PROBLEM XIII. (Plate III. Fig. 1.) (593) Given the latitude of the place, the sun's declination and altitude, to find the azimuth. EXAMPLE I. In latitude 51° 32′ North the sun's altitude was observed to be 50°, and his declination 23° 28′ North; required his azimuth from the North. In the triangle szN, we have given SN=66° 32′ the co-declination, or sun's polar distance; for DS is the sun's declination North. sz=40 O the co-altitude of the sun; for cs is his altitude above the horizon нo. ZN 38 28 the co-latitude of the place, or what on the elevation of the pole wants of 90°. To find the angle szN, measured by the arc of the horizon co, the sun's distance from the North. Call the co-latitude ZN the base, bisect zN in t, and draw the perpendicular sf. Then (484) tanzN=zt = 19° 14' : tan (SN+sz) = 53 16 :: tan (SN-sz) = 13 16 : tan ft = 42 10 Take the difference between ft and half the complement of latitude. Then Radius, sin of 90° : tan (ft-zt)=22° 56′ :: tan of the sun's altitude=cs=50° 10.00000 9.62645 10.07619 9.70264 : sin of azimuth from the east or west -30° 17′ Hence 90° +30° 17′ 120° 17′ the sun's azimuth from the North. (594) A slight review of Figure I. Plate III. will naturally point out to us the following observations: If the latitude of the place and the declination of the sun be of the same name, the sun is less than 90° from the elevated pole; therefore the co-declination is the polar distance. If the sun or star be in the equinoctial, it is 90° from the pole, and forms a quadrantal triangle. If the declination be of a contrary name with the latitude, the distance of the object from the elevated pole will be equal to the declination increased by 90°. (595) The preceding solution is on the supposition that the sun's declination, as taken from the Nautical Almanac, remains the same in all places, and at all times of the day. But this cannot be strictly true, for the declination is subject to a daily variation, and is adapted to the meridian of Green wich. An azimuth will not be materially affected by this change, but it will make a considerable difference when we want to find the time of the day. EXAMPLE II. Suppose the sun's altitude at London, latitude 51° 32′ North, to be 1° 14' when the declination is 20° 11' South, what is the azimuth from the North? Answer. 125° 34'. EXAMPLE III. Given the latitude of the place 51° 32′ North, the sun's altitude 25°, his declination 4° 47′ South; required the azimuth from the North. Answer. 137° 18'. EXAMPLE IV. Suppose in latitude 51° 32′ North, the altitude of the star Arcturus in Bootis was 44° 30′, when its declination was 20° 1' N.; required its azimuth from the North at that time. Answer. 117° 44'. EXAMPLE V. In latitude 48° 51' North, when the sun's declination is 18° 30′ North, and his altitude 52° 35'; what is his azimuth from the North? Answer. 134° 36'. (596) 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 for refraction, &c. (597) 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 aoa at right angles to it. Set off the polar distance from P to p, draw pp through the centre c, and from the point 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 IR 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 of ZN, and from the north if it fall on the right. See Leybourn's Mathematical Repository, old series, vol. i. page 283., second edition, 1799, and vol. ii. pages 26 and 27. See also Dr. Mackay's Navigation, first edition, 1804, page 200. PROBLEM XIV. (Plate III. Fig. 1.) (598) 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 19° 39′ North; required the hour of the day. In the triangle sZn, we have given SN=70° 21′ the sun's distance from the elevated pole. sz 51 41 the sun's co-altitude. ZN 38 28 the co-latitude. = To find the angle zNs, measured by the arc ED, the distance between the meridian which the sun is on at the time of observation, and the meridian of the place of observation. Since Az 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. Num. in col. ris., Requisite Tables, 3h 30m 4.59230 Log sine of half the hour angle 26° 14′ 42′′ 9.64563 The hour angle 2 52 29 24=3h 29m 57s 36t. 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 logarithmic sine of half the required angle (281). EXAMPLE II. In latitude 39° 54' North, longitude 80° 39′ 45′′ West of Greenv ich, suppose the altitude of the sun's lower limb to be 15° 40′ 57′′ on the 9th of May, 1840, at 5h 34m 21s P. M. per watch; required the correct apparent and mean time at that place, and the error of the watch. 1. To find the time by acct. at Greenwich, and equation of time. May 9d 5h 34m 21s time by watch. Eq. of time May 9, 1840 =-3 475 *The XVIth of the Requisite Tables, or the XXXVth, XXXVIth, and XXXVIIth 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 middle time, is the logarithmic sine of half the elapsed time, increased by the logarithm of 2, and the index of the sum diminished by 5. The THIRD, called column of rising, is a table of logarithmic versed sines with their indices diminished by 5. Thus for 1 hour of time, or 15 degrees. cosec 15 10.58700 sin 15° 9.41300 vers 15° 8.53243 = † Table IV. Nautical Almanac. Table VI 4. To find the correct time. In the triangle szN, Complement of latitude Nz=50° 6′ 0′′ cosec sum sz=74 6 25 SN 72 25 37 2 | 196 38 2 reject indices. •11511 cosec ⚫02076 (Subtract from App. time) 5 34 21 Time per watch 28 Watch too slow. (599) 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, 1840, in latitude 45° 37′ North, and longitude 169° 59′ 30′′ West, suppose the following altitudes of the sun's lower limb to be observed: App. time P. M. per watch. 2h 53m 32s 2 54 30 2 55 36 2 56 46 411 40 24 Mean 2 55 6 P. M. |