Nine geometricall exercises, for young sea-men and others that are studious in mathematicall practices: containing IX particular treatises, whose contents follow in the next pages. All which exercises are geometrically performed, by a line of chords and equal parts, by waies not usually known or practised. Unto which the analogies or proportions are added, whereby they may be applied to the chiliads of logarithms, and canons of artificiall sines and tangents. By William Leybourn, philomath.

SECTION I.

BEfore I come to the Resolving of such Problems as principally appertain to Navigation, which are such as concern Longitude, Latitude, Rhumb and Distance; I shall shew how the Solution of plain Triangles may be made applicable to the ta∣king of Heights and Distances, and so (in the first place) pro∣pose and work severall Nauticall Questions, which to the indu∣strious

Mariner will be both delightfull and profitable, and give occasion to him to invent and put in Practice others of his own contrivance.

QUESTION I. There are two Ships set sail from the Port A, the one saileth di∣rectly North 24 Centesms, (or 4 Leagues and ⅘ parts of a League,) and the other directly East 32 Centesms, (or 6 ⅖ Leagues;) I demand how the two Ships bear one from the other, and also how far they are asunder.

[illustration] geometrical diagram

DRAW a right Line A B, and upon A raise the Perpen∣dicular A C: let the Point A represent the Port from whence the two Ships set sail: then, because the first Ship sailed 24 Centesms North, from a Scale of equal parts take 24 Cent. and set them from A to C; so shall C be the place of the first Ship. Then, because the other Ship sailed directly

East, which is a Quadrant or Quarter of the Compass distant from the North, therefore the Angle at A must be a right Angle: And because the second Ship sailed East 32 Cent. take 32 Cent. from the same Line of equal parts, and set them upon the Line A B, from A unto B; so shall B be the place of the second Ship.

Now first, To know how these two Ships bear one from a∣nother, Draw the Line C B, and measure the Quantity of the Angle at B, which you shall find to be 33 degr. 45 min. which is three Points from the West Northerly, that is the N. W. by West Point of the Compass; and so doth the second Ship B bear from the first Ship C.—Again, find the quantity of the Angle at C, which you shall find to be 56 degr. 15 m. which is five Points from the South Easterly, that is the S. E. by East Point of the Compass; and so do the two Ships bear one from the other.—Then for the Distance that the two Ships are from each other, Take in your Compasses the di∣stance between B and C, which measure upon your Scale of equal parts, and you shall find it to contain 40 Centesms or 8 Leagues; and so far asunder are the two Ships B and C.

The Bearing of the Ships one from the other is found by the first Case of Right-angled plain Triangles by this Ana∣logie.

As the Distance that the first Ship sailed is to the Distance that the second Ship sailed;

So is the Radius to the Tangent of the Angle that the first Ship bears to the second. The Complement whereof is the bearing of the second Ship to the first.

The Distance of the Ships from each other is found by the seventh Case, by the following Analogie. Having found the Angle C, the bearing of the first Ship from the second, say,

As the Bearing of the first Ship is to the Distance that the se∣cond Ship sailed;

So is the Radius to the Distance of the two Ships.

QUEST. II. A Ship at A discovers an Island at C, lying from her directly East, but she sails from A towards B 32 Cent. or 6 ⅖ Leagues directly South; but her Compass coming to some mischance, that use cannot be made of it, she again at B discovers the same Island, and sails upon an unknown Point of the Compass directly upon the Island, and touches upon it, having sailed 8 Leagues.—I demand upon what Point of the Compass the Ship sailed from B to C, and also how far off the Island was from A, where it was first discovered.

DRAW a Line C A, representing a Line of East and West, and upon A erect a Perpendicular A B, and from A to B set off 32 Cent. or 6 ⅖ Leagues, the distance that the Ship sailed from A to B. Then take out of your Scale of equal

[illustration] geometrical diagram

parts 40 Cent. or 8 Leagues, the distance that the Ship sailed from B to the Island; and setting one foot of the Compasses in B, with the other describe an obscure Arch of a Circle m m, crossing the East and West Line in C: so is C the place of the Island.

Now first, to find upon what Point of the Compass the Ship sailed from B to the Island, you must find the quantity of the Angle at B, (either by your Line of Chords, or Pro∣tracting Quadrant,) and you shall find it to contain 33 degr. 45 min. which is three Points from the North Easterly, that is N. E. by N. and upon that Point did the Ship sail from B to the Island at C.—Then, to know how far the Island C was from A, where it was first discovered, Take in your Com∣passes the length of the Line A C, and measure it upon your Scale; so shall you find that to contain 24 Cent. or ⅘ Leagues: and so far distant was the Island from A.

The Point of the Compass that the Ship sailed upon from B to C may be found by the second Case of Right-angled plain Triangles, by this Analogie.

As the Distance which the Ship sailed from B to C is to the Radius;

So is the Distance sailed between A and B to the Co-sine of the Point that the Ship sailed upon from B to C.

The Distance that the Ship was from the Island, when first discovered, may be found by the fifth Case of Right-angled plain Triangles, by the following Analogie.

(1.) As the Distance that the Ship sailed from B to C is to the Radius;

So is the Distance that the Ship sailed from A to B to the bear∣ing of the Island from B.

(2.) As the Radius is to the Distance that the Ship sailed from C to B;

So is the Sine of the Rhumb that the Ship sailed upon from B to C to the Distance of the Island from A.

QUEST. III. There are two Ports at A and B which are distant 6 ⅖ Leagues, and lie directly North and South of each other; from whence two Ships set sail, both for the Port C: the Ship at B sails away upon a South-W. by South Point; and the Ship at A sails directly West.—I demand how many Leagues either of the Ships had sailed when they met at the Port C, and also how the Port C did bear from that at B.

[illustration] geometrical diagram

DRAW a right Line A B, and upon it set off 32 Cen∣tesms, or 6 ⅖ Leagues. Now because the Ship at B steered a S. W. by S. Course, which is three Points from the South-Westerly, therefore upon the Point B protract an An∣gle of 33 degr. 45 min. and draw the Line B C.—Then, because the Ship at A steered a Westerly Course, which is a Quarter from the North, upon the Point A protract an Angle

of 90 degr. and draw the Line A C, cutting the former Line B C in C.—Now to know how many Leagues each Ship sailed, take in your Compasses the length of the Line B C, and measuring it upon your Scale, you shall find it to contain eight Leagues; and so many did the Ship that came from B sail. Also take the length of the Line A C in your Compasses, and measuring that upon your Scale, it will be found to contain 24 Centesm. or 4 ⅘ Leagues; and so much did the Ship that came from A sail. Now to know how the Port at C did bear from that at B, find the quantity of the Angle at C, which you shall find to be 56 degr. 15 min. that is, five Points from the East Northerly, namely, N. E. by N. and so did the Port C bear from B.

The finding of the Distance that each Ship sailed may be done by the third Case of Right-angled plain Triangles by this Analogie.

As the Distance of the two Ports A and B is to the bearing of the Port C from B;

So is the Sine of the Rhumb that the Ship sailed upon from B to C to the Distance that the Ship sailed from A to C;

And so is the Radius to the number of Leagues that the Ship sailed from B to C.

QUEST. IV. A Ship at C discovers a Point of Land at A bearing from her S. S. E. but she shapes a Course E. by S. and sails away 8 Leagues to B, and at B she discovers the same Point of Land bearing from her W. S. W.—I demand how far the Ship was from Land being at C and B.

DRAW a Line C B containing 40 Cent. or 8 Leagues, and upon C protract an Angle of 56 degr. 15 min. or five Points, which is the difference the Point of Land did bear

from the Ship being at C, and the Point upon which he sailed from C to B; and draw a right Line C A.—Then upon the Point B protract an Angle of 33 degr. 45 min. which is the difference of the Ship's bearing from C and A, she being at B, namely, W. S. W. and draw the Line B A, cutting the Line C A, before drawn, in A.

Now to find how far the Ship was from Land being at C, measure the Line CA upon your Scale of equal parts, and you shall find it to contain 24 Centesms, or 4 ⅘ Leagues: and so far was the Ship from the Land, when she was at C. Also measure the length of the Line B A, and you shall find that to contain 32 Cent. or 6 ⅖ Leagues: and so far from Land was the Ship being at B.

To find these Distances by the Canon of Sines and Table of Logar. you may doe it by the fourth Case of Right-angled Triangles, by this Analogie.

As the Radius is to the Distance that the Ship sailed from C to B;

So is the Bearing of the Ship, being at C, to her Distance from Land, being at B.

Or,

The Bearing of the Sip, she being at B, to her Distance from Land at C.

QUEST. V. A Ship being at A discovers two other Ships at C and B; the Ship at C bears from her directly East, and the other Ship at B bears from her directly South. The Ship at A sails directly South 32 Cent. to B, and being at B, steers away upon an un∣known Course to C 40 Cent. or 8 Leagues.—I demand upon what Point the Ship failed from B to C,—and also how far C is distant from A.

DRAW a right Line A B, for the bearing of the Ship B from the Ship A, which was direct South. Also from A draw another Line A C, for the bearing of the Ship C from the Ship A, which was directly-East. Now because between the South and the East is 90 degr. or one Quarter of the Compass, therefore upon the Point A protract an Angle of 90 degr. drawing the Lines A C and A B at right Angles. This done, take 32 Cent. out of your Scale of equal parts, which is the distance that the Ship sailed South from A to B. Then take from the same Scale 40 Cent. which is the distance that the Ship sailed from B to C upon an unknown Point. And with this distance, setting one foot of the Compasses in B, with the other describe an obscure Arch of a Circle m m, cutting the Line A C in the Point C, and draw the Line C B.—Now to find upon what Point of the Compass the Ship sailed from B to C, find the quantity of the Angle at B, which you shall find to contain 33 degr. 45 min. that is three Points from the North Easterly, namely, N. E. by N. and upon that Point did the Ship sail from B to C.—Then to find how far C is distant from A, Take the Line C A in your Compasses, and measuring it upon your Scale, you shall find it to contain 24 Cent. or 4 ⅘ Leagues: and so far is C distant from A.

The Point upon which the Ship sailed from B to C may be found by the second Case of Right-angled Triangles, by this Analogie.

As the Distance that the Ship sailed from B to C is to the Ra∣dius;

So is the Distance that the Ship sailed from A to B to the Co-sine of the Rhumb from the Meridian.

Then for the Distance of C from A.

As the Radius is to the Distance that the Ship sailed from B to C;

So is the Rhumb from the Meridian that the Ship sailed upon from B to C to the Distance of C A.

QUEST. VI. Two Islands at A and C are discovered by a Ship at B, the Island A bears from the Ship at B N. N. W. and the Island at C bears N. by E. from B; the Ship being at B sails away N. N. W. to the Island A, and having sailed 32 Cent. touches upon the Island, and being there findes that the Island C bears from the Island A E. N. E.—I demand how far the Ship at B was from the Island C, and also how far the two Islands were asunder.

DRAW a Line A B, and upon it set 32 Cent. which is the Distance that the Ship sailed from B to the Island at A. And because the Island A did bear from B N. N. W. and the Island at C N. by E. which are three Points, or 33 degr. 45 min. asunder, upon the Point B protract an Angle of 33 d. 45 min. and draw the Line B C.—Then because the Island at C bears from the Island at A E. N. E. which is eight Points, or 90 degr. from N. N. W. upon the Point A pro∣tract an Angle of 90 degr. and draw the Line A C, cutting the Line B C in C.

Now to find the Distance of the Ship being at B from the Island C, take the Line C B in your Compasses, and applying it to your Scale, you shall find it to contain 40 Cent. or 8 Leagues; and so far was the Ship at B from the Island at C. And to find the Distance of the Islands one from the other, take C A in your Compasses, and measure it upon your Scale, you shall find it to contain 24 Cent. or 4 ⅘ Leagues; and so far distant were the Islands one from the other.

The Distance from A to C may be found by the sixth Case of Right-angled plain Triangles, by this Analogie.

As the Co-sine of the Rhumb that the Ship sailed upon from B to A is to the Distance that the Ship sailed from B to A;

So is the Radius to the Distance of the Ship at B from the Island at C.

Then for the Distance of the two Islands, by the fourth Case say,

As the Radius is to the Distance C B;

So is the Sine of the Difference between the bearing of the two Islands from B to the Distance of the two Islands C and A.

QUEST. VII. Two Ships set out from one and the same Port A; the Ship C sails 24 Cent. or 4 ⅘ Leagues directly East, and the Ship B sails away 32 Cent. or 6 ⅖ Leagues directly South.—When they have thus sailed, I demand how far the two Ships are from each other.

DRAW a right Line A B, and set off upon it 32 Cent. the Distance that the Ship sailed from A to B South. —Then because the other Ship sailed directly East, which is 90 degr. from the South, upon the Point A erect the Perpen∣dicular A C, and upon it set off 24 Cent. or 4 ⅘ Leagues from A to C, which was the Distance the other Ship sailed East. —Then draw the Line C B, which being taken in your Compasses, and measured upon your Scale, will be found to contain 40 Cent. or 8 Leagues. And so far are the two Ships from each other.

This Distance, by the seventh Case of Right-angled plain Triangles, may be found by this Analogie.

(1.) As the Distance that the Ship sailed from A to B is to the Distance that the Ship sailed from A to C;

So is the Radius to the Tangent of the Angle at B.

(2.) As the Sine of the Angle at B is to the Distance C A; So is the Radius to the Distance C B.

QUEST. VIII. Two Ships set sail from the Port at K; the one sails 3 77/100 Leagues upon the S. W. Point towards M, the other sails 8 Leagues upon the West Point towards L.—I demand how many Leagues the Ships at M and L are asunder, and also how the Ship at M bears from the Port K, and the other Ship at L.

DRAW a right Line K L, and by help of your Scale set off upon it 8 Leagues, the Distance that the Ship sailed from K to L upon the West-Point. Then because the other Ship sailed 3 77/100 Leagues from K towards M upon the S. W. Point, which is 45 degr. or 4. Points from the West, therefore upon the Point K protract an Angle of 45 degr. and draw the Line K M, setting off upon it from K to M 3 77/100 Leagues, the Distance that the Ship sailed from K to M, and draw the Line M L.

[illustration] geometrical diagram

Now to know, First, how far distant the Ships at M and L are from each other, take in your Compasses the length of the Line M L, which applie to your Scale, and you shall find it to contain 6 62/100 Leagues.—And, Secondly, to find how the Ship at M bears from the Port K and the other Ship at L, you must find the quantity of the Angle at M, which you will find to be 112 degr. 30 min. that is, eleven Points. Now because the Course from K to M was S. W. therefore the Ship at M bears from the Port K N. E. And seeing that the Angle at M is 112 degr. 30 min. or eleven Points; therefore eleven Points counted from the N. E. Point is the Bearing of the Ship at M from that at L, which is W. N. W.

The Distance of the Ships M and L may be found by tho fifth Case of Oblique-angled plain Triangles; and the Bear∣ings by the second Case.

QUEST. IX. There are three Ships, K, L, and M: the Ship K is distant from the Ship L 8 Leagues; the Ship at L is distant from that at M 6 62/100 Leagues; and the Ship at M is distant from that at K 3 77/100 Leagues; and they lie directly North and South.—I demand how the Ship at M bears to that at L, and how that at L bears to that at K.

DRAW a right Line, and out of your Scale take 8 Leagues, and set them thereon from K to L, for the Di∣stance of the Ships at K and L. Then take 3 77/100 Leagues, the Distance of the Ships K and M, out of your Scale; and setting one foot of the Compasses in K, with the other describe the obscure Arch of a Circle o o. Again, take 6 62/100 Leagues from your Scale, which is the Distance that the Ship L was from the Ship M; and setting one foot of the Compasses in L, with the other describe the obscure Arch of a Circle n n, crossing

the former Arch in the Point M. Then draw the Lines M K and M L; so have you their true Positions.

Now to find their Bearing one from another; forasmuch as the Ships M and K did lie North and South of each other, find the quantity of the Angle at M, which is 112 degr. 30 min. that is, eleven Points from the South Eastward, (or 3 Points from the East Northward,) either of which will be the N.E. by E. Point: and so doth the Ship M bear from that at K. And for the Bearing of that at K from that at L, finde the quantity of the Angle at L, which will be 22 degr. 30 min. or two Points; so two Points from the S. W. by W. Point Southward is S. W. by S. and so doth the Ship L bear to that at K.

[illustration] geometrical diagram

The Bearings of the Ships from each other may be found by the third Case of Oblique-angled plain Triangles, by the Analogie in that Case set down.