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The Ship is in Latitude 34° 20' N. the Departure is 47.4. W

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N PLANE SAILING the Earth was considered as a Plane, representing a Bowling Green, having the Meridians parallel to

each other, and consequently the Degrees of Longitude equal in all Places; but this cannot be true, as the Earth is a Globe or Sphere; for

As the Meridians are Circles on the Terraqueous Globe, meeting . in the Poles, (as may be seen in the following Figure) it is obvious, that any two of these Circles must recede more at greater Distances from the Poles; and at equal Distances from each Pole, or at the Equator, the Distance between the Meridians is greatest.

The true Place of a Ship at Sea depends upon its Distance from the Equator, and some noted Meridian ; and since the Meridional Distance, that is, the Distance between any two Meridians, varies in every Latitude, it is therefore convenient this Distance should be reckoned in a fixed Latitude; and where the Degrees are of the same Magnitude with those of the Meridian, which can be no where but on the Equator, where 60 Geographical Miles make a Degree.

The Circumference of all Circles are in direct Proportion to each other, as their Radii; and fince the Earth turns once round its Axis in 24 Hours, every Point upon its Surface must describe Circles parallel to the Equator: Hence it follows, that the Circumference of any Parallel of Latitude in Miles, is to the Circumference of the Equator in Miles, as the Co-Sine of that Latitude is to Radius; and, that the Breadth of a Degree in any Parallel of Latitude, is to the Breadth of a Degree upon the Equator, as the Sine Complement of that Latitude is to Radius.

By the last Proportion was the following Table calculated, which shews the Breadth of a Degree of Longitude in every Latitude; and may be made to answer for any Degrees and Minutes by taking proportional Parts. * *

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The above Table shews how many Miles answer to a Degree of Longitude at every Degree of Latitude.

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Hence it follows that:

As Co-Sine of any Paral. of Lat. As Radius, or Sine 90°
Is to the Dist. run in Miles in that | * | Is to the Diff of Long. in Mile
Lat. 2. R So is Co-Sine of any Paral. of Lat:
So is the Radius, or Sine of 90° F | To the Dist. in Miles between any
To the Diff. of Long, in Miles. Two Merid. in that Paral. of Lat.

From what has been said arises the Solution of the following Problems :

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The Diff'rence of Longitude between two Places both in one Parallel of Latitude being given, to find the Distance between them.

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With the Sine of 90° in your Compasses taken from the Plane Scale, and with one Foot in P describe the Arch E Q, and upon it fet off the Difference of Longitude 210 Miles, and draw the Lines P F. and PQ to represent the two Meridians; then will E. Q represent the Equator, and P the Pole. Again, with the Sine Com. of the Latitude 49° 30' viz. 40° 30' in your Compasles, taken from the Line of Sines on the Plane Scale, and with one Foot in P. describe an Arch, and *.e Distance between the Points, where it cuts the two Meridians, being measured upon the same Scale of equal Parts that the Difference of Longitude was, will be the Departure

136.4 Miles. Or,

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Or, thus: Draw the Meridian A B, and with the Chord of 60 in your Compasses describe an Arch, and upon it set off the Complement of the Lat. 40° 30' (taken from the Line of Chords) and set it off. upon the Arch as a Course in Plain Sailing, and draw the Line A C as a Distance, which make equal to the Difference of Longitude 210 Miles; then will the Departure C, D be the Distance 136.4 Miles, as before ; this last Method is preferable to the former, as we are not confined to any particular Scale. Reverse this Problem, and suppose the Distance sailed in any Parallel of Latitude given, to find the Difference of Longitude. With the Sine Com. of Latitude in your Compasses describe an Arch, upon which set off the Departure 136.4 Miles, and through the Points where it cuts the Arch draw the Lines P E and P Q = then with the Sine of 90° in the Compasses, and one Foot in the former Centre P, describe an Arch to cut P E and P Q_i then E Q being measured upon the small Scale of equal Parts that the Departure was, will be the Difference of Longitude 21o Miles.

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To find the Departure. As Radius - — 90° fo.cooco Is to the Difference of Longitude — 210 2. 22222 Sois Co-Sine Latitude — — 49°.30' .# 54 12. 13476 1o.ooooo To the Distance or Departure – 136.4 2.13476

By G U N T E R.

* The Extent from Radius to Sine Com. Lat. 40°33' on the Line of Sines, will reach from the Difference of Longitude 210 to the Distance 136.4 on the Line of Numbers.”

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Find the Sine Com, of the Latitude among the Degrees, and in the Distance Column the Difference of Longitude, opposite to which, in the Column of Departure, is the Distance required; but as the Co-Latitude is 40° 30', therefore,

For 4o Degrees you’ll find 135
For 41 Degrees you'll find 137.7
The Sum is 272.7
Half is the Distance required 136.3
*-

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