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Erample IV.

When the sun’s declination is 19° 18' 20" N. what are his right ascension and longitude 2 First, to find the longitude, employ equa. 15. that is, sin l = sin d cosec i. To log sin d.... 19° 8' 20".... 9.5156873 Add log cosec i. 23° 27' 49".... 10-8999353

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Here since the declination is north, or positive, sin l is positive, and lies, therefore, either in the first or second quadrant, l is, therefore, either 55° 25'43" (that is, 1° 25° 25'43") as above expressed, or its supplement 124° 34' 17", that is, 4° 4° 34' 17".

To find the right ascension, take equa. 11,

that is, cos a = cos l sec d. * To log cos l .. 55° 25'43"..... 9.753914 Add log sec d. 19° 8' 20" .... 10.0246938

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This reduced to time at the rate of 15° to an hour, gives 3" 32° 20', for the right ascension corresponding to the longitude 55° 25' 48". If the other longitude had been taken, the resulting right ascension would have been 8" 27" 40°, the supplement of the former to 12 hours.

The days that correspond to these in the Nautical Almanac for 1816, are May 16 and July 27.

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PRoBLEM II.

6. To investigate formulae that shall be applicable to inquiries in reference to the times of risings and settings of the heavenly bodies, the azimuths, the duration of twilight, &c. Let HzR be the meridian of the place of observation, Hv R the horizon, z the zenith, So Z s” s P the elevated pole, s the place == of the sun, a fixed star, or other heavenly body, sos s” part of the circle it appears to describe about the pole P during the earth’s diurnal rotation, zsv part of the vertical circle on which the body is at any proposed time, and Ps a portion of a great circle passing through the pole and the place of the body. Then Psis = 90° – d the co-declination; Pz = 90° - l, the co-latitude; zs = n, the zenith distance, or = 90° — a the co-altitude; zPs = P the horary or polar angle between the two meridians ZP, SP; and SZP = z, the azimuth, measured also by the arc VR. Then the formula we may first employ in the present investigation, is equa. 2 of chap. vi. which suited to the case before us, is cos zs = cos Ps cos Pz + sin Ps sin Pz cos P, ... (1*) or adopting the characters just specified cos n = sin d sin l + cos d cos l cos P ...... (1.) Suppose P = 0, or the body in the meridian, then cos P = rad = 1, and the fundamental equation becomes, cos Zs = cos Ps cos Pz + sin Ps sin Pz = cos (Ps Pz) whence zs = + (Ps — Pz) = + Ps + Pz; ps = Pzitzs; and PR or 180° - PR = Rs” + Ps" or Hs’ + Ps'. . l ) (2.

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is 6 hours from the meridian, then its cosine will vanish, and equa. 1, becomes cos n or sin a = sin disin l . . . . . . . . (3.) from which the altitude, a, becomes known. In the same circumstances theorem 10, of right angled spherical triangles, gives cot szP = cot Ps sin Pz = tan d cos l .... (4.) by which the azimuth in that case may be determined. 8. When the body s is in the horizon, or s and v coincide, we have zs = 90°, or cos n =0; hence 0 = sin d sin l + cos d cosol cos P, sin d sin

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— tan d tan l ... (5.) Here it is evident that when the declination and latitude are both of the same kind, cos P is negative, or P is greater than 90°; that is, the time occupied by the heavenly body in passing from the horizon to the meridian, or from the meridian to the horizon, exceeds 6 hours. If the declination and latitude are of different kinds, cos P will be positive, and the time of passage from the horizon to the meridian, or from the meridian to the horizon, less than 6 hours. Thus, this theorem will serve to determine the times of the risings and settings of the sun, or other heavenly bodies, disregarding the changes of declination, and the effects of parallax and refraction. To find the azimuth when s is in the horizon, we have from the principles of quadrantal triangles,

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COS Z E — = . . . sin PZ coS 9. If the body s be upon the prime vertical, then Pzs = 90°, and the formulae for right angled triangles give cos P = tan zP cot sp = cotl tan d .... (7.)

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10. When the parallel circle so s” which a heavenly body describes in consequence of the diurnal rotation can have a vertical zsv drawn to touch it, which it may while the declination of that body exceeds the latitude of the place, then the body when at the place of contact of the two circles will move vertically, and with the greatest apparent rapidity. In that case the spherical triangle Psz will be right angled at s, and we shall have from the formulae for right angled triangles in chap. vi.

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11. Returning to the first equation
cos Zs = cos Ps cos PZ + sin Ps sin Pz cos P

it may evidently be opo to determine the altitude, or the zenith distance, when the latitude, declination, and horary angle, are known. The same may also be readily effected by an auxiliary angle, as explained in case 2 of oblique spherical triangles, equa. 12 and 13.

If we wish to find the hour from the meridian, by the observed zenith distance, we have from the above equation coS ZS - coS PS coS PZ

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a theorem fitted to logarithmic use. A like theorem will obviously serve for the determination of the azimuths, viz.

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12. Since, according to the determinations of astronomers, the crepusculum or twilight commences and terminates when the sun is 18° below the horizon, either previously to his rising, or subsequently to his setting, it follows that equa. 12 will serve to compute the time from noon to the beginning or end of twilight, if zs the zenith distance be assumed equal to 90°,+18°, or 108°. Or, to determine the duration of twilight, regarding the effect of horizontal refraction, we, have, these formulae: . - - cos P’ = — sin 18° sec l sec d tan l tan d, for the hour angle for the commencement of the morning, or the end of the evening twilight. cos P = — sin 34' sec l sec d – tan l'tan d, for the hour'angle from noon to sunrise or sunset, including refraction. And to (P’ - P) = duration of twilight..... (15.) These theorems may be employed in the solution of a variety of questions: a few are subjoined.

Example I.

At Cambridge, in north latitude 52°12'35", when does the sun rise and set, on the 4th of May, 1816, and what is the azimuth at rising and setting? The sun's declination on the proposed day at noon, is 16° 0'46" north; and the formulae to be employed are the 5th and 6th. For the first, viz. cos P = — tan d tan l, we have,

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