right ascension by the following theorems similarly deduced, viz. cos rs = sin P'p sin p ́s cos pp's + cos PP' cos p ́s, ;} (6.) =- -tan L sin i secl + tan l cos i These equations are obviously analogous to those we have just deduced for the latitude and longitude, the only difference being that here the obliquity enters the formulæ negatively. Hence by taking another subsidiary angle ', so that sin l tan p' = tan L 4. As for the angle of position PSP'p, it is easily found from the relation between the sines of angles and the sines of their opposite sides; for from hence we have 5. We have also, from the same consideration, cos a cos à = cos L cos l.... (10.) And when L = 0, as is always the case with the sun, we have cos @= = cos i secd.... (11.) tan a cos d (12.) tan / cosi .... The preceding formulæ have been deduced upon the supposition that the heavenly body has not gone beyond the first quadrant of right ascension from the vernal equinox. But they are applicable to all positions by simply regarding the mutations of signs in the several sines, cosines, tangents, &c. according to the arcs to which they refer. The right ascensions and longitudes, being reckoned from the first point of Aries through their respective circles, will at once indicate by their attendant signs, + or - "2 to which quadrant they belong. And as to the declinations and latitudes, they being regarded as positive when toward the elevated pole, will be regarded as negative when towards the contrary pole. Thus, with us, north latitudes and declinations will be positive, south latitudes and declinations, negative. Example I. The right ascension of Aldebaran being 67° 40′ 30′′, its declination 16° 8' 20" N. Required the longitude, latitude, and angle of position. Here, taking so that tan = ing see 4, cosec 4, and sec d, for in equa. 3, 4, and 9, they become sin L tan and sin p sin a tan d' and substitut Cos sin and cos d sin d cos (4 + i) sec, tan a sin (4+ i) cosec 4, the log. operations corresponding to which will be as follows: From log sin a..67° 40′ 30′′.... 9.9661625 ... Remains tan..72° 37′ 46′′....10.5047081 Then, to find L the latitude, add together, .... .72° 37′ 46′′.... 10·5249824 Here, because cos (4+ i) being in the second quadrant is negative (chap. iv. art. 4), and the other terms are positive, the product is negative, and therefore the latitude is south. ...... Next, to find the longitude, add together, cosec The sum tan l .... • 68° 29′ 28′′ 10-4044092 .... Here all the terms being positive, their product is positive; conseq. the longitude is in the first quad. Lastly, for p the angle of position, add together, Log sin i.... 23° 27′ 49′′.... 9.6000647 cos l.... 68° 29′ 28′′ 9.5642090 sec d 16° 8′ 20′′.... 10.0174616 ... The sum sin p... 8° 44′ 26′′.... 9:1817353 Example II. What are the latitude and longitude of the moon, when her right ascension is 304° 21', and declination 22° 57'.S? Ans. Longitude 10' 1° 21′ 54′′, latitude 3° 8′ 46′′ S. Example HI. When the longitude of the moon is 17° 41′ 23′′, and the latitude 3° 49′ 57′′ S., what are the right ascension and declination? Ans. Right ascension 36° 36′, declination 10° 28′ N. Example IV. When the sun's declination is 19° 18′ 20′′ N. what are his right ascension and longitude? First, to find the longitude, employ q. 15. To log sin d....19° 8' 20".... 9.5156873 .... 10.8999353 The sum is log sin l... 55° 25′ 43′′.... 9.9156226 Here since the declination is north, or positive, sin l is positive, and lies, therefore, either in the first or second quadrant, lis, 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.. 55° 25′ 43′′.... 9.7539143 Add log sec d 19° 8' 20" .... 10.0246938 The sum is log cos a ..53° 5′ 6′′.... 97786081 This reduced to time at the rate of 15° to an hour, gives 3h 32" 20, for the right ascension corresponding to the longitude 55° 25′ 43′′. If the other longitude had been taken, the resulting right ascension would have been 8h 27m 40s, 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. Example V. Given the sun's longitude 6' 8° 9' 36", to find the right ascension, declination, and time in 1816. Ans. 12h 29m 58', right ascension, 3° 14′ 24′′ S. declin.. time Oct. 1, 1816, at noon. PROBLEM II. 6. To investigate formula 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. H S' Z s' S" R Let HZR be the meridian of the place of observation, HVR the horizon, z the zenith, P the elevated pole, s the place of the sun, a fixed star, or other heavenly body, s's 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 Ps is = 90°-d the co-declination; PZ=90° – 1, the co-latitude; zs=n, the zenith distance, or = 90° a the co-altitude; ZPSP 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 = PZ±zs; and PR or 180° - PRRS" + PS" or нS' + PS'.. S (2.) = mer. alt. from north or south+ declin.. sum. of mer. alts. in a circumpolar star . 7. If the horary angle r = 90°, as it is when the body |
