REDUCTION TO THE MERIDIAN. 1. To find the Greenwich Time. To the time by watch apply the error for app. time at ship; add if slow, subtract if fast. Then apply the diff. of long. in time; add if E., subtract if W.; the result is the app. time at ship; to which apply the long. in time; if E. subtract, if W. add; the result is the app. time at Greenwich. 2. To find the “ Time from Noon.” If P. M. the minutes and seconds of the app. time at ship; but if A. M. the hours, minutes, and seconds of the app. time at ship, subtracted from 24h. 00m. 00s., will be the time from noon. 3. Take out the dec. from page 1, and correct it as usual. 4. Correct the obs. alt. for index error and Table IX. 5. Take out the log. rising of the time from noon (Table XXIX), log. cosine of the cor. dec., and the log. cosine of the lat. by acc., the sum of these logs. (rejecting the tens) is the common log. of the natural number. 6. From Table XXVI take out the nat. sine of the true alt. to 5 places of figures, and add the nat. number to it; the sum is the nat. cosine (Table XXVI) of the meridian zenith dist. ; always to be named opposite to the bearing of the Sun. 7. Put the cor. dec. under the meridian zenith dist.; if they are the same names add, but if different names subtract. The result is the true latitude. EXAMPLE I. 1870, June 3rd, P. M. at ship, in latitude by account 47° 30' N., longitude 35° 15' W., the observed altitude of the Sun's l. l. South of the observer near the meridian 64° 2' 10", height of the eye 18 feet, when a watch shewed 2d. 20h. 17m. 10s., which was slow on apparent time at ship 4h. 9m. 178., and diff. long. to the East 20 miles. Find the true latitude. d. h. m. S. Time by watch 2 20 17 10 D. long. 20' E. Long. 35° 15' W. Slow + 4 9 17 4 4 Dec. 22° 19' 43' N. 52 Diff. for 1 hour + 18:44 2.8 Cor. dec. 22 20 35 14752 3688 51 632 m. S. Time from noon 27 47 Log. rising 2.86566 47 30 Cosine 9.82968 True Dec. 22 21 Cosine 9.96608 EXAMPLE II. 1870, April 12th, A. M. at ship, in latitude by account 43° 25' N., longitude 125° 35' E., the observed altitude of the Sun's l. 1. near the meridian was 55° 12 44", South of the observer, index error 1' 10", height of the eye 16 feet, time by watch 11d. 22h. 59m. 56s., which had been found to be 40m. 4s. slow on apparent time at ship, but since the error was determined the ship has made 13 miles d.long to the West. Required the latitude by reduction to the meridian. d. h. m. s. Time by watch 11 22 59 56 D.long. 13'W. Long. 125° 35'E. Error + 40 4 4 4 6,0)50,2 20 11 23 40 00 52 8 22 20 D.long:52 h. m. 8. App. ship time 11 23 89 08 App. time at ship 23 39 08 Long. - 8 22 20 24 00 00 App. Gr. time il 15 16 48 Time from noon 20 52 IIourly diff. + 55.04 Dec. 8° 20' 45" N. 15.3 Cor. + 14 2 16512 27520 True dec. 8 34 47 5504 60842-112 14 2 Obs. alt. 55 12 44 1 10 55 11 34 11 30 Cor. m. S. 11.5 Time from noon 20 52 Log rising 2.61712 0 True dec. 8° 35' Cosine 9.99511 Lat by acc. 43° 25' Cosine 9.86116 11.5 6 Nat. number 297....2.47339 Nat. Cosine 82594 True dec. 8 35 N. Cor. + True alt. 55 23 04 Latitude 42 54 North. LATITUDE BY STAR. 1. Put down the observed alt., and correct it for index error; subtract the cor. for dip. (Table V.), and for refraction (Table IV.); the result is the true altitude. 2. Subtract the true alt. from 90° 00'00" for the Z. dist., which is named opposite to the bearing. 3. Put the dec. under the Z. Dist. ; add them, if the same names, subtract if different names; the result is the latitude. EXAMPLE. 1870, January 1st, the observed meridian altitude of the star Spica, bearing S., was 44° 25' 30", index error -- 2 45", height of the eye 12 feet. Required the latitude. Obs. alt. 44° 25' 30" 2 45 44 22 45 Dip. 3 19 44 19 26 Ref. 0 59 90 00 00 28 46 S. 41 TO CORRECT COURSES FOR DEVIATION. Allow the deviation the same way as the variation, that is, Easterly deviation to the right, and Westerly deviation to the left. As the deviation is given in degrees, you must convert the given course into degrees before applying the deviation. EXAMPLE. Correct the two following courses for deviation from the table given at page 42. SSW. 1 W.. .S. 28° 70 W. N.W. .N. 45° 00' W. Deviation 4 2 left. Deviation.... 7 52 left. Corrected course ..S. 2+ 5 W. Corrected course N. 52 52 W. EXPLAIN THE DEVIATION OF THE COMPASS, AND SHOW HOW IT MAY BE FOUND ? Deviation of the Compass, or local attraction in a ship, is the effect of the iron of the ship on the Compass or Compasses; it different on each course steered, and the deviation of one ship will not do for any other. The deviation generally is the greatest when the ship’s head is E. or W., and least when N. or S. To find it take the bearing of the centre of the ship from on shore with a compass, and at the same time take the bearing from on board the ship to the compass ashore; the bearings will read just opposite; if not, the difference is the deviation. In correcting courses, allow it always the same way as the variation ; that is, Easterly deviation to the right; Westerly to the left. MASTERS ARE REQUIRED TO CORRECT COURSES FOR THE DEVIATION OF THE COMPASS BY THIS TABLE OF N.N.W., W.S.W. } W.E.N.E. & E., W. by S., S.E., W. N., N. 80° E., S. 45° W., N. 11° 15' W., E. 20° N., S. 50° E., S. 90° E. CURRENT SAILING. Given the ship's position, the port, the set and drift of the current, and the ship's rate by the log.,—to show by a figure (or diagram) the course to be steered to fetch her port. RULE. Draw a line from the ship's position to the port, and another from the ship's position in the direction of the set of the current, and lay off on it the drift in any given time. Open a pair of compasses to the extent of the ship's rate in the given time, and with one leg on the current point describe a circle, cutting the line joining the ship's position to the port; then draw a straight line through the ship's position parallel to the straight line joining the current point to the point where the circle cuts the direct course to the port, this will be the course required. EXAMPLE. A current on the starboard beam sets three knots per hour, while the ship's rate is ten knots per hour; required the course to be steered to fetch her port. SET OF THE B 3' A Let A be the ship's position ; the current will set her to B (three miles) in an hour. Open the compasses to 10, the ship's rate, and with one leg on B describe a circle, cutting the direct course to her port in C. Through A draw a straight line parallel to BC. This line is the course to be steered. |