« PreviousContinue »
In parallel sailing the ship is supposed to be kept on a parallel of latitude, as TS, fig. p. 3. The course will evidently be due east or due west, and the distance between two places as T and S, will be the arc TS between the two meridians passing through the places.
To find the course and distance, having given the latitude of the two places, and their longitudes.
(1.) Find the difference longitude.
(2.) The course is evidently due east or due west, according as the longitude in is to the east or west of longitude from.
(3.) To find the distance. Add together log. cos. latitude and log. difference longitude; the sum (rejecting 10 in index) is the log distance, which find in the table.
100. Required the course and distance from A to B.
Required the compass course and distance from A to B.
Variation of the compass two points E, and deviation as
Required the true course and distance from A to B, in
each of the following examples:
Answers. Course and dist. East 144-2′
To find the longitude in, having given the course and distance, and latitude and longitude from.
Add together log. sec. lat. and log. distance, the sum (rejecting 10 in the index) will be the log. difference longitude. Find the natural number thereof, and turn it into
*In this example it is evident we must modify the general rule; for the diff. long. is never considered to be greater than 180°. When, therefore, the above rule gives the diff. long. greater than 180°, subtract it from 360°, and apply thereto a contrary letter to the one directed by the rule; the result will be the diff. long. to be used.
degrees, and mark it E. or W. according as the course is E. or W. Apply difference longitude to longitude from, and (Rule f.)
thus find longitude in.
The latitude in is the
same as the latitude from.
107. Sailed from A due east 1000 miles, required the latitude and longitude in.
Required the latitude and longitude in, in each of the
The preceding rules are the principal ones used in navigation. It would be easy for the mathematical student to make for himself others, by means of the relations between the several terms course, dist., dep., &c., as shown by the fig. p. 151, in the author's volume of Astronomical Problems: he would find then no difficulty in solving problems similar to the following:
Sailed from A, in long. in 3° 10' W., 300 miles due east, and altered my longitude 10 degrees; required the latitude and longitude in.
... lat. in 60°, and long. in = 6° 50′ E.
Wishing to make a small island, I took the ship to windward of it in the same latitude with the island, namely, 50° 48′ N. The longitude of the ship by chronometer was 20° 35′ W., and the long. of the island was 23° 50′ W. What was my distance from the island?
In this example of parallel sailing we have given lat. 50° 48', and diff. long. 3° 15', or 195', to find distance.
To find the place of the ship at noon, that is, its latitude and longitude, having given the latitude and longitude at the preceding noon, the compass courses, and distances run in the interval, the deviation of the compass for each course on account of local attraction, the variation of the compass, the leeway, the velocity and direction of current (if any) &c., constitutes what is called the Day's Work.
Rule VII. (the Day's Work).
(1.) Correct each course for variation (Rule g), deviation (Rule 2), and leeway (Rule 7); thus get the true courses, and arrange the same in a tabular form, as in the example, p. 34. Add together the hourly distances sailed on each course, and insert the same in table opposite the true course.
(2.) Take out of the traverse table the true difference latitude and departure for each course and distance, putting
them down in the columns headed with the same letters as in course. Previously to opening the traverse table, fill up the columns of true difference latitude and departure not wanted, by drawing horizontal lines; this will frequently prevent mistakes.
(3.) If the ship does not sail from a place whose latitude and longitude are known, her bearing and distance from some near object, as a church-spire, &c., must be ascertained, and also its latitude and longitude. Then the ship is supposed to sail from this known object to her anchorage, her course being the opposite to the bearing of the object from the ship. This course must be corrected like the rest for variation and deviation, and inserted in the table as an actual course, with the distance of the object as a distance.
(4.) If a current sets the ship in any ascertained direction, and with a known velocity, these also may be conceived to be an independent course and distance, and must be corrected for variation, and should be for deviation also, if the latter correction is appreciable, which is rarely the
(5.) To find the latitude in. The quantities in the four columns of true difference latitude and departure being added up separately, the difference between the north difference of latitude and south difference of latitude, with the name of the greater, will give the true difference of latitude, made at the end of the day. The departure is found in a similar manner. Apply true difference latitude to latitude from, so as to obtain the latitude in.
(6.) To find the longitude in. Add together log. sec. mid. lat. and log. departure, the result (rejecting 10 in the index) is the log. difference longitude. Find this in table, and thus the longitude in is found.*
The following example, worked out in detail, will perhaps
* Or thus :-To find diff. long., add together log. M.D. lat. and log. dep., and from the sum subtract log. T. D. lat.; the remainder is the log. diff. long., which find in the tables.