PROBL. V. Both Latitudes and the Meridian Distance of two Places being given, To find the Difference of Longitude, and Course and Distance on the True Sea-Chart.
THis is a most useful Problem, when the Mariner hath cast up his Traverse: Sup∣pose a Ship sail upon the S. W. Quarter of the Compass, from Latitude 51 deg. 22 min. unto Latitude 13 deg. 10 min. and the Departure from the first Meridian to the Westward 865 Leagues.
You must find first the Difference of Latitude inlarged, as is before-directed in the first Problem 934 5/10.
As the true Difference of Latitude AB 764 Leagues | 288309 |
Is to the Meridian-distance or Departure BE 865 Leagues | 293701 |
So is the Difference of Latitude inlarged AC 934 5/10 Leagues | 297057 |
To the Difference of Longitude in Leagues 1058 CD | 590758 |
302449 |
EXtend the Compasses from AB the true Difference of Latitude 764 Leagues, to BE 865 Leagues Meridian-distance; the same Extent will reach from AC. 934 5/10 Leagues the Difference of Latitude inlarged, to the Difference of Longitude 1058 Leagues; which laid off upon the Parallel-Line from C to D, is the Point and Place of the Ship in Mr. Wright's or Mercator's Chart.
As the true Difference of Latitude 764 Leagues AB | 288309 |
Is to the Meridian-distance 865 Leagues BE | 1293701 |
So is the Radius 90 deg. | 10, |
To the Tangent of the Course 48 deg. 33 min. at A | 1005392 |
EXtend the Compasses from AB 764 Leagues, to BE 865; the same Distance will reach from 90 deg. to the Tangent of 48 deg. 33 min. that is, 4 Points and above a Quarter from the South Westward, that is, S. W. ¼ Westerly, the Course the Ship hath kept.
As the Sine of the Course at A, 48 deg. 33 min. | 987479 |
Is to the Radius 90 deg. | 10 |
So is the Departure from the Meridian 865 Leagues | 1293701 |
To the Distance sailed AE 1154 2/10 Leagues | 306222 |
EXtend the Compasses from the Sine of 48 deg. 33 min. at A, to the Sine of 90 deg. the same Extent will reach from 865 Leagues BE, to 1154 Leagues AE, the Distance sailed.