sin alt. = sin decl. sin lat. + cos decl. cos lat. cos ACX ... (1)
which equation determines the hour-angle ACX shown by the bead.
To determine the hour-angle of the sun at the same moment, let fig. 10 represent the celestial sphere, HR the horizon, P the pole, Z the zenith and S the sun.
From the spherical triangle PZS, we have
cos ZS = cos PS cos ZP + sin PS sin ZP cos ZPS
but ZS = zenith distance = 90° - altitude
ZP = 90° - PR = 90°- latitude
PS = polar distance = 90° - declination,
therefore, by substitution
sin alt. = sin decl. sin lat. + cos decl. cos lat. cos ZPS ... (2)
and ZPS is the hour-angle of the sun.