CHAP. XXII. How to find the Arches and Angles which are requisite in a North Decliner Recliner, and a South Decliner Incliner.

I Could not pass by this Example of the North Recliner Decliner, and South In∣cliner Decliner, although it is shewed in the Fundamental Diagram; but it may be too obscure, and harder to be apprehended by the Industrious Practitio∣ner there; therefore I would advise him to draw a Scheme of the Dial, as was shewed Chap. 15. & 18. or draw the Fundamental Diagram in Paper, and with a small Needle prick the Hour-lines, Horizon, and Meridians; Aequinoctial, and the Tro∣picks; and then you have a Figure ready to be stamped with a little Charcoal-dust as often as you have occasion: Or if you apprehend your Work in any manner, the Figure following, or the like, may serve your turn, to shew you the Angles you are to find, and Arches for the making of your Dial. I shall be short in this, and refer you to Chap. 15. & 18.

The Circle ESWN is our Horizon, as before; NS our Meridian, FLC the Plane, ZL the Reclination thereof, FC the Base or Horizontal Line of the Plane, AEN the Vertical of the Plane, cutting it right at L, and cutting the Pole thereof at H: for N is the Pole of a Plane erect upon FC; but the Pole of the Reclining Plane FLC is H; SE or nN the Declination of the Plane.

Now you see your three Triangles all adjoyning in this Scheme, viz. FSO and ORP rectangled at S and R, and PZH obtuse angled at Z.

It is true, That the two first may do the Work, and so we will be brief. Observe, you are to find as followeth.

• 1. The Arch between the Horizon and the Plane FO.
• 2. The Arch of the Meridian between the Horizon and the Plane SO.
Page  34
• 3. The Arch of the Meridian between the Pole and the Plane PO.
• 4. The Angle of Inclination between the Meridian and the Plane FOS.
• 5. The Angle of Inclination between both Meridians OPR.
• 6. The Substile Distance from the Meridian OR.
• 7. The Height of the Stile above the Plane PR.

1. To find the Arch of the Meridian between the Horizon and the Plane, it is thus.

As the Sine of 90 at S	1000000
To the Sine of Reclination at F 45	984948
So is the Co-tangent of Declination FS 45	1000000
To the Complement Tangent of FO 35 deg. 16 min.	984948

Or thus: Extend the Compasses from the Sine of 90, to the Sine of 45; the same Extent will reach from the Tangent of 45 deg. to the Sine of 35 deg. 16 min. as be∣fore; which substracted from 90 deg. there remains 54 deg. 44 min. the Arch of the Plane between the Horizon and Meridian FO.

2. To find the Arch of the Meridian between the Pole and the Plane, first find the Arch of the Meridian between the Horizon and the Plane SO thus.

As the Sine of 90 at S	1000000
To the Sine of the Arch of the Plane between the Horizon and Plane FO 54. 44	991194
So is the Tangent of Reclination at F 45	1000000
To the Tangent of the Arch of the Meridian between the Horizon and Plane SO 39 deg. 14 min.	991194

Or thus: Extend the Compasses from the Sine of 90, to the Sine of 54 deg. 44 m. the same Extent will reach from the Tangent of 45 deg. to the Tangent of SO 39 d. 14 min. by the Tables found before: which 39 deg. 14 min. taken out of 90 deg.Page  35 there remains 50 deg. 46 min. the Arch of the Meridian between the Zenith and the Plane OZ; which being added to the Complement of the Latitude ZP, there will be 89 deg. 16 min. for the Arch of the Meridian between the Pole and the Plane PO.

3. For the Angle between the Meridian and the Plane FOS or ZOL, it is

As the Sine of the Arch FO 54 deg. 44 min.	991194
To the Radius or Sine of 90 deg. S	1000000
So is the Sine of the Declination FS 45 deg.	98494•
To the Sine of the Angle at O 60. 0	993754

Or, by Gunter's Rule, Extend the Compasses from the Sine of 54 deg. 44 min. to the Sine of 90 deg. the same Extent will reach from the Sine of 45 deg. O, to the Sine of 60 deg. 0, the Angle between the Meridian and the Plane.

4. Before we find the Inclinations of both Meridians, or the Difference of Longi∣tude or Angle at P, we will find the Substiler Distance OR.

As the Sine of the Angle at R 90 deg.	1000000
To the Co-sine of the Angle at O 60 deg.	969897
So is the Tangent of the Side PO 89 deg. 16 min.	1189279
To the Tangent of the Substile OR 88 deg. 32 min.	1159176

Extend the Compasses from the Sine of 90 deg. to the Sine of 30 deg. the same Extent will reach from the Tangent of 89 deg. 16 min. to the Tangent of the Substile 88 deg. 32 min.

5. Now for the Inclination of both Meridians, or Difference of Longitude, or Angle at P, it is thus. For,

As the Sine of the Arch PO 89 deg. 16 m.	999996
To the Radius or Sine of 90 at R	1000000
So is the Substiler Distance OR 88 deg. 32 min.	999985
To the Sine of the Angle of Inclinations between both Meridians 88 deg. 45 min. Difference of Longitude	999989

Which converted into Hours, by allowing 15 deg. to one Hour, 4 Minutes to a De∣gree will be 5 ho. 55 min. as you may find by the Line of Inclination and Hours on your Scale.

6. For the Height of the Pole above the Plane, or Stiles Height,

As the Radius or Sine of 90 deg.	1000000
To the Sine of the Arch PO 89 deg. 16 min.	999996
So is the Sine of the Angle at O 60 deg.	993753
To the Sine of the Stiles Height PR 60 deg.	993749

the Height of the Pole above the Plane or Stiles Height. And thus if you observe what was laid before in the South Decliners Recliners, and now for the North Decli∣ners Recliners Incliners, you have the Propositions for any sorts of Reclining Decli∣ning Inclining Dials, and how to find the Arches and Angles, as before, fitting for the making of them. Now we will proceed and draw the Dial.