In the preceding problems the earth has been considered as a plane, we now proceed to the solution of problems in which it is necessary to advert to the spherical form of the earth, which we must do in all problems in which longitude is concerned.

In parallel sailing the distance of two meridians on a given parallel, the latitude of that parallel, and the distance of the meridians at the equator, or the difference of their longitudes, are the quantities whose relations are the object of computation.

It has already been shewn, that if the base of a right angled plane triangle represent the distance of two meridians, on a parallel whose latitude is the measure of the acute angle at the base of the triangle, the hypothenuse will represent the distance of the meridians at the equator, or the difference of their longitudes.

Hence by the solution of a right angled triangle, when any two of these quantities are given, the other may readily be found; it may however be convenient to recollect the following proportions, viz. 1.

rad : diff long : : cos lat : mer dist;
cos lat : mer dist : : rad : diff long ;

diff long : rad :: mer dist : cos lat ; 2. cos any lat : mer dist in that lat : : cos any other lat : mer dist in that other lat.




1. If a ship sail from Cape Finisterre west 196 miles, required her longitude ?

BY CONSTRUCTION. Make the angle A = 42° 54', the latitude of Cape Finisterre, and from a scale of equal parts take A B = 196, the given C meridian distance ; let B C, perpendicular to A B meet A Cin C; then A C measured on the scale of equal parts will be found to be 268, the required difference of longitude.

B Dist.



To the angle A, nearly 43o, in Table 3. and A B, 196, in the lat column, corresponds A C, 263 in the dist column.


Extend on the line of sines from 47° 6', the complement of the latitude, to radius, 90°, and that extent will reach on the line of numbers from 196, the meridian distance, towards the right to 268, the difference of longitude.

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2. If a ship sail E 126 miles from the North Cape, in Lapland, and then due N till she arrives in lat 73° 26' N, how far must she sail W to reach the meridian of the North Cape?



Make A B = 126; the given distance sailed from the North Cape, and the < BAC = 71° 10', the lat of the North Cape, and draw BC perpendicular to AB; then A C will be the dif long. On A C describe a semicircle, and from A draw A D, making the angle CAD= 73° 26', the latitude of which the ship sailed W, meeting the semicircular arc in D, then A D will be the required distance lli.



In Table 3. with the argie C AB, nearly 5 l°, and A B 126 in the lat column, the diff long A C is found = 337 nearly in the dist column. Then with C A D nearly 73.1°, and 387 in the dist colunın, D A the required distance is found = 110 nearly in the lat column.

Note. As the given numbers are too great to be found in their respective columns, the half of each number may be sought for, and the corresponding result doubled to obtain the required number.


Extend from 15° 50', the complement of the latitude of the North Cape, to 16° 34', the complement of the other given latitude on the line of sines, and that extent will reach in the same direction on the line of numbers from 126, the given meridian distance in the latitude of the North Cape to 111.3, the corresponding meridian distance in

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1. How far must a ship sail W on the parallel of 60° to change her longitude 20 ?

Answer, 60 miles. 2. If a ship sail E from Cape Race 212 miles, required her longitude

Answer, 47° 54' W. 3. Two places in lat 50° 12' diff in long 34° 48', required their meridian distance ?

Answer, 1336 miles. 4. If a ship sail from the Naze, in Norway, due W 78 miles, required her longitude ?

Answer, 4° 36' E. 5. How far must a ship sail W from the Cape of Good Hope that her course to James Town, St. Helena, may be due north ?

Answer, 1201 miles. 6. A ship in south latitude sails on a parallel 189 miles, and another in north latitude sails on a parallel 230 miles, they have both made the same diff long, 267 miles, required their difference of latitude ?

Answer, 75° 27', or 4527 miles. 7. Two ships in the same north latitude, but 126 miles distant from each other, sail S 248 miles, and their distance from each other is then 200 iniles, required the latitude left, and the latitude arrived at ? Answer, lat left 83° 2' N, and lat arrived at 78° 54'.

8. The master of a West Indiaman, bound for Ireland, in long 16° 12' W, determines to take the shortest distance to the parallel of Cape Clear, when he is arrived in that parallel what will be his distance from the Cape ?

Answer, 246 miles. 9. How far must a ship sail E from Cape Horn, to reach the meridian of the Cape of Good Hope ?

Answer, 2884 miles. 10. In what latitude will a ship's diff long be three times the distance which she sails upon a parallel ? Answer, 10° 31' 44".

11. I sailed E from Tinmouth 123, and then N 156 miles, how far must I sail W to reach the meridian of Tinmouth?

Answer, 115 miles. 12. On what parallel of latitude must a ship sail that her distance and difference of longitude may be equal ?

Answer, she must sail on the equator.


By PLANE SAILING, as we have seen, we can correctly determine the relations which connect the nautical distance, difference of latitude, departure, and course, and their respective magnitudes ; and by parallel sailing, as we have also seen, the distance sailed on a parallel, the latitude of the parallel, and the corresponding difference of longitude, may be determined from each other.

MIDDLE LATITUDE and MERCATOR's Sailing are two different methods of determining the relation between a ship's change of place, and her difference of longitude, when she sails on an oblique rhumb.

In MIDDLE LATITUDE SAILING, the departure computed by plane sailing is considered as a meridian distance in the middle latitude, and the difference of longitude, on that supposition, is computed as in parallel sailing. For, the departure being taken as the base of a right angled plane triangle, and the middle latitude as the acute angle adjoining the base, the hypothenuse of the triangle is nearly equal to the difference of longitude.

This method of deducing the difference of longitude is slightly erroneous, because the departure is not strictly equal to the meridian distance in the middle parallel ; but it will scarcely produce any material error in computing the difference of longitude for an ordinary day's run ; and in any case when the course is nearly east or west, it is safer to compute the difference of longitude by this method than by Mercator's sailing; for a small mistake in the course by account will, under such circumstances, produce an error of much greater importance in the difference of longitude computed by Mercator's sailing, than

any that can arise from the trilling error in the assumption on which middle latitude sailing' is founded.

Besides the proportions which will readily suggest themselves from a consideration of the figure, the following, which have been previously deduced, are here again stated.

cos mid lat : dist :: sin course : diff long,

diff lat : diff long : : cos mid lat : tan course. These proportions may be varied by inversion, alternaţion, &c. so as to make any term that may be required the last in the proportion,

In MERCATOR's Sailing the difference of longitude is computed by considering the meridians as parallel lines, and of course all the parallels of latitude as equal to the equator. But the elementary parts of the meridian are also conceived to be increased in the same proportion as their parallels of latitude are ; so that the proportion between the elementary parts of the meridians and their parallels remains unaltered. The increased meridian between the two distinction to the actual distance of the parallels, which is called the proper difference of latitude. The meridional difference of latitude is obtained by taking the difference of the meridional parts corresponding to each latitude from Table 3. when the latitudes are of the same name, but their sum, when the latitudes are of different names.

Then if a triangle be constructed similar to that formed by the difference of latitude, departure, and nautical distance, in plane sailing, having the side adjacent to the course equal to the meridional difference of latitude, the side opposite to the course will be the difference of longitude; and the different parts of these triangles may be computed from each other. It may be useful in practice to recollect the following proportions :

prop diff lat : dep : : mer diff lat : diff long,
mer diff lat : rad : : diff long : tan course.


1. Required the course and distance from the east point of St. Michael's, Azores, to the Start ?

Lat of Start ... 50° 13' N

= 3495
lat of St. Michael's 37 49 N

= 2454
diff lat 12 24 = 1744 miles 1041 mer diff lat

1)88 02
mid lat 44 1

Long of Start

3° 38' W long of St. Michael's

25 11 W
diff long 21 33 W

1293 miles diff long.


Construction of the figure. Draw the meridian line A D, and at the point A make an angle BAC equal to the complement of the middle latitude, 45° 59'; from A on A B lay off 1293 miles, the diff long; from B let fall the perpendicular BC on AD, then B C will be the mer dist in the mid lat, or the departure nearly.

B Produce A C until C D be equal to 744 miles, the diff lat, and join D B, which will be the distance, and the angle D will be the course.

Hence D B, measured on a line of equal parts, will be 1179 miles, and 2 D the course N 51° E.

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