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 Entry: DIAL