Mathematical Instruments

## Section I.

This Figure represents an Astronomical Quadrant upon it’s Pedestal, with it’s Limb curiously divided diagonally, and furnished with a fixed and moveable Telescope.

This Quadrant may be moved round horizontally, by turning a perpetual Screw fitted into the Pedestal: For as this Screw is turned about by means of a Key, at the same time it causes the Axis A to turn, by the falling in of its Threads between the Teeth of a strong thick Circle on the said Axis.

Behind the Quadrant is fixed, at Right Angles to its Plane, a strong thick Portion of a Circle greater than a Semicircle, having one Semi-circle of the outside thereof cut into Teeth. There is likewise another strong thick Portion of a Circle something greater than a Semi-circle behind the Quadrant, which is moveable upon two fixed Studs, at Right Angles to the former Portion; so that the Plane of this Portion may be parallel, inclined, or at Right Angles to the Plane of the Quadrant. On the side of this Portion, which is made flat next to the other fixed Portion, is a contrivance with a Screw and perpetual Screw, such that in turning the Screw the Threads of the perpetual Screw may be locked in between the Teeth of the fixed circular Portion; and by this means the Quadrant fixed to any Point, according to the direction of the Plane of the fixed Portion. And when the Quadrant is to be moved but a small matter in the aforesaid Direction, this may be done by turning the perpetual Screw with a Key.

The Outside of the abovementioned moveable circular Portion is cut into Teeth, and about the Center thereof the Axis A is moveable, according to the Direction of the Plane of the said Portion. In this Axis Aides a little Piece carrying a perpetual Screw, whose Threads, by means of a Trigger, may be locked in between the Teeth of the moveable circular Portion. And so when the Axis is set in the Pedestal, the Quadrant may be fixed to any Point, according to the Direction of the Plane of the said moveable Portion.

Therefore by these Contrivances the Quadrant may be readily fixed to any required Situation, for observing Celestial Phænomeua, without moving the Pedestal.

There is a Piece sliding on the Index, upon which the moveable Telescope is fastened, carrying a Screw and perpetual Screw; so that when the Telescope and Index are to be fixed upon any Point in the Limb of the Instrument, this may be done, by means of the Screw which locks the Threads of the perpetual Screw in between some of the Teeth cut round the curve Surface of the Limb of the Instrument: and when the Index and Telescope is to be moved a very minute Space backwards or forwards along the Limb, this is done by means of a Key turning a small Wheel fastened upon the aforenamed Piece, which is cut into a certain Number of Teeth, and whose Axis is at Right Angles to the Plane of the Quadrant; for this Wheel moves another (having the same Number of Teeth as that) which is at the end of the Cylinder whereon the perpetual Screw is: and by this means the perpetual Screw is turned about; and so the Index and Telescope may be moved a very minute Space backwards or forwards along the Limb. Note, The Number of Teeth the Curve Surface of the Limb is divided into, must be as great as possible, and the Threads of the perpetual Screw falling between them very fine; for the Exactness of the Instrument very much depends upon thus.

These Quadrants are commonly two Feet Radius, and all Brass, except the Pedestal and the perpetual Screws; the Telescopes have each two Glasses and Cross-hairs in their Foci; and for the Manner of dividing their Limbs, &c. See our Author’s Quadrants.

## Section II. Concerning a Micrometer.

This Micrometer is made of Brass: ABCg is a rectangular Brass Frame, the Side AB being about 3 Inches long, and the Side BC, as likewise the opposite Side Ag, are about 6 Inches; and each of these three Sides are $$\frac{1}{8}$$ of an Inch deep. The two opposite Sides of this Frame are screwed to the circular Plate, which we shall speak of by and by.

The Screw P having exactly 40 Threads in an Inch, being turned round, moves the Plate GDEF, along two Grooves made near the Tops of the two opposite Sides of the Frame; and the Screw Q having the same Number of Threads in an Inch as P, moves the Plate RNMY along two Grooves made near the bottom of the said Frame, in the same direction as the former Plate moves, but with half the Velocity as that moves with. These Screws are both at once turned, and so the said Plates moved along the same way, by means of a Handle turning the perpetual Screw S, whose Threads fall in between the Teeth of Pinions on the Screws P and Q. Note, Two and a half Revolutions of the perpetual Screw S, moves the Screw P exactly once round.

The Screw P turns the Hand a, fastened thereto over 100 equal Divisions made round the Limb of a circular Plate, to which the abovenamed two opposite Sides of the Frame are screwed at Right Angles. The Teeth of the Pinion of the Screw P, whose Number are 5, takes into the Teeth of a Wheel, on the backside of the circular Plate, whose Number are 25. Again, On the Axis of this Wheel is a Pinion of two, which takes into the Teeth of another Wheel moving about the Center of the circular Plate, without side the same, having 50 Teeth. This last Wheel moves the lesser Hand b once round the abovenamed circular Plate, in the $$\frac{1}{100}$$ part of the time the Hand a is moving round: for because the Number of Teeth of the Pinion on the Screw P, are 5, and the Number of Teeth of the Wheel this Pinion moves round, are 20; therefore the Screw P moves four times round in the same time the said Wheel is moving once round. Again, Since there is a Pinion of two takes into the Teeth of a Wheel, whose Number are 50, therefore this Wheel with go Teeth will move once round in the same time that the Wheel of 20 Teeth hath moved twenty-five times round; and consequently the Screw P, or Hand a, must move a hundred times round in the same time as the Wheel of 50 Teeth, or the Hand b, hath moved once round.

It follows from what hath been said, that if the circular Plate W, which is fastened at Right Angles to the other circular Plate, be divided into 200 equal Parts, the Index x to which the Handle is fastened, will move five of these Parts in the same time that the Hand a has moved one of the hundred Divisions round the Limb of the other circular Plate: and so by means of the Index x, and Plate W, every fifth Part of each of the Divisions round the other Plate may be known.

Moreover, Since each of the Screws P and Q have exactly 40 Threads in an Inch; therefore the upper Plate GDEF will move 1 Inch, when the Hand a hath moved forty times round, the four thousandth part of a Inch, when the said Hand hath moved over one of the Divisions round the Limb, and the twenty thousandth part of an Inch, when the Index x hath moved one part of the 200 round the Limb of the circular Plate W; and the under Plate RNMY, half an Inch, the two thousandth part of an Inch, and the ten thousandth part of an Inch the same way, in the said respective times.

Hence, if the under Plate, having a large round Hole therein, be fixed to a Telescope, so that the Frame may be moveable together with the whole Instrument, except the said lower Plate, and the strait smooth Edge HI, of the fixed narrow Plate ABIH, as likewise the strait smooth Edge DE of the moveable Plate GDEF, be perceivable thro’ the round Hole in the under Plate, in the Focus of the Object-Glass; then when the Handle of the Micrometer is turned, the Edge HI of the narrow Plate ABIH, fixed to the Frame, and DE of the moveable Plate, will appear thro’ the Telescope equally to accede to, or recede from, each other. And so these Edges will serve to take the apparent Diameters of the Sun, Moon, &c. the manner of doing which is thus: Suppose in looking at the Moon thro’ the Telescope, you have turned the Handle ’till the two Edges DE and HI are opened, so as to just touch or clasp the Moon’s Edges; and that there was twenty-one Revolutions of the Hand a to compleat that Opening. First say, As the focal Length of the Object-Glass, which suppose 10 Feet, is to Radius, So is 1 Inch to the Tangent of an Angle subtended by 1 Inch in the Focus of the Object-Glass, which will be found 28 Min. 30 Sec. Again, Because there are exactly 40 Threads of the Screws in one Inch, say, If forty Revolutions of the Hand a give an Angle of 28 Min. 38 Sec. what Angle will twenty-one Revolutions give? The Answer will be 15 Min. 8 Sec. and such was the Moon’s apparent Diameter, and so may the apparent Diameters of any distant Objects be taken.

It is to be observed, that the Divisions upon the top of the Plate GDEF, are Diagonal Divisions of the Revolutions of the Screws, with Diagonal Divisions of Inches against them; and so as the said Plate Aides along, these Diagonals are cut by Divisions made on the Edge of the narrow Plate KL, fixed to the opposite Sides of the Frame by means of two Screws. These Diagonal Divisions may serve to count the Revolutions of the Screws, and to shew how many there are in an Inch, or the Parts of an Inch.

## Section III. Of the Construction of Gunter’s Quadrant.

This Quadrant, which is partly a Projection, that is, the Equator, Tropicks, Ecliptick, and Horizon, are stereographically projected upon the Plane of the Equinoctial, the Eye being supposed to be placed in one of the Poles, may be thus made.

About the Center A describe the Arc CD, which may represent either of the Tropicks. Again, Divide the Semidiameter AT so in E, that AE being Radius, AT may be the Tangent of 56 Deg. 46 Min. half the Sun’s greatest Declination above the Radius or Tangent of 45 Deg. To do which, say, As the Tangent of 56 Deg. 46 Min is to 1000; So is Radius to 655: therefore if AT be made 1000 equal Parts, AE, the Radius of the Equator, will be 655 of those Parts. And if about the Center A, with the Distance AE, the Quadrant EF be described, this will serve for the Equinoctial.

Now to find the Center of the Ecliptick, which will be somewhere in the left Side of the Quadrant AD (representing the Meridian) you must divide AD so in G, that if AF be the Radius, AG may be the Tangent of 23. Deg. 30 Min. the Sun’s greatest Declination; therefore if AF be 1000, AG will be 434. And if about the Center G, with the Semidiameter GD, an Arc ED be described, this will be $$\frac{1}{4}$$ of the Ecliptick. And to divide it into Signs and Degrees, you must use this Canon, viz. As Radius is to the Tangent of any Degree’s distance from the nearest Equinoctial Point, So is the Co-sine of the Sun’s greatest Declination to the Tangent of that Degree’s Right Ascension, which must be counted on the Limb from the Point B, by which means the Quadrant of the Ecliptick may be graduated.

As, for Example, The Right Ascension of the first Point of ♉︎ being 27 Deg. 54 Min. lay a Ruler to the Center A, and 27 Deg. 54 Min. on the Limb, from B towards C, and where it cuts the Ecliptick, will be the first Point of ♉︎; and so for any other.

The Line ET, between the Equator and the Tropick, which is called the Line of Declination, may be divided into 23 Deg. 30 Min. in laying off from the Center A, the Tangent of each Degree added to 45 Deg, the Line AE being supposed the Radius of the Equinoctial. As suppose the Point for 10 Degrees of Declination be to be found, add 5 Deg. (half 10.) to 45 Deg. and the Sum will be 50 Deg. the Tangent of which will be (supposing the Radius 1000) 1192: therefore laying 1192 Parts from A, or 192 from E, and you will have a Point for 10 Degrees of Declination; and so for others.

Most of the principal Stars between the Equator and Tropick of Cancer, maybe put on the Quadrant by means of their Declination, and Right Ascension. As suppose the Wing of Pegasus be 13 Deg. 7 Min. and the Right Ascension 358 Deg. 34 Min. from the first Point of Aries. Now if about the Center A, you draw an occult Parallel thro’ 13 Deg. 7 Min. of Declination, and then lay a Ruler from the Center A thro’ 1 Deg. 26 Min. (the Complement of 358 Deg. 34 Min. to 360 Deg.) in the Limb BC, the Point where the Ruler cuts the Parallel, will be the Place for the Wing of Pegasus, to which you may set the Name, and the Time when he comes to the South.

There being Space sufficient between the Equator and the Center, you may there describe the Quadrant, and divide each of the two Sides furthest from the Center into 100 Parts; so shall the Quadrant be generally prepared for any Latitude. But before the particular Lines can be drawn, you must have tour Tables fitted for the Latitude the Quadrant is to serve in.

First, A Table of Meridian Altitudes for the Division of the Circles of Days and Months, which may be thus made: Consider the Latitude of the Place, and the Sun’s Declination for each Day of the Year; if the Latitude and Declination be both North, or both South, add the Declination to the Complement of the Latitude; if they be one North and the other South, substract the Declination from the Complement of the Latitude, and you will have the Meridian Altitude for that Day. As, in the Latitude of 51 Deg. 32 Min. North, whose Complement is 38 Deg. 28 Min. the Declination on the 10th of June will be 23 Deg. 30 Min. North; therefore add 23 Deg. 30 Min. to 38 Deg. 28 Min. and the Sum will be 61 Deg. 58 Min. the Meridian Altitude on the 10th of June. Again, The Declination on the 10th of December will be 23 Deg. 30 Min. South; wherefore take 23 Deg. 30 Min. from 38 Deg. 28 Min. and the Remainder will be 14 Deg. 58 Min. the Meridian Altitude on the 10th of December. And in this manner may the Meridian Altitude for each Day in the Year be found, and put in a Table.

Your Table being made, you may inscribe the Months and Days of each Month on the Quadrant, in the Space left below the Tropick. As, Laying a Ruler upon the Center A, and 16 Deg. 42 Min. the Sun’s Meridian Altitude on the 1st of January, in the Limb BC, you may draw a Line for the end of December and beginning of January. Again, Laying a Ruler to the Center A, and 24 Deg. 34 Min. the Sun’s Altitude at Noon the end of January, or first of February, on the Limb, and you may draw a Line for that Day. And so of others.

Now to draw the Horizon, you must find its Center, which will be in the Meridian Line AC; and it the Point H be taken such, that if AH be the Tangent Complement of the Latitude, viz. of 38 Deg. 28 Min. AF being supposed Radius; or if AF be supposed 1000, and AH 776 of those Parts, then will H be the Center of the Horizon. Therefore if about the Center H, with the Distance HE, an Arc be described cutting the Tropick TD, the said Arc will represent the Horizon.

The next thing done, must be to make a Table for the Division of the Horizon, which may be done by this Canon, viz. As Radius is to the Sine of the Latitude, So is the Tangent of any Number of Degrees in the Horizon (which will be not more than 40 in our Latitude) to the Tangent of the Arc in the Limb which will divide the Horizon.

As in our Latitude, 7 Deg. 52 Min. belong to 10 Deg. of the Degrees of the Horizon; therefore laying a Ruler to the Center A, and 1 Deg. 52 Min. in the Limb BC, the Point where the Ruler cuts the Horizon, will be 10 Deg. in the Horizon; and so of the rest. But the Lines of Distinction between every 5th Degree are best drawn from the Center H.

The third Table for drawing the Hour-Lines, must be a Table of the Sun’s Altitude above the Horizon at every Hour, especially when he comes to the Equator, Tropicks, and other intermediate Declinations. If the Sun be in the Equator, and so have no Declination, as Radius to the Co-sine of the Latitude, so is the Co-sine of any Hour from the Meridian to the Sine of the Sun’s Altitude at that Hour.

But if the Sun be not in the Equator, you must say, As the Co-sine of the Hour from the Meridian is to Radius, So is the Tangent of the Latitude to the Tangent of a 4th Arc. Then consider the Sun’s Declination, and the Hour proposed; it the Latitude and Declination be both alike, and the Hour fall between Noon and Six, substract the Declination from the aforesaid 4th Arc, and the Remainder will be a 5th Arc.

But if the Hour be either between Six and Midnight, or the Latitude and Declination unlike, add the Declination to the 4th Arc, and the Sum will be a 5th Arc. Then as the Sine of the fourth Arc is to the Sine of the Latitude, so is the Co-sine of the 5th Arc to the Sine of the Altitude sought.

Lastly, You may find the Sun’s Declination when he rises or sets, at any Hour, by this Canon, viz. As Radius is to the Sine of the Hour from Six, So is the Co-tangent of the Latitude to the Tangent of the Declination.

As in our Latitude you will find, that when the Sun rises at five in the Summer, or seven in the Winter, his Declination is 11 Deg. 36 Min. whence you will find the Sun’s Meridian Altitude in the beginning of ♋︎ will be 61 Deg. 58 Min. in ♊︎ 58 Deg. 40 Min. in ♉︎ 49 Deg. 58 Min. in ♈︎ 38 Deg. 30 Min. &c. but the beginning of ♋︎ and ♍︎ is represented by the Tropick TD, drawn thro’ 23 Deg. 30 Min. of Declination, and the beginning of ♈︎ and ♎︎ by the Equator EF. Now if you draw an occult Parallel between the Equitor and the Tropick, at 11 Deg. 30 Min. of Declination, it shall represent the beginning of ♉︎, ♍︎, ♏︎, and ♓︎. If you draw another occult Parallel thro’ 20 Deg. 12 Min. of Declination, it will represent the beginning of ♊︎, ♑︎, ♐︎ and ♒︎.

Then lay a Ruler from the Center A thro’ 61 Deg. 58 Min. of Altitude in the Limb BC, and note the Point where it crosses the Tropick of ♋︎. Then move the Ruler to 58 Deg.

40 Min. and note where it crosses the Parallel of ♊︎; then to 49 Deg. 58 Min. and note where it crosses the Parallel of ♉︎; and again to 38 Deg. 28 Min. noting where it crosses the Equator; and a Line drawn thro’ these Points will represent the Line of 12 in the Summer while the Sun is in ♈︎, ♉︎, ♊︎, ♋︎, ♑︎, or ♍︎. In like manner, if you lay a Ruler to A and 26 Deg. 58 Min. in the Limb, and note the Point where it crosses the Parallel of ♓︎; then move it to 18 Deg, 16 Min. and note where it crosses the Parallel of ♒︎. And again, to 14 Deg, 58 Min. noting where it crosses the Tropick of ♑︎; the Line drawn thro’ these Points shall shew the Hour of twelve in the Winter. And in this manner may the rest of the Hour-Lines be drawn, only that of seven from the Meridian in Summer, and five in the Winter, will cross the Line of Declination, at 11 Deg. 35 Min. and that of eight in the Summer, and four in the Winter, at 21 Deg. 38 Min.

The fourth Table for drawing of the Azimuth Lines must also be made for the Altitude of the Sun above the Horizon, at every Azimuth, especially when the Sun comes to the Equator, Tropicks, and some other intermediate Declinations.

If the Sun be in the Equator, and so has no Declination, as Radius to the Co-sine of the Azimuth from the Meridian; so is the Tangent of the Latitude to the Tangent of the Sun’s Altitude at the Azimuth in the Equator.

If the Sun be not in the Equator, as the Sine of the Latitude is to the Sine of the Declination, so is the Co-sine of the Sun’s Altitude at the Equator, at a given Azimuth, to the Sine of a 4th Arc.

Now when the Latitude and Declination are both alike in all Azimuths, from the Prime Vertical to the Meridian, add this 4 th Arc to the Arc of Altitude at the Equator But when the Azimuth is above 90 Degrees distant from the Meridian, take the Altitude at the Equator from this 4th Arc When the Latitude and Declination are unlike, take the said 4th Arc from the Arc of Altitude at the Equator, and then you will have the Sun’s Altitude for a proposed Azimuth.

Lastly, When the Sun rises or sets upon any Azimuth, to find his Declination, say, As Radius to the Co-sine of the Latitude, So is the Co-sine of the Azimuth from the Meridian, to the Sine of the Declination.

Now a Table being made according to the arforesaid Directions, if you would draw the Line of East or West, which is 90 Degrees from the Meridian, lay a Ruler to the Center A and 30 Deg 38 Min. numbered in the Limb from C towards B, and note the Point where it crosses the Tropick of ♋︎, then move the Ruler to 26 Deg. 10 Min. and note where it crosses the Parallel of ♊︎; then to 14 Deg. 45 Min. and note where it crosses the Parallel of ♉︎; then to 0° and 0°, and you will find it cross the Equator in the Point F; then a Line drawn thro’ these Points will be the East and West Azimuth. And so may all the other Azimuths be drawn.

These Lines being thus drawn, if you set two Sights upon the Line AC, and at the Center A hang a Thread and Plummet, with a Bead upon the Thread, the foreside of the Quadrant will be finished.

## Section IV. The Use of Gunter’s Quadrant.

### Use I.The Sun’s Place being given, to find his Right Ascension, and contrariwise.

Let the Thread be laid upon the Sun’s Place in the Ecliptick, and the Degrees which it cuts in the Limb, will be the Right Ascension sought.

For example; Suppose the Sun’s Place be the 4th Degree of ♊︎, the Thread laid on this Degree will cut 62 Deg. in the Limb, which is the Right Ascension required. But if the Sun’s Place be more than 90 Deg. from the beginning of Aries, the Right Ascension will be more than 90 Deg. And in this Case the Degrees cut by the Thread must be taken from 180, to have the Right Ascension.

Now if the Sun’s Right Ascension be given, to find its Place, lay the Thread on the Right Ascension, and it will cross the Sun’s Place in the Ecliptick.

### Use II.The Sun’s Place being given, to find his Declination, and contrariwise.

Lay the Thread, and set the Bead to the Sun’s Place in the Ecliptick; then move the Thread to the Line of Declination, and there the Bead will fall upon the Degrees of the Line of Declination sought.

For example; Let the Sun’s Place be the 4th Degree of ♊︎, the Bead being first set to this Place, move the Thread to the Line of Declination, and there you will find the Sun’s Declination 21 Deg. from the Equator.

Now the Sun’s Place being sought, in having the Declination given, you must first lay the Thread and Bead to the Declination, and then the Bead moved to the Ecliptick will give the Sun’s Place sought.

### Use III.The Day of the Month being given, to find the Sun’s Meridian Altitude, and contrariwise.

Lay the Thread to the Day of the Month, and the Degrees which it cuts in the Limb will be the Sun’s Meridian Altitude.

Suppose the Day given be May the 15th, the Thread laid upon this Day will cut 59 Dew. 30 Min. the Meridian Altitude sought.

Again, If the Thread be let to the Meridian Altitude, it will fall upon the Day of the Month.

As, suppose the Sun’s Meridian Altitude be 59 Deg. 30 Min. the Thread set to this Altitude falls upon the 15th Day of May, and the 9th of July; and which of those two is the true Day, may be known by the Quarter of the Year, or by another Day’s Observation: for it the Sun’s Altitude be greater, the Thread will fall upon the 16th of May, and the 8th of July; and if it prove lesser, then the Thread will fall on the 14th of May, and the 10th of July; whereby the Question is fully answered.

### Use IV.The Sun’s Altitude being given, to find the Hour of the Day, and contrariwise.

Having put the Bead to the Sun’s Place in the Ecliptick, observe the Sun’s Altitude by the Quadrant; and then if the Thread be laid over the same in the Limb, the Bead will fall upon the Hour required. For example; Suppose on the 10th of April, the Sun being then in the beginning of Taurus, I observe his Altitude by the Quadrant to be 36 Deg. place the Bead to the beginning of Taurus in the Ecliptick, and afterwards lay the Thread over 36 Degrees of the Limb; then the Bead will fall upon the Hour-Line of 9 and 3: and so the Hour is 9 in the Morning, or 3 in the Afternoon. Again, If the Altitude be near 40 Degrees, the Bead will fall half way between the Hour-Line of 9 and 3, and the Hour-Line of 10 and 2. Wherefore it must be either half an Hour past 9 in the Morning, or half an Hour past two in the afternoon; and which of these is the true Time of the Day, may be known by a second Observation: For if the Sun rises higher, it is Morning, and if it becomes lower, it is Afternoon.

Now to find the Sun’s Altitude by having the Hour given, you must lay the Bead upon the Hour given (having first rectified or put it to the Sun’s Place) and then the Degrees of the Limb cut by the Thread, will be the Sun’s Altitude sought.

Note, The Bead may be rectified otherwise, in bringing the Thread to the Day of the Month, and the Bead to the Hour-Line of 12.

### Use V.To find the Sun’s Amplitude either for setting, when the Day of the Month or the Sun’s Place is given.

Let the Bead rectified for the time, be brought to the Horizon; and there it will shew the Amplitude sought. If, for example, the Day given be the 4th of May, the Sun will then be in the 4th Degree of Gemini. Now if the Bead be rectified and brought to the Horizon, it will there fall on 35 Deg. 8 Min. and this is the Sun’s Amplitude of rising from the East, and of his setting from the West.

### Use VI.The Day of the Month or Sun’s Place being given, to find the Ascensional Difference.

Rectify the Bead for the given time, and afterwards bring it to the Horizon; then the Degrees cut by the Thread in the Limb will be the ascensional Difference. And if the ascensional Difference be converted into time, in allowing an Hour for 15 Degrees, and four Minutes of an Hour for one Degree, then we shall have the time of the Sun’s rising before six in the Summer, and after six in the Winter: and consequently the Length of Day and Night may be known by this means.

### Use VII.The Sun’s Altitude being given, to find his Azimuth, and contrariwise.

Rectify the Bead for the time, and observe the Sun’s Altitude. Then bring the Thread to the Complement of that Altitude, and so the Bead will give the Azimuth sought upon or among the Azimuth Lines.

And to find the Altitude by having the Azimuth given, having rectified the Bead to the Time, move the Thread ’till the Bead falls on the given Azimuth; then the Degrees of the Limb cut by the Thread, will be the Sun’s Altitude at that time.

### Use VIII.The Altitude of any one of the five Stars on the Quadrant being given, to find the Hour of the Night.

First, Put the Bead to the Star, which you intend to observe, and find how many Hours he is from the Meridian by Use IV. then from the Right Ascension of the Star, substract the Sun’s Right Ascension converted into Hours, and mark the Difference: for this Difference added to the observed Hour of the Star from the Meridian, will shew how many Hours the Sun is gone from the Meridian, which is in effect the Hour of the Night.

For Example; The 15th of May, the Sun being in the 4th Degree of Gemini, I set the Bead to Arcturus, and observing his Altitude, find him to be in the West, about 52 Deg. high, and the Bead to fall upon the Hour-Line of two after Noon; then the Hour will be 11 Hours 50 Min. past Noon, or 10 Minutes short of Midnight. For 62 Deg. the Sun’s right Ascension, converted into Time, makes 4 Hours 8 Min. which if we take out of 13 Hours 58 Min. the right Ascension of Arcturus, the Difference will be 9 Hours 50 Min. and this being added to two Hours, the observed Distance of Arcturus from the Meridian, shews the Hour of the Night to be 11 Hours 50 Min.

Thus have I briefly shewn the Manner of solving several of the chief and must useful Astronomical Problems, by means of this Quadrant. As for the Manner of taking Altitudes in Decrees, as likewise the Use of the Quadrant, see our Author‘s Quadrant.

There are other Quadrants made by Mr Sutton long since; one of which (being in my Opinion the best) is a Stereographical Projection of $$\frac{1}{4}$$ of those Circles, or quarter of the Sphere between the Tropicks, upon the Plane of the Equinoctial, the Eye being in the North Pole.

The said quarter on the Quadrant, is that between the South part of the Meridian, and Hour of Six, which will leave out all the outward Part of the Almicanters between it and the Tropick of Capricorn; and instead thereof, there is taken in such a like Part of the depressed Parallels to the Horizon, between the same Hour of Six and Tropick of Capricorn, for the Parallels of Depression have the same Respect to the Tropick of Capricorn, as the Parallels of Altitude have to the Tropick of Cancer, and will produce the same Effect.

This Projection is fitted for the Latitude of London: and those Lines therein that run from the Right-hand to the Left, are Parallels of Altitude; and those which cross them, are Azimuths. The lesser of the Circles that bounds the Projection, is one fourth of the Tropick of Capricorn, and the other one fourth of the Tropick of Cancer. There are also the two Eclipticks drawn from the same Point in the left Edge of the Quadrant, with the Characters of the Signs upon them; as likewise the two Horizons from the same Point. The Limb is divided both into Degrees and Time, and by having the Sun’s Altitude given, we may find the Hour of the Day to a Minute by this Quadrant.

The Quadrantal Arcs next the Center contain the Calendar of Months; and under them in another Arc is the Sun’s Declination: so that a Thread laid from the Center over any Day of the Month, will fall upon the Sun’s Declination that Day in this last Arc, and on the Limb upon the Sun’s right Ascension for that same Day. There are several of the most noted fixed Stars between the Tropics, placed up and down in the Projection; and next below the Projection is the Quadrat and Line of Shadows, being only a Line of natural Tangents to the Arcs of the Limb; and by help thereof, the Heights of Towers, Steeples, &c. may be pretty exactly taken.

Now the Manner of using this Projection in finding the Time of the Sun’s rising or setting, his Amplitude, Azimuth, the Hour of the Day, &c. is thus: Having laid the Thread to the Day of the Month, bring the Bead1 to the proper Ecliptick (which is called rectifying it), and afterwards move the Thread, and bring the Bead to the Horizon: then the Thread will cut the Limb in the Time of the Sun’s rising or setting, before or after Six. And at the same Time the Bead will cut the Horizon in the Degrees of the Sun’s Amplitude. Again, Suppose the Sun’s Altitude on the 24th of April be observed 45 Degrees, What will the Hour and Azimuth then be? Having laid the Thread over the 24th of April, bring the Bead to the Summer Ecliptick, and then carry it to the Parallel of the Altitude 45 Degrees: and then the Thread will cut the Limb at 55 Deg. 15 Min. and so the Hour will be either 41 Min. past Nine in the Morning, or 19 Min. past Two in the Afternoon. And the Bead among the Azimuths shews the Sun’s Distance from the South to be 50 Deg. 41 Min.

Note, If the Sun’s Altitude be less than that which it hath at six o’Clock, on any given Day; then the Operation must be performed among those Parallels above the upper Horizon, the Bead being rectified to the Winter Ecliptick.

There are a great many other Uses of this Quadrant, which I shall omit, and refer you to Collins’s Sector upon a Quadrant, wherein it’s Description and Use, together with those of two other Quadrants, are fully treated of.

1. 1The Bead is a little round Leaden or Brass Shot, with a small Hole thro’ it, that moves stiffly up an down, so as to remain at any Point or Part of the Thread of a Plummet fastened to the Centre of the Quadrant.