short runs and days' works; but in calculating large distances across distant parallels, it is liable to error. The principle on which the calculations of Middle Latitude Sailing are founded, is this:-Instead of calculating the difference of longitude corresponding to the departure made on each of the small surfaces, into which we have supposed the sphere to be divided, and adding them together, the whole departure (or sum of the meridian distances) is calculated, and the longitude deduced therefrom by the rules of Parallel Sailing, using for the latitude the arithmetical mean between the latitude sailed from and that arrived at. On this supposition, we have the two first of he following theorems for calculating the departure from the difference of longitude, or the difference of longitude from the departure, which are the same as Theorems III. and IV. of Parallel Sailing, except in writing departure for distance, and middle latitude for latitude: the other theorems are easily obtained by combining the two first with the common theorems of Plane Sailing; observing that the middle latitude is half the sum of the two latitudes, if they are of the same name, or half their difference if of contrary names. This method may be rendered perfectly accurate by applying to the middle latitude a correction taken from the table following Case VII. of this article We shall, however, in the following examples, make the calculations without applying this correction, because, in most cases in practice, it is of but little importance. THEOREM I. As radius is to the cosine of the middle latitude, so is the difference of longitude to the departure. THEOREM II. As the cosine of the middle latitude is to the radius, so is the departure to the difference of longitude. Now, by Case I. of Plane Sailing, the radius is to the sine of the course, as the distance sailed is to the departure, and, if we combine this analogy with Theorem II., we shall have THEOREM III. As the cosine of the middle latitude is to the sine of the course, so is the distance sailed to the difference of longitude. By Case II. of Plane Sailing, we have this analogy; As radius is to the tangent of the course, so is the difference of latitude to the departure; by combining this with Theorem II., we have THEOREM IV. As the cosine of the middle latitude is to the tangent of the course, so is the difference of latitude to the difference of longitude. Whence we easily deduce the following, THEOREM V. As the difference of latitude is to the difference of longitude, so is the cosine of the middle latitude to the tangent of the course. By means of the preceding theorems, we have formed the following table, which contains all the rules necessary for solving the various cases of Middle Latitude Sailing. 68 MIDDLE LATITUDE SAILING MIDDLE LATITUDE SAILING SOLUTIONS. Radius diff. of long. :: cosine middle lat. : departure. Difference of lat. radius :: departure: tangent course Radius departure: cotangent course: diff. of latitude. We shall now proceed to illustrate these rules, by working an example in every case, CASE I. The latitudes and longitudes of two places given, to find their bearing and distance. Required the bearing and distance between Cape Cod light-house, in the latitude of 42° 3′ N., longitude 70° 4′ W., and the island of St. Mary (one of the Western Islands), in the latitude of 36° 59′ N., and longitude 25° 10′ W. Cape Cod. A Longitude... 70° 4′ W. 44 54 Diff. of long. 2694 miles с DepartureQ Diff. Longitude Midd. Lat. S St. Mary. Draw the east and west line DC; with the chord of 60° describe the arc QS about the centre D, to cut DC in Q; upon this arc, set off, from Q to S, the middle latitude 39° 31'; through D and S draw the line DB, which make equal to the difference of longitude 2694 miles; from B let fall upon DC the perpendicular BC; continue this towards A, making AC equal to the difference of latitude 304 miles;† join AD, and it is done. For by this method of construction, on the principles before explained, A will be the situation of Cape Cod, D the situation of St. Mary; CD will be the departure, which, being measured, B The correction of this quantity, in the table at the end of Case VII., is 3' additive, making i 39° 34', which can be used instead of 39° 31', if great accuracy be required. If the place A be to the southward of D, the line AC should be set off upon the line CB, from C will be found to be 2078 miles; the distance will be represented by AD, which, being measured, will be found to be 2102 miles, and the course from Cape Cod to St. Mary will be represented by the angle CAD equal to 81° 41′; therefore the course will be S. 81° 41' E., or E. & S., nearly. Note. The course is put S. 81° 41′ E. because St. Mary, being in a less northern latitude than Cape Cod, is to the southward of it; it is also to the eastward of Cape Cod, because it is in a less western longitude. Extend from the radius, or 90°, to 50° 29', the complement of the middle latitude, on the line of sines; that extent will reach from the difference of longitude 2694, to the departure 2078, on the line of numbers. 2dly. Extend from the difference of latitude 304, to the departure 2078, on the line of numbers; that extent will reach from radius, or 45°, to the course 81° 41', on the line of tangents. 3dly. Extend from the course 81° 41', to the radius 90°, on the line of sines; that extent will reach from the departure 2078, to the distance 2102 miles, on the line of numbers. BY INSPECTION. RULE. Look for the middle latitude, as if it was a course in Plane Sailing, and the difference of longitude in the distance column, opposite to which, in the column of latitude, will stand the departure; having the difference of latitude and departure, the course and distance are found (as in Case VI. Plane Sailing) by seeking in Table II., with the difference of latitude and departure, until they are found to agree in their respective columns; opposite to them will be found the distance in its column, and the course will be found at the top of that table, if the departure be less than the difference of latitude, otherwise at the bottom. Thus, with one tenth of the difference of longitude 269.4 or 269, I enter Table II., and opposite to it, in the distance column of the tables of 39° and 40°, I find 209.1, and 206.1 in the latitude column; now, the middle latitude being nearly 3940, I take the mean of these, 207.6, for the departure, which being multiplied by 10, gives the whole departure 2076. Again, I enter Table I. with one tenth of the departure 207.6, and one tenth of the difference of latitude 30.4, and find that they agree nearly to a course of 7 points, and a distance of 210, which, multiplied by 10, gives the sought distance, 2100 miles, nearly. CASE II. Both latitudes ard departure from the meridian given, to find the course, distance, ana difference of longitude. A ship in the latitude of 49 57' N., and longitude of 15° 16′ W., sails south-westerly till her departure is 194 miles, and latitude in 47° 18' N. Required the course, distance, and longitude in. B a Mid.Lat Diff. Long Diff. Lat. Draw the meridian ACD, on which take AC equal to the difference of latitude 159 miles; draw CB perpendicular to AC, and make it equal to the departure 194 miles; about B, as a centre, describe an arc ab, on which set off the middle latitude 48° 38'; through B and b draw the line BD, meeting ACD in D; join AB, and it is done; for AB will be the distance sailed, which, being measured, will be found equal to 250.8 miles; BD will be the difference of longitude, equal to 293.5 miles; and the angle CAB will represent the course from the meridian, 50° 40'. 1st. The extent from the difference of latitude 159, to the departure 194, on the line of numbers, will reach from radius, or 45°, to the course 50° 40', on the line of tangents. 2dly. The extent from 50° 40′ to radius, or 90°, on the line of sines, will reach from the departure 194, to the distance 251, on the line of numbers. 3dly. The extent from the complement of middle latitude 41° 22′, to radius, or 90°, On the line of sines, will reach from the departure 194, to the difference of longitude 294, on the line of numbers. BY INSPECTION. RULE. With the difference of latitude and departure, find the course and distance (as in Case VI. of Plane Sailing), by seeking in Table II. until the difference of latitude and departure are found to correspond, against which, in the distance column, will be the distance; and if the departure be less than the difference of latitude, the course will be found at the top of that table, otherwise at the bottom. Then take the middle latitude as a course, and find the departure in the latitude column; the number corresponding in the distance column will be the difference of longitude. In the present example, with the difference of latitude 159, and the departure 194, we find that the nearest numbers to these are 158.0 and 195.1, standing together *The correction of this latitude in the table at the end of Case VII. is about 1', making the corrected over 51°, against the distance 251; whence the course by inspection is S. 51° W., and the distance 251. Then, taking as a course 49° (which is the nearest to the middle latitude 48°38′), seek for the departure 194 in the latitude column; the nearest number is 194.2; opposite to this, in the distance column, is 296, for the difference of longitude; this value differs a little from that found by logarithms, owing to the miles of middle latitude neglected; for if we were also to find the difference of longitude for the middle latitude 48°, and proportion for the minutes, the result would come out nearly the same as by logarithms. CASE III. me latitude, course, and distance given, to find the difference of latitude and difference of longitude. A ship in the latitude of 42° 30′ N., and longitude 58° 51′ W., sails S. E. by S 300 miles. Required the latitude and longitude in. BY PROJECTION. Draw the meridian ADE (as in Case I. Plane Sailing); upon A, as a centre, describe an arc with the chord of 60°, and upon it set off, from where it cuts AD, the course S. E. by S., or 3 points; through that point of the arc, and the point A, draw the line AC, which make equal to the distance 300 miles; from C let fall upon AD the perpendicular CD; then will CD be the departure 166.7 miles, and AD the difference of latitude 249.4 miles. Hence we obtain the latitude arrived at, and the middle latitude; draw the line CE, making an angle with DC of 40° 26' equal to the middle latitude; and the distance CE will be the difference of longitude 219 miles; hence the longitude is easily obtained. To find the difference of latitude. As radius 8 points.... Is to the distance 300 . So is cosine course 3 points.... 9.91985 BY LOGARITHMS. To find the departure. 1st. The extent from radius 8 points, to the complement of the course 5 points, en the line marked SR, will reach from the distance 300, to the difference of latitude 249, on the line of numbers. 2dly. The extent from radius 8 points, to the course 3 points, on the line SR, will reach from the distance 300, to the departure 167, on the line of numbers. 3dly. The extent from the complement of middle latitude 49° 34', to radius 90°, on the line of sines, will reach from the departure 167, to the difference of longitude 219, on the line of numbers. *The correction of this latitude in the table at the end of Case VII. is 2', making the corrected middle atitude 40° 28. The logarithm of the departure was found by the preceding canon to le 2.22186, differing a little from the logarithm of 166.7. |