Then the shadow of the line AB will obviously fall on the line XII ... XII at apparent noon, on the line I ... I at one hour after noon, on II ... II at two hours after noon, and so on. If now the cylinder be cut by any plane MN representing the plane on which the dial is to be traced, the shadow of AB will be intercepted by this plane and fall on the lines AXII AI, AII, &c.
The construction of the dial consists in determining the angles made by AI, AII, &c. with AXII; the line AXII itself, being in the vertical plane through AB, may be supposed known.
For the purposes of actual calculation, perhaps a transparent sphere will, with advantage, replace the cylinder, and we shall here apply it to calculate the angles made by the hour-line with the XII o'clock line in the two cases of a horizontal dial and of a vertical south dial.
| Fig. 2. |
Horizontal Dial.—Let PEp (fig. 2), the axis of the supposed transparent sphere, be directed towards the north and south poles of the heavens. Draw the two great circles, HMA, QMa, the former
horizontal, the other perpendicular to the axis Pp, and therefore coinciding with the plane of the equator. Let EZ be vertical, then the circle QZP will be the meridian, and by its intersection A with the horizontal circle will determine the XII o'clock line EA. Next divide the equatorial circle QMa into 24 equal parts ab, bc, cd, &c. ... of 15° each, beginning from the meridian Pa, and through the various points of division and the poles draw the great circles Pbp, Pcp, &c. ... These will exactly correspond to the equidistant generating lines on the cylinder in the previous construction, and the shadow of the style will fall on these circles after successive intervals of 1,2, 3, &c., hours from noon. If they meet the horizontal circle in the points B, C, D, &c., then EB, EC, ED, &c. ... will be the I, II, III, &c., hour-lines required; and the problem of the horizontal dial consists in calculating the angles which these lines make with the XII o'clock line EA, whose position is known. The spherical triangles PAB, PAC, &c., enable us to do this readily. They are all right-angled at A, the side PA is the latitude of the place, and the angles APB, APC, &c., are respectively 15°, 30°, &c., then
tan AB = tan 15° sin latitude,
tan AC = tan 30° sin latitude,
&c. &c.
These determine the sides AB, AC, &c., that is, the angles AEB, AEC, &c., required.
The I o'clock hour-line EB must make an angle with the meridian EA of 11° 51' on a London dial, of 12° 31' at Edinburgh, of 11° 23' at Paris, 12° 0' at Berlin, 9° 55' at New York and 9° 19' at San Francisco. In the same way may be found the angles made by the other hour-lines.
The calculations of these angles must extend throughout one quadrant from noon to VI o'clock, but need not be carried further, because all the other hour-lines can at once be deduced from these. In the first place the dial is symmetrically divided by the meridian, and therefore two times equidistant from noon will have their hour-lines equidistant from the meridian; thus the XI o'clock line and the I o'clock line must make the same angles with it, the X o'clock the same as the II o'clock, and so on. And next, the 24 great circles, which were drawn to determine these lines, are in reality only 12; for clearly the great circle which gives I o'clock after midnight, and that which gives I o'clock after noon, are one and the same, and so also for the other hours. Therefore the hour-lines between VI in the evening and VI the next morning are the prolongations of the remaining twelve.