Sine Rumbs, or 36d. 15m. on the Sines) the fame Extent Chap. V. will reach from the Distance 987 to Difference of Latitude $20 Miles or Minutes, or 13d. 40m. which added to the Latitude failed from, finds the Latitude come to 63d. 40m. then to find the Meridional Difference of Latitude, extend the Compaffes upon the Meridian Line from 5cd. om. to 63d. 40m. that Extent applied to the equal Parts, accounting (as before directed) every 10 Degrees for 600, and each Degree for 60 Miles or Minutes, will be found to be 1522, the Meridional Difference of Latitude. Then for Difference of Longitude, proceed according to the Propertion in Mercator's Sailing Trigonometrical, Cafe 1. vis. Extend from Radius to the Tangent of the Course, viz. 3. on the Tangent Rumbs, or 33d. 45m. on the Tangents, the fame Extent will reach from the Meridional Difference of Latitude 1522 to the Difference of Longitude 1017, and fo in others.

In Cafe II. where both Latitudes, and Courfe is given, the Distance is found, as in Cafe II. of Plane Sailing Inftrumental, and the Difference of Longitude as in Cafe 1. hereof, which is fo eafy, that it needs no Example.

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

A Ship in Latitude 50 North, fails N.N.W. till her Difference of Longitude be 7 Degrees, or 420 Minutes, I

demand as above.

Extend from the Tangent of the Courfe, two Points to Radius, the fame Extent will reach from 420 the Difference of Longitude, to 1014 the Meridional Difference of Latitude.

Take the Meridional Difference of Latitude 1014 from the Scale of Equal Parts, then with that Extent, and one Foot in Lat. 50 on the Meridian Line, the other will reach to 59d. 40m. the Latitude come to. the Distance extend from Sine Complement of the Courfe Then for (6 on the Sine Rumbs, or 67: 30 on the Sines) to Radius,

the

the fame Extent will reach from the proper Difference of Latitude 580 to 628, the Distance required.

SECT. II.

Parallel and Middle Latitude Sailing Inftrumental.

TH HE Proportion for folving all the reft of the Cafes both in Mercator and Parallel Sailing, and also in Middle Latitude, may be fo easily deduced from the Proportions laid down in the Trigonometrical Part, that I need fay no more about them; only in the Inftrumental Operation, where a Tangent is required, the Point of the Compass will fometimes fall beyond the End of the Line; as suppose the Proportion were, as Tangent 21d. 30m. to Tangent 37d. 20m. fo Tangent 42d. 40m. to a Tangent required. And here, if you extend from the firft Term 21d. 30m. to Tangent 37d. 20m. the fame Extent will reach from 42d. 40m. to beyond the End of the Scale; but to remedy this, extend from Tangent 21d. 30m. to Tangent 37d. 20m, then with that Extent, and one Foot in Tangent 45, extend the other back towards the Left-hand, and where-ever it lights, keep it fixed and clofe the other to the third Term, viz. Tangent 42d. 40m. then with this Extent, and one Foot in Tangent 45, the other will fall upon Tangent 6od. 45m. the Tangent required.

As for Traverfe Sailing, it being compounded of feveral Questions in Plane Sailing, I fhall not heed to fpeak of it in this Place, fuppofing that by what is already laid down in this Section of Inftrumental Navigation, the Learner will be able to make Application, as Occafion requires, without any further Inftructions.

CHAP.

CHAP. VI.

Navigation New Modell'd;

OR,

The WHOLE ART Performed

BY A

NEW METHOD.

SECT. I.

Rules and Grounds of the Method.

IN order to the right Understanding of this new Method of Trigonometry, I fhall proceed according to the ufual Manner, and fhall, for the Help of Memory, lay down fome fundamental Rules or Axioms, upon which the whole Operation depends, and by which all the Cafes in Plane Trigonometry, both Right and Oblique, may be folved, without any Book, Table, or Inftrument whatfover. But before I come to the Axioms, I fhall premife, that whenever a Side and an Angle is given, to find another Side, (which is the firft and most useful Cafe in Navigation) there muft first be a Number found, which I call the Natural Radius, not only because it is the Original, from whence the Solutions are deduced, but alfo because being found, it pro

duces

duces the fame Anfwer in Natural Numbers, that the Radius, or Sine of 90, produces in a Sinical Proportion; and this Natural Radius is thus found.

METHOD the First,

Take the Angle whofe oppofite Side is either given or fought, and divide four times the Square of its Complement to 90 Degrees, by 300 added to three times the faid Complement, and then the Quotient added to the faid Angle is the Natural Radius required; and this Rule is univerfally true in all Angles from 0 to 90,

METHOD the Second.

But because in Angles under 45, the Complements are above 45, and their Squares amount to greater Numbers than the Squares of the Complements of the Angles above 45; therefore to render the Work as easy, and the Contrivance as useful as poffible, I fhall fhew another Way to find the Natural Radius for all Angles under 45, and the Rule is,

Divide three Times the Square of the Angle (whose oppofite Side is given or sought) by 1000, the Quotient added to 57 3, (that is 57 3) the Sum is the Natural Radius required.

This being premised, the Rules are these ;

RULE the First. In Right-angled Triangles,

An Angle and a Side given, to find another Side,

The Natural Radius bears always the fame Proportion to the Hypotenufe that the Angle (by which the Natural Radius was found) bears to its oppofite Side.

Therefore if the Angles and Hypotenuse be given, it is, As Natural Radius to Hypotenufe, fo the Angle to its oppofite Side. But if the Angles and a Leg be given, then it is, As the Angle to its oppofite Side, fo Natural Radius to the Hypotenuse,

RULE

RULE the Second. In Right-angled Triangles.

Two Sides given, to find a Third.

The Hypotenufe is equal in Power to the two Legs: That is, the Square of the Hypotenufe is equal to the Square of both Legs added together; of which fee more in Plane Sailing Arithmetical.

RULE the Third. In Right-angled Triangles.

The Hypotenuse and a Leg given, to find the other Leg.

Multiply the Sum of the Hypotenufe, and given Leg by their Difference: The Square Root of the Product is the other Leg required.

RULE the Fourth. In Right-angled Triangles.

Three Sides given, to find an Angle,

Add half the longer Leg to the Hypotenuefe: Then As that Sum to 86, fo the fhorter Leg to its oppofite Angle.

RULE the Fifth. In Oblique Triangles.

Three Sides given, to find where the Perpendicular muft fall.

Multiply the Sum of the two fhorteft Sides by their Difference, and divide the Product by the third Side, which is the greatest, and upon which the Perpendicular is to fall: The Quotient added to the greateft Side, or fubtracted from it, fhall be double the greater or leffer Segment, on each Side of the Perpendicular.

Another Way.

Add the Squares of the biggest and leaft Sides together, and from their Sum fubtract the Square of the

middle