Institutio mathematica, or, A mathematical institution shewing the construction and use of the naturall and artificiall sines, tangents, and secants in decimal numbers, and also of the table of logarithms in the general solution of any triangle, whether plain or spherical, with their more particular application in astronomie, dialling, and navigation / by John Newton.

CHAP. II.

THE ART OF SHADOWS: Commonly called DIALLING. Plainly shewing out of the Sphere, the true ground and reason of making all kinde of Dials that any plain is capable of.

Problem 1.

How to divide diverse lines, and make a Chord to any proportion given.

FOrasmuch as there is continuall use both of Scales and Chords in drawing the Scheams and Dials following, it will be necessary first to shew the making of them, that such

as cannot have the benefit of the skilful ar∣tificers labour, may by their own pains sup∣ply that defect.

Draw therefore upon a piece of paper or pastboard a streight line of what length you please, divide this line into 10 equal parts, and each 10 into 10 more, so is your line divided into 100 equal parts, by help where of a line of Chords to any proportion may be thus made.

First, prepare a Table, therein set down the degrees, halves, and quarters, if you please, from one to 90. Unto each degree and part of a degree joyn the Chord proper to it, which is the naturall sine of halfe the arch doubled, by the 19th. of the second of the first part: if you double then the natu∣rall sines of 5. 10. 20. 30. degrees, you shall produce the Chords of 10. 20. 40. 60. degrees: Thus 17364 the sine of 10 de. being doubled, the sum will be 34••28, the Chord of 20 deg. and so of the rest as in the Table following.

De	Chord
1	17
2	35
3	52
4	70
5	87
6	105
7	122
8	139
9	157
10	175
11	192
12	209
13	226
14	244
15	261
16	278
17	296
18	313
19	330
20	347
21	364
22	382
23	398
24	416
25	432
26	450
27	466
28	384
29	501
30	••18
31	534
32	551
33	568
34	585
35	601
36	618
37	635
38	651
39	668
40	684
41	700
42	717
43	733
44	749
45	765
44	781
47	797
48	813
49	830
50	845
51	861
52	876
53	892
54	908
55	923
56	939
57	954
58	970
59	984
60	1000
61	1015
62	1030
63	1045
64	1060
65	1074
66	1089
67	1104
68	1118
69	1133
70	1147
71	1161
72	1176
73	1190
74	1204
75	1217
76	1231
77	1245
78	1259
79	1273
80	1286
81	1299
82	1312
83	1325
84	1338
85	1351
86	1364
87	1377
88	1389
89	1402
90	1414

This done, proportion the Radius of a circle to what extent you please, make AB equal thereto, in the middle whereof, as in C, erect the perpendicular CD, and draw the lines AD and BD, equal in length to your line of equal parts, so have you made an equiangled Triangle, by help whereof and the Table aforesaid, the Chord of any arch proportionable to this Radius may speedily be obtained.

As for example. Let there be required the Chord of 30 deg. the number in the Table answering to this arke is 518, or in proportion to this Scale 52 almost, I take therefore 52 from the Scale of equal parts, and set them from D to E and F, and draw the line EF, which is the Chord desired. Thus may you finde the Chord of any other arch agreeable to this Radius. Or if your Radius be either of a greater or lesser ex∣tent, if you make the base of your Triangle AB equal thereunto, you may in like man∣ner finde the Chord of any arch agreeable to any Radius given. Only remember that if the Chord of the arch desired exceed 60 deg. the sides of the Triangle AD and DB must be continued from A and B as far as need shall require. In this manner is made the line of Chords adjoyning, an∣swerable

to the Radius of the Fundamental Scheme.

[illustration]

And in this manner may you finde the Sine, Tangent or Secant of any arch pro∣portionable to any Radius, by help of the Canon of Naturall Sines, Tangents and Secants, and the aforesaid Scale of equall parts, as by example may more plainly ap∣pear.

Let there be required the sine of 44 de∣grees in the table of natural sines, the num∣ber answering to 44 degrees is 694. I take therefore with my compasses 69 from my Scale of equal parts, and set them from D to G and H; so is the line GH the sine of 44 degrees, where the Radius of the circle is AB.

Again, if there were required the tangent of 44 degrees, the number in the table is 965; and therefore 96 set from D to K and L shall give the tangent required; and so for any other.

Your Scales being thus prepared for the Mechanicall part, we will now shew you how to project the Sphere in plano, and so proceed to the arithmeticall work.

Probl. 2.

The explanation and making of the funda∣mental Diagram.

THis Scheme representeth to the eye the true and natural situation of those circles of the Sphere, whereof we shall have use in the description of such sorts of Dials as any flat or plane is capable of. It is therefore necessary first to explain that, and the making thereof, that the Sy∣metry

of the Scheme with the Globe being well understood, the representation of eve∣ry plane therein may be the better concei∣ved.

Suppose then that the Globe elevated to the height of the Pole be prest flat down in∣to the plane of the Horizon, then will the outward circle or limbe of this Scheme NESW represent that Horizon, and all the circles contained in the upper Hemi∣sphere of the Globe may artificially be con∣trived, and represented thereon, as Azi∣muths, Almicanters, Meridians, Parallels, Equator, Tropicks, circles of position, and such like, the which in this Diagram are thus distinguished.

The letter Z represents the Zenith of the place, and the center of the horizontal cir∣cle, NZS represents the meridian, P the pole of the world elevated above the North part of the Horizon N here at London, 51 degrees 53 minutes, or centesmes of a de∣gree, the complement whereof PZ 38 de∣grees, 47 minutes, and the distance between the Pole and the Zenith; EZW is the prime vertical, DZG and CZV any other intermediate Azimuths, NOS a circle of position, EKW the Equator, the distance whereof from Z is equal to PN, the height

of the Pole, or from S equal to PZ, the complement thereof, HBQX the Tro∣pick or parallel of Cancer, LFM, the Tro∣pique of Capricorn, the rest of the circles intersecting each other in the point P, are the meridians or hour-circles, cutting the Horizon and other circles of this Diagram in such manner as they do in the Globe it∣self.

[illustration]

Amongst these the Azimuths onely in this projection become streight lines, all the rest remain circles, and are greater or lesser, according to their natural situation

in the Globe, and may be thus described.

Open your compasses to the extent of the line AB in the former Probleme, (or to a∣ny other extent you please) with that Ra∣dius, or Semidiameter describe the hori∣zontal circle NESW, crosse it at right an∣gles in Z with the lines NZS and EZW.

That done, seek the place of the Pole at P, through which the hour circles must pass, the Equinoctial point at K, the Tropiques at T and F, the reclining circle at O, and the declining reclining at A; all which may thus be found.

The Zenith in the Globe or Materiall Sphere is the Pole of the Horizon, and Z in the Scheme is the center of the limbe, representing the same, from which point the distance of each circle being given both wayes, as it lyeth in the Sphere, and set upon the Azimuth, or streight line of the Scheme proper thereunto, you may by help of the natural tangents of half their arch∣es give three points to draw each circle by, for if the naturall tangents of both distan∣ces from the Zenith be added together, the half thereof shall be the Semidiameters of those circles desired.

The reason why the natural tangent of half the arches are here taken, may be

made plain by this Diagram following. Wherein making EZ the Radius, SZN is a tangent line thereunto, upon which if you will project the whole Semicircle SWN, it is manifest, by the work, that every part of the lines ZN or ZS can be no more then the tangent of half the arch desired, because the whole line ZN or ZS is the tang. of no more then half the Quadrant, that is, of 45 degrees, by the 19th. of the second Chapter of the first Part; and there∣fore WEA is but half the angle WZA and WEB is but half the angle WZB.

[illustration]

Now then if EZ or Radius of the fun∣damental

Scheme be 1000, ZP shal be 349, the natural tangent of 19 degrees, 23 mi∣nutes, 50 seconds, the half of 38 degrees, 47 minutes, the distance between the North pole and the Zenith in our Latitude of 51 degrees, 53 minutes, or centesmes of a de∣gree. And the South pole being as much under the Horizon as the North is above it, the distance thereof from the Zenith must be the complement of 38 degrees, 47 mi∣nutes to a Semicircle, that is, 141 degrees, 53 minutes; and as the half of 38 degrees, 47 minutes, viz. 19 degrees, 23 minutes, 50 seconds is the quantity of the angle PEZ, and the tangent thereof the distance from Z to P, so the half of 141 degrees, 53 minutes, viz. 70 degrees, 76 minutes, 50 seconds must be the measure of the angle in the circumference between the Zenith and the South, the tangent whereof 2866 must be the distance also, and the tangents of these two arches added together 3215, is the whole diameter of that circle, the half whereof 1607, that is, one Radius, and neer 61 hundred parts of another is the Semidiameter or distance from P to L in the former Scheme, to which extent o∣pen the compasses, and set off the distance PL, and therewith draw the circle

WPE for the six of the clock hour.

The Semidiameters of the other circles are to be found in the same manner: the distance between the Zenith and the Equi∣noctiall is alwayes equal to the height of the Pole, which in our Latitude is 51 degr. 53 min. and therefore the half thereof 25 degrees, 76 minutes, 50 seconds is the mea∣sure of the angle WEB, and the natural tangent thereof 483, which being added to the tangent of the complement 2070, their aggregate 2553 will be the whole diameter of that circle, and 1277 the Radius or Se∣midiameter by which to draw the Equi∣noctiall circle EKW.

The Tropique of Cancer is 23 degrees, 53 minutes above the Equator, and 66 degrees 47 minutes distant from the Pole, and the Pole in this Latitude is 38 degrees 47 min. distant from the Zenith, which being sub∣stracted from 66 degrees 47 minutes, the distance of the Tropique of Cancer from the Zenith, will be 28, the half thereof is 14, whose natural tangent 249 being set from Z to T, giveth the point T in the Meridian, by which that parallel must passe; the distance thereof from the Zenith on the North side is TN 90 degrees, and substracting 23 degrees, 53 minutes, the

height of the Tropique above the Equator, from 38 degrees, 47 minutes, the height of the Equator above the Horizon, their dif∣ference is 14 degrees, 94 minutes, the di∣stance of the Tropique from N under the Horizon; and so the whole distance there∣of from Z is 104 degrees, 94 minutes, the half whereof is 52 degrees, 47 minutes, and the natural tangent thereof 1302 ad∣ded to the former tangent 249, giveth the whole diameter of that circle 1551, whose half 776 is the Semidiameter desired, and gives the center to draw that circle by.

The Tropique of Capricorn is 23 degrees, 53 minutes below the Equator, and there∣fore 113 degrees 53 minutes from the North pole, from which if you deduct, as before, 38 degrees, 47 minutes, the distance of the Pole from the Zenith, the distance of the Tropique of Capricorn from the Zenith will be 75 degrees, 6 minutes, and the half thereof 37 degrees, 53 minutes, whose na∣tural tangent 768 being set from Z to F, giveth the point F in the Meridian, by which that parallel must pass: the distance thereof from the Zenith on the North side is ZN 90 degrees, as before; and adding 23 degrees, 53 minutes, the distance of the Tropique from the Equator to 38 degrees,

47 minutes, the distance of the Equator from the Horizon, their aggregate is 62 de∣grees, the distance of the Tropique from the Horizon, which being added to ZN 90 degrees, their aggregate is 152 degrees, and the half thereof 76 degrees, whose natural tangent 4011 being added to the former tangent 768, giveth the whole di∣ameter of that circle 4.779, whose half 2.389 is the Semidiameter desired, and gives the center to draw that circle by.

The distance of the reclining circle NOS from Z to O is 40 degrees, the half thereof 20, whose naturall tangent 3.64 set from Z to O, giveth the point O in the prime ver∣tical EZW, by which that circle must pass; the distance thereof from the Zenith on the East side is ZE 90 degrees, to which adding 50 degrees, the complement of the former arch, their aggregate 140 degrees is the di∣stance from Z Eastward, and the half there∣of 70 degrees, whose natural tangent 2747 being added to the former tangent 364, their aggregate 3111 is the whole diameter of that circle, and the half thereof 1555 is the Semidiameter desired, and gives the center to draw that circle by.

The distance of the declining reclining circle DAG from the Zenith is ZA 35 deg.

the half thereof 17 degrees, 50 minutes, whose natural tangent 315 being set from Z to A, giveth the point by which that circle must passe, and the natural tangent of 7•• degr. 50 min. the complement thereof 317•• being added thereto is 3486, the whole di∣ameter of the circle, and the half thereof 1743, the Semidiameter desired, and giveth the center to draw that circle by.

The streight lines CZA or DZG are put upon the limbe by help of a line of Chords 30 degrees distant from the Cardi∣nal points NESW, and must crosse each other at right angles in Z, representing two Azimuths equidistant from the Meridian and prime verticall.

Last of all, the hour-circles are thus to be drawn; first, seek the center of the six of clock hour-circle, as formerly directed, making ZE the Radius, and is found at L upon the Meridian line continued from P to L, which cross at right angles in L with the line 8 L 4, extended far enough to serve the turn, make PL the Radius, then shall 8 L 4 be a tangent line thereunto, and the natural tangents of the Equinoctiall hour arches, that is the tangent of 15 degrees 268 for one hour, of 30 degr. 577 for two, hours, of 45 degrees 1000 for three hours

of 60 deg. 1732 for four hours, and 75 deg. 3732 for five hours set upon the line from L both wayes, that is, from L to 5 and 7, 4 and 8, and will give the true center of those hour-circles: thus, 5 upon the line 8 L 4 is the center of the hour-circle 5 P 5, and 7 the center of the hour-circle 7 P 7; and so of the rest.

The centers of these hour-circles may be also found upon the line 8 L 4 by the natu∣rall secants of the same Equinoctiall arch∣es, because the hypothenuse in a right an∣gled plain triangle is alwayes the secant of the angle at the base, and the perpendicular the tangent of the same angle: if there∣fore the tangent set from L doth give the center, the secant set from P shall give that center also. The Scheme with the lines and circles thereof being thus made plain, we come now to the Art of Dialling it self.

Probl. 3.

Of the severall plains, and to finde their scituation.

ALL great Circles of the Sphere, pro∣jected upon any plain, howsoever si∣tuated, do become streight lines, as any one may experiment upon an ordinary bowle thus. If he saw the Bowle in the midst, and joyne the two parts together again, there will remain upon the circum∣ference of the Bowle, some signe of the for∣mer partition, in form of a great Circle of the Sphere: now then, if in any part of that Circle the roundnesse of the bowle be taken off with a smoothing plain or other∣wise, as the bowle becomes flat, so will the Circle upon the bowle become a streight line; from whence it follows, that the houre lines of every Diall (being great Circles of the Sphere) drawn upon any plain super∣ficies, must also be streight lines.

Now the art of Dialling consisteth in the artificiall finding out of these lines, and their distances each from other, which do continually varie according to the situation of the plain on which they are projected.

Of these plains there are but three sorts.

• 1. Parallel to the Horizon, as is the Ho∣rizontal only.
• 2. Perpendicular to the Horizon, as are all erect plains, whether they be such as are direct North, South, East or West, or such as decline from these points of North, South, East, or West.
• 3. Inclining to the Horizon, or rather Reclining from the Zenith, and these are direct plains reclining and inclining North and South, and reclining and inclining East and West, or Declining-reclining and inclining plains.

To contrive the houre lines upon these severall plains, there are certain Spherical arches and angles, in number six, which must of necessity be known, and divers of these are in some Cases given, in others they are sought.

• 1. The first is an arch of a great Circle perpendicular to the plain, comprehended betwixt the Zenith and the plain, which is the Reclination, as ZT, ZK, and ZF, in the fundamental Diagram.
• 2. The second is an arch of the Horizon betwixt the Meridian and Azimuth passing by the poles of the plain, as SV or NC in the Scheme.
• 3. The third is an arch of the plain be∣twixt

• the Meridian and the Horizon, pre∣scribing the distance of the 12 a clock houre from the horizontal line, as PB in the Scheme of the 11th. Probl.
• 4. The fourth is an arch of the plain be∣twixt the Meridian and the substile, which limits the distance thereof from the 12 a clock houre line, as ZR in the Scheme.
• 5. The fifth is an arch of a great Circle perpendicular to the plain, comprehended betwixt the Pole of the World; and the plain, commonly called the height of the stile, as PR in the Scheme.
• 6. The last is an angle at the Pole betwixt the two Meridians, the one of the place, the other of the plain (taking the substile in the common sense for the Meridian of the plain) as the angle ZPR in the fun∣damental Scheme.

The two first of these arches are alwayes given, or may be found by the rules fol∣lowing.

To finde the Inclination or Reclination of any plain.

If the plain seem to be level with the Ho∣rizon, you may try it by laying a ruler thereupon, and applying the side of your Quadrant AB to the upper side of the ruler,

so that the center may hang a little over the end of the ruler, and holding up a threed and plummet, so that it may play upon the center, if it shall fall directly upon his le∣vel line AC, making no angle therewith, it is an horizontal plain.

[illustration]

If the plain seeme to be verticall, like the wall of an upright building, you may try it by holding the Quadrant so that the threed may fall on the plumb line AC, for then if the side of the Quadrant shall lie close to the plain, it is erect, and a line

drawn by that side of the Quadrant shall be a Verticall line, as the line DE in the figure.

If the plain shall be found to incline to the Horizon, you may finde out the quan∣tity of the inclination after this manner. Apply the side of your Quadrant AC to the plain, so shall the threed upon the limbe give you the inclination required.

Suppose the plain to be BGED, and the line FZ to be verticall, to which applying the side of your Quadrant AC, the threed upon the limbe shall make the angle CAH the inclination required, whose comple∣ment is the reclination.

To finde the declination of a plain.

To effect this, there are required two ob∣servations, the first is of the horizontal di∣stance of the Sun from the pole of the plain, the second is of the Suns altitude, thereby to get the Azimuth: and these two obser∣vations must be made at one instant of time as neer as may be, that the parts of the work may the better agree together.

• 1. For the horizontal distance of the Sun from the pole of the plain, apply one edge of the Quadrant to the plain, so that the other may be perpendicular to it, and

• the limbe may be towards the Sun, and hold the whole Quadrant horizontal as neer as you can conjecture, then holding a threed and plummet at full liberty, so that the shadow of the threed may passe through the center and limb of the Quadrant, ob∣serve then what degrees of the limb the shadow cuts, counting them from that side of the Quadrant which is perpendicular to the horizontal line, those degrees are cal∣led the Horizontal distance.
• 2. At the same instant observe the Suns altitude, by this altitude you may get the Suns Azimuth from the South, by the 10th. Probleme of the first Chapter hereof.

When you make your observation of the Suns horizontal distance, marke whether the shadow of the threed fall between the South, and the perpendicular side of the Quadrant, or not, for,

• 1. If the shadow fall between them, then the distance and Azimuth added together do make the declination of the plain, and in this case the declination is upon the same coast whereon the Suns Azimuth is.
• 2. If the shadow fall not between them, then the difference of the distance and Azi∣muth is the declination of the plain, and if the Azimuth be the greater of the two, then

• the plain declineth to the same coast where∣on the Azimuth is, but if the distance be the greater, then the plain declneth to the con∣trary coast to that whereon the Suns Azi∣muth is.

Note here further, that the declination so found, is alwayes accounted from the South, and that all declinations are num∣bered from North or South, towards East or West, and must not exceed 90 deg.

• 1. If therefore the number of declination exceed 90, you must take its complement to 180, and the same shall be the plains decli∣nation from the North.
• 2. If the declination found exceed 180 deg. then the excesse above 180, gives the plains declination from the North, towards that Coast which is contrary to the Coast whereon the Sun is.

By this accounting from North & South, you may alwayes make your plains decli∣nation not to exceed a Quadrant or 90 de. And as when it declines nothing, it is a full South or North plain, so if it decline just 90, it is then a full East or West plain.

These precepts are sufficient to finde the declination of any plain howsoever situated, but that there may be no mistake, we will adde an Example.

1 Example.

Let the horizontall distance from the pole of the plains horizontal line represent∣ed in the last diagram by RZ the line of shadow, be 24 degrees, and let the Suns A∣zimuth from the South be 40 deg. describe the circle BCMP, the which shall repre∣sent the horizontal circle, and draw the di∣ameter BAC representing the horizontal

[illustration]

line of the plain, and the diameter MP representing the poles of the plains hori∣zontal line; then by a line of Chords set off your horizontal distance 24 degrees (found by observation in the afternoon) from P to

G, and from G to S set off the suns Aizmuth 40 degrees, so shall the point S represent the South, N the North, E the East, and W the West.

Now because the line of shadow AG, fal∣leth between P the pole of the plains hori∣zontal line, and S the South point, there∣fore according to the former direction, I adde the horizontal distance PG 24 deg. to the Suns Azimuth GS 40 deg, and their aggregate is PS 64 deg. the declination sought; and in this case it is upon the same coast with the sun, that is West, according to the rule given, and as the figure it selfe sheweth, the East and North points being hid from our sight by the plain it selfe; this therefore is a South plain declining West 64 degrees.

2. Example.

Let the horizontal distance taken in the afternoon by observation, be 67 degrees, and the Suns Azimuth from the South 42 degr be given, then draw, as before, the Circle BCMP, and from P to H set off the hori∣zontal distance 67 deg. from H to S the suns Azimuth, 42 deg. Now then, because the South point doth fall between P the pole of the plains horizontal line, and H,

[illustration]

the horizontal distance, I deduct the Suns Azimuth HS 42 degrees, from HP the ho∣rizontal distance, and their difference is SP 25 degrees, the declination sought; and because the horizontal distance is greater than the Azimuth, therefore the declinati∣on is contrary to the Coast of the Sun. This then is a South plain declining East 25 de∣grees.

To finde a Meridian line upon an Horizontal plain.

If your plain be levell with your Hori∣zon, draw thereon the Circle BCMP, then holding a threed and plummet, so as the

shadow thereof may fall upon the center, and draw in the last diagram the line of shadow HA: then if the Suns Azimuth shall be 50 deg. and the line of shadow taken in the afternoon, set off the 50 deg. from H to S, and the line SN shall be the Meridian line desired.

Probl. 4.

To draw the houre lines upon the Horizontal plain.

THis plane in respect of the Poles thereof, which lie in the Vertex and Nadir of the place may be called vertical, in respect of the plane it self, which is parallel to the Horizon, horizon∣tal, howsoever it be termed, the making of the Dial is the same, and there is but one onely arch of the Meridian betwixt the pole of the world and the plane required to the artificiall projecting of the hour-lines thereof, which being the height of the pole above the horizon (equal to the height of the stile above the plane) is alwayes given, by the help whereof we may presently pro∣ceed to calculate the hour distances in man∣ner following.

This plane is represented in the funda∣mental

Diagram by the outward circle ESWN, in which the diameter SN drawn from the South to the North may go both for the Meridian line, and the Meridian circle, Z for the Zenith, P for the pole of the world, and the circles drawn through P for the hour-circles of 1, 2, 3, 4, &c. as they are numbred from the Meridian, and limit the distance of each hour line from the Meridian upon the plane, according to the arches of the Horizon, N 11, N 10, N 9, &c. which by the severall Triangles SP 11, SP 10, SP 9, or their verticals NP 11, NP 10, NP 9 may thus be found; because every quarter of the Horizon is a∣like, you may begin with which you will, and resolve each hours distance, either by the small Triangle NP 11, or the verticall Triangle KP 11. In the Triangle PN11, the side PN is alwayes given, and is the height of the pole above the horizon, the which at London is 51 deg. 53 min. and the angle at P is given one hours distance from the Meridian, whose measure in the Equino∣ctiall is 15 deg. & the angle at N is alwayes right, that is 90 deg. wherefore by the first case of right angled spherical Triangles, the perpendicular N 11 may thus be found.

As Radius 90,	10.000000
To the tangent of NP11, 15d.	9.428052
So is the sine of PN 51.53.	9.893725
To the tangent of N11, 11.85.	9.321777

Which is the distance of the hours of 1 and 11, on each side of the Meridian, thus in all respects must you finde the distance of 2 and 10 of clock, by resolving the tri∣angle NP10, and of 3 and 9 of clock, by resolving the triangle NP9; and so of the rest: in which, as the angle at Pincreafeth which for 2 hours is 30 degrees, for 3 hours 45 degr. for 4 hours 60 degr. for 5 hours 75 degr. so will the arches of the Horizon N10, N9, N8, N7, vary proportionably, and give each hours true distance from the Meridian, which is the thing desired.

Probl. 5.

To draw the hour-lines upon a direct South or North plane.

EVery perpendicular plane, whether direct or declining, lieth in some Azi∣muth or other; as here the South wall or plane doth lie in the prime vertical or Azimuth of East and West, represented in

the fundamental Diagram by the line EZW, and therefore it cutteth the Meri∣dian of the place at right angles in the Ze∣nith, and hath the two poles of the plane seated in the North and South intersection of the Meridian and Horizon; and be∣cause the plane hideth the North pole from our sight, we may therefore conclude, (it being a general rule that every plane hath that pole depressed, or raised above it, which lieth open unto it) that the South pole is elevated thereupon, and the stile of this Diall must look downwards thereunto, erected above the plane the height of the Antartick Pole, which being an arch of the Meridian betwixt the South pole and the Nadir, is equall to the opposite part there∣of, betwixt the North pole and the Zenith; and therefore the complement of the North pole above the horizon.

Suppose then that P in the fundamental Scheme, be now the South pole, and N the South part of the Meridian, S the North; then do all the hour-circles from the pole cut the line EZW, representing the plane unequally, as the hour-lines will do upon the plane it self, and as it doth appear by the figures set at the end of every hour line in the Scheme. Now having already the

poles elevation given, as was in the ho∣rizontal, there is nothing else to be done, but to calculate the true hour-distances upon the line EZW from the meridian SZN; and then to proceed, as formerly, and note that because the hours equidistant on both sides the meridian, are equal upon the plane, the one half being found, the other is also had, you may there∣fore begin with which side you will.

In the triangle ZP11, right angled at Z, I have ZP given, the complement of the height of the pole 38 deg. 47 min. the which is also the height of the stile to this Diall, and the angle at P15 degrees one hours distance from the meridian upon the Equator to finde the side Z11, for which by the first case of right angled sphericall triangles, the proportion is, as before.

As the Radius 90,	10.000000
To the sine of PZ 38.47.	9.793863
So is the tangent of ZP11, 15d.	9.428052
To the tangent of Z11, 9.47.	9.221915

And thus in all respects must you finde the distance of 2 and 10, of 3 and 9; and so forward, as was directed for the houres in the horizontal plane.

The North plane is but the back side of the South, lying in the same Azimuth with it, & represented in the Scheme by the back part of the same streight line EZW, what∣soever therefore is said of the South plane may be applied to the North; because as the South pole is above the South plane 38 degr. 47 min. so is the North pole under the North plane as much, and each stile must respect his own pole, onely the meridian upon this plane representeth the mid∣night, and not the noon, and the hours a∣bout it 9, 10, 11, and 1, 2, 3, are altoge∣ther uselesse, because the Sun in his great∣est northern declination hath but 39 degr. 90 min. of amplitude in this our Latitude; and therefore riseth but 22 min. before 4. in the morning, and setteth so much after 8 at night; neither can it shine upon this plane longer then 35 min. past 7 in the morning, and returning to it as much before 5 at night, because then the Sun passeth on the North side of the prime vertical, in which this plane lieth, and cometh upon the South.

Now therefore to make this Dial, is but to turn the South Dial upside down, and leave out all the superfluous hours between 5 and 7, 4 and 8, and the Diall to the

North plane is made to your hand.

The Geometricall projection.

To project these and the Horizontal Di∣als, do thus: First, draw the perpendicu∣lar line CEB, which is the twelve of clock hour, crosse it at right angles with 6C6, which is the six of clock hour; then take with your compasses 60 deg. from a line of Chords, and making C the center draw the circle 6E6, representing the azimuth in which the plane doth lie; this done, take from the same Chord all the hour distan∣ces, and setting one foot of your compasses

[illustration]

in E, with the other mark out those hour distances before found by calculation, both

wayes upon the circle 6E6; streight lines drawn from the center C to those pricks in the circle are the true hour-lines desired.

Having drawn all the hour-lines, take from the same line of Chords the arch of your poles elevation, or stile above the plane, and place it from E to O, draw the prickt line COA representing the axis or heighth of the stile, from any part of the meridian draw a line parallel to 6C6, as is BA, & it shall make a triāgle, the fittest form to support the stile at the true height; let the line 6C6 be horizontal, the triangular stile CBA erected at right angles over the 12 of clock line, and then is the Diall perfected either for the Horizontal, or the direct North and South planes.

Probl. 6.

To draw the hour-lines upon the direct East or West planes.

AS the planes of South and North Dials do lie in the Azimuth of East and West, and their poles in the South and North parts of the meridian; so do the planes of East and West Dials lie in the South and North azimuth, and their poles in the East and West part of the Ho∣rizon,

from whence these Dials receive their denomination, and because they are parallel to the meridian line in the funda∣mental Scheme SZN, some call them me∣ridian planes.

And because the meridian, in which this plane lieth, is one of the hour-circles, and no plane that lieth in any of the hour cir∣cles can cut the axis of the world, but must be parallel thereunto; therefore the hour lines of all such planes are also parallel each to other, and in the fundamental Scheme may be represented in this man∣ner.

Let NESW in this case be supposed to be the Eq••inoctiall divided into 24 equall parts, and let the prickt line E 8. 7. paral∣lel to ZS be a tangent line to that circle in E, straight lines drawn from the center Z thorow the equal divisions of the limbe, intersecting the tangent line, shall give points in 4, 5, 6, 7, 8, 9, 10, 11, thorow which parallels being drawn to the prime vertical, or 6 of clock hour line EZW, you have the hour-lines desired, which may for more certainties sake be found by tangents also; for making ZE of the for∣mer Scheme to be the Radius, and E 8. 7. a tangent line, as before; then shall the na∣turall

tangent of 15 degr. 268 taken from a diagonal scale equal to the Radius, and set both wayes from E upon the tangent line E 8. 7. gives the distance of the houres of 5 and 7, the tangent of 30 degr. the di∣stance of the hours of 4 and 8, and the tangent of 45 degr. the distance of the hours of 3 and 9, &c. from the six of clock hour, as before; and is a general rule for all Latitudes whatsoever.

The Geometricall projection.

Proceed then to make the Diall, and first draw the horizontal line BA upon any part

[illustration]

thereof, as at A, draw two obscure arches DBG and FCE; and with that line of Chords, with which the arches were drawn

set off 38 deg. 47 min. the height of the E∣quator at London from B to D, and from C to E, set off likewise 51 deg. 53 min. the height of the Pole from B to G, and from C to F, and draw the streight line DAE, re∣presenting the Equinoctial, as is manifest by the angle BAD 38 deg. 47 min. which the Horizon makes with the Equator: and the streight line FAG representing the Ax∣is of the World, as is manifest by the angle FAC 51 deg. 53 min. which the Pole and Horizon make, and this wil be also the six of clock houre, or substile of this Diall, see∣ing the plain it selfe lieth in the Meridian, 90 deg. distant. And because the top of the Stile (which may be a streight pin fixed in the point A) doth give the shadow in all plains that are parallel to the Axis, it will be necessary to proportion the stile to the plain, that the hour lines may be enlarged or contracted according to the length there∣of, the which is done in this manner. Let the length of the plain from A be given in some known parts, then because the extream houre of the East Dial is 11, in the West 1, reckoning 15 deg. to every houre from six, the arch of the Equator will be 75 deg. and therefore in the right angled plain triangle AHE, we have given the base AE, which

is the length of the plain from A, and the angle AHE 75 deg. to find the perpendi∣cular HA, for which (by the 1 Case of right angled plain Triangles) the proportion wil be,

As the Radius, 90	10.000000
To the Base AE, 3.48	2.541579
So the Tangent of AEH, 15	9.428052
To the perpendicular HA, 93	1.969631

The length of the stile being thus propor∣tioned to the plain, make that the Radius of a Circle, and then the Equator DAE shall be a Tangent line thereunto, and therefore, the naturall Tangent of 15 deg. being set upon the Equinoctiall DAE both wayes from A, shall give the points of 5 and 7: the Tangent of 30 deg. the points of 8 and 4, &c. through which streight lines be∣ing drawn parallel to the six a clock houre, you have at one work made both the East and West Dials, only remember that because the Sun riseth before 4 in Cancer, and set∣teth after 8, you must adde two houres be∣fore six in the East Diall, and two houres after six in the West, that so the plain may have as many houres as it is capable of.

The West Dial is the same in all respect

with the East, only the arch BD, or the height of the Equator, must be drawn on the right hand of the center A for the West Dial, and on the left for the East, that so the houre lines crossing it at right angles, may respect the Poles of the world to which they are parallel.

Probl. 7.

To draw the houre-lines upon a South or North erect plain declining East or West, to any decli∣nation given.

EVery erect plain lieth under some A∣zimuth or other, and those only are said to decline which differ from the Meridian and Prime Vertical. The decli∣nation therefore being attained by the rules already given, (or by what other means you like best) we come to the calculation of the Diall it selfe, represented in the fundamentall Scheme by the right line GZD, the Poles whereof are C and V, the declination from the South Easterly NC, or North Westerly SV, 25 deg. supposing now S to be North, and N South; W East, and E the West point, the houre circles proper to this plain are the black lines pas∣sing

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through the Pole P, and crossing upon the plain GZD, wherein note generally that where they run neerest together, there∣abouts must the sub-stile stand, and alwayes on the contrary side to the declination, as in this example declining East, the stile must stand on the West side (supposing P to be the South Pole) between Z and D, the reason whereof doth manifestly appear; be∣cause the Sun rising East, sendeth the sha∣dow of the Axis West, and alwayes to the opposite part of the Meridian wherein he is, wherefore reason enforceth, that the morn∣ing houres be put on the West side of the Meridian, as the evening houres are on the East, and from the same ground that the substile of every plain representing the Me∣ridian thereof, must alwayes stand on the contrary side to the declination of the plain and that the houre-lines must there run neerest together, because the Sun in that position is at right angles with the plain. For the making of this Diall three things must be found.

• 1. The elevation of the Pole above the plain, represented by PR, which is the height of the stile, and is an arch of the Meridian of the plain, between the Pole of the world and the plain.
• ...

• 2. The distance of the substile from the Meridian, represented by ZR, and is an arch of the plain between the Meridian and the substile.
• 3. The angle ZPR, which is an arch between the substile PR the meridian of the plain, and the line PZ the meridian of the place, and these are thus found.

Because the substile is the Meridian of the plain, it must be part of a great circle passing through the pole of the world, and crossing the plain at right angles, therefore in the supposed right angled triangle PRZ, (for yet the place of R is not found) you have given the base PZ 38 deg. 47 min. and the angle PZR the complement of the de∣clination 65 deg. and the supposed right angle at R, to finde the side PR, which is the height of the stile as aforesaid, but as yet the place unknown: wherefore by the 8 Case of right angled Spherical Triangles the analogie is,

As the Radius,	10.00000
To the sine of PZ, 38.47	9.793863
So the sine of PZR, 65	9.957275
To the sine of PR, 34.32	9.751138

Secondly, you may finde ZR the distance of the substile from the meridian, by the 7 case

of right angled Spherical Triangles.

As the Radius, 90	10.000000
To the Co-sine of PZR, 65	9.629378
So is the tangent of PZ, 38.47	9.900138
To the tangent of ZR, 18.70	9.529516

These things given, the angle at P be∣tween the two meridians may be found by the 9 Case of right angled Sphericall Tri∣angles, for the proportion is,

As the Radius,	10.000000
To the Co-sine of PZ, 38.47	9.893725
So the Tangent of PZR, 65	10.331327
To the Co-tang. of RPZ, 30.78	10.225052

Having thus found the angle between the Meridians to be 30 deg. 78 min. you may conclude from thence, that the substile shall fall between the 2d. & third houres distance from the Meridian of the place, and there∣fore between 9 and 10 of the clock in the morning, because the plain declineth East from us, 9 of the clock being 45 deg. from the Meridian, and 10 of the clock 30 deg. distant, now therefore let fall a perpendi∣cular between 9 and 10, the better to inform the fancie in the rest of the work, and this

shall make up the Triangle PRZ before mentioned and supposed, which being found we may calculate all the houre distances by the first case of right angled sphericall Triangles. For,

• As the Radius,
• Is to the sine of the base PR;
• S•• is the Tangent of the angle at the per∣pendicular, RP 9,
• To the tangent of R 9 the perpendicular

The angle at P is alwayes the Equinocti∣all distance of the houre line from the sub∣stile, and may thus be sound: If the angle between the Meridians be lesse than the houre distance, substract the distance of the substile from the houre distance; if greater substract the houre distance from that, and their difference shall give you the Equino∣ctiall distance required.

Thus in our Example, the angle between the Meridians was found to be 30 deg. 78 m. and the distance of 9 of the clock from 1•• is three houres, or 45 deg. if therefore I substract 30 deg. 78 min. from 45 deg. the remainder will be 14 deg. 22 min. the di∣stance of 9 of the clock from the substile. Again, the distance of 10 of the clock from the Meridian is 30 deg. and therefore

if I substract 30 deg. from 30 deg. 78 min. the distance of 10 of the clock from the substile will be 78 centesms or parts of a de∣gree: the rest of the houres and parts are easily found by a continual addition of 15 deg. for every houre, 7 deg. 50 min. for half an houre, 3 deg. 75 min. for a quarter of an houre, as in the Table following you may perceive, the which consists of three columns, the first containeth the houres, the second their Equinoctiall distances from the substile, the third and last the houre arches, computed by the former proportion in this manner.

As the Radius, 90	10.000000
Is to the sine of PR, 34.32	9.751136
So is the tang. of RP 9, 14.22	9.403824
To the tangent of R 9, 8.13	9.154960

H	Aequ.	Arches
4	89 22	88 61
5	74 22	63 38
6	59 22	43 43
7	44 22	28 75
8	29 22	17 50
9	14 22	8 13
	merid.	substil
10	00 78	00 44
11	15 78	9 05
12	30 78	18 56
1	45 78	30 08
2	60 78	45 23
3	75 78	65 80
4	90 78	88 61

The Geometricall Projection.

Having calculated the hour distances, you may thus make the Diall; Draw the Horizontall line ACB, then crosse it at right angles in C, with the line CO 12. Take 60 degrees from a Chord, and ma∣king C the Center, draw the Semicircle AOB, representing the azimuth GZD in the Scheme, in which the plane lieth; up∣on this circle from O to N set off the di∣stance of the substile from the Meridian, which was found before to be 18.70. upon the West side of the Meridian, because this plane declineth East, then take off the same Chord the severall hour-distances, as they are ready calculated in the table, viz. 8.13. for 9, 17.50. for 8: and so of the rest; and set them from the substile upon the circle RNO, as the Table it self di∣recteth; draw streight lines from the cen∣ter C to these several points, so have you the true hour lines, which were desired: and lastly, take from the same Chord the heighth of the stile found to be 34. 32. which being set from N to R, and a streight line drawn from C through R representing the axis, the Diall is finished for use.

In applying it to any wall or plane, let

[illustration]

ACB be horizontal, CO perpendicular, and the side or axis of the stile, CR point∣ing downwards, erected over the substile line CN; so have you fitted a Diall for a∣ny South plane declining East 25 degrees.

Nay, thus have you made four Dials in one, viz. a South declining East and West 25 degrees, and a North declining East or West as much; to make this plainly ap∣pear, suppose in the fundamental Scheme if N were again the North part of the ho∣rizon, P the North pole, and that GZD were a North declining West 25 degrees, then do all the hour-circles crosse the same plane, as they did the former; onely DZ which was in the former the East side will now be the West: and consequently the afternoon hours must stand where the fore∣noon hours did, the stile also, which in the East declining stood between 9 and 10, must now stand between 2 and 3 of the af∣ternoon hours. And lest there should be yet any doubt conceived, I have drawn to the South declining East 25, the North de∣clining West as much; from which to make the South declining West, and North declining East, you need to do no more then prick these hour lines through the pa∣per, and draw them again on the other

side, stile and all; so shall they serve the turn, if you place the morning hours in the one, where the afternoon were in the other.

APPENDIX.

To draw the hour lines upon any plane de∣clining far East or West, without re∣spect to the Center.

THe ordinary way is with a Beam∣compasse of 16, 18, or 20 foot long, to draw the Diall upon a large floor, and then to cut off the hours, stile and all, at 10, 12, or 14 foot distance from the cen∣ter, but this being too mechanical for them that have any Trigonometrical skill, I o∣mit, and rather commend the way follow∣ing; by help whereof you may upon half a sheet of paper make a perfect model of your Diall, to what largenesse you please, without any regard at all to the Center.

Suppose the wall or plane DZG, on which you would make a Diall to decline from N to C, that is from the South Easter∣ly 83 degrees, 62 min. set down the Data,

and by them seek the Quaesita, according to the former directions.

The Data or things given are two.

• 1. PS the poles elevation 51 degrees, 53 minutes.
• 2. SA, the planes declination southeast 83 deg. 62 min.

The Quaesita or things sought are three,

• 1. PR the height of the stile 3 degrees 97 minutes.
• 2. ZR, the distance of the substile from the meridian 38 deg. 30 min.
• 3. ZPR, the angle of the meridian of the plane with the meridian of the place 85 degrees, which being found, according to the former directions, the substile line must fall within five degrees of six of the clock, because 85 degrees wanteth but 5 of 90, the distance of 6 from 12.

Now there∣fore make a table, according to this ex∣ample following, wherein set down the houres from 12, as they are equidistant from the meridian, and unto them adjoyn their Equinoctial distances, and write Me∣ridian and substile between the hours of 6 and 7, and write 5 degrees against the hour of 6, 10 degrees against the hour of 7, and to the Equinoctial distances of each hour adde the natural tangents of those distan∣ces,

as here you see. So is the Table prepa∣red for use, by which you may easily frame•• the Diall to what greatnesse you will, after this manner.

Hours	Equ. dist.	Tang.
4	8	35 0	700
5	7	20 0	364
6	6	5 0	087
		Meridiā	Substile
7	5	10 0	176
8	4	25 0	166
9	3	40 0	839
10	2	55 0	1.428
11	1	70 0	2.747
12	12	85 0	11.430

The Geometricall projection.

Proportion the plane BCDE, whereon you will draw the Diall to what scantling you think fit. Let VP represent the hori∣zontal line, upon any part thereof, as at P, make choice of a fit place for the perpen∣dicular stile (though afterwards you may use another forme) neer about the upper part of the plane, because the great angle between the two Meridians maketh the substile, which must passe thorow the point

P, to fall so near the 6 of clock hour, as that there may be but one hour placed a∣bove it, if you desire to have the hour of 11 upon the plane, which is more useful then 4, let P be the center, and with any Chord (the greater the better) make two obscure arches; one above the horizontal line, the other under it, and with the same Chord set off the arch of 51.70. which is the angle between the substile and horizon, and is the complement of the angle be∣tween the substile and meridian, and set it from V to T both wayes, then draw the streight line TPT, which shall be the sub∣stile of this Diall.

This done, proportion the length of PO the perpendicular stile to what scantling you will, and from a diagonal Scale fitted to the Radius of your intended perpendi∣cular stile, set off 69, the natural tangent of 3 degrees 97 min. the height of the stile found by calculation from P to H. Then by a scale proportional to the Radius PO, and at the point H draw the Equinoctial KH11, cutting the substile at right angles; which if rightly drawn, will cut the horizontal line at 6 of the clock, and make an angle of 38 deg. 30 min. with the horizon, equal to the distance of the substile from the Me∣ridian,

[illustration]

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upon this Equinoctial line making HO the Radius, set off 364, the natural tangent of 20 degrees from H upwards for the 5 of clock hour, and 2747 the natural tangent of 70 degrees, from H downwards for the 11 of clock hour, if these two hour distances fit not the plane to your liking, make PO greater or lesser, as you see cause, for according to this, the distance of H from P, (by which the Equinoctial line must be drawn) the length of HO, and the width of all the hour lines must vary pro∣portionably, but if they fit the plane, then by your scale proportioned to the Radius HO, and the help of the natural tangents set the hours upon the Equinoctial, after this manner: In the right angled plain triangle HG11, having the perpendicular HG equal to HO given in some known parts: as suppose 206, that is 2 inches and 6 parts of an inch, and the angle HG11, 70 degrees, the base H11 may be found by the first case of right angled plain trian∣gles: for,

As the Radius 90	10.000000
Is to the perpendicular HG 206,	2.313867
So is the tangent of HG11, 70.	10.438935
To the base H11, 566.	2.752802

Which is 5 inches, and 66 hundred parts for the distance of 11 a clock from the point H, and will be the same with those points set off by the natural tangents in the Table. Having done with this Equinocti∣all, you must do the like with another: to finde the place whereof, it will be necessa∣ry first to know the length of the whole line from H the Equinoctial to the center of the Diall in parts of the perpendicular stile PO, if you will work by the scale of inch∣es, or else the length in natural tangents, if you will use a diagonall Scale: first therefore, to finde the length thereof in inch-measure, we have given in the right angled plain triangle HOP, the base OP, and the angle at O to finde HP, and in the triangle OP center. We have given the perpendicular OP, and the angle PO center the complement of the former, to finde H center: wherefore, by the first case of right angled plain triangles:

As the Radius 90	10.000000
Is to the base OP 206;	2.313867
So is the tang. of HOP 3.97.	8.841364
To the perpendicular PH14.	1.155231

Again,

As the Radius 90,	10.000000
Is to the perpend. OP 206,	2.313867
So is the tang. PO center 86.3.	11.158636
To the base P center 2972	3.472403

Adde the two lines of 014 and 2972 to∣gether, and you have the whole line H cen∣ter 2986 in parts of the Radius PO, viz. 29 inches, and 86 parts; out of this line abate what parts you will, suppose 343, that is, 3 inches and 43 parts, and then the re∣mainer will be 2643. Now if you set 343 from H to I, the triangle IO center will be equiangled with the former, and I center being given, to finde LO, the proportion is;

As H center the first base 2986, co. ar.	6.524911
Is to HO, the first perpend. 206.	2.313867
So is I center the 2d. base 2643,	3.422097
To IO the 2d. perpend. 182,	2.260875

Having thus found the length of IO to be one inch, and 82 parts; make that the Radius, and then NT4 shall be a tangent line thereunto, upon which, according to this new Radius, set off the hour-distances

before found, and so have you 2 pricks, by which you may draw the height of the stile OO, and the hour-lines for the Dial.

The length of H center in natural tan∣gents, is thus found, HP 069 is the tangent line of the angle HOP 3 deg. 97 min. and by the same reason P center 14421 is the tangent line of PO center 86.3. the com∣plement of the other, and therefore these two tangents added together do make 14490, the length of the substile H center, that is, 14 times the Radius, and 49 parts, out of which substract what number of parts you will, the rest is the distance from the second Equinoctial to the center in na∣tural tangents; suppose 158 to be sub∣stracted, that is, one radius, and 58 parts, which set from H to T, in proportion to the Radius HO, and from the point T draw the line NT4 parallel to the former Equi∣noctial, and there will remain from T to the center 1291. Now to finde the length of LO, the proportion, by the 16th. of the second, will be

As H center 1449, co. ar.	6.838932
Is to HO 321,	2.506505
So is T center 1291,	3.110926
To TO 286,	2.456363

Now then if you set 286 from T to O in the same measure, from which you took HO, then may you draw ONO, and the tangents in the Table set upon the line NT in proportion to this new radius TO, you shall have two pricks, by which to draw the hour-lines, as before.

Probl. 8.

To draw the hour lines upon any direct plane, reclining or inclining East or West.

HItherto we have only spoken of such planes, as are either parallel or per∣pendicular to the horizon, all which except the horizontal, lie in the plane of some azimuth or other. The rest that fol∣low are reclining or inclining planes, ac∣cording to the respect of the upper or ne∣ther faces of the planes, in those that re∣cline, the base is a line in the plane, paral∣lel to the Horizon or Meridian, and al∣wayes scituate in some azimuth or other:

thus the base of the East and West recli∣ning planes lie in the Meridian, or South and North azimuth, and the poles thereof in the prime vertical, but the plane it self in some circle of position (as it is Astrolo∣gically taken) which is a great circle of the Sphere, passing by the North or South in∣tersections of the meridian and horizon, and falling East or West from the Zenith upon the prime vertical, as much as the poles of the plane are elevated and depres∣sed above and under the horizon. And this kinde of plane rightly conceived and represented in the fundamental Scheme by NOS, is no other but an erect declining plane in any Countrey, where the pole is elevated the complement of ours: for if you consider the Sphere, it is apparent, that as all the azimuths, representing the de∣cliners, do crosse the prime vertical in the Zenith, and fall at right angles upon the horizon, so do all the circles of position, representing the reclining and inclining East or West planes crosse the horizon in the North and South points of the Meridi∣an, and fall at right angles upon the prime vertical. From which analogie it commeth to passe, that making a Diall declining 30 degr. from the Meridian, it shall be the

same that a reclining 30 degr. from the Zenith, and contrary, onely changing the poles elevation into the complement there∣of, because the prime vertical in this case is supposed to be the horizon, above which the pole is alwayes elevated the comple∣ment of the height thereof above the ho∣rizon.

And therefore the poles elevation and the planes reclination being given, which for the one suppose to be, as before, 51 deg. 53 min. and the other, that is, the reclina∣tion 35 degrees towards the West, we must finde (as in all decliners) first the height of the pole above the plane, which in the fundamental diagram is PR, part of the meridian of the plane between the Pole of the world and the plane. 2. The distance thereof from the meridian of the place, which is NR part of the plane betwixt the substile and the meridian. 3. The angle betwixt the two meridians NPR, by which you may calculate the hour distances, as in the decliners.

First, therefore in the supposed triangle NPR (because you know not yet where R shall fall) you have the right angle at R the side opposite PN 51 degr. 53 min. and the angle at N, whose measure is the recli∣nation

ZO 35 degr. to finde the side PR, the height of the stile, or poles elevation above the plane, wherefore, by the eighth case of right angled spherical triangles, the analogie is

As the Radius 90,	10.000000
Is to the sine of PN 51.53.	9.893725
So is the sine PNR 35.	9.758591
To the sine of the side PR 26.69.	9.652316

Secondly, you may finde the side NR, which is the distance of the substile from the meridian, by the seventh case of right angled spherical triangles; for

As the Radius 90,	10.000000
Is to the ••••sine of PNR 35.	9.913364
So is the tangent of PN 51.53.	10.099861
To the tangent of NR 45.87.	10.013225

Thirdly, the angle at P between the two meridians m•••• be found by the ninth case of right angle•• spherical triangles.

As the Radius ••••	10.000000
Is to the co-sine ••••N, 51.53.	9.793864
So is the tangent ••f ••NR 35.	9.845227
To the co-tangent of RPN 66.46.	9.639091

The angle at P being 66 deg. 46 min. the perpendicular PR must needs fall some∣what neer the middle between 7 and 8 of the clock; if then you deduct the Equino∣ctial distance of 8, which is 60, from 66 deg. 46 min. the Equinoctial distance of 8 of the clock from the substile will be 6 deg. 46 min. again, if you deduct 66 degr. 47 min. from 75 deg. the distance of 7 from the Meridian, the Equinoctiall distance of 7 from the substile will be 08. deg. 53 min. the rest are found by the continual additi∣on of 15 deg. for an hour: thus, 15 degr. and 6 degr. 47 min. do make 21 deg. 47 min. for 9 of the clock; and so of the rest. And now the hour distances upon the plane may be found by the first case of right angled spherical triangles: for

As the Radius 90	10.000000
Is to the sine of PR 26.69.	9.652404
So is the tangent of RP. 8, 6.46.	9.053956
To the tangent of R 8, 2.91.	8.706360

These 2 deg. 91 min. are the true distance of 8 of the clock from the substile. And there is no other difference at all in calcu∣lating the rest of the hours, but increasing the angle at ••, acccording to each hours

Equinoctial distance from the substile.

The Geometrical Projection.

Having calculated the hour distances, you shall thus make the Diall; let AD be the base or horizontal line of the plane pa∣rallel to NZS, the meridian line of the Scheme. And ADEF the plane reclining 35 degr. from the Zenith, as doth SON of the Scheme▪ through any part of the plane, but most convenient for the houres, draw a line parallel to the base AD, which shall be GO 12, the 12 of the clock hour representing NZS of the Scheme; because the base AD is parallel to the meridian, take 60 degrees from a Chord, and make G the center, and draw the circle PRO, representing the circle of position NOS in the Scheme in which this plane lieth; from the point O to R Westerly in the East reclining, & Easterly in the West reclining, set off the distance of the substile and me∣ridian formerly found to be 45 degrees, 87 min. and draw the prickt line GR for the substile, agreeable to PR in the Scheme, GO in the Diall representing the arch PN, and OR in the plane the arch NR in the Scheme. From the point R of the substile both wayes set off the hour distan∣ces,

by help of the Chord, for 8 of the clock 2 degr. 91 min. and so of the rest; and draw streight lines from the center G through those points, which shall be the true hour lines desired. Last of all, the height of the stile PR 26 degr. 69 min. being set from R to P, draw the streight line

[illustration]

GP for the axis of the stile, which must give the shadow on the diall, Erect GP at the angle RGP perpendicularly over the substile line GR, and let the point P be di∣rected to the North pole, GO12 placed in the Meridian, the center G representing the

South, and the plane at EF elevated above the horizon 55 degrees; so have you finish∣ed this diall for use, onely remember, be∣cause the Sun riseth but a little before 4, and setteth a little after 8, to leave out the hours of 3 and 9, and put on all the rest.

And thus you have the projection of four Dials in one; for that which is the West recliner is also the East incliner 〈◊〉〈◊〉 you take the complements of the recliners ••ours un∣to 12, and that but from 3 in the afternoon till 8 at night: again, if you draw the same lines on the other side of your ••••per, and change the houres of 8, 7, 6, &c. into 4, 5, 6, &c. you have the East recliner, and the complement of the East recliners hours from 3 to 8 is the West incliner: one∣ly remember, that as the stile in the West recliner beholds the North, and the plane the Zenith; so in the East incliner, the stile must behold the South, and the plane the Nadir.

Probl. 9.

To draw the hour-lines upon any direct South reclining or inclining plane.

AS the base of East and West reclining or inclining planes do alwayes lie in the meridian of the place, or pa∣rallel

thereunto, and the poles in the prime vertical; so doth the base of South and North reclining or inclining planes lie in the prime vertical or azimuth of East and West, and their poles consequently in the Meridian. Now if you suppose the circle of position, (which Astrologically taken is fixed in the intersection of the meridian and horizon) to move about upon the ho∣rizon, till it comes into the plane of the prime vertical, and being fixed in the inter∣section thereof with the horizon, to be let fall either way from the Zenith upon the meridian, it shall truly represent all the South and North reclining and inclining planes also, of which there are six varieties three of South and three of North recli∣ning; for either the South plane doth re∣cline just to the pole, and then it becom∣meth an Equinoctial, because the poles of this plane do then lie in the Equinoctiall; some call it a polar plane, or else it recli∣neth more and less then the pole, and con∣sequently the poles of the plane above and under the Equinoctiall, somewhat differing from the former. In like manner, the North plane reclineth just to the Equi∣noctial, and then becometh a polar plane, because the poles of that plane lie in the

poles of the world; some term it an Equi∣noctiall plane. Or else it reclineth more or lesse then the Equinoctial, and conse∣quently the poles of the plane above and under the poles of the world, somewhat differing from the former.

Of the Equinoctiall plane.

The first of these six varieties which I call an Equinoctial plane, is in the fundamen∣tal Scheme, & also in this, represented by the six of clock hour-circle EPW, wherein you may observe out of the Scheme it self

[illustration]

that none of the other hour circles do cut the same, and therefore (as in the 5 Probl.)

you may conclude, that the hour-lines thereof have no center to meet in, but must be parallel one to another, as they were in the East and West Dialls.

And because this Diall is no other but the very horizontall of a right Sphere, where the Equinoctial is Zenith, and the Poles of the world in the Horizon; there∣fore it is not capable of the six of clock hour (no more then the East and West are of the 12 a clock hour) which vanish upon the planes, unto which they are parallel: and the twelve a clock hour is the middle line of this Diall (because the Meridian cut∣teth the plane of six a clock at right an∣gles) which the Sun attaineth not, till he be perpendicular to the plain. And this in my opinion, besides the respect of the poles, is reason enough to call it an Equinoctiall Diall, seeing it is the Diall proper to them that live under the Equinoctiall.

This Diall is to be made in all respects as the East and West were, being indeed the very same with them, onely changing the numbers of the hours: for seeing the six of clock hour in which this plane lieth crosseth the twelve of clock hour at right angles, in which the East and West plane lieth, the rest of the hour-lines will have equall respect

unto them both: so that the fifth hour from six of the clock is equal to the fift hour from twelve; the four to the four; and so of the ••est. These analogies holding, the hour di∣stances from six are to be set off by the natu∣ral tangents in these Dials, as they were from twelve in the East and West Dials.

The Geometricall Projection.

Draw the tangent line DSK, parallel to the line EZW in the Scheme, crosse it at right angles with MSA the Meridian line, make SA the Radius to that tangent line, on which prick down the hours; and that there may be as many hours upon the plane as it is capable of, you must proportion the stile to the plane (as in the fifth Problem) after this manner: let the length of the plane from A be given in known parts, then because the extream hours upon this plane are 5 or 7, reckoning 15 degrees to every hour from 12, the arch of the Equator will be 75 degrees: and therefore in the right angled plain triangle SA ♎, we have given the base A ♎, the length of the plane from A, and the angle AS ♎ 75 degrees, to finde the perpendicular SA; for which, as in the fifth Chapter, I say;

As the Radius 90,	10.000000
Is to the base A ♎ 3.50.	2.544068
So is the tangent of A ♎ S 15	9,428052
To the perpendicular AS 94	1.972120

[illustration]

At which height a stile being erected over the 12 a clock hour line, and the hours from

12 drawn parallel thereunto through the points made in the tangent line, by setting off the natural tangents thereon, and then the Diall is finished.

Let SA 12 be placed in the meridian, and the whole plane at S raised to the height of the pole 51 degr. 53 min. then will the stile shew the hours truly, and the Diall stand in its due position.

2. Of South reclining lesse then the pole.

This plane is represented by the prickt circle in the fundamental Diagram ECW, and is intersected by the hour circles from the pole P, as by the Scheme appeareth, and therefore the Diall proper to this plane must have a center, above which the South pole is elevated; and therefore the stile must look downwards, as in South direct planes; to calculate which Dials there must be given the Poles elevation, and the quantity of reclination, by which to finde the hour distances from the meridian, and thus in the triangle PC 1, having the poles elevation 51 degr. 53 min. and the reclina∣tion 25 degr. PC is given, by substracting 25 degr. from PZ 38 degr. 47 min. the complement of the poles height, the angle CP 1 is 15 degrees, one hours distance, and

the angle at C right, we may finde C 1, by the first case of right angled spherical trian∣gles: for,

As the Radius 90,	10.000000
Is to the sine of PC 13.47.	9.367237
So is the tangent of CP 1. 15.	9.428052
To the tangent of C 1 3.57.	8.795289

And this being all the varieties, save one∣ly increasing the angle at P, I need not re∣iterate the work.

3. Of South reclining more then the pole.

This plane in the fundamental Scheme is represented by the prickt circle EAW, of which in the same latitude let the reclina∣tion be 55 degrees, from which if you deduct PZ 38 deg. 47 min. the complement of the poles height, there will remain PA 16 deg. 53 min. the height of the north pole above the plane, and instead of the triangle PC 1, in the former plane we have the triangle PA 1, in which there is given as before the angle at P 15 deg. & the height of the pole PA 16 deg. 53 min. and therefore the same proportion holds: for,

As the Radius 90,	10.000000
Is to the sine of PA 16.53.	9.454108
So is the tangent of A 15.	9.428052
To the tangent of A 1. 4.36.	8.882160

The rest of the hours, as in the former, are thus computed, varying onely the angle at P.

The Geometricall Projection.

These arches being thus found, to draw the Dials true, consider the Scheme, where∣in so oft as the plane falleth between Z and P, the Zenith and the North pole, the South pole is elevated; in all the rest the North; the substile is in them all the meridian, as in the direct North and South Dials; in which the stile and hours are to be placed, as was for them directed: which being done let the plane reclining lesse then the pole, be raised above the horizon to an angle e∣qual to the complement of reclination, which in our example is to 65 degr. and the axis of the plane point downwards; and let all planes reclining more then the pole have the hour of 12 elevated above the ho∣rizon to an angle equal to the complement of the reclination also, that is in our exam∣ple, to 35 deg. then shall the axis point up to the North pole, and the Diall-fitted to the plane.

Probl. 10.

To draw the hour-lines upon any direct North reclining or inclining plane.

THe direct north reclining planes have the same variety that the South had; for either the plane may recline from the Zenith just to the Equinoctial, and then it is a Polar plane, as I called it before, be∣cause the poles of the plane lie in the poles of the world; or else the plane may re∣cline more or lesse then the Equinoctial, and consequently their poles do fall above or under the poles of the world, and the houre lines do likewise differ from the for∣mer.

Of the Polar plain.

This place is well known to be a Circle divided into 24 equall parts, which may be done by drawing a circle with the line of Chords, and then taking the distance of 15 degrees from the same Chord, drawing streight lines from the center through those equall divisions, you have the houre-lines desired. The houre-lines being drawn, erect a streight pin of wier upon the center, of wh•••• length you please, and the Diall is

finished: yet seeing our Latitude is capa∣ble of no more then 16 houres and a halfe, the six houres next the South part of the Meridian, 11, 10, 9, 1, 2, and 3, may be left out as uselesse. Nor can the reclining face serve any longer then during the Suns aboad in the North part of the Zodiac, and the inclining face the rest of the year, be∣cause this plain is parallel to the Equino∣ctial, which the Sun crosseth twice in a year. These things performed to your liking, let the houre of 12 be placed upon the Meridi∣an, and the whole plain raised to an angle equall to the complement of your Latitude, the which in this example is 38 deg. 47 min. so is this Polar plain and Diall rectified to shew the true houre of the day.

2. Of North reclining less then the Equator.

The next sort is of such reclining plains as fall between the Zenith and the Equator, and in the Scheme is represented by the pricked circle EFW, supposed to recline 25 degrees from the Zenith, which being added to PZ 38 deg. 47 min. the comple∣ment of the poles elevation, the aggregate is PF, 63 deg. 47 min. the height of the Pole or stile above the plane. And there∣fore

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in the triangle PF1, we have given PF, and the angle at P, to finde F1, the first houres distance from the Meridian upon the plain, for which the proportion is,

As the Radius, 90,	10.000000
Is to the sine of PF, 63.47	9.951677
So is the tangent of FP1, 15	9.428052
To the Tangent of F1, 13.48	9.379729

In computing the other houre distances there is no other variety but increasing the angle at P as before we shewed.

3. Of North reclining more then the Equator.

The last sort is of such reclining plain; as fall between the Horizon and Equator, represented in the fundamental Scheme by the prickt circle EBW, supposed to recline 70 deg. And because the Equator cutteth the Axis of the world at right angles, all planes that are parallel thereunto have the height of their stiles full 90 deg. above the plane: and by how much any plane re∣clineth from the Zenith, more then the E∣quator, by so much less then 90 is the height of the stile proper to it, and therefore if you adde PZ 38 deg. 47 min. the height of the Equator, unto ZB 70 deg. the reclination

of the plain, the totall is PB 108 deg. 47 mi. whose complemenc to 180 is the arch BS, 71 deg. 53 min. the height of the pole above the plain. To calculate the houre lines thereof, we must suppose the Meridian PFB and the houre circles P1, P2, P3, &c. to be continued till they meet in the South pole, then will the proportion be the same as before.

As the Radius, 90,	10.000000
To the sine of PB, 71.53	9.977033
So is the tangent of 1PB, 15	9.428052
To the tangent of B1, 14.27	9.405085

And so are the other houre distances to be computed, as in all the other planes.

The Geometricall Projection.

The projection of these planes is but lit∣tle differing from those in the last Probl. for the placing the hours and erecting the stile, they are the same, and must be eleva∣ted to an angle above the horizon equall to the complement of their reclinations, which in the North reclining lesse then the Equa∣tor is in our example 65 degrees, and in this plane the houres about the meridian, that is, from 10 in the morning till 2 in the after∣noon, can never receive any shadow, by

reason of the planes small reclination from the Zenith, and therefore needlesse to put them on. In the North reclining more then the Equator, the plane in our example must be elevated 120 degr. above the horizon, and the stiles of both must point to the North pole.

Lastly, as all other planes have two faces respecting the contrary parts of the hea∣vens; so these recliners have opposite sides, look downwards the Nadir, as those do to∣wards the Zenith, and may be therefore made by the same rules; or if you will spare that labour, and make the same Dials serve for the opposite sides, turn the centers of the incliners downwards, which were up∣wards in the recliners; and those upwards in the incliners which were downwards in the recliners, and after this conversion, let the hours on the right hand of the meri∣dian in the recliner become on the left hand in the incliner, and contrarily; so have you done what you desired: and this is a general rule for the opposite sides of all planes.

Probl. 11.

To draw the hour-lines upon a declining reclining, or declining inclining plane.

DEclining reclining planes have the same varieties that were in the for∣mer reclining North and South; for either the declination may be such, that the reclining plane will fall just upon the pole, and then it is called a declining Equino∣ctial; or it may fall above or under the pole, and then it is called a South declining cast and west recliner: on the other side the declination may be such, that the recli∣ning plane shall fall just upon the interse∣ction of the Meridian and Equator; and then it is called a declining polar; or it may fall above or under the said interse∣ction, and then it is called a North decli∣ning East and West recliner. The three va∣rieties of South recliners are represented by the three circles, AHB falling between the pole of the world and the Zenith: AGB just upon the pole; and AEB be∣tween the pole and the horizon: and the particular pole of each plane is so much elevated above the horizon, (upon the a∣zimuth)

DZC, crossing the base at right angles) as the plane it self reclines from the Zenith, noted in the Scheme, with I, K, and L.

[illustration]

1. Of the Equinoctiall declining and reclining plane.

This plane represented by the circle AGB, hath his base AZB declining 30 degrees from the East and West line EZW equal to the declination of the South pole thereof 30 degrees from S the South part of the Meridian Easterly unto D, reclining from the Zenith upon the azimuth CZD the quantity ZG 34 degrees, 53 min. and

Passeth through the pole at P. Set off the reclination ZG, from D to K, and K shall represent the pole of the reclining plane so much elevated above the horizon at D, as the circle AGB representing the plane declineth from the Zenith Z, from P the pole of the world, to K the pole of the plane, draw an arch of a great circle PK, thereby the better to informe the fancie in the rest of the work. And if any be desirous, to any declination given, to fit a plane reclining just to the pole: or any reclination being given, to finde the decli∣nation proper to it, this Diagram will sa∣tisfie them therein: for in the Triangle ZGP, we have limited,

First, the hypothenusal PZ 38 degrees, 47 min.

Secondly, the angle at the base PZG, the planes declination 30 degrees. Hence to finde the base GZ, by the seventh case of right angled spherical triangles, the pro∣portion is;

As the Radius 90,	10.000000
To the co-sine of GZP 30;	9.937531
So the tangent of PZ 38.47.	9.900138
To the tangent of GZ 34.53.	9.837669

the reclination required.

If the declination be required to a recli∣nation given, then by the 13 case of right angled spherical triangles, the proportion is

As the Radius 90,	10.000000
To the tangent of ZG 34.53.	9.837669
So the co-tangent of PZ 38.47.	10.099861
o the co-sine of GZP 39.	9.937530

And now to calculate the hour-lines of this Diall, you are to finde two things: first, the arch of the plane, or distance of the meridian and substile from the horizon∣tal line, which in this Scheme is PB, the intersection of the reclining plane with the horizon, being at B. And secondly, the di∣stance of the meridian of the place SZPN, from the meridian of the plane PK, which being had, the Diall is easily made.

Wherefore in the triangle ZGP, right angled at G, you have the angle GZP given 30 degrees, the declination; and ZP 38 degr. 47 min. the complement of the Pole; to finde GP: and therefore, by the eighth case of right angled spherical trian∣gles, the proportion is:

As the Radius, 90	10.000000
To the sine of ZP, 38.47	9.793863
So is the sine of GZP, 30	9.698970
To the sine of GP, 18.12	9.492833

Whose complement 71 deg. 88 min. is the arch PB desired.

The second thing to be found is the di∣stance of the Meridian of the place, which is the houre of 12 from the substile or meri∣dian of the plane, represented by the angle ZPG, which may be found by the 11 Case of right angled sphericall Triangles, for

As the Radius, 90	10.000000
Is to the sine of GP, 18.12	9.492833
So is the co-tang. of GZ, 34.53	10.162379
To the co-tang. of GPZ, 65.68	9.655212

Whose complement is ZPK 24 deg. 32 min. the arch desired.

Now because 24 deg. 32 min. is more then 15 deg. one houres distance from the Me∣ridian, and lesse then 30 deg. two houres di∣stance, I conclude that the stile shall fall between 10 and 11 of the clock on the West side of the Meridian, because the plain de∣clineth East: if then you take 15 deg, from 24 deg. 32 min. there shall remain 9 deg. 32 min. for the Equinoctiall distance of the

11 a clock houre line from the substile, and taking 24 deg. 32 min. out of 30 deg. there shall remain 5 deg. 68 min. for the distance of the houre of 10 from the substile: the rest of the houre distances are easily found by continual addition of 15 deg. Un∣to these houre distances joyn the naturall tangents as in the East and West Dials, which will give you the true distāces of each houre from the substile, the plane being pro∣jected as in the 5 Pro. for the east & west di∣als, or as in the 8 Prob. for the Equinoctial, according to which rules you may propor∣tion the length of the stile also, which being erected over the substile, and the Diall placed according to the declination 30 deg. easterly, and the whole plain raised to an angle of 55 deg. 47 min. the complement of the reclination, the shadow of the stile shall give the houre of the day desired.

2. To draw the houre lines upon a South re∣clining plain, declining East or West, which passeth between the Zenith and the Pole.

In these kinde of declining reclining plains, the South pole is elevated above the plane, as is clear by the circle AHB re∣presenting the same, which falleth between the Zenith and the North pole, and there∣fore

hideth that pole from the eye, and forceth you to seeke the elevation of the contrary pole above the plain, which not∣withstanding maketh the like and equall angles upon the South side objected to it, as the North pole doth upon the North side, (as was shewed in the 7 Prop.) so that either you may imagine the Scheme to be turned about, and the North and South points changed, or you may calculate the houres as it standeth, remembring to turn the stile upwards or downwards, and change the numbers of the houres, as the nature of the Diall wil direct you.

In this sort of declining reclining Dials, there are four things to be sought before you can calculate the houres.

• 1 The distance of the Meridian from the Horizon.
• 2 The height of the pole above the plain.
• 3 The distance of the substile from the Meridian.
• 4 The angle of inclination between the Meridian of the plane, and the meri∣dian of the place.

1 The distance of the Meridian from the Horizon, is represented by the arch OB, to finde which, in the right angled Triangle HOZ, we have HZ the reclination 20 deg.

and the angle HZO the declination, to find HO, the complement of OB, for which, by the first case of right angled sphericall tri∣angles, the analogie is,

As the Radius, 90	10.000000
o the sine of HZ, 20	9.534051
o is the tangent of HZO, 30	9.761439
o the tangent of HO, 11.17	9.295490

Whose complement 78 deg. 83 min. is OB, the arch desired.

2. To finde the height of the pole above the plane, there is required two operations, the first to finde OP, and the second to finde PR; OP may be found by the 3 Case of right angled Sphericall Triangles, for,

As the Radius, 90	10.000000
Is to the co-sine of HZP. 30	9.937531
So is the co-tang. of HZ, 20.	10.438934
To the co-tangent of ZO, 22.80	10.376465

Which arch being found, and deducted out of, ZP 38 deg. 47 min. there resteth PO 15 deg. 67 min.

Then may you finde PR, by the triangles HZO, and PRO both together, because the sines of the hypothenusals and the sines of the perpendiculars are proportional, by the first of the 7 Chap. of Triangles.

Therefore,

As the sine of ZO, 22.80	9.588289
Is to the sine of ZH, 20	9.534052
So is the sine of PO, 15.67	9.431519
To the sine of PR, 13.79	9.377282

The height of the stile desired.

3 The distance of the substile from the Meridian may be found by the 12 Case of right angled sphericall triangles, for

As the co-sine of PR, 13 78	9.987298
Is to the Radius, 90	10.000000
So is the co-sine of PO, 15.67	9.983551
To the co-sine of OR, 7.41	9.996253

The arch desired.

4. The angle of inclination between the Meridians, may be found by the 11 Case of right angled Spherical triangles, for,

As the Radius, 90	10.000000
Is to the sine of PR, 13.79.	9.377241
So is the co-tang. of OR 7.51	10.879985
To the co-tang. of OPR, 28.93	10.257226

Now as in all the former works, the angle P between the two Meridians being 28 deg. 93 min. which is more then one houres di∣stance from the Meridian, and lesse then

two, you may conclude that the substile must stand between the first & second hours from the Meridian or 12 of the clock West∣erly, because the declination is easterly: and 28 deg. 93 min. being deducted out of 30 deg. there resteth 1 deg. 7 min. for the di∣stance of 10 of the clock from the substile; again, deducting 15 deg. from 28 deg. 93 min. there resteth 13 deg. 73 min. the di∣stance of the 11 a clock houre line from the substile, the rest are found by continuall ad∣dition of 15 deg. as before.

And here the true houre distances may be found by the first case of right angled Sphe∣ricall triangles, for,

As the Radius, 90	10.000000
Is to the sine of PR 13.79	9.377240
So is the tangent of RP, 11.15	9.428052
To the tangent of R 11, 3.66	8.805292

And so proceed with all the rest.

3. To draw the houre lines upon a South reclining plain, declining East or west, which passeth between the Pole and the Horizon.

In this plain represented by the circle of reclination AFB, the North pole is ele∣vated above the plane, as the South pole

was above the other, and the same four things that you found for the former Diall must also be sought for this; in the finding whereof there being no difference, save on∣ly deducting ZP from ZO, because ZO is the greatest arch, as by the Scheam appea∣reth: to calculate the houres of this plane needeth no further instruction.

Probl. 12.

To draw the houre lines upon a polar plain, declining East or west, being the first variety of North declining reclining planes

AS in the South declining recliners, there are three varieties, so are there in the North as many: for either the plane reclining doth passe by the intersecti∣on of the Meridian and Equator, and then it is called a declining Polar, which hath the substile alwayes perpendicular to the Meridian; or else it passeth above or under the intersection of the Meridian and Equa∣tor, which somewhat differeth from the for∣mer. I will therefore first shew how they lie in the Scheam, and then proceed to the particular making of the Dials proper to them.

[illustration]

1. Of the Polar declining reclining plane.

This plane is in this diagram represented by the circle AGB, ZG is the reclination, ZAE the distance of the Equator from the Zenith, the declination NC, K the pole thereof. Here also as in the last Probl. there may be a reclination found to any declina∣tion given, and contrary, by which to fit the plane howsoever declining, to passe through the intersection of the Meridian and Equa∣tor, by the 7 and 13 Cases of right angled sphericall triangles.

As the Radius, 90	10.000000
To the co-sine of GZAE, 60	9.698970
So is the tangent of ZAE, 51.53	10.099861
To the tangent of ZG, 32.18	9.798871

The reclination desired.

And,

As the Radius, 90	10.000000
To the tangent of GZ, 32.18	9.798831
So is the co-tangent of ZAE, 51.53	9.900138
To the co-sine of GZAE, 60	9.698969

The declination.

And now to calculate the houre lines of this Dial, you must finde, first, the distance of the Meridian from the Horizon, by the 8 Case of right angled Spherical triangles.

As the Radius, 90	10.000000
Is to the sine of ZAE, 51.53	9.893725
So is the sine of GZAE, 60	9.937531
To the sine of AEG, 42.69	9.831256

Whose complement 47 deg. 31 min. is AAE the arch desired.

2. You must finde RP, the height of the pole above the plane, by the 2 Case of right angled Sphericall Triangles, for

As the Radius, 90	10.000000
Is to the sine of AEZG, 60	9.937531
So is the co-sine of ZG, 32.11	9.927565
To the co-sine of ZAEG, 42.87	9.865••9••

Which is the height of the pole above the plane, AER being a Quadrant, PR must needs be the measure of the angle at AE.

3. Because in all decliners (whose planes passe by the intersection of the Meridian and Equinoctiall) the substile is perpendicular to the Meridian, therefore you need not seek AER, the distance between the substile and Meridian, which is alwayes 90 deg. and fal∣leth upon the 6 a clock houre.

4. Lastly, the arch AER, which is the distance of the substile from the Meridian: being 90 degrees, the angle at P opposite thereunto must needs be 90 also: from whence it followes, that the houres equidi∣stant from the six of the clock hour in Equi∣noctial degrees shall also have the like di∣stance of degrees in their arches upon the plane, and so one half of the Diall being calculated, serves for the whole; these things considered, the true hour-distances may be found, by the first case of right an∣gled spherical triangles: for,

As the Radius,	10.000000
Is to the sine of PR 42.87.	9.832724
So is the tangent of RP 5. 15 d.	9.428052
To the tangent of R 5▪ 10.34.	9.260776

The which 10 degr. 34 min. is the true distance of 5 and 7 from the substile or six of the clock hour, and so of the rest.

The Geometrical projection of this plane needs no direction; those already given are sufficient, according to which this Di∣all being made and rectified by the decli∣nation and reclination given, it is prepared for use.

2. To draw the hou•••• lines upon a North re∣clining plane, declining East or West, which cutteth the meridian between the Zenith and the Equinoctial.

All North reclining planes howsoever declining, have the North pole elevated a∣bove them, and therefore the center of the Diall must be so placed above the plane, that the stile may look upwards to the pole, neither can it be expected that the plane be∣ing elevated above the horizon Southward, should at all times of the year be enlight∣ened by the Sun, except it recline so far from the Zenith, as to intersect the Meridian be∣tween the horizon and the Tropique of Ca∣pricorn; this plane therefore reclining but 16 degrees from the Zenith, and declining 60 cannot shew many hours, when the Sun

is in his greatest Northern declination, part∣ly by reason of the height of the plane a∣bove the horizon, and partly by reason of the great declination thereof, hindring the Sun-beams from all the morning houres, which may be therefore left out as useless.

In this second variety, the plane repre∣sented by the Circle AMB in the last Dia∣gram, cutteth the Meridian at O between the Zenith and the Equator, ZM being the reclination, 16 deg. ZAE the distance of the Equator from the Zenith, 51 deg. 53 m. and the declination NC 60 as before.

As in the former, so in this Diall, the same four things are again to be found be∣fore you can calculate the houre distances thereof. The first is the distance of the Me∣ridian from the Horizon, represented in this plain by the arch A••. The second is PR, the height of the pole above the plane. The third is ••R, the distance of the substile or Meridian of the plane, from the Meridian of the place. The fourth is the angle ••PR between the two Meridians: all which, and the houre distances also, being to be found according to the directions of the last Probl. there needeth no further instruction here.

3 To draw the houre lines upon a North reclining plane, declining East or West, which cutteth the Meridian between the Equator and Horizon.

The last variety of the six declining recli∣ners, represented by the circle ALB, and cutteth the Meridian at H, between the E∣quator and the Horizon, ZL being the re∣clination, 54 deg. the declination NC, 60 deg. as before; and hence the four things mentioned before must be sought ere you can calculate the houre distances.

• 1 The distance of the Meridian and Ho∣rizon, represented by AH.
• 2 RH the substile. or Meridian of the plane from the Meridian of the place.
• 3 PR, the height of the pole above the plane.
• 4 HPR, the angle between the two Meridians.

In finding whereof the propor∣tions are still the same, though the triangles are somewhat altered, for when you have found ZH, it is to be added to ZP to finde PH, both which together do exceed a Qua∣drant, therefore the sides PN must be con∣tinued to X, then is PX the complement of PH to a semicircle, and if RB be continued

••o X also, RX may be found by the 12 Case of right angled spherical triangles as before, whose complement is RH, the distance of the substile from the Meridian; and hence the angle at P must be found in that trian∣gle also, though the proportion be the same, there being no other variety, I think it need∣lesse to reiterate the work.

The Geometrical Projection.

There is so little difference between the South & North declining reclining planes, that the manner of making the Dials for both may be shewed at once: Let the exam∣ple therefore be a Diall for a South plane declining East 30 deg. reclining 20 deg.

First, draw the horizontal line ACB in the middle of the plane, because the stile of this Dial must looke downwards to the South pole, and because the plane declineth East, therefore the morning houres must stand on the West side of the Meridian, and the di∣stance of the Meridian and Horizon 78 deg. 83 min. must be set upon the circle ADBF, from A to E, and there draw the line CE for the 12 a clock houre, from E reckon 7 deg. 51 min. the distance of the substile and Meridian Westwards to D, and draw the prickt line CD for the substile: from

the point of the substile at D, set off the houre distances, as of 3 deg. 66 min. for 11, and so of the rest: unto every prick draw streight lines from the center C, so have you

[illustration]

all the houres truly drawn. Last of all, see off from D the height of the stile DE, and draw the line CE for the axis, which being

erected over the substile, CD, the Diall is fit for use, but must be placed in its due po∣sition by the declination and reclination thereof.

And thus have you made four Dials at once, or at least, this Dial thus drawn may be made to serve four sorts of planes, for first, it serves for a South declining East 30 deg. reclining 20 deg. and if you prick the houre lines through the paper, and draw them on the other side stile and all, this Diall will then be fitted for a South plane declining West 30 deg. reclining 20 deg. only re∣member to change the houres, that is to say, instead of writing 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, from A, the west side of the East declining plane, you must write, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7. Again, if you turn the Zenith of your Dial downwards, the South declining East reclining shall in all respects serve for a North declining west inclining as much; and the South declining West reclining, will likewise serve for a North declining East inclining; and there∣fore there needs no further direction either to make the one, or calculate the other.