NAVIGATION BY THE Mariners Plain Scale new plain'd: OR, A TREATISE OF Geometrical and Arithmetical Navigation; Wherein Sayling is performed in all the three kindes by a right Line, and a Circle divided into equal parts.

Containing

- 1. New Ways of keeping of a Reckoning, or Platting of a Traverſe, both upon the Plain and Mercators Chart, without drawing any Lines therein, with new ways for meaſuring of the Courſe and Diſtance in each Chart.
- 2. New Rules for Eſtimating the Ships way through Currents, and for Correcting the Dead Reckoning.
- 3. The refutation of divers Errors, and of the Plain Chart, and how to remove the Error committed thereby, with the Demonſtration of Mercators Chart from Proportion, and how to ſupply the Meridian-line of it Geometrically, albeit there is added to the Book a Print thereof from a Braſs-Plate to go alone, or with the Book; as alſo a Table thereof made to every other Centeſm.
- 4. A new eaſie Method of Calculation for Great Circle-ſayling, with new Projections, Schemes and Charts, giving the Latitudes of the Arch without Calculation.
- 5. Arithmetical Navigation, or Navigation performed by the Pen, if Tables were wanting, with excellent new eaſie ways for raiſing of a Table of Natural Sines, which ſupplies the want of all other Tables.

By John Collins of London, Pen-man, Accomptant, Philomathet.

London, Printed by Tho Johnſon for Francis Coſsinet, and are to be ſold at the Anchor and Mariner in Tower-ſtreet, as alſo by Henry Sutton Mathematical Inſtrument-maker in Thread needle ſtreet, behinde the Exchange. 1659.

[THE AZIMVTH COMPASS AND PLAINE SCALE.:

THe Reſtitution and Eſtabliſhment of your Honorable Society, promiſeth and preſageth no leſs then the future felicity of this Nation, the more or leſs diſemboguings of Nilus could not be more infallible Signals of Egypts Exuberance or Indigence, then the greater or leſſer enlargement of your Trade, is of the increaſe or decay of our Fame, Riches, and Strength.

You by penetrating thoſe Oriental Coaſts, will to our Clime in great meaſure reſtore the Golden Age; many poor being employed by your noble Adventures and Returns, will eſcape the miſeries of complaint and want.

Your Navy, as at preſent, it doth employ many able and experienced Sea-men, who poſſibly might otherwiſe languiſh under many diſcouragements, ſo we may aſſuredly promiſe to our ſelves the further encreaſe of their number in your Service.

That their Art and Knowledge might likewiſe be advanced and rendred more facil, as it ſhould be the deſire of all, ſo it hath been my particular aim and endeavor in the enſuing Treatiſe, which by that experience I have had amongſt them, I gueſs not to be unſuitable to their deſires and ſpare hours, and though intended chiefly for their benefit, yet may be of general uſe to every perſon, but eſpecially to thoſe that are Mathematically inclined. And now before whom can I more juſtly proſtrate my Endeavors in this kinde, then Your Honored Selves, whoſe renowned Employments will exerciſe the choiceſt Reſults of this nature, which by how much the more ſerviceable they are rendred unto you, by ſo much the more will they become the Objects of Applauſe and Deſire in others; Accept therefore this Offſpring of ſome ſpare Hours improved, more with an intent for the Advancement of the Publique good, then for any private benefit.

I ſhall conclude this my humble Addreſs with a Temporal and Spiritual Wiſh, viz. That the Encreaſe of your Treaſures may anſwer your Hazards and Deſires, and that your Vertues and Graces may exceed your Treaſures in this Life; and in that to come, may your Glories as far tranſcend BOTH, as heaven does earth: And for this you have not onely the earneſt Wiſhes, but the Cordial Prayers of

Your Honours moſt humble Servitor to be commanded, JOHN COLLINS.

HAving formerly ſpent ſome part of my time at Sea, I perceived the Genius and Inclination of moſt Mariners to be affected with drawing ſuch Paper Schemes and Delineations of the Sphere, as might not onely repreſent the ſame to their fancy, but likewiſe ſuit the reſolution of ſuch Spherique queſtions, as commonly they had occaſion of; and finding a complaint amongſt them of the want of a good Treatiſe of that kinde, I have often wiſhed that ſome of our able Artiſts would diſplay their endeavours to that purpoſe, wherein failing of my deſire, and conſidering that ſome mean performances of this nature have formerly found good Acceptance with Seamen, as being ſuitable to the meaneſt Capacity, and performed by a Scale of ſmall bulk and price, I thought fit to impart my Obſervations of that kinde, and certainly to the moſt learned it will be matter of uſe and delight, to ſee that whatſoever they can perform by exact Calculation, if their Tables be free from error, may be confirmed and performed without them, by the ſole aſſiſtance of a pair of Compaſſes and a bare Ruler, and what great eaſe enſues, both in Navigation and Dyalling, &c. by skill in carrying on of Proportions Geometrically, is ſcarcely as yet expected or apprehended by thoſe who are ſtrangers thereto.

If thou art juſt entring upon the Mathematicals, this Treatiſe may ſerve as an Introduction, in which after thou haſt read to page 20 of the firſt Part, thou mayeſt apply thy ſelf to the Spherical Definitions in the ſecond Part, and there proceed, or elſe where, as ſhall be moſt ſuitable to thy deſires. Thou wilt eaſily perceive that the right Angled Caſes from page 57 to 63, in the ſecond Book, and alſo from page 2 to 11 in the third Book, might have been contrived in one general Scheme for each Book, but not ſo perſpicuous to be underſtood: If this finde acceptance it will encourage me to be more laborious hereafter, and if thou reap benefit thereby, I have attained my deſire, and that thou mayeſt ſo do, is the hearty wiſh of thy Well-willer and Friend,

John Collins.

I fear'd, good Friend, your Works would ſole employ

O ur Braſs, not Wood, which time might putrifie,

H owe're I ſee your time not ſpent in vain,

N or Labor loſt the Scale in making plain.

C apacious, uſeful, long before neglected,

O bſcur'd and tedious, therefore diſreſpected.

L et thoſe that work in Braſs thy Worth admire,

L et them applaud thee, they have their deſire

I n Wood, if any will be doing too,

N or can they leſs in imitation do,

S o they'l have work, and Students they will gain

The uſe o'th rough-hewn Scale now its made plain:

And I for thee an Anagram have choſe,

S o good a Work as this NO HIL INCLO'S.

N oted are the Accounts thou didſt put forth,

I ngenious are thy Quadrants of like worth:

L et Dyalling Geometrical ne're fail,

L eſt Mariners forget to uſe the Scale;

O h let not Momus with his inked thumbs,

C ome near to ſlur thy Works, or try the Rumbs.

N ever deſiſt, but let's have more of thine,

H ere's but a Tangent, but let's have a Sine,

O r boſom full of thy induſtrious toyl,

I t will inform the weak, enrich our Soyl.

Your loving Friend, Sylvanus Morgan.

In the Proportional Part.

- GEometrical Definitions. Page 1, 2
- To raiſe Perpendiculars. 2, 3, 4
- To draw a line parallel to another Line. 5
- To bring three points into a Circle. 6
- To finde a right line equal to the Arch of a Circle. 9, 10
- Chords, Sines, Tangents, Secants, Verſed Sines, &c. defined. 11, 12, 13
- The Scale in the Frontiſpiece deſcribed. 14
- Plain Triangles both right and obliqued Angled reſolved by Protraction. from p 15 to 25
- Proportions in Sines reſolved by a Line of Chords. p 21 to 25
- Proportions in Tangents alone ſo reſolved. p. 25, 27, 28, 29, 30
- Proportions in Sines and Tangents reſolved by a Line of Chords. p. 26, 31, 32, 33

Particular Schemes fitted from Proportions to the Caſes of Oblique Angled Sphoerical Triangles.

- To finde the Azimuth. p. 34, 35
- As alſo the Amplitude. p. 36
- The Azimuth Compaſs in the Frontiſpiece deſcribed. p. 38
- The Variation found by the Azimuth Compaſs. p. 39
- To finde the Hour of the Day. p. 40
- As alſo the Azimuth and Angle of Poſition, p. 41
- To finde the Suns Altitudes on all Hours. p. 43, 46
- Alſo the Diſtances of places in the Arch of a great Circle. p. 44
- To finde the Suns Altitudes on all Azimuths p. 48
- The Latitude, Declination and Azimuth, given to finde the Hour. p. 50 to 54
- ☞ To finde the Amplitude with the manner of meaſuring a Sine to a leſſer Radius. p. 55
- To get the Suns Altitude by the ſhadow of a Thread or Gnomon. p. 56

The Contents of the Treatiſe of Navigation.

- OF the Imperfections and Uncertainties of Navigation. p. 1 to 5
- To meaſure a Courſe and Diſtance on the Plain Chart. p. 7, 8, 9
- Of the quantity of a degree, and of the form of the Log-board. p. 9, 10
- A Reckoning kept in Leagues, how reduced by the Pen to degrees and Centeſmes. p. 11, 12
- Of a Traverſe-Quadrant. p. 13
- A Traverſe platted on the Plain Chart, without drawing Lines thereon. p. 14 to 18
- A Scheme, with Directions to finde what Courſe and Way the Ship hath made through a Current. p. 18 to 21
- Divers Rules for Correcting of the Dead Reckoning from p. 21 to 33
- Of the errors of the Plain Chart. p. 33
- And how ſuch Charts may be amended. p. 34
- To finde the Rumbe between two places. p. 35
- Proportions having one tearm, the middle Latitude how far to be truſted to. p. 35 to 38
- To finde the Rumbe between two places by a Line of Chords onely. p. 39 to 42
- The Meridian-line of Mercators Chart ſupplied generally by a line of Chords. p. 42 to 47
- The Meridian-line divided from the Limbe of a Quadrant, with the uſe thereof in finding the Rumbe. p. 48 to 51
- The error committed by keeping of a Reckon•ng on the Plain Chart removed. p. 52 to 54
- Of the nature of the Rumbe on the Globe. p. 55 to 57
- Mereators Chart Demonſtrated from Proportion. p. 58 to 60
- Objections againſt it anſwered. p. 60 to 63
- To finde the Rumbe between two places in the Chart. p. 64
- Diſtances of places how meaſured on that Chart. p. 65 to 71
- Another Traverſe-Quadrant fitted for that Chart, with a Traverſe platted thereby, without drawing lines on the Chart. p. 71 to 78
- To meaſure a Courſe and Diſtance in that Chart, without the uſe of Compaſſes. p. 79
- Of Sailing by the Arch of a great Circle. p. 81
- To finde the Latitudes of the great Arch by the Stereographick Projection. p. 82 to 83
- Of a Tangent Projection from the Pole for finding the Latitudes of the great Arch p. 84 to 88
- With a new Method of Calculation raiſed from it p. 89, 90
- And how to meaſure the Diſtance in the Arch, and the Angles of Poſition. p. 91
- Another Tangent Projection from the Equinoctial for finding the Latitudes of the Arch. p. 93 to 100
- And how to finde the Vertical Angles, and Arkes Latitudes Geometrically. p. 100 to 102
- To draw a Curved-line in Mercators Chart reſembling the Arch, with an example for finding the Courſes and Diſtances in following the Arch. p. 102 to 104
- The Dead Reckoning caſt up by Arithmetick. p. 106 to 108
- A brief Table of Natural Sines, Tangents and Secants for each point of the Compaſs and the quarters. p. 107
- The difference of Longitude in a Dead Reckoning found by the Pen. p. 109
- That a Table of Natural Sines ſupplyes the want of all other Tables p. 110
- Many new eaſie Rules and Proportions to raiſe a Table of Natural Sines. p. 111 to 113
- And how by having ſome in ſtore to Calculate any other Sine in the Quadrant at command. p. 114
- Of the contrivance of Logarithmical Tables of Numbers, Sines and Tangents, and how the want of Natural Tables, and of a Table of the Meridian-line, are ſupplied from them. p. 117
- The Sides of a Plain Triangle being given, to Calculate the Angles without the help of Tables two ſeveral ways. p. 118, 119
- An Inſtance thereof in Calculating a Courſe and Diſtance. p. 119

1## CHAP. I. Containing Geometrical Definitions.

A Point, is an imaginary Prick void of all length, breadth or depth.

A Line, is a ſuppoſed Length without breadth or depth, the ends or limits whereof are Points.

An Angle, derived from the word Angulus in Latine, which ſignifieth a Corner, is the inclination or bowing of two lines one to another, and the one touching the other, and not being directly joyned together.

If the Lines which contain the Angle be right Lines, then is it called a Right lined Angle.

A right Angle, when a right Line ſtanding upon a right Line, maketh the Angles on either ſide equal, each of theſe Angles are called Right Angles, and the Line erected is called a Perpendicular Line unto the other.

An obtuſe Angle, is that which is greater then a right Angle.

An acute Angle, is that which is leſs then a right Angle, when tvvo Angles are both acute or obtuſe, they are of the ſame kinde, othervviſe are ſaid to be of different affection.

An Angle, is commonly denoted by the middlemoſt of the three Letters ſet to the ſides, including the ſaid Angle.

The quantity of an Angle is meaſured by the arch of a Circle, deſcribed upon the point of Concurrence or Interſection, where the two ſides incloſing the ſaid Angle, meet.

By the complement of an Arch or Angle, is meant the remainder of that Arch taken from 90^{d}, unleſs it be expreſſed the complement thereof to a Semicircle of 180^{d}.

A Circle, is a plain Figure contained under one Line, which is called the Circumference thereof, by ſome the Perimeter, Periphery, or Limbe, a portion or part thereof is called a Segment.

The Center thereof is a Point in the very midſt thereof, from which Point all right lines drawn to the Circumference are equal,2 if divers Circles are deſcribed upon one and the ſame Center, they are ſaid to be Concentrick, if upon divers Centers, they are in reſpect of each other ſaid to be Excentrick.

The Diameter of a Circle is a right Line drawn through the Center of any Circle, in ſuch ſort that it may divide the whole Circle into two equal parts.

The Semidiameter of a Circle, commonly called the Radius thereof, is juſt one half of the Diameter, and is contained betwixt the Center and one ſide of the Circle.

A Semicircle is the one medeity or half of a whole Circle, deſcribed upon the Diameter thereof.

And a Quadrant is juſt the one fourth of a whole Circle.

All Circles are ſuppoſed to be divided into 360 equal parts, called Degrees, conſequently a Semicircle contains 180^{d}, and a Quadrant 90^{d}, and ſo much is the quantity of a right Angle.

A Minute is the ſixtieth part of a degree, being underſtood of meaſure, but in time a Minute is the ſixtieth part of an hour, or the fourth part of a degree, 15 degrees anſwering to an hour, and 4 Minutes to a degree. A Minute is marked thus 1′, a ſecond is the ſixtieth part of a Minute, marked thus 1″.

Where two lines or arches croſs each other, the point of meeting is called their Interſection, or their common Interſection.

[illustration]

Let it be required to raiſe a Perpendicular upon the end of the Line C D, upon C as a Center, deſcribe the arch of a Circle as D G, prick the Extent of the Compaſſes from D to G, and upon G as a Center, with the ſaid Extent, deſcribe the ark D E G, in which prick down the Extent of the Compaſſes twice, firſt3 from D to E, and then from E to F, then joyn the points F D with a right line, and it ſhall be the Perpendicular required.

Upon G with the Extent of the Compaſſes unvaried, deſcribe a ſmall portion of an Ark near F, then a Ruler laid over the Points C and G, will cut the ſaid Ark at the Point F, from whence a line drawn to D, ſhall be the Perpendicular required.

In this latter caſe the Extent C F becomes the Secant of 60^{d} to the Radius C D, which Secant is always equal to the double of the Radius.

As to the ground of the former Way, if the three Points C G F were joyned in a right line, it would be the Diameter of a Semicircle, an Angle in the circumference whereof made by lines, drawn from the extreamities of the Diameter, as doth the Angle C D F, is a right Angle by 31 Prop. 3 Book of Euclid.

Otherwiſe:

[illustration]

Set one foot of the Compaſſes in the Point A, and opening the other to any competent diſtance, let it fall in any convenient point at pleaſure, as at D, then retaining that foot in D, without altering the Compaſſes, make a mark in the line A C, as at E: Now if you lay a Ruler from D to E, and by the edge of it from D, ſet off the Extent of the Compaſſes, it will find the point B, from whence draw the line A B, and it ſhall be the Perpendicular required: Thus upon a Dyal we may raiſe a Perpendicular from any point or part of a Line, without drawing any razes to deface the Plain.

Otherwiſe upon the Point D, having ſwept the Point E, with4 the ſame Extent draw the touch of an Arch on the other ſide at B, then laying the Ruler over D and E as before, it will interſect the former Arch at the Point B, from whence a line drawn to A, ſhall be the Perpendicular required.

The converſe of this Propoſition would be from a Point aſſigned, to let fall a Perpendicular on a line underneath.

To be done by drawing a line from the Point B, to any part of the given line A C, as imagine a ſtreight line to paſs through B D E, and to finde the middle of the ſaid line at D, where ſetting one foot of the Compaſſes with the Extent D E, draw the touch of an Arch on the other ſide the Point E, and it will meet with the ſaid line A C at A, from whence a line drawn to B, ſhall be the Perpendicular required.

[illustration]

In the Scheme annexed, let A B be the line given, and let it be required to raiſe a Perpendicular in the Point C.

Firſt ſet one foot of the Compaſſes in the point C, and open the other to any diſtance at pleaſure, and mark the given line therewith on both ſides from C, at the point A and B, then ſetting one foot of the Compaſſes in the point A, open the other to any competent diſtance beyond C, and draw a little Arch above the line at D, then with the ſame diſtance ſet one foot in B, and with the other croſs the Arch D with the arch E, then from the Point of Interſection or croſſing, draw a ſtreight line to C, and it ſhall be the Perpendicular required.

The converſe will be, to let a Perpendicular fall from a Point upon a given Line: Let the Point given be the Interſection of the two Arks D, E, ſetting down one foot of the Compaſſes there, with any Extent draw two Arks upon the line underneath, as at B5 and A, divide the diſtance between them into halfs, as at C, and from the given Point to the Point C draw a line, and it ſhall be the Perpendicular required; and if this be thought troubleſom, upon the Points A and B with any ſufficient Extent, you may make an Interſection underneath, and lay a Ruler to that and the upper Interſection, and thereby finde the Point C.

[illustration]

In this Figure, let the Line A B be given, and let it be required to draw the Line C D parallel thereto, according to the diſtance of A C.

Firſt open the Compaſſes to the diſtance required, then ſetting one foot in A, or further in, with the other draw the touch of an Arch at C, then retaining the ſame extent of the Compaſſes, ſet down the Compaſſes at B, and with the former extent draw the touch of an Arch at D, then laying a Ruler to the outwardmoſt edges of theſe two Arks, draw the right line C D, which will be the Parallel required.

This Propoſition will be of frequent uſe in Dyalling, now the drawing of ſuch pieces of Arks as may deface the Plain, may be ſhunned; for having opened the Compaſſes to the aſſigned Parallel diſtance, ſet down one foot oppoſite to one end of the line propoſed A B, ſo as the other but juſt turned about, may touch the ſaid Line, and it will finde one Point: Again, finde another Point in like manner oppoſite to the other end of the Line at B, and through theſe two Points draw a right Line, and it ſhall be the Parallel required.

This way, though it be not ſo Geometrical as the former, yet in other reſpects may be much more convenient, and certain enough.

6### Prop. 4. To draw the Arch of a Circle through any three Points, not lying in a ſtreight Line.

[illustration]

In the Figure adjoyning let A B C be the three Points given, and let it be required to draw a Circle that may paſs through them all.

Set one foot in the middle Point at B, and open the Compaſſes to above one half of the diſtance of the furtheſt point therefrom, or to any other competent extent, and therewith draw the obſcure Arch D E F H with the ſame extent, ſetting one foot in the point C, draw the Arch F H: Again, with the ſame extent, ſetting one foot in the point A, draw the Arch D E, then laying a Ruler to the Interſections of theſe Arches, draw the lines D G and G H, which will croſs each other at the point G, and there is the Center of the Circle ſought; where ſetting one foot of the Compaſſes, and extending the other to any of the three points, deſcribe the Arch of a Circle, which ſhall paſs through the three points required.

[illustration]

And if they be multiplied the other way, from D to H, and from C to E, then the lines E H and F G being produced, find the ſame point I, without continuing either of the lines firſt given, and with much more certainty.

An Oblong, a Rectangle, a right angled Parralellogram, or a Long Square, are all words of one and the ſame ſignification, and ſignifie a flat Figure, having onely length and breadth, the four Angles whereof are right Angles, the oppoſite ſides whereof are equal.

In Proportions the product of two tearms or numbers, are called their Rectangle or Oblong, becauſe if the ſides of a flat be divided into as many parts as there are units in each multiplyer, lines ruled over thoſe parts, will make as many ſmall ſquares as there are units in the Product, and the whole Figure it ſelf will have the ſhape of a long Square.

[illustration]

8A Rhombus (or Diamond) is a Figure with four equal ſides, but no right Angle.

But a Rhomboides (or Diamond-like Figure) is ſuch a Figure whoſe oppoſite Sides and oppoſite Angles onely are equal, either of theſe Figures are commonly called Oblique Angled Parralellograms: Thus either of the Figures A B, F E, or B C, E D are Rhomboides, or Oblique Angled Parralellograms.

This foregoing Figure is much uſed in Dyalling, thereby to ſet off the Hour-lines: Admit the Sides A B and B E were given, and it were required on both ſides of B E to make two oblique Parralellograms, whoſe oppoſite ſides ſhould be equal to the lines given, this may be done either by drawing a line through the point E, parallel to A B C, and then make F E, E D and B C, each equal to A B, and through thoſe points draw the ſides of the Parralellogram, or continue A B, and make B C equal thereto, and with the extent B E upon A and C, draw the croſs of an Ark at F and D: Again, upon E, with the extent A B, draw o•her Arks croſſing the former at F, and thoſe croſſes or interſections limit the extreamities of the ſides of the Parralellogram.

A line drawn within a four-ſided Figure from one corner to another, is called a Diagonal-line.

A Parralellipipedon, is a ſolid Figure, contained under ſix fourſided figures, whereof thoſe which are oppoſite are parallel, and is well repreſented by two or many Dice ſet one upon another, or by the Caſe of a Clock-weight.

Let the given Circle be B D C, divide the upper Semicircle B D C into halfs at D, and the lower Semicircle into three equal parts, and draw the lines D E, D F, which cut the Diameter at G and H, and make G I equal to G H, then is the length D I a little more then the length of the quadrant B D, neither doth the exceſs amount unto one part of the Diameter B C, if it were divided into five thouſand, and four times the extent D I will be a little more then the whole circumference of the Circle.

9
[illustration]

Let C D be an arch of a given Circle leſs then a Quadrant, whereto it is required to finde a right line equal.

Divide the Arch C D into halfs at E, and make the right line F G equal to the Chord C D, and make F H equal to twice C E, and place one third part of the diſtance between G and H, from H to I, and the whole line F I will be nigh equal in length to the Arch C D, but ſo near the truth, that if the line F I were divided into 1200 equal parts, one of thoſe parts added thereto would make it too great, albeit the Ark C D were equal to a Quadrant, but in leſſer Arches the difference will be leſs, and if the given Arch be leſs then 60 degrees, or one ſixth part of the whole Circle, the line found will not want one ſix thouſandth part of its true length; But when the given Arch is greater then a quadrant, it may be found at twice, thrice, or four times by former Directions: Theſe two Propoſitions are taken out of Hugenius de magnitudine Circuli, Page 20, 21.

10
[illustration]

In Dyalling, to ſhun drawing of Lines on a Plain, it may be of frequent uſe to prick off an Angle by Sines or Tangents in ſtead of Chords, it will therefore be neceſſary to define theſe kinde of Lines.

[illustration]

2. The right Sine of an Arch is half the Chord of twice that Arch: thus G F being the half of G L, is the Sine of the Arch, G A half of the Arch G A L, whence it follows that the right Sine of an Arch leſs then a quadrant, is alſo the right Sine of that Arks reſidue from a Semicircle, becauſe, as was ſhewed above, the Chord of an Ark is the ſame both to an Ark leſſer then a quadrant, and to its complement to a Semicircle.

What an Arch wants of a quadrant, is called the Complement thereof: thus the Arch D G is the Complement of the Arch A G, and H G is the Sine of the Arch D G, or which is all one, it is the Coſine of the Arch A G, and the Line H G being equal to C F, it follows that the right Sine of the Complement of an Arch is equal to that part of the Diameter, which lieth between that Arch and the Center.

From the former Scheme alſo follows another Definition of a right Sine, as namely, that it is a right Line falling from one end of any Arch perpendicularly upon the Radius, drawn to the other end of the ſaid Arch, ſo is G F perpendicular to C A, being the Sine of the Arch G A; likewiſe A I falls perpendicularly on C G, therefore by the ſame definition is alſo the Sine of the ſaid Arch.

3. The Verſed Sine of an Arch is that part of the Diameter which lieth between the right Sine of that Arch and the Circumference: thus F A is the Verſed Sine of the Arch G A, and F B the Verſed Sine of the Arch B D G.

4. If unto one end of an Arch there be drawn a Radius, and to the other end a right line from the Center, cutting the Circle, and if from the end of the Radius a Perpendicular be raiſed till it meet with the Line cutting the Circle, that Perpendicular is the Tangent of that Arch: thus A E is the Tangent of the Arch G A, and D M is the Cotangent of the ſaid Ark, namely, it is the Tangent of the Arch H G, which is the Complement of the former Arch.

125. The aforeſaid right Line cutting the Circle, is called the Secant of the ſaid Arch: thus C E is the Secant of the Arch G A, and C M is the Coſecant of the ſaid Arch, for it is the Secant of the Arch D G.

6. Aſſigning the Radius C A to be an unit with Ciphers at pleaſure, to define or expreſs the quantities of theſe reſpective lines, in relation to the Arches to which they belong, were to make a Table of natural Sines, Tangents and Secants; of which at large ſee Mr. Newtons Trigonometria Brittanica, for abbreviate ways, and ſomething we ſhall adde about it in the Arithmetical part of Navigation.

7. The Tables being made, their chief uſe was to work the Rule of Three, or Golden Rule Arithmetically, by multiplying the ſecond and third tearms of any Proportion, and dividing by the firſt, and thereby to reſolve all Propoſitions relating either to Plain or Spherical Triangles, which in lines is performed by drawing a line parallel to the Side of a Triangle, and where four tearms either in Sines, Tangents, Secants, Verſed Sines, are expreſſed, as the firſt to the ſecond, ſo is the third to the fourth, it implyes a Proportion, and that the ſecond and third tearm are to be multiplyed together, and the Product divided by the firſt.

The proportion or reaſon of two numbers, or reference of one to the other, is meaſured by the Quotient, the one being divided by the other.

A Proportion is then ſaid to be direct, when the third tearm bears ſuch proportion to the fourth, as the firſt to the ſecond: four numbers are ſaid to be proportional when as often as the firſt and ſecond are the one contained in the other, ſo often are the third and fourth the one contained in the other.

Reciprocal Proportion is when the fourth tearm bears ſuch Proportion to the third, as the firſt doth to the ſecond.

A Proportion is ſaid to hold alternately, when the ſecond and third tearms thereof change places, and inverſly, when the order of the tearms are ſo altered, that one of the three firſt tearms ſhall become the laſt.

Divers Proportions are expreſſed in this following Book, which if the Reader would apply to Tables, he muſt underſtand that when a Side or an Angle is greater then a Quadrant, that in ſtead of the13 Sine, Tangent, or Secant of ſuch an Ark, he muſt take the Sine, Tangent, or Secant of that Arks complement to a Semicircle.

That the words Coſine or Cotangent of an Arch given or ſought, ſignifie the Sine or Tangent of the Complement of the Arch given or ſought.

That the Coſine or Cotangent of an Arch greater then a Quadrant, is the Sine or Tangent of the exceſs of that Arch above a Quadrant, or 90 degrees.

8. What trouble the Ancients were at in reſolving of Proportions by Multiplication and Diviſion, is wonderfully abbreviated by an admirable Invention called Logarithmes, where by framing and ſubſtituting other numbers in ſtead of the former, Multiplication is changed into Addition, and Diviſion into Subſtraction; of which alſo ſee the former Book Trigonometria Brittanica.

9. What may be performed by either of the former kindes of Tables, may alſo with a Line of Chords and equal parts be performed, but not ſo near the truth, without them, and that either by projecting or repreſenting the Sphere on a Flat or Plain, as we have handled in the ſecond and third Part, or by Protracting and Delineating of ſuch Proportions as may be wrought by the Tables: and this in ſome meaſure is the intended ſubject of our ſubſequent Diſcourſe.

10. Therefore before we proceed any further, it will be neceſſary to deſcribe ſuch Lines as are upon the Scale, intended to be treated of.

1. The firſt or uppermoſt line is a line of a Chords, numbred to 180 degrees, and is called the Leſſer Chord, being a double Scale, the undermoſt ſide whereof being numbred with half thoſe Arks, is a line of Sines and is called the greater Sines; at the end of this Scale ſtands another little Scale, which is called the leſſer Sines, being numbred with 90 degrees, and both theſe Scales ſeem to be one continued Scale.

2. The ſecond Scale is another line of Chords, called the greater Chord, being fitted to the ſame Radius with the greater Sine, and numbred to 90 degrees.

Annexed thereto is a ſingle Line called the line of Rumbes or14 Points of the Compaſs, numbred from 1 to 8, in which each Rumbe is divided into quarters, having pricks or full points ſet thereto.

3. The third Scale is a line of equal parts or leagues, divided into ten greater diviſions, and each of thoſe parts into ten ſmaller diviſions, and each of thoſe ſmaller diviſions into halfs.

Annexed thereto is another Scale of ſix equal parts, each of which parts is ſub-divided as the former.

How to make a Line of Chords or Sines from the equal diviſions of the Circle, is ſpoke to in the ſecond Book, page 2. The Scale thus deſcribed is indeed a double Scale, for it hath two Lines of Chords, two Lines of Sines, and two Lines of equal Parts upon it, and this rather for conveniency then neceſſity, whereas indeed one of each kinde had been ſufficient, yea, the Sines might have been wholly ſpared; for throughout theſe Treatiſes nothing is more required of neceſſity, then a Circle divided into equal parts, as in the third Book, page 4. and a right Line divided into equal parts, which we ſuppoſe in every mans power to do if he have Compaſſes. The Schemes throughout theſe Books are fitted either to the Radius of the greater or leſſer Chord before deſcribed, that is to ſay, they are drawn with 60 degrees of the one or the other of thoſe Lines of Chords.

THis Chapter, from the very Title of it, will ſeem to many to be unneceſſary, if not impertinent, as being in it ſelf ſo eaſie that any perſon that knows any thing can perform, however I thought fit to adde it for methods ſake, that it might be ſaid that all the uſual Caſes of Triangles are here performed with Scale and Compaſſes, which poſſibly ſome that are meer Beginners may be ignorant of.

If a Figure be made of three right Lines, ſo joyned that oppoſite to each there be an angular point or corner, it is called a right lined Triangle.

If it have three equal ſides, it is called an Equilateral Triangle.

If it have but two equal ſides, it is called an Iſoceles, or Equi•rural Triangle.

15If all the ſides be unequal, it is called A Scalenon Triangle, or Oblique plain Triangle.

If it have one right angle, it is called A right angled Triangle, that ſide which ſubtends the right Angle, is called the Hipotenuſal, the other ſides are indifferently called Sides, or Legs, or one of them the Baſe, and the other the Perpendicular, thoſe parts of the Six of a Triangle which are given or known, are tearmed the Data, and thoſe unknown or ſought, the Queſita.

[illustration]

Admit a Ship ſayl South-eaſt and by South ninety Leagues, In this caſe the Courſe is the Angle given, and the diſtance is the Hipotenuſal.

Having drawn E N and N S at right Angles one to another, and deſcribed the Quadrant E A with 60^{d} of the Chords, therein ſet off A B equal to the Courſe from the ſouth, being 3 points16 or 33^{d} 45′ of the greater Chord, and this is called ſetting off of an Angle, and draw N B, wherein ſet off the diſtance 90 Leagues to D, the neareſt diſtance from D to N A meaſured on the equal parts, ſhews the ſide ſought, if B N A be the Angle given, which in this example is D C 50 Leagues, and ſo much is the Ships departure from the Meridian or Separation.

But if B N E were the Angle given, the neareſt diſtance from D to E N meaſured on the equal parts, is the ſide ſought, in this Example D F 74 Leagues, and eight tenths more of another League, and ſo much is the difference of Latitude, called the Variation: Now to perform this without letting fall or drawing the prickt Perpendicular Lines D C, D F, which are added onely for illuſtration, will be of excellent uſe, as ſhall follow hereafter.

Admit the difference of Latitude were the given Side, and the Diſtance the Hipotenuſal, and it were required to finde the departure from the Meridian.

Prick out of the equal parts the difference of Latitude from N to C 74, 8 Leagues, and from the point C raiſe the Perpendicular C D, then take the diſtance out of the equal parts ninety Leagues, and ſetting one foot in N, the other will ſomewhere meet with C D, as at D, and the line D C meaſured on the equal parts, is the Departure from the Meridian, to wit, 50 Leagues in this example, being the Side ſought.

Admit the Diſtance 90 Leagues were the given Hipotenuſal, and the Difference of Latitude N C n were the other Side, and that it were required to finde the Courſe, which is the Angle ſought: Having proceeded ſo far as is directed in the ſecond Caſe, draw N D, and upon N as a Center, deſcribe the Arch B A, which meaſured on the Chords or Rumbes, ſheweth the Courſe from that Coaſt of the world that N C repreſents, to be 3 points or 33 degrees 45 minutes.

17#### 4. To finde a Side. Given a Side and one acute Angle.

Admit the Side given were the Difference of Latitude N C, and the Angle were the Courſe B A from the Meridian, and that it were required to finde the Departure from the Meridian D C.

In this caſe ſet off the Difference of the Latitude from the equal parts from N to C 74, 8 Leagues, and raiſe the Perpendicular D C. Upon N deſcribe the Arch E B A, and therein out of the Rumbes, having ſet off the Courſe B A, draw the Line N B produced, and it meets with D C produced at D, and the Side D C meaſured on the equal parts is 50 Leagues, the Departure from the Meridian, as before.

In the former Triangle let D C repreſent a Tower of unknown height, perpendicular to the Horizon, and let N C be a diſtance meaſured off from it, 74 yards and eight tenths more, the line N C making right Angles with D C then ſtanding at N. If by a Quadrant the Altitude of D C were found to be 33^{d} 45′, upon N as a Center, deſcribe an Arch with 60^{d} of the Chords, and therein ſet off 33^{d} 45′ from A to B, and draw the line N B produced, till it concur with D C at D, then the line N D meaſured on the equal parts, ſhews the Diſtance between the eye and the top of the Tower, which is the Hipotenuſal ſought, and in this Example is 90 yards, and the line D C there meaſured, ſhews the height of the Tower 50 yards.

Admit in the former Triangle that D C being the Departure from the Meridian, namely 50 Leagues, and C N the difference of Latitude, to wit, 74, 8 Leagues were given, and it were required to finde the Diſtance N D: Having drawn D C and N C at right Angles one to another, and therein prickt down from the equal parts the Variation and Separation, the diſtance between the Points N D meaſured on the equal parts, ſheweth the Hipotenuſal ſought, in this Example 90 Leagues.

18#### 7. To finde an Angle. Given both the Sides.

If it were required to finde the Courſe by what in the former Caſe was given: Having proceeded ſo far as is there expreſt, draw N D, and upon N as a Center with 60^{d} of the Chords, draw the Arch B A, which Arch meaſured on the Chords, ſheweth the Courſe from that coaſt of the world that N A repreſents, in this Example 3 points, or 33^{d} 45′ from the Meridian, and the Angle N D C is the complement thereof; and the Ark might as well have been deſcribed on D as a Center, if the Sides D N and D C had been produced far enough.

[illustration]

Let the quantities of the two given Sides be 46, 6 and 30.

Prick off that next the Angle given from A to B, and upon A as a Center, deſcribe the Ark D F, and therein prick down 30 degrees 58 minutes for the Angle given, and draw the line A F C, then from the end of the Side prickt off, prick the other Side from B to C, or E, and ſo the Angle B C A or B E A, is the Angle ſought, but which of the two cannot poſſibly be determined, unleſs the affection be alſo given, to wit, whether it be obtuſe or19 acute, though ſome of our Writers affirm it may be determined by drawing the Triangle as true as you can.

Then upon the angular Point C, deſcribe an Arch with 60^{d} of the Chords, and meaſure the ſaid Arch in the Chords, continuing the Sides if need be, and it ſhews the quantity of the ſaid Angle to be 53^{d} 6′, and the Complement thereof to 180^{d}, being 126^{d} 54′ is the meaſure of the Angle B E A, becauſe the Angle B E C is equal to the Angle B C E.

How to frame ſuch Triangles whoſe Sides ſhall be all whole numbers, is ſhewed in our Engliſh Ramus, page 155, 156.

In this caſe alſo, unleſs the quality of the Angle oppoſite to the greateſt Side be determined, the third Side will be doubtful, to wit, it may be either A E or A C, which extents meaſured on the equal parts, ſhew the Side accordingly, and the third Angle to be meaſured, as before.

This Mr. Norwood doth not make a Caſe, becauſe an Angle muſt be firſt found and determined before the third Side can be found, and then it will be reſolved by the following Caſe.

In this Caſe the third Angle is alſo given, as being the complement of the ſum of the two former Angles to a Semicircle. As if there were given the Side A B in the former Triangle, the Angle B A C, and the Angle A C B.

Firſt prick off the Side A B, then ſubſtract the ſum of the Angles B A C, and A C B, from 180^{d} or a Semicircle, and there remains the Angle A B C, then upon A as a Center, ſet off upon the Arch D F the meaſure of the Angle A.

Alſo upon B as a Center, the meaſure of that Angle muſt be ſet off, and lines drawn through the extreamities of thoſe Arks will meet, as at C, the point that limits the two Sides A C, and B C, which are to be meaſured on the equal parts: By this Caſe if A C were the Wall of a Town, and B a Fort ſhooting into the ſaid Town, the diſtance of the ſaid Fort might be found from any part of the Town wall, without going out to meaſure it; for firſt, with any whole circle or a compaſs, obſerve the Arch between A B20 and A C, and meaſure the diſtance C A, again at A obſerve the Arch between A B and A C, and protract as in this Example, and you may meaſure the diſtance between B and any point in the line A C; And ſo if B were a Tower or Mark on the land, and A C repreſent the Ships diſtance in her courſe, by obſerving how B bears both at A and C, the Ships diſtance therefrom might be meaſured.

Thus if there were given the two Sides A B and A C, with the Angle A between them, the ſaid Angle muſt firſt be ſet off in the Arch D F, then a line joyning the extreamities of the two Sides, as doth B C, is the third Side, which being firſt found, upon the angular Points C and B, with 60^{d} of the Chords Arks muſt be drawn, which being limited by the two Sides (produced when it is neceſſary) are the meaſures of the Angles ſought.

If the three Sides be joyned in a Triangle which is eaſily done, firſt pricking down any one of the Sides, and from its extreamities with the other Sides deſcribe two Arks which will interſect at the Point where the other two Sides concur, then will the three angular Points be given, upon which Arks being deſcribed between the given Sides, are the meaſures of the Angles ſought.

7. Three Angles being given, are not ſufficient to finde any one of the Sides.

LEt the Proportion be:

As the Sine of 19^{d} to the Sine of 25^{d}, So is the Sine of 31^{d}.

To what Sine? to wit, the Sine of 42^{d}.

[illustration]

Otherwiſe:

Place the extent wherewith the Ark F was deſcribed from V to E, and draw the line E G juſt touching the extreamity of the ſaid Ark, then with the extent V A, one foot of the Compaſſes reſting in V, croſs the aforeſaid line at G, and a ruler over V and G will finde the Point D in the Limbe, as before.

Here obſerve, that every Proportion without the Radius, may be made into two ſingle Proportions, with the Radius in each, thus:

- As the Radius, Is to one of the middle tearms,
- So is the other middle tearm, To a fourth Proportional.

Again:

- As the firſt tearm, Is to the Radius,
- So is the fourth Proportional before found, To the true Proportional ſought.

From which conſideration the former Scheme was contrived.

22#### Two other general ways for working Proportions in Sines.

Let the Proportion be:

As the Sine of 70^{d} to the Sine of 50^{d}, So is the Sine of 35^{d}. To what Sine? Anſwer: 27^{d}, 50′.

[illustration]

Having drawn the Quadrant D E with 60^{d} of the Chords, and its two Radii D C, C E at right Angles in the Center, prick down from the Chords one of the middle tearms, to wit, 35^{d} from D to H, and draw a line into the Center, and upon the ſaid line from the Center to A, prick down out of the Sines the other middle tearm, to wit, 50^{d}, and through the Point A draw the line A B parallel to D C, then count the firſt tearm from D to G 70^{d}, and draw a line from the Center which paſſeth through A B at F, and the extent C F meaſured on the Sines, is 27^{d} 50′ the fourth Proportional, and thus the firſt tearm may be varied as much as you pleaſe.

Otherwiſe:

Place the Sine of the firſt tearm, to wit, of 70^{d}, which in this Example is the neareſt diſtance from G to D C ſo from the Center, that it may croſs A B produced when need requires: In this Example it croſſeth it at I, a ruler over the Center and the point I,23 cuts the Limbe at K, and the Arch D K being 27^{d} 50′, is the meaſure of the fourth Proportional, as before.

When it cannot be there placed, to wit, as when the Sine of the firſt tearm is ſhorter then C B, the fourth Proportional is more then the Radius, and the Arch of the firſt tearm being counted from D towards E, a line from the Center meeting with A B produced, ſhall be the Secant of the fourth Proportional to the common Radius of the Quadrant.

The Demonſtration of this Scheme is inferred from varying the Proportion, which may ſtand thus, by changing the two firſt tearms into their complements, and altering their order they become Secants.

As the Secant of 55^{d} to Secant 20^{d}, So is the Sine of 50^{d} to the Sine of 27^{d} 50′.

If C B be made Radius, then is C A the Secant of 55^{d} to the ſame Radius, and it is alſo by conſtruction the Sine of 50^{d} to the common Radius, then from the proportion of equality, becauſe the firſt tearm is equal to the third, the ſecond tearm C F being the Secant of 20^{d} to the leſſer Radius, is alſo the Sine of the fourth Proportional to the common Radius, this ſuitable to the former of thoſe directions, the latter carries on this Proportion:

As the Sine of 50^{d} to Sine of 70^{d}, So is the Secant of 55^{d} to the Secant of 62^{d} 10′.

[illustration]

Thus from the former Scheme make C A the Secant of 55^{d} to the leſſer Radius, and C H the Sine of 50^{d} to the common Radius, and joyne A H; alſo make C F the Secant of 20^{d}, and draw F G parallel to A H, then becauſe A C is equal24 to H C, therefore F C ſhall be equal to G C, wherefore the Sine of the fourth Proportional is equal to the Secant of the ſecond Proportional, which is the foundation of the former Scheme; which form of Operation holds generally, though by reaſon of excurſions we ſhall purſue other forms of Operation more ſpeedy.

Let the Proportion be:

As the Sine of 30^{d} to the Sine of 18^{d}, So is the Sine of 23^{d}. To what Sine? Anſwer is, 14^{d}.

If a Line of Sines be wanting, it may be obſerved that a Line of Chords will ſupply the defect thereof, by doubling the three firſt tearms, and you will for the fourth tearm finde the Chord of twice the Arch ſought, the half whereof is the Arch ſought.

The three tearms of the former Proportion doubled, are 60^{d}, 36^{d}, 46^{d}.

[illustration]

It will be convenient to deſcribe the Ark C always with the leaſt of the three tearms, and this kinde of Operation is beſt when the two firſt tearms of the Proportion are fixed.

If the Proportion had been of the leſs to the greater:

As the Sine of 36^{d} is to the Sine of 60^{d}, So is the Sine of 14^{d} to the Sine of 23^{d}.

In this caſe the Chord of 28 degrees muſt be ſo entred on the line A B, that on foot reſting thereon, the other turned about ſhould but juſt touch the line A C, the foot of the Compaſſes would reſt at D, and the extent D A, being the Chord of 46 degrees, the half thereof 23 degrees, is the fourth Proportional ſought.

This way was chiefly added, to ſhew the manner of Proportioning out of any line to a leſſer Radius, for here A B being the common Radius, and the extent wherewith the ark C was deſcribed another leſſer Radius, the neareſt diſtance from D to A C, ſhal be the Chord of 46 degrees to the ſaid leſſer Radius.

Let the Proportion be:

As the Tangent of 40^{d}: Is to the Tangent of 50^{d} ∷ So is the Tangent of 20^{d}: To the Tangent of 27^{d} 20′.

Double the three firſt tearms, and they are 80^{d}, 100^{d}, 40^{d}.

[illustration]

Example. Let the Proportion be:

As the Sine of 20 degrees, Is to the Sine of 10 degrees,

So is the Tangent of 31 degrees, To the Tangent of 17 degrees.

[illustration]

Having drawn a Circle and its Diameter, as before, prick any extent from C to H, and from A to I, and draw the lines A H and C I, which will be parallel, then double the three firſt tearms of the Proportion, and they are 40^{d} 20^{d} ∷ 62^{d}.

Prick the Chord of 40 degrees from A to B, and the Chord of 20 degrees from C to D, a ruler over B and D cuts the Diameter at E, and the Segment A E bears ſuch Proportion to E C, as the Sine of 20 degrees doth to the Sine of 10 degrees, then prick the Chord of 62 degrees from A to F, a ruler over F and E, cuts the Circle at G, and the Arch C G meaſured on the Chords, is 34 degrees, the half whereof 17 degrees, is the Tangent ſought.

In like manner when a Sine is ſought, the Diameter muſt be divided, according to the Ratio of the two firſt tearms.

27If the Proportion were:

As the Tangent of 17 degrees, Is to the Tangent of 31 degrees,

So is the Sine of 10 degrees, To a fou•th Proportional, to wit, the Sine of 20 degrees.

Here A H and C I being drawn parallel to each other, if C G be 34 degrees, the double of the firſt tearm and A F 62 degrees, the double of the ſecond tearm, a ruler over G and F cuts the Diameter at E, and the Segment C E bears ſuch Proportion to E A as the Tangent of 17 degrees doth to the Tangent of 31 degrees, then pricking the Chord of 20 degrees the double of the third tearm from C to D, a ruler over D and E cuts the other line at B, and the extent A B being the Chord of 40 degrees, the half thereof 20 degrees, is the fourth Proportional Sine ſought.

In the performance whereof, it is required to take out one Tangent by help of the Chords, and to meaſure another, and ſo to regulate theſe two tearms, as they may both be leſſe then the Radius, thereby to ſhun Excurſion, which requires knowledge in varying of Proportions.

Example:

As Radius, to Tangent 67 degrees, So is the Tangent of 70 degrees to a fourth Tangent. The anſwer is: 79 degrees 30 minutes.

A Proportion wholly in Tangents may be all changed into their Complements, without altering the order of their places, the former Porportion ſo changed, is:

As the Radius is to the Tangent of 27 degrees, So is the Tangent of 20 degrees to the Tangent of 10 degrees 30 minutes.

Operation.

[illustration]

Otherwiſe:

Any two Tangents may be changed into their complements, if the other two tearms of the Proportion do onely change places.

The former Proportion thus changed, is:

As the Tangent of 63 degrees: To the Radius ∷ So is the Tangent of 20 degrees: To the Tangent of 10 degrees 30 minutes.

This Proportion is alſo wrought by the former Scheme.

The former Proportion lyes evident, imagine a Perpendicular raiſed from the point A till it meet with D C, which being continued, is the Secant of the ark A D, it then lyes:

As the Radius C A to the ſaid Perpendicular or Tangent:

So is the other Tangent L C, equal to E C to L F, the Tangent ſought.

The other Proportion lyes evident alſo.

Imagine a Perpendicular erect from the point B, till it meet with C D continued, then:

29As the ſaid Perpendicular the Tangent of 63 degrees: Is to its Baſe B C, the Radius ∷ So is the Perpendicular L C to its Baſe L F.

Theſe are Equiangled Triangles, for the Angle C F Z, is equal to the Angle D C B: what hath been ſaid is ſufficient for any Proportion whatſoever, if where the Radius is not ingredient, it be brought in according to former Directions.

Example.

As the Tangent of 40^{d}: To the Tangent of 50^{d} ∷

So is the Tangent of 62^{d} 40′: To the Tangent of 70^{d}.

If this Proportion be changed into the complements, it will be:

As the Tangent of 50 degrees: To the Tangent of 40 degrees ∷

So is the Tangent of 27 degrees 20 minutes: To the Cotangent of the Arch ſought, namely, to the Tangent of 20 degrees.

[illustration]

Draw the Quadrant A C B, and from A to D out of a Line of Chords, prick off the firſt tearm 50 degrees, and draw the line D C, then ſetting one foot in A, draw the Arch H C, and therein prick off 27^{d} 20′ from C to I, a ruler laid from A to I, will30 help you to the Tangent of 27 degrees, 20 minutes C E, then either finde the Point F by drawing a line through E parallel to A C, through which Point F draw the parallel M F, or take E C and enter it, ſo that one foot reſting in D C, the other turned about may juſt touch A C, and then take the neareſt diſtance from F to B C, which place from C to M, and through the Points F and M draw a right line, then prick off 40 degrees the other middle tearm from A to L, a ruler laid from the Center to L will find the Point N, place M N from C to G, a ruler from A to G, will finde the Arch K C, which meaſured on the Chords, is 20 degrees, the fourth Proportional ſought in the latter Proportion, the complement whereof 70 degrees, is the fourth Proportional in the former Proportion.

The foregoing Scheme doth alſo repreſent a Proportion of the leſs to the greater.

As the Tangent of 20 degrees, to the Tangent of 40 degrees,

So is the Tangent of 27 deg. 20 min. to the Tangent of 50 degrees.

Having pricked off 40 degrees to L, draw a Line from it into the Center, then enter the Tangent of 20 degrees, ſo as one foot reſting in L C, the other turned about may juſt touch A C, and it will finde the Point N, through which draw a line parallel to B C, and therein prick down M F the third Tangent, then lay a ruler from the Center over F, and it will cut the Circle at D, the Arch A D is the meaſure of the fourth Proportional, and by changing in many Caſes the places of the ſecond and third tearm, and admitting the work to be a little without the Quadrant, it may be performed after the ſame manner, onely the parallel F M will be more remote from B C, and poſſibly the Operation thereby rendred more certain.

If the Proportion were:

As the Tangent of 5 degrees, To the Tangent of 10 degrees,

So is the Tangent 65 degrees, To the Tangent of 77 degrees.

This Proportion being changed wholly into their Complements would not be inconvenient; or if the two latter tearms were changed into their Complements, it would be:

As the Tangent of 10 degrees, To Tangent of 25 degrees,

So is Tangent 5 degrees, To Tangent of 13 degrees.

Either way might be commodious enough.

31### Proportions in Sines and Tangents protracted from Equality of Proportion.

Example:

As the Sine of 10 degrees, To the Sine of 20 degrees,

So is the Tangent of 59 degrees, To the Tangent of 73 degrees.

Which by altering the two latter tearms into their Complements, and altering the place of the firſt and ſecond tearm, will ſtand thus:

As the Sine of 20 degrees, To the Sine of 10 degrees,

So is the Tangent of 31 degrees, To the Tangent of 17 degrees.

And the Radius might be introduced into either of theſe Proportions ſeveral ways, from whence might iſſue divers Varieties of Operation, either of which Proportions may alſo be performed with as much facility without the Radius.

Example of the latter.

[illustration]

32Again, when a Sine is ſought, I ſay the former Scheme ſerves to finde it, if the Proportion were:

As the Tangent of 31^{d}: Is to the Sine of 20^{d} ∷

So is the Tangent of 17^{d}: To the Sine of 10^{d}.

If the three firſt tearms were given, the Scheme would finde E G to be the meaſure of the fourth Proportional, the Sine ſought.

To ſum or difference two Arks would be eaſily done, by applying the Chord of the leſſer Ark both upwards and downwards: And thus to ſolve with a Line of Chords all the Caſes of Sphaerical Triangles, will be a matter of no difficulty, but to preſcribe Protractions for many particular queſtions, without intimation of the Proportion applyed, were but to miſlead and puzzle the Reader, as hath been the vain affectation of ſome.

Example: Let a Tangent be ſought, and let the Proportion be:

As the Sine of 40^{d}: Is to the Sine of 25^{d} ∷

So is the Tangent of 39^{d}: To another Tangent, to wit, 28^{d}.

[illustration]

Having drawn the Quadrant D L C, place the Sine of 40 degrees from C to A, and the Sine of 25 degrees to B, and draw33 B K and A I, parallel to C L, then prick the third tearm 39^{d} in the Limbe from D to E, and draw C E, it cuts the Line paſſing through the ſecond tearm at F, place the extent B F upon the line paſſing through the firſt tearm from A to G, a rule•over C and G cuts the Limbe at H, and the Arch D H being 28^{d}, is the fourth Proportional ſought.

If a Sine were ſought. Example.

Let the Proportion be:

As the Tangent of 28 degrees: Is to the Tangent of 39 degrees ∷

So is the Sine of 25 degrees To what Sine? 40 degrees.

Place 28 degrees and 39 degrees, the two Tangent Tearmes in the Limbe at H and E, and draw Lines into the Center, then place the Sine of 25 degrees the third tearm from C to B, and draw B K parallel to C L, and thorough the point where it cuts the line of the ſecond tearm of the Proportion, as at F, draw a line parallel to C D, as is G M, and mind where it croſſeth the line H G drawne thorough the firſt tearm of the Proportion, as at G, ſo is the extent G M the Sine of the fourth Proportional ſought, to wit, 40 degrees, as it will be found being meaſured on the Sines.

By the like reaſon Proportions in Tangents alone, or in equal parts and Tangents, might have been protracted, for the extents A C, B C, might as well have been Tangents or equall parts, as Sines.

THe particular Example ſhall be to finde the Suns Azimuth, the Proportion is:

As the Coſine of the Altitude: Is to the Secant of the Latitude ∷

So is the difference of the Verſed Sines of the Ark of Difference between the Latitude and Altitude, and of the Suns Diſtance from34 the Elevated Pole: To the Verſed Sine of the Azimuth from the North in this Hemiſphere.

Example:

- Let the Latitude be 51
^{d}32′. - The Altitude — 41 34.
- The Suns declination 23 31 North.

[illustration]

Having firſt drawn the quadrant B F C, with the Radius, or 60 degrees of the Chords, then prick the Chord of the Latitude 51 degrees 32 minutes from B to L, and draw C L produced: prick the Chord of the Atitude, to wit, 41 degrees 34 minutes, from B to A, prick the Chord of the Declination 23 degrees 31 minutes in ſummer, from F upwards to D, but in winter downwards in the arch of the quadrant continued, and draw D E parallel to C F, the neareſt diſtance from A to C F, is the Coſine of the Altitude, which place on the Latitude line, from C to H, ſo is C H the ſecant of the Latitude, the neareſt diſtance from H to C F being Radius, which extent place from C to K.

Then place the arch A L from B to G, and the neareſt diſtance from G to E D is the difference of the verſed Sines of the third tearm of the Proportion; it is alſo by reaſon of the Proportion of equality, the verſed Sine of the Azimuth from the North, which35 place from C to M, then becauſe it is greater then the Radius C K, the Azimuth is more then 90 degrees from the North, and K M is the Sine of the Azimuth from the Eaſt or Weſt, wherewith on the point K deſcribe the ark O, and a ruler from the Center touching it, cuts the Limbe at Q, and the arch F Q being 15 degrees, is the Suns Azimuth to the Southwards of Eaſt or Weſt, whereas if the point M had fallen as much on the other ſide K, it had been ſo much to the Northwards of it.

This is an excellent Scheme, in regard it requires the drawing of no new Lines till the Declination vary.

The Coſine of the Altitude may be multiplied, &c. being doubled, it reacheth to I, the Radius, to wit, C K doubled reacheth to L, the extent C M being doubled, reacheth to N, with L N deſcribe the Ark P, and a ruler from the Center, touching its extremity, finds the Point Q in the Limbe, as before.

Note alſo, that the Coſine of the Altitude, and the difference of the verſed Sines aforeſaid, may be taken from a line of Verſed or Natural Sines on a Ruler of any Radius big enough, and therewith proceed as if they were taken from the Scheme.

Moreover, the extent C R is the Sine of the Amplitude to the ſame Radius, wherewith the quadrant was drawn, and the extent E R is the Sine of the Aſcenſional difference, or of the time of Sun riſing or ſetting from Six, the Radius to which Sine being D E.

Example:

- Latitude 51
^{d}32′. - Declination 13
^{d}North. - Altitude 31
^{d}18′.

[illustration]

36Having with 60 degrees of the Chords, deſcribed a Semicircle A D B, draw the Diameter A C B, perpendicular thereto from the Center raiſe C D, prick off the Latitude from A to L, and draw the Latitude line L M parallel to D C, prick off the height from D to H, and place the Sine thereof being the nearſt diſtance from H to D C, from L to N place the Chord of the Suns Polar diſtance, to wit 77 degrees, from N to P, and ſetting one foot on M, with the extent M P, deſcribe an Ark, then take the Coſine of the Altitude being the neareſt diſtance from H to A C, and upon C as a Center, deſcribe another Ark which croſſeth the former at S, a Ruler from the Center over S, cuts the Limbe at V, and the Arch A V being 110^{d}, is the Suns Azimuth from the North.

The like Operation holds for South Declination, onely then the Chord of the Polar diſtance is greater then 90^{d} by the Suns declination.

If the Sun have Depreſſion, the Sine of it muſt be pricked upward above L in the Latitude Line, and then as before.

Thus when three Sides are given to finde an Angle, the complements of thoſe Sides may bear the names of Latitude, Altitude and Declination as here, and the Solution will be the ſame.

Prick 77^{d} the Chord of the Suns Polar diſtance from L to P, and ſetting one foot in M, with the extent M p, croſs the Limbe at K, and the arch D K being 21^{d} 12′, is the meaſure thereof.

In following Schemes we ſhall finde the hour from Noon from the Point A, alſo the Azimuth from the South may be found from the ſame Point, and poſſibly with more convenience in regard the Interſections may not happen ſo oblique, and that upon this conſideration, that whatſoever Altitude the Sun hath in any Signe upon any Azimuth from the north, he hath the like Depreſſion in the Oppoſite Sign upon the like Azimuth counted from the South.

Wherefore retaining the former conſtruction, L N the Sine of the Altitude, which in the former Scheme was placed downwards, in this following Scheme place upwards.

37
[illustration]

This is a Scheme of much worth, in regard it requires the drawing of no new Lines till the Latitude vary. In South Latitudes when the Altitude is very great, the interſections of the prickt Arks will fall near the Center, in that caſe let the Altitude and Latitude change Names, and fit the Scheme thereto: Here note, that the three points M C S are the angular points of the Sides of a plain Triangle, & if thoſe Sides be doubled (doubling the ſine M C outward beyond M) the interſection at S which now happens within the Semicircle, would happen without and beyond it: The foundation of this Scheme ſhall afterwards be ſuggeſted.

It is ſuppoſed to be made of a ſquare Board, on which there is a Circle deſcribed which is cut out through: the Line N S repreſents a thread, upon the Point N as a Center placed in the Circumference, there moveth a Labell or Triangle repreſented by N S A, the Side N A is ſuppoſed to ſtand Perpendicular, and to have a Slit in it, the line S A is to be a thread extended from S to A, the other ſide of which Triangle reſembles a Moveable Toung or Labell, the Center being in the Circumference, every degree is twice as large as it would be, if it were at the Center, wherefore the quadrants S E, S W, are numbred with 45^{d} on each ſide the38 Line N S, but are not divided with concentrick Circles and Diagonals, nor can they be with truth when ever the Center is placed in the Circumference, and this I call an Azimuth Compaſs, becauſe though it be not ſo, yet it ſupplies the uſe of one, and if a right line be continued from N to E, and made a line of Sines; alſo a Tangent of 45^{d} put through the Limbe, it, or an Azimuth Compaſs, is rendred general, without the uſe of Paper-draughts, as I have ſhewed in the Uſes of the ſmalleſt Quadrant in my Treatiſe, The Sector on a Quadrant, Page 277 to 284. where the Reader will meet with ready Proportions for Calculating the Suns Azimuth or true Coaſt, not before publiſhed.

The Uſe of the ſaid Azimuth Compaſs at Sea, is readily to apply it to any Compaſs in the Ship, and thereby finde the true Coaſt of the Suns bearing by that Compaſs to which it is applyed, and conſequently the Variation thereof, and by my own experience I have often found, that by the thread which paſſeth through the Diameter of the ſaid A•imuth Compaſs, it may very well by the View be placed over the Meridian line of the Compaſs, and then turning the moveable Label towards the Sun, ſo that the ſhadow of the thread may paſs thorough the Slit, the tongue of the Label, amongſt the graduations of the Limbe, ſhewes how the Sun bears by the ſaid Compaſs, in which the toucht wires is ſuppoſed to be preciſely under the Flower-deluce, & when the Sun is more then 45^{d} from the Meridian either way, the thread in the•iameter of the Circle muſt be placed by the view over the Eaſt or Weſt point of the Compaſs, and the Suns bearing accordingly reckoned from thence.

Then admiting that the bearing of the Sun by the Compaſs and his true Azimuth or Coaſt of bearing, be found either by Calculation or the former Schemes, the Variation of the ſaid Compaſs from the North (which all Needles are lyable unto) with the Coaſt thereof may thus be found.

Example:

Let the bearing of the Sun by the Compaſs be 55^{d} Eaſt-ward from the South, and his true coaſt in the Heavens be 43^{d} ¾ from the South Eaſt-wards.

[illustration]

Then admit it were required to ſteer the Ship away N E by E, being the fifth point from the North Eaſt-ward, it is deſired to know how ſhe muſt wind or ſteer by that Compaſs. Out of the Scale of Rumbes place five point from B to R, then meaſure the extent N R on the Rumbes, and it ſheweth four points, whence we may conclude th•t to make good the former Courſe, the Ship muſt be ſteered North-eaſt by this Compaſs.

The re•dieſt wa•for fin•ing the Variation, is by thoſe Sea Rings deſcribed by M•. Wright, but thoſe are chargeable, are but in few ſhips, fixed but to one Compaſs, reſerved for the Ship maſters own peculiar Ob•ervations, ſo that the common Mariners can have no practiſe thereby.

40### A Scheme for finding the Hour.

Example.

- Latitude 51
^{d}22′ North. - Declination 23
^{d}31′ North. - Altitude 10
^{d}28′, Complement 79^{d}32′.

[illustration]

Having deſcribed the Semicircle, and divided it into two Quadrants by the line D C, prick as before, the Latitude 51^{d} 32′ from A to L, and draw L M parallel to D C: Prick off the Declination 23^{d} 31′ from D to E, prick the Sine thereof being the neareſt diſtance from E to D C in Winter, or South Declination from L to Q upwards, but in Summer or North Declination O N downwards, and with the Coſine of the Declination being the neareſt diſtance from E to A C, upon C as a Center, deſcribe the Ark G W, ſo is the Scheme prepared for that Declination both North and South.

The Suns height being 10^{d} 28′, its Complement is 79^{d} 32′, prick the Chord thereof from Q to T, and ſetting one foot upon M, with the extent M T draw the arch I, a ruler from the Center over that Interſection, finds the point K, and the Arch A K being 30^{d}, the hour is either 10 in the Morning, or 2 in the Afternoon.

Prick the Chord of 90^{d} from Q to O, and with the extent M O upon M as a Center, croſs the ark G W at P, a ruler from the Center over P, finds the point R in the Limbe, and the ark D R41 being 33^{d} 12′, in time about 2 hours 13′, is the time of the Suns riſing or ſetting from Six to that Declination both North and South.

Let the Altitude be 9^{d} 30′, its Complement is 81^{d} 30′, prick the Chord thereof from N to H, and with the extent M H upon M as a Center, croſs the arch G W as at S, a ruler from the Center over that Interſection, finds the point V in the Limbe, and the ark D V being 15^{d}, the true time of the day is either five in the Morning, or ſeven in the Afternoon.

Having found the Hour firſt, then the Azimuth and Angle of Poſition may be eaſily found from the Proportion of the Sines of Sides to the Sines of their oppoſite Angles, as in the following Scheme.

Example:

- Latitude 51
^{d}32′. - Declination 13
^{d}North. - Altitude 37
^{d}18′.

[illustration]

By the former Directions the Hour will be found to be 45^{d} from noon, either 9 in the morning, or 3 in the afternoon, the Interſection whereof happens at e, through e, draw e F parallel to A B, and prick the Altitude 37^{d} 18′ from D to H, and draw H C, alſo joyn e C and make C O equal to C M, and through the point O, draw O Q parallel to A B, ſo is the extent C Q the Sine of42 the Angle of Poſition, and the extent C P the Sine of the Azimuth from the Meridian.

With the neareſt diſtance from H to C B, ſetting one foot in C, croſs the parallel e F at F, a ruler from the Center cuts the Limbe at I, and the arch B I is the Suns Azimuth either from the North or South, in this Caſe 60^{d} from the South.

With the former extent croſs the parallel O Q at G, a ruler from the Center cuts the Limbe at K, and the arch B K being 33^{d} 34′, is the meaſure of the Angle of Poſition, and this work might have been performed on the other ſide D C; but to avoid confuſion when the Doubts about Oppoſite Sides and Angles may be removed, and when not, as when a double anſwer is to be given, I have ſhewed in a Treatiſe, Entituled, The Sector on a Quadrant, page 139, 140. And how to find the points I or K without drawing the lines e F or O G, and that by help of a croſs or Interſection like that at e, which may either happen within or without the outward Circle, the Reader may attain from the laſt Scheme for finding the Amplitude.

[illustration]

The Converſe of the former Scheme for finding the Hour, will finde the Suns Altitudes on all Hours, and the Diſtances of Places in the Arch of a great Circle.

43Example.

- Latitude 51
^{d}32′. - Declination 23
^{d}31′ North. - Hour 75
^{d}from Noon, that is either 7 in the Morning, or 5 in the Afternoon.

Having drawn the Semicircle, its Diameter, and by a Perpendicular from the Center divided it into two quadrants, and therein having prickt off A L the Latitude, and thorough the ſame drawn L M produced and parallel to D C, therein from L to M and Q prick off the Sine of the Declination.

Then prick off the Hour from Noon from A to R, and laying the Ruler from the Center, draw the line R E, and with the Coſine of the Declination, namely, the neareſt diſtance from F to D C, draw the Arch G E, and transfer the diſtance between the points M, and E from M to H.

Laſtly, the diſtance between the points N and H, is the Chor•of the Suns diſtance from the Zenith for that Hour, namely, 62^{d} 37′, the Complement whereof 27^{d} 23′ is the Altitude ſought.

Moreover, the diſtance between H and Q, is the Chord of the Suns diſtance from the Zenith for the winter declination, namely, 99^{d} 30′, which being greater then a quadrant, argues the Sun to have 9^{d} 30′ of Depreſſion under the Horizon, and ſo much is his Altitude for the hours of 5 in the Morning, or 7 in the Afternoon when his Declination is 23^{d} 31′ North.

Let the Hour from Noon be either 10 in the Morning or 2 in the Afternoon, prick off 30^{d} from A to K, and from the Center draw the line K G, place the diſtance M G from M to O, ſo is the diſtance O N the Chord of 36^{d} 16′, the Complement whereof 53^{d} 44′ is the Summer Altitude for that Hour, and the diſtance O Q is the Chord of 79^{d} 32′, the Complement whereof 10^{d} 28′ is the Winter Altitude for that Hour.

Alſo for the Diſtances of places in the ark of a great Circle, the Caſe in Sphaerical Triangles is the ſame with that here reſolved.

So if there were two places, the one in North Latitude 51^{d} 32′, the other in North Latitude 23^{d} 31′, the difference of Longitude between them being 75^{d}, their diſtance by the former Scheme will be found to be 62^{d} 37′; but if the latter place were in as much South Latitude, then their diſtance would be 99^{d} 30′.

44### Another Example for finding the Diſtances of Places in the Arch of a great Circle.

Example.

Let the two Places be according to the Sea-mans Calendar.

- Iſle of Lobos, Longitude 307
^{d}41′. Latitude 40^{d}21′ South. - Lizard — 18 30. Latitude 50 10 North.
- Difference of Longitude 289 11. Complement 70
^{d}49′.

[illustration]

Having deſcribed a Circle, make AE M 70^{d} 49′, M I the Latitude of the Iſland, B I the Sine thereof falling Perpendicularly on C M, AE L the Latitude of the Lizard, L A the Sine thereof, make A E equal to the extent A B, and prick B I from L upwards to H, when the places are in different Hemiſpheres, but downwards to K, when in the ſame Hemiſphere, and the extent H E or K E, is the Chord of the Ark of diſtance between both places in this Example

- H E is 109
^{d}, 41′. - K E is 48
^{d}, 57′.

45### Demonſtration.

This Scheme I firſt met with in a Map made in Holland, the foundation whereof was long ſince laid by Copernicus and Regiomontanus, who from a right lined plain Triangle, happening at the Center of the Sphere, have preſcribed a Method of Calculation for finding an Angle when three Sides are given: Here we ſhall illuſtrate the Converſe, how from two Sides and the Angle comprehended, to find the third Side.

From any two points in the Sphere, ſuppoſe Perpendiculars to fall on the plain of the Equator here repreſented by AE L Q M, which Perpendiculars are the Sines of the Latitudes of thoſe two Points, and the diſtance of the points in the Plain of the Equinoctial from the Center of the Sphere, are the Coſines of thoſe Latitudes, & the angle at the Center between thoſe points in the plain of the Equator, is equal to the arch of the Equinoctial between the two Meridians, paſſing through the ſuppoſed Points in the Sphere: now a right line extended in the Sphere between any two Points, is the Chord of the Ark of diſtance between thoſe Points.

Underſtand then, that the three Points A, C, B, limit the Sides of a righ•lined Triangle in the Plain of the Equator, whereof the Angle A C B is at the Center, then the extent A B is placed from A to E, if then we draw D E G perpendicular to A Q, and place B I from E to G and D, the extent L G ſhall be the Chord of the third Side when the places are in different Hemiſpheres, and the extent L D is the Chord of their diſtance when they are in the ſame Hemiſphere.

And if the extent E D, E G, be placed from L to H and K, the line D E G need not be drawn, becauſe the extent L G and L D, if it be rightly conceived, are the two very Points at firſt ſuppoſed in the Sphere, the extent A B as to matter of Calculation, being one Side of a plain Triangle right angled. The ſum or difference of the Sines of both Latitudes the other Side, and the Chord of the diſtance is the Hipotenuſal, or third Side ſought.

Thus the Ancients by Calculation, and we by Protraction, having two Sides and the Angle comprehended given, find the third Side; or having three Sides given, find the Angle oppoſite to that Side which in the Scheme is meaſured by a Chord, as by reſult from the three Sides there will be got the two extents A E46 and C B, and conſequently the Interſection at B, and thence the Angle AE M, which before was inſiſted on in finding the Azimuth and Hour, by the like reaſon the Diſtances of Stars may be found from their Longitudes and Latitudes, or from their Declinations and right Aſcenſions. Divers other Schemes from other Proportions might be added for finding the Hour and Azimuth, &c. which which I am loth to trouble the Reader withal.

I ſhall adde another Scheme for this purpoſe, which carries on the ſame Proportion, by which this Caſe is uſually Calculated: The Proportions are expreſſed in a Treatiſe, The Sector on a Quadrant, page 127.

Example.

- Latitude — 51
^{d}32′. - Declination 23 31.
- Hour — 60 from Noon.

[illustration]

Having drawn a Semicircle and the Radius Z D, prick the Latitude from C to L, and draw L A parallel to C E, and making A L Radius, place the Sine of 30^{d} from A to 30^{d}: how to do this, is ſhewed in the Second Part, a ruler over D and 30^{d} cuts the Limbe at B, from B to F ſet off 66^{d} 29′ the Suns diſtance from the Elevated Pole, and draw a line into the Center, and make D K equal to D 30^{d}, and draw H K parallel to C E, and it is the parallel of Altitude required, the meaſure whereof being the Arch C H, is 36^{d} 42′.

Two Sides with the Angle comprehended to find both the other angles, & then the third ſide. Schemes may be fitted to Proportions47 which at two Operations find both thoſe Angles, firſt according to the firſt Direction for operating Proportions in the Sines and Tangents, and then by another operation, the third Side may be found: alſo this Caſe is performed otherwiſe in the Geometrical Dyalling.

Example.

- Latitude — 51
^{d}32′. - Declination 23 31 North.
- Azimuth 30 deg. from the Vertical or Point of Eaſt and Weſt.

[illustration]

Having deſcribed a Semicircle, and divided it into halfs by the Perpendicular H C, prick the Latitude from S to L, and draw L A parallel to S N, and proportion out the Sine of 30^{d} to the Radius L A, and prick it from A to 30^{d}, upon which point with the Sine of the Declination, deſcribe a prickt Circle, and a ruler from the Center juſt touching the outward extreamities of it, cuts the Limbe at T and K: Here note, that H C is the Horizontal line and any Ark counted from H towards K is Altitude, but towards N is Depreſſion; ſo in this Example the Arch H K being 49^{d} 56′, is the Altitude for 30^{d} of Azimuth to the Southwards of the Eaſt and Weſt for North Declination, and the Arch H T being 6^{d} 36′, is the Suns Depreſſion under the Horizon for the ſame Azimuth, when his Declination is 23^{d} 31′ South.

It is alſo his Altitude for 30^{d} of Azimuth to the North-wards of the Eaſt or Weſt, when his Declination is 23^{d} 31′ North, becauſe48 if we place A 30^{d} as much on the other ſide the Vertical at M, and with the Sine of the Declination deſcribe an Ark at f, a ruler from the Center will cut the Limbe as much on this ſide H, as T is on the other ſide.

If the prickt Circle had hapned wholly on one ſide C H, a ruler from the Center touching its extreamities, and cutting the outward Circle in two points, that neareſt unto H had been the Altitude to the aſſigned Azimuth for South Declination, and the remoteſt for North Declination.

And if one Side of the prickt Circle happens below S C, as it may do both for the Sun or Stars, when their Declination is more then the Latitude of the place, then the Quadrant H S. muſt be continued to a Semicircle, and H C muſt be alſo produced, and the Altitude counted above the other end of the ſaid Horizontal Line.

If the ruler ſhould touch the prickt Circle near the Center, let it be noted that C 30^{d} is the Radius, and the extent that deſcribed the prickt Circle, is a Sine of an Ark to that Radius; whence it followes, that a right angled Triangle may be framed, and the anſwer given in the Limbe, as we have often done before. See the laſt Scheme for finding the Amplitude.

The Demonſtration of this Scheme, is founded not onely on the Analemma, as I have ſaid in the Second Part, but likewiſe on thoſe Proportions, whereby the Suns Altitudes are Calculated on all Azimuths.

The firſt Proportion finds the Equinoctial / Altitude proper to the given Azimuth, and is:

As the Radius: To the Cotangent of the Latitude ∷

So is the Sine of any Azimuth from the Vertical:

To the Tangent of the Equinoctial Altitude ∷

In the former Scheme L A is the Radius of a line of Sines, it is alſo the Cotangent of the Latitude, if we make C A Radius.

Then becauſe of the Proportion of Equality in the two firſt tearmes, it followes that the ſine of the Azimuth from the Vertical A 30^{d}, to the Radius L A, is alſo the Tangent of the Equinoctial Altitude, making C A Radius, wherefore a ruler over C 30^{d} cuts the Limbe at B, and the Ark H B being 21^{d} 40′, is the Equinoctial Altitude ſought.

The next Proportion is:

As the ſine of the Latitude:

Is to the Coſine of the Equinoctial Altitude ∷

So is the ſine of the Declination:

To the ſine of a fourth Ark ∷

Get the ſum and diffe•ence of the Equinoctial Altitude, and of this fourth Ark, the Sum is the Summer Altitude for Azimuths from the Vertical toward the South in this Hemiſphere.

The difference when this fourth ark is

- leſſer
- greater

then the Equinoctial Altitude, is the

- Winter
- Summer

Altitude for Azimuths from the vertical towards the

- South.
- North.

In the former Scheme if C 30^{d} be Radius, 30^{d} A is the ſine of the Equinoctial Altitude, and C A is the Coſine thereof, which is alſo the ſine of the Latitude to the common Radius, then by reaſon of the Proportion of equality in the two firſt tearms, it follows that the ſine of the Declination to the common Radius being the third tearme, is alſo the ſine of the fourth Arch ſought, C 30^{d} being Radius, and the ſumming or differencing of the two Arks, is performed by deſcribing the prickt Circle, and laying a ruler from the Center touching its Extreamities, after this m•nner a Scheme is contrived from Proportions, to find the third Side, having two Sides with an angle oppoſite to one of them given.

This of all other Caſes is the moſt difficult either on the Analemma or from Proportions to ſuit Schemes unto, however we ſhall adde a particular Queſtion, with the Proportions and Scheme fitted thereto:

- Let there be given the Latitude,
- The Suns Declination,

And the Azimuth to the Sun or Stars from the Eaſt or Weſt, and let it be required to finde the Hour from Noon,

50The Proportion is:

As the Tangent of the Azimuth from Eaſt or Weſt:

Is to the ſine of the Latitude ∷

So is the Radius:

To the Cotangent of the firſt Ark ∷

Again:

As the Cotangent of the Declination:

Is to the Coſine of the firſt Ark ∷

So is the Cotangent of the Latitude:

To the Coſine of the ſecond Ark ∷

The Declination being North, if the Azimuth and Angle of Poſition be both acute, the Sum of the firſt and ſecond Ark is the Hour from Noon, if of different kindes, their difference.

But when the Declination is South, if the Azimuth be to the South-wards of the Eaſt or Weſt, the complement of the Sum of the former Arkes to a Semicircle, is the Hour from Noon; but if the Azimuth be to the Northwards of the Eaſt or Weſt, the complement of their difference is the Hour from Noon.

When the Angle of Poſition will be acute, and when obtuſe. See my Treatiſe, The Sector on a Quadrant, page 117.

Theſe Proportions we ſhall carry on in the Scheme following.

Example.

- Latitude 51
^{d}32′ North. - Declination 23
^{d}31′ - Azimuth from the Vertical 30
^{d}

[illustration]

51Upon C as a Center, deſcribe a Semicircle, draw the Diameter N C S, and the Radius C E W perpendicular thereto, ſet off the Declination from S to D, and the Latitude to L, drawing lines into the Center.

Set off the Azimuth being 30^{d} from the Vertical from E to A, the neareſt diſtance from A to N C, prick on the Latitude line from C to I, and draw I Q parallel to C E, and where it interſects the Declination line (which may ſometimes fall on the other ſide of L) ſet K; alſo draw I H parallel to S C, and H B equal to the neareſt diſtance from A to C E, then with K Q upon the point B deſcribe an Ark, and on both Sides of it lay a ruler from the Center juſt touching the outward Extreamitie thereof, and you will finde the points R and T in the Limbe.

If the Azimuth be to the Northwards of the Eaſt, and Angle of Poſition acute, the Ark N T is the Hour from Noon, in this Example 110^{d} 16′, but if the Azimuth be to the Southwards of the Vertical, and Angle of Poſition acute, or to the Northwards and the Angle of Poſition obtuſe, the Ark S R is the Hour from Noon, in the former of theſe Caſes 37^{d} 26′.

If the Azimuth be to the Southwards of the Eaſt or Weſt, the Ark S T is the Hour from Noon, in this Caſe 69^{d} 44′.

If it be to the Northwards, then is the Ark N R the Hour from Noon, to wit, 142^{d} 34′, we ſpeak in reference to the Northern Hemiſphere, in the other Hemiſphere let the words North and South change places.

Thus when we have two Sides with an Angle oppoſite to one of them, we may finde the Angle included by this Scheme, calling one of thoſe Sides the Colatitude, the other the Polar diſtance, and the Angle given the Azimuth, and that ſought the Hour; and thus for the Sun or ſuch Stars as have two Altitudes on every Azimuth, we may finde the reſpective times when they ſhall be on that Azimuth, by turning the Stars Hour into common time, as I have ſhewed in a Treatiſe called, The Sector on a Quadrant.

52
[illustration]

Here we have retained the former Scheme, but onely continued ſome of the lines thereof, making the Coſine of the Azimuth, to wit, the neareſt diſtance from A to N C to be Radius, equal to C I the ſine thereof, to wit, the neareſt diſtance from A to C E, will become the Tangent, which is equal to H B or C R, and making C I Radius, the extent C H equal to I Q, is the ſine of the Latitude, then the firſt Proportion lyes plain:

As C R the Tangent of the Azimuth from the Vertical:

Is to R B the ſine of the Latitude ∷

So is C S the Radius:

To S G the Cotangent of the firſt Ark ∷

Wherefore the point B is in a right line from the Center with the firſt Ark.

Then making C B Radius, H B is the ſine of the firſt Ark, and C H the Coſine, then the ſecond Proportion lyes thus:

As F H the Cotangent of the Declination:

Is to H C the Coſine of the firſt Ark ∷

So is M K equal to H I the Cotangent of the Latitude:

To M C equal to K Q, the Coſine of the ſecond Ark to the Radius C B. Laſtly, the ſumming and differencing of theſe Arks is performed by deſcribing an Ark with K Q upon the Point B.

53### Another Scheme for the ſame purpoſe, in which the Operations are Sinical, and the Latitude, Declination and Azimuth, the ſame as before.

[illustration]

Having deſcribed a Semicircle, and divided it into two Quadrants by the Radius C W, prick the Declination from W to D, the Azimuth to A, drawing Lines into the Center, alſo the Latitude from W to L.

Then place the Coſine of the Latitude (being the neareſt diſtance from L to N C) from C to I, and draw I K parallel to N C, then place the extent C O ſo at Q as a ſine in the Limbe, that it may fall Perpendicular on C E, as doth Q E, wherein make E B equal to K O, and upon the Point B as a Center, with the ſine of the Azimuth, being the neareſt diſtance from A to C W, deſcribe the prickt Ark, and lay a ruler from the Center touching its extreamities on both Sides, and cutting the Limbe at R and T, and the Arkes

- S T
- S R
- N T
- N R

are agreeable in order to the four Cautions given in the former Scheme.

This Scheme firſt findes the Angle oppoſite to the other of the given Sides, whereof C O is the ſine, and then we have two Angles with a Side oppoſite to one of them given, to finde the third Angle, which is the Angle included, and the Proportions54 are of the ſame kind with thoſe for finding the Altitudes on all Azimuths, which may be found in this Scheme, if we place C I in the Limbe as is Q E, and in it make the extent E B equal to I K, then an Ark deſcribed after the ſame manner, with the neareſt diſtance from D to C W, ſhall in like kind give the Suns Altitudes on all Azimuths, C W being the Horizontal Line.

In page 60 of a Treatiſe of Dyalling by the Plain Scale or Line of Chords, for the more eaſie drawing the Parallels of Declination on the Horizontal Projection of the Sphere, I have intimated that a ready way ſhould be ſhewed for finding the Amplitude, and although that before preſcribed be very ready, yet I ſhall adde one more from the common Proportion:

As the Coſine of the Latitude: Is to the Sine of the Declination ∷

So is the Radius: To the Sine of the Amplitude ∷

Example.

- Latitude — 62
^{d}35′. - The complement is 27
^{d}25′. - Declination — 23
^{d}30′.

[illustration]

55Having drawn the Quadrant C B F, double the complement of the Latitude, and it is 54^{d} 50′, prick the Chord thereof from C to R, then double the Declination, which will be 47^{d}, with the Chord thereof upon the point R, deſcribe the prickt Ark G, a ruler laid from the Center juſt touching that Ark, cuts the Limbe at A, and the Arch B A being 60^{d}, is the meaſure of the Amplitude.

Here note, That the extent wherewith G was drawn is the Sine of the Amplitude, if you make C R Radius, and when the Amplitude falls to be large, it may be better found the other way by help of the Coſine, thus: Prick the extent R G from C to D, then make C L and D E equal to C R, ſo is C E the Coſine of the Amplitude, with which extent upon the Point L, deſcribe the Ark H, and a ruler from the Center juſt touching it, cuts the quadrants Limbe at A, as before.

Upon the Point E, with the extent C D, deſcribe a little Ark neare N: Again, upon the point D with the extent C E, deſcribe another Ark croſſing the former at N, then a ruler laid over the Center and the croſs at N, will cut the Limbe at A, as before; Note this well, becauſe it is wholly omitted in the Second Book,☝ page 18, where it ſhould rather have been handled; and if you will double the extent C D, and the Radius C R, you will finde the point E twice as far out towards B, and the interſection at N twice as remote from the Center, as now it is, and by the like reaſon when a line is to be drawn unto the Limbe from a Point that happens near the Center, take the neareſt diſtance from the point, firſt to C F, and triple it in the ſaid line, then the neareſt diſtance to C B, and triple it in that line, and find a point of interſection more remote from the Center, through which and the Center the line required is to be drawn, and the ſaid extents may be doubled, or often multiplyed.

How to perform this, I have ſhewed in the Second Part, but there the Shadow doth not immediately give the height, whereto notwithſtanding a Gnomon may be fitted.

54Imagine the former Quadrant to lye flat upon the Plain of the Horizon, and draw the line F K perpendicular to C F of what length you pleaſe, which is ſuppoſed to be done upon a ſquare piece of Wood.

Then to fit the Gnomon, prick the Chord of 45 degrees from F to M, and draw C K, then imagine the Triangle C K F to be a Gnomon or Cock ſtanding perpendicularly upright over the line C E F, if then you turn this Quadrant ſo about towards the Sun, that the ſhadow of the ſtreight edge of the Cock may fall in the line F K, the ſhadow of the ſlope edge of the Cock (which edge may be ſupplied by a Thread) will happen in the Limbe, and there ſhew the Suns Height required, if counted from B towards F. This Conceit is taken from Clavius de Aſtrolabio.

FINIS.

THE MARINERS PLAIN SCALE NEW PLAIN'D: OR, A Treatiſe ſhewing the ample Uſes of a Circle equally divided, or of a Line of CHORDS and equal Parts, Divided into Three Books or Parts.

Being contrived to be had either alone, or with the other Parts.

I. The firſt containing Geometrical Rudiments, and ſhewing the Uſes of a Line of Chords, in reſolving of all Proportions relating to Plain or Spherical Triangles, with Schemes ſuited to all the Caſes derived from Proportions, and the full Uſe of the Scale in Navigation, according to the particulars in the following Page, and in finding the Variation of the Compaſs.

II. The ſecond ſhewing the Uſes of a Line of Chords, in reſolving all the Caſes of Spherical Triangles, by Projecting the Sphere Orthographically, or laying down the Sphere in right Lines, commonly called, The Drawing or Delineating of the Analemma.

With an everlaſting Almanack in two Verſes, &c.

III. The third part ſhewing the Uſes of a Line of Chords, in reſolving all the Caſes of Spherical Triangles Stereographically that is on the Circular Projection with Dyalling, from three Shadows of a Gnomon on a regular Flat, or by two Shadows, &c.

To which may alſo