PARALLEL SAILING.

In Plane Sailing it was observed that the meridians are considered as being all parallel to each other, and the length of a degree on the meridian and parallel every where equal; which supposition will give just conclusions, so far as the course, distance, difference of latitude, and departure, are concerned; because a ship, when sailing on a rhumb, makes equal angles with the meridians: but as the earth is a sphere, or globe, and the meridians meet at the Poles, it is evident that the distance between any two meridians must vary in every latitude, their greatest distance being at the Equator, on which the difference of longitude is measured: hence the difference of longitude always exceeds the departure, or meridian distance (except on the Equator, where they are the same), in proportion as the given places are situated further from the Equator.

PARALLEL SAILING is the method of finding the distance between two places in the same latitude when their difference of longitude is known, or of finding the difference of longitude answering to the meridian distance or departure made good, when a ship sails due east or west.

This sailing is particularly useful in making small or low islands, in which case it is usual to run into the latitude, and then steer due east or

west.

P

R

The principles upon which Parallel Sailing depends, may be thus illustrated: Let PAEC represent a section of one-fourth part of the earth, through the centre c, and one of the Poles P; then PA E will be part of a meridian, PC the polar, and E c the equatorial, semi-axis; also let PBO represent part of another meridian, A and B two places in the same parallel, being equally distant from the Equator EOQ; then will A B be the meridian distance, and Eo their difference of longitude; the arches AE or BO will measure their latitude, and A P or BP their co. latitude; A R, the radius of the parallel A B S, will be the sine of the arch A P, the co. latitude; or co. sine of A E the latitude of A or B. Now the angles AR B and E C O being equal, the arches A B and E o are similar; and as circles and similar arches of circles are in direct ratio to their radii, therefore

That is, As radius,

E

EC (or A C): EO :: AR: A B.

Is to difference of longitude,
So is co. sine of latitude,

And, AR

AB :: EC (or A c): E O.
That is, As co. sine of latitude

Is to meridian distance,
So is radius

To difference of longitude.

Also, E O EC: AB AR. That is, As difference of longitude

Is to radius,

So is meridian distance

To co. sine of latitude.

Hence, if a triangle, as A B C (see figure in Case I.), be so constructed that the longest side a c may represent the difference of longitude in miles, the base A B the meridian distance, and the angle opposite to it A C B the co. latitude, consequently the other angle B A C equal to the latitude, and any two of these parts be given, the other may be found by Trigonometry.

CASE I.

The Difference of Longitude between two Places, both in one Parallel of Latitude given, to find their Distance.

EXAMPLE. A ship in latitude 36° 58′ N., and longitude 20° 25′ W., is bound to St. Mary's, one of the Western Islands, in the same latitude, and in longitude 25° 13′ W., what distance must she run to arrive at the Island?

10.00000

Is to the difference of longitude A c 288... 2. 45939
So is the co. sine of latitude 36° 58'

......

9.90254

12.36193
10.00000

To the departure or distance A B 230. 1... 2. 36193

M

BY INSPECTION.

Seek for the complement of the latitude 53° among the degrees in the Traverse Table, as if it were a course, and for the difference of longitude 288 in one of the distance columns of that page, opposite to which, in the departure column, will be found 230, the departure or distance required.

BY GUNTER'S SCALE.

Extend from 90° to the co. latitude 53° 2' on the line of sines; that extent will reach from the difference of longitude 288 to the distance 230.1 on the line of numbers.

CASE II.

The Distance between two Places both in the same Parullel of Latitude given, to find the Difference of Longitude.

EXAMPLE. A ship from latitude 49° 32′ N., and longitude 10°16′ W. sails due West 118 miles: required her present longitude.

BY CONSTRUCTION.

Draw the line A B, which make equal to the

given distance 118, and make the angle CAB

90° 00' Lat....49 32

Diff. of Long.

equal to the latitude 49° 32′(Prob. XII.Geom.); Co. lat. 40 28
upon в erect the perpendicular в C, cutting
the line a C in c (Prob. II. Geom.); then
will the line A c measure 182, the difference
of longitude required.

Lat. 49° 32'

Co. Lat. 40° 28

Look for the co. latitude 40° in the Table, as a course, and for the distance 118 in one of the departure columns, opposite to which, in the

way between 40° and 41°, look again in the page with 41° at the top, for the distance 118 in one of the departure columns, and opposite, in the distance column, will be found 180; then half the sum of this and 184, found before, will be 182, the difference of longitude.

BY GUNTER'S SCALE.

Extend from the co. latitude 40° 28′ to radius 90° on the line of sines; that extent will reach from the distance 118 to the difference of longitude 182 miles.

CASE III.

The Difference of Longitude and Distance between two Places in the same Parallel of Latitude given, 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 she has sailed on.

BY CONSTRUCTION.

Draw the line A в, and make it equal to the distance 156; on в erect the perpendicular в 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 arch cutting вc in c, and draw the line A c; then the angle C A B will measure 60° 13' (Prob. XIII. Geom.), the latitude required.

BY CALCULATION.

A Dist. 156 m.

Seek in the several pages of the Table till half the difference of longitude and distance, viz. 157 and 78, (the whole exceeding the limits of the Table) are found opposite each other in the distance and departure columns, which will give the co. latitude 30° at the top of the page, and consequently the latitude required will be 60°.

BY GUNTER'S SCALE.

Extend from the difference of longitude 314 to the distance 156 on the line of numbers; that extent will reach from 90° to the complement of the latitude 29° 47′ on the line of sines : hence the latitude required is 60° 13'

EXAMPLES FOR EXERCISE.

1. A ship having taken her departure from North Cape, New Zealand, in latitude 34° 24′ S., and longitude 173° 10' E., being bound to Port Jackson, sails due West until she arrives in longitude 163° 35′ E.: required her distance run.

Answer. Distance 474. 4 miles.

2. A ship from Buchanness, in latitude 57° 29' N., and longitude 1° 47′ W., sails due East 125 miles: required her present longitude. Answer. Longitude in 2° 6' E.

3. A ship in latitude 32° 22′ N., and longitude 52° 20′ W., sails West 365 miles required her distance from the Island of Bermuda, in the same latitude, and in longitude 64° 43′ W.

Answer. Distance of ship from Bermuda 262. 6 miles.

4. A ship in latitude 60° N., and longitude 22° 30′ W., sails West 200 miles required her present longitude.

Answer. Longitude 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 latitude 36° 19′ S., and longitude 56° 42′ W., how far must she sail due East to arrive at the meridian of the Cape of Good Hope, in longitude 18° 24′ E.?

Answer. 3631 miles.

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

Answer. In latitude 70° 32.

7. A ship from longitude 81° 36′ W., sails West 310 miles, and then finds by observation she is in longitude 91° 50′ W.; on what parallel of latitude has she sailed?

Answer. In latitude 59° 41'.

8. Suppose a ship from latitude 35° 30′ N., and longitude 6° 15′ W., sails West 250 miles, North 525 miles, and then East 250 miles required her present latitude and longitude.

Answer. Latitude 44° 15′ N., and longitude 5° 33′ W.

9. A ship from latitude 49° 32′ N., and longitude 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.