Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

tiont, to find his amplitude, ascensional difference, and the time of his rising and setting.

In the right-angled spherical triangle abs or Agc. 1. AB or ag=the sun's, or star's, ascensional difference. 2. bs or gc=the sun's, or star's, declination. 3. As or ac=the amplitude.

4. BAS, measured by the arc go; or cag, measured by the arc HÆ,=the complement of the latitude.

5. ASB or ACB=the angle formed between the horizon and the meridian passing through the sun.

Any two of the above five quantities being given, the rest may be found.

(T) The same things may be found from the right-angled triangle son or chs.

For, in the triangle son, so is the complement of As ; sn is the complement of bs; on is the complement of the arc go, which measures the angle Bas; and the angle sno, measured by the arc Bg, is the complement of the ascensional difference. In the same manner the triangle chs may be compared with the triangle agc.

EXAMPLE I.

Given the latitude of Lordon 51°.32' N. on the longest day when the sun has 230.28' N. declination; required the sun's amplitude, ascensional difference, time of rising, setting, the length of the day and night, &c.

1. In the triangle Ass to find the amplitude As.

*

Rad x sine bs=sine Bas x sine As. ::: (10+ log sine Bs) – log sine bas= or, Sine BAS=38°.28': rad::sine Bs=239.28': sine as= 39o.48'.

=log

sine As.

+ This problem may be applied to the rising or setting of a planet, or of a star. But where great exactness is required, the declination of the sun or planet must be calculated as near to the time of rising and setting as possible; especially for the moon on account of her swift and irregular motion. Likewise, the declination of the sun near the equinoxes alters considerably in the compass of an hour. Hence what is generally called a parallel of the sun's declination, and drawn as such in the projection of the sphere, is not strictly a parallel circle, but a spiral line,

When the latitude of the place and the declination of the object have the same name, the right ascension diminished by the ascensional difference leaves the oblique ascension, and increased by the same gives the oblique descension.

But when the latitude of the place and declination have contrary names; the right ascension increased by the ascensional difference gives the oblique ascension ; and being diminished by the ascensional difference leaves the oblique descension,

viz. cosine of the latitude, is to radius; as the sine of the sun's declination is to the sine of the amplitude.

(U) This part of the problem is useful in navigation, for finding the variation of the compass.- It is evident, by comparing the triangle als with the triangle agc, that the amplitude is always of the same name with the declination, whether north or south.

2. To find the ascensional difference AB.

Rad x sine AB=cot BAS X tang Bs. ::: (log cot Bas+log tang bs)—10=log sine AB. or, Rad : cot Bas=38°.28'::tang. Bs=23°.28' : sine AB=330.7'. viz. Radius, is to the tangent of the latitude; as tangent of the sun's declination, is to the sine of the ascensional difference.

(W) By comparing the triangle ABS with the triangle AGC, it is evident that when the sun's declination and the latitude of the place are of the same name, the ascensional difference in time subtracted from six hours gives the time of rising, and added to six hours gives the time of sun-setting: but when the latitude of the place and the sun's declination are of contrary names, the ascensional difference added to six hours gives the time of sun-rising, and subtracted gives the time of setting.

In the present case the sun rises at 3h.47'.32", and sets at gh.12.28"; double the time of rising gives the length of the night, and double the time of setting gives the length of the day.

(X) When the sun's declination is equal to, or greater than the complement of the latitude; viz. when go is equal to or exceeds go (a circumstance that can never happen in latitudes lower than 660.32') the parallel of declinations will not cut the horizon Ho, and consequently the sun will never set at these times. The same observations hold true with stars whose co-declination or polar distance cn, is equal to, or less than, the latitude of the place, or elevation of the pole on, and in the same hemisphere.

EXAMPLE II.

Required the amplitude, time of rising, culminating, and setting of Arcturus at Greenwich, latitude 510.28.40" N. on the first of December 1822, its right ascension being 146.7'.32" t, and declination 20°.6'.50" N.

+ Sec Table VIII, at the end of the Book.

The sun's right ascension at noon December 1st (Naut. Alm.) being 16h 28'.16", and on December 20, 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=380.316.20" : rad :: sinebs=20°.6'.50" : sineas= 330.30'.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 = 270.23.21". Then 270.23'.21" x 4 = 19.49'.33", and 6h. + 16.49'.33"=7h. 49'.33" the arc BADÆ, or half the time of the star's continuance above the horizont. 3. Find the time of the *'s culminating, Prob. VII. (D. 273.) Then, From the time of the *'s culm. (D. 273.)=214.35'.23". Take the semi-diurnal arc

= 76.49'.33".

*

Time of the *'s rising, December 1st - = 13h.45'.50".

And, To the time of the *'sculminating (D.273.)=216.35'.23". Add the semi-diurnal arc

= 76.49'.33".

Sum 295.24.56".

Then 295.24'.56". -- 24h=5h.24'.56". time of setting, De

cember 2d.

† The amplitude of this star on the same day, in the year .1796, was 330.45.46", 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

(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=390.48' N. his declination= 239.28' N.; to find the latitude of the place, time of sun rising and setting, and the length of the day and night.

-The latitude= 51°.32' N.

Sun rises at 3h.47.32". Answer.

Sun sets at 8h.12.28".

Length of day=165.24'.56" length of night=7h.35'.4". 2. Required at what time Sirius, the Dog star, will rise, culminate, and set at Greenwich lat. 510.28'.40" N. on the 18th of December 1822.

The right ascension of Sirius being 6h.37'.18", and declination 16o.28.39" S; the sun's right ascension at noon (Naut. Alm.) being 17h.42.55", and on December 19th at noon 176.471.21".

Ascensional difference=219.48'.43"=19.27.13".

Semi-diurnal arc=45.32'.47".
Answer.(Sirius will be on the meridian at 12.52'.

rise at 8h.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 154.41'.29", and on the 21st November, at noon, 15h.45.41".

Ascensional difference=21°.19'.19"=1b.25'.17".
Semi-diurnal arc 76.25'.17".

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

rises at

56.16'.44". evening. sets at 8h. 7.18". next morning. The amplitude

= 260.30'.53". towards the N. 4. Given the latitude=510.32 N., the sun's amplitude

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

Sun's declination=230.28'. N.

Ascensional diff. = 33°. 7.
Answer.

Sun rises at 36.47.32".
Sun sets at 8h.12.48".

PROBLEM II. (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 510.32' North, when the sun has 19o.39' North declination; required his altitude and azimuth at 6 o'clock.

1. To find the altitude ds, in the triangle ADş.

Rad x sine ds=sine A x sine As.

:: (log sine A +log sine as) -10=log sine ps. or, Rad :siness=190.39':: sinesad=519.32': sineps=150.16',

2. To find ad the co-azimuth, in the triangle Ads.

Rad x cos A=cot As x tang Ad.

(10+log cos A)- log cot as=log tang AD. or, Cot as = 19o.39' : rad::cos SAD = 512.32' : tang AD =

« ΠροηγούμενηΣυνέχεια »