Though this method of finding the difference of latitude and departure by logarithms is accurate, yet the calculations may be more easily made by the tables of difference of latitude and departure, as in Case I. Plane Sailing. Again, the southing being greater than the northing, subtract the northing from it, and the remainder 65.3 shews how far the ship is to the southward of her first place. To find the direct course or bearing of To find the distance of the Island. Block Island from the ship. As the diff. lat. 65.3 Is to radius 450 So is the departure 14.2 To tang. course 12° 16' 1.81491 As sine of course 12° 16′ 10.00000 Is to the departure 14,2 1.15229 So is radius 90° 9.33738 To the distance 66.8 BY INSPECTION. 9.32728 1.15229 10.00000 1.82501 Which, because the difference of latitude is southerly, and the departure westerly, is S. 12° 16′ W. Whence Block Island Find the course and distance by Case VE bears from the ship N. 12° 16' E. or N. by of Plane Sailing. E. 1° 1' E. EXAMPLE II. A ship from Mount-Desert Rock, in the latitude of 43° 52′ N. sails for Cape Cod in the latitude of 42° 5' N. its departure from the meridian of Mount-Desert Rock being supposed to be 84 miles west; but by reason of contrary winds, she is obliged to sail on the following courses, viz. South 10 miles, W. S. W. 25 miles, S. W. 30 miles, and W. 20 miles. Required the bearing and distance of the two places, the course and distance sailed by the ship, and the bearing and distance of her intended port? C towards F upon the meridian, and this point represents the place of the ship after sailing her first coure; continue setting off the various courses and distances as in the last example, viz. W. S. W. 25 miles, S. W. 30 miles, and West 20 miles to the point A; then will A represent the place of the ship after sailing these courses. Join CE, AC, AE; draw AB perpendicular to the meridian CF, and AD parallel thereto : then will AC 76,2 miles be the distance made good, AE=69,1 miles the distance of Cape Cod from the ship; CE the distance of the two places=136 miles; ACB=57° 36′, the course made good; EAD=16° 34' the course to Cape Cod, and ECF the course from Mount-Desert Rock to Cape Cod=38° 8', &c. BY LOGARITHMS. To find the bearing and distance of the two places by Case VI. Plane Sailing. To find the distance. To find the bearing. As diff. of lat. 107 Is to radius 45° So is departure 84 To tang. course 38° 8' 2.02938 As radius 90° 10.00000 Is to diff. of lat. 107 1.92428 So is sec. course 38° 8' 9.89490 To the distance 136 10.00000 2.02938 10.10426 2.13364 Whence the course from Mount-Desert Rock to Cape Cod is S. 38° 8' W. distance 136 miles. The same may be found by the scale or by inspection. Subtract the difference of latitude made good by the ship 40,3 miles, fromt the whole difference of latitude 107 miles, and there remain 66,2 miles, which is the difference of latitude between the ship and Cape Cod. In the same manner by subtracting the ship's departure, 64,3 miles, from the whole departure, 84 miles, there remain 19,7 miles for the departure between the ship and Cape Cod. With this difference of latitude, 66,2, and departure, 19,7, the bearing of Cape Cod is found, by Case VI. Plane Sailing, S. 160 34' W. and its distance 69,1 miles. All the preceding calculations may be made by logarithms, by the scale, or by inspection. But we shall leave them to exercise the learner; and for the same purpose shall add the following example. EXAMPLE III. A ship in the latitude of 37° 10′ N. is bound to a port in the latitude of 35° 0' N. which lies 180 miles west of the meridian of the ship; but by reason of contrary winds she sails the following courses, viz. S. W. by W. 27 miles-W. S. W. W. 30 miles-W. by S. 25 miles-W. by N. 18 milesS. S. E. 32 miles-S. S. E. 4 E. 27 miles-S. by E. 25 miles-S. 31 miles, and S. S. E. 39 miles. Required the latitude the ship is in, and her departure from the meridian, with the course and distance to her intended port? TRAVERSE TABLE. IN Plane Sailing the earth is considered as an extended plane, but this supposition is very erroneous, because the earth is nearly of a spherical figure, in which the meridians all meet at the poles, consequently the distance of any two meridians measured on a parallel of latitude (which distance is called the meridian distance) decreases in proceeding from the equator to the poles. To illustrate this, let PB represent the semiaxis of the earth, B the centre, P the pole, PCA a quadrant of the meridian, AB the radius of the equator, and CD (parallel thereto) the radius of a parallel of latitude. Then it is evident that CD will be the co-sine of AC or the co- C sine of the latitude of the point C, to the radius AB: now if the quadrantal arch PCA be supposed to revolve round Al the axis PB, the point A will describe the circumference of the equator, and C the circumference of a parallel of latitude; and the former circumference will be to the latter as AB to CD (as may easily be deduced D * Instead of putting the course S. S. E. 32 miles, and S. S. E. 39 miles, you might make one entry only, calling it S. S. E. 71 miles. from Art. LV. Geometry;) that is, as radius to the co-sine of the latitude or the point C: hence it follows, that the length of any arch of the equator intercepted between two meridians, is to the length of a corresponding arch of any parallel intercepted between the same meridians, as radius is to the co-sine of the latitude of that parallel. Hence we obtain the following theorems. THEOREM I. The circumference of the equator is to the circumference of any other parallel of latitude, as radius is to the co-sine of that latitude. THEOREM II. As the length of a degree of the equator is to the meridian distance corresponding to a degree on any other parallel of latitude, so is radius to the co-sine of that parallel of latitude. THEOREM III. As radius is to the co-sine of any latitude, so are the miles of difference of longitude between two meridians (or their distance in miles upon the equator) to the distance of these two meridians on that parallel of latitude in miles. THEOREM IV. As the co-sine of any latitude is to radius, so is the length of any arch on that parallel of latitude (intercepted between two meridians) in miles to the length of a similar arch on the equator, or miles of difference of longitude. THEOREM V. As the the co-sine of any latitude is to the co-sine of any other latitude, so is the length of any arch on the first parallel of latitude in miles, to the length of the same arch on the other in miles. By means of Theorem II. the following Table was calculated, which shows the meridian distance corresponding to a degree of longitude in every latitude: and may be made to answer for any degree or minute by taking proportional parts. The following Table shews for every degree of latitude how many miles distant the two meridians are, whose difference of longitude is one degree. D. L. MILES. D. L. MILES. D. L. MILES. D. L. MILES. D. L. MILES. 60 30. 00 78 12. 47 88 61 29. 09 79 11. 45 7 59. 55 25 54. 38 43 43. 9 59. 26 27 53. 46 45 42. 43 63 27. 24 81 9. 39 8. 35 7. 31 6. 27 36 67 23. 44 85 5. 25 57 68 22. 48 86 4. 19 69 21. 50 87 3. 14 70 20. 52 88 2. 09 1. 05 7119. 53 89 When a ship sails east or west on the surface of the earth supposed to be spherical, she describes a parallel of latitude, and this is called Parallel Sailing. In this case, the distance sailed (or departure) is equal to the distance between the meridians sailed from and arrived at in that parallel, and it is easy, by Theorem IV. (preceding) to find the difference of longitude from the distance, or the distance from the difference of longitude, as will appear plain by the following examples. CASE I. The difference of longitude between two places in the same parallel of latitude being given, to find the distance between them. Suppose a ship in the latitude of 49° 30' north or south, sails directly easi or west until her difference of longitude be 3° 30′, required the distance sailed? BY PROJECTION. Take the sine of 90° from the Plane Scale, and with one foot of the compasses on P (Fig. 1) as a centre, describe the arch EQ; with the difference of longitude 210 miles in the compasses, and one foot in E, as a centre, describe an arch cutting EQ in Q; join PE, PQ. Take the sine of the complement of the latitude 40° 30' in your compasses, and with one foot in P, as a centre, describe the arch FG, cutting PE, PQ, in FG; then the length of the chord FG being measured on the scale of equal parts will be the departure 136.4 miles. Or this projection may be made in the following manner. Draw AD (Fig. 2) of an indefinite length, make the angle DAC equal to the latitude 49° 30', and AC equal to the difference of longitude 210 miles; draw CD perpendicular to AD; then will the line AD be the distance or departure required. To the distance or departure 136,4 BY GUNTER. The extent from radius to the complement of the latitude 40° 30′ on the line of sines, will reach from the difference of longitude 210 to the distance 136,4 on the line of numbers. BY INSPECTION. Find the latitude among the degrees in Table II. and in the distance column the difference of longitude, opposite to which in the column of latitude will be the distance required. In the present example the latitude is 49° 30′, and as the table is only calculated to single degrees, I find the numbers in the tables of 490 and 500, and take the mean of them; the former is 137,8, the latter 135,0, the mean of which is the sought distance or departure 136,4. CASE II. The distance between two places on the same parallel of latitude given, to find their difference of longitude. Suppose a ship in the latitude of 49° 30' N. or S. and long. 36° 40′ W. sails directly west 136,4 miles; required the difference of longitude, and longitude in ? L |