PROBLEM I. The Sun's Place being given, to find his De­clination.

COUNT what Parallel from the Equinoctial doth cut the Place of the Sun in the Ecliptick, for that is the Declination required. So in ♉ 00° 00′ you will find the 11° 30′ Parallel cuts the Ecliptick.

PROB. II. Contrarily, the Declination being given, to find the Sun's Place.

Count 11° 30′ among the Parallels, and that will cut 00° 00′ of ♉.

PROB. III. The Place of the Sun being given, to find his Right Ascension.

Look what Meridian, reckoning from the Center, doth pass through the Suns place, for that in Degrees is the Rig [...] Ascension required. So will the Right Ascension for [...] beginning of ♉ be found 27° 54′. So likewise f [...] [...] beginning of ♏; to which adding 180, because [...] Southern Sign, the Right Ascension will be found 207° [...]

PROB. IV. The Elevation of the Pole, and Degree of the Ecliptick being given, to find, 1st. The Or­tive Latitude. 2dly. The Rising and Set­ting of the Sun. 3dly. The Semidiurnal Arch.

Place the Finitor to the Latitude, then see what degree thereof doth cross the Sun's Parallel, for the Degree in the Ecliptick given, for that is the Ortive Latitude sought, being reckoned from the Center, or East point in the Instrument. So in the Latitude of 49 I find the Ortive Latitude about 38 in the Tropick of Cancer. 2dly, The Meridian there passing, shews the Sun's Rising to be ve­ry near four in the Morning. And 3dly. The Semidiur­nal Arch to be 8 hours, and so the longest day there 16 hours.

PROB. V. The same things being given, to find the Ascensional Difference.

The Horizon, as before, being set to the Latitude, count from the Center to the Meridian that cuts it in the Sun's Place for that day; that is to say, in the Sun's Parallel for the place of the Eccliptick given. So shall you find the Ascensional Difference for the beginning of ♉ in the fore­said Latitude of 49° to be 13° 32′.

PROB. VI. To find the Oblique Ascension.

First, find the Right Ascension by the third Problem, and the Ascensional Difference by the preceding one. In Northern Signs subduct the Ascensional Difference out of the Right Ascension, but in Southern Signs add it. So shall the Summe, or Difference give you the Oblique Ascen­sion required. So in the beginning of ♉ I find the Sun's Right Ascension 27 54, and in the Latitude of 49° 00′, his Ascentional Difference 13° 32′, their Difference is 14° 22′, which is the Oblique Ascension for the beginning of ♉. So the Oblique Ascension for the beginning of ♏, will be found 221° 26′ by adding 13° 32′ to 207° 54′, the Right Ascension before found.

PROB. VII. To find the Oblique Ascension of any other Point in the Ecliptick, not reckoning from the Aequinoctial points, by which you may know whether the said Sign doth ascend Right, or Obliquely.

By the preceding Problem find the Oblique Ascension for the beginning and end of ♉, or any other Arch pro­pounded, then subducting the lesser out of the greater, the residue is the Oblique Ascension required. So in the La­titude of 49° 00′ I desire to know what degrees of the Aequinoctial do in an Oblique Sphere ascend between the beginning and end of ♉, I find the oblique Ascension for the beginning of ♉ to be 14° 22′, and of the latter end 32° 45′, their Difference is 18° 23′, the Oblique Ascen­tion required. Where note, because the remaining Summe is in the Example less than the given Arch, the said Sign doth ascend Obliquely, contrarily if it had been greater.

PROB. VIII. To sind the Hour of the day, the Altitude of the Sun being first observed.

Set the Ruler to the Latitude, and remove the Vertical till it cross the Sun's Parallel in the Altitude observed, the Me­ridian there passing shews the hour of the day. So in the beginning of ♉ in the Latitude of 52° 00′ I observed the ☉ Altitude 27° 18′, and the 60th. Meridian doth cross the Sun's Parallel in that Altitude, therefore it was then [Page 5]eight of the Clock in the Morning. Thus also may you make a Table of Altitudes for every hour of the day, by re­moving the Vertical from hour to hour, and observing the Degrees of Altitude cut off by the several Meridians.

PROB. IX. To find what Degree of the Ecliptick is in the Meridian at any hour given.

Find the Right Ascension by the third Problem, then turn the hour given from the Meridian into degrees and minutes of the Aequinoctial, add unto these the Right Ascen­sion found, then examine what point in the Ecliptick doth answer to those Degrees of Right Ascension, which is done by following the Meridian, which shews the Right Ascen­sion up to the Ecliptick, for that part of the Ecliptick is in the Meridian.

So in the beginning of ♉, and at the hour of nine in the Morning, I find the Right Ascension 27° 54′, the hours from the Meridian last past 21° 00′ in degrees 315°, to which I add 27° 54′, it makes 342° 54′ which is the Right Ascen­sion for the 11th. Degree of ♓, or near, which was the de­gree of the Ecliptick then in the Meridian, which will be ♓ 11° 30′.

PROB. X. To find the Sun's Azimuth at any Altitude given.

Place the Finitor to your Latitude, and remove the Ver­tical till it intersect the Sun's Parallel in the Altitude given, then applying it over in the Aequinoctial-line, the Meridian which intersects the Vertical in the given Altitude, shews [Page 6]the Sun's Azimuth, or Distance from the South or East, as you shall please to account it. So in the Latitude of 48° 50′, the Sun's Altitude being 31° 40′ in the Tropick of ♋. Then moving the Vertical till it cross the Parallel of ♋ in that Altitude, I there find it, and then moving the Horizon over in the terms of the Aequinoctial, I find that 88° 2′ Me­ridian doth cross the Vertical in the Altitude given, which is the Azimuth from the South. Here you are to take no­tice, that those which were before Meridians are now be­come Vertical Circles in this manner of working.

PROB. XI. To make an Horizontal DIAL.

Set the Horizon or Finitor to the Latitude of that place for which you desire to make your Dial, and then observe the several Angles made by the Meridians upon the Hori­zon, which express upon your Plain, and let your Style make an Angle equal to the Latitude of your Place for which you make your Dial. This needs no Example.

PROB. XII. To make an Ʋpright Vertical DIAL.

Set the Ruler to your Latitude, and the moveable Verti­cal in the Center of your Instrument, then observe the se­veral Angles made between the 12 of Clock line and the other Meridians, not upon the Horizon, as you did before; but upon the Vertical, and those shall be the hour spaces required. So in the Latitude of 49° the angle of 11 will be 9° 57′, and also for 1; 20° 44′ for 10h. and 2h. 33° 16′ for 9h. and 3h. and so the rest.

PROB. XIII. To make a Ʋpright Declining DIAL.

For the placing the Style, and drawing the Hour-lines there must necessarily be resolved a Sperical Rect-angled Triangle, in which there will be given, beside the Right An­gle, the Complement of your plains Declination (which I all­ways count from the Pole of the Plain to the North or South Points) and thirdly, the Complement of your Lati­tude; by help whereof you may find, 1st. The Elevation of the Pole above the Plain. 2dly. The distance of the Substyler from the Vertical in your Plain. And 3dly. The Difference of Longitude, or the Angle comprehended between the Meridian of the Plain, and Meridian of the Place. All which will be made plain by the Figure following: in which,

The Scheme following represents a South Plain Declining from the North Westward 55° 30′ in the North Latitude 46° 12′.

[Page 8]

Let ENWS be the Horizon of the place and the letters as they are in their order, mark, the East, North, and West and South points. Let then a plain BC be given, the pole whereof will be D, the Meridian of the Plain which passeth allwayes through P, the pole of the world, and D the pole of the plain DPR, the Declination DZN the Comple­ment thereof NZC, the Triangle to be resolved PZR, in which are given;

• 1st. PZR the Complement of the Declination: 34° 30′.
• 2dly. PZ the Complement of the Latitude 43° 48′.
• 3dly. PRZ the Right Angle.

Things required are,

1st. PR the Elevation of the Pole above the Plain.

2dly. ZR the distance of the Substylar from the Verti­cal, or the angle made between the Substylar and the Me­ridian, or line of 12h.

3dly. ZPR, the Difference of Longitude, or the angle comprehended between the Meridian of the Plain, and of the Place.

[Page 9]Now these things are thus found at one Operation by this Instrument.

1st. Reckon from the Aequinoctial towards the Pole, the Complement of the Declination of your Plain, which in our Example will be 34° 30′, at that point place the Ho­rizontal Ruler, then from the Center of your Instrument towards the Limb, reckon upon the Horizontal Ruler, the side given PZ the Complement of your Latitude, and the Parallel there intersecting being reckoned from the Aequinoctial, shall be PR the Elevation of the Pole above your Plain.

2dly. The Instrument standing in the same situation, count from the Limb of your Instrument towards the Cen­ter among the Meridians, the Complement of your Latitude, for the Meridian, there intersecting the Horizon, shall, upon the Degrees of the Horizon give you ZR, the Distance of the Substylar from the Meridian of the Place.

3dly. the Parallel of Declination in the common Intersecti­on of the foresaid Meridian with the Horizon, being num­bred from the Pole of the World, doth give you the Angle sought, ZPR, or the difference of Longitude. But because it may some times fall out, that the said Intersection will be more then 23° 30′, and in little Instruments the Parallels not drawn beyond the Tropicks, which are 23°½ from the Aequinoctial: in that case you may thus help your self: at the common Intersection make a little spot of ink or red Ocre with your pen, and then putting the Horizon back to the terms of the Aequinoctial, run the movable Vertical to the said mark, the degrees of which shall shew you the Angle sought, counted from the Pole of the world; or otherwise place one foot of your compasses in the said Intersection, and take the nearest Distance between that and the Aequi­noctial, which being applyed over, either in the limb, or Vertical, shall give from the Aequinoctial the Angle requi­red.

PROB. XIV. To draw the Hour-lines upon a Reclining Plain, whose Face looketh directly toward the North or South.

Let a Plain be given, reclining from the Zenith toward the North 20° in the Latitude of 49° 00′. Here I set the Finitor to the Latitude, and the moveable Vertical to the Degree of Reclination, reckoning from the Zenith to­wards that Pole towards which the Reclination is. As in our Example of 20° Reclination towards the North Pole, the Meridians there intersecting denote the Horary Angles, and the Degrees between the Pole and the Vertical, that is to say, in our Example 22° 00′ is the Elevation of the Pole above the Plain. This is very little different from the way of making an upright Vertical Dial, and therefore needs no Example.

PROB. XV. In any Spherical Triangle whatever, having two sides given with the Angle comprehend­ed to find the rest.

In the foregoing Triangle BAC let BA be 60, AC 50, and the angle BAC 30.

[Page 13]Reckon on the Limb from the Aequator toward the North Pole, and on that side of your Instrument that is on your Right Hand, one of the sides given, Viz. AC 50°, and there place the Horizontal Ruler, and then among the Parallels from the North count the other side AB 60°, and among the Meridians from the Left Hand, there count the Angle given, Viz. 30° BAC, & at the common Intersection of that Meridian with the Parallel, place the Moveable Ver­tical; this being done, apply the Horizontal Ruler over in the terms of the Aequinoctial, so shall the Degree in the Vertical, which was before observed, viz. 26½, in the common Inter­section of the Meridian with the Parallel, give you among the Parallels, reckoning from the Pole the Base 26°½, and among the Meridians, from the utmost on the Right Hand towards the Left you shall find the Angle C. 103. Now for the Angle B, count AB 60° in the Limb, and there place the Horizon, and count the side AC 50° among the Pa­rallels, and among the Meridians towards the Right Hand, count the Angle given 30°, and apply the Vertical to the common Intersection, then apply the Horizon as before over in the Aequinoctial; so among the Meridians from the utmost on the Right Hand towards the Left, you shal in the common Intersection find the Angle unknown B 59½ So have you now in your Triangle the three sides and two Angles, the third Angle may be found either by letting fal a Perpendicular, or continuation of the sides to a Semicircle

PROB. XVI. To find when the Twilight begins and ends.

The Twilight in the Morning begins when the Sun is di­stant 18 degrees from the Horizon, & continues till it Rise; and at Night begins at Sun Set, and continues till he is 18 degrees below the Horizon, and the [...] Darkness begins.

To find this, you must know the Place of the Sun, and [Page 14]his Declination. The Sun's Place may be known by most Almanacks, or by a Scale, which may be annexed to any convenient Place in the Instrument. The 2d. to wit, the ☉'s Declination, is found by the first Problem.

Example First. If the ☉ be in the Aequinoctial, he hath no Declination; if you therefore place the Finitor to the Latitude of your Place, move your Vertical till the 18th. degree thereof crosses the Aequinoctial, and mark what Me­ridian that is, counting from the Center of your Planisphere, which in the Latitude of 52 degrees, you will find to be the 30th. Meridian; so that you may conclude the Twilight begins two hours before ☉ Rise, and at Night ends two hours after ☉ Set.

But if the ☉ have Declination, you must observe that Me­ridian which crosseth the 18th. degree of the Vertical in the Parallel of Declination.

Example. Suppose the ☉ to be in ♊ 00° 00′ in the La­titude of 52° 00′. When the ☉ hath 20° of Declination, and observe the Intersection made by the 18th. degree of the Vertical, and the 20th. Parallel of Declination, and there you may observe the 33 Meridian, or thereabout, from ☉ Rising, passing which sheweth the Twilight will then end, to wit, 2 hours and 12′ before the ☉ should rise, but at this time you will find the ☉ is up, which teaches you there is then no Twilight at all. So in the Winter, the ☉ Sets so much before 8h. that it will not come to be 18 de­grees below the Horizon till ☉ Rise, and will continue so till the ☉ comes back to ♌, there will be no dark time in that Latitude, if the Sky be not Cloudy.

PROB. XVII. Of Spherical Rect-angled Triangles in all their Varieties.

I have in the third and [...]ourth Problems foregoing, shewed you the Solution of some Questions, wherein the Triangle [Page 15]hath been Rect-angled and Spherical. I shall now handle the Use of them more fully, and shew you how three parts, that is, two beside the Rectangle, which is always known, being given; the other parts sought, either Sides or Angles may be by this Planisphere many ways concluded, as will appear in Practice.

1. In the Scheme.

2 Or,

The Data in the Tri­angle in our following Geniture are 3° 00′ ☉ Long. 1½ ☉ Declin. with the Rect. and the Angle at C will be found 66° 32′, which will be demonstrated by the Scheme, Nº 4.

The Proportions in Trigonometry are thus: BC Rad. BA.C Scheme 1. Or, CA. Scheme 2.

PC Rad. PD.C. That is, As the Cosine of the Declination given is to the Radius, so is the Cosine of the greatest Declination to the Angle sought.

In the Triangle BAC Rect-angled, at A the Longitude, BC is 60° 00′; the Right Ascension may be found by Pro­blem 3d. 57.48, and the Angle at B 23° 30′. Now to find the Declination, do thus: Choose out the 57° 48′ Me­ridian, and then move the Horizontal Ruler till the 60th. degree from the Center meets the 57° 48′ Meridian, which you will find will be 20° 12′ the Declination sought, if you count the Parallels from the Aequinoctial till the Interse­ction is made in the 57° and 48′ Meridian: or if you have the Declination given, bring down the Horizontal Ruler to the Aequinoctial, and you will find the 57° 48′ of the Ruler will be cut by the 57° 48′ Meridian.

[Page 16]But we still want the Angle C; to obtain this, we must turn the Triangle, and as in the annex­ed Scheme,

call BA 20° 12′ and AC 57° 48′: choose now the 20° 12′ Meri­dian, and remove the Horizontal Ruler till the Longitude given, Viz. 60° doth intersect the 20° 12′ Meridian, then count from the Aequinoctial upon the Limb the Degrees cut off by the Ruler, which you will find to be 77° 43′, for the Acute Angle at C, and its Supple­ment 113° 17′ will be the Obtuse An­gle. So you have all the Angles and Sides of your Triangle.

Though I have a very low opinion of Astrology, especially as our Genethliacall-men, or rather Fortune-Tellers use it; yet that I may shew the excellent use of this Planisphere, and because it may be useful to know the figure of the Heavens in Eclipses, and the Ingress of the Sun into the Aequinoctial points: I shall now shew you how to place the 12 Houses with the Signs to them belonging. In this I shall take for Example ageniture long since past, and not agreeable to any Example before given, that by variety of Operations the Reader may better comprehend the use of the Instrument.

PROB. XVIII. How to erect a Figure of the Heavens.

First for the time propounded you must first find the Suns place, that is, in what Degree of the Ecliptick the Sun is in at the time proposed; and afterward, in the following Ex­ample; I shall teach you to find the Medium-Coeli, or that de­gree of the Ecliptick that is in the Meridian.

PROB. XIX. In order to find the Cusps of the other Houses, divers Spherical Triangles are to be Re­solved.

We had before shewed how to find the Angle between the Meridian and the Eccliptick, to be 66°½, by Prob. 7. which we shall now find another way.

The Degree culminating, or the Medium-coeli was ♓ 26½, and consequently its Distance from ♈ was 3°½; count there­fore 3°½ among the Meridians, from the utmost towards the Center, then remove the Horizontal Ruler downward to­wards the Antartick Pole, the Fiducial Edge of the Vertical remaining fixt at the Center, till such time as the Vertical shall cross the 3½ Meridian in the Artick Circle, that is, in the 23½ Parallel from the Pose, and in the Limb you shall meet the 66°½, which is the Angle between the Meridian and the Eccliptick.

2dly. Things standing all in this Posture, look from the extremity of the Vertical Ruler to the 3½ Meridian, there you will find 1½ crossing, which is the Declination of the Mid-Heaven Southward; take this out of 37° the Comple­ment of your Latitude, there rests 35° 30′, which is the Al­titude of the Mid-Heaven, or the Degree culminant, because the Declination is Southward, otherwise it must have been added.

3dly. Number 35° 30′ from the extremity of the Verti­cal, and this will, among the Parallels from the Pole, shew about 42°½, or thereabout, which is the Angle made be­tween the Eccliptick and the Horizon.

4thly. Count among the Meridians from the utmost to that point, and you shall find the 61st. Meridian there, which is the degree of the Eccliptick between the Meridian and the Horizon.

[Page 19] Lastly, You shall find the same Degrees of the Eccliptick, (Viz. 61) if you count the 35½ or thereabout, from the Center for those Degrees of the Ruler will shew you 61, as before.

Thus we have obtained,

All these things will be made plain by the following SCHEMES.

1. The Degree Culminant, that is, the Mid-Caeli was be­fore found between 26 and 27° of ♓, neer the 27°.

2. The Ascendant was found between 25 and 26° of ♋.

[Page 20]

3. The Angle between the Meridian and Eccliptick repre­sented in the Scheme, No. 2, by the angle at L, may be found by the Precept of the 15th. Problem, and will be 66½, and the Triangle BAC in the Scheme, No. 4, is equipollent to that wrought by the Analemma, where you have the same Data, Viz. The Right Angle at A, the greatest Declinati­on at B, the present Declination at the Parallel 1°½, and the third Meridian, and where they cross, will be 66°½ from the Aequinoctial. The Proportion in Trigonometry will be, As Cosine of the Declination given is to the Radius, so will the Cosine of the Declination be to tht Angle sought. Sch. No. 5.

As BC.BAC ∷ Rad. to C.

4. In the Scheme, No. 1. I seek the Angle at 11, which is the An­gle between the Meridian and Cir­cle of Position for the 9th. & 11th. Houses. As HAE Sin. to DAE Tang. ∷ Rad. to the Tang. of the Angle at H. which will be found 44.47.

In the Scheme, No. 1. in the Rect-angled Triangle PRO Rect-angled at R.PR, the Elevation of the Pole above that Circle of Position, is found by this Proportion.

[Page 21]As Rad. to the Lat. So the Angle at ☉ will be to the side PR.

Rad. to OP ∷ POR.PR. But this Work will be shortned by the Table annexed, where you have the Eleva­tion of the Pole given to several Latitudes.

6. I seek the Altitude of the Mid-Caeli in the Scheme, No. 2. HL. There are given HAE the Complement of the Latitude. 2dly. 1°½ the Declination of the Mid-Coeli in our Example. Now because this is a Southern Sign, I take 1°½ from 37° the Complement of my Latitude, the Residue will be 35° 30′, the Altitude of the Mid-Coeli in our Example.

7. I still want No. 2. LG, which is the Degree of the Eccliptick between the Meridian and the Circle of Position. In the Oblique-Angled Triangle HLG you have given HL, the height of the Mid-Caeli, with the two Angles at O and L, by which LG may be found by Trigonometry, ac­cording to the Rules delivered by Artists, by letting fall a Perpendicular either within or without the Triangle, when two Angles and the adjacent side is given. In this manner you may set a Figure of the 12 Houses, which will be near as the Figure adjoyning.

[Page 22]