CHAPTER XVII. LONGITUDE BY CHRONOMETER. ART. 82. Longitude by Chronometer. The time by chronometer when corrected for its original Error and accumulated Rate gives the time at Greenwich, whilst the time at ship can be calculated from an observed Altitude of a Celestial Object and the known Latitude. The difference converted into Arc is the Longitude. The Longitude is E. when Greenwich time is less than ship time, and W. when greater. The following are required in the Formula : (a) The true Altitude. (b) The Polar Distance. (c) The Latitude. The Polar Distance is measured from the elevated Pole, which N S FIG. 134. is the N. Pole in N. Lat., and the S. Pole in S. Lat. It is therefore 90°- Dec. when the Dec. is of the same name as the Lat., and 90° + Dec. when of a different name. ART. 83. To find the Hour E Angle of a Celestial Object, the Altitude, Latitude, and Polar Distance being known.-Let O be the object, then PO is its polar distance (p), ZO its zenith distance (90° - a), and PZ the co-latitude (90° - l); the angle ZPO is the hour angle. Therefore, in the spherical triangle PZO, three sides are given to find an angle ZPO. Let the letters a, l, p, and h stand for altitude, latitude, polar distance, and hour angle respectively; which is the formula used for computing the hour angle in the longitude by chronometer and other time problems. ART. 84. To find the Time when the Sun or other Celestial Object is on the Prime Vertical.-Let O be an object on the prime vertical; then PO is its polar distance, PZ is the co-latitude, and the angle P is the hour angle. PZO is a spherical triangle, right angled at Z, of which the circular parts are PZ and ZO, and the complements of PO and the angles P and O. N FIG. 135. Taking complement of P as middle part, PZ and the com plement of PO are adjacent parts: then sin (comp. of P) = tan PZ x tan (comp. of PO) that is, cos H.A. cot lat. x tan dec. In the case of the Sun, the H.A. is the interval from apparent noon. For any other object, add its R.A. to its H.A. west to find the R.A. of meridian, from which subtract the sun's R.A. to find the time. ART. 85.-The following rules must be observed in Sun chronometers : (1) Find the correct G.M.T. by applying the error when last found, and accumulated rate, to the time shown by chronometer. As in other problems, great care must be taken to make the time shown by the chronometer agree with the rough Greenwich time deduced from D.R. long. and ship time; both as regards the date and hours. (2) Correct declination, equation of time, and altitude. Find polar distance; and look at the heading of the equation of time column on page I., to see how it is to be applied. (3) Add together true alt., lat., and pol. dist.; find the half sum, and from it subtract the alt.; this gives the chron. remainder. Add the secant lat., cosec pol. dist., cos half sum, and sin remainder, reject 10's, and look out result in Table XXXI., Norie; this is the hour angle. NOTE 1. Should the pol. dist. exceed 90°, use sec dec. for cosec pol. dist. NOTE 2.-The lat. used must be the lat. at observation; if not given, it must be deduced from the lat. at noon. (4) If P.M. at ship, the hour angle is the apparent time at ship of the same date as given in question. If A.M. at ship subtract the hour angle from 24h and date the day before day in question for app. time at ship. To the A.T.S. apply equation of time for M.T.S. (5) The difference between the M.T.S. and M.T.G. is the longitude in time, which must be expressed in arc, for long. at obs. named west when G.M.T. is best, and east when G.M.T. is least. Examples. (1) August 8th, about 3h 50m P.M. at ship, in lat. 56° 48' N., long. D.R. 7° W.; when a chron. showed 4" 29m 34', the obs. alt. of the Sun's L.L. was 30° 57' 10"; ind. cor., -2' 10"; height of eye, 20 ft. Find the longitude, the chron. having been found 7" 46"-2 fast of G.M.T. on July 18th, and losing daily 2'7. Approx. time at ship, Aug. 84 3h 50m elapsed time 21.2 days Long. W. acc. loss 57.24 Construction of Fig. 136.-Draw circle NESW representing rational horizon, and NZS meridian. Measure NP = ZQ = 57° lat.; set off dec. 16 to north, and draw EQW for equinoctial, and dd parallel of declination. With centre Z and radius = zen. dist. 59° draw an arc cutting dd in X for position of sun, draw hour circle PXB, and vertical circle ZXA. = In triangle PZX; given PZ = co-lat., PX = pol. dist., ZX = zen. dist. Req. P Hour angle. = (2) September 1st, about 9" 15" A.M., in lat. at noon 46° 1' 30" S., D.R. long. 27° E.; when a chron. showed 7h 19" 43′, and had been found to be fast 3 115 on May 3rd, and slow 12' on June 27th, the alt. of Sun's L.L. was 24° 14' 40"; ind. cor., -3' 15"; eye, 29 ft. Find long. at sights; and if run from obs. to noon was S. 70° E. 30 miles, find long. at noon. |