CONSTRUCTION.

Draw the Line A B to reprefent the Meridian of the Lizard, and with the Chord of 60° describe an Arch, upon which fet off the Comp. of Mid. Lat. 58° 26', as a Courfe in Plane Sailing, taken from the Line of Chords, and draw the Line A C as a Distance, equal to the Diff. of Long. 3263; from C, let fall the Perpendicular CD, and that will be the Departure 2780 Miles.

Again fet off the Diff. of Lat. 2205 from A to E; upon E, erect the Perpendicular E G, which make equal to the Departure D C, 2780; draw the Line G A, and it is done, for the AngleG A E will be the Course 51° 35', and Á G the Distance 3549 Miles, the fame as before.

Hence it will be eafy to conftruct any Cafe in Middle Latitude Sailing; by confidering the Comp. of Mid. Lat. as a Course in Plane Sailing, and the Diff. of Long. as the Distance; and the Perpendicular let fall on the Meridian will be the Departure.

Then having the Difference of Latitude and the Departure given, the Course and Distance is found the fame as in Cafe the VI. in Plane Sailing.

By this Method are the following Cafes in Middle Latitude Sailing conftructed, being in my Opinion preferable to any other, as the Learner is not confined to any particular Scale of equal Parts, and has only to make Use of the Line of Chords.

Hence the direct Course from the Lizard to the East End of Barbadoes is S. 51 35 W. or, nearly S. W. by W. half W. Distance 3549 Miles, or 1183 Leagues.

By GUNTER.

First, Extend from Radius or 90° to the Co-Sine Mid. Lat. 58° 26' on the Line of Sines, that Extent will reach from the Diff. of Long. 3263 to 2780 the Dep. on the Line of Numbers.

Secondly, Extend from the Diff. of Lat. 2205 to the Dep. 2780 on the Line of Numbers, that Extent will reach from Radius, or 45° to the Course 51° 35' cn the Line of Tangents.

Thirdly, Extend from Radius or 90° to the Sine Course 51° 35′ on the Line Sines, that Extent will reach from the Departure 2780 to the Distance 3549 on the Line of Numbers.

By INSPECTION in the Tables of Difference of Latitude and Departure.

RULE. Find the Degrees of the Complement of Middle Latitude in the Tables of Difference of Latitude and Departure, as if it was a Course in Plane Sailing, and the Difference of Longitude in the Distance Column, as a Distance, in the Departure Column correfponding to these will be the Departure; with the Difference of Latitude and this Departure, find the Course and Distance, as in Cafe VI. in Plane Sailing.

But when the Difference of Longitude happens to be too large, as in this Cafe, divide it by any convenient Number, obferving, to multiply the Departure, when found by the fame Number you divided by.

Here the Difference of Longitude 3263 being too large, I divide it by 12, which gives 272, and the Complement of Middle Latitude is 58.26, but fince the Tables are only calculated to fingle Degrees I look for a 12 of the Difference of Longitude, viz. 272 in the Distance Column over 58 and 59 Degrees, and the Departures correfponding to thefe are 230,7 and 233,1 their Sum is 463,8 half is 231,9, which multiplied by 12 gives 2782 for the Departure.

Now having the Difference of Latitude and Departure the Courfe and Distance is found, as in Cafe VI. in Plane Sailing.

CASE II.

Both Latitudes and the Departure from the Meridian given, to find the Courfe and Distance, and Difference of Longitude.

A Ship in Latitude 49° 57' N. and Longitude 5° 24′ W. fails. South Wefterly, till her Departure is 789 Miles, and she be in Latitude 39° 20' N. I demand the Courfe, Diftance, and Longitude

fhe is in?

Draw the Meridian A D, from A to D fet off the Difference of Latitude 637 Miles, and on D, erect the Perpendicular D G, which make equal to the Departure 789 Miles. Draw the Line AG, and that will be the Distance 1014 Miles, and the Angle DAG the Course 51° 05'.

Again, draw E K parallel to A D, making the Distance from AD equal to the Departure D G 789, take the Complement of the Middle Latitude 45° 22′ in your Compaffes from the Line of Chords, and fet that off, on the Arch on the oppofite Side of the Meridian A D, through where that cuts the Arch, draw the Line A E, to cut the LineK E in E, from E, let fall the Perpendicular E B, and it is done, for A E will be the Difference of Longitude 1109 Miles.

To find the Difference of Longitude it will be,

First. The Extent from the Difference of Latitude 637 to the Departure 789 on the Line of Numbers, will reach from Radius, or 45° backward to 51° 05′ the Courfe on the Line of Tangents.

Secondly. The Extent from 51° 5' to Radius or 90° on the Line of Sines will reach from the Departure 789 to the Distance 1014 on the Line of Numbers.

Thirdly. The Extent from the Complement of Middle Latitude, 45° 22' to Radius or 90° on the Line of Sines, will reach from the Departure 789, to the Difference of Longitude 1109, on the Line of Numbers,

By INSPECTION.

RULE. With the Difference of Latitude and Departure, find the Courfe and Distance, as in Cafe VI. in Plane Sailing.

Secondly. Taking the Comp. of Mid. Lat. as a Course and the Departure in its Column, and the Distance correfponding to these, will be the Difference of Longitude.

Thus, taking a Tenth of the Difference of Latitude 637 and Departure 789, viz. 63.7 and 78.9 the nearest Numbers to these are 63.6

and 78.5, ftanding together over 51°, against Distance 101, which multiplied by 10 gives 1010, whence the Course by Inspection is S. 51° W. and the Distance 1010.

Taking 45° 22' or 45° as a Courfe, and a Tenth of the Departure 78.9 in its Column, the nearest is 78. 5, in the Distance Column ftands III which multiplied by 10 gives 1110 for the Difference of Longitude nearly as before.

One Latitude, Courfe, and Distance, given, to find the Difference of Latitude and Difference of Longitude.

A Ship in Latitude 42° 30′ N. and Longitude 18° 31' W. fails S. E. by S. 591 Miles, or 197 Leagues, I demand the Latitude and Longitude the Ship is in?

By PROJECTION,

With the Course and Distance, find the Difference of Latitude and Departure, as in Cafe I. in Plane Sailing: viz. draw the Meridian A D, and on A describe an Arch with the Chord of 60°, and upon it fet off the Course S. E. by S. or 3 Points, through where that cuts the Arch draw the Line A C, making it equal to the Distance 591, from C, let fall the Perpendicular CD, then will CD be the Departure, and A D the Difference of Latitude 491 Miles.

Draw the Line E F Parallel to A D, making the Distance from it equal to the Departure.

Take the Com. of Mid. Lat, 51° 36′ from the Line of Chords in your Compaffes, and fet it off on the Arch on the other Side of the Meridian A B, and through where that cuts the Arch, draw the Line A E, to cut the Line E F in E, from E let fall the Perpendicular E D and it is done, for A E will be the Difference of Longitude, 419 Miles.