EXAMPLE II. Suppose the sun's altitude at London, latitude 51°.32'North, to be 10.14' when the declination is 20°.11'South, what is the azimuth from the North ? Answer. 1250.34'. EXAMPLE III. Given the latitude of the place 51°.32' North, the sun's altitude 25°, his declination 4o.47' South; required the azimuth from the North ? Answer. 137°16'. EXAMPLE IV. Suppose in latitude 510.32' North, the altitude of the star Arcturus in Bootes was 440.30', when its declination was 20°. 16' N.; required its azimuth from the North at that time. Answer. 1179.8'. EXAMPLE V. In latitude 480.51' North, when the sun's declination is 18o. 30' North, and his altitude 520.35'; what is his azimuth from the North? Answer. 1349.36'.8". (D) This problem is not only useful for finding the variation of the compass, but it will be found very convenient for determining the true meridian, and the four cardinal points. For, if with a line representing the vertical on which the sun is, an angle be made equal to the supplement of the azimuth from the North, the meridian will be obtained: provided care has been taken to correct the sun's altitude by refraction. (E) An azimuth may be constructed by a scale of chords, sufficiently accurate for Nautical purposes; Thus, With the chord of 60° describe a circle (Plate IV. Fig. 2.), set off the complement of the latitude from z to P, and the complement of the altitude from z to a, draw zon through the centre c and aca at right angles to it. Set off the polar distance from P to p, draw pp through the centre c, and from the point i where it intersects aoa draw Ir parallel to zon, and through c the centre draw cs parallel to aoa. With c as a centre, and the distance oa, cross the line in in m; through m draw cn, then sn, measured on a scale of chords will be the azimuth from the south in North latitude) if the point i fall on the left-hand of zn, and from the north if it fall on the right-hand. See the Mathematical Repository, vol. i. page 283, second edition, 1799, and vol. ïi. page 26 and 27. See also Dr. Mac PROBLEM XIV. (Plate III. Fig. 1.) (F) Given the latitude of the place, the sun's declination and altitude, to find the hour of the day. EXAMPLE I. At London, latitude 51°.32' North, the altitude of the sun's centre was 38°.19', and his declination 19o.39' North, required the hour of the day? In the triangle szn, we have given To find the angle zns, measured by the arc Æd, the distance between the meridian which the sun is on at the time of obseryation, and the meridian of the place of observation. Since Æz is equal to the latitude on; it follows, that if the latitude and declination be both north or both south, the difference between them is equal to the meridional zenith distance; if the one be north and the other south, their sum is the meridional zenith distance. Hence the following general method of solution : Nat cos of merid zenith dist (51°.32' - 190.39') 31°.53=84913 Nat sine of the sun's altitude 38. 19=62001 Num.in col.ris., Requisite Tables, 35.30 4:59230 Or, Increase the index by 5+ Log sine of half the hour angle 26°.14'.40"=9•64563 2 The hour angle 52.29.20=35.29'.57".20". (G) This solution is analogous to Problem 8th in the Requisite Tables published by order of the Commissioners of longitude; or to Problem VII. Chap. II. Book III. of Dr. Mackay's Navigation. If you have the Requisite Tables * by you, the sum of the three logarithms above may be looked for in the column of rising, and the hours and minutes answering thereto will be the time from noon. Or, increase the index by 5, and the sum is the log. versed sine of the same angle. "Or, the-log. versed sine of an arc increased by the logarithm of 5 (with 9 for its index), will give double the logarithmical sine of half the required angle. (P. 131.) EXAMPLE II. In latitude 39o.54' North, longitude 80°.39'.45" West of Greenwich, suppose the altitude of the sun's lower limb to be 150.40%.57" on the 9th of May 1822, at 5h.34'21" P.M. per watch. Required the error of the watch. 1. To find the time at Greenwich, nearly. 80°.39',45" x 4=5h.23.39" the difference of time between the clocks at Greenwich and the place of observation. Hence, 5h.34.21" +5h.22.39"=109.57' P.M. the time at Gree: .wich. 2. To correct the O's declin. 3. To correct the O's altı yde. O's decl. May 9th 1822 = 170.16'.51" || O's observed alt. =15- 40.57 Ditto May 10th =170.32.47" Refraction t 8'.17" * The XV Ith of the Requisite Tables, or the XXXV, XXXVI, and XXXVII Tables in Dr. Mackay's Navigation. : The XVIth of the Requisite Tables is divided into three columns. The first column, called half elapsed time, is only a table of secants with 10 rejected from the index ; the time being reduced to degrees. The second, called mildle time, is the logarithmical sine of half the elapsed time, increased by the logarithm of 2, and the index of the sum diminished by 5. I he third, called column of rising, is a table of logarithmical versed sines with their indices diininished by 5. Thus for 1 hour of time, or 15 degrees. co-sec of 15°= 10:58700, sine of 15° –9.41300, ver, sine of 15o=8.53243 -10 log. of 2 =0.30103, -5 Log. elap. time 0.58700 sum 9.71403 5 3.53243 logarith. Log. mid. time 4•71403 Proceed in the same manner for any other time. 4. To find the true time. In the triangle szn, reject indices. Complement of latitude nz=50°. 6. 0" cosecant - 11811 Complement of altitude Sz=74. 6. 20 Polar distance Sn=72. 35. 53 cosecant - 02035 (H) In the practical application of this problem, it will be proper to take several altitudes of the sun within a minute or two of each other, and to note the corresponding times per watch. The mean of these altitudes, and of the times, will be more correct than a single altitude and time, and will in a great measure counteract the errors arising from the imperfection of the instrument. EXAMPLE III. On the 4th of March 1822, in latitude 450.37' North, and longitude 169o.59'.30" West, suppose the following altitudes of the sun's lower limb to be observed : Times per watch P.M. A watch P.M. Altitudes of the sun. 2h 53.32' 24°.55 2. 54. 30 24. 48 2. 55. 36 24. 40 2. 56. 46 24. 31 4 | 11. 40. 24 4 | 98. 54 Mean 2. 55. 6. P.M. Mean 24. 43. 30 Required the apparent time of observation, and the error of the watch. 18 The sun's declination at Greenwich at noon being 6o.31'.59" S. (Naut. Alm.) and on the 5th 6°.8'.53" S. also the O's semidiameter 16.9". Answer. The time at Greenwich is 14h.15'.4", the O's declination at that time = 6°.18'.16" S. true altitude of the O's centre=240.57.45"; hence the apparent time of observation = 2h.55'.50", and the watch is 44" too slow. EXAMPLE IV. On the 15th of September 1822, in latitude 33.56' South, and longitude 84°.6.15" West, suppose the mean time per watch of several altitudes of the sun's lower limb to be sh.11'.5' A.M. and the mean of the altitudes themselves 24°.43'; required the error of the watch ? The O's declination at noon (Naut. Alm.) on the 16th being 22.46'.24" N. and on the 17th 20.23.11", also the O’s semidiameter 15'.56". Answer. The time at Greenwich on the 16th is 1h.47.30" P.M. the O's declination at that time=20.44'.40" N. true alt. of the O's centre=240.57'.2"; hence the apparent time is 3h.48'. 54" from noon, or 86.11.6" A.M. and the watch is right. PROBLEM XV. (Plate III. Fig. 1.) (I) Given the latitude of the place, the declination and altitude of a known fixed star, to find the hour of the night when the observation was made. EXAMPLE I. In latitude 70.45' South, and longitude 30°.18' East of Greenwich, on the 7th of September 1822, suppose the altitude of the star Procyon, when east of the meridian, to be 28°.12'; required the true time of the day? The declination of Procyon being 59.40'.29' N. Procyon's alt =(28o.12! – refraction 1.47"=)28°.10.13" Co-lat NZ =82o.15' Cosec •00399 Co-alt sz =619.49.47" Polar dist Sn =950.40'.29" Cosec ·00213 |