The sector on a quadrant, or A treatise containing the description and use of four several quadrants two small ones and two great ones, each rendred many wayes, both general and particular. Each of them accomodated for dyalling; for the resolving of all proportions instrumentally; and for the ready finding the hour and azimuth universally in the equal limbe. Of great use to seamen and practitioners in the mathematicks. Written by John Collins accountant philomath. Also An appendix touching reflected dyalling from a glass placed at any reclination.

General Proportions for the Hour.

The Proportion selected for the Hour is,

• As the Radius,
• Is to the Cosine of the Latitude,
• So is the Cosine of the Declination,
• To a fourth:

Namely, the difference of the Sines of the Meridian Altitude, and of the Altitude at 6.

Again,

1.

• As that fourth,
• Is to the Radius,
• So in Summer, is the difference; but in VVinter, the sum of the Sines of the Suns Altitude or Depression at 6,
• To the Sine of the Hour from 6 towards Noon or Midnight, according as the Altitude or Depression is greater or less then the Altitude or Depression at 6.

2.

• And so is the difference of the Sines of the Meridian, and proposed Altitude,
• To the Versed Sine of the Hour from Noon:
• And so is the sum of the Sines of the Midnight Depression, and given Altitude,
• To the Versed Sine of the Hour from Midnight.

3.

• And so is the Sine of the Altitude,
• To the difference of the Versed Sines of the Semidiurnal Ark, and of the Hour sought.

By the first Proportion, the hour may be found generally, either in the equal Limb, or Line of Sines.

By the second Proportion, it may be found generally, either in the Versed Sines of 90^d, or 180^d.

By the third Proportion, it may be found in the Line of Versed Sines issuing from the Center in many cases. I shall add a brief Application of all three ways.

The first Work will be to find the point of Entrance.

Example, For the Latitude of Nottingham, 53^d.

Lay the thred to the Declination, admit 20 in the Limb, counted from the left edge, and from the Latitude in the Line of Sines, counted towards the Center from 90^d; take the nearest distance to the thred, the said extent measured from the Center, will fall upon 34 25′, and there will be the point of Entrance; let it be recorded, or have a mark set to it.

If the Suns Declination be North, the Meridian Altitude in that Latitude, will be 57^d, the said extent will reach from the Sine there∣of, to the Sine of the Suns Altitude, or Depression at 6, to that Declination, namely, to 15^d 51′: which may also be found with∣out

the Meridian Altitude, by taking the distance from 20^d in the Sines, to the thred laid over the Arch 53^d, counted from the right Edge, and by measuring that extent from the Center, the point thus found, I call the Sine point.

Thirdly, If the respective distances between the Sine point, and the Sine of the given Altitude, be taken and entred upon the point of Entrance, laying the thred to the other foot, according to near∣est distance, the hour may be found all day for that Declination, when it is North in the equal Limb.

Example, For the Latitude of Nottingham, to the former Declination, being North.

When the Sun hath

• 11^d 31′
• 20 17

of Altitude, the Hour in each Case will be found half an hour from 6, to the lesser Altitude be∣yond it, towards Midnight; to the greater, towards Noon.

And when the Altitude is 38^d 19′, the time of the day will be either half an hour past 8 in the morning, or half an hour past 3 in the Afternoon.

An Example for the Latitude of Nottingham, when the Declination is as much South.

Let the Altitude be 10^d 6′, In this case add the Sine thereof to the Sine of 15^d 51′, the whole extent will be equal to the Sine of 26^d 39′; Enter this Extent upon the point of Entrance at 34^d 25′ laying the thred to the other foot, according to nearest distance, and the time of the day found in the Limb, will be either half an hour past 9 in the morning, or half an hour past 2 in the afternoon.

An Example for working the second Proportion.

The Summer Meridian Altitude is 57^d, if the given Altitude be 46^d 11′, take the distance between the Sines of these two Arks, and entring this extent upon the point of Entrance, lay the thred to the other foot, according to nearest distance, it will in the Versed

Sine of 90^d, shew the Hour from Noon to be 37^d 30′, that is, ei∣ther half an hour past 10 in the morning, or half an hour past 1 in the Afternoon.

And when the Hour falls near Noon, we may double or triple the extent of the Compasses, and find an Ark in the Limb, which if counted from the other edge, and the thred laid over it▪ will give answer in the Versed Sines doubled or tripled accordingly.

A third Example.

If the Altitude were 3^d 15′, in this case the distance between it and the Meridian Altitude being greater then the distance of the point of Entrance from the Center, the hour must be found from Midnight; add the Sine thereof to the Sine of 17^d, the Winter Meridian Altitude, the whole extent will be equal to the Sine of 20 25′; Enter the said extent upon the point of Entrance▪ as before, and in the Versed Sine of 90^d, the hour will be found to be either half an hour past 4 in the Morning, or half an hour past 7 in the Evening.

Examples for working the third Proportion.

Take the Sine of 30^d, and enter it upon the point of Entrance, laying the thred to the other foot, according to nearest distance, and there keep it; then take the nearest distance to it from the Sine of 57, the Meridian Altitude; and the said Extent prick upon the Line of Versed Sines on the left edge, and it will reach to 118^• 54′, set a mark to it. Lastly, the nearest distance from the Sine of each respective Altitude to the thred, being pricked from the said mark, will reach to the Versed Sine of the hour from Noon, for North Declinations.

So when the Sun hath 24^d 48′ of Altitude, the Hour from 7: 17 Noon will be found to be—

• 75^d
• 105

A Winter Example for that Declination.

The nearest distance from the Sine of 17^d, the Winter Meridian Altitude, to the thred, will reach to the Versed Sine of 61^d 6′, the Complement of the former to a Semicircle, at which set a mark; then if the Altitude were —

• 12^d 30′
• 14: 26

the nearest distances to the thred prickt from the latter mark, would shew the hours to these Altitudes to be

• 2 hours
• 1 ½ hour

from Noon

This last Proportion in some cases will be inconvenient, being liable to excursion in Latitudes more Northwardly.

Two sides with the Angle comprehended, to find the third side.

• As the Radius,
• Is to the Sine of one of the Includers,
• So is the Sine of the other Includer,
• To a fourth.

Again,

• As the Radius,
• Is to the Versed Sine of the Angle included,
• So is that fourth,
• To the difference of the Versed Sines of the third side, and of the Ark of difference between the two including sides,

And

• so is the Versed Sine of the Included Angles Complement to 180.
• To the difference of the Versed Sines of the third side, and of the sum of the two including sides.

Another Proportion for finding it in Sines, elsewhere delivered.

By the former Proportion, having the advantage both of lesser and greater Versed Sines, we may find the side sought, either in the line of Sines, or in the line of Versed Sines on the the left edge▪ issu∣ing from the Center.

The Converse of the Proportion that found the Hour, will find the Suns Altitudes on all Hours.

• As the Radius,
• Is to the Cosine of the Latitude,
• So is the Cosine of the Declination,
• To a fourth Sine.

Namely, The difference of the Sines of the Suns Meridian Altitude, and of his Altitude at 6 in Summer, but the sum of the Sines of his Depression at 6, and Winter Meridian Altitude, here∣by we may obtain the point of Entrance and Altitude, or Depres∣sion at 6, as before, and let them be recorded, then it holds,

• As the Radius,
• Is to the Sine of the Hour from 6,
• So is that fourth Sine,
• To the difference of the Sines of the Suns Altitude at 6, and of his Altitude sought; But in Winter, To the sum of the Sines of his Depression at 6, and of the Altitude sought.

Hereby we may find two Altitudes at a time.

Lay the thred to the Hour in the Limb, and from the point of Entrance, take the nearest distance to it, the said Extent being set down at the Altitude at 6, shall reach upward to the greater Alti∣tude, and downward, to the lesser Altitude.

Example.

Admit the hour to be 5 and 7 in the morning, the Altitudes thereto for 20 North Declination for the Latitude of Notting∣ham, will be found to be 7^d 17′, and 24^• 48′

If the Hour be more remote from 6 then the time of Rising, we may find a Winter Altitude to as much South Declination, and a Summer Altitude, to the said North Declination.

Thus if the Hour be 45^d from 6, that is either 9 in the morning, or 3 in the afternoon, the nearest distance from the point of En∣trance to the thred, will reach from the Sine of 15^d 51′, the Alti∣tude

at 6 upwards, to the Sine of 42^d 18′, the Summer Altitude to that Declination: But downwards, it reaches beyond the Center: In this case measure, that extent from the Center, and take the di∣stance between the inward foot of the Compasses, and the Alti∣tude at 6▪ which measured on the Sines, will be found to be 7^d 17′ for the Winter Altitude to that Hour.

So if the hour were 60^d from 6, that is either 10, or 2, the Sum∣mer Altitude would be found to be 49^d 42′, and the Winter Alti∣tude 12^d 30′.

And this may be found in the Versed Sines on the left edge, ac∣counted as a Sine each way from the middle, if use be made of the lesser Sines, instead of the Limb, in finding the point of Entrance, as also, in laying it to the Sine of each hour from 6, in which case the Compasses will alwaies find two Altitudes at once; for when they fall beyond the midst of the said Line, it shews the Winter Altitudes counted from thence towards the end of the said Versed Sines.

Having found the fourth Sine, which gives the point of Entrance as before, the Altitudes on all hours may be found by the Versed Sines of 90^d in the Limb, the Proportion will be,

• As the Radius,
• Is to the Versed Sine of the Hour from Noon,
• So is the fourth abovesaid,
• To the difference of the Sines of the Meridian Altitude, and of the Altitude sought.

But for hours beyond 6, the Proportion will be,

• As the Radius,
• Is to the Versed Sine of the Hour from Midnight,
• So is the fourth abovesaid,
• To the sum of the Sines of the Suns Depression at Midnight (equal to his Winter Meridian Altitude,) and of his Altitude sought,

Hereby also we may find two Altitudes at once.

Operation.

Lay the thred to the Versed Sine of the Hour from Noon, and from the point of Entrance at 34^d 25′, take the nearest distance to

it, the said Extent shall reach from the Summer Meridian Altitude, accounted in the Sines to the Altitude sought, also from the Win∣ter Meridian Altitude, to the Altitude sought.

Example.

• Latitude of Nottingham is 53^d, Complement—37^•
• Suns Declination,— 20
• Sum being the Summer Meridian Altitude— 57^d
• Difference being Winter Meridian Altitude 17

If it were required to find the Altitudes for the hours of

• 11 and 1 The Extents so taken out will find the Summer Alti∣tudes to be— 55^• 00′ And the Winter Altitudes to the same hours and Declination— 15^• 51^•
• 10 and 2 The Extents so taken out will find the Summer Alti∣tudes to be— 49, 42 And the Winter Altitudes to the same hours and Declination— 12, 30
• 9 and 3 The Extents so taken out will find the Summer Alti∣tudes to be— 42, 18 And the Winter Altitudes to the same hours and Declination— 7, 17
• 8 and 4 The Extents so taken out will find the Summer Alti∣tudes to be— 33, 47 And the Winter Altitudes to the same hours and Declination— 00, 32

But for hours more remote from the Meridian then 6, as admit for 5 in the morning, or 7 at night, which is 75^d from the North Meridian; lay the thred to the said Ark in the Versed Sine of 90^d, and the distance from the point of Entrance to it, shall reach from the Sine of 57^d, the Meridian Altitude, to the Sine of 24^d 48′, the Summer Altitude for the Hour 75^d from Noon, and if that Extent be pricked from the Winter Meridian Altitude, it will reach beyond the Center, in which case, enter that Extent upon the Line of Sines, and take the distance between the point of limitation and 17^d, which will (being measured) be found to be the Sine of 7^d 17′, the Altitude belonging to the hour 105 from Noon.

In like manner, the Altitudes for the hours

• 97^d 30 from Noon that is 82^d 30′ from Midnight, will be 11^d 31′ and for the like hours from Noon 20^d 17′
• 112, 30 from Noon that is 67, 30 from Midnight, will be 3: 15 and for the like hours from Noon 29: 19

In like manner, it might have been found in the Versed Sines is∣suing from the Center, if in finding the point of Entrance, and in laying the thred to the Versed Sine of the Hour, we make use of the lesser Sines, and of the Versed Sine of 180^d in the Limb.

For the Azimuth.

Two of the former Proportions may be conveniently applied to other sides, for finding the Azimuth universally.

• As the Radius,
• Is to the Cosine of the Latitude,
• So is the Cosine of the Altitude,
• To a fourth Sine.

Get the sum of the Altitude and Colatitude; or, which is all one, the sum of the Latitude and Colatitude; and if it exceeds a Qua∣drant, take its Complement to a Semicircle: This fourth Sine is equal to the difference of the Sines of this Compound Ark, and of another Ark to be thereby found, called the latter Ark.

Then it holds,

• As the fourth Sine,
• Is to the Radius,
• So in Summer is the difference, but in Winter, the sum of the Sines of this latter Ark, and of the given Declination,
• To the Sine of the Azimuth from the Vertical.

When the latter Ark is more then the Declination, the Azimuth will be found from the Vertical towards the Noon Meridian, other∣wise towards the Midnight Meridian, and in winter, always towards the Noon Meridian.

For such Stars as come to the Meridian between the Zenith and the elevated Pole, the fourth Ark will never exceed the Stars decli∣nation, and their Azimuth will be alwaies found from the Vertical towards the Meridian they come to, above the Horizon.

Example for the Latitude of Nottingham.

• Complement of the Latitude is— 37^d
• Altitude is 40^•—40
• Sum —77

Let the Declination be 20^d North.

To find the point of Entrance, take the nearest distance to the thred laid over 50 in the Limb, counted from right edge from the Sine of 37^d, the said Extent measured from the Center, falls upon the Sine of 27^d 26′, and there will be the point of Entrance; the said Extent prickt from 77^• in the Sines, will reach to the Sine of 30 51′, where the Sine point falls.

Lastly, The distance between the Sine point, and the Sine of 20^d being entred at the point of Entrance, and the thred laid to the other foot, the Azimuth will be found in the equal Limb to be 21^d 48′ from the East or West Southwards, because the Sine point fell beyond the Declination.

Another Example for that Latitude, the Declination being 20^d South Altitude. 12^d 30′

The point of Entrance will fall at the Sine of 36^•

The Sine point may be found without summing or differencing of Arks, by taking the nearest distance from the Sine of the La∣titude, to the thred laid over the Altitude, counted in the Limb from the right edge; which Extent being added to the Sine of 20^d the Declination, the whole Extent will be equal to the Sine of 31^d, this being entred on the point of Entrance, and the thred laid to the other foot, the Azimuth will be found to be 61^d 14′ from the East or West Southwards.

A third Example for the Latitude of London, 51^d 32′.

Let it be required to find the Azimuth of the middlemost Star in the great Bears tail, Declination is 56^d 45′, let the Altitude be 44^d 58′.

The nearest distance from the Sine of 38^d 28′ to the thred laid over the Altitude counted from the right edge, will find the point of Entrance to be at the Sine of—26^d 6′.

The nearest distance from the Sine of 51^d 32′ to the thred laid over the Altitude, counted from the right edge, need not be known, but the distance between that Extent, and the Sine of 56^d 45′, the Stars Declination being entred on the point of Entrance, will find the Azimuth of that Star, by laying the thred to the other foot, to be 40^d from the East or West Northwards.

Thus we find it the general way, and so it will also be found by the fitted particular Scale; for the Hour, the point of Entrance, and Sine point, vary not till the Declination change; but for the Azi∣muth, they vary to every Altitude.

To find the Azimuth in the Versed Sines.

• As the fourth, found by the former Proportion; namely, where the point of Entrance hapned,
• Is to the Radius,
• So is the difference of the Versed Sines of the Polar distance, and of the Ark of Difference between the Altitude and the Latitude,
• To the Versed Sine of the Azimuth from Midnight Meridian.

This finds the Angle it self in the Sphere.

• And so is the difference of the Versed Sines of the Polar distance, and of the Ark of residue of the sum of the Latitude and Alti∣tude taken from a Semicircle.
• To the Versed Sine of the Azimuth from Noon Meridian. This finds the Complement of the Angle in the Sphere to a Semi∣circle.

The Proportion to find it from Midnight Meridian, the third term being express'd in Sines, will be thus.

Get the sum of the Altitude and Colatitude, and when it exceeds a Quadrant, take its Complement to a Semicircle, the Ark thus found, is called the Compound Ark. Then it holds,

• As the fourth found before.
• Is to the Radius,
• So in Summer Declinations, is the difference, but in Winter De∣clinations, the sum of the Sines of the Suns or Stars declination, and of the compound Ark,
• To the Versed Sine of the Azimuth from the Midnight Meridian of the place.

Use this Proportion alwaies for the Sun or Stars, when they come to the Meridian between the Zenith and elevated Pole.

And to find it from the Noon Meridian,

Get the difference between the Altitude and Colatitude, and then it holds,

• As the fourth Sine found before,
• Is to the Radius,
• So is the sum of the Sines of the said Ark of Difference, and of the Suns Declination,
• To the Versed Sine of the Azimuth from the Noon Meridian, in Summer only, when the Suns Altitude is less then the Colati∣tude.

In all other cases,

• So is the difference of the said Sines,
• To the Versed Sine of the Azimuth, as before, from Noon Meri∣dian.

If by the former Proportion it be required to find the Azimuth in the Versed Sine of 90^d, a difference of Versed Sines taken out of the Line of Versed Sines on the left edge must be doubled, and being taken out of the Line of Sines, as sometimes representing the former, sometimes the latter half of a Versed Sine, needs not be doubled.

Example:

• Latitude of Nottingham — 53^d
• Altitude of the Sun —4
• Ark of difference — 49
• ☉ Declination 20 North, the Polar distance is—70

The Point of entrance will fall at the Sine of 36^d 54′

And the difference of the Versed Sines of 49^• and 70••, equal to the distance between the Sines of 41^• and 20 being entred at the Point of entrance, and the Tbread laid to the other foot will lye over 61^d 30′ of the Versed Sine of 90^•, and so much is the Suns Azimuth from the North.

Another Example for finding it from the South when the Alti∣tude is more then the Colatitude.

• Altitude—47^d
• Colatitude—37 of Nottingham.
• difference—10

The Point of entrance will fall at the sine of 24^d 14′ found by taking the nearest extent from sine of 37^d to the Thread lying over 43^d of the Limb the Coaltitude.

Then the distance between the sines of 10^d, the Ark of difference as above, and the sine of 20 the Suns North Declination being entred at the Point of entrance, and the Thread laid to the other foot, will shew 53^d 55′ in the Versed sine of 90^d for the Suns Azi∣muth from the South.

A third Example when the Altitude is less then the Cola∣titude in Summer.

• Complement Latitude 37^d of Nottingham.
• Altitude—34
• difference— 3

The Point of entrance will fall at the sine of 29^d 55′, and the sum of the sines of 3^d, and of 20^d the Suns declination supposed North, is equal to the sine of 23^d 13′: Which Extent entred at 29^d 55′, the Point of entrance, and the Thread laid to the other foot according to nearest distance, it will intersect the Versed sine of 90^d at the Ark of 77^d 57′, and so much is the Suns Azimuth from the South.

And if there were no Versed sines in the Limbe, find an Ark of the equal Limbe, and enter the sine of the said Ark down the Line of sines from the other end, and you may obtain the Versed sine of the Ark sought.

More Examples need not be insisted upon, having found the Point of entrance, the distance between the Versed sines of the Base or side subtending the angle sought, and of the Ark of dif∣ference between the two including sides, being taken out of the streight Line of Versed Lines on the left edge, and entred at the Point of entrance, laying the Thread to the other foot shews in the Versed Sine of 180^d in the Limb the angle sought; and if the said distance or Extent be doubled, and there entred it shews the angle sought in the Versed Sine of 90^d, when the Angle is less then a Quadrant, when more, the distance between the Versed Sines of the Base and the sum of the Legs, will find the Comple∣ment of the angle sought to a Semicircle without doubling in the Versed Sine of 180^d in the Limb, with doubling in the Versed Sine of 90^d.

Lastly, Three sides, viz. all less then Quadrants, or one of them greater, generally to find an angle in the equal Limb, the Pro∣portion will be,

• As the Radius,
• Is to the Cosine of one of the including sides:
• So is the Cosine of the other Includer,
• To a fourth Sine.

Again,

• As the Sine of one of the Includers,
• To the Cosecant of the other:
• ...

• So when any one of the sides is greater then a Quadrant is the sum, but when all less, the difference of the fourth Sine, and of the Cosine of the third side,
• To the Cosine of the angle sought.

If any of the three sides be greater then a Quadrant, it subtends an Obtuse angle, the other angles being Acute; But when they are all less then Quadrants, if the 4^th Sine be less then the Cosine of the third side, the angle sought is Acute, if equal thereto, it is a right angle, if greater an Obtuse angle.

From the Proportion that finds the Hour from six, we may educe a single Proportion applyable to the Logarithms without natu∣ral Tables for Calculating the Hour of the day to all Altitudes, By turning the third Tearm, being a difference of Sines or Ver∣sed Sines into a Rectangle, and freeing it from affection.

The two first Proportions to be wrought are fixed for one De∣clination; The first will be to find the Suns Altitude or Depressi∣on at six.

The second will be to find half the difference of the Sines of the Suns Meridian Altitude, and Altitude sought, &c. as before de∣fined, the Proportion to find it is,

• As the Secant of 60^d,
• To the Cosine of the Declination:
• So is the Cosine of the Latitude,
• To the Sine of a fourth Arch.

Lastly, To find the Hour.

Get the sum and difference of half the Suns Zenith distance at the hour of six; and of half his Zenith distance to any other pro∣posed Altitude or Depression.

Then,

• As the Sine of the fourth Arch,
• Is to the Sine of the sum:
• So is the Sine of the difference,
• To the Sine of the hour from six towards Noon or Mid∣night,

according as the Altitude or Depression was greater or lesser then the Altitude or Depression at six.

Observing that the Sine of an Arch greater then a Quadrant, is the Sine of that Arks Complement to a Semicircle.

Of the Stars placed upon the Quadrant below the Projection.

ALL the Stars placed upon the Projection are such as fall be∣tween the Tropicks and the Hour may be found by them with the Projection, as in the Use of the small Quadrant: Which may also be found by the fitted particular Scale, not only for Stars within the Tropicks, but for all others without, when their Alti∣tude is less then 62^d, and likewise their Azimuth may be thereby found when their Declination is not more then 62^d.

For other Stars without the Tropicks, they may be put on be∣low the Projection any where in such an angle that the Thread laid over the Star shall shew an Ark in the Limb, at which in the Sines the Point of entrance will always fall; And again, the same Star is to be graved at its Altitude or Depression at six in the Sines, and then to find the Stars hour in that Latitude whereto they are fitted, will always for Northern Stars be to take the di∣stance in the Line of Sines between the Star and its given Altitude, and to enter that Extent at the Point of entrance, laying the Thread to the other foot according to nearest distance, and it gives the Stars hour in the equal Limb from six, which may also be found in the Sines by a Parrallel entrance, laying the Thread over the Star.

Example.

Let the Altitude of the last in the end of the great Bears Tail be 63^d, take the distance between it and the Star which is graved at 37^d 30′ of the Sines, the said Extent entred at the Sine of 23^d, the Ark of the Limb the Thread intersects when it lies over the said Star, and by laying the Thread to the other foot you will find that Stars hour to be 46^d 11′ from six towards Noon Meridian, if the Altitude increase, and in finding the true time of the night, the Stars hour must be always reckoned from the Meridian it was last upon; in this Example it will be 5 minutes past 9 feré.

Of the Quadrant of Ascensions on the backside,

This Quadrant is divided into 24 Hours with their quarters and subdivisions, and serves to give the right Ascension of a Star, as in the small Quadrant to be cast up by the Pen.

It also serves to find the true Hour of the night with Compasses.

First having found the Stars hour, take the distance on the Qua∣drant of Ascensions in the same 12 hours between the Star and the Suns Ascension (given by the foreside of the Quadrant) the said Extent shall reach from the Stars hour to the true hour of the night, and the foot of the Compasses always fall upon the Qua∣drant; Which Extent must be applyed the same way it was taken, the Suns foot to the Stars hour.

Example.

If upon the 30^th of December the last in the end of the Bears Tail were found to be 9 hours 05′ past the Meridian it was last up∣on, the true time sought would be 16 minutes past 3 in the morn∣ing.

Another Example for the Bulls Eye.

Admit the Altitude of that Star be 39d, that Stars hour as we found it by the Line of Versed Sines was 3 ho 3′ from the Meridian, if the Altitude increase, then that Stars hour from the Meridian it was last upon was 57 minutes past 8—8^h:57′

If this Observation were upon the 23^d of October, the Complement of the Suns Ascension would be— 9: 30

• The Ascension of that Star is —4: 16
• The true time of the night would be forty— 10: 43

three minutes past ten.

The distance between the Star and the Suns Ascension being applyed the same way, by setting the Sun foot at the Stars hour will shew the true time sought.

When the Star is past the Meridian, having the same Altitude, the Stars hour will be 3′ past 3, and the true time sought, will be 49′ past 4 in the next morning.

[illustration]

The Geometrical Construction of M^r Fosters Circle.

THe Circle on the Back side of the Quadrant, whereof one quarter is only a void Line, is derived from M. Foster's Trea∣tise of a Quadrant, by him published in An••o 1638. the founda∣tion and use whereof being concealed, I shall therefore endeavour to explain it.

Upon the Center H describe a circle, and draw the Diameter A C, passing through the Center, and perpendicularly there∣to, upon the point C, erect a Line of Sines C I, whose Radius shall be equal to the Diameter A C, let 90^d of the Sine end at I; I say then, if from the point A, through each degree of that Line of Sines, there be streight lines drawn, intersecting the Quadrant of the circle C G, as a line from the point D doth intersect it at B the Quadrant C G, which the Author calls the upper Quadrant, or Quadrant of Latitudes shall be constituted, and if C I be con∣tinued as a Secant, by the same reason the whole Semicircle C G A may be occupied; hence it will be necessary to educe a ground of calculation for the accurate dividing of the said Quadrant, and that will be easie; for A C being Radius, the Sine C D doth also represent the Tangent of the Angle at A, therefore seek the natu∣ral Sine of the Ark C D in the Table of Natural Tangents, and the Ark corresponding thereto, will give the quantity of the Angle D A C, then because the point A falls in the circumference of the Circle, where an Angle is but half so much as it is at the Center, by 31 Prop. 3. Euc. double the Angle found, and from a Quadrant divided into 90 equal parts, and their subdivisions, by help of a Ta∣ble so made, may the Quadrant of Latitudes be accurately divided: but the Author made his Table in page 5. without doubling, to be graduated from a Quadrant divided into 45 equal parts.

Again, If upon the Center C, with a pair of Compasses, each degree of the line of Sines be transferred into the Semicircle C G A it shall divide it into 90 equal parts; the reason whereof is plain,

because the Sine of an Arch is half the chord of twice that Arch, and therefore the Sines being made to twice the Radius of this cir∣cle, shall being transferred into it, become chords of the like Arch, to divide a Semicircle into 90 equal parts.

Again, upon the point A, erect a line of Tangents of the same Radius with the former Sine, which we may suppose to be infinitely continued, here we use a portion of it A E.

If from the point C, the other extremity of the Diameter lines be drawn, cutting the lower Semicircle (as a line drawn from E intersects it at F) through each degree of the said Tangent, the said lower Semicircle shall be divided into 90 equal parts; the reason is evident a line of Tangents from the Center shall divide a Qua∣drant into 90 equal parts, and because an Angle in the circumfe∣rence is but half so much as it is in the Center, being transferred thither, a whole Semicircle shall be filled with no more parts.

The chief use of this Circle, is to operate Proportions in Tan∣gents alone, or in Sines and Tangents joyntly, built upon this foundation, that equiangled plain Triangles have their sides Pro∣portional.

In streight lines, it will be evident from the point D to E, draw a streight line intersecting the Diameter at L, and then it lies as C L to C D; so is A L to A E: it is also true in a Circle, provided it be evinced, that the points B L F fall in a streight line.

Hereof I have a Geometrical Demonstration, which would re∣quire more Schemes, which by reason of its length and difficulty, I thought fit at present not to insert, possibly an easier may be found hereafter: As also, an Algebraick Demonstration, by the Right Honourable, the Lord Brunkard, whereby after many Al∣gebraick inferences it is euinced, that as L K is to K B ∷ so is L N to •• F: whence it will follow, that the points B, L, F, are in a right line.

If a Ruler be laid from 45^d of the Semicircle, to every degree of the Quadrant of Latitudes, it will constitute upon the Diame∣ter, the graduations of the Line Sol, whereby Proportions in Sines might be operated without the other supply.

From the same Scheme also follows the construction of the streight line of Latitudes, from the point G, at 90^• of the Quadrant of Latitudes, draw a streight Line to C, and transfer each degree of the Quadrant of Latitudes with Compasses, one foot resting up∣on C into the said streight line, and it shall be constituted.

To Calculate it.

The Line of Latitudes C G bears such Proportion to C A as the Chord of 90^d doth to the Diameter, which is the same that the Sine of 45^d bears to the Radius, or which is all one, that the Radius bears to the Secant of 45d, which Secant is equal to the Chord of 90d; from the Diagram the nature of the Line of La∣titudes may be discovered.

Any two Lines being drawn to make a right angle, if any Ark of the Line of Latitudes be pricked off in one of those Lines retai∣ning a constant Hipotenusal A C, called the Line of Hours, equal to the Diameter of that Circle from whence the Line of Latitudes is constituted, if the said Hipotenusal from the Point formerly pricked off, be made the Hipotenusal to the Legs of the right an∣gle formerly pricked off, the said Legs or sides including the right angle shall bear such Proportion one to another, as the Radius doth to the sine of the Ark so prickt off; and this is evident from the Schem, for such Proportion as A C bears to C D, doth A B bear to B C, for the angle at A is Common to both Triangles, and the angle at B in the circumference is a right angle, and conse∣quently the angle A C B will be equal to the angle A D C, and the Legs A C to C D bears such Proportion by construction, as the Radius doth to the Sine of an Ark, and the same Proportion doth A B bear to B C, in all cases retaining one and the same Hypote∣nusal A C, the Proportion therefore lies evident.

• As the Radius, the sine of the angle at B,
• To its opposite side A C, the Secant of 45d:
• So is the sine of the angle at A,
• To its opposite side B C sought.

Now the quantity of the angle at A was found by seeking the natural Sine of the Ark proposed in

the Table of natural Tangents; and having found what Ark an∣swers thereto, the Sine of the said Ark is to become the third Tearm in the Proportion.

But the Cannon prescribed in the Description of the small Qua∣drant is more expedite then this, which Mr Sutton had from Mr Dary long since, for whom, and by whose directions he made a Quadrant with the Line Sol, and two Parrallel Lines of Sines up∣on it, as is here added to the backside of this Quadrant.

Of the Line of Hours, alias, the Diameter or Proportional Tangent.

This Scale is no other then two Lines of natural Tangents to 45d, each set together at the Center, and from thence beginning and continued to each end of the Diameter, and from one end thereof numbred with 90d to the other end.

This Line may fitly be called a Proportional Tangent, for wher∣soever any Ark is assumed in it to be a Tangent, the remaining part of the Diameter is the Radius to the said Tangent.

So in the former Schem, if C L be the Tangent of any Ark, the Radius thereto shall be A L.

[illustration]

In the Schem annexed, let A B be the Radius of a Line of Tan∣gents equal to C D, and also parralel thereto, and from the

Point B to C draw the Line B C, and let it be required to divide the same into a Line of Proportional Tangents: I say, Lines drawn from the Point D to every degree of the Tangent, A B shall divide one half of it as required from the similitude of two right angled equiangled plain Triangles, which will have their sides Proportional, it will therefore hold, As C F, To C D: So F B, To B E, and the Converse, As the second Tearm C D, To the fourth B E: So is the first C F, To the third F B, and therefore C F bears such Proportion to F B, as C D doth to B E, which is the same that the Radius bears to the Tangent of the Ark proposed.

If it be doubted whether the Diameter wil be a double Tangent or the Line here described such a Line, a Proportion shall be given to find by Experience or Calculation, what Line it will be; for there is given the Radius C D, and the Tangent B E, the two first Tearms of the Proportion, with the Line C B the sum of the third and fourth Tearms, to find out the said Tearms respectively; and it will hold by compounding the Proportion,

• As the sum of the first and second Tearm,
• Is to the second Tearm:
• So is the sum of the third and fourth Tearm,
• To the fourth Tearm,

that is,

• As C D + B E,
• Is to B E:
• So is C F + F B = C B,
• To F B,

see 18 Prop. of 5 of Euclid, or page 18 of the English Clavis Mathematicae, of the famous and learned Mr Oughtred.

After the same manner is the Line Sol, or Proportional Sines made, that being also such a Line, that any Ark being assumed in it to be a Sine, the distance from that Ark to the other end of the Diameter, shall be the Radius thereto.

A Demonstration to prove that the Line of Hours and Latitudes will jointly prick off the hour Di••tances in the same angles as if they were Calculated and prickt off by Chords.

[illustration]

Draw the two Lines A B and C B crossing one another at right angles at B, and prick off B C the quantity of any Ark out of the line of Latitudes, and then fit in the Scale of Hours; so that one end of it meeting with the Point C, the other may meet with the other Leg of the right angle at A, from whence draw A E parra∣lel to B C; So A B being become Radius, B C is the Sine of the Arch first prickt down from the line of Latitudes; from the Point B through any Point in the line of Proportional Tangents, at L draw the Line B L E, and upon B with the Radius B A draw the Arch A D, which measureth the Angle A B E to the same Radius: I say, there will then be a Proportion wrought, and the said Arch measureth the quantity of the fourth Proportional, the Proportion will be,

• As the Radius,
• To the Sine of the Ark prickt down from the Line of Latitudes:
• So is any Tangent accounted in the Scale, beginning at A,
• To the Tangent of the fourth Proportional;

in the Schem it lies evident in the two opposite Triangles L C B and L A E, by con∣struction equiangled and consequently their sides Proportional.

Assuming A L to be the Tangent of any Ark, L C becomes the Radius, according to the prescribed construction of that Line, it then lies evident,

• ...

• As L C the Radius,
• To C B the Sine of any Ark,
• So is L A, the Tangent of any Ark,
• To A E, the Tangent of the fourth Proportional.

Namely, of the Angle A B E, and therefore it pricks down the Hour-lines of a Dyal most readily and accurately: the Proportion in pricking from the Substile being alwaies,

• As the Radius,
• To the Sine of the Stiles height,
• So the Tangent of the Angle at the Pole,
• To the Tangent of the Hour-line from the Substile.

Uses of the Graduated Circle.

To work Proportions in Tangents alone.

In any Proportion wherein the Radius is not ingredient, it is sup∣posed to be introduced by a double Operation, and the Poportion will be,

• As the first term,
• To the second,
• So the Radius
• to a fourth.

Again,

• As the Radius
• is to that fourth,
• So is the third Term given,
• To the fourth Proportional sought.

In illustrating the matter, I shall make use of that Theorem•• for varying of Proportions, that the Tangents of Arches, and the Tangents of their Complements are in reciprocal Proportions.

As Tangent 23^d, to Tangent 35^d, So Tangent 55^d to the Tan∣gent of 67^d.

In working of this Proportion, the last term may be found to the equal Semicircle, or on the Diameter.

1. In the Semicircle.

Extend the thred through 23^d on the Diameter, and through 3•• in the Semicircle, and where it intersects the Circle on the oppo∣site

side, there hold one end of it, then extend the other part of it over 55 in the Diameter, and in the Semicircle, it will intersect 67^d for the term sought.

2. On the Diameter.

Extend the thred over 23^d in the Semicircle, and 35^d on the Di∣ameter, and where it intersects the void circular line on the oppo∣site side, there hold it, then laying the other end of it over 55d in the Semicircle, and it will cut 67d on the Diameter.

If the Radius had been one of the terms in the Proportion, the operation would have been the same, if the Tangent of 45d had been taken in stead of it.

To work Proportions in Sines and Tangents joyntly.

1. If a Sine be sought, the middle terms being of a different species.

Extend the thred through the first term on the Diameter, being a Tangent, and through the Sine, being one of the middle terms, counted in the unequal Quadrant, and where it intersects the Op∣posite side of the Circle hold it, then extend the thred over the Tangent, being the other middle term counted on the Diameter, and it will intersect the graduated Quadrant at the Sine sought.

Example.

If the Proportion were as the Tangent of 14^d to the Sine of 29^d So is the Tangent of 20^d to a Sine, the fourth Proportional would be found to be the Sine of 45^d.

2. If a Tangent be sought, the middle terms being of several kinds, Extend the thred through the Sine in the upper Quadrant, being the first term, and through the Tangent on the Diameter, being one of the other middle terms, holding it at the Intersection of the Circle on the opposite side, then lay the thred to the other middle term in the upper Quadrant, and on the Diameter, it shews the Tangent sought.

Example.

If the Suns Amplitude and Vertical Altitude were given, the Proportion from the Analemma to find the Latitude would be,

• As the Sine of the Amplitude
• to Radius,
• So is the Sine of the Vertical Altitude,
• To the Cotangent of the Latitude

• Let the Amplitude be—39^d 54′
• And the Suns Altitude being East or West—30 39′

Extend the thred through 39 54′, the Amplitude counted in the upper Quadrant, and through 45^d on the Diameter, holding it at the intersection with the Circle on the Opposite side, then lay the thred over 30^d 39′, the Vertical Altitude, and it will inter∣sect the Diameter at 38^d 28′, the Complement of the Latitude sought.

But Proportions derived from the 16 cases of right angled Spherical Triangles, having the Radius ingredient, will be wrought without any motion of the thred.

An Example for finding the Suns Azimuth at the Hour of 6.

• As the Radius
• to the Cosine of the Latitude,
• So the Tangent of the Declination,
• To the Tangent of the Azimuth, from the Vertical towards Mid∣night Meridian.

Extend the thred over the Complement of the Latitude in the upper Quadrant, and over the Declination in the Semicircle, and on the Diameter. it shews the Azimuth sought.

So when the Sun hath 15^d of Declination, his Azimuth shall be 9^d 28′ from the Vertical at the hour of 6 in our Latitude of London.

Another Example to find the time when the Sun will be due East or West.

Extend the thred over the Latitude in the Semicircle, and over the Declination on the Diameter, and in the Quadrant of Latitudes it shews the Ark sought.

The Proportion wrought, is,

• As the Radius
• to the Cotangent of the Latitude,
• So is the Tangent of the Declination,
• To the Sine of the Hour from 6.

Example.

So when the Sun hath 15^• of North Declination, in our Lati∣tude of London, the Hour will be found 12^d 18′ from 6 in time 49⅕′ past 6 in the morning, or before it in the afternoon.

Another Example to find the Time of Sun rising.

• As the Cotangent of the Latitude,
• to Radius,
• So is the Tangent of the Declination,
• To the Sine of the Hour from 6 before or after it.

Lay the thred to the Complement of the Latitude in the Semi∣circle, and over the Declination on the Diameter, and in the Qua∣drant of Latitudes, it shews the time sought in degrees, to be con∣verted into common time, by allowing 15^• to an hour, and 4′ to a degree.

So in the Latitude of London, 51^d 32′ when the Sun hath 15^• of Declination, the ascensional difference or time of rising from 6, will be 19^d 42′, to be converted into common time, as before.

By what hath been said, it appears, that the Hour and Azimuth may be found generally by help of this Circle and Diameter.

For the performance whereof, we must have recourse to the Proportions delivered in page 123. whereby we may alwaies find the two Angle adjacent to the side on which the Perpendicular fal∣leth,

which may be any side at pleasure; for after the first Pro∣portion wholly in Tangents is wrought, to find either of those Angles, will be agreeable to the second case of right angled Sphe∣rical Triangles, wherein there will be given the Hypotenusal, and one of the Legs, to find the adjacent Angle, only it must be sug∣gested, that when the two sides that subtend the Angle sought, are together greater then a Semicircle, recourse must be had to the Opposite Triangle, if both those Angles are required to be found by this Trigonometry, otherwise one of them, and the third An∣gle may be found by those directions, by letting fall the perpendi∣cular on another side, provided the sum of the sides subtending those Angles be not also greater then a Semicircle; or, having first found one Angle, the rest may be found by Proportions in Sines only.

[illustration]

IN the Triangle ☉ Z P, if it were required to find the angles at Z and ☉, because the sum of the sides ☉ P and Z P are less then a Semicircle they might be both found by making the half of the Base ☉ Z the first Tearm in the Proportion, and then because the angles ☉ Z are of a different affection, the Perpendicular would fal without on the side ☉ Z continued towards B, as would be evinced

by the Proportiod, for the fourth Ark discovered, would be found greater then the half of ☉ Z; hence we derived the Cannon in page 124, for finding the Azimuth; Whereby might also be found the angle of Position at ☉; so if it were required to find the angles at ☉ and P, the sides ☉ Z and Z P being less then a Semicircle the Perpendicular would fall within from Z on the side ☉ P, as would also be discovered by the Proportion, for the fourth Ark would be found less then the half of ☉ P.

But if it were required to find both the angles at Z and P, in this Case we must resolve the Opposite Triangle Z B P, because the sum of the sides ☉ Z and ☉ P are together greater then a Se∣micircle, and this being the most difficult Case, we shall make our present Example. The Proportion will be,

• As the Tangent of half Z P,
• Is to the Tangent of the half sum of Z B and P B:
• So is the Tangent of half their difference,
• To a fourth Tangent.

That is,

• As Tangent 19^d 14′,
• Is to the Tangent of 86^d 30′,:
• So is the Tangent of 9^d 30′,
• To a fourth.

Operation. Extend the Thread through 19d 14′ on the Semicircle, and 9d 30′ on the Diameter, and hold it at the Intersection on the oppo∣site side the Semicircle, then lay the Thread to 86d 30′ in the Se∣micircle, and it shews 82d 44′ on the Diameter for the fourth Ark sought.

Because this Ark is greater then the half of Z P, we may con∣clude that the Perpendicular B A falls without on the side Z P con∣tinued to A.

• fourth Ark— 82^d 44′
• half of Z P is — 19 14
• sum— 101 58 is Z A
• difference — 63 30 is P A

Then in the right angled Triangle B P A, right angled at, A we have P A and B P the Hypotenusal, to, find the angle B P A, equal to the angle ☉ Z P. The Proportion is

• ...

• As the Radius,
• Is to the Tangent of 13d, the Complement of B P:
• So is the Tangent of P A 63d 30′,
• To the Cosine of the angle at P.

Extend the Thread through 13d on the Diameter, and through 63d 30′ in the Semicircle counted from the other end, and in the upper Quadrant, it shews 27d 35′ for the Complement of the angle sought.

And letting this Example be to find the Hour and Azimuth in our Latitude of London, so much is the hour from six in Winter when the Sun hath 13d of South Declination, and 6d of Altitude, in time 1 ho 50⅓ minutes past six in the morning, or as much be∣fore it in the afternoon.

To find the Azimuth.

Again, in the Triangle Z A B right angled at A, there is given the Leg or Side Z A 101d 58′, and the Hipotenusal Z B 96d, to find the angle B Z P; here noting that the Cosine or Cotangent of an Ark greater then a Quadrant is the Sine or Tangent of that Arks excess above 90d, and the Sine or Tangent of an Ark grea∣ter then a Quadrant, the Sine or Tangent of that Arks Comple∣ment to 180d, it will hold,

• As the Radius,
• To the Tangent of 6d:
• So is the Tangent 78d 2′,
• To the Sine of 29d 44′,

found by extending the Thread through 78d 2′ on the Semicircle, counted from the other end, alias, in the small figures, and in the Quadaant it will intersect 29d 44′; now by the second Case of right angled Sphoerical Triangles, the angle A Z B will be Acute, wherefore the angle ☉ Z B is 119d 44′ the Suns Azimuth from the North, the Complement being 60d 16′ is the angle A Z B, and so much is the Azimuth from the South.

To work Proportions in Sines alone.

THat this Circle might be capacitated to try any Case of Sphoe∣rical Triangles, there are added Lines to it, namely, the Line Sol falling perpendicularly on the Diameter from the end of the Quadrant of Latitudes, whereto belongs the two Parrallel Lines of Sines in the opposite Quadrants, the upermost being extended cross the Quadrant of Latitudes.

The Proportion not having the Radius ingredient, and being of the greater to the less.

Account the first Tearm in the line Sol, and the second in the upper Sine extending the Thread through them, and where it inter∣sects the opposite Parrallel hold it; then lay the Thread to the third Tearm in the line Sol, and it will intersect the fourth Propor∣tional on the upper Parrallel.

• As the Sine of 30^d,
• To the sine of any Arch:
• So is the Cosine of that Arch,
• To the sine of the double Arch and the Converse.

By trying this Canon, the use of these Lines will be suddenly attained.

Example.

• As the sine of 30^d,
• To the sine of 20^d:
• So is the sine of 70^d,
• To the sine of 40^d.

But if it be of the less to the greater, the answer must be found on the Line Sol.

Account the first Tearm on the upper Sine, and the second in the Line Sol, and hold the Thread at the Intersection of the opposite Parrallel, then lay the Thread to the third Tearm on the uper Parrallel, and on the line Sol it will intersect the fourth Proportio∣nal if it be less then the Radius.

But Proportions having the Radius ingredient, will be wrought without any Motion of the Thread.

• As the Cosine of the Latitude,
• To Radius:
• So is the sine of the Declination,
• To the sine of the Amplitude.

So in our Latitude of London, when the Declination is 20^d 12′ the Amplitude will be found to be 33^d 42′.

Extend the Thread through 38d 28′ on the line Sol. and through the Declination in the upper Sine, and it will intersect the opposite Parrallel Sine at 33d 42′, the Amplitude sought.

The use of the Semi-Tangent and Chords are passed by at pre∣sent.

The line Sol is of use in Dyalling, as in Mr Fosters Posthuma, page 70 and 71, where it is required to divide a Circle into 12 equal parts for the hours, and each part into 4 subdivisions for the quarters, and into such parts may the equal Semicircle be divided; that if it were required to divide a Circle of like Radius into such parts, it might be readily done by this.

Of the Line of Hours on the right edge of the foreside of the Quadrant.

This is the very same Scale that is in the Diameter on the Back∣side, only there it was divided into degrees, and here into time, and placed on the outermost edge; there needs no line of Latitudes be fitted thereto, for those Extents may be taken off as Chords from the Quadrant of Latitudes, by help of these Scales thus placed on the outward edges of the Quadrants may the hour-lines of Dyals be prickt down without Compasses.

To Draw a Horizontal Dyal.

[illustration]

FIrst draw the line C E, for the Hour-line of 12, and cross it with the Perpendicular A B, then out of a Scale or Quadrant of La∣titudes set of C B and C A, each equal to the Stiles height, or La∣titude of the place, then place the Scale of 6 hours on the edge of the Quadrant, whereto the Line of Latitudes was fitted, one ex∣tremity of it at A, and move the Quadrant about, till the other end or extremity of it will meet with the Meridian line C E; then in regard the said Scale of Hours stands on the very brink or out∣ward most edge of the Quadrant, with a Pin, Pen, or the end of a black-lead pen, make marks or points upon the Paper or Dyal against each hour (and the like for the quarters, and other lesser

parts) of the graduated Scale, and from those marks draw lines into the Center, and they shall be the hour-lines required, without drawing any other lines on the Plain, the Scale of Hours on the Quadrant is here represented by the lines A E, and E B, the hour lines above the Center, are drawn by continuing them out through the Center.

And those that have Paper prints of this line, may make them serve for this purpose, without pricking down the hour points by Compasses, by doubling the paper at the very edge or extremity of the Scale of Hours.

Otherwise to prick down the said Dial without the Line of La∣titudes and Scale of hours in a right angled Parallellogram.

Having drawn C E the Meridian line, and crossed it with the perpendicular C A B, and determining C E to be the Radius of a∣ny length, take out the Sine of the Latitude to the same Radius, and prick it from C to A and B, and setting one foot at E, with the said Extent sweep the touch of an Arch at D and F, then take the length of the Radius C E, and setting down one foot at B, sweep the touch of an Ark at D, intersecting the former, also setting down the Compasses at A, make the like Arch at F, and through the points of Intersection, draw the streight lines A F, B D, and F E D, and they will make a right angled Parallellogram, the sides whereof will be Tangent lines.

To draw the Hour-lines:

Make E F, or E D Radius, and proportion out the Tangents of

• 15^d and prick them down from E to 1 and 11 and draw lines
• 30 and prick them down from E to 2 and 10 and draw lines

through the points thus found, and through the points F and D, and there will be 3 hours drawn on each side the Meridian line.

Again, make A F or B D Radius, and proportion out the Tan∣gent of 15^d, and prick it down from A to 5, and from B to 7.

Also proportion out the Tangent of 30^d, and prick it down from A to 4, and from B to 8, and draw lines into the Center, and so the Hour-lines are finished, and for those that fall above the 6 of clock

line, they are only the opposite hours continued, after the like man∣ner are the halfs and quarters to be prickt down.

Lastly, By chords prick off the Stiles height equal to the Lati∣tude of the place, and let it be placed to its due elevation over the Meridian line.

Of Ʋpright Decliners.

DIvers Arks for such plains are to be calculated, and may be found on the Circle before described.

1. The Substiles distance from the Meridian.

By the Substilar line is meant, a line over which the Stile or cock of the Dyal directly hangeth in its nearest distance from the Plain, by some termed the line of deflexion, and is the Ark of the plain between the Meridian of the Plain, and the Meridian of the place. The distance thereof from the Hour-line of 12, is to be found by this Proportion.

• As the Radius,
• To the Sine of the Plains Declination,
• So the Cotangent of the Latitude,
• To the Tangent of the Substile from the Meridian.

2. For the Angle of 12 and 6.

An Ark used when the Hour-lines are pricked down from the Meridian line in a Triangle or Parallellogram, (and not from the Substile,) without collecting Angles at the Pole.

• As the Radius,
• Is to the Sine of the Plains Declination,
• So is the Tangent of the Latitude,
• To the Tangent of an Ark, the Complement whereof is the Angle of 12 and 6.

3. Inclination of Meridians.

Is an Ark of the Equinoctial, between the Meridian of the plain, and the Meridian of the place, or it is an Angle or space of time elapsed between the passage of the shaddow of the Stile from the Substilar line into the Meridian line, by some termed the Plains dif∣ference of Longitude; and not improperly, for it shews in what Longitude from the Meridian where the Plain is; the said Plain would become a Horizontal Dyal, and the Stiles height shews the Latitude, this Ark is used in calculating hour distances by the Tables and in pricking down Dyals by the Line of Latitudes, and hours from the Substile.

• As the Radius,
• Is to the Sine of the Latitude,
• So the Cotangent of the Plains Declination,
• To the Cotangent of the Inclination of Meridians.

Or,

• As the Sine of the Latitude
• to Radius,
• So is the Tangent of the Plains Declination,
• To the Tangent of Inclination of Meridians.

4. The Stiles height above the Substile.

• As the Radius,
• Is to the Cosine of the Latitude,
• So is the Cosine of the Plains Declination,
• To the Sine of the Stiles height.

Or the Substiles distance being known,

• As the Radius,
• To the Sine of the Substiles distance from the Meridian,
• So is the Cotangent of the Declination,
• To the Tangent of the Stiles height.

Or, The Inclination of Meridians being known.

• As the Radius,
• To the Cosine of the Inclination of Meridians,
• So is the Cotangent of the Latitude,
• To the Tangent of the Stiles height.

5. Lastly, For the distances of the Hour-lines from the Substilar Line.

• As the Radius,
• Is to the Sine of the Stiles height above the Plain,
• So is the Tangent of the Angle at the Pole,
• To the Tangent of the Hours distance from the Substilar Line.

By the Angle at the Pole, is meant the Ark of difference between the Ark called the Inclination of Meridians, and the distance of any hour from the Meridian, for all hours on the same side the Substile falls, and the sum of these two Arks for all hours on the other side the Substile.

These Proportions are sufficient for all Plains to find the like Arks, without having any more, if the manner of referring De∣clining Reclining Inclining Plains to a new Latitude, and a new Declina∣tion in which they shall stand as upright Plains, be but well explain∣ed, for East or West Reclining Inclining Plains, their new Latitude is the Complement of their old Latitude, and their new Declination, is the Complement of their Reclination Inclination, which I count always from the Zenith, and upon such a supposition, taking their new Latitude and Declination, those that will try, shall find that these Proportions will calculate all the Arks necessary to such Dials.

So if an Upright Plain decline 25^d in our Latitude of Lon∣don from the Meridian.

• The Substiles distance from the Meridian is—18^d 34′
• The Angle of 12 and 6 is—62: 00
• The Inclination of Meridians is—30: 47
• The Stiles height is —34: 19

To Delineate the same Dial from the Substile by the Line of Latitudes, and Scale of hours in an Equicrutal Triangle.

To Draw an Ʋpright Decliner.

[illustration]

An Ʋpright South Plain for the Latitude of London, Declining 25^d Eastwards.

TO prick down this Dial by the line of Latitudes, and Scale of Hours in an Isoceles Triangle.

Draw C 12 the Meridian Line perpendicular to the Horizontal line of the Plain, and with a line of Chords, make the Angle

F C 12, equal to the Substiles distance from the Meridian, and draw the line F C for the Substile; Draw the line B A perpendi∣cular thereto, and passing through the Center at C, and out of the line of Latitudes on the other Quadrants, or out of the Quadrant of Latitudes on this Quadrant, set off B C and C A each equal to the Stiles height, then fit in the Scale of 6 hours, proper to those Latitudes, so that one Extremity meeting at A, the other may meet with the Substilar line at F.

Then get the difference between 30^d 47′, the inclination of Me∣ridians, and 30^d the next hours distance lesser then the said Ark, the difference is 47′ in time, 3′ nearest then fitting in the Scale of hours as was prescribed.

Count upon the said Scale,

Hour.	Min.
0	3	from F to	10
1	3	11
2	3	12
3	3	1
4	3	2
5	3	3

And make points at the termi∣nations with a pin or pen, & draw lines from those points into the Center at C, & they shall be the true hour-lines required on this side the Substile.

Again, Fitting in the Scale of Hours from B to F, count from that end at B the former Arks of time.

Ho	Min
00,	03	from B to	4
1,	3	5
2,	3	6
3,	3	7
4,	3	8
5,	3	9

And make Points at the Termi∣nations, through which draw Lines into the Center, and they shall be the hour Lines requi∣red on the other side the Sub∣stile.

The like must be done for the halfs and quarters, getting the difference between the half hour next lesser (in this Example 22^d 30′) under the Ark called the inclination of Meridians, the diffe∣rence is 1^d 17′ in time 33′ nearest to be continually augmented an

hour at a time, and so prickt off as before was done for the whole hours.

By three facil Proportions, may be found the Stiles height, the Inclination of Meridians, and the Substiles distance from the Plains perpendicular, for all Plains Declining, Reclining, or Inclining, which are sufficient to prick off the Dyal after the manner here described, which must be referred to another place.

If the Scale of hours reach above the Plain, as at B, so that B C cannot be pricked down, then may an Angle be prickt off with Chords on the upper side the Substile, equal to the Angle F C A, on the under side, and thereby the Scale of hours laid in its true situation, having first found the point F on the under side.

To prick down the former Dyal in a Rectangular ☉blong, or long square Figure from the Substile.

Having set off the Substilar F C, assume any distance in it, as at F to be the Radius, and through the fame at right Angles, draw the line E F D, then having made F C any distance Radius, take out the Sine of the Stiles height to the same Radius, and entring it at the end of the Scale of three hours, make it the Radius of a Tangent, and proportion out Tangents to

• 3′ and set them off from F to 10
• 1 hour 3 and set them off from F to G
• 2 3 and set them off from F to H

Again, Take out the Tangents of the Complement of the first Ark, increasing it each time by the augmentation of an hour, namely

• 57′ and prick them from F to I and from the points
• 1 ho. 57 and prick them from F to K and from the points
• 2 57 and prick them from F to E and from the points

thus found, draw lines into the Center.

Then for the other sides of the Square, make C F the Radius of the Dyalling Tangent of 3 hours, and proportion out Tangents to the former Arks, namely,

• 3′ and prick them from B to P Also to the latter Arks, 57′ and prick them from A to—N
• 1 ho. 3 and prick them from B to O Also to the latter Arks. 1 h. 57 and prick them from A to—M
• 2 3 and prick them from B to L Also to the latter Arks. 2 57 and prick them from A to—D

and draw lines from these terminations into the Center, and the Hour-lines are finished; after the same manner must the halfs and quarters be finished.

And how this trouble in Proportioning out the Tangents may be shunned without drawing any lines on the Plain, but the hour-lines, may be spoke to hereafter, whereby this way of Dyalling, and those that follow, will be rendred more commodious.

Lastly, the Stile may be prickt off with Chords, or take B C, and setting one foot in F, with that Extent sweep the touch of an oc∣cult Arch, and from C, draw a line just touching the outward ex∣tremity of the said Arch, and it shall prick off the Angle of the Stiles height above the Substile.

To prick off the former Dyal in an Ob∣lique Parallellogram, or Scalenon alias unequal sided Triangle from the Meri∣dian.

[illustration]

First, In an Oblique Parallellogram.

DRaw CE the Meridian line and with 60^d of a line of Chords, draw the prickt Arch, and therein from K, contrary to the Coast of Declination, prick off 62^d, the angle of 12 and 6, and

draw the line C D for the said hour line continued on the other side the Center, and out of a line of Sines, make C E equall to 65^d the Complement of the Declination; then take out the sine of 38^d 28′ the Complement of the Latitude, and enter it in the line D C, so that one foot resting at D, the other turned about, may but just touch the Meridian line, the point D being thus found, make C F equall to C D, and with the sides C F and C E make the Pa∣rallellogram D G F H, namely, F H and G D equal to C E: and E G and E H equal to D C. And where these distances (sweeping occult arches therewith) intersect will find the points H and G limiting the Angles of the Parallellogram.

Then making E H or C D Radius, proportion out the Tangents of

• 15^d and prick them down from E to 1 and 11 and
• 30 and prick them down from E to 2 and 10 and

draw lines into the Center through those points, and the angular points of the Parallellogram at H and G, and there will be 6 hours drawn, besides the Meridian line or hour line of 12. Then making D G Radius, proportion out the Tangent of 15^d, and prick it down from D upwards to 5, and downward to 7, also proporti∣on out the tangent of 30^d and prick it from D to 8, and from F to 4, and draw lines into the Center, and so the hour lines are fi∣nished; after the same manner are the halfs and quarters to be proportioned out and pricked down: and if this Work is to be done upon the Plain it selfe, the Parallel F H will excur above the plain, in that case, because the Parallel distance of F H from the Meridian, is equal to the parallel distance of D G the space G. 8. may be set from H to 4, and so all the hour lines prickt down.

To prick down this Dyal in a Scalenon, or unequal sided tria∣ngle from the Meridian, from E to D draw the streight line D E, and from the same point draw another to F, and each of them (the former hour lines being first drawn) shall thereby be divi∣ded into a line of double tangents, or scale of 6 hours, such a one as is in the Diameter of the Circle on this quadrant, or on the right edge of the foreside; and therefore by helpe of either of them lines, if it were required to prick down the Dyal, it might be done by Proportioning them out, take the extent D E, and prick it from one extremity of the Diameter in the Semicircle on the qua∣drant,

and from the point of Termination draw a line with black Lead to the other extremity, (which will easily rub out again ei∣ther with bread or leather parings) and take the nearest distance from

• 15 of the Diameter to the said line, and the said extents
• 30 of the Diameter to the said line, and the said extents
• 45 of the Diameter to the said line, and the said extents

• shall reach from, E to 11 and from D to 7
• shall reach from, E to 10 and from D to 8
• shall reach from, E to 9 and from D to 9

and the like must be done for the line E F, entring that in the Se∣micirle as before; or without drawing lines on the quadrant, if a hole be drilled at one end of the Diameter, and a thred fitted into it, lay the thred over the point in the Diameter, and take the nearest distances thereto.

Lastly, from a line of Chords, prick off the substilar line, and the stiles height as we before found it.

This way of Dyalling in a Parallellogram, was first invented by John Ferrereus a Spaniard, long since, and afterwards largely hand∣led by Clavius, who demonstrates it, and shews how to fit it in∣to all plains whatsoever, albeit they decline, recline, or incline, without referring them to a new Latitude; the Triangular way is also built upon the same Demonstration, and is already published by Mr Foster in his Posthuma, for it is no other then Dyalling in a Parallellogram, if the Meridian line C E be continued upwards, and C E set off upwards, and lines drawn from the point, so found to D and E, shall constitute a Parallellogram.

An Advertisement about observing of Altitudes.

IMagine a line drawn from the beginning of the line Sol, to the end of the Diameter, and therein suppose a pair of sights placed with a thred and bullet hanging from the begining of the said line, as from a Center; I say the line wherein the sights are placed, makes a right angle with the line of sines on the other side of Sol, and so may represent a quadrant, the equal Limbe whereof is either re∣presented by the 90^d of the equal Semicirle, or by the 90^d of the Diameter and thereby an Altitude may be taken. Now to make an Isoceles equicrural, or equal legged triangle made of three streight Rulers, the longest whereof will be the Base or Hipotenusal line; thus to serve for a quadrant to take Altitudes withal, will be much cheaper, and more certain in Wood, then the great Arched woo∣den framed quadrants. Moreover, the said Diameter line supplies all the uses of the Limbe, from it may be taken off Sines, Tangents, or Secants, as was done from the Limbe; and therein the Hour and Azimuth, found generally by helpe of the line of Sines on the left edge, as is largely shewed in the uses of this quadrant, besides its uses in Dyalling, onely when such an Instrument is made apart, it will be more convenient to have the line of Sines to be set on the right edge, and the Diameter numbred also by its Complements; this Diameter or double Tangent, or Hipotenusal line being first divided on, all the other lines may from it, by the same Tables that serve to graduate them from the equal Limbe be likewise inscribed: and here let me put a period to the uses of this quadrant.

Gloria Deo.

[illustration]

The Description of an Universal small Pocket Quadrant.

THis quadrant hath only one face.

On the right edge from the Center is placed a line of sines divided into degrees and half degrees up to 60 d. afterwards into whole degrees to 80 d.

On the left edge issueth a line of 10 equal parts, from the Center being precise 4 inches long, each part being divided into 10 subdivisions and each subdivision into halfs.

These two lines make a right angle at the Center, and be∣tween them include a Projection of the Sphere for the Lati∣tude of London.

Above the Projection are put on in quadrants of Circles a line of Declinations 4 quadrants for the dayes of the moneth, above them the names of 5 Stars with their right Ascensions graved against them, and a general Almanack.

Beneath the Projection are put on in quadrants of Circles a particular sine and secant, so called, because it is particular to the Latitude of London.

Below that the quadrat, and shadowes.

Below that a line of Tangents to 45 d.

Last of all the equal Limbe.

On the left edge is placed the Dialling scale of hours 4 In∣ches long, outwardmost on the right edge a line of Latitudes fitted thereto.

Within the line of sines close abutting thereto is placed a small scale called the scale of entrance beginning against 52 d. 35 min. of the sines numbred to 60 d.

The line of sines that issueth from the Center should for a particular use have been continued longer to wit to a secant of 28 d. because this could not be admitted, the said secant is placed outward at the end of the scale of entrance towards

the Limbe, and as much of the sine as was needful placed at its due distance, at the other end of the scale of entrance.

Of the uses of the said quadrant.

THe Almanack hath been largely spoke to in pag. 12, and 13, also again in the uses of the Horizontal quadrant pag. 11, 12.

The quadrat and shadowes from pag. 35 to 44.

The line of Latitudes and scale of hours pag. 250. Again, from page to 262 to 274, also the line of sines, equal parts, and Tangents, in other parts of the Booke.

The use of the Projection.

THis projection is only fitted for finding the hour in the limb, and not the Azimuth, all the Circular lines on it are parallels of Altitude or Depression except the Eclip∣tick and Horizon, the Ecliptick, is known by the Characters of the signes, and the Horizon lyeth beneath it, being numbred with 10, 20, 30, 40.

The parallels of Altitude are the Winter parallels of Stofler's Astrolable, and are numbred from the Horizon upwards to∣wards the Center, the parallels of Depression which supply the use of Stoflers Summer Altitudes are numbred downwards from the top of the Projection towards the limb.

To find the time of Sun rising, and his Amplitude.

LAy the thread over the day of the moneth, and set the Bead to the Ecliptick, then carry the thread and Bead to the Horizon, and the thread in the limb, shewes the time of rising, and the Bead on the Horizon the quantity of the Amplitude.

Example.

So on the second of August the Suns Declination being 15 d▪ his Amplitude will be 24 d. 35 min. and the time of rising 41′ past 4 in the morning.

To find the Hour of the Day.

HAving taken the Suns Altitude and rectified the Bead as before shewed, if the Sun have South Declination bring the Bead to that parallel of Altitude on which the Suns height was observed, amongst those parallels that are numbred upward towards the Center, and the thread in the limb sheweth the time of the day.

Eample.

So when the Sun hath 15 d. of South Declination, as about 28th of January, if his Altitude be 15 deg. the time of the day will be 39 m. past 2 in the afternoon, or 21 m. past 9 in the morning.

But in the Summer half year bring the Bead to lye on those parallels that are numbred downwards to the limb, and the thread sheweth therein the time of the day sought.

Example.

If on the second of August his Declination being 15 d. his Altitude were 40 d. the true time of the day would be 8 m. past 9 in the morning, or 52 m. past 2 in the afternoon.

If the Bead will not meet with the Altitude given amongst those parallels that run donwnwards towards the right edge, then it must be brought to those parallels that lye below the Horizon downward towards the left edge, and the thread in the Limb shewes the time of the day before six in the morning or after it in the evening in Summer.

Example.

When the Sun hath 15 d. of North Declination as on the second of August if his Altitude be 5 d. the time of the Day will be 44 m. before 6 in the morning, or after it in the evening.

Of the general lines on this quadrant.

THe line of sines on the right edge is general for finding either the hour or the Azimuth in the equal limb, or in the said line of sines, as I have largely shewed in page 231 for finding the hour, also for finding the Suns Altitudes on all hours, as in page 234, for finding the Azimuth from page 237 to pag. 2.9.

Though this quadrant hath neither secants nor versed sines as the rest have, yet both may be easily supplyed, let it be re∣quired to work this Proportion.

• As the Co-sine of the declination,
• Is to the secant of the Latitude,
• So is the difference of the sines of the Suns Meridian and given Altitude,
• To the versed sine of the hour from noon,

before or after six the hour may be found from midnight by the proportions in page 230.

Let the Radius of the sines be assumed to represent the se∣cant of the Latitude, the Radius to that secant will be the cosine of the Latitude, then lay the thread to the complement of the Declination in the limb counted from right edge, and take the nearest distance to it. I say that extent shall be the cosine of the Declination to the Radius of the Secant enter this at 90 d. of the line of sines laying the thread to the other foot according to nearest distance, then in the sines take the distance between the Meridian Altitude and the given Altitude, and enter that extent so upon the sines that one foot resting thereon, the other turned about may just touch the thread the distance between the resting foot and the Center is equal to the versed sine of the ark sought and being measured from the end of the line of sines towards the Center shewes the ark sought.

Example.

When the Suns Declination is 15 d. North if his Altitude were 35 d. 21 m. the time of the day would be found 45 d. from noon, that is 9 in the morning or 3 in the afternoon.

Of finding the Azimuth generally.

THough this may be found either by the sines alone in the e∣qual limbe as before mentioned, or by versed sines as was instanced for the hour, see also page 239, 240, 241, yet where the Sun hath vertical Altitude or Depression, as in places with∣out the Tropicks towards either of the Poles, it may be found most easily in the equal limb by the joynt help of sines and tangents by the proportions in page 175.

First, find the vertical Altitude as is shewed in page 174. Then for Latitudes under 45 d.

Enter in Summer Declinations the difference, but in Winter Declinations the sum, of the sines of the vertical Altitude, and of the proposed Altitude once done the line of sines from the Center, and laying the thread over the Tangent of the Latitude take the nearest distance to it, then enter that Extent at the complement of the Altitude in the Sines, and lay the thread to the other foot, and in the limb it shewes the Azimuth from the East or West.

Example. For the Latitude of Rome to witt 42 d. If the Sun have 15 d. of North Declination his vertical Altitude is 22 d. 45 m. If his given Altitude ••e 40 d. the Azimuth of the Sun will be 17 d. 33 m. to the Southward of the West. If his declina∣tion were as much South and his proposed Altitude 18 d. his Azimuth would be 41 d. 10 m. to the Southwards of the East or West. For Latitudes above 45.

If we assume the Rad. of the quadrant to be the tangent of the Latit. the Rad. to that Tang. shall be the co-tangent of the Latit. wherefore lay the thread to the complement of the Latitude in the line of Tangents in the limb, and from the complement of the Altitude in the Sines take the nearest distance to it, I say

that extent shall be cosine of the Altitude to the lesser Radius which measure from the Center, and it finds the Point of en∣trance whereon enter the former Sum or difference of sines as before directed, and you will find the Azimuth in the equal limb.

Or if you would find the answer in the sines, enter the first extent at 90 d. laying the thread to the other foot, then enter the Sum or difference of the Sines of the vertical and given Al∣titude, so between the scale and the thread, that one foot turned about may but just touch the thread, the other resting on the sines, and you will find the sine of the Azimuth sought.

Example. For the Latitude of Edinburg 55 d. 56 m. If the Sun have 15 d. of Declination, his vertical Altitude or depres∣sion is 18 d. 14 m. the Declination being North, if his proposed Altitude were 35 d. the Azimuth of the Sun would be 28 d. to the South-wards of the East or West.

But if the Declination were as much South, and the Altitude 10 d. the Azimuth thereto would be 46 d. 58 m. to the South-wards of the East or West.

The first Operation also works a Proportion to witt.

• As the Radius
• Is to the cotangent of the Latitude.
• So is the cosine of the Altitude,
• To a fourth sine.

I say this 4th sine beares such Proportion to the Radius as the cosine of the Altitude doth to the tangent of the Latitude, for the 4th term of every direct Proportion beares such Pro∣portion to the first terms thereof, as the Rectangle of the two middle terms doth to the square of the first term.

But as the rectangle of the co-tangent of the Latitude and of the cosine of the Altitude is to the square of the Radius, So is the cosine of the Altitude is to the tangent of the Latitude, or which is all one, So is the co-tangent of the Latitude, To the secant of the Altitude, as may be found by a common division of the rectangle, and square of the Radius by either of the terms of the said rectangle, by help of which notion I first found out the particular scales upon this quadrant.

All Proportions in sines and Tangents may be resolved by the sine of 90 d. and the Tangent of 45 d. on this quadrant if what hath been now wrote, and the varying of Proportions be understood, as in page 72 to 74 it is delivered.

Because the Projection is not fitted for finding the Azimuth there are added two particular scales to this quadrant, namely, the particular sine in the limb, and the scale of entrance abutting on the sines fitted for the Latitude of London.

Lay the thread to the day of the moneth, and it shewes the Suns Declination in the scales proper thereto.

Then count the Declination in the Limbe laying the thread thereto, and in the particular sine, it shewes the Suns Altitude or Depression being East or West.

To find the Suns Azimuth.

FOr North Declinations take the distance between the sines of the vertical Altitude and given Altitude, but for South Declinations adde with your compass the sine of the given Al∣titude to the sine of the vertical Altitude, enter the extent thus found, at the Altitude in the scale of entrance laying the thread to the other foot according to nearest distance, and in the equal limb it shewes the Azimuth sought from the East or West, or it may be found in the sines by laying the thread to that arch in the limb that the Altitude in the scale of entrance stands against in the sines, and entring the former extent paralelly between the thread and the sines.

Example. So when the Sun hath 13 d. of Declination his vertical Altitude or Depression is 16 d. 42 m. If the Declination were North and his Altitude 8 d. 41 m. his Azimuth would be 10 d. to the North-wards of the East or West. But if it were South and his Altitude 12 d. 13 m. the Azimuth would be 40d to the South-wards of the East or West.

By the same particular scales the hour may be also found.

To find the time of Sun rising or setting.

TAke the sine of the Declination, and enter it at the Declination in the Scale of entrance and it shewes the time sought in the equal lim••e from six.

Example. When the hath 10 d. of Declination the Ascensional diffe∣rence is 49 m. which added to, or substracted from six shewes the time of rising and setting.

To find the hour of the day for South Declination.

IN taking the Altitude, mind what Ark in the particular Sine the thread cut, adde the Sine of that Ark to the Sine of the Declination, and enter that extent at the Declination in the Scale of entrance laying the thread to the other foot according to nearest distance, and in the equal limb it shewes the hour from six. So if the Declination were 13 d. South and the Suns Altitude 14 d. 38 m. the thread in the particular Sine would cut 18 d. 49 m. and true time of the day would be 9 in the morning or 3 in the morning.

To find the time of the day for North Declination.

HAving observed what Ark the thread in taking the Altitude hung over in the Particular Sine take the distance between the Sine of the said Arke, and the Sine of the Declination and entring that extent at the De∣clination in the Scale of entrance the thread in the limbe shewes the hour from six.

Example. If the Suns Declination were 23 d. 31 m. North, and his Al∣titude 39 d. the Arch in the particular Sine would be 53 d. 32 m. and the time of the day would be about 3 quarters past 3 in the afternoon, or a quar∣ter past 8 in the morning.

When the Altitude is more then the Latitude the thread will hang over a Secant in the particular Scale, this happens not till the Sun have more then 13 d. of North Declination, in this case take the distance be∣tween the Secant before the beginning of the Scale of entrance, and the Sine of the Declination at the end of the same and enter it as before.

Example. The Suns declination being 23 d. 31 m. North if his Altitude were 55 d. 29 m. the thread in the particular Scale would hang over the Secant of 18 d. 11 m. and the true time of the day would be a quarter past 10 in the morning, or 3 quarters past 1 in the afternoon. The Proportions here used are expressed in Page 193.

The Stars hour is to be found by the projection by rectifying the Bead to the Sar and then proceed as in finding the Suns hour, afterwards the ••ue time of the night is to be found as in page 32.

ERRATA.

In the Treatise of the Horizontal quadrant, pag. 43 line 6 for the 3 January read the 30th. In the Reflex Dialling pag. 5, adde to the last line these words, As Kircher sheweth in his Ars Anaclastica.

FINIS.