288 
TRIGONOMETRY 
the diagram is one position of it, coinciding with the equator), 
and it was called iKTrjfiopo9 kvkXos (‘the circle in six parts’) 
because the highest point of it above the horizon corresponds 
to the lapse of six hours; the second, called the hour-circle, is 
the circle represented by any position, as BSQA, of the circle 
of the horizon as it revolves round BA as axis. 
The problem is, as above stated, to find the position of the 
sun at a given hour of the day. In order to illustrate 
the method, it is sufficient, with A, v. Braunmuhl, 1 to take the 
simplest case where the sun is on the equator, i.e. at one of 
the equinoctial points, so that the hectemoron circle coincides 
with the equator. 
Let S be the position of the sun, lying on the equator MSG, 
P the pole, MZA the meridian, BCA the horizon, BSQA the 
hour-circle, and let the vertical great circle ZSV be drawn 
through S, and the vertical great circle ZQG through Z the 
zenith and 0 the east-point. 
We are given the arc SC = 90° — t, where t is the hour- 
angle, and the arc MB = 90° — 0, where 0 is the elevation of 
the pole; and we have to find the arcs (the sun’s altitude), 
z 
VC, the ‘ascensional difference’, SQ and QC. Ptolemy, in 
fact, practically determines the position of S in terms of 
certain spherical coordinates. - 
Draw the perpendiculars, Si 1 to the plane of the meridian, 
SH to that of the horizon, and SE to the plane of the prime 
1 Braunmiihl, Gesch. der Trigonometric, i, pp. 12, IB.