Mathematical Instruments
Book II.

# Of the Constructions and Uses of the English Sector.

The principal Lines that are now generally put upon this Instrument, to be used Sector-wise, are the Line of Equal Parts, the Line of Chords, the Line of Sines, the Line of Tangents, the Line of Secants, and the Line of Polygons.

The Line of equal Parts, called also the Line of Lines, is a Line divided into 100 equal Parts; and if the Length of the Legs of the Sector is sufficient, it is again subdivided into Halfs and Quarters: they are placed on each Leg of the Sector, on the same Side, and are numbered by 1, 2, 3, &c. to 10, which is very near the End of each Leg. These Lines are denoted by the Letter L; it is divided into the same Number of equal Parts, as the same Line on the Sector described by our Author. And here note, that this 1 may be taken for 10, or for 100, 1000, 10000, &c. as Occasion requires; and then 2 will signify 20, 200, 2000, 20000, &c. and so of the rest.

The Line of Chords is a Line divided after the usual way of the Line of Chords, From a Circle, whose Radius is nearly equal in Length to the Legs of the Sector, beginning at the Center, and running towards the End thereof. It is numbered with 10, 20, &c. to 60, and to this Line, on each Leg, is placed the Letter C.

The Line of Sines is a Line of natural Sines, divided from a Circle of the same Radius, as the Line of Chords on the Sector was; these are also placed upon each Leg of the Sector, and numbered with the Figures 10, 20, 30, &c. to 90; at the End of which, on each Leg, is set the Letter S.

The Line of Tangents, is a Line of natural Tangents, divided from a Circle, and is placed upon each Leg of the Sector, and runs to 45 Degrees. It has the Numbers 10, 20, &c. to 45 placed upon it, with the Letter T for Tangent.

There is likewise another small Line of Tangents, divided from a Radius, of about two Inches, and is placed upon each Leg of the Sector; it begins at 45, which stands at the Length of the Radius from the Center, and runs to about 75 Degrees or farther, having the Numbers 45, 50, &c. to 75, with the Letter t set thereto. The Use of this Line (as hereafter shall be shewn) is to supply the Defect in the great Line.

The Line of Secants is only a Line of natural Secants, divided from a Circle of about two Inches Radius. These are placed upon each Leg of the Sector, beginning, not from the Center, but at two Inches Distance therefrom, and run to 75 Degrees. To these are set the Numbers 10, 20, &c. to 75 with the Letter S at the End there of for Secants.

Finally, the Line of Polygons, denoted by the Letter P on each Leg of the Sector, is divided in the same manner as the Line of Polygons on the French Sector; only there the Number 3, for an equilateral Triangle, is the first Polygon, and here the Number 4 for a Square.

These are the principal Lines that are now put upon this Sector, to be used Sector-wise.

The other Lines, that are placed near and parallel to the outward Edges of the Sector, on both Faces thereof, and which are to be used, as on Gunter’s Scale, are,

1st, The Line of artificial Sines, numbered (as per Fig.) with 1, 2, 3, 4, 5, on one of the Legs, and with 6, 7, 8, &c. to 90, on the other Leg; which last Numbers, as they appear in the Figure, must be set backwards; to the End, that when the Sector is quite opened, they may become forwards. This Line is denoted by the Letter S, signifying Sines.

2dly, The Line of artificial Tangents, placed next below the Line of artificial Sines is numbered, on one Leg, 1, 2, 3, 4, 5, and on the other 6, 7, 8, &c. to 45. which last are likewise set backwards; but the Numbers 80, 70, 60, 50, placed at the Divisions 10, 20, 30, 40, which signify their Complements, are set forwards.

3dly, Near the Edges, on the other Face of the Sector, is a Foot divided into 12 Inches, numbered 1, 2, 3, &c. and each Inch into 20 equal Parts. There is set to it In. signifying Inches.

4thly, and Lastly, Next to that is placed Gunter’s Line of Numbers, denoted by the Letter N, as per Figure.

There are also some other Lines placed sometimes upon the vacant Spaces of the Sector, as the Lines of Hours, Latitudes, and Inclinations of Meridians; which are no otherwise used than if they were placed upon common Scales.

All the aforesaid Lines, except the small Lines of Tangents, Secants, and the Line of Polygons, are furnished with Parallels, and the Divisions marked by unequal Lines, that the Eye may the better distinguish them.

## Section I. Of the general Use and Foundation of the Sector.

The Excellency of this Instrument above the common Scales, or Rules, is, that it may be made to fit all Scales and Radius’s; for by the Sector you may divide a Line, not exceeding it's Length when quite opened, into any Number of equal Parts; also from the Line of Chords, Sines, Tangents, &c. placed on the Sector, as before directed, you may have a Line cf Chords, Sines, Tangents, &c. to any Radius, betwixt the Breadth and Length of the Sector when opened; by which Contrivance a Sector is made almost an universal Instrument. The Invention and Contrivance of this Instrument arose from a premeditate Consideration of Prop. IV. Lib. 6. Eucl. where it is demonstrated that similar Triangles have their homologous Sides proportional.

For let the Lines AB, AC, represent the Legs of the Sector; and AD, AE, two equal Sections from the Center: then, I say, if the Points C, B, also D, E, are joined, the Lines CB, DE, per Prop. II. Lib. 6. Eucl. will be parallel; therefore because the Lines DE, CB, are parallel, the Triangles ADE, ACB, per Schol Prop. IV. Lib. 6. Eucl. will be similar; and consequently from the said Prop. IV. the Sides AD, DE, AB, BC, are proportional: that is, As AD to DE, so is AB to BC; whence if AD be the half, or a third part of the Side AB, DE will be a half, or a third part of the Parallel CB; the like Reason holds of all other Sections: whence you see that if AD be the Chord, Sine, or Tangent of any Number of Degrees, to the Radius AB, DE will be the Chord, Sine, or Tangent of the same Number of Degrees to the Radius BC.

Now the Lines found out by the Sector, are of two Sorts, viz. Lateral, or Parallel: Lateral are such as are found upon the Sides of the Sector, as AB, AC: Parallel are the Lines that run from one Leg of the Sector to the other, in equal Divisions from the Center, as DE, CB.

And here note, that the innermost of the Parallels, is the true divided Line, and therefore in using the Compasses, you must set them upon the innermost Line, both in lateral and parallel Entrance.

And further note, that the Lines are placed on this Sector, different from those that are placed upon Sectors formerly made; for instead of putting the same Lines at equal Distances from the inwards Edges of both Legs of the Sector, they are put at unequal Distances, as may be seen in the Figure: where, upon one Leg the Line of Chords is innermost, upon the other line of Tangents is inner most; that is, the innermost Line of Chords and Tangents are equally distant from the inward Edge, and so are the outermost Line of Chords and Tangents. The Benefit of the Contrivance is this, When you have set the Sector to a Radius for twixt 60 and 60 of the Chords, 90 and 90 of the Sines, also 45 and 45 of the Tangents, are all equal, which is the Reason that the Chords run but to 60 Degrees.

## Section II. Of the general Use of the Lines of Chords, Sines, Tangents, and Secants, on the Sector.

By disposing and placing these Lines, as before directed, on this Instrument, we have Scales to several Radius’s; that is, having a Length, or Radius given (not exceeding the Length of the Sector when opened), we can by the Sector find the Chord, Sine, &c. thereto; for which Property, this Instrument is often of great Use.

For Example; Suppose the Chord, Sine, or Tangent of 10 Degrees, to a Radius of three Inches, be required: take that three Inches, and make it a Parallel between 60 and 60 on the Line of Chords, then, as I have already said, the same Extent will reach from 45 to 45, on the Line of Tangents; also on the other Side of the Sector, the same Distance of three Inches, will reach from 90 to 90 on the Line of Sines: so that if the Lines of Chords be set to any Radius, the Lines of Sines and Tangents are also set to the same. Now the Sector being thus opened, if you take the parallel Distance between 10 and 10 on the Line of Chords, it will give the Chord of 10 Degrees. Also if you take the parallel Distance on the Line of Sines between 10 and 10, you will have the Sine of 10 Degrees. Lastly, if you take the parallel Extent on the Line of Tangents, between 10 and 10, it will give you the Tangent of 10 Degrees.

If the Chord, or Tangent of 70 Degrees, had been required; then for the Chord you must take the parallel Distance of half the Arc proposed, that is, the Chord of 35 Degrees, and repeat that Distance twice on the Arc you lay it down on, and you will have the Chord of 70 Degrees; and for finding the Tangent of 70 Degrees to the aforesaid Radius, you must make use of the small Line of Tangents: for the great one running but to 45 Degrees, the Parallel of 70 cannot be taken on that, therefore take the Radius of three Inches, and make it a Parallel between 45 and 45 on the small Line of Tangents; and then the parallel Extent of 70 Degrees on the said Line, is the Tangent of 70 Degrees to 3 Inches Radius.

If you would have the Secant of any Arc, then take the given Radius, and make it a Parallel between the Beginning of the Line of Secants, that is 0 and 0; so the parallel Distance between 10 and 10, or 70 and 70, on the said secant Line, will give you the Secant of 10, or 70 Degrees, to the Radius of three Inches.

After this manner may the Chord, Sine, or Tangent of any Arc be found, provided the Radius can be made a Parallel between 60 and 60 on the Line of Chords, or between the small Tangent of 45, or Secant of 0 Degrees. But if the Radius be so large, that it cannot be made a Parallel between 45 and 45 on the small Line of Tangents, then there cannot be found a Tangent of any Arc above 45 Degrees, nor the Secant of no Arc at all to such a Radius, because all Secants are greater than the Radius, or Semi-diameter of a Circle.

If the Converse of any of these things be required; that is, if the Radius be sought, to which a given Line is the Chord, Sine, Tangent, or Secant of any Arc, suppose of 10 Degrees; then it is but making that Line (if it be a Chord) a Parallel on the Line of Chords between 10 and 10, and the Sector will stand at the Radius required; that is, the parallel Extent between 60 and 60, on the said Chord-Line, is the Radius.

And so if it be a Sine, Tangent, or Secant, it is but making it a Parallel between the Sine, Tangent, or Secant of 10 Degrees, according as it is given; then will the Distance of 90 and 90 on the Sines, if it be a Tangent, the Extent from 45 to 45 on the Tangents, and if it be a Secant, the Extent or Distance between 0 and 0, be the Radius.

Hence, you see, it is very easy to find the Chord, Sine, Tangent, or Secant to any Radius.

## Section III. Of the Use of the Sector in Trigonometry.

### Use I.The Base AC of the right-lined right-angled Triangle ABC being given 40 Miles, and the Perpendicular AB 30: to find the Hypothenuse BC.

Open the Sector, so that the two Lines of Lines may make a right Angle (by Use VI. of our Author’s) then take, for the Base, AC, 40 equal Parts upon the Line of Lines on one Leg of the Sector; and for the Perpendicular AB, 30 equal Parts on the Line of Lines upon the other Leg of the Sector. Then the Extent from 40 on one Line, to 30 on the other, taken with your Compasses, will be the Length of the Hypothenuse BC; and applying it on the Line of Lines, you will find it to be 50 Miles.

### Use II.The Perpendicular AB of the right-angled Triangle ABC being given 30 Miles, and the Angle BCA 37 Degrees; to find the Hypothenuse BC.

Take the given Side AB, and set it over, as a Parallel, on the Sine of the given Angle ACB; then the parallel Radius will be the Length of the Hypothenuse BC, which will be found 50 Miles, by applying it on the Line of Lines.

### Use III.The Hypothenuse BC being given, and the Base AC; to find the Perpendicular AB.

Open the Sector, so that the two Lines of Lines may be at right Angles; then lay off the given Base AC on one of these Lines from the Center; take the Hypothenuse BC in your Compasses, and setting one Foot in the Term of the given Base AC, cause the other to fall on the Line of Lines on the other Leg of the Sector, and the Distance from the Center to where the Point of the Compasses falls, will be the Length of the Perpendicular AB.

### Use IV.The Hypothenuse BC being given, and the Angle ACB; to find the Perpendicular AB.

Take the given Hypothenuse BC, and make it a parallel Radius, and the parallel Sine of the Angle ACB will be the Length of the Side AB.

### Use V.The Base AC, and Perpendicular AB being given, to find the Angle BCA.

Lay off the Base AC on both Sides of the Seder from the Center, and note it’s Extent; then take the Perpendicular AB, and to it open the Sector in the Terms of the Base AC: so the Parallel Radius will be the Tangent of BCA.

### Use VI.In any right-lined Triangle, as ABC, the Sides AC, and BC, being given, one 20 Miles, and the other 30, and the included Angle ACB 110 Degrees, to find the Base AB.

Open the Sector, so that the two Lines of Lines may make an Angle equal to the given Angle ACB of 110 Degrees; then take out the Sides AC, CB, of the Triangle, and lay them off from the Center of the Sector on each of the Lines of Lines, and take in your Compasses the Extent between their Terms, or Ends, and that will be the Length of the sought Side AB, which will be found 41$$\frac{1}{2}$$ Miles.

### Use VII.The Angles CAB, and ACB, being given, and the Side CB: to find the Base AB.

Take the given Side CB, and turn it into the parallel Sine of it’s opposite Angle CAB, and the parallel Sine of the Angle ACB, will be the Length of the Base AB.

### Use VIII.The three Angles of a Triangle, as ABC, being given, to find the Proportion of the Sides AB, AC, BC.

Take the lateral Sines of the Angles ACB, CBA, CAB, and measure them in the Line of Lines, for the Numbers belonging to those Lines Will give the Proportions of the Sides.

### Use IX.The Three Sides AC, AB, CB, being given, to find the Angle ACB.

Lay the Sides AC, CB, on the Lines of Lines of the Sector from the Center, and let the Side AB be fitted over in their Terms; so shall the Sector be opened in those Lines, to the Quantity of the Angle ACB.

### Use X.The Hypothenuse AC, of the right-angled Spherical Triangle ABC, being given, suppose 43 Degrees, and the Angle CAB, 20 Degrees, to find the Side CB.

As Radius is to the Sine of the given Hypothenuse 43 Degrees, So is the Sine of the given Angle CAB 20 Degrees, to the Sine of the Perpendicular CB.

Take either the lateral Sine of the given Angle CAB, 20 Degrees, and make it a parallel Radius; that is, take 20 Degrees from the Center on the Line of Sines, in your Compasses, and set that Extent from 90 to 90; then the parallel Sine of 43 Degrees, the given Hypothenuse, will, when measured from the Center on the Line of Sines, give 13 Deg. 30 Min. Or take the Sine of the given Hypothenuse AC, 43 Degrees, and make it a parallel Radius; and the parallel Sine of the given Angle CAB, taken and measured laterally on the Line of Sines, will give the Length of the Perpendicular CB, 13 Deg. 30 Min. as before.

### Use XI.The Perpendicular BC given, and the Hypothenuse AC, to find the Base AB.

As the Sine Complement of the Perpendicular BC, is to Radius, So is the Sine Complement of the Hypothenuse AC, to the Sine Complement of the Base required.

Make the Radius a parallel Sine of the given Perpendicular BC, viz. 76 Deg. 30 Min. and then the parallel Sine of the Complement of the given Hypothenuse, viz. 47 Degrees, measured laterally on the Line of Sines, will be found 49 Degrees, 25 Minutes: therefore the Complement of the required Base, will be 49 Degrees, 25 Minutes; and consequently the Base will be 40 Degrees, 35 Minutes.

The Use of the Sector in the Solution of the before-mentioned Cases of Trigonometry, being understood, it's Use in solving the other Cases, which I have omitted, will not be difficult.

Note, The several Uses of the Line of Lines, and Line of Polygons, on this Sector, are the same as the Uses of these Lines upon the French Sector, which see.

I now proceed to give some of the particular Uses of the Sector in Geometry, Projection of the Sphere, and Dialling.

## Section IV.

### Use I.To make any regular Polygon, whose Area shall be of a given Magnitude.

Let it be required to find the Length of one of the Sides of a regular Pentagon, whose superficial Area shall be 125 Feet, and from thence to make the Polygon.

Having extracted the square Root of $$\frac{1}{5}$$. Part of 125 (because the Figure is to have 5 Sides) which Root will be 5; make the Square AB, whose Side let be 5 Feet: then by means of the Line of Polygons (as directed by our Author in USE I. of the Line of Polygons) upon any right Line, as CD, make the Isosceles Triangle CGD so, that CG, being the Semi-diameter of a Circle, CD may be the Side of a regular Pentagon inscribed in it, and let fall the Perpendicular GE. Now continuing the Lines EG, and EC, make EF equal to the Side of the Square AB; and from the Point F, draw the right Line FH parallel to GC; then a mean Proportional between GE, and EF, will be equal to half the Side of the Polygon sought, which doubled, will give the whole Side. Now having found the Length of the whole Side, you must, upon the Line expressing it’s Length, make a Pentagon (as directed by our Author in USE II. of the Line of Polygons), which will have the required Magnitude.

### Use II.A Circle being given, to find the Side of a Square equal to it.

Let EF be the Diameter of the given Circle, which divide into 14 equal Parts, by means of the Line of Lines (as directed by our Author in the Use of the Line of equal Parts), then EP, which is 12.4 of those Parts, will be the Side of the Square sought. Note, 12.4 is the square Root of 11 × 14.

### Use III.A Square being given, to find the Diameter of a Circle equal to it.

Let AB be one Side of the given Square, which divide into 11 equal Parts, by means of the Line of Lines on the Sector; then continue the said Side, so that AG may be 12.4; that is, 1.4 of those Parts more, and the Line AG, will be the Diameter of a Circle, equal to the Square whose Side is AB.

### Use IV.The transverse and conjugate Diameters of an Ellipsis being given, to find the Side of a Square equal to it.

Let AB, and CD, be the transverse and conjugate Diameters of an Ellipsis: first, find a mean Proportional between the transverse and conjugate Diameters, which let be the Line EF; then divide the said Line EF, into 14 equal Parts, 12 and $$\frac{4}{10}$$ of which, will be EG, the Side of the Square equal to the aforesaid Ellipsis.

### Use V.To find the Magnitude of two right Lines which shall be in a given Ratio; about which, an Ellipsis being described, in taking them for the transverse and conjugate Diameters, the Area of the said Ellipsis, may be equal to a given Square.

Let the given Proportion that the transverse and conjugate Diameters are to have, be as 2 to 1; then divide the Side AB of the given Square, into 11 equal Parts. Now as 2 is to 1 (the Terms of the given Proportion), So is 11 × 14 = 154 to a fourth Number; the square Root of which being extracted, will be a Number to which, if the Line AG is taken equal (supposing one of those 11 Parts the Side of the Square is divided into, to be Unity), the said Line AG, will be the conjugate Diameter sought. Then to find the transverse Diameter, say, As 1 is to 2, So is the conjugate Diameter AG, to the transverse Diameter sought. To work the first of the said Proportions by the Line of Lines on the Sector, set 1 over as a Parallel on 2; then the parallel Extent of 154 taken, and laterally measured on the Line of Lines, will give 77, the fourth Proportional sought. In the same manner may the latter Proportion be worked.

### Use VI.To describe an Ellipsis, by having the transverse and conjugate Diameters given.

Let AB, and ED, be the given Diameters: take the Extent AC, or CB, between your Compasses, and to that Extent, open the Legs of the Sector so, that the Distance between 90 and 90 of the Line of Sines, may be equal to it: then may the Line AC be divided into a Line of Sines, by taking the parallel Extents of the Sine of each Degree, on the Legs of the Sector, between your Compasses, and laying them off from the Center C; the Line AC being divided into a Line of Sines (I have only divided it into the Sine of every 10 Degrees) from every of them raise Perpendiculars both ways. Now to find Points in the said Perpendiculars, through which the Ellipsis must pass, take the Extent of the semi-conjugate Diameter CE, between your Compasses; and then open the Sector so, that the Points of 90 and 90, on the Lines of Sines of the Sector, may be at that Distance from each other. This being done, take the parallel Sines of each Degree, of the Lines of Sines of the Sector, and lay them off, on those Perpendiculars drawn thro’ their Complements, in the Line of Sines AC, both ways from the said Line AC, and you will have two Points in each of the Perpendiculars thro’ which the Ellipsis must pass.

As for Example, the Sector always remaining at the same Opening, take the Distance from 80 to 80, on the Line of Sines, between your Compasses, and letting one Foot in the Point 10, on the Line AC; with the other make the Points a and b, in the Perpendicular passing thro’ that Point: then the Points a and b, will be the two Points in the said Perpendicular, thro’ which the Ellipsis must pass. All the other Points, in this manner, being found, if they are joined by an even Hand, there will be described the Semi Ellipsis DAE. In the same manner may the other half of the Ellipsis be described.

### Use VII.The Bearings of three Towers, standing at ABC, to each other being given, that is, the Angles ABC, BCA, and CAB; and also the Distances of each of the from a fourth Tower standing between them, as at D, being given; that is, BD, DC, and AD being given: to find the Distances of the Towers at ABC from each other; that is, to find the Lengths of the Sides AB, BC, AC, of the Triangle ABC.

Having drawn the Triangle EFG similar to ABC, divide the Side EG in the Point H; So that EH may be to HG, as AD is to DC; which may be done by taking the Sum of the Lines AD and DC between your Compasses, and setting that Extent over as a Parallel on the Line of Lines of the Sector, upon the Side EG of the Triangle, laterally taken on the Line of Lines; for then the parallel Extent of AD will give the Length of EH, and consequently the Point H will be had.

In like manner must the Side EF (or FG) be divided so in I, that EI may be to IF, as AD is to DB (or FG must be so divided, that the Segments must be as BD to DC).

Again, having continued out the Sides EG, EF, say, As EHHG is to HG, So is EH + HG to GK; and as EIIF is to IF, So let EI + IF be to FM, which Proportions may easily be worked by the Line of Lines on the Sector. This being done, bisect HK and IM, in the Points LN; and about the said Points, as Centers, and with the Distances LH and IN describe two Circles intersecting each other in the Point O; to which, from the Angles EFG, draw the right Lines EO, FO, and OG, which will have the same Proportion to each, other, as the Lines AD, BD, DC. Now if the Lines EO, FO, and GO are equal to the given Lines AD, BD, DC, the Distances EF, FG, and EG will be the Distances of the Towers sought. But if EO, OF, OG are lesser than AD, DB, DC, continue them out so, that PO, OR, and OQ be equal to them; then the Points P, Q, R being joined, the Distances PR, RQ, and PQ will be the Distances of the Towers sought. Lastly, If the Lines EO, OF, OG, are greater than AD, DB, DC, cut off from them Lines equal to AD, BD, DC, and join the Points of Section by three right Lines; then the Distances of the said three right Lines, will be the sought Distances of the three Towers.

Note, If EH be equal to HG, or EI to IF, the Centers L and N, of the Circles, will be infinitely distant from H and I; that is, in the Points H and I there must be two Perpendiculars raised to the Sides EF, EG, instead of two Circles, ’till they intersect each other: But if EH be lesser than HG, the Center L will fall on the other Side of the Base EG continued; understand the same of EI, IF.

### Use VIII.To project the Sphere Orthographically upon the Plane of the Meridian.

Let the Radius of the Meridian Circle, upon which the Sphere is to be projected, be AE; then divide the Circumference of the said Circle into four equal Parts in E, P, Æ, S, and draw the Diameters , PS; the former of which will represent the Equator, and the latter PS, the Hour-Circle of 6, as also the Axis of the World; P being the North-Pole, and S the South-Pole. Then must each Quarter of the Meridian be divided into 90 Degrees, by making the Extent from 60 to 60 of the Line of Chords, on the Sector, equal to the Radius of the Meridian Circle; and taking the parallel Extent of every Degree, and laying them off from the Equator towards the Poles; in which if 23 Deg. 30 Min. be numbered, (viz. the Sun’s greatest Declination) from E to ♋︎ Northwards, and from Æ to ♑︎ Southwards, the Line drawn from ♋︎ to ♑︎ will be the Ecliptick, and the Lines drawn Parallel to the Equator, through ♋︎ and ♑︎, will be the Tropicks.

Now if each Semidiameter of the Ecliptick be divided into Lines of Sines (by making the Distance of the Points of 90 and 90, on the Line of Sines of the Sector, equal to either of the Semidiameters, and taking out the parallel Extent of each Degree, and laying them off both ways from the Center A), the first 30 Degrees, from A towards 25, will stand for the Sign Aries; the 30 Degrees next following for Taurus; the rest for ♊︎, ♋︎, ♌︎, &c. in their Order.
If, again, AP, AS, are divided into Lines of Sines, and have the Numbers 10, 20, 30, &c. to 90 set to them, the Lines drawn thro’ each of these Degrees, parallel to the Equator, will represent the Parallels of Latitude, and shew the Sun’s Declination.

If, moreover, AE, are divided into Lines of Sines, and also the Parallels, and then there is a Line carefully drawn thro’ each 15 Degrees; the Lines so drawn will be Elliptical, and will represent the Hour-Circles; the Meridian PES the Hour of 12 at Noon; that next to it, drawn thro’ 75 Degrees from the Center, the Hours of 11 and 1; that which is drawn thro’ 60 Degrees from the Center, the Hours of 10 and 2, &c.

Then with respect to the Latitude, you may number it from E, Northwards, towards Z, and there place the Zenith (that is, make the Arc EZ 51 Deg. 32 Min. for London); thro’ which, and the Center, the Line ZAN being drawn, will represent the vertical Circle passing thro’ the Zenith and Nadir East and West; and the Line MAH, crossing it at right Angles, will represent the Horizon. These two being divided, like the Ecliptick and Equator, the Lines drawn thro’ each Degree of the Radius AZ, parallel to the Horizon, will represent the Circles of Altitude, and the Divisions in the Horizon, and it’s Parallels will give the Azimuths, which will be Ellipses.

Lastly, If thro’ 18 Degrees in AN, be drawn a right Line IK, parallel to the Horizon, it will show the Time of Day-breaking, and the End of Twilight. For an Example of this Projection, let the Place of the Sun be the last Degree of ♉︎, the Parallel passing thro’ this Place is LD, and therefore the Meridian Altitude will be ML; the Depression below the Horizon at Midnight HD; the semidiurnal Arc LC; the seminocturnal Arc CD; the Declination Ab; the ascensional Difference bC; the Amplitude of Ascension AC: The Difference between the End of Twilight, and the Break of Day, is very small; for the Sun’s Parallel hardly crosses the Line of Twilight.

If the Sun’s Altitude be given, let a Line be drawn for it parallel to the Horizon; so it shall cross the Parallel of the Sun, and there shew both the Azimuth and the Hour of the Day. As suppose the Place of the Sun being given, as before, the Altitude in the Morning was found, 20 Degrees, the Line FG, drawn parallel to the Horizon thro’ 20 Degrees in AZ, would cross the Parallel of the Sun in ☉︎; wherefore F☉︎ shews the Azimuth, and L☉︎ the Quantity of the Hour from the Meridian, which is about half an Hour past 6 in the Morning, and about half a Point from the East. The Distance of two Places may be also shewn by this Projection, in having their Latitudes and Difference of Longitude given.

For suppose a Place in the East of Arabia hath 20 Degrees of North Latitude, whose Difference of Longitude from London, by an Eclipse, is found to be five Hours and an half: Let Z be the Zenith of London, and the Parallel of Latitude for that other Place be LD, in which the Difference of Longitude is L☉︎; wherefore ☉︎ representing the Position of that Place, draw thro’ ☉︎ a Parallel to the Horizon MH, crossing the vertical AZ about 70 Degrees from the Zenith; which multiplied by 69, the Number of Miles in a Degree, gives 4830 Miles, the Distance of that Place from London.

### Use IX.To project the Sphere Stereographically upon the Plane of the Horizon; suppose for the Latitude of 51 Degrees, 32 Minutes.

Draw a Circle of any Magnitude at pleasure, as NE, SW, representing the Horizon; in which draw the two Diameters, WE, NS, eroding one another at right Angles, which will be the Representations of two great Circles of the Sphere crossing each other at right Angles in the Zenith. Let N represent the North, E the East, S the South, and W the West Part of the Horizon.

Note, In all these Projections, the Eye is commonly supposed to be in the Under-pole of the primitive Circle, projecting that Hemisphere which is opposed to the Eye, which will all fall within the primitive Circle; but that Hemisphere in which the Eye is, will all fall without the primitive Circle, and will run out in an infinite annular Plane, in the Plane of the Projection, and consequently cannot all of it be projected by Scale and Compass.

1. But now let us begin with projecting the Equinoctial. And here we must first determine the Line of Measures, in which the Center of this Circle will be; and this will be done by determining in what Points a Plane, perpendicular to the primitive Circle, will cut the Horizon, whether in the North and South, East and West, or in what other intermediate Points such a Plane shall cut it. The Pole of the World, in this Projection, is elevated 51 Deg. 32 Min. and consequently the Equinoctial, on the Northern Part of the Horizon, will fall below the Horizon, and it is the Southern Part which here must be projected, or which will fall within the primitive Circle; that Plane, whose Intersection with the Horizon shall produce the Line of Measures, will be the Plane of a Meridian passing thro’ the North and South Parts of the Horizon: wherefore NS will be the Line of Measures, in which the Center of the projected Equinoctial must fall; and since it is the Southern Part of the Equinoctial which we are to project, it’s Center will be towards the North.

To find whereabouts in the Line of Measures the said Center will fall, you must first open the Legs of the Sector, so that the Distance from 45 Degrees to 45 Degrees, on the Lines of Tangents, is equal to the Radius of the primitive Circle; then take the parallel Extent of the Tangents of 38 Deg. 28 Min. the Height of the Equinoctial above the Plane of the Horizon, and lay it off from Z to n, and n will be the Center of the projected Equinoctial; and the Secant of the same, 38 Deg. 30 Min. will give it’s Radius, with which the Circle WQE must be described, which is the Representation of that part of the Equinoctial which is above our Horizon, for the Latitude of 51 Deg. 32 Min.

2. We will next project the Ecliptick, which being a great Circle of the Sphere, must cut the Equinoctial at a Diameter’s Distance; that is, in E, W, the East and West Points of the Horizon, and consequently will have the same Line of Measures with that of the Equinoctial, viz. NS. Now let us consider whether the Center of the Ecliptick falls towards the North, or towards the South of the Horizon; and this will easily be determined, by considering that the Equinoctial is elevated above the Southern Part of the Horizon 38 Deg. 28 Min. and the Northern Part of the Ecliptick, or the Northern Signs, are elevated above the Equinoctial 23 Deg. 30 Min. which in all, make 62 Degrees, which is lesser than 90 Deg. So that it must fall towards the South, and consequently the Center must be Northwards, and will be found (the Sector remaining open as before), by setting off the Tangent of 62 Deg. from z to b, and the Secant of 62 Deg. will give it’s Radius; with which the Circle WCE, the Representation of the Northern half of the Ecliptick, must be described.

The Southern Part of the Ecliptick is likewise, for the most part, projected on the horizontal Projection, and made to fall within the primitive Circle; but this cannot be, the Globe remaining fixed: for that part of the Ecliptick, which is below the Horizon, will be thrown out of the primitive Circle; so that it cannot be projected, unless the Globe be supposed to be turned round, and by that means the Southern Part of the Ecliptick to be brought above the Horizon; but such a Revolution of the Sphere, where it makes any Alteration, is scarce allowable: however, I shall shew how it is usually projected.

The same Line of Measures NS remains still, and the Circle must fall to the South, and consequently it’s Center to the North of the Horizon; therefore nothing remains but to find it’s Elevation above the Horizon. The Northern Part of the Ecliptick falls 23 Deg. 30 Min. nearer the Zenith than the Equinoctial does; therefore the Southern Part, being brought above the Horizon, must be 23 Deg. 30 Min. nearer the Horizon than the Equinoctial: So that 23 Deg. 30 Min. being taken from 38 Deg. 28 Min. there remains 15 Deg. for the Distance of that part of the Ecliptick above the Horizon. It will be represented by WeE, which is described by setting off the Tangent of 15 Degrees for the Center, and taking the Secant of the same for the Radius.

3. NS produced will also be the Line of Measures for all Parallels of Declination, and Parallels of Latitude: for the Poles of lesser Circles being the same as those of the great Circles, to which they are parallel, it is manifest that the same Plane, which is at right Angles to the Equinoctial and Horizon, will also be at right Angles to all lesser Circles parallel to the Equinoctial, and the same will hold as to Circles parallel to the Ecliptick: But NS is the Line of Measures of the Equinoctial and Ecliptick, and consequently must be the Line of Measures of all Circles parallel to either of them; therefore the Centers of such lesser Circles will be in NS produced, if there be Occasion. Now to project them, for Instance, the Tropick of Cancer; consider, in this Position of the Sphere, what will be it’s nearest and greatest Distance from the Zenith, or the Pole of the primitive Circle, which you will find to be 28 Degrees; for the Equinoctial being elevated 38 Deg. 28 Min. above the Horizon, and the Tropick of Cancer being 23 Deg. 30 Min. from the Equinoctial, which, being added together, gives 62 Deg. which substracted from 90 Deg. leaves 28 Deg. it’s Distance from the Zenith on the South-side of the Horizon; therefore the Half-Tangent of 28 Deg. or the Tangent of 28 Deg. set from Z to C, will give one Extremity of it’s projected Diameter: Then the Distance from the Zenith to the Pole, being 38 Deg. 28 Min. and from the Pole to the Tropick of Cancer 66 Deg. 30 Min. the Sum of these, viz. 104 Deg. 58 Min. will be it’s greatest Distance from the Zenith; the Half-Tangent of which, set from Z to a, will give the other Extremity of it’s projected Diameter: therefore having got Ca the Diameter, bisect it, and describe the Circle ♋︎C♋︎.

The Tropick of Capricorn may be described in the same manner: for the Distance of the Equinoctial and the Zenith being 51 Deg. 32 Min. if to this be added 23 Deg. 30 Min. you will have 55 Deg. 2 Min. equal to the nearest Distance of the Tropick of Capricorn, on the South-side of the Horizon; the Half-Tangent of which being set from Z to e, will give one Extremity of it’s Diameter. Then the Distance between the Zenith and the Pole, viz. 38 Deg. 28 Mm and the Distance between the Pole and the Equinoctial, which is 90 Deg. and the Distance between the Equinoctial and the Tropick of Capricorn, which is 23 Deg. 30 Min. being all added together, will give the greatest Distance of the Tropick of Capricorn, from the Zenith, viz. 152 Deg. 2 Min. the Semi-tangent of which being set from Z towards the North will give the other Extremity of the Diameter. Bisect the Diameter found in e, and describe the Circle ♑︎C♑︎, which is the Representation of so much of the Tropick of Capricorn, as falls within the primitive Circle.

4. The Polar Circle is a 3 Deg. 30 Min. from the Pole; but the Pole being elevated, on the North-side the Horizon, 51 Deg. 32 Min. and 51 Deg. 32 Min. added to 23 Deg 30 Min. whose Sum is 75 Deg. 2 Min. is lesser than 90 Deg. so that it does not pass beyond the Zenith; therefore 75 Deg. 2 Min. taken from 90 Deg. leaves 15 Deg. which is the nearest Distance of the Polar Circle from the Zenith: And the Half-Tangent of 15 Deg. set from Z to v, will give one Extremity of it’s projected Diameter, and then, 15 Deg. added to 47 Deg. equal to 62 Deg. will be it’s greatest Distance from the Zenith: the Half-Tangent of which Distance, set from Z to P, will give the other Extremity of it’s projected Diameter; so that it’s Diameter vp being found, it is but bisecting it, and the Circle may be described.
5. I shall now shew how to project the Hour-Circles. And First, a Line of Measures must be determined, in which their Centers shall be, if possible; but you may easily discover it impossible for one Line of Measures to serve them all; for they are differently inclined to the Horizon, and so the Plane of no one great Circle can be at right Angles to the Horizon and all the Hour-Circles; therefore the Plane of a great Circle at right Angles to the Horizon, and one of them, must be found: which is possible, because the Hour-Circles being all at right Angles to the Plane of the Equinoctial, their Poles will be all found in this Circle; but the Poles of all great Circles, being 90 Degrees distant from their Planes, the Hour-Circle of 12, and the Hour-Circle of 6, must of necessity pass thro’ each other’s Poles, and so will be at right Angels to one another: But the Hour-Circle of 12 is at right Angles to the Horizon, and intersects it in NS; therefore the Line NS will be the Line of Measures, in which the Center of the Hour-Circle of 6 will be, and it’s Center will be towards the South Parts of the Horizon, because all the Hour-Circles pass thro’ the Pole which falls towards the North, the Elevation of this Circle above the Horizon being the same with that of the Pole, viz. 51 Deg. 32 Min. then take the Tangent of 51 Deg. 32 Min. and set it from Z to K; and upon the Center K, and with the Secant of the same Elevation, describe WPE, which is the Circle required.

The Point P, where NS, WE, intersect one another, is the Representation of the Pole of the World; for NS being the Representation of the Hour-Circle of 12, the projected Pole must be somewhere in this Line; but it must be somewhere in WE, which is likewise the Projection of an Hour-Circle: therefore it must be in that Point where these two projected Circles intersect one another, that is, in the Point P; P is the Point thro’ which all the Hour-Circles must pass in the Projection.

In order to draw the rest of the Hour-Circles, we must have recourse to a Secondary Line of Measures, which may thus be determined: To PK, at the Point K, erect DB at right Angles, and produce the Circle WPE, ’till it meet the Line DB, in the Points D and B; and the Line DB will be the secondary Line of Measures in which the Centers of all the Hour-Circles will be found; for let the Hour-Circle of 6, DPB, be considered as the primitive Circle, in whose Under-Pole (which will be in the Equinoctial) K, let the Eye be placed; then DB will be the Representation of the Equinoctial, for it passing thro’ the Eye will be projected into a right Line: but the Equinoctial is at right Angles to the Hour-Circles, both the primitive and all the rest; therefore it will be the secondary Line of Measures, upon this Supposition, upon which will be all their Centers. In order to find which, set the Sector to the Radius PK, then take off parallel-wise the Tangents of 15 Deg. 30 Deg. 45 Deg. the Elevations of the Hour-Circles above the Hour-Circle at 6, and set them both ways, from K to r, from K to s, from K to r, &c. then upon those Centers, and with the Secants of the same Elevations, describe the Circles PP, PQ, and PT, which will be the Hour-Circles; for they are all great Circles of the Sphere, passing thro’ the Pole P, and make Angles with one another of 15 Deg. or are 15 Deg. distant from each other: and the Portions of those Circles which fall within the primitive Circle NESW, as HPh, are the Representations of those Halves of the Hour-Circle, which are above our Horizon in our Latitude.

6. In like manner the Circles of Longitude may be drawn, by determining the secondary Line of Measures RS, in which all their Centers will be; and this Line will be determined after the same manner with DB above, and the Circles of Longitude drawn as before the Meridians were drawn: for the Line NS will be the Line of Measures, with respect to one of them passing thro’ E and W, the East and West Points of the Horizon. In order to draw this Circle, consider it’s Elevation above the Horizon, which will be found by considering the Distance of the Pole of the Ecliptick, from the Pole of the World, which will be 28 Deg. 2 Min. the Elevation of this Circle above the Horizon. Set the Tangent of 28 Deg. 2 Min. from Z to Q, and with the Secant of the same Distance, describe the Circle WpE; to pQ, at the Point Q, erect RS at right Angles, which will be the secondary Line of Measures. In this Line from Q (the Sector being set to pQ), set off the Tangents of 24 Deg. 40 Deg. according to the Number of Circles you have a Desire to draw, from Q to x, from Q to y, &c. and with the Secants of 20 Deg. 40 Deg. &c. describe the Circles of Longitude, MPm, &c.
7. The Representations of Azimuths, in this Projection, will be all right Lines, and any Number of them may be drawn, making any assigned Angles with one another, if the Limb be divided into it’s Degrees by help of the Sector, and thro’ these Degrees be drawn Diameters to the primitive Circle.
8. All Parallels of Altitude, in this Projection, will be Circles parallel to the primitive Circle, and may be easily drawn, by dividing a Radius of the primitive Circle, into Half-Tangents, and describing upon the Center Z, thro’ the Points of Division, concentrick Circles. I shall omit drawing of them, left the Scheme be too much perplexed.

### Use X.To project the Sphere Stereographically upon the Plane of the Solstial Colure for the Horizon of 51 Deg. 32 Min.

Draw the Circle HBOC, representing the primitive Circle; and the Diameter HO, representing the Horizon: Set off the Chord of 51 Deg. 32 Min. from O to P, having first set the Sector to the Radius of the Circle, which will give the Polar Point, and draw the Diameter Pp, representing the Hour-Circle of 6.

1. The Equinoctial may be represented, by drawing the Diameter EQ at right Angles to the Diameter Pp.
2. Set off 23 Deg. 30 Min. from the Chords, from E to ♋︎, and from Q to ♑︎︎, which will represent the Ecliptick.
3. The Tropicks of Cancer and Capricorn may be drawn thus: Take the Secant of 66 Deg. 30 Min. the Distance of each of them from their respective Poles, and set it both ways, from the Center A in Pp produced, which will give the two Points ee the Centers of the two Circles, and their Radii will be the Tangents of the same 66 Deg. 30 Min.
4. The Polar Circles, as also all other Parallels of Declination, may be drawn in the same manner.
5. The Line of Measures for the Azimuths will be HO, and the Line of Measures for the Almacanters will be BC.
6. ♋︎, ♑︎︎, or the Ecliptick, will be the Line of Measures for the Circles of Longitude, and the Line of Measures for the Circles of Latitude will be NS, all of which may be easily drawn from what is said in the precedent Use.
7. The Ecliptick may be divided into it’s proper Signs in this Projection, by setting off the Tangents of 15 Deg. 30 Deg. 45 Deg. both ways from A.

### Use XI.To draw the Hour-Lines upon an erect direct South Plane, as also on an Horizontal Plane.

First, draw the indefinite right Line CC, for the Horizon and Equator, and cross it at right Angles in the Point A, about the middle of the Line, with the indefinite right Line AB, serving for the Meridian, and the Hour Line of 12. then take out 15 Deg. from the Line of Tangents, on the Sector (the Sector being set to a parallel Radius lesser than the Extent from 45 Deg. to 45 Deg. of the lesser Lines of Tangents, when the Sector is quite opened), and lay them off in the Equator on both Sides from A, and one Point will serve for the Hour of 10, and the other for the Hour of 2. Again, Take out the Tangent of 30 Deg. (the Sector being opened to the same Radius), and lay it off on both Sides the Point A in the Equator, and one of these Points will serve one for the Hour of 10, and the other for the Hour of 2. In the same manner, lay off the Tangent of 45 Deg. for the Hours of 9 and 3, the Tangent of 60 Deg. for the Hours of 8 and 4, and the Tangent of 75 Deg. for the Hours of 7 and 5; But note, because the greater Tangents on the Sector run but to 45 Deg. therefore you must set the parallel Radius of the lesser Tangents, when you come above 45 Deg. to the Extent of the Radii of the greater Tangents.

Now if you have a mind to set down the Parts of an Hour, you must allow 7 Deg. 30 Min. for every half Hour, and 3 Deg. 45 Min. for one quarter. This done, you must consider the Latitude of the Place in which the Plane is, which suppose 51 Deg. 30 Min. then if you take the Secant of 51 Deg. 32 Min. off from the Sector, it remaining opened to the parallel Radius of the lesser Tangents, and set it off from A to V, this Point V will be the Center of the Plane; and if you draw from V, right Lines to 11, 10, 9, &c. and the rest of the Hour Points, they will be the required Hour Lines.

But if it happen, that some of these Hour Points fall out of the Plane, you may thus remedy yourself, by means of the larger Tangents.

At the Hour Points of 3 and 9, draw occult Lines parallel to the Meridian; then the Distances DC, between the Hour Line of 6, and the Hour Points of 3 and 9, will be equal to the Semi-diameter AV; and if they be divided in the same manner as the Line AC is divided, you will have the Points of 4, 5, 7, and 8, with their Halves and Quarters.

For take out the Semi-diameter AV, and make it a parallel Radius, by fitting it over in the Tangents of 45 and 45; then take the parallel Tangent of 15 Deg. and it will give the Distance from 6 to 5, and from 6 to 7. The Sector remaining thus opened, take out the parallel Tangent of 30 Deg. and it will give the Distance from 6 to 4, and from 6 to 8: the like may be done for Halves and Quarters of Hours.

The Hour Points may be otherwise denoted thus: Having drawn a right Line for the Equator, as before, and a (Turned the Point A for the Hour of 12, cut off two equal Lines A10, and A2, then upon the Distance between 10 and 2, make an equilateral Triangle, and you will have B for the Center of the Equator, and the Line AB, will give the Distance from A to 9, and from A to 3. This done, take out the Distance between 9 and 3, and this will give the Distance from B to 8, and from 8 to 7, and from 8 to 1: and again, from B to 4, and from 4 to 5, and from 4 to 11; so have you the Hour Points: and if you take out the Distances B1, B3, B5, &c. the Points may be found not only for the Half-Hours, but for the Quarters.

In the same manner are the Hour Lines drawn on a Horizontal Plane, only with this Difference, that AH is the Secant of the Complement of the Latitude, and the Hour Lines of 4, 5, 7, 8, are continued thro’ the Center.

### Use XII.To draw the Hour Lines upon a Polar Plane, as also on a Meridian Plane.

In a Polar Plane, the Equator may be also the same with the Horizontal Plane, and the Hour Points may be denoted as before, in the last Use: but the Hour Lines must be drawn parallel to the Meridian.

In a Meridional Plane, the Equator will make an Angle with the Horizontal Line, equal to the Complement of the Latitude of the Place; then may you assume the Point A, and there cross the Equator with a right Line, which will serve for the Hour Line of 6: then the tangent at 15 Deg. being laid off in the Equator on both Sides from 6, will give the Hour Points at 5 and 7; and the Tangent of 30 Deg. the Hour Points of 8 and 4; the Tangent of 45 Deg. the Hour Points of 3 and 9; the Tangent of 60 Deg. the Hour Points of 2 and 10: and lastly the Tangent of 75 Deg. will give the Hour Points of 1 and 11; and if right Lines are drawn thro’ these Hour Points, crossing the Equator at right Angles, these shall be the Hour Lines required.

### Use XIII.To draw the Hour Lines upon a vertical declining Plane.

First draw AV the Meridian, and AE the Horizontal Line, crossing one another in the Point A; then take out AV, the Secant of the Latitude of the Place, which suppose 51 Deg. 32 Min. and prick it down on the Meridian from A to V. Now because the Plane declines, which suppose 40 Deg. Eastward, you must make an Angle of the Declination upon the Center A, below the Horizontal Line, on the left Side of the Meridian, because the Plane declines Eastwards; for if it had declined Westward, the said Angle must have been made on the right Side of the Meridian. This being done, take AH, the secant Complement of the Latitude, out of the Sector, and prick it down in the Line of Declination from A to H, as was done for the Semi-diameter in the Horizontal Plane: then draw an indefinite right Line thro’ the Point A, perpendicular to AH, which will make an Angle, with the Horizontal Line, equal to the Plane’s Declination, and will be as the Equator in the Horizontal Plane. Again, take the Hour Points out of the Tangents, as in the last Problem, and prick them down in this Equator on both Sides, from the Hour of 12 at A; then lay your Ruler, and draw right Lines thro’ the Center H, and each of these Hour Points, and you will have all the Hour Lines of an Horizontal Plane, except the Hour of 6, which is drawn thro’ H perpendicular to HA. Lastly, you must note the Intersections that these Hour Lines make with AE, the Horizontal Line of the Plane, and then if right Lines are drawn thro’ the Center V, and each of these Intersections, they will be the Hour Lines required.

The Hour Points may be pricked down otherwise, thus: Take out the Secant of the Plane’s Declination, and prick it down in the Horizontal Line from A to E, and thro’ E draw right Lines parallel to the Meridian, which will cut the former Hour Lines of 3 or 9, in the Point C; then take out the Semi-diameter AV, and prick it down in those Parallels from C to D, and draw right Lines from A to C, and from V to D; the Line VD will be the Hour of 6: and if you divide those Lines AC, DC, in the same manner as DC is divided in the Horizontal Plane, the Hour Points required will be had.

Or you may find the Point D, in the Hour of 6, without knowing either H or C; for having pricked down AV in the Meridian Line, and AE in the Horizontal Line, and drawn Parallels to the Meridian thro’ the Points at E, take the Tangent of the Latitude out of the Sector, and fit it over in the Sines of 90 Deg. and 90 Deg. and the parallel Sine of the Plane’s Declination, measured in the same Tangent Line, will there shew the Complement of the Angle DVA, which the Hour Line of 6 makes with the Meridian: then having the Point D, take out the Semi-diameter VA, and prick it down in those Parallels from D to C; so shall you have the Lines DC, AC, to be divided, as before.

Thus have you the Use of the Sector applied in resolving several useful Problems. I might have laid down many more Problems in all the practical Parts of Mathematicks, wherein this Instrument is useful; but what I, and our Author have said of this Instrument, will, I believe, be sufficient to shew Persons skilled in the several practical Parts of Mathematicks, the Manner of using this Instrument therein.

For the Uses of the Lines of Numbers, Artificial Sines, and Tangents, as also the Lines of Latitude, Hours, and Inclination of Meridians; See Uses of Gunter’s Scale.