To find the difference of longitude.
Cosine middle latitude 52° 20' 9:89849
: Departure 54

1'73239
:: Radius

10'00000

: Difference of longitude 68.2

Or 1° 8' E.

1.83390

By the Traverse Table.

With distance 80 and departure 53'5 (the nearest in the table to 54) are found course 42°, and difference of latitude 59'4. And with comiddle latitude 52 as course, and departure 53:6, is found distance 68, which is the difference of longitude.

Solution by Mercator's Sailing.

.Construction.

A

Construct the triangle DAC as before, DB and find the present latitude. From the two latitudes and their correspondent meridional parts, find the meridional difference of latitude, which set off from D southward to E. Draw EB to the eastward, and parallel to CA. · Continue DA till it intersect EB in B.

EB will be the difference of longitude.

Computation.

1 The present latitude and course are found by plane sailing, as has already been done.

Lat.

To find the difference of longitude. Proper diff. lat. 59 ár. co. 8.22893 : Mer. diff. lat. 75

1.87506 :: Departure 54.

173239

: Dif. long

68.6

1.83638

By the Traverse Table.

With' distance 80, and departure 53'5, are found course 42° and difference of latitude 59*4. And under the same course with meridional difference of latitude 75, instead of proper difference of latitude, is found departure 68:6, or difference of longitude.

6. A ship in latitude 29° 40' S. sails 250 miles in the NW. quarter, and then finds her latitude by observation to be 26° 13' S. Her course, departure, and difference of longitude are required to be found.

Construction.

Draw the meridian NS, and from

N a point D in the same, set off to the northward DC=207 miles, the difference of latitude. From C, the latitude come to, draw the parallel of latitude CA, unlimited toward A westward.

From D, with the distance 250m. describe an arc cutting CA in A, and join DA.

From A draw the line AE, making the angle CAE = 27° 56', the middle latitude, and intersecting NS in E.

The angie ADN will be the course, the line AC the de. parture, and the line AE the difference of longitude.

D

Its comp. 34 06 or the angle of course.

With the distance 250, and nearest difference of latitude 207*3, is found course 34°, and departure 139.8. And with the comiddle latitude 62° as course, and departure 139.5, is found distance 158, which is the difference of longitude.

Solution by Mercator's Sailing.

Construction.

Construct the triangle DAC, as B before. Find the meridional difference of latitude, and set it off northward from D to E. Draw EB to the westward of E, and parallel to CA. Continue DA till it intersect EE in B.

EB will be the difference of longitude.

Computation. The course and departure are found by plane sailing, as before.

Lat. sailed from 20° 40' S. Mer. parts 1865
Lat. come to 26 13 S. Mer. parts 1631

Mer. diff. lat. 234

To find the difference wide.

Radius
: Tangent course 34° C
:: Mer. diff. lat. 234

10'00000
9.83062
2*36921

: Diff. long. 158-4

2'19983 By the Traverse Table. Course and departure being found, as before, the meridional difference of latitude 234, taken as difference of latitude under the same course, gives departure or difference of longitude 1583.

7. From a port in latitude 53° 7' N. a ship sails ESE. till her departure becomes 112 miles. It is required to determine the distance sailed, and the difference of longitude made.

Solution by Middle Latitude Sailing.

Construction.
Draw the meridian NS, and from a

N
point D in the same draw DA on a
SSE. course. Draw the parallel of lat- en
itude Da to the eastward = 112 miles,
the departure. Draw likewise the line
aA parallel to NS, and intersecting DA
in A. Through A draw the parallel
AC intersecting NS in C. DC will be
the difference of latitude, and DA the
distance.

Find the latitude, to which she is come, and thence the middle latitude.

From A draw the line AE, making the angle CAE the middle latitude, and the said line will be the difference of longitude.

Computation.

1

3