MIDDLE LATITUDE SAILING BY LOGARITHMS For ordinary purposes the course and distance found by inspection is quite satisfactory; but when the distance between the two points is too great or an exact result is wanted, it is necessary to use logarithms and is the method usually required at examinations. A general description of the various operations is not given here; but in its place an individual explanation follows each example. The following table contains the necessary rules for the solution of any problem by middle latitude sailing: 36° 51′ 00′′ N. EXAMPLE NUMBER I Both latitudes and longitudes to find the course and distance: A ship is in latitude 36° 51′ 30′′ N. and longitude 70° 55′ 45′′ W. Find the course and distance to Barnegat light in latitude 39° 45′ 52′′ N. and longitude 74° 06′ 24′′ W. EXPLANATION Find the difference of latitude by subtracting the latitude of the ship from that of Barnegat light, convert it into miles and name it north because Barnegat light is north of the latitude of the ship. Find the difference of longitude by subtracting the longitude of the ship from that of Barnegat light, convert it into miles and name it west because Barnegat light is west of the longitude of the ship. Add together the latitude of the ship and that of Barnegat light. Then divide by 2, which will give the middle latitude. To the logarithm of the difference of longitude 190'.6 add the logarithm cosine of the middle latitude 38° 18′ 26′′, which gives the logarithm of the departure 149′.6 and from that subtract the logarithm of the difference of latitude 174'.9 and the result will be the logarithm tangent of the course N. 40° 32′ W. To the logarithm secant of the course N. 40° 32′ W. add the logarithm of the difference of latitude 174'.9, which gives the logarithm of the distance 230'.1. EXAMPLE NUMBER 2 Both latitudes, one longitude and departure to find the course and distance, and longitude in: A ship left a point in latitude 37° 55′ 25′′ N. and longitude 74° 56′ 22′′ W. and arrived in latitude 35° 57′ 25′′ N., the departure being 216'.5 E. Find the course and distance and longitude of the ship. ind the difference of latitude by subtracting the latitude of the ship from that of the point left, convert it into miles and name it south because the ship has sailed south. Add together the latitude of the point left and that of the ship. Then divide by 2, which will give the middle latitude. From the logarithm of the departure 216'.5 subtract the logarithm cosine of the middle latitude 36° 56′ 25′′, which will give the difference of longitude 270'.9, which is subtracted from the longitude of the point left because the ship has sailed east and by doing so has decreased the longitude. From the logarithm of the departure 216.5 subtract the logarithm of the difference of latitude 118'.0, which will give the logarithm tangent of the course S. 61° 24′ E. From the logarithm of the departure 216'.5 subtract the logarithm sine of the course S. 61° 24′ E. and the result will be the logarithm of the distance 246′.6. EXAMPLE NUMBER 3 One latitude, one longitude, course and distance to find the latitude and longitude of the ship: A ship sailed S. 65° E. for 362'.0 from a point in latitude 35° 15′ 17′′ N. and longitude 75° 31′ 16′′ W. Find the latitude and longitude in. To the logarithm of the distance 362′.0 add the logarithm cosine of the course S. 65° E., which gives the logarithm of the difference of latitude 153'.0 and is subtracted from the latitude left because the ship has sailed south. To the logarithm of the distance 362'.0 add the logarithm sine of the course S. 65° E., which gives the logarithm of the departure 328'.1, |