The Third Book of the CrossStaff
Those Planes wherein a Line may be drawn parallel to the Axis of the World are called Polar Planes, because that Line pointeth unto the Poles; and these Planes are always parallel to some one of the Hourcircles. If they be parallel to the Hour of 6, they are called Direct Polar Planes; if to the Hour of 12, they are called Meridian Planes; and both these are described before: if to any other of the Hourcircles, they are then called by the name of Polar Declining Planes, because their inclining to the Pole, and declining from the Vertical.
These kind of Planes may be known in this sort: First, consider the Inclination of the Plane to the Horizon, which in these parts of the World must always be Northward, and more than the Latitude of the Place. Then find the Declination from the Vertical. These two being known, if the proportion hold,
 As the Sine of 90 gr. to the Cosine of the Declination:
 So the Tangent of the Declination, to the Tangent of the Latitude.
It is then a Polar declining Plane; otherwise not.
For example: In our Latitude of 51 gr. 30 m. a Plane is proposed declining from the Vertical 65 gr. 40 m. and inclining Northward 71 gr. 51 m. the upper Face being open to Southeast, and the lower to the Northwest. If I number those 65 gr. 40 m. in the Horizon of the Fundamental Diagram from E unto Q, and draw the Line HCQ, it shall represent the Horizontal Line of the Plane; then crossing it at Right Angles with the Plane BZD drawn through the Zenith, I number 71 gr. 51 m. for the Inclination from D unto R, and there draw the Circle HRQ: this Circle so drawn shall represent the Plane proposed; and because it also passeth through the Pole, it is therefore a Polar Plane. But for further trial, I extend the Compasses from the Sine of 90 gr. To the Sine of 24 gr. 20 m. the Complement of the Declination, and I find the same Extent to reach from the tangent of 71 gr.51 m. the Inclination proposed, unto the tangent of 51 gr. 30 m. which is the true Latitude of the Place; and therefore it is a Polar Plane.
Again, I number the Inclination of 71 gr. 51 m. in the Circle BZD from Z unto M, so this point M will fall at the meeting of BZD with the Equator, and being 90 gr. From the Plane R, it shall be the Pole of this Plane; and the Circle drawn though M and P will be the proper Meridian of this Plane. The Meridian MP here falling on the Hour of 8 doth give MPZ, the Angle of the Inclination of Meridians, to be 4 Hours, or 60 Degrees; then crossing the Plane at the point P, it shews that the Substylar should be CP, and be placed at the Hour of 8. But because P is the Pole, and CP the Axis of the World wherein all Hourcircles do meet, and so there would be no distinction between the Axis, the Substylar, and the Hourlines, I now suppose the Plane in a parallel to the Circle HRQ, according to the distance that I would have between the Axis of the Style and the Substylar, then will the Style be parallel to the Plane, as appears in the Fundamental Diagram.
Here then the Style will be parallel to the Plane, and the Hourlines parallel one to the other, as in the meridian and Direct Polar Planes. Yet that we may better know how to draw the Hourlines, and where to place the Style, we are to consider,
In a Meridian Plane, the Ark between the Horizon and the Pole, which represents the Ark between the Horizon and the Hourlines, is always equal to the latitude of the Place; in a direct Polar it is an Ark of 90 gr. In these Declining Polars it is greater than the latitude, and yet less than 90 gr. This Ark here is represented by PQ, and may be known by resolving the Triangle QNP, or PRZ.
 As the Sine of 90 gr. To the Cosine of the Latitude:
 So the Sine of the Declination, to the Cosine of the Ark between the Horizon and the Pole.
Extend the Compasses from the Sine of 90 gr. Unto the Sine of 38 gr. 30 m. the Complement of the Latitude, the same extent will reach from the Sine of 65 gr. 40 m. the Declination proposed, unto the Sine of 34 gr. 34 m. whose Complement is 55 gr. 26 m. the Ark of the Plane required between the Horizon and the Pole.
 Or, As the Cosine of the Inclination to the Horizon, to the Sine of 90 gr.
 So the Cotangent of the Declination, to the Tangent of the Ark between the Horizon and the Pole.
And so extending the Compasses from the Sine of 18 gr. 9 m. the Complement of the Inclination to the Tangent of 24 gr. 20 m. the Complement of the Declination, the same extent doth reach from the Sine of 90 gr. unto the Tangent of 55 gr. 26 m. And such is QP the Ark of the Plane between the Horizon and the Pole, the measure of the Angle QCP between the Horizontal Line and the Substylar.
The Substylar in a Direct Polar Plane is always the same with the Hourline of 12; in a Meridian Plane it is the same with the Hourline of 6; in these Declining Polars it must be placed between 12 and 6, according to the Inclination of the Meridian of the Plane to the Meridian of the Place, which is here represented by MPZ, the Complement of the Angle RPZ, and thus known.
 As the Sine of 90 gr to the Sine of the Latitude:
 So the Tangent of the Declination of the Plane, to the Tangent of the Inclination of Meridians.
Extend the Compasses from the Sine of 90 gr. to the Sine of 51 gr. 30 m. the latitude of the Place, the same extent will reach from the Tangent of 65 gr. 40 m. the Declination proposed, unto the tangent of 60 gr. and such is the Angle of Inclination between the Meridian of the Place and the proper Meridian of the Plane, which resolved into Time, doth make four Hours; and so the Substylar must here be placed upon the Hour of 8 in the Morning.
This Angle being known, the rest of the Angles at the Pole are easily gathered: For if the Hour of 12 be 60 gr. distant from the meridian of the Plane, the Hour of 1 will be 75 gr. and the Hour of 11 will be 45 gr. distant, and the rest of the Hours, as in the Table following. Then coming to the Plane,
 I draw an occult Line HQ, wherein I make choice of a Center at H, and describe an occult Circle for the Horizon of the Plane.
 I find a Chord of 55 gr. 26 m. and inscribe it into this Circle from Q unto B, according to the situation of the Plane; so the Line HB shall be the Meridian of the Plane, and therefore the Substylar; and the Line AC, crossing it at Right Angles, shall be the Equator.
 I consider the length of the Plane, and how many Hours I am to draw upon it, that so I may proportion the Height of the Style; and I find by the Fundamental Diagram, and the former table, that it will contain all the Hours from Sunrising until it be past 1 after Noon: and therefore the Meridian of the Plane falling on the Hour of 8 in the morning, there will be four Hours on the one side, and five on the other side of the Substylar. But in all Polar Planes the height of the Style above the Substylar must be equal to the distance of the third Hour from the Substylar, of about 4/7 of the fourth Hour, or little that ¼ of the fifth Hour, and thereupon I allow the height of this Style to be equalt o CB, which you may suppose to be 10 Inches.
 Because the Equator AC is a Tangentline, in respect of the Radius BC, and the parts thereof are such as belong to the Angles between the Meridian of the Plane and the Hourlines, which Angles are set down in the Table following, I may find the length of each several tangent in this manner.
 As the Tangent of 45 gr. is to the Tangent of the Hour:
 So the parts of the Radius, to the Parts of the Tangentline.
The Angle ABC between the Meridian of the Plane and the Hour of 12, the Meridian of the Place, is 60 gr. in the former table, and the Radius BC is supposed to the 10 Inches; whereupon I extend the Compasses from the Tangent of 45 gr. unto the tangent of 60 gr. the same extent will reach from 10 in the Line of Numbers, unto 17.32, which shews the length of the Tangent AC, between the Substylar and the Hour of 12, to be 17,32 cent. The like reason holds for the rest of the Hours.
These Lengths being thus found and set down in the Table, I take out 17 Inches 32 cent, and prick them in the Equator from C unto A for the Hour of 12, and 37 Inches 32 cent and prick them down for the Hour of 1: And so the rest of the Hourpoints.
 This done, if I draw Right Lines through each of these Points, crossing the Equator at Right Angles, they shall be the Hourlines required: And if I set the Style over the Substylar, so as the edge of it may be parallel to the Plane, and the height of it be 10 Inches, equal to the former Radius BC, it shall represent the Axis of the World, and be truly placed for casting of the Shadow upon the Hourlines in this Declining Polar Plane.
Latitude 
51 
30 
Declinat. 
65 
40 
Inclinat. 
71 
51 
Diff. Merid. 
60 
0 
Dist. Subst. 
55 
20 
Hours 
Ang. Po. 
Tang. 
Gr. 
M. 
In. 
Par. 
4 
60 
00 
17 
31 
5 
45 
00 
10 
00 
6 
50 
00 
5 
77 
7 
45 
00 
2 
68 
8 
Merid. 
Substyl. 
9 
15 
00 
2 
68 
10 
30 
00 
5 
77 
11 
45 
00 
10 
00 
12 
60 
00 
17 
32 
1 
75 
00 
37 
32 
2 
90 
00 
Infint. 
