This Scale is commonly made of Box, and sometimes of Brass, exactly two Foot long (though there are others but a Foot long, which are not so exact) about an Inch and \(\frac{1}{4}\) broad, and of a convenient Thickness.

The Lines that are put on one Side of it are the Line of Numbers, marked on the Scale Numbers; the Line of artificial Sines, marked Sines; the Line of artificial Tangents, marked Tangents; the Line of artificial versed Sines, marked V. S. signifying Versed Sines; the artificial Sines of the Rhumbs, marked S. R. signifying the Sines of the Rhumbs; the artificial Tangents of the Rhumbs, marked T. R. signifying Tangents of the Rhumbs; the Meridian-Line in Mercator’s Chart, marked Merid. signifying Meridian-Line; and equal Parts, marked E. P. signifying equal Parts.

There are commonly placed on these Scales, that are but a Foot long, the Lines of Latitudes, Hours, and Inclinations of Meridians.

On the Back-side of this Scale are placed all the Lines that are put upon a Plain-Scale.

The Lines of artificial Sines, Tangents, and Numbers are so fitted on this Scale, that, by means of a Pair of Compasses, any Problem, whether in right-lined, or spherical Trigonometry, may be solved by them very expeditiously, with tolerable Exactness; and therefore the Contrivance of these Lines on a Scale is extremely useful in all Parts of Mathematicks that Trigonometry hath to do with; as Navigation, Dialling, Astronomy, &c.

Construction of the Line of Numbers.

The Construction of the Line of Numbers is thus: Having pitched upon it’s Length, which, on Gunter's Scale, let be 23 Inches, take exactly half that Length, which will be the Length of either of the Radius’s; then take that half Length, and divide into 10 equal Parts, one of which diagonally subdivide into 100 equal Parts, that is, make a Diagonal Scale of 1000 equal Parts of the aforesaid Half-Length, which may easily be done from our Author’s 8th Use.

Now having drawn, on Gunter’s Scale, three Parallels, for better distinguishing the Divisions of the Line of Numbers, and made a Mark for the Beginning of it, half an Inch from the Beginning-end of the Scale, look in a Table of Logarithms lor the Number 200, and against it you will find 2.301030; and rejecting the Characteristick 2, and also the three last Figures 030, because the Length of the Radius is divided but into 1000 equal Parts, take 301 of those 1000 Parts in your Compasses, and lay off that Distance from the Beginning of the Line, at the End of which write 2 for the first Prime; Again, to find the Division for the second Prime, look, in the Table of Logarithms for the Number 300, and against it you will find 2.477121; and rejecting the Characteristick 2, and the three last Figures 121, as before, take 477 from your Diagonal Scale, and lay off that Distance from 1 at the Beginning, at the End of which write 3 for the second Prime. In this manner proceed for all the Primes of the first Radius to 1, which will be the whole Length of your Diagonal Scale, or 1000 equal Parts. And because each of the Primes of the second Radius are at the same Distance from 1, at the End of the first Radius, as the same Primes, on the first Radius, are distant from 1 at the Beginning, the Primes on the second Radius are easily found.

The Divisions of the Tenths, between each of the Primes in both Radius’s, are found thus: Look in the Table of Logarithms for 110, and against it you will find 2.041393, and rejecting the Characteristick 2, and the three last Figures, there will remain 41; which taken from the aforesaid Diagonal Scale of 1000, will give the first Tenth in the first Prime. Again, look in the Table for 120, and against it you will find 2.079181; and by rejecting the Characteristick, and the three last Figures, there will remain 79; which taken from the Diagonal Scale, will give the second Tenth in the first Prime. Proceed thus for all the Tenths in the first Primes of both Radius’s. And to find the Tenths in the second Primes of both Radius’s, look in the Table for the Number 210, and against it you will find 2.322219, whence rejecting as before, you will have 322, which laid off from the Beginning of the first Prime, will give the first Tenth in the second Prime. Again, to find the second Tenth in the second Prime, look for the Number 220, and against it you will find 2.342423, whence by rejecting, as before, you will have 342 for the second Tenth, in the second Prime. In like manner may the Tenths in all the Primes of both Radius’s be found.

To find every two Centesms in the first Prime of the second Radius, look for the Number 102 in the Table of Logarithms, and against it you will find 2.008600, and by rejecting, as in the Table of Logarithms, and against it you will find 2.008600, and by rejecting, as at first, you will have 8 for the second Centesm. Again, look in the Table for 104, and proceed as before, and you will have 17 for the third Centesm. In like manner you may have every second Centesm in the first, and also the Second Primes of the second Radius.

Note, In bisecting every of the two Centesms in the first Prime, Centesms will be had. Note also, That the third, fourth, and fifth Primes, cannot be divided into every two Centesms, but only into every five, because of the Smallness of the Divisions.

Construction of the Line of artificial Sines and Tangents.

The Line of artificial Sines on Gunter’s Scale, is nothing but the Logarithms of the natural Sines, translated from the Tables of artificial Sines and Tangents, almost in the same Manner as the Logarithms of the natural Numbers were; the Method of doing which is thus:

Having drawn three Parallels under the Line of Numbers for distinguishing the Divisions of the Line, and marked a Point exactly half an Inch from the Beginning-end of the Scale, representing the Beginning of the Line of Sines, look in the Tables of artificial Sines and Tangents, for the Sine of 40 Minutes, which is the first Subdivision of the Line, and it will be found 8.065776: then rejecting the Characteristick 8, and the three 1 last Figures 776, as in the Construction of the Line of Numbers, the 65 remaining, must be taken on the same Scale of 1000 Parts, as served before for the Line of Numbers; this 65 laid off from the Beginning of the Line of Sines, will give the Division on the Line of Sines for 40 Minutes. Again, To make the next Division, which is for 50 Minutes, seek in the Table for the Sine of 50 Minutes, which will be found 8.162681; then rejecting the Characteristick 8, and the three last figures 681, take the Remainder 162, from your Scale of 1000 Parts, and lay it off from the Beginning of the Line, and that will give the Division for 50 Minutes. Moreover, to make the Division for 1 Degree, seek the Sine of 1 Degree, which is 8.241855, and rejecting as before, take the Remainder 241 from the Scale of 1000, and lay it off from the Beginning on the Line of Sines, which will give the Division for 1 Degree. Proceed thus for the other Degrees and Minutes to 90; only take notice, that when you come to 5 Degrees, 50 Minutes, the Parts to be taken off the Scale are more than 1000, and, consequently, longer than the Scale itself. In that Case you must make a Mark in the Middle of the Line of Sines; from which lay off all the Parts found above 1000, for the Degrees and Minutes: As, to make the Division for 6 Degrees, the Sine of which is 9.019235, the Parts to be taken off the Scale will be 1019' therefore lay off 19 from the middle Point, representing 1000, and the Division for 6 Degrees will be had. Proceed in the same manner for the Line of artificial Tangents, ’till you come to 45 Degrees, whose Length is equal to Radius; and the Divisions for the Degrees and Minutes above 45, which should go beyond 45, are set down by their Complements to 90. For Example, the Division of 40 Degrees hath it’s Complement 50 set to it, because the proper equal Parts taken off the Scale of 1000 to make the Division, for the Tangent of 50 Deg. will be as much above 1000 (which are the equal Parts for the Tangent of 45 Degrees, to be laid off from the middle of the Line of Tangents) as the equal Parts for the Division of the Tangent of 40 Degrees wants of 1000; an Example of which will make it manifest: The Tangent of 40 Degrees is 9.924813, and by rejecting the Characteristick, and the three last Figures, the Parts of 1000, viz. 924 taken from 1000, and there remains 76, which are the Parts that the Tangent of 40 Degrees is distant from the Tangent of 45 Degrees. Again, the Tangent of 50 Degrees is 10.076186, and by rejecting the Characteristick, and the three last Figures, the Parts 76 above 1000, for the Division of the Tangent of 50 Degrees, which must fall beyond 45 Degrees, are equal to the Parts that the Division of 40 Degrees wants of 1000. Understand the same for the Tangent of any other Degree, or Minute, and it’s Complement: the Reason of this is, because Radius is a mean Proportional between any Tangent and it’s Complement.

The Construction of the artificial Sines of the Rhumbs, and quarter Rhumbs, is deduced from a Consideration that the first Rhumb makes an Angle of 11 Deg. 15 Min. with the Meridian; the second, 22 Deg. 30 Min. the third, 33 Deg. 45 Min the fourth, 45 Deg. &c. therefore to make the Division on Gunter’s Scale, for the first Rhumb, take the Extent of the artificial Sine of 11 Deg. 15 Min. on the Scale, and lay it off upon the Line drawn to contain the Divisions of the Line of Rhumbs, and that will give the Division for the first Rhumb. Again; take the Extent, on the Line of artificial Sines, of the Sine of 22.30 Min. and lay it off in the same manner as before, and you will have the second Rhumb: proceed thus for all the other Rhumbs. The Divisions for the half Rhumbs, and quarter Rhumbs, are also made in the same manner: the Divisions of the artificial Tangents of the Rhumbs, are made in the same manner as the Divisions of the artificial Sines of the Rhumbs, by taking the artificial Tangents of the several Angles that the Rhumbs and quarter Rhumbs make with the Meridian.

The Construction of the Line of artificial versed Sines.

This Line, which begins at about 11 Deg. 45 Min. and runs to 180 Deg which is exactly under 90 of the Line of Sines (though on the Scale they are numbered backwards; that is, to the versed Sine of each 10 Degrees above 20, are set the Numbers of their Complements to 182, for a Reason hereafter shewn), may be thus made, by means of the Table of Sines, and the aforesaid equal Parts. Suppose the Division for the versed Sine of 15 Degrees be to be made. Take half 15 Degrees, which will be 7^{d}. 30^{m}; the Sine of which doubled will be 18.231396, and by substracting the Radius therefrom, you will have 8.231396; and rejecting the three last Figures, and the Characteristick, there will remain 231; this 231 taken from your Scale of 1000, and laid off from a Point directly under the Beginning of the Line of Sines, will give the Division for the versed Sine of 15 Degrees, at which is set 165, viz. the Complement of 15^{d} to 180^{d}. Again; to make the Division for 20 Degrees; twice the Sine of 10 Degrees (it’s half) will be 18.470340; from which, substracting Radius, and rejecting the Characteristick, and the three last Figures, you will have 479; which taken from your Scale, and laid off from the Beginning of the Line, will give the Division for the versed Sine of 20 Degrees. And in this manner may the Line of versed Sines be divided to 180 Degrees, by observing what I have said in the Construction of the Line of Sines.

The Manner of projecting the Lines of Numbers, artificial Sines and Tangents, in Circles, and Spirals of any Number of Revolutions.

Suppose the Circle BC is to be divided into a Line of Numbers of but one Radius; first, divide the Limb into 1000 equal Parts, beginning from the Point G; then take 301 of those Parts, which suppose to be at p, and lay a Ruler from the Center A, on the said Point p, and that will cut the Periphery of the Circle BC in the Point for the Log. of 2. Again, Take 477 Parts upon the Limb, and a Ruler laid from the Center upon the laid Division, will cut the Circle BC in the Point for the Log. of the Number 3: and thus by taking the proper Parts upon the Limb, from the Point G, which were before directed to be used in dividing this Line upon the Scale; and laying a Ruler from the Center, may the Line of Numbers be projected upon the Circle BC. And in the same manner may the Lines of artificial Sines and Tangents be projected, from the Sine of 5^{d} H 45^{m}, and Tangent of 5^{d} 42^{m}, to the Sine of 90^{d}, and the Tangent of 45^{d}, by taking (as before directed in the Construction of the straight Lines of Sines and Tangents) the Parts of 1000 for the Degrees and Minutes, and laying them off upon the Limb from the Point G, and then laying a Ruler from the Center, which will divide the Circles into Lines of Sines and Tangents.

Now to project a Line of Numbers upon the Spiral or having four Revolutions, or Turns; first, divide the Limb into 1000 equal Parts, beginning from the Point G; then take 301, which is the Log. of the Number 2 (when the Characteristick, and the three last Figures are rejected), and multiply it by 4, because the Spiral hath four Revolutions, and the Product is 1204: then if 204 of the Parts of 1000, be taken upon the Limb from G top, and a Ruler be laid from the Center A to p, it will cut the second Revolution of the Spiral in the Point for the Number 2. Again; having multiplied 477, the Log. of the Number 3, by 4, the Product will be 1908; whence taking 908 Parts from the Point G on the Limb, to the Point q, lay a Ruler from A to q, and that will cut the second Revolution of the Spiral, in the Point for the Number 3. Moreover, multiply 602 by 4, and the Product will be 2408; whence take 408 Parts upon the Limb from G, and laying a Ruler from A, it will cut the third Revolution of the Spiral in the Point, for the Number 4: and in thus proceeding may the Spiral be divided into a Line of Numbers, whole Beginning is at the Point C, and End at the Point B. This being understood, it will be no difficult Matter to project the Sines and Tangents in a Spiral of any Number of Revolutions.

In using either the Circular or Spiral Lines of Numbers, Sines, and Tangents, there is an opening Index placed in the Center A, confiding of two Arms; the one called the antecedent Arm, and the other the consequent Arm; then three Numbers, Sines, or Tangents being given, to find a fourth. If you move the antecedent Arm to the first, and open the other Arm to the second (the two Arms keeping the same Opening), and afterwards the antecedent Arm be moved to the third, the consequent Arm will fall upon the fourth required.

But, Note, That as many Revolutions of the Spiral as the second Term is distant from the first, so many Revolutions will the fourth Term be distant from the third.

Of the Meridian Line.

The Meridian Line, on Gunter’s Scale, is nothing but the Table of Meridional Parts in Mercator’s Projection transferred on a Line, which may be done in the following Manner, by help of the Line of equal Parts set under it, and a Table of Meridional Parts.

Take any one of the large Divisions of the aforesaid Line of equal Parts, whose Length Fig. 9, let be AB, and divide it into six equal Parts upon some Plane; at the Points AB raise the Perpendiculars AC, BD, equal to AB, and compleat the Parallelogram ABDC; divide the Sides AC, BD, into ten equal Parts, and the Side DC into six, draw the Diagonals AF, 10, 20, &c. as per Figure, and you will have a Diagonal Scale, by which any part of the aforesaid Division under 60 may readily be taken.

Now to make the Divisions of the Meridian Line, look in the Table of meridional Parts for 1 Degree, and against it you will find 60: and rejecting the last Figure, which in this Case is 0, take six equal Parts from the aforementioned Diagonal Scale, and lay it off on the Meridian Line, which will give the Division for one Degree. Again, to find the Division for 2 Degrees, seek in the Table of Meridional Parts, for the Parts against 2 Degrees, and they will be found 120: whence rejecting the last Figure (which always must be done), take 12 from your Scale, and lay it off from the Beginning of the Meridian Line, and the Division for 2 Degrees will be had. Moreover, to find the Division for 11 Degrees, you will find answering to it 664; and rejecting the last Figure, the Remainder will be 66, which must be laid off from the Beginning of the Meridian Line to have the Division for 11 Degrees. But because 66 cannot be taken from the Diagonal Scale, you must take only 6 from it; and for the 60, take it’s whole Length, or else lay off the 6 from the End of the first Division of the Line of equal Parts, and the Division for 11 Degrees will be had. In this manner may the Meridian Line be divided into Degrees, and every thirty Minutes, as it is upon the Scale.

There are several other ways of dividing this Meridian Line, but let this suffice.

The Use of this Line is to project a Mercator’s Chart.

Upon the End A, of the Diameter of the Circle, erect a Line of Sines at right Angles, of the Length of the Diameter; then from the Point B, the other End of the Diameter, draw right Lines to each Degree of that Line of Sines, cutting the Quadrant AC. Now having drawn the Chord-Line AC, which is to be the Line of Latitudes, set one Foot of your Compasses upon the Point A, and with the other transfer the Intersections made by the Lines drawn from B, on the Quadrant, to the Chord-Line AC, by means of which it will be divided into a Line of Latitudes. Or the Line of Latitudes may be made by this Canon, viz. As Radius is to the Chord of 90 Deg. So is the Tangent of any Degree, to another Tangent, the natural Sine of whose Arc, taken from a Diagonal Scale of equal Parts, will give the Division, for that Degree, on the Line of Latitudes, and so for any other Degree.

Again, To graduate the Line of Hours, draw the Tangent GH equal to the Diameter AB, and parallel thereto; then divide each of the Arcs of half the Quadrants AK, KB, into three Parts, for the Degrees of every Hour from 12 to 6, which must again be each subdivided into Halfs, Quarters, &c. then, if thro’ each of the aforesaid Divisions and Subdivisions, Lines be drawn from the Center, cutting the Tangent Line GH, they will divide the said Line into a Line of Hours.

As for the Line of Inclination of Meridians, usually put upon Scales, it is nothing but the Line of Hours numbered with Degrees instead of Time; and the Lines of the Style’s Height, and Angle of 12 and 6, sometimes put upon Scales, are made from Tables of the Style’s Height, &c. and no otherwise used.

Whence the Line of Hours is but two Lines of natural Tangents to 45 Degrees, each set together at the Center, and from thence Beginning and continued to each End of the Diameter, and from one End thereof, numbered with 90 Deg. to the other End; and may otherwise be thus divided: Let AB be the Radius of a Line of Tangents, CD another Radius equal and parallel thereto, and CB the Diameter to either of the said Radius’s, which is to be divided into a Line of Hours. Now if right Lines are drawn from the Point D, to every Degree of the Tangent-Line AB, those Lines will divide GB, half of the Line of Hours, as required; and Lines drawn from the Point A, to every Degree of the Tangent CD, will divide the other half of CB: therefore from the similar Triangles CDF, EFB, it will be as the Radius CD is to the Tangent EB of any Arc under 45: So is CF to FB; that is, As Radius is to the Tangent of any Arc under 45 Degrees, So is Radius + the Cotangent of the said Arc to 45 Degrees, to Radius − the said Cotangent, as in Fig. 12.

As the Radius AB, to the Tangent of BC of any Arc, So is AB + EG, to AB − EG: for call AB, r; and BC, b; and from the Point C, draw CF parallel to EG, and make BD equal to AB. Then DF (=FC) = \(\frac{\sqrt{rr-2rb+bb}}{2}\), and AF = \(\frac{\sqrt{rr+2rb+bb}}{2}\): Whence As AF : \(\left(\frac{\sqrt{rr+2rb+bb}}{2}\right)\) : FC \(\left(\frac{\sqrt{rr-2rb+bb}}{2}\right)\) :: AB (r) : EG \(\left(\frac{\sqrt{rr-rb}}{r+b}\right)\), therefore it will be AB (r) :: BC (b) :: AB + EG \(\left(\frac{\sqrt{2rr}}{r+b}\right)\) : AB − EG \(\left(\frac{\sqrt{2rb}}{r+b}\right)\).

Thus having given the Construction of the Lines on Gunter’s Scale, I now proceed to shew their manner of using; but, Note, These Lines are also put upon Rulers to slide by each other, and therefor called Sliding-Gunters, so that you may use them without Compasses; but any Person that understands how to use them with Compasses, may also by what I have said of Everard’s and Coggeshall’s Sliding-Rules, use them without.

Use of the Lines of Numbers, Sines, and Tangents.

Use I.The base of a right-angled right-lined Triangle being given 30 Miles, and the opposite Angle to it 26 Degrees, to find the Length of the Hypothenuse.

As the Sine of the Angle, 26 Degrees, is to the Base, 30 Miles, So is Radius to the Length of the Hypothenuse. Set one Foot of your Compasses upon the 26th Degree of the Line of Sines, and extend the other to 30 on the Line of Numbers; the Compasses remaining thus opened, set one Foot on 90 Degrees, or the End of the Line of Sines, and cause the other to fall on the Line of Numbers, which will give 68 Miles and about a half, for the Length of the Hypothenuse sought.

Use II.The Base of a right-angled Triangle being given 25 Miles, and the Perpendicular 15, to find the Angle opposite to the Perpendicular.

As the Base 25 Miles is to the Perpendicular 15 Miles, So is Radius to the Tangent of the Angle sought; because if the Base is made Radius, the Perpendicular will be the Tangent of the Angle opposite to the Perpendicular. Extend your Compasses on the Line of Numbers, from 15, the Perpendicular given, to 25, the Base given, and the same Extent will reach the contrary way, on the Line of Tangents, from 45 Degrees, to 31 Degrees, the Angle sought.

Use III.The Base of a right-angled Triangle being given, suppose 20 Miles, and the Angle opposite to the Perpendicular 50 Degrees, to find the Perpendicular.

As Radius is to the Tangent of the given Angle 50 Degrees, So is the Base 20 Miles to the Perpendicular sought. Extend your Compasses on the Line of Tangents, from the Tangent of 45 Degrees, to the Tangent of 50 Degrees, and the same Extent will reach on the Line of Numbers the contrary way, from the given Base 20 Miles, to the required Perpendicular, about 23\(\frac{3}{4}\) Miles.

Note, The Reason why the Extent on the Line of Numbers was taken from 20 to 23\(\frac{1}{4}\) forwards, is, because the Tangent of 50 Degrees (as I have already mentioned in the Construction of the Line of Tangents) should be as far beyond the Tangent of 45 Degrees, as it’s Complement 40 Degrees wants of 45 Degrees.

Use IV.The Base of a right-angled Triangle being given, suppose 35 Miles, and the Perpendicular 48 Miles; to find the Angle opposite to the Perpendicular.

As the Base 35 Miles is to the Perpendicular 48 Miles, So is Radius to the Tangent of the Angle sought. Extend your Compasses from 35, on the Line of Numbers, 1048; the same Extent will reach the contrary way on the Line of Tangents, from the Tangent of 45 Degrees, to the Tangent of 36 Degrees 5 Minutes, or 53 Degrees 55 Minutes; and to know which of those Angles the Angle sought is equal to, consider that the Perpendicular of the Triangle is greater than the Base; therefore (because both the Angles opposite to the Perpendicular and Base together make 90 Degrees) the Angle opposite to the Perpendicular will be greater than the Angle opposite to the Base, and consequently the Angle 53 Degrees 53 Minutes, will be the Angle sought.

Use V.The Hypothenuse of a right-angled Spherical Triangle being given, suppose 60 Degrees, and one of the Sides 20 Degrees; to find the Angle opposite to that side.

As the Sine of the Hypothenuse 60 Degrees is to Radius, So is the Sine of the given Side 20 Degrees, to the Sine of the Angle sought. Extend your Compasses, on the Line of Sines, from 60 Degrees to Radius, or 90 Degrees, and the same Extent will reach on the Line of Sines the same way, from 20 Degrees, the given Side, to 23 Degrees 10 Minutes, the Quantity of the Angle sought.

Use VI.The Course and Distance of a Ship given; to find the Difference of Latitude and Departure.

Suppose a Ship fails from the Latitude of 50 Deg. 10 Min. North, S. S. W. 48.5 Miles: As Radius is to the Distance failed 48.5 Miles, So is the Sine of the Course, which is two Points, or the second Rhumb, from the Meridian, to the Departure. Extend your Compasses from S, on the artificial Sine Rhumb-Line, to 48.5 on the Line of Numbers; the same Extent will reach the same way from the second Rhumb, on the Line of artificial Sines of the Rhumbs, to the Departure Westing 18.6 Miles. Again, as Radius is to the Distance failed 48-5 Miles, So is the Co-Sine of the Course 67 Deg. 30 Min. to the Difference of Latitude. Extend your Compass from Radius, on the Line of Sines, to 48.5 Miles on the Line of Numbers; the same Extent will reach the same way, from 67 Deg. 30 Min. on the Line of Sines, to 44.8 on the Line of Numbers; which converted into Degrees, by allowing 60 Miles to a Degree, and substracted from the given North-Latitude 50 Deg. 10 Min. leaved the Remainder 49 Deg. 25 Min. the present Latitude.

Use VII.The Difference of Latitude and Departure from the Meridian being given; to find the course and Distance.

A Ship, from the Latitude of 59 Deg. North, fails North-Eastward ’till me has altered her Latitude 1 Deg. 10 Min. or 70 Miles, and is departed from the Meridian $7.5 Miles; to find the Course and Distance.

As the Difference of Latitude 70 Miles is to Radius, So is the Departure 57.5 Miles to the Tangent of the Course 39 Deg. 20 Min. or three Points and a half from the Meridian. Extend your Compasses from the fourth Rhumb, on the Line of artificial Tangents of the Rhumbs, to 70 Miles on the Line of Numbers: the same Extent will reach from 57.5 on the Line of Numbers, to the third Rhumb and a half on the Line of artificial Tangents of the Rhumbs. Again; As the Sine of the Course 39 Deg. 20 Min. is to the Departure 57.5 Miles, So is Radius to the Distance 90.6 Miles. Extend your Compasses from the third Rhumb and a half, on the artificial Sines of the Rhumbs, to 57.5 Miles on the Line of Numbers, and that Extent will reach from the Sine of the eighth Rhumb, on the Sines of the Rhumbs, to 90.6 Miles on the Line of Numbers.

Use of the Line of Versed Sines.

The three Sides of an oblique Spherical Triangle being given, to find the Angle opposite to the greatest Side.

Suppose the Side AB be 40 Degrees, the Side BC 60 Degrees, and the Side AC 96 Degrees, to find the Angle ABC. First add the three Sides together, and from half the Sum substract the greater Side AC, and note the Remainder; the Sum will be 196 Degrees, half of which is 98 Degrees; from which substracting 96 Degrees, the Remainder will be two Degrees.

This done, extend your Compasses from the Sine of 90 Degrees, to the Sine of the Side AB 40 Degrees; and applying this Extent to the Sine of the other Side BC 60 Degrees you will find it to reach to a fourth Sine about 34 Degrees. Again; from this fourth Sine extend your Compasses to the Sine of half the Sum, that is, to the Sine of 72 Degrees, the Complement of 98 Degrees to 180, and this second Extent will reach from the Sine of the Difference 2 Degrees to the Sine of 3 Deg. 24 Min. against which, on the Versed Sines, stands 151 Deg. 50 Min. which is the Quantity of the Angle sought.

That the Reason of this Operation may appear, it is demonstrated in most Books of Trigonometry, that as Radius is to the Sine of AB, So is the Sine of BC to a fourth Sine; then As this fourth Sine is to Radius, So is the Difference of the versed Sines of AC and AB + BC to the versed Sine of the Complement of the Angle ABC to 180 Degrees. It is also demonstrated, that As Radius is to the Sine of half the Sum of any two Arcs, so is the Sine of half their Difference to half the Difference of the versed Sines of these two Arcs whence, if the Sine of AB be called a, the Sine of BC, b, and the Sine of AC, c, the fourth Sine in the first Analogy will be hand; in saying as \(r:a::b:\frac{ab}{r}\). Now to get the Difference of versed Sines of AC, and AB + BC, let us call the Sine of \(\frac{\class{fig-label}{\mathrm{AB}}+\class{fig-label}{\mathrm{BC}}+\class{fig-label}{\mathrm{AC}}}{2}p\), and the Sine of \(\frac{\class{fig-label}{\mathrm{AB}}+\class{fig-label}{\mathrm{BC}}-\class{fig-label}{\mathrm{AC}}}{2}q\), then as \(r:p::q:\frac{pq}{r}\), which last Term will be half the difference of the versed Sines of AC, and AB + BC: therefore, if we again say, as \(\frac{ab}{r}:r::\frac{2pq}{r}:\frac{2rpq}{ab}\) this last Term will be the versed Sine of the Complement of the Angle ABC: To find which at two Operations, you must say, As \(r:a::b:\frac{ab}{r}\); then as \(\frac{ab}{r}:p::q:\frac{rpq}{ab}\); which last Term, multiplied by 2, will be the versed Sine Complement sought. But to avoid multiplying by 2, the versed Sines on Scales are fitted from this Proportion, viz. As Radius is to half the Sine of an Arc, so is half the Sine of the same Arc, to half the versed Sine of that Arc.

Use the Line of Latitudes and Hours.

These Lines are conjointly used, in readily pricking down the Hour-Lines from the Substyle, in an Isosceles Triangle, on any kind of upright Dials, having Centers in any given Latitude; that is, by means of them there will be this Proportion worked, viz. As Radius is to the Sine of the Style’s Height, So is the Tangent of the Angle at the Pole, to the Tangent of the Hour-Lines Distance from the Substyle.

Now suppose the Hour-Lines are to be pricked down upon an upright Declining-Plane, declining 25 Deg. Eastwards: First draw C12 the Meridian, perpendicular to the Horizontal Line of the Plane, and make the Angle FC12 equal to the Substyle’s Distance from the Meridian, and draw the Line FC for the Substyle. This being done, draw the Line BA perpendicular to the said Substyle, passing thro’ the Center C; then out of your Line of Latitudes set off CA, CB, each equal to the Style’s Height, and fit in the Hour-Scale, so that one End being at A, the other may meet with the Substyle Line at F.

Now get the Difference between 30 Deg. 47 Min. the Inclination of Meridians, and 30 Degrees, the next Hour’s Distance letter than the said 30 Deg. 47 Min. and the Difference is 47 Minutes, that is, 3 Minutes in time; then count upon the Line of Hours,

Hours.

Min.

0

3

from F to

10

1

3

from F to

11

2

3

from F to

12

3

3

from F to

11

4

3

from F to

2

5

3

from F to

3

And make Points at the Terminations, to which drawing Lines from the Center C, they shall be the Hour-Lines on one Side.

Again, fitting in the Hour-Scale from B to F, count from that End at B, the former Arcs of Time.

Hours.

Min.

0

3

from B to

4

1

3

from B to

5

2

3

from B to

6

3

3

from B to

7

4

3

from B to

8

5

3

from B to

9

And make Points at the Terminations, thro’s which draw Lines from the Center C, and they will be the Hour-Lines on the other Side the Substyle.

You must proceed thus for the Halfs and Quarters, in getting the Difference between the Half-Hour next lesser (in this Example 22 Deg. 30 Min.) under the Arc of Inclination of Meridians; the Difference is 1 Deg. 17 Min. which in time is 33 Minutes, to be continually augmented an Hour at a Time, and so be pricked off, as before was done for the whole Hours.

If the Hour-Scale reach above the Plane, as at B, so that BC cannot be pricked down; then may an Angle be made on the upper Side of the Substyle, equal to the Angle FCA on the under Side, and thereby the Hour-Scale laid in it’s due Position, having first found the Point F on the Substyle.

That the Reason of the conjoint Use of these Lines, in pricking off the Hour-Lines from the Substylar-Line may appear; let us suppose AC to be the Substylar-Line, A the Center of a Dial, BA a Portion of the Line of Latitudes, at right Angles to AC, and BC the Line of Hours fitted thereto. Now if CD be the Quantity of any Arc taken on the Line of Hours, and a right Line be drawn from the Center A through the Point D, the Angle FAC will be the same, as that found by saying, As Radius is to the Sine of the Number of Degrees pricked off upon the Line of Latitudes (that is, to the Sine of the Style’s Height), From A to B; So is the Tangent of that Number of Degrees pricked off from C to D on the Line of Hours (that is, the Tangent of the Angle at the Pole), to another Tangent, whose Arc will be equal to FAC (that is, to the Tangent of the Distance of the Hour-Line AF from the Substyle).

Now to prove this, it is evident, from the Construction of the Line of Latitudes, that as the Radius BC is to the Sine BG of an Arc; So is AC to AB: whence if AC be supposed Radius, BA is the Sine of the Arc pricked down from the Line of Latitudes.

Again, from the Nature of the Line of Hours; if CD be taken for the Tangent of am Arc, BD will be the Radius thereto. This being evident, let CE be the Tangent of the Angle FAC, then the Triangles BAD, DEC, will be similar; whence as the Radius BD is to BA, the Sine of an Arc; so is CD, the Tangent of an Arc, to EC, the Tangent of the Angle FAC.