58 BY GUNTER. Extend from the difference of latitude 271 to the departure 406 on the line of numbers, that extent will reach from radius to 56° 17' the course on the line of tangents. 2dly. For the distance we must consider it as radius (unless there is a line of secants on the scale) and extend from the course 56° 17′ to the radius or 90° on the line of sines, that extent will reach from the departure 406, to the distance 488 on the line of numbers. BY INSPECTION. Seek in the tables till the given difference of latitude and departure are found together in their respective columns, then against them will be the distance in its column, and the course will be found at the top of that table if the departure be less than the difference of latitude, otherwise at the bottom. Thus with half the difference of latitude 135.5, and half the departure 208, enter the tables, and these numbers will be found to correspond nearly to 5 points or N. W. by W. course, and a distance equal to 244 miles, which being doubled gives the sought distance, 488. Questions to exercise the learner in the foregoing Rules. Question I. A ship in 20 10′ south latitude, sails N. by E. 89 leagues; what latitude is she in, and what is her departure? Answer. Latitude in 2° 12′ N. and departure 17,36 leagues. Question II. A ship sails S. S. W. from a port in 41° 30' north latitude, and then by observation is in 360 57' north latitude; I demand the distance run and departure? Answer. Distance run 98,5 leagues, departure 37,7 leagues. Question III. A ship sails S. S. W. W. from a port in 2° 30′ south latitude, until her departure be 59 leagues; I demand her distance run and latitude in? Answer. Distance run 125,2 leagues, latitude in 80 1' south. Question IV. If a ship sail 360 miles south westward from 21° 59′ south latitude, until by observation she be in 24° 49' south latitude, what is her course and departure? Answer. The course is S. W. by W. half W. or S. 61° 49′ W. and her departure from the meridian is 317,3 miles. Question V. Suppose a ship sails 354 miles north eastward from 20 9′ south latitude, until her departure be 150 miles, what is her course and latitude in? Answer. Her course is N. 25° 4' E. or N. N. E. E. nearly, and she is in latitude 3° 12′ N. Question VI. Sailing between the north and the west, from a port in 1° 59' south latitude, and then arriving at another port in 4° 8' north latitude, which is 209 miles to the westward of the first port-I demand the course and distance from the first port to the second? Answer. The course is N. 29° 40′ W. or N. N. W. W. nearly, and the distance of the ports is 422,4 miles, or 140,8 leagues. Question VII. Four days ago we were in lat. 3° 25′ S. and have since that time sailed in a direct course N. W. by N. at the rate of 3 miles an hour; required our present latitude and departure? Answer. Latitude in 7° 14′ N. Departure 426,7 miles. Question VIII. A ship in the latitude of 3° 52′ S. is bound to a port bearing N. W. by W. W. in the latitude of 4° 30' N. how far does that port lie to the westward, and what is the ship's distance from it? Answer. The port lies 939,2 miles to the westward, and the direct distance 1065 miles. Question IX. A ship from the latitude of 48° 17' N. sails S. W. by S. until she has depressed the north pole 2 degrees; what direct distance has she sailed, and how many miles has she got to the westward? Answer. Distance run 144,3 miles, and has got to the westward 80,2 miles. TRAVERSE SAILING. A TRAVERSE is an irregular track which a ship makes by sailing on several different courses; these are reduced to a single course by means of two or more cases of Plane Sailing, either by geometrical construction, or by arithmetical calculation.* The geometrical construction is performed as follows: Describe a circle with the chord of 60°, to represent the compass, and lay off on its circumference the various courses sailed. From the centre, upon the first course set off the first distance, and mark its extremity: through this extremity, and parallel to the second course, draw the second distance of its proper length; through the extremity of the second distance, and parallel to the third course, draw the third distance of its proper length; and thus proceed till all the distances are drawn. A line, drawn from the extremity of the last distance to the centre of the circle, will represent the distance made good; a line, drawn from the same point, perpendicular to the meridian, will represent the departure; and the part of the meridian intercepted between this and the centre, will represent the difference of latitude. The arithmetical calculation to work a traverse is as follows; Make a traverse table consisting of six columns; title them, Course, Dist. N. S. E. W. begin at the left side, and write the given courses and distances in their respective columns. Find the difference of latitude and departure for each of these courses, by Gunter's Scale, or by Tables I. or II. (as in Case I. Plane Sailing) and write them in their proper columns; that is, when the course is southerly, the difference of latitude must be set in the column S. when northerly in the column N. The departure, when westerly, in the column W. and when easterly in the column E. Add up the columns of northing, southing, easting, and westing; take the difference between the northing and southing, and also between the easting and westing; the former difference will be the difference of latitude, which will be of the same name as the greater; and the latter will be the departure, which will be also of the same name as the greater. With this difference of latitude and departure, the course and distance made good are to be found as in Case VI. Plane Sailing. EXAMPLE I. Suppose a ship takes her departure from Block Island, in the latitude of 41° 10' N. the middle of it bearing N. N. W. distance by estimation 5 leagues, and sails S. E. 34, W. by S. 16, W. N. W. 39, and S. by E. 40 miles; required the latitude she is in, and her bearing and distance from Block Island? This method of reducing compound courses to a single one is perfectly accurate in sailing on a plane, and is nearly so in sailing a short distance on the spherical surface of the earth; and though in this case it is liable to a small error in high latitudes, yet in general the rule is sufficiently accurate for reducing the several courses and distances sailed in one day to a single course and distance. Let L represent the middle of Block-Island; draw the meridian LM, and on Las a centre, with a chord of 600, describe a circle to represent the compass, on which mark the various courses sailed, and the bearing of the land at the time of taking the departure; opposite to this bearing draw the S. S. E.D line LA, which make equal to 15 miles, the estimated distance of the land; then will A represent the place of the ship at the time of taking the departure: through A draw AB= 34 miles parallel to the S. E. line; then will B be the place of the ship after sailing her first course; in like manner draw BC-16 miles parallel to the W. by S. line: CD= BY PROJECTION. B =40 miles parallel to the S. by E. line; then will E represent the place of TO FIND THE SAME BY LOGARITHMS. To find the difference of latitude. As radius 90° For departure. Is to distance 15 So is co-sine course 2 points 10.00000 As radius 90° 10.00000 1.17609 9.58284 To Diff. lat. 13.9 1.14171 To departure 5.7 Second course S. E. 34 miles. 0.75893 For departure. Is to co-sine course 45° So is distance 34 10.00000 As radius 90° To diff. latitude 24 1.38097 To departure 24 1.38097 Third course W. by S. 16 miles. For difference of latitude. As radius 90° For departure. 10.00000 As radius 90° 10.00000 Is to co-sine course 78° 45' 9.29024 Is to sine-course 78° 45' 9.99157 1.20412 So is distance 16 1.20412 To diff. latitude 3.1 0.49436 To departure 15.7 1.19569 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 45° So is the departure 14.2 To tang. course 12° 16' 1.S1491 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. 9.32729 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 VI. 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? Let C represent Mount-Desert Rock, draw the meridian CF, which make equal to 107 miles the difference of latitude between the two places; and perpendicular thereto the line FE equal to the departure 84 miles, then is E the place of Cape Cod. With the chord of 60° sweep about the centre C, a circle E. S. W. to represent the compass, & upon it note the various courses sailed. The first course being South the distance 10 miles is set off from Join 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. 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. 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. |