A history of Greek mathematics - Wilbourhall.org

THE ANALEMMA OF PTOLEMY 289
vertical ; and draw FG perpendicular to BA, and ET to OZ.
Join HG, and we have FG = SH, GH = FS = ET.
We now represent SF in a separate figure (for clearness'
sake, as Ptolemy uses only one figure), where B'Z' A' corresponds
to BZA, P' to P and O'M' to OM. Set off the arc
P'S' equal to 08 (= 90° -t), and draw S'F' perpendicular
to O'M'. Then S'M'= 8M, and S'F r = SF; it is as if in the
original figure we had turned the quadrant MSG round MO
till it coincided with the meridian circle.
In the two figures draw IFK, I'F'K' parallel to BA, B'A\
and LFG, L'F'G' parallel to OZ, O'Z'.
Then (1) arc Zl — arc ZS = arc (90° — SV), because if we
turn the quadrant ZSV about ZO till it coincides with the
s '
Z
meridian, S falls on /, and V on B. It follows that the
required arc SV — arc B'l' in the second figure.
(2) To find the arc VC, set off G'X (in the second figure)
along G'F' equal to FS or F'S', and draw O'X through to
meet the circle in X'. Then arc ^'X'=arc VC; for it is as if
we had turned the quadrant BVG about BO till it coincided
with the meridian, when (since G'X = FS = GH) H would
coincide with X and V with X'. Therefore 5Fis also equal
to B'X'.
(3) To find QG or ZQ, set off along TF' in the second figure
T'Y equal to F'S\ and draw O'Y through to Y' on the circle.
Then arc B'Y' = arc QG: for it is as if we turned the prime
vertical ZQG about ZO till it coincided with the meridian,
when (since T'Y=S'F'= TE) E would fall on 7, the radius
OEQ on O'YY' and Q on Y'.
(4) Lastly, arc BS = arc BL = arc B'U, because 8, L are
1523.2
u