For 4, 5, 6, 7, 8, 9, 10, 11 in the Morning, in the East-Dial. Must be changed to 8, 7, 6, 5, 4, 3, 2, I in the Afternoon in the Woft Dial: Which is all the difference. CHAP. VIII. To make an Erect Dial, declining from the South, Eastward, or West-ward; 30 degrees in the Latitude of 51 deg. 30 min. I. By the Globe. Fig. LV. T Quadrant of HE Globe being Rectified to the Latitude of the place, the Fig. Altitude in the Zenith, the Index of the Hour-Cir- LVL cle at 12, and the Equinoctial Colure brought under the Meridian; 1. Count the Declination of the Plain upon the Horizon, from the East or West-points thereof (according as the Plain declines) towards the South: namely, 30 degrees; and to that point of the Horizon bring the Quadrant of Altitude, and there keep it. 2. Turn the Globe about till the Index of the Hour-wheel cuts 11 of the Clock, or rather (as I said before) till 15 degrees of the Equinoctial have passed the Meridian, and then shall you find the Equinoctial Colure to cut the Quadrant of Altitude at 9 deg. 50 min. if you count the degrees from the Zenith point downwards. 3. Turn the Globe farther about, till 30 degrees of the Equinoctial be past the Meridian, and then shall yon find the Colure to cut the Quadrant of Altitude at 18 deg. 14 min. counted from the Zenith downwards as before. 4. Do the like with all the rest of the Hours, and you shall find that at the several 15 degrees of the Equinoctial, the Equinoctial Colure will cut such degrees of the Quadrant of Altitude as are expressed in this Table, if you count them from the Zenith downwards, as is before directed. This done; the o 5. Bring the Quadrant of Altitude to ther fide of the Meridian, and set it to 30 degrees, the Plains declination, counted from the Eaft or Weft-points Northward, as you did : Bbb 2 Hours from Noon. 12 11 I 10 2 93 [Hour-di- : d. m. 0000 0950 84 34 56 7 5 44 56 5749 5 7175 37 before 1 Fig.LVI before towards the South, which will be in the juft opposite point of the Horizon to which it was before; and also, bring the Equinoctial Colure under the Meridian. Then, 6. Turn the Globe about (the contrary way to what you did before) till 15 degrees of the Equinoctial be past the Meridian, and then shall you find the Equinoctial Colure to cut at 12 deg. 23. min of the Quadrant of Altitude counted from the Zenith. And fo continuing turning the Globe about till 10, 45, and 60 degrees of the Equinoctial have paffed the Meridian, you shall find the Equinoctial Colure to cut the Quadrant of Altitude at fuch degrees as are expressed in this Table. 12 23 The Hour-distances upon the Plain being thus attained, there are two other requifites in all upright declining Dials also to be found by the Globe, before the Dial can be finished. Namely, 1. The distance of the Sub-ftile from the Meridian. 2. The height of the Pole above the Plain, or the height of the Stile above the Sub-ftile. To find both which, Bring the Equinoctial Colure to the Plains declination 30 degrees counted upon the Horizon from the South-Eastward; and the Quadrant of Altitude to 30 degrees counted in the Horizon from the Eaft-Northward: So shall the Quadrant cut the Colure at Right Angles. And The number of degrees of the Quadrant contained between this Intersection and the Zenith (which here is 21 deg. 41 min.) is the diftance of the Sub-ftile from the Meridian. And the degrees of the Colure contained between this Interfection and the Pole (which here is 32 deg. 37 min.) is the height of the Pole above the Plain. II. The Geometrical projection of this Dial, in order to the Trigonometrical Calculation of the Hour distances and other Requifites belonging to such an Erect Declining Plains. I. Upon the Point Q, as a Centre, with 60 de. of a Scale of Chords, defcribe a Circle, representing your Dial Plain: And cross it with two Diameters ZQN for the Vertical; and H QO for the Horizontal Line of the Plain. 2. Set 2. Set 30 deg. the Plains Declination, from N to c, if the Plain Fig. Decline Eastward, as in this Example; or from Nto e, if Westward: LVI. Then lay a Ruler from Z to c, and it will cut the Horizontal Line of the Plain in K, so have you three Points Z, K and N, by which to defcribe the Arch ZKN, reprefenting the Meridian of the Place. And to find the Pole thereof, fet 90 de: from c, to d, and then a Ruler laid from Z to d, will cut the Horizontal Line HO, in W, which is the Pole of the Meridian Circle ZKN: and the West-point of the Horiz. 3. Set 51 deg. 30 m. the Latitude of the Place, from O to a, and from N to b: Then a Ruler laid from W to a, will cross the Meridian in P, the Pole of the World: And laid from W to b, it will cross the Meridian in Æ, the point through which the Equinoctial Circle is to pass: And now you have two points W and Æ, through which the Equinoctial Circle æ Æ æ may be defcribed (by the XXI Geometrical Problem, Lib. 1.) 4. Through P, the Pole of the World, and Q, the Pole of the Plain, draw the right Line PQ, for the Axis of the World, and Sub-ftilar Line of your Dial: And in this Line (extended) will the Center of the Equinoctial Circle æ Æ æ be found. 5. From P, the Pole of the World, lay a Ruler to Æ, the interfection of the Meridian and Equinoctial, and it will cut the Plain B: At this point B, begin to divide the Semicircle HNO of the Plain, into 12 equal Parts, at the points, &c. 6. FromQ, (the Pole of the Plain) lay a Ruler to every of the points, &c. and it will cross the Æquinoctial Circle æ Ææ, in the the Points ***, &c. ?. 7. Lay a Ruler to P (the Pole of the World) and every of the points ***, &c. and it will cut the Primitive Circle reprefenting the Dial Plain, in the points N, 9, 10, 11, &c. on the Weft fide, and N, 1, 2, 3,&c, on the East fide of the Meridian. 8. Lastly, Lines drawn from the Centre Q through these points, shall be the true Hour-lines of an Erect Plain Declining from the South-Eastward 30 de. in the Latitude 51 de. 30 m. And now, Concerning the other Requisites belonging to this Erect Decliing Plain. These are all of them reprefented to the Eye in the Scheme of the Projection of the Plain: Wherein, by the interfection of the several Circles there is conftituted a Right-angle Spherical Triangle ZTP, in which, The |