dow 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.

_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

[Illustration: FIG. 2.]

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 deg. 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 deg., 30 deg., &c., then

tan AB = tan 15 deg. sin _latitude_,
tan AC = tan 30 deg. sin _latitude_,