, or 1.44'.21" time of rising December 2d.

The astronomical day begins at noon, and is counted forward to 24 hours, Or the succeeding noon, when the next day begins, being 12 hours later than the civil day, which commences at the preceding midnight; thus, December 1st, at 21.34.13" astronomical time, is December 2d, at 9.34.13" in the morning, according to civil reckoning.

21.34.13" 7.49′.52′′=29.24'.5", or 5.24'.5", the time of setting December 2d, in the afternoon.

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.

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Length of day=16".24′.56′′lengthofnight=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 1813, the right ascension of Sirius being 6.36′.54′′, and declination 16°.27.51", S.

The sun's right ascension at noon being 17.43'.40', and on December 19th at noon 17".48′.7".

Answer.

Ascensional difference=1h.27.10".

Sirius will be on the meridian, at 12.50'.51"
rise at 8.18.1" in the evening.

set at 5.23.41" 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 1813. The right ascension of Aldebaran being-4".25′.12′′, declination=16°.7′.27′′ N.

The sun's right ascension at noon being 15".42'.12", and on the 21st November, at noon, 15".46'.24".

Ascensional difference 1.25'.10".
Aldebaran culminates at 12".40'.46" at night

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5.15.36" evening

8h. 5′.56" next morning =26°.28′.53" towards the N.

4. Given the latitude 51°.32' N., the sun's amplitude= 39°.48' N. of the east; required the sun's declination, ascensional difference, time of rising, setting, &c.

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(P) The latitude of the place, and the sun's (or a star's) de

clination

clination being given, to find the altitude and azimuth, &c. at 6 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 6 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.

(Q) 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, is 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 510.32′ North, when the sun has 19°.39' North declination; required his altitude and azimuth at 6 o'clock. 1. To find the altitude DS.

Radx sine DS=sine AX sine as.

Rad: sine as 19°.39′ :: sine SAD=51°.52′ : sine Ds=15°.16′.

2. To find AD the co-azimuth.

*

Rad x cos Acot AS X tang AD.

Cot As=19°.39' : rad::cos SAD=51°.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′′ Ñ., and what will its altitude and azimuth be on the first of April 1813, when its right ascension is 14.7.8" and declination 20°.9'.48" N.

The sun's right ascension on the firstof April being 0.41′.50′′, and on the second of April 0".45.28′′.

1. To find the time of the star's culminating. (N. 252.) Right ascension of the sun 1st April 1813 at noon 0.41'.50' Right ascension of the sun 2d April 1813 at noon 0.45′.28

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:

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0. 41'.50"

13.25'.18"

24.3'.38" 24". :: 13.25.18" 13h.23.16" true time of the star's culminating or passing the meridian. Consequently (Q. 254.) it will be upon the 6 o'clock hour circle at (13.23.16′′ 6) 7.23.16" in the evening, in the eastern hemisphere; and at (13.23.16"+6") 19".23.16", or 7".23'.16" the next morning, in the western hemisphere.

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2. To find the altitude Ds.

*

Rad x sine Ds=sine SAD X sine As.

Rad sine As=20°.9′.48′′ :: sine SAD=51°.28′.40": sine DS= 15°.38'.44".

3. To find the azimuth ▲ from the east.

*

Rad x cos SAD=cot AS X tang AD.

Cot As 20°9′.48": rad :: cos SAD=51°.28.40′′ : tang AD= 13°.12′.30". Hence оD=76°.47′.30" the azimuth from the north.

(R) 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 6 o'clock hour circle; and the difference of the times it arrives there, may be considered equal to the diurnal difference of the sun's right ascension.

PRACTICAL EXAMPLES.

1. At what time will Aldebaran appear upon the six o'clock hour circle at Greenwich, latitude 51°.28'.40" N., and what will its altitude and azimuth be on the 20th November 1813, when its right ascension is 4".25'.12", and declination 16°.7.27" N. The sun's right ascension on the 20th November being 15h.42′.12′′ and on the 21st 15h.46′′.24′′.

Answer. Aldebaran culminates at 12".40′.47′′ at night, therefore it is on the 6 o'clock hour circle in the eastern hemisphere

at

at 6.40.47" in the evening; and in the western hemisphere at 6.40′.47′′ the next morning.

Its altitude at each time being 12°.32'.57" and the azimuth 79°.47′.36" 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, latitude 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 of February 1813, when its right ascension is 9.58'.24" and declination 12°.52.41" North.

The sun's right ascension at noon on the same day being 21.19′.25′′, and on the 7th February 21".23′25′′?

Answer. Regulus culminates at 12.36′52′′ at night, is on the 6 o'clock hour circle, at 6".36.52" in the evening in the eastern hemisphere; and at 61.36′.52′′ the next morning in the western hemisphere. Its altitude=10°.2′.30", and its azimuth =81°.48′.50" from the north.

6. At what time will Castor appear upon the six o'clock hour circle at Greenwich, latitude 51°.28'.40" N., and what will its altitude and azimuth be on the first of December 1813, when its right ascension is 7.22.39", and declination 32°.17.14" N. The sun's right ascension on the first of December at noon being 16".29.2, and on the second 16.33'.21".

Answer. Castor culminates at 14.50′.56", will be on the six o'clock hour circle at 8.50'.56" in the evening, in the eastern hemisphere, and at 8.50.55" next morning in the western hemisphere.

Altitude 24°.42′.10′′, azimuth—68°.31′.18′′ from the north.

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

The latitude of a place, and the declination of the sun (or of a star) being given; to find the altitude, and the time when it will be due east and west.

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