In each of the following examples the difference of longitude is required: 275. The method of parallel sailing will apply correctly enough for all practical purposes to cases where the course is nearly East and West (true). In latitudes not higher than 5°, when the distance does not exceed 300 miles, the departure may be used at once for the difference of longitude, the resulting error scarcely exceeding one mile. 276. Given the difference of longitude of two places on the same parallel, to find their distance as measured along the parallel. RULE LXVIII. To the log. of the diff. of long. add the cosine of lat.; the sum (neglecting 10) is log. of the distance required. EXAMPLE. Ex. 1. Required the distance between St. Abb's Head, in latitude 55° 55′ N., longitude 2° 10' W., and Uraniberg in the same latitude, but in longitude 12° 52′ E. 277. Given the meridian distance and difference of longitude, to find the latitude. RULE LXIX. From the log. of meridian distance (adding 10 to the index) subtract the log. of diff. long.; the remainder is the log. cosine of the latitude. EXAMPLE. Ex. 1. From a place in longitude 3° 12′ W., a ship sails due East 246 miles, and by observation is found to be in longitude 4o 8′ E.: required the latitude of the parallel on which she sailed. Since the diff. long. given, 440, exceeds the distance given in the Traverse Table, its half is taken, and also the half of the meridian distance 246, these are 220 and 123 respectively. Entering the Tables with 220 as distance, and 123 as diff. lat., we find, on searching the Table, these quantities, in their respective columns, on the page with 56° at the bottom; hence the latitude sought is 56°. EXAMPLES FOR PRACTICE. 1. Required the compass course and distance from A to B. Given lat. A 52° 15′ S.; var. 14 points W.; long. A 37° 30′ W. B 48 18 W. 3. Define a great circle and a small circle of a sphere, giving an example of each. connection is there between the tropic of Cancer and the Arctic Circle ? 4. Required the compass course and distance from A to B. Lat. A 28° 40′ N.; var. 14 points W.; long. A 2° 20′ E. B 28 40 N.; dev. 8° 50′ E.; B 4 10 E. What 5. In what latitudes are the lengths of a degree of longitude 30 and 20 miles respectively? 6. In travelling 35 nautical miles on the parallel of 55° 25′ N., how much do I change my longitude? 7. Find the true course and distance from A to B. Lat. A 54° 25' S.; long. A 1° 30′ B. B 54 25 S.; B 9 15 W. MIDDLE LATITUDE SAILING. 278. Middle Latitude Sailing is a method founded on the principle of parallel sailing, converting Departure into Difference of Longitude, and the Difference of Longitude into Departure, when the ship's course lies obliquely across the meridian, that is, when besides departure she makes difference of latitude. Suppose a ship, in going on the same course, from latitude 40° to latitude 44°, make 100 miles departure: this departure, if made good altogether in latitude 40°, would give 130.5 difference of longitude by Rule LXVI, page 194; and again, if made good in latitude 44°, it would give 139 difference of longitude. Now, since the ship has sailed between these two parallels, and not on either of them exclusively, her real difference of longitude must be between 130-5 and 139, and therefore we may conclude it to be not far from that which would result from a departure made good altogether in the middle parallel; hence the name Middle Latitude Sailing. Middle latitude sailing, then, is founded on the consideration that the arc of the parallel of middle latitude of two places intercepted between their meridians is nearly equal to the departure. If we conceive the ship to sail along this middle parallel, we may apply the principle of parallel sailing to the cases in point. In parallel sailing, as has been shown, the departure (or distance) and difference of longitude are connected by the relation, dep. diff. of long. X cos. lat. When the ship's course lies obliquely across the meridian, making good a difference of latitude, a modification of this formula gives the formula for middle latitude sailing, dep. (nearly) = diff. of long. X cos. mid. lat.; or in logarithms, log. dep. log. diff. of long. + log. cos. mid. lat. IO. Middle latitude sailing has thus the same two cases as parallel sailing, and accordingly the rules for inspection and computation already given, Rule LXVI, page 194, apply equally to this sailing, observing merely to read middle latitude for latitude. 279. To find the latitude and longitude in, the course and distance from a known place being given, by Traverse Table and Middle Latitude. By working a Traverse the difference of latitude and departure are obtained. Hence, by applying the difference of latitude to the latitude from, we have the latitude in. The middle latitude is then found, and the solution of the problem completed by the aid of the formula above, No. 278, note, viz. : 1o. the diff. of long. applied to the long. from giving the long. in, hence RULE LXX. With the given course and distance enter the Traverse Table, and take out true difference of latitude and departure (see Rule LXIII, page 184). 2o. With difference of latitude and latitude from, find latitude in (see Rule XLIV, page 108). 3°. Get the middle latitude, as directed, Rule XLV, page 108. 4°. With the middle latitude as course, look in the difference of latitude column for the departure, the corresponding distance at the top is the difference of longitude. 5°. With difference of longitude and longitude from get longitude in, as in Rule XLVII, page 110. NOTE.-When the departure to be looked for as difference of latitude at the middle latitude, is beyond the limits of the Table, one-half, one-third, &c., must be used, and the resulting difference of longitude multiplied by the divisor, in order to get the whole difference of longitude. Ex. 1. EXAMPLES. A ship from lat. 52° 6' N., long. 35° 6′ W., sailed S.W. by W., 256 miles: required her latitude and longitude in. Course S. 5 pts. W. Distance 250 miles.} give diff. lat. 142′2, and dep. 212-9 (see Rule LXIII, page 184). 5 38 Diff. long. 337'7 log. 2.528525 5° 38′ W. Long. from 35 6 W. Long. in (See Rule XLVII, page 110). 40 44 W. Explanation. The difference of latitude and departure is found as described in Rule LXIII, page 184. The latitude in is found by Rule XLIV, page 108; and thence the middle latitude, by adding the latitude from and latitude in together, and divided by 2 (see Rule XLV, page 108). The departure exceeding the limits of the Tables, the half is taken. Then with middle latitude as a course, and half the departure, in difference of latitude column, half the difference of longitude is found in the distance column. This being doubled (as half the departure was taken) and divided by 60, gives the difference of longitude expressed in degrees and minutes. The ship is in West longitude, sailing West, add difference of longitude to longitude left to obtain longitude in (Rule XLVII, page 110). This Method by Inspection is the usual case at sea of Working the Day's Work. Ex. 2. A ship from lat. 48° 27′ S., and long. 29° 12′ W., sails S.E. by S., 22.5 miles : required the latitude in, also the longitude in. Course S.E. by S. 3 pts.; then 3 pts. and dist. 22'5 give diff. lat. 18'7, Ex. 3. A ship from the Lizard, in lat. 49° 57′ N., sails between the South and West until diff. lat. is 126'7, and dep. 102·6: required the latitude come to, and difference of longitude. Ex. 4. Lat. from 59° 0' N., long. from 3° 33′ E., the ship sailed between S. and E., making diff. lat. 817 and dep. 172°7. - 9 6,0)17,1 D. long. for mid. lat. 58° is 326 " 59, 335 Diff. for 1° of mid. lat. 9 D. long. for mid. lat. 58° is 326 + 2.8 D. long. for mid. lat. 58° 19′ is 328.8 Remark. When the mid. lat. is high and between two whole degrees, and also the dep. great as in this example, the diff. long. is best found by calculation. Ex. 5. Sailed from A, in lat. 50° 48′ N., long. 1° 10′ W., S. 41° E., 275 miles. Entering Traverse Table II with dist. 275 miles, and course 41°, the true diff. lat. is 207′5, or 3° 27'5 S.; applying this to lat. from, the lat. in is 47° 20′5 N. The corresponding dep. is taken out at the same opening, which is 180′4. The mid. lat., or half sum of lat. from and lat. in, is 49° to the nearest degree. The dist. corresponding to 49° as a course, and 180'4 in diff. lat. column, is found to be 275', in degrees 4° 35′ E., which is the diff. long. Applying this to the long. from, 1° 10′ W., we have the long. in 3° 25′ E. EXAMPLES FOR PRACTICE. In each of the examples following, the latitude and longitude arrived at are required to be found, having given the latitude and longitude from, with the course and distance sailed. |