Meteorology By Aristotle

Next let the horizon be ABG but let H have risen above the horizon.
Let the axis now be HI. The proof will be the same for the rest as
before, but the pole I of the circle will be below the horizon Ag
since the point H has risen above the horizon. But the pole, and the
centre of the circle, and the centre of that circle (namely HI) which
now determines the position of the sun are on the same line. But since
KH lies above the diameter AG, the centre will be at O on the line
KI below the plane of the circle AG determined the position of the
sun before. So the segment YX which is above the horizon will be less
than a semicircle. For YXM was a semicircle and it has now been cut
off by the horizon AG. So part of it, YM, will be invisible when the
sun has risen above the horizon, and the segment visible will be smallest
when the sun is on the meridian; for the higher H is the lower the
pole and the centre of the circle will be.

In the shorter days after the autumn equinox there may be a rainbow
at any time of the day, but in the longer days from the spring to
the autumn equinox there cannot be a rainbow about midday. The reason
for this is that when the sun is north of the equator the visible
arcs of its course are all greater than a semicircle, and go on increasing,
while the invisible arc is small, but when the sun is south of the
equator the visible arc is small and the invisible arc great, and
the farther the sun moves south of the equator the greater is the
invisible arc. Consequently, in the days near the summer solstice,
the size of the visible arc is such that before the point H reaches
the middle of that arc, that is its point of culmination, the point
is well below the horizon; the reason for this being the great size
of the visible arc, and the consequent distance of the point of culmination
from the earth. But in the days near the winter solstice the visible
arcs are small, and the contrary is necessarily the case: for the
sun is on the meridian before the point H has risen far.