EXAMPLE I. 1870, January 26th, in longitude 96° 45′ W. the observed meridian altitude of the Sun's 1. 1. was 47° 46′ 20′′, bearing S., index error + 2′ 4′′, height of the eye 16 feet. Required the latitude EXAMPLE II. 1870, February 27th, in longitude 155° E., the observed meridian altitude of the Sun's 1. 1. was 57° 46′ 10′′, bearing N., index error 4' 20", height of the eye 18 feet. Required the latitude. d. h. m. S. 27 00 00 00 4 Long. in time 10 20 00 E. 6,0)62,0 10 20 App. time at Green. 26 13 40 00 Add together the log Secant of the latitude and the common log of the departure, rejecting 10 from the index; the number corresponding will be the d. long. EXAMPLE. In latitude 26° 45', the departure made good was 16 miles. Required the d.long. by parallel sailing. Lat. 26° 45'. . . . Secant 0.049159 Dep. 16 ......Log. D.long. 17.92.. 1.204120 1.253279 COURSE AND DISTANCE BY MERCATOR'S SAILING. Find the d.lat and d.long., observing that if the given. latitudes and longitudes are the same names, the difference must be taken; but if they are different names, the sum; reduce both the d.lat. and d.long. into miles; then from Table III. take meridional parts for each of the given latitudes, adding or subtracting, as you do the given latitudes. EXAMPLE. Required the course and distance, by Mercator's sailing, from A to B. M.D.lat. 1629 D.long. 796 log. +10..12-900913 Course 26°2′ Secant 0.046463 ...... 3.211921 T. d.lat. 1125 ......3·051152 Course 26° 2' Tang. 9.688992 Dist. 1252....3.097615 Distance 1252 miles. Course S 26° 2′ W. The course is always named N. or S. and E. or W. according to the way in which the ship has to sail. Should the longitude of the two places be great, and opposite names, the sum exceeding 180° 00', subtract the sum from 360° 00' to obtain the difference of longitude, and change the name. EXAMPLE. Required the d. long. between A and B. Longitude B. 175° 50′ W. 351 40 W. 360 00 8 20 E. 60 D. long. 500 miles E. TIDES. 1. Take from the Nautical Almanac (pages 496 and 497) the establishment at London Bridge and the given place, the difference always to be taken between them; marking this difference, (add) if the place is greater than London Bridge, and (subtract) if less. This is called the constant difference. 2. Enter pages 494 and 495 of the Almanac, and take the second tide opposite the day preceding the given date under the month, adding or subtracting the constant difference as the case requires. There are four Cases. CASE I. Should the result be between 12 and 24 hours, subtract 12 hours, and the remainder will be the A. M. tide; and the first tide opposite the given date and under the month will give the P. M. tide. CASE II. Should the result exceed 24 hours, take 24 hours from it, and the remainder will give the P. M. tide; and the first tide opposite the preceding date will give the A. M. tide. (3.) 1870, March 27th. Find the A. M. and P. M. tides at Shoreham. *If this tide had not reached 12 hours, there would have been no A. M. tide. Try March 27th, at Sunderland. CASE III. In seeking the tide, should the second tide opposite the day preceding the given date be a blank thus use this rule : If the constant diff. is + (additive), take the first tide opposite the preceding date for the A. M. tide, and the first tide opposite the given date for the P. M. tide. If the constant diff. is (subtractive), take the first tide opposite the given date, increased by 24 hours, for the A. M. tide, and the second time opposite the given date for the P. M. tide. (4.) 1870, April 13th. Find the A. M. and P. M. tides at Hull. (5.) 1870, April 13th. Find the A. M. and P. M. tides at CASE IV. Should the result be less than 12 hours, then the first tide opposite the given date, increased by 24 hours, will give the A. M. tide, and the second tide opposite the given date the P. M. tide. *If this tide had been more than 12 hours, there would have been no P. M. tide. Try June 12th, at Chatham, |