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.

The Doctrine of RIGHT-LINED TRIANGLES applied to Practice in NAVIGATION: Whereby Sundry Nauticall Questions are resolved; and many Problems of Sailing, both by the Plain and Merca∣tor's Chart, performed by Protraction, by Calculati∣on, and also wrought upon the Chart it self. The Ninth EXERCISE.

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.

PROBLEMS Of Sailing by the Plain Sea-Chart. SECTION II.

AMONG Sea-men there are Three principal waies of Sailing most in Use and Practice: Two whereof are Rectilineal, performed by Right Lines, the Third is Sphericall or Cir∣cular, performed by Arches of great Cir∣cles of the Sphere.

Of the Two first, the one is called Plain Sai∣ling, or, Sailing by the Plain Sea-Chart.

The other is called Mercator's Sailing, or, Sailing by Merca∣tor's Chart.

These two Charts are both of them composed of Right Lines, yet differ both in their Construction and Use, though not so much in their Use as in their making or composition.

The Plain Sea-Chart consisteth of Meridians and Parallels, which are drawn in all parts equal from the Aequinoctial towards either of the Poles; which is erroneous, as hereafter shall be dis∣covered.

Mercator's Chart hath the Degrees of Longitude in every Parallel of Latitude equal to those in the Aequinoctial, as the Plain Chart hath: But the Degrees of Latitude do increase more and more (as they grow nearer the Poles) in such a Pro∣portion as every Parallel of Longitude doth decrease.

The way of Sailing by the Plain Sea-Chart is much in use, nay too much, considering the Errours that it leads Sea-men into; though they are not so easily discovered in short as in long Voi∣ages, nor in places near the Aequinoctial as those nearer the Poles. But, I suppose, it is more used, for the ease there is in Projecting of this Chart, more then in that of Mercator's: other∣wise I know not why that should be so as it is embraced, and the other (I mean that of Mercator's) so much neglected; which comes so near to the Spherical way of Sailing, that there is an insensible difference between them. But I shall bring them face to face, that the ingenious Sea-man may see their Difference, and thereby abandon Errour, and embrace the Truth. For in the fol∣lowing Problems I shall perform the same thing by both Charts, by which the Errours may more palpably be discovered. And to retain the Method which I have observed in all the foregoing EXERCISES, I shall shew how these Nauticall Problems may be Trigonometrically performed by the Tables of Loga∣rithms, and Canons of Artificial Sines and Tangents. And to be∣gin this Exercise, the first thing that I shall propose unto you is

The Making of the Plain Sea-Chart.

A Sea-Chart may be made either general, or particular. A General Sea-Chart is that whose Degrees of Latitude pro∣ceed from the Aequinoctial to either Pole, which in the com∣mon Sea-Chart may be done; but it will be egregiously false, as the Degrees of Latitude grow nearer the Pole, as I have al∣ready declared.—A Particular Sea-Chart is such a one as is made properly for one particular Navigation: as if your

whole Navigation were not to exceed the Latitudes of 48 and 60 degr. of Latitude, and not to differ in Longitude above 8 degrees.

Now to project or make such a Chart; First, draw a right Line A B, representing the Meridian, and cross it at right Angles in the Point A with another right Line A D, representing the Parallel of your least Latitude, namely, of 48 degr.— Secondly, consider what Distance you will have your Parallels of Longitude and Latitude to be, (for in this Chart they are both equal,) whether an Inch, 2, 3, or 4 Inches, (for the lar∣ger the better.) But in this Example I have made them one∣ly half an Inch. I take therefore half an Inch out of an exact Scale, and run it up upon the Meridian Line A B, from A to 49, from 49 to 50, from 50 to 51, &c. till I come to my greatest Latitude, which is here supposed to be 60 degr.—Thirdly, run the same Distance of half an Inch from A towards D, upon the Line A D, eight times, because the Difference of Longitude in your whole Navigation will not exceed 8 degrees.— Fourthly, draw the Line C D, parallel to A B, and B C, par∣allel to A D, and run the same Distances upon the Line B C as are upon the Line A D, and the same upon C D as are upon the Line A B.—Fifthly, from each Degree of Latitude in the Line A B draw to the like Degree of Latitude in the Line C D a right Line, as 49, 49; 50, 50; 51, 51; 52, 52; &c. till you have drawn all your Parallels of Latitude.—Sixthly, for your Meridians, they are to be drawn in like manner as were the Parallels of Latitude, all of them equidistant, and par∣allel to your first Meridian A B, as the Lines 1, 1; 2, 2; 3, 3; &c. And by this means have you the Meridians and Parallels drawn.

The grand Divisions, or whole Degrees, being thus set upon your Chart, we now come to sub-divide them. And for the dividing of the Degrees of the Aequinoctial at the top and bottom of your Chart, let each of them be divided into 5 or 10 parts, and each of those parts sub-divided into 5 or 10

more less parts, according as Quantity will permit; for eve∣ry one of them is supposed to be divided into 100 or 1000 parts.

For the dividing of the Degrees of Latitude; they may be di∣vided as those of Longitude were, into 100 parts. But some∣times each Degree is subdivided into 60 Minutes, or English Miles, or into 20 Leagues.—Now I have divided the Degrees of Latitude in this Chart each of them into 5 parts, by which means it is capable of the Numeration either by Miles, Leagues, Centesms, or 100 parts.—For if you count by 60 minutes, or miles, then every of those Divisions will be 12 minutes, or miles; if by 20 Leagues, then every Division will contain 4 Leagues; and if by Centesms or 100 parts, then every of them is 20 Centesms. And thus much concerning the Making or Projecting of this Chart. I now come to shew

Some Uses of the Plain Sea-Chart.

THE Problems that are to be resolved by (or upon) the Sea-Chart are chiefly such as concern Longitude, Lati∣tude, Rhumb or Course, and Distance.

Longitude is the Distance of a Place from some known Meri∣dian to that Place, and is alwaies counted upon the Aequinoctial.

Latitude is the Distance of any Place from the Aequinoctial, counted upon that Meridian Circle which passeth over that Place.

Rhumb or Course is the Angle that a Ship in his Sailing makes with the Meridian, and is discovered in the general by the Magneticall Needle, which alwaies respecteth the North; and (though not directly, yet) its Variation being often observed, and the Chart rectified thereby, (as I have before shewed how it may be done by severall means) is the best help that Navigators yet have to steer their Course by.

Distance is the number of Leagues, Miles, or Centesms, that any Ship hath sailed.

Raising of the Pole is when a Ship sails from a lesser to a greater Latitude.

Depressing of the Pole is sailing from a greater to a lesser Latitude.

These Terms thus explained, I will proceed to Practice, as followeth.

PROBL. I. How to set any Place upon your Chart according to its Longitude and Latitude.

IF the two Places lie under one and the same Parallel, dif∣fering not at all in Latitude, but onely in Longitude, then the Course leading from the one to the other is directly East or West. As E and F are two Places lying under the Par∣allel of 50 degr. of Latitude, and differ in Longitude 5 ½ deg. Lay a Ruler to 5 ½ degr. both at the top and bottom of the Chord, and where the Ruler crosseth the Parallel of 50 degr. as at F, there is your other Place upon the Chart. So E and F lie in 50 degr. of Latitude, and differ in Longitude 5 ½ degr.

But if the two Places to be set upon the Chart differ onely in Latitude, and lie under the same Meridian as G F, then the Course leading from the one to the other is directly North or South, and the Difference of Latitude of F and G is 2 degr. G lying in the Latitude of 48 degr. and F in the Latitude of 50 degr.

But if the Places to be set upon the Chart differ both in Longitude and Latitude, as A and F, then the Course lea∣ding from the one to the other is upon some other Point of the Compass, so far distant from the Meridian as is the quanti∣ty of the Angle E A F, which here is 70 degr. 1 min. that is upon the E. N. E. Point 2 degr. 31 min. Easterly.

This Angle may be found either by Protraction by your Line of Chords, or it may be protracted by your Protracting

Quadrant, which in all these Operations upon the Chart is best, for that it avoids the drawing of Arches of Circles upon your Chart or Blank. So then if you were to protract the Angle E A F by your Protracting Quadrant; Lay the Centre A of your Quadrant upon the Point A in your Chart, and the Meridian Line of the Quadrant A B upon the Meridian Line of your Chart; then will the Line A C of the Quadrant lie upon the Parallel A D of your Chart: and the Angle that you are to protract being 70 degr. 1 min. by the edge of your Quadrant make a small Mark or Prick with your Needle, and from A through that Point draw a right Line, which will be the Line A F.

[illustration] geometrical diagram

And in the same manner as you set any Place upon your Chart, you may find in what Latitudes and difference of Lon∣gitude any Places already set upon your Chart are in.

PROBL. II. Any Places being set upon the Chart, to find in what La∣titudes they are, and also how they differ in Longitude.

LET the Points Q and R upon the Chart be two Places, and I would know in what Latitudes they lie. First, through the Point Q draw a Line parallel to the Line B C of your Chart; and also through the Point R draw another Line parallel to A C. The Line that is drawn through Q

shoots upon the Latitude of 58 degr. 36 min. and the Line passing through R cuts the Meridian of the Chart on either side at the Latitude of 57 degr. 16 min. And under those two Latitudes are the two Places Q and R.

Then to find their Difference of Longitude; Take in your Compasses the Distance between R and S, and measuring it upon the bottom of the Chart, it will reach from A to 4 degr. 24 min. And such is the Difference of Longitude of the two Places Q and R.

PROBL. III. Having the Rhumb, and the Distance that the Ship hath run upon that Rhumb, to find the Difference of Lon∣gitude and Latitude.

The Analogie or Proportion.

As the Radius is to the Distance run;

So is the Sine of the Rhumb to the Difference of Longitude:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Rhumb being 70 degr. 1 min. that is E. N. E. 2 degr. 31 min. Easterly, and the Distance run 117 Leagues, the Dif∣ference of Longitude will be found to be 5 ½ degr. and the Difference of Latitude 2 degr.

Ʋpon the Chart.

UPON the Point A protract an Angle of 70 d. 1 m. as the Angle E A F, and draw the Line A F, which is the Rhumb upon which the Ship sailed. Upon this Line set 117, the number of Leagues that the Ship sailed from A to F. Then through the Point F draw the Line F E parallel to A D. So shall E F be the Difference of Longitude, 5 d. and an half, and A E the Difference of Latitude, 2 degr.

PROBL. IV. The Difference of Latitude and the Rhumb being given, to find the Distance run and the Difference of Lon∣gitude.

The Analogie or Proportion.

As the Co-sine of the Rhumb is to the Difference of Latitude; So is the Radius to the Distance run:

And

So is the Sine of the Rhumb to the Difference of Longitude.

So the one Latitude being 48 degr. and the other 50 degr. the Difference is 2 degr. and the Rhumb being E. N. E. 2 deg. 31 min. Easterly, the Distance run will be found to be 117 Leagues, and the Difference of Longitude 5 ½ degr.

Ʋpon the Chart.

SET the Difference of Latitude 2 degr. from A to E, and draw the Line E F parallel to A D. Then upon the Point A protract the Angle of the Rhumb 70 degr. 1 min. E. N. E. 2 degr. 31 min. Easterly, and draw the Line A. F, cutting the other Line E F in F. Then taking in your Compasses the length of the Line A F, and measuring it upon the side of the Chart, you shall find it to contain 117; which is the number of Leagues the Ship sailed: And the Line E F, being so measured, will contain 5 ½ degr. the Difference of Longitude.

PROBL. V. Having the Difference of Longitude and the Rhumb gi∣ven, to find the Distance run and Difference of Latitude.

The Analogie or Proportion.

As the Sine of the Rhumb is to the Difference of Longitude; So is the Radius to the Distance run:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Rhumb being E. N. E. 2 degr. 31 min. Easterly, and the Difference of Longitude 5 ½ degr. the Distance run will be found to be 117 Leagues, and the Difference of La∣titude 2 degr.

Ʋpon the Chart.

UPON the Point A protract an Angle of the Rhumb 70 degr. 1 min. and draw the Line A F. Then the Difference of Longitude being 5 ½ degr. count 5 ½ degr. upon the bottom of your Chart from A to G, and upon the Point G raise a Perpendicular G F, cutting the Line A F before drawn in F. Then the Line A F, being measured upon the Side of your Chart, will be found to contain 117 Leagues, the Distance run: And F G, there also measured, will be found to be 2 degr. the Difference of Latitude.

PROBL. VI. The Distance that the Ship hath run, and the Difference of Latitude, given, to find the Rhumb and Difference of Longitude.

The Analogie or Proportion.

As the Distance run is to the Radius;

So is the Difference of Latitude to the Co-sine of the Rhumb:

And

So is the Sine of the Rhumb to the Difference of Longitude.

So the distance run being 117 Leagues, and the Difference of Latitude being 2 degr. the Rhumb will be found to be E. N. E. 2 degr. 31 min. Easterly, and the Difference of Lon∣gitude 5 ½ degrees.

Ʋpon the Chart.

SET the Difference of Latitude 2 degr. upon your Chart from A to E, and draw the Line E F parallel to A B. Then out of the Side of your Chart take the Distance run, 117 Leagues; and setting one foot of the Compasses in A, turn the other about till it cross the Line E F, which it will doe in F. Then F E, being measured upon the bottome of your Chart, will contain 5 ½ degr. the Difference of Longitude. And by your Line of Chords or Protracting Quadrant find the Quantity of the Angle E A F, which will be 70 degr. 1 min. the E. N. E. Point 2 degr. 31 min. Easterly.

PROBL. VII. The Distance that the Ship hath run, and the Difference of Longitude, being given, to find the Rhumb and Difference of Latitude.

The Analogie or Proportion.

As the Distance run is to the Radius;

So is the Difference of Longitude to the Rhumb:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Difference of Longitude being 5 ½ degr. and the Di∣stance that the Ship hath run 117 Leagues; the Rhumb will be found to be E. N. E. 2 degr. 31 min. Easterly, and the Difference of Latitude 2 degr.

Ʋpon the Chart.

COUNT the Difference of Longitude upon the bot∣tome of the Chart from A to G, and upon the Point G raise the Perpendicular G F. Then take out of the Side of your Chart the Distance run, 117 Leagues, and setting one foot of the Compasses in A, with the other cross the Perpen∣dicular F G in the Point F. Now if you take F G in your Compasses, and measure it on the Side of your Chart, you shall find it to contain 2 degr. for the Difference of Latitude; and the Angle E A F, being measured by your Chord or Quadrant, will be 70 degr. 1 min. that is the E. N. E. Point 2 d. 31 m. Easterly for the Rhumb.

PROBL. VIII. The Difference of Longitude and Difference of Latitude being given, to find the Rhumb and the Distance run.

The Analogie or Proportion.

As the Difference of Latitude is to the Radius;

So is the Difference of Longitude to the Tangent of the Rhumb:

And

As the Sine of the Rhumb is to the Difference of Longitude;

So is the Radius to the Distance run.

So the Difference of Longitude being 5 ½ degr. and the Difference of Latitude 2 degr. the Rhumb will be found to be E. N. E. 2 degr. 31 min. Easterly, and the Distance upon the Rhumb 117 Leagues.

Ʋpon the Chart.

COUNT the Difference of Latitude from A to E, and draw the Line E F parallel to A D. Also count the Dif∣ference of Longitude from A to G, and upon the Point G raise the Perpendicular G F, cutting the Line E F in the Point F. Then take in your Compasses the length of the Line A F, and measuring it upon the Side of the Chart, you shall find it to contain 117 Leagues, the Distance that the Ship hath run. And if by your Line of Chords, or Quadrant, you find the Quantity of the Angle E A F, it will be the Rhumb, which you may find to be E. N. E. 2 degr. 31 min. Easterly, or 70 degr. 1 min.

PROBL. IX. The Rhumb that a Ship hath sailed upon, and the num∣ber of Leagues she hath sailed upon that Rhumb, being given, to know how much she hath raised or depressed the Pole.

The Analogie or Proportion.

As the Radius is to the Distance run;

So is the Co-sine of the Rhumb from the Meridian to the Difference of both Latitudes.

So the Rhumb being E. N. E. 2 degr. 31 min. Easterly, that is, 70 degr. 1 min. and the Distance that the Ship hath sailed upon that Rhumb 117 Leagues, the Pole will be found to be raised 2 degr.

Ʋpon the Chart.

UPON the Point A, the lesser Latitude, protract an Angle of 70 degr. 1 min. and draw the Line of the Rhumb A F, and out of the Side of your Chart take 117 Leagues, (the Distance the Ship sailed) and set them upon the Rhumb from A to F. Then through the Point F draw the Line E F parallel to A D, cutting the Meridian of your Chart in E, which is 2 degr. from A: so that the Ship hath raised the Pole 2 degrees.

PROBL. X. The Longitude and Latitude of the Place from whence you came, with the Rhumb and Distance sailed, be∣ing given, to find the Longitude and Latitude of the Place to which you are come.

The Analogie or Proportion.

As the Radius is to the Distance run;

So is the Sine of the Rhumb from the Meridian to the Diffe∣rence of Longitude:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Latitude of the Place from whence you came being 52 degr. and the Longitude 35 degr. the Rhumb upon which you have sailed N. E. by N. 33 degr. 45 min. and the Distance which you have sailed upon that Rhumb 96 2/10 Leagues; you shall find the Difference of Longitude to be 2 degr. 40 min. and the Difference of Latitude 4 degr. So that the Place to which you are come is in the Latitude of 56 degr. and in the Longitude of 37 degr. 40 min.

Ʋpon the Chart.

THE Place from whence you came being in the Latitude of 52 degr. and in the Longitude of 35 degr. is repre∣sented by H. The Rhumb you have sailed upon being N. E. by N. 33 degr. 45 min. upon the Point H protract an Angle of 33 degr. 45 min. and draw the Line H K for the Rhumb. Then out of the Side of your Chart take 96 2/10 Leagues, which is so much as the Ship sailed, and set that upon the Rhumb-Line from H to K, and through the Point K draw the Line K L

parallel to B C, (or perpendicular to A B,) and it will cut the Line A B in L. So K L, being measured on the bottom of your Chart, will be found to contain 2 degr. 40 min. the Dif∣ference of Longitude; which added to 35 degr. the Longitude you came from, gives 37 degr. 40 min. for the Latitude you are in. Also the Line H L, being measured on the Side of your Chart, will be found to contain 4 degr. And such is the Difference of Latitude, which added to 52 degr. the Latitude from whence you came, gives 56 degr. the Latitude in which you are.

PROBL. XI. The Longitude and Latitude of the Place from whence you came, the Rhumb upon which you sailed, and the Latitude of the Place to which you are come, being given, to find the Distance and Difference of Lon∣gitude.

The Analogie or Proportion.

As the Difference of Latitude is to the Radius;

So is the Tangent of the Rhumb from the Meridian to the Difference of Longitude:

And

As the Sine of the Rhumb is to the Difference of Longitude;

So is the Radius to the Distance run.

So the Latitude of the Place from whence you came being 52 degr. and the Longitude 35 degr. and the Rhumb upon which you sailed the third from the Meridian N. E. by N. 33 degr. 45 min. you shall find the Distance run to be 96 Leagues 2/10, and the Difference of Longitude 2 degr. 40 min.

Ʋpon the Chart.

UPON your Chart assign H for your Place from whence you came, in the Latitude of 52 degr. and Longitude 35 degr. Upon this Point H protract the Angle of the Rhumb 33 d. 45 m. N. E. by N. and draw the Rhumb-Line H K. Then the Latitude of the Place where you are being found by ob∣servation (or being otherwise given) to be 56 degr. draw a Line quite cross your Chart at the 56^th degree of Latitude, as the Line 56. 56 in the Chart crossing the Rhumb-Line in the Point K. So K L, being measured at the bottome of your Chart, will be found to contain 2 degr. 45 min. which added to 35 degr. the Longitude you came from, makes 37 degr. 40 min. And that is the Longitude in which you are. In like manner measure H K upon the Side of your Chart, and you shall find it to contain 96 2/10 Leagues. And so much hath the Ship run upon that Point N. E. by N.

PROBL. XII. The Latitude of two Places, and the Difference of Longi∣tude between them, being known, to find what Rhumb leadeth from one to the other, and how many Leagues distant they are asunder.

The Analogie or Proportion.

As the Difference of Latitude is to the Radius;

So is the Difference of Longitude to the Tangent of the Rhumb:

And

As the Sine of the Rhumb is to the Difference of Longitude;

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

[illustration] geometrical diagram

The Plaine See Chart.

page 172.

So the Latitude of one of the Places being 50 degr. and the other 52 degr. 30 min. and the Difference of Longitude 6 ½ degrees; the Rhumb will be found to be 67 degr. 23 min. and the Distance upon the Rhumb 6 ½ degr. or 120 Leagues.

Ʋpon the Chart.

UPON the Point of the greater Latitude at N 52 deg. 30 min. draw a Line N M, parallel to A D, upon which Line set 6 degr. the Difference of Longitude of the two Places (being taken from the bottom of the Chart) from N to M. Then from the Point M draw the Line to E, the lesser Latitude, 52 degr. which Line, taken in the Compasses and measured upon the Side of the Chart, will be found to contain 6 ½ degr. or 130 Leagues. Also the Angle N E M, being measured by your Chord, or Protracting Quadrant, will be found to contain 67 degr. 23 min. which is the Rhumb leading from one to the other, namely, short of the E. N. E. Point 7 degr. or, N. E. by E. 11 degr. 8 min. Easterly.

PROBLEMS Of Sailing by Mercator's Chart. SECTION III.

PROBL. I. How to make a Sea-Chart according to MERCATOR's Projection, by your Line of Chords.

UPON a piece of thick and smooth Paper (or ra∣ther Past-board) draw a right Line, as A B, and upon the Point A, with 60 degr. of your Line of Chords, describe the Quadrant A C D, which divide into 90 equal Parts or Degrees, as here in this Figure there is onely every fifth Degree. —This done, upon the Point D erect the Perpendicular D F, (in which you must be very exact.) And from the Point A (through each Degree of the Quadrant) draw right Lines, as A 10, 10; A 20, 20; A 30, 30; A 40, 40; &c. to∣wards 90. till they touch the Line D F.—Then with your

Compasses, one foot being placed in A, extend the other to 60 degr. in the Line D F, and with that Distance describe the Arch F H G B. Again, extend the Compasses from A to 55 in the Line F D, and keeping one foot in A, with the other describe 55 K; so shall the Point K be the Point of 55 degr. in the Line A B.—Also, extend the Compasses from A to 50 in the Line F D, and draw the Arch 50, 50. Doe so with 45, 40, 35, &c. till you come to the beginning of the Degrees at D. So shall these Arch-lines, by meeting with the Line A B, divide that part of it D B into unequal parts, at 10, 20, 30, 40, 50, 60, and so forward. But this Figure is sufficient for Example.

[illustration] geometrical diagram

Now from this Line A B, being thus unequally divided, you may divide the Meridian-line of a Sea-Chart according to Mercator's Projection of any bigness, so that the Distance be∣tween Degree and Degree in the Aequinoctial be less then the Distance A D, which is here two Inches. And if a Chart were made that the Aequinoctial Degrees were two Inches di∣stant, and it passed upon a smooth Board, many Nauticall Con∣clusions might be wrought upon it very exactly. Being thus far prepared, I will now shew you how, from the Line A B,

To divide the Meridian Line of a Sea-Chart.

A Sea-Chart, according to this Projection, may be made either General, or Particular. I call that a General Sea-Chart, whose Line E H, in the following Figure, represents the Aequinoctial, as the Line E H there doth the Parallel of 49 degr. and so I will make the Chart following to contain all Latitudes between 49 degr. and 57 degr. whose Difference of Longitude exceedeth not 8 degr.

Now to project such a Chart, having drawn the Line E F for the Meridian, and crossed it at right Angles with another Line representing the Parallel of 49 d. parallel thereto draw another Line F G, representing the Parallel of 57 degr. and another Meridian G H, parallel to F E. So shall you have made the Parallelogram E F G H.

This done, consider how far distant you would have your Degrees of Longitude upon the Aequinoctial each from other, as suppose (and as in this Chart I have made them to be) half an Inch. Take half an Inch out of a Line of Inches, and run that Distance along the Line E H from E to 1, from 1 to 2, from 2 to 3, &c. And also doe the like upon the Line F G, at the top of the Chart, drawing the Lines 1, 1; 2, 2; 3, 3; &c.

Now for the Dividing of the Meridians E F and H G, re∣pair to the foregoing Figure, taking in your Compasses the Distance that is between Degree and Degree of the Aequino∣ctial,

which in our Example is half an Inch. With this Di∣stance, set one foot of the Compasses in the Point D, and with the other describe the Arch m m; by the very Edge whereof draw the Line A G: so is your Figure prepared to divide the Meridian-line of a Sea-Chart whose Degrees of Longitude are half an Inch distant.

Now in respect that your first Parallel of Latitude E H in your Chart is drawn for 49 degr. your next Parallel must be 50 degr. Wherefore set one foot of your Compasses upon 50 degr. in the Line A B, and with the other take the nearest Distance to the Line A G: that is done by turning the Com∣passes about till the moveable foot do onely touch the Line A G; which when it so doth, that Distance at which your Com∣passes then are, being set upon the Meridian of your Chart, will reach from 49 degr. to 50, which being set upon your Chart, on both sides thereof, from 49 draw the Line 50. 50 will give you the Parallel of 50 d. of Latitude. In like man∣ner for the Parallel of 51 degr. Set one foot of the Compasses in 51 degr. upon the Line A B of the former Figure, and with the other take the least Distance to the Line A G: this Distance set upon the Meridian of your Sea-Chart, on both sides thereof, will reach from 50 to 51; and there draw the Parallel 51, 51.—Likewise for the Parallel of 52 degr. Set one foot of the Compasses in 52 degr. in the Line A B, taking the nearest Distance to the Line A G: that Distance set upon the Meridian of your Sea-Chart, on both sides thereof, will reach from 51 to 52; and there draw the Parallel of 52, 52. Doe thus with all the Degrees, as 53, 54, 55, 56, and 57. So shall the Meridians of your Chart E F and H G be divided into whole Degrees.

For the Sub-divisions of these Degrees, they may be divi∣ded each of them into equal parts, as the Divisions at the top and bottome of the Chart ought to be; but the Degrees of the Meridian, as they grow higher, they ought still to grow greater. But the Difference is so small, that it cannot produce

any considerable Errour, though the Sub-divisions be all made equal between Degree and Degree. You may therefore di∣vide them either into 60 Minutes or English Miles, or into 20 Leagues, or into 100 parts of Degrees, as you shall best like of.

But if you would make a Chart that the Distance between De∣gree and Degree upon the Aequinoctial should be an Inch, or any other Distance less then A D in the foregoing Figure; take that Di∣stance (as suppose an Inch) in your Compasses, and setting one foot in D, with the other describe the Arch o o, and draw the Line A H onely to touch the Arch o o. The least Distance taken from each Degree to this Line A H shall give you the Distance of the Degrees upon the Meridian of a Sea-Chart, whose Distance of Degrees up∣on the Aequinoctial are an Inch from each other.

Your Chart being thus prepared, I will now come to shew you how to resolve severall Problems upon it.

PROBL. II. To find how many Leagues do answer to one Degree of Longitude in every severall Latitude.

UPON the two edges of your Protracting Quadrant there are two Lines, the one divided into 20, the other into 60 equal parts.

Take therefore the least Distance from the Complement of the Parallel's distance from the Aequator, (or the Complement of the given Latitude:) this Distance, being measured upon the edge that is divided into 20, shall shew you what number of Leagues make one Degree of Longitude in that Parallel of Latitude. And the same Distance, being measured upon the other edge that is divided into 60, will give so many of our Miles, or so many Minutes of the Aequinoctial, or any other

great Circle, as are answerable to one Degree of Longitude in that Latitude.

Example. Let it be required to find how many Leagues do answer to one Degree of Longitude in the Latitude of 18 d. 12 min.

Set one foot of your Compasses in 71 degr. 48 min. the Complement of the given Latitude, and with the other take the nearest Distance to the side of the Quadrant which is di∣vided into 20: that Distance, measured upon the Line 20, will reach from the beginning thereof to 19: and so many Leagues do answer to one Degree of Longitude in the Lati∣tude of 18 degr. 12 min.

Or, If you take the least Distance from 18 degr. 12 m. the Latitude it self, in the Limb of the Quadrant, to that edge which is divided into 60, that Distance will also reach to 19 upon the Line 20, as before.

And the same Distance, being measured upon the Line 60 of the Quadrant, will give you 57 parts: and so many Mi∣nutes of the Aequator are answerable to one Degree of Lon∣gitude in the Parallel of 18 degr. 12 min. of Latitude.

So likewise in the Latitude of 25 degr. 15 min. if you take the least Distance from the Complement thereof, or from the Latitude it self, to the edges of the Quadrant, you shall find that Distance to reach 18 in the Line of 20: and so many Leagues do answer to one Degree of Longitude in the Lati∣tude of 25 degr. 15 min. or unto 54 in the Line of 60: and so many Minutes of the Aequator do answer to one Degree of Longitude in that Parallel of Latitude.

The Analogie or Proportion.

As the Radius is to the Co-sine of the Latitude;

So is

• 20 Leagues to the num∣ber of Leagues
• 60 Minutes to the num∣ber of Minutes

which answer to one Degree of Longitude in that Lati∣tude.

PROBL. III. By the Latitude of two Places and their Distance, to find the Rhumb.

The Analogie or Proportion.

As the Distance upon the Rhumb is to the Radius;

So is the Difference of Latitudes to the Co-sine of the Rhumb from the Meridian.

Thus if the Places given were one in the Latitude of 50 d. and the other in the Latitude of 55 degr. and the Distance up∣on the Rhumb 6 degr. or 120 Leagues; the Rhumb leading from one to the other will be found to be the third from the Meridian, namely, N. E. by N. 33 degr. 45 min.

Ʋpon the Chart.

LET A represent the Place in the Latitude of 50 degr. and C that in 55 degr. whose Distance from A to C is 6 degr. Take 6 degr. out of the Meridian-line, by setting one foot as much below the lesser Latitude as above the grea∣ter, which will be from K in the Latitude of 49 ½ degr. to L in the Latitude of 55 ½; either of which are half a Degree above and under the two given Latitudes. Take this Distance K L in your Compasses, and setting one foot in A, (the lesser Latitude) with the other cross the Parallel of the greater La∣titude 55 degr. in the Point C, and draw a right Line from A to C. So shall the quantity of the Angle B A C, being found (either by your Chord or Quadrant,) shew you the Inclina∣tion of the Rhumb to the Meridian to be 33 degr. 45 min. the N. E. by N. Point.

Note, That in the Propositions following, the Difference of Lon∣gitude must always be taken out of the Aequator, and measured

thereupon also. But the Difference of Longitude and Distance upon the Rhumb must alwaies be measured upon, and taken out of, the Meridian Line of your Chart. And hereafter I shall call them the proper Difference, and proper Distance.

PROBL. IV. The Longitude and Latitude of two Places being gi∣ven, to find the Rhumb.

The Analogie or Proportion.

As the proper Difference of Latitude is to the Radius;

So is the Difference of Longitude to the Tangent of the Rhumb from the Meridian.

Thus if the Places should lie one in the Latitude of 50 deg. and the other in the Latitude of 55 degr. and the Difference of Longitude between them were 5 degr. 30 min. the Rhumb leading from one Place to the other will be found to be the third from the Meridian N. E. by N. 33 degr. 45 min.

Ʋpon the Chart.

THE Meridians and Parallels being drawn through the two Places at A and C, and a straight Line from A to C, for the Rhumb, by your Chord or Quadrant find the quanti∣ty of the Angle B A C, which you will find to be 33 d. 45 m. or the third Rhumb from the Meridian N. E. by N.

But if this Rhumb were to be found by the Common Sea-Chart, it would be found to be above 47 degr. that is, N. E. 2 degr. Easterly, that is, one whole Point and 2 degr. more Easterly then it should be.

PROBL. V. The Latitude of two Places and the Rhumb being given, to find the Difference of Longitude.

The Analogie or Proportion.

As the Radius is to the Tangent of the Rhumb from the Meri∣dian;

So is the proper Difference of Latitudes to the Difference of Longitude.

Thus the Latitude of one Place being 50 degr. and the other 55 degr. and the Rhumb leading from one to the other being the third from the Meridian, the Difference of Longi∣tude will be found to be 5 ½ degr.

Ʋpon the Chart.

LET a Meridian be drawn through A, and a Parallel of Latitude through C. Then upon the Angle A protract the Angle of the Rhumb 33 degr. 45 min. So the Distance B C upon the Parallel, being measured upon the bottome of the Chart, will be found to contain 6 degr. 30 min.

But if this Difference of Longitude were to be found by the Plain Sea-Chart, the Difference of Longitude would be found to be but 3 degr. 20 min. which is more then 3 degr. less then the truth; a vast Difference. And yet this Errour would be yet greater, if either the Latitude be greater, or the Rhumb farther from the Me∣ridian.

PROBL. VI. The Difference of Longitude of two Places, the Latitude of one of them, and the Rhumb leading from one to the other, given, to find the Latitude of the other Place.

The Analogie or Proportion.

As the Radius is to the Co-tangent of the Rhumb from the Meridian;

So is the Difference of Longitude to the proper Difference of Latitude.

Thus if the Latitude of one of the Places were 50 degr. the Rhumb leading from that to the other N. E. by N. 33 d. 45 min. and the Difference of Longitude between the two Places were 5 degr. 30 min. the Latitude of the other Place will be found to be in 55 degr.

Ʋpon the Chart.

LET A B and D C be two Meridians drawn through A and C, at 5 ½ d. the Difference of Longitude, and a Par∣allel of Latitude through A, crossing the Meridian C D in D. Then upon the Point A protract an Angle equal to the Rhumb from the Meridian given 33 degr. 45 min. So the Line C D, being measured upon the Meridian from A, the given Latitude, 50 degr. will reach to 56 degr. the proper Difference of Latitude. So that the other Place lies in the Latitude of 56 degr.

But if this Difference of Latitude were to be found by the Plain Sea-Chart, this Difference of Latitude would be found to be 8 d. 13 min. and the Latitude sought would be found to be 58 degr.

13 min. above three Degrees more then the truth. As by the Tri∣angle for that purpose drawn upon the Plain Sea-Chart, marked with T V E, may appear.

PROBL. VII. Having the Latitude of one Place, the Rhumb leading from that Place to another unknown, and the Distance upon the Rhumb from the first to the second Place, to find the Difference of Longitude of the two Places.

The Analogie or Proportion.

As the Radius is to the Sine of the Rhumb from the Meridian;

So is the proper Distance upon the Rhumb to the Difference of Longitude.

Thus if the two Places were one in the Latitude of 50 degr. and the other in a greater Latitude, but unknown; the proper Distance upon the Rhumb leading from one place to the other being 6 degr. and the Rhumb N. E. by N. 33 degr. 45 min. the Difference of Longitude will be found to be 5 ½ degr.

Ʋpon the Chart.

THrough the Point A in the Latitude of 50 degr. let be drawn a Meridian A B, and a Parallell A D; and upon the Point A protract an Angle equal to the Rhumb from the Meridian 33 degr. 45 min. Then take with the Compasses 6 degres, the proper Distance upon the Rhumb, out of the Meridian-line, (having respect to the Latitude of the Places) as from K to L, and set that Distance upon the Rhumb from A to C. Then through C draw another Meridian C D, cros∣sing the Parallel drawn through A in the Point D. So the Line A D, being measured at the bottom of the Chart, will be found to contain 5 ½ d. the Difference of Longitude sought.

But if this Difference of Longitude had been to be found by the Common Sea-Chart, it would be found to have been onely 3 d. 20 min. which is 2 degr. 10 min. less then the truth; as in the Plain Chart may be seen, where the third Rhumb from the Me∣ridian cuts the Parallel of 55 degr. of Latitude in 3 degr. 20 m. of Longitude at the Point X.

PROBL. VIII. The Difference of Longitude between two Places, the Rhumb leading from one Place to the other, and the Latitude of one of the Places, being given, to find their Distance.

The Analogie or Proportion.

As the Sine of the Rhumb from the Meridian is to the Diffe∣rence of Longitudes;

So is the Radius to the proper Distance of the two Places up∣on the Rhumb.

Thus, if the Latitude of one Place were in 50 degr. the other in a greater Latitude unknown, the Difference of Lon∣gitude between the two Places 5 ½ degr. and the Rhumb N. E. by N. 33 degr. 45 min. from the Meridian; the proper Distance upon the Rhumb will be found to be 6 degrees.

Ʋpon the Chart.

LET two Meridians, A B and C D, be drawn through A and C, according to the Difference of Longitude, and a Parallel of Latitude through A, crossing the Meridian C D in the Point D. Then upon the Point A protract an Angle of 33 degr. 45 min. the quantity of the Rhumb from the Meri∣dian, and draw the Line A C crossing the Meridian C D in C. So the Distance C D, being taken in the Compasses, and

measured upon the Meridian-line of the Chart, (respect be∣ing had to the Latitude of the Places) that is, so much above the greater Latitude as below the lesser Latitude, you will find it to contain 6 degr.

But if this settting of the Compasses so much above one La∣titude as below another seem difficult, it may be thus other∣wise done.—For, the Rhumb Line being drawn, it will cut the Meridian C D in C: so a Parallel drawn through C will cut the Meridian A B in B: so is B the Latitude of the second Place, viz. 55 degr. Then divide the Distance be∣tween the two Latitudes A and B in two equal parts in the Point M; also divide the Rhumb-Line A C in two equal parts in N: then take the Distance N C or N A, and setting one foot of the Compasses in M, the other will reach to L above the greater Latitude, and from M to K as much below the lesser Latitude, namely, 30 min. or half a Degree on either side; so that between K and L are contained 6 degr. and that is the proper Distance upon the Rhumb.

But if this Distance were to be found by the Plain Chart, it would be almost 10 degr. or 197 Leagues, which is 77 Leagues more then in truth it should be. As may appear, if you measure the Line A L in the Plain Chart, upon the Side thereof.

PROBL. IX. The Difference of Longitude, and Distance of two Places, with the Latitude of one of the Places, being given, to find the Rhumb that leads from one to the other.

The Analogie or Proportion.

As the proper Distance upon the Rhumb is to the Difference of Longitude;

So is the Radius to the Sine of the Rhumb from the Meridian.

Thus, if one of the Places lay in the Latitude of 50 degr. and the other in a greater Latitude, but unknown; the Dif∣ference of Longitude between them 5 ½ degr. and their pro∣per Distance upon the Rhumb 6 degr. the Inclination of the Rhumb to the Meridian which leadeth from one Place to the other will be found to be 33 degr. 45 min. that is the N. E. by N. Point.

Ʋpon the Chart.

LET the Meridians A B and D C be drawn through A and C, and through A a Parallel of Latitude A D. Then open the Compasses (having respect to the Latitudes) from K to L, the quantity of 6 degr. in the Meridian; and setting one foot of that Extent in A, with the other foot cross the Meridian C D in C, and draw the right Line A C for the Rhumb. Lastly, by your Chord or Quadrant find the quantity of the Angle B A C, 33 degr. 45 min. and that is the Rhumb required N. E. by N.

But if you were to find this Rhumb by the Plain Sea-Chart, it would be found almost the E. N. E. Point within 1 degr. 30 min. differing from truth very near 3 whole Points to the Eastward.

PROBL. X. The Longitude and Latitude of two Places being given, to find the Distance upon the Rhumb.

The Analogie or Proportion.

As the proper Difference of Latitudes is to the Radius;

So is the Difference of Longitudes to the Tangent of the Rhumb from the Meridian:

And

As the Sine of the Rhumb from the Meridian is to the Diffe∣rence of Longitude;

So is the Radius to the proper Distance upon the Rhumb.

Thus, the two Places being one in the Latitude of 50 degr. the other in the Latitude of 55 degr. and the Difference of Longitude between them being 5 ½ degr. the proper Distance upon the Rhumb will be found to be 6 degr.

Ʋpon the Chart.

DRAW the Meridians A B and C D, the Difference of Longitude between them being 5 ½ degr. and through A and B draw two Parallels B C and A D, and then the Line for the Rhumb leading from the one to the other A C. So A C, being taken in the Compasses, and measured upon the Meridian-line of the Chart, with this Condition, that at the resting of the Compasses upon the Meridian-line, one foot be so many Degrees above the greater Latitude as the other foot is below the lesser Latitude; so will the feet of the Compasses rest in the Points K and L, one being 30 min. below the lesser Latitude, and the other 30 min. above the greater.

But if this Distance upon the Rhumb were to be found by the Plain Chart, it would be found to be almost 7 degr. 15 min. or 245 Leagues, which is 25 Leagues more then it should be.

PROBL. XI. The Latitude of two Places and their Distance upon the Rhumb being given, to find their Difference of Lon∣gitude.

The Analogie or Proportion.

As the proper Distance upon the Rhumb is to the Radius;

So is the proper Difference of Latitudes to the Co-sine of the Rhumb from the Meridian:

And

So is the Sine of the Rhumb from the Meridian to the Diffe∣rence of Longitude.

Thus, if one of the Places be in the Latitude of 50 degr. and the other in 55 degr. and their proper Distance upon the Rhumb 6 degr. or 120 Leagues; their Difference of Longi∣tude will be found to be 5 degr. 30 min.

Ʋpon the Chart.

DRaw A D and B C, two Parallels of Latitude, through 50 degr. and 55 degr. which were the two given Lati∣tudes. Then out of the Meridian Line take the proper Distance upon the Rhumb (having respect to both Lati∣tudes) from K to L: the Compasses being opened to this Distance, one foot being set in A, the lesser Latitude, the other will cross the Parallel of the greater Latitude in C. So the Distance B C, being measured at the bottome of the Chart from E, will reach to 5 degr. 30 min. And such is the Diffe∣rence of Longitude between the two Places.

But if this Difference of Longitude were to be found by the Plain Chart, it would be but 3 degr. 20 min. which is no less then 2 d. 10 min. less then the truth; as by the Triangle T V. E drawn upon the Plain Chart may appear.

PROBL. XII. The Difference of Longitude of two Places, their Distance upon the Rhumb, and the Latitude of one of the Pla∣ces, being given, to find the Difference of Latitudes.

The Analogie or Proportion.

As the proper Distance of the two Places upon the Rhumb is to the Radius;

So is the Difference of Longitudes to the Inclination of the Rhumb to the Meridian:

And

So is the Co-sine of the Rhumb from the Meridian to the Dif∣ference of Latitudes.

Thus, the Difference of Longitudes being 5 ½ degr. their pro∣per Distance upon the Rhumb 6 degr. and the Latitude of one of the Places 50 d. the Difference of Latitudes will be found to be 5 d.

Ʋpon the Chart.

THrough the given Latitude A draw a Meridian AB, and a Parallel A D, and upon the Parallel set the Difference of Longitude 5 ½ d. taken from the bottom of the Chart, from A to D, and through D draw the Meridian D C. Then out of the Meridian-line take the proper Distance upon the Rhumb, 6 d. from K to L, and setting one foot of the Compasses in A, with the other cross the Meridian C D in C: so a Parallel of Latitude drawn through C will be the Parallel of 55 d. So is 55 d. the Latitude of the other Place, and 50 being taken from 55, leaves 5 d. for the Difference of Latitudes required.

Which Difference, had it been to be found by the Plain Chart, would have been but 2 d. 25 m. that is, 2 d. 35 m. less then the truth; as by the Triangle T V E upon the Plain Chart may appear.

PROBL. XIII. The Latitude of two Places and their Difference of Lon∣gitudes being given, to find the Rhumb leading from one to the other, and also how many Degrees distant they are asunder.

THIS Proposition is already performed in the Example of the two Places A and B; but for Variety I will take two other Places, and onely shew the manner of working upon the Chart.

Suppose then two Places, one (as before) in the Latitude of 50 d. the other in the Latitude of 52 degr. 30 min. whose Dif∣ference of Longitudes is 6 degr.

Ʋpon the Chart.

Through the two given Latitudes 50 d. and 52 ½, at A and O draw two Parallels, O P and A D, upon which set the Dif∣ference of Longitudes from O to P, and from A to Q, 6 degr. Then draw the Line A P, which shall be the Line of the Rhumb leading from one Place to the other: wherefore, by your Chord or Protracting Quadrant find the quantity of the Angle O A P, which shall be the Inclination of the Rhumb to the Me∣ridian, and will be found to be 56 d. 15 m. that is the N. E. by E. Point; which was the First thing that was required.

Then to find the proper Distance upon the Rhumb; Take the Line A P in your Compasses, and measure it upon the Me∣ridian-line, so that one foot may be above the greater Latitude so much as the other is below the lesser; and you will find the Compass-points to rest in E and S, E being one whole Degree below the lesser Latitude, and S one Degree above the grea∣ter. So that there is intercepted between E and S 4 ½ degr. And that is the proper Distance upon the Rhumb; which was the Second thing required.

But if this Problem had been wrought upon the Plain Chart, the Rhumb from the Meridian would be found to be 67 d. 23 m. that is, within 7 m. of the 6^th Rhumb; which is more then the truth by 11 d. 8 m.

PROBL. XIV. A Ship set sail from the Latitude of 50 degr. upon the fifth Rhumb N. E. by E. after that she had made 36 Leagues of way upon that Rhumb, the wind changing, she was constrained to sail 50 Leagues upon the 7^th Rhumb E. by N. I would know in what Longitude and Latitude the Ship is.

Ʋpon the Chart.

THE Rhumb-Line A P being drawn, set off thereupon 36 Leagues (which was the way that the Ship made upon the fifth Rhumb before the Wind changed) from A to T, (which Distance must be taken out of the Meridian-line by opening the Compasses from 50 d. to 51, 48. or better, to as much below 50 d. as above 51 d.) So shall the Point T be the Place that the Ship was in when the Wind altered. So a Paral∣lel drawn through T upon the Chart will cut the Meridian at V in 51 d. and in that Latitude the Ship was. Now to find in what Longitude she was; Take in your Compasses the Line T V, and measure it at the bottom of the Chart, you shall find it will reach from E to 2 d. 21 m. And in that Longitude the Ship then was.

This done, upon the Point T (where the Wind changed, and drove the Ship 2 Points more Eastwardly, namely, upon the E. by N. Point) protract an Angle of 22 d. 30 m. namely, the An∣gle P T X, which is the Rhumb upon which the Ship sailed 50 Leagues after the Wind changed. Therefore take 50 Leagues out of the Meridian-line, and set them from T to X. So shall X be the Place that the Ship was in after she had sailed 50 Leagues upon the E. by N. Point; which, by drawing a Parallel through K, will be found in the Latitude of 51 d. 30 m. and by drawing of a Meridian through K also, it will be found to be in the Lon∣gitude of 6 degr. 16 min.

But if these Courses had been protracted according to the Plain Sea-Chart, the Point T would fall in the Latitude of 51 degr. and the Point X in the Latitude of 51 degr. 30 m. But the Longitude of T would be onely 1 d. 30 m. and the Longitude of X in 3 d. 57 min. Both these Longitudes being added, make but 5 d. 27 m. for the Difference of Longitude between X and the first Meridian; whereas by the other Chart it is 6 d. 16 m. So that the Ship at X is 33 m. Westward of the Place to which she was bound.

These Differences, which I have observed to be between the Plain and Mercator's Chart, may be seen by comparing the Scheme of the two Charts together.