The Third Book of the CrossStaff
In Planes neither Equinoctial nor Polar, the Equator will be a Right Line, the Tropicks and any other Parallels of Declination will be Conical Sections, some of them parabolic, some eliptical, but most of them hyperbolical.
To find the Points of Intersection of these Parallels with the Hourlines, we are to consider,
 First, the length of the Axis of the Style in Inches and parts of Inches.
 Secondly, The height of the Style above the Plane.
 Thirdly, The Angles at the Pole between the proper Meridian and the Hourcircles.
These being known, will help us to find, first, the Angle between the Axis and the Hourlines on the Plane; and then the distance between the Center and the Parallels. Both these may be represented in this manner,
Let the Triangle ABC be made equal to the Style belonging to you Plane, AC the Substylar, BC the Axis of the Style, AB the length of the Style perpendicular to the Plane. Then having drawn the Line BD perpendicular to the Axis on the Center B, and any Semidiameter BD, describe an occult Ark of a Circle, and therein inscribe a Chord of 23 gr. 30 m. from D unto T, on either side of the Line, with such other intermediate Declinations as you intend to describe on the Plane; so the Perpendicular BD shall be the Equator, and BT the Tropicks, and the other intermediate Lines the Parallels of Declination. Wherefore you may take out the distance CV from the Center of the Equator, and prick it down on the Substylar of your Plane from the Center at C unto V; so the Line drawn through ♈, perpendicular to your Substylar, shall be the Equator of your Plane.
That done, take the distance of each Hourline between the Center and the Equator of your Plane, and prick them down in the Equator of this Figure, from the Center at C, noting the place where they cross the Equator, with the Number belonging to the Hour, and drawing the Hourlines from C, through the Lines of Declination.
Or, having the Sector, you may draw an occult Line CE, perpendicular to the Axis BC, and therein prick down the Tangent of the height of the Style above the Plane, from C unto E: Then draw the Line EF parallel to the Axis, crossing the Substylar produced in the point F; this Line EF will be the Line of Sines upon the Sector, and therein you may prick down the Sines of the Complement of the Angles at the Pole from E toward F, and draw the Hourlines by those Points through the Lines of Declination; so the Angles at C, between the Axis BC and those Hourlines, shall be the Angles between the Axis of your Style and the Hourlines in your Plane; and the several Distances between the Point C and the Lines of Declination, shall give you the like Distances between the Center and the Parallels of Declination upon the Hourlines in your Plane. Upon this ground followeth,
Consider the height of the Style above the Plane, and the length of the Substylar between the Center and the Place which you intend for the Tropick. If it be the Tropick which is farthest from the Center, add 113 gr. 30 m. if the nearer Tropick, add 66 gr. 30 m. unto the height of the Style, the Remainder unto 180 gr. Shall give you the Altitude of the Sun above the Plane, when it cometh to that Tropick. As in our latitude, the height of the Style above an Horizontal Plane is 51 gr. 30 m. add unto this 113 gr. 30 m. the sum is 165 gr. Which being taken out of 180 gr. The remainder will be 15 gr. And such is the Altitude of the Sun above this Plane when it cometh to be in the WinterTropick: But if you add 66 gr. 30 m. unto 51. gr. 30 m. the remainder to 180 gr. Will be 62 gr. And such is the Altitude of the Sun in the SummerTropick. Then,
 As the Sine of 66 gr. 30 m. to the Sine of the Suns Altitude:
 So the Length of the Substylar Line, to the Length of the Axis of the Style.
As in the first Examples of the Declining Vertical, where the height of the Style was found to be 34 gr. 33 m. and is here represented before, pag.31 by the Angle BC♋; add to this height 113 gr. 30 m. for the Angle CB♋, the sum will be 148 gr. 3 m. and the remainder to 180 gr. will be 31 gr. 57 m. and such is the Angle B♋C of the Altitude of the Sun above the Plane, when it cometh to be in the Tropick of ♋, which is here the farthest Tropick from the Center.
Then supposing the length of the Substylarline between the Center and the Place which is fit for the farthest Tripick, to be about 21 Inches, extend the Compasses from the Sine of 66 gr. 30 m. unto the Sine of 31 gr. 57 m. the same extent will reach in the Line of Numbers from 21 unto 12.11, and so the length of the Axis of the Style should be 12 inch. 11 cent. Or it may suffice to make it just 12 Inches, as a more easie ground for the rest of the Work.
But if it were required to proportion the Style unto the Plane, so as it may cast the Shadow to the full length of the Substylarline at all times of the Year, you may then consider the Sun in the Tropick, which is to be set nearest unto the Center, and add 66 gr. 30 m. unto 34 gr. 33 m. so the remainder unto 180 gr. will be 78 gr. 57 m. And if you extend the Compasses from the Sine of 66 gr. 30 m. unto the Sine of 78 gr. 57 m. the same extent will reach in the Line of Numbers from 21 unto 22.47 for the length of the Axis of the Style.
The Style of a Plane neither Equinoctial nor Polar, may be either a small Rod of Iron set parallel to the Axis of the World, or perpendicular to the Plane, or else a thin Plate of Iron or Brass, made in form of a Rectangle Triangle BAC, with the Base BC parallel to the Axis of the World, the Side AB perpendicular to the Plane, and the Side AC the same with the Substylarline; wherein knowing BC, and the Angle BAC,
 As the Sine of 90 gr. to the Length of the Axis:
 So the Sine of the height of the Style, to the Length of the Perpendicular Side:
 And so the Cosine of the Height of the Style, to the Length of the Substylar side.
Thus in the former Example. The length of the Axis being supposed to be 12 Inches, and the height of the Style 34 gr. 33 m. Extend the Compasses from the Sine of 90 gr. (or else from the Sine of 5 gr. 45 m.) unto 12 in the Line of Numbers, the same extent will reach from the Sine of 34 gr. 33 m. unto 6.80 in the Line of Numbers, for the length of the perpendicular Side; and from the Sine of 55 gr. 27 m. unto 9.88 for the length of the Substylar side.
This is here represented by C♈, and may be found by resolving the rectangle Triangle CB♈.
 As the Sine of the height of the Style, is to the Sine of 90 gr.
 So the Length of the Axis, To the Distance of the equator from the Center.
Extend the Compasses from the Sine of 55 gr. 27 m. unto the Sine of 90 gr. the same extent will reach in the Line of Numbers from 12 unto 14.17. Wherefore if you take 14 inch. 57 cent. and pricking them down on your Substylarline from C unto ♈, draw a Line through ♈, crossing the Substylar at Right Angles, the Line drawn shall be the Equator.
These Angles made by ♈ and the Hourlines are Complements of those which are at C, between BC the Axis and those several Hourlines, and depend upon the Angles at the Pole, between the proper Meridian and the Hourcircles.
 As the Sine of 90 gr. to Cosine the Angle at the Pole:
 So the Cotangent of the Height of the Style, to the Tangent of the Angle between the Equator and the Hourline.
In our Example the height of the Style is 34 gr. 33 m. and the proper Meridian falleth to be the same with the Circle of the second Hour after Noon; whereupon the Angle at the Pole, between this proper Meridian, and the Circles of the Hour of 1 on the one side, and 3 on the other side, will be 15 gr. So between this Meridian and the Hourcircles of 10 and 4 the Angle will be 30 gr. &c. as in the Table.
Hour. 
An. Po. 
Ar. Po. 
Ar. Eq. 
C ♈ 
C ♋ 
C ♑ 
Gr. 
M. 
Gr. 
M. 
Gr. 
M. 
In. 
P. 
In. 
P. 
In. 
P. 
Substyl. 
0 
0 
0 
0 
55 
27 
14 
57 
20 
80 
1 
21 
1 
3 
15 
0 
8 
38 
54 
30 
14 
74 
21 
36 
11 
25 
12 
4 
30 
0 
18 
8 
51 
30 
15 
33 
23 
44 
11 
40 
11 
5 
45 
0 
19 
33 
45 
45 
16 
75 
29 
06 
11 
76 
10 
6 
60 
0 
44 
30 
36 
0 
20 
00 
50 
84 
12 
77 
9 
7 
75 
0 
64 
42 
20 
36 
34 
10 
Infinit. 
15 
82 
8 
8 
90 
0 
90 
0 
0 
0 
Infinit. 

27 
60 
If then it be required to find the Angle which the Hourline of 4 after noon doth make with the Plane of the Equator, that is the Angle C4B, contained between the Hourline C4 and the Line B4, drawn from the top of the Style unto the Intersection of the Hourline of 4 with the Equator.
Extend the Compasses from the Sine of 90 gr. Unto the Sine of 60 gr. The Complement of the Angle at the Pole, the same extent will reach from the Tangent of 55 gr. 27 m. the Complement of the height of the Pole, unto the Tangent of 51 gr. 30 m. and such is the Angle C4B in the Diagram.
Or in Crosswork, if it were required to find the Angle C9B, look into the table for the Hour of 9, and there you shall find the Angle at the Pole to be 75 gr. And if you extend the Compasses from the Sine of 90 gr. Unto the tangent of 55 gr. 27 m. the same extent will reach from the Sine of 15 gr. The Complement of 75 gr. Unto the Tangent of 20 gr. 36 m. and such is the Angle C9B, made at the Equator between the Line B9, drawn from the top of the Style, and the Hourline C9, drawn from the center. The like reason holdeth for the rest, which may be found and set down in a Table: Then you may either draw these Angles at C in the former Figure more perfectly, and thence finish your Work, or else proceed,
The Distances between the Center and the Parallels of Declination may be found by resolving the Triangles made by the Axis BC, the Lines of Declination, and the Hourlines. For having the Angles at the Equator, and knowing the Declination of the Parallel, if the Parallel shall fall between the Equator and the Center, add the Declination unto the Angle at the Equator: or if it shall fall without the Equator, take the Declination out of the Angel at the Equator, so shall you have the Angle at the Parallel. Then,
 As the Sine of the Angle at the Parallel, to the Cosine of the Declination:
 So the length of the Axis of the Style, to the Distance between the Center and the Parallel.
Thus in our Example, the Angle at the Equator belonging to the Hour of 4 afternoon was found to be 51 gr. 30 m. if you would find the distance between the Center and the Equator, extend the Compasses from the Sine of 51 gr. 30 m. unto the Sine of 90 gr. The Complement of the declination, the same extend will reach in the Line of Numbers from 12 unto 15.33, and such is the distance upon the Hourline of 4 between the Center and the Equator.
If you would find the distance upon this Hourline between the Center and the inner Tropick, whose Declination is known to be 23 gr. 30 m. add the declination to the Angle at the Equator, so the Angle at the Parallel will be 75 gr. Wherefore extend the Compasses from the Sine of 75 gr. Unto the Sine of 66 gr. 30 m. the Complement of the declination, the same extent will reach in the Line of Numbers from 12 unto 11.40, and such is the length of the Hourline of 4 between the center and the Tropick of ♑.
If you would find the distance upon this Hourline between this Center and the Tropick of ♋, which is here farthest from the Center, take the Declination out of the Angle at the Equator, so the Angle at the parallel will be 28 gr. wherefore extend the Compasses from the Sine of 28 unto the Sine of 66 gr. 30 m. the same extent will reach in the Line of Numbers from 12 unto 23.44, and such is the distance between the Center and the Tropick of ♋ upon the Hourline of 4. The like reason holdeth for all the rest, which may be gathered and set down in a Table.
That done, and the Equator drawn as before, if you would draw the Tropick of ♋, look into the Table, and there finding under the Title C♋ the distance of the Substylar between the center and the parallel of ♋ to be 20 inch. 80 cent. take 20 inch. 80 cent. out of the Line of Inches, and prick them down in the Substylar of your Plane from C unto ♋.
Or if either the Center fall without your Plane, or the extent be too large for your Compasses, you may prick down the difference between C♈ and C♋: as here the distance C♈ between the Center and the Equator is 14.57, the distance C♋ 20.80, the difference 6.23. Therefore taking ♈ 6 inch. 23 cent. prick them down on the Substylar from 6 unto C♋, and you shall have the same Intersectuion of the Tropick and the Substylar as before: And the like reason holdeth for pricking down the rest of these Distances on their several Hourlines.
Then having the Points of Intersection between the Hourlines and the Parallel, you may joyn them all in a crooked Line, without making of any Angles, the Line so drawn shall be the Tropick required. And after this Manner you may draw any other Parallel of Declination, whereof you have Examples in most of the former Diagrams.
