
Meteorology
H, K there will have been drawn lines with the same ratio to one
another, not only to the circumference MN but to another point as
well, which is impossible. Since then D cannot bear that ratio to
any line either lesser or greater than IM (the proof being in either
case the same), it follows that it must stand in that ratio to MI
itself. Therefore as MI is to IK so IH will be to MI and finally MH to
MK.
If, then, a circle be described with I as pole at the distance MI it
will touch all the angles which the lines from H and K make by their
reflection. If not, it can be shown, as before, that lines drawn to
different points in the semicircle will have the same ratio to one
another, which was impossible. If, then, the semicircle A be
revolved about the diameter HKI, the lines reflected from the points
H, K at the point M will have the same ratio, and will make the
angle KMH equal, in every plane. Further, the angle which HM and MI
make with HI will always be the same. So there are a number of
triangles on HI and KI equal to the triangles HMI and KMI. Their
perpendiculars will fall on HI at the same point and will be equal.
Let O be the point on which they fall. Then O is the centre of the
circle, half of which, MN, is cut off by the horizon. (See diagram.)
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
