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. COS.

To find the Departure or Distance

= cos. lat., .. dep. = D. long. x cos. lat.
D. Long.

Log. dep. = log. D. long. + L cos. lat. - 10
Difference of longitude 288' ......log. 2.459392
Latitude 36° 58' ..

9-902539 Departure or distance 230'I miles. .log. 2.361931 By Inspection.Enter Traverse Table with the lat. 37° as a course, and the difference of longitude 288 in the distance column, then the distance (or departure) 230 will be found in the difference of latitude column.

Ans. Distance 230.1 miles. NOTE.-By the above Formula the meridian distance in miles and decimal of a mile, answering to a degree of longitude on any parallel of latitude, may be found.

Example.Find the number of miles contained in a degree of longitude in lat. 48o.

Naut. miles in a degree of long. at the equator (60')..log. 1.778151
Latitude 48°

..cos. 9.825511
Naut. miles in a degree of long. in lat. 48°; 40.15..log. 1.603662

Given the Distance between two Places, both on the same Parallel of Latitude,

to find their Difference of Longitude Example.--A ship from lat. 49° 32' N., and long. 10° 16' W. sails due west 118 miles. Required her present longitude.


Dift. of Long

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Draw the line A B, which make equal to the given distance 118, and make the angle CAB equal to the lat. 491° (Prob. XII. Geom.); from B erect the perpendicular B C, cutting the line A C in C (Prob. II. Geom.); then will the line AC measure 182, the difference of longitude required.


or Dep. 118 m.
To find the D. Long.

To find the Longitude in
D. long.

= sec. lat.
Longitude left

10° 16' W, .:. D. long dist. X sec. lat. D. long. 182'


2 W. Log. D. long. = log. dist. Longitude in

13 18 W + L sec. lat. Ιο Distance 118 miles log. 2071882 Latitude 49° 32' sec. 10*187752 D. long. 181.8' log. 2'259634


By Inspection.-Look for the lat. 50° in the Traverse Table, as a course, and for the distance 118 in the difference of latitude column, opposite to which in the distance column will be found 184 ; but as the latitude is nearly half way between 49° and 50°, look with 49° as a course, and the distance 118 in the difference of latitude column, opposite to which, in distance column, will be found 180; then half the sum of 180 and 184 will be 182, the difference of longitude.

Ans. Long. in 13° 18' W. Given the Difference of Longitude and Distance between two Places on the same

Parallel of Latitude, to find the Latitude of that Parallel Example.—A ship sails due east 156 miles, and then finds she has altered her longitude 314 miles. Required the latitude of the parallel on which she has sailed.



Draw the line A B, and make it equal to the distance 156; from B erect the perpendicular B C, and with an extent in the compasses equal to the difference of longitude 314, set one foot in A, and with the other describe an arc cutting B C in C, and draw the line A C; then the angle C AB will measure 601° (Prob. XIII. Geom.), the latitude required.

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To find the Parallel of Latitude sailed on:


= cos. lat.

D. long
Log. dep. + 10 — log. D. long. = L cos. lat. .
Distance or departure 156 miles log. ( + 10) 12.193125
D. long. 314

log. 2.496930
Latitude 60° 13'

cos. 9.696195

By Inspection.-Seek in the Traverse Table till the difference of longitude and distance, viz., 314 and 156, are found opposite to each other in the distance and difference of latitude columns, which will give the lat. 60° at the bottom of the page.

Examples for Practice 1. A ship having taken her departure from North Cape, New Zealand, in lat. 34° 24' S., and long. 173° 10' E., being bound to Port Jackson, sails due west until she arrives in long. 163° 35' E. Required her distance run.

Ans. Distance 474.4 miles.


2. A ship from Buchanness, in lat. 57° 29' N., and long. 1° 47' W., sails due east 125 miles. Required her present longitude.

Ans. Long. in 2° 6' E.

3. A ship in lat. 32° 22' N., and long: 52° 20' W., sails west 365 miles. Required her distance from the Island of Bermuda, in the same latitude, and in long. 64° 43' W.

Ans. Distance of ship from Bermuda 262.6 miles.

4. A ship in lat. 60° N., and long. 22° 30' W., sails west 200 miles. Required her present longitude.

Ans. Long. in 29° 10' W.

5. If a ship take her departure from Cape St. Antonio (at the entrance to the River Plate), which lies in lat. 36° 19' S., and long. 56° 42' W., how far must she sail due east to arrive at the meridian of the Cape of Good Hope, in long. 18° 24' E. ?

Ans. 3,631 miles.

6. In what parallel of latitude is the departure or meridian distance onethird the difference of longitude ?

Ans. In lat. 70° 32'.

7. A ship from long. 81° 36' W., sails west 310 miles, and then finds by observation she is in long. 91° 50' W. On what parallel of latitude has she sailed ?

Ans. In lat. 59° 41'.

8. Suppose a ship from lat. 35° 30' N., and long. 6° 15' W., sails west 250 miles, north 525 miles, and then east 250 miles. Required her present latitude and longitude.

Ans. Lat. 44° 15' N., and long. 5° 33' W.

9. A ship from lat. 49° 32' N., and long. 21° 56' W., sails N.W. by N. 20 miles, S.W. 40 miles, N.E. by E. 60 miles, S.E. 55 miles, W. by S. 41 miles, and E.N.E. 66 miles. Required her present latitude and longitude.

Ans. Lat. 49° 32' N., and long. 20° 8' W.

10. At what rate per hour is Greenwich Observatory, in lat. 51° 28' 38", carried around the earth's axis ?

Ans. 560-54 geographical miles per hour.

11. Two ships in lat. 34° 20' N. are 307 miles apart : if they both sail due north, how many miles will they be apart when they arrive in lat. 46° 34' N.?

Ans. 255.6 miles.

12. Two ships in lat. 50°18' N. are distant apart 256 miles : they sail due south until their distance apart is 356 miles. In what latitude have they arrived, and how many miles have they sailed on their south course ?

Ans. Lat. arrived in 27° 20' N.; distance sailed 1,3771 miles.

MIDDLE LATITUDE SAILING When a ship sails due north or south she keeps on the same meridian, and therefore does not change her longitude, and her distance run is the difference of latitude : consequently her place is easily determined by the latitude left, and difference of latitude. Again, when a ship sails due east or west, her difference of longitude is found by the latitude in, and the departure or meridian distance, as already explained in Parallel Sailing ; but when she sails upon any other course she changes both her latitude and longitude. Now, the difference of longitude cannot be inferred either from the departure, considered as a meridian distance in the latitude left, or that come to; for in the greater latitude it would give the difference of longitude too much, and in the less latitude too little : for example, in lat. 50°, a departure of zoo miles is the equivalent of 155'-6 difference of longitude, while in lat. 48°, the same departure makes only 149' 5 diff. long. ; the departure is therefore accounted a meridian distance in the mean of the two latitudes, and then the difference of longitude is found as in Parallel Sailing; hence this method, which is compounded of plane and parallel sailing, is called MIDDLE LATITUDE SAILING.

The middle latitude is half the sum of the two latitudes when they are of the same name; this is arithmetically correct, and an assumption sufficiently accurate for short distances; but when precision is required a small correction must be applied to the middle latitude, since the true middle latitude is always a little nearer to the pole than the mean of the latitude left and latitude arrived at.

Workman's Table* (see Norie's Tables) for the correction of the middle latitude is to be entered at the top with the difference of the two latitudes, and at the side with the middle latitude : under the former, and opposite the latter, is the correction to be added to the middle latitude to obtain the true middle latitude.

When the difference of latitude is under 2°, the correction may be neglected; when it is between 2 and 3°, add a correction of 1'.

This method of sailing, although not strictly accurate, especially in high latitudes, approaches sufficiently near to the truth for a day's run; it is accurate in low latitudes, and generally when the ship makes a course greater than 45o

Middle latitude should be used only when the latitudes are of the same name. If of different names, and the distance small, the departure may be assumed equal to the difference of longitude, since the meridians are sensibly parallel near the equator. But if the distance is great, the two portions of the track on opposite sides of the equator should be calculated separately.

By referring to the figures connected with the examples worked out, it will be seen that in middle latitude sailing the figure illustrative of the method consists of two triangles having a common side, one part of which properly appertains to plane sailing, and the other part to parallel sailing, with the departure as the base or connecting link between them.

* This Table was published by Workman in 1805 under the sanction of Dr. Maskelyne, then Astronomer Royal. The Table is not as well known as it ought to be, and, as it removes the only objection to the method of finding the difference of longitude by middle latitude sailing, it is introduced here to the notice of the navigator.

7 of Lat. 265m?


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In middle latitude sailing the departure is taken to be the sum of all the various meridian distances (or departures) corresponding to the indefinitely small parts of the rhumb line between the latitude left and laiitude arrived at, and the assumption is that this

departure = the meridian distance in the middle latitude. Given the Latitudes and Longitudes of two Places, to find the True Course

and Distance from the one Place to the other Required the true course and distance from Cape St. Vincent, in lat 37° 3' N., and long. 9° 1' W., to Funchal, Madeira, in lat. 32° 38' N., and long. 16° 56' W. Lat. Cape St. Vincent 37° 3' N.-37° 3' N. Long. C. St. Vincent 9° 1' W. Lat. Funchal

32 38 N.-32 38 N. Long. Funchal 16 56 W. Diff. lat. 4 25 S. 2) 69 41 Sum

D. long. 7 55 W. 60 Mid. lat. 34 50

бо In miles 265

475' BY CONSTRUCTION Draw the line A D to represent the meridian of Cape St. Vincent; make the angle ADC equal to the co-mid. lat. 55° (Prob. XII. Geom.), and from D to C lay off the difference of longitude 475; from C draw the line B C perpendicular

to A D (Prob. III. Geom.); make B A equal to
the difference of latitude 265, and draw the
line A C: then will B C represent the departure
390, the angle B A C the course 552°, or 5 points
nearly; and A C the distance 471 miles.

To find the Course

To find the Distance

= cos. mid. lat.

D. lat. .:. Dep. = d. long. X cos. mid. lat.

Mid. Lat. 34:50
Diff. of Long. 475 m.

= sec. CO.

D. long

.:. Dist.

d. lat. X sec. co. Also- Dep.

Log. dist. = log. d. lat.
= tan. co.
D. lat.

+ L sec. co. Therefore

D. lat. 265 m. log. 2.423246 D. long x cos. mid. lat.

Course 55° 48' sec. 10*250199

tan. co. Diff. lat.

Distance 471.5 m. log. 2-673445 L tan. course log. d. long. + L cos, mid. lat.

log. d. lat. D. long. 475' log. 2•676694 Mid. lat. 34° 50' COS. 9:914246

12:590940 D. lat. 265 m. log. 2423246 Course S. 55° 4S' W. tan. 10.167694

ΙΩ. .

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