Vertical South Dial.—Let us take again our imaginary transparent sphere QZPA (fig. 4), whose axis PEp is parallel to the earth's axis. Let Z be the zenith, and, consequently, the great circle QZP the meridian. Through E, the centre of the sphere, draw a vertical plane facing south. This will cut the sphere in the great circle ZMA, which, being vertical, will pass through the zenith, and, facing south, will be at right angles to the meridian. Let QMa be the equatorial circle, obtained by drawing a plane through E at right angles to the axis PEp. The lower portion Ep of the axis will be the style, the vertical line EA in the meridian plane will be the XII o'clock line, and the line EM, which is obviously horizontal, since M is the intersection of two great circles ZM, QM, each at right angles to the vertical plane QZP, will be the VI o'clock line. Now, as in the previous problem, divide the equatorial circle into 24 equal arcs of 15° each, beginning at a, viz. ab, bc, &c.,—each quadrant aM, MQ, &c., containing 6,—then through each point of division and through the axis Pp draw a plane cutting the sphere in 24 equidistant great circles. As the sun revolves round the axis the shadow of the axis will successively fall on these circles at intervals of one hour, and if these circles cross the vertical circle ZMA in the points A, B, C, &c., the shadow of the lower portion Ep of the axis will fall on the lines EA, EB, EC, &c., which will therefore be the required hour-lines on the vertical dial, Ep being the style.
| Fig. 4. |
There is no necessity for going beyond the VI o'clock hour-line on each side of noon; for, in the winter months the sun sets earlier than 6 o'clock, and in the summer months it passes behind the plane of the dial before that time, and is no longer available.
It remains to show how the angles AEB, AEC, &c., may be calculated.
The spherical triangles pAB, pAC, &c., will give us a simple rule. These triangles are all right-angled at A, the side pA, equal to ZP, is the co-latitude of the place, that is, the difference between the latitude and 90°; and the successive angles ApB, ApC, &c., are 15°, 30°, &c., respectively. Then
tan AB = tan 15° sin co-latitude;
or more simply,
tan AB = tan 15° cos latitude,
tan AC = tan 30° cos latitude,
&c. &c.
and the arcs AB, AC so found are the measure of the angles AEB, AEC, &c., required.
In this ease the angles diminish as the latitudes increase, the opposite result to that of the horizontal dial.