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REDUCTION TO THE MERIDIAN.
To the time shewn by watch add the original error if slow, and subtract if fast; to the time thus found, apply the d.long in time, adding if the ship has sailed E., and subtracting if W. This is the true apparent time at ship; to it apply the longitude in time to obtain Greenwich time, as in preceding problems.
The apparent time at ship is the time from noon, when it is P. M.; but if the apparent time is A. M., it must be taken from 24 hours to find the time froin noon ; for example, if the following is the apparent time at ship:
With the true Greenwich time take out of the Nautical Almanac the declination, and correct it as in the preceding problems, which will give the true declination ; correct the observed altitude for index error, and correction from Table IX ; then use the following formula :
In taking out the Natural sine of the true altitude, only use the first five figures.
Time from noon ............ Log. rising (Table XXIX)......... Lat. by acct. ............ Cosine (Table XXV.)............. True declination ............ Cosine (Table XXV.) ........
(Table XXIV.) Nat. number ............
T. alt. ...... Nat. sine .... (Table XXVI.) Z. dist. ............... Nat. cosine ............ (Table XXVI.) T. dec.
EXAMPLE.—1864, April 12th, A. M. at ship, in latitude by account 43° 25' N., longitude 125° 35' E., the observed altitude of the Sun's 1.1. 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. 22b. 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.
MERIDIAN ALTITUDE OF A STAR.
Correct the observed altitude for index error, (Dip, Table V.) (Refraction, Table IV.,) the result will be the true altitude. Take the true altitude from 90°, as in meridian altitude of the sun; which will give the zenith distance; add or subtract the declination, and the result will be the latitude.
EXAMPLE.—1864, 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.
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 33. SSW. 1 W. ...... S. 28° 7' W. NW. .............N. 45° 00' W. Deviation .......... 4 2 left. Deviation ...... 7 52 left. Corrected course S. 24 5 W. Corrected course N. 52 52 W.
Given the ship's position, the port, and set and drift of the current, to show by a figure (or diagram) the course to be steered to fetch the port.
RULE.-Join the port and the ship, draw a line from the ship equal in length to the drift of the current, and exactly opposite in direction to the set of the current.
Draw a line from the port equal in length to this life, parallel to it, and in the same direction.
A line drawn from the ship to the extremity of the line last drawn will give the course to be steered.
The remaining line may be then drawn to complete the parallelogram, when the course to be steered will be seen by its diagonal.
EXAMPLE.—Let S represent the ship's position, P the port, and the arrow the set and drift of the current.
From S draw S A, equal to the arrow in length, but in the opposite direction. From P draw P B parallel to S A, equal to S A, and in the same direction. Join S B. Then S B is the course to be steered. The line A B completes the parallelogram SPBA.
EXPLAIN THE DEVIATION OF THE
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 is 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