CASE I. Course and distance sailed given, to find the difference of latitude and departure from th meridian. A ship from the latitude of 49° 57′ N., sails S. W. by W. 244 miles; required the latitude she is in, and her departure from the meridian sailed from. BY PROJECTION. Draw the line CA, to represent the meridian of the place C, from whence the ship sailed. With the chord of 60° in your compasses, and one foot in C, as a centre, describe the compass W. S. E. Take 5 points in your compasses from the line of rhumbs on the plane scale, and set it off on the are, from S. towards W., for the course; through this point and C draw the line CB, and make it equal to the distance 244; draw BA parallel to the east and west line EW, to cut the meridian in A.. Then will CA be the difference of latitude 135.6, and AB the departure 202.9. BY LOGARITHMS. By making the distance radius. W Distance 244 Courses pts. Departure Diff. Lat Now, as the ship is in north latitude sailing southerly, From the latitude left.... Take the difference of latitude 135.6. Gives the latitude in....... And the departure from the meridian is 202.9 miles. * BY GUNTER. 49° 57' N. 47 41 N. Extend from radius or 8 points to 5 points on the line marked SR; that extent will reach from the distance 244, to the departure 202.9, on the line of numbers. 2dly. Extend from radius or 8 points to 3 points, the complement of the course, on the line SR; that extent will reach from the distance 244, to the difference of latitude 135.6, on the line of numbers. Thus may all the operations be performed in the several cases of Navigation. By this case are calculated the tables of latitude and departure (Tables I. and II.) for every degree, point, and quarter point of the mariner's compass, to the distance of 300 miles. By the inspection of these tables, a day's work may be calculated in a much more expeditious manner than by logarithms or by Gunter's scale. In consequence of this facility, the method by inspection is generally used at sea in preference to every other method. BY INSPECTION. Find the given course at the top or bottom of the tables, either among the points or degrees, and in that page, against the distance taken in its column, will stand the difference of latitude and departure in their columns.† It must be observed, that, in using these tables, the names Dist. Lat. Dep. must be found at the top if the course is found there, but if the course is found at the bottom, those names must be found at the bottom. Thus the course S. W. by W. or 5 points, is found at the bottom of the table of difference of latitude and departure for points; and against 244 in the distance column stands 135.6 for the difference of latitude, or 202.9 for the departure. When the course is given in points, make use of the lines marked sine rhumbs, and tangent rhumbs, on the upper side of the scale; when in degrees, make use of the lines marked sine and tangent. When the distance is too great to be found in the tables, you must divide it by 2, 3, 4, or any convenient number; the numbers corresponding to the quotient being multiplied by the divisor will give CASE II. Course and difference of latitude given, to find the distance run, and departure from the meridian. A ship runs S. E. by E. from 1° 45′ north latitude, and then, by observation, is in 0° 31' south latitude; required her distance and departure. In this case, as the ship has crossed the equator, the sum of the two latitudes, 1° 45 and 0° 31', is the difference of latitude, 2° 16' 136 miles. BY PROJECTION. Draw BC equal to 136, and BA making an angle with BC equal to the course 5 points, or 56° 15'; draw CA perpendicular to BC to cut BA in A, and it is done; for CA will be the departure equal to 203.5, and AB the distance equal to 244.8. By making the distance AB radius * 10.00000 As cosine course 5 points..... 9.74474 So is tangent course 5 points... 10.17511 Hence the ship's distance run is 244.8 miles, and her departure from the meridian is 203.5 easterly. BY GUNTER. Extend from radius or 4 points to the course 5 points on the line marked TR; that extent will reach from the difference of latitude 136, to the departure 203.5, on the line of numbers. 2dly. Extend from the complement of the course 3 points to the radius 8 points on the line SR; that extent will reach from the difference of latitude 136, to the distance 244.8, on the line of numbers. BY INSPECTION. Find the course among the points or degrees, and the difference of latitude m its column, against which will stand the distance and departure in their columns. CASE III. Course and departure from the meridien given, to find the distance and difference of latitude. If a ship sails N. E. by E. 3 E. from a port in 3° 15′ south latitude, until she depart from her first meridian 203 miles, required the distance sailed, and the latitude she is in. BY PROJECTION. B Departure 203 Draw the meridian AB, upon which erect the perpendicular BC, and set off thereon the departure 203, easterly from B to C; with the chord of 60° on C, as a centre, describe an arc, and set off thereon the complement of the course; through this point and C draw the line CA, cutting the meridian in the point A ; then AC measured on the same scale before used, gives the distance 224.6, and AB 96, the difference of latitude. Diff. Lat. A 25 19 Distance 6441 Course By making BC radius, you would have, radius: difference of latitude :: scant course : distance at this canon would not do for a common scale on which there is no line of secants. ing is to be observed in the following cases The same Extend from radius or 4 points to the complement of the course 2 points on the line marked TR; that extent will reach from the departure 203, to the difference of latitude 96, on the line of numbers. 2dly. Extend from the course 53 points to radius on the line SR; that extent will reach from the departure 203, to the distance 224.6 miles, on the line of numbers. BY INSPECTION. Find the course, either among the points or degrees, and the departure in its column, against which will stand the distance and difference of latitude in their respective columns. Thus with the course 53 points, and departure 203, we find 224.6 for the distance, and 96.0 for the difference of latitude. CASE IV. Distance and difference of latitude given, to find the course and departure Suppose a ship sails 244 miles, between the south and the east, from a port in 2o 52 south latitude, and then, by observation, is in 5° 08′ south latitude; what course has she steered, and what departure has she made? the remainder, From the latitude by observation 5° 08′, take 2° 52′, the latitude left; Draw the meridian AB=136; upon which erect the perpendicular BC; take 244 in your compasses, and with one foot on A, as a centre, describe an arc cutting BC in C; join A and C ; then will BC be the departure 202.6, and the angle BAC the course, equal to 56° 08′, or 5 points, nearly. Hence the course is S. E. by E., and the departure 202.6. BY GUNTER. The extent from the distance 244, to the difference of latitude 136, on the line of numbers, will reach from radius or 90° to 33° 52', the complement of the course on the line of sines. And the extent from radius, to 56° 08′ on the line of sines, will reach from the distance 244, to the departure 202.6, on the line of numbers. BY INSPECTION. Seek in the tables till against the distance, taken in its column, is found the given |