A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

.
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
290 TRIGONOMETRY
both in the plane LSHG at right angles to the meridian
therefore arc SQ — arc UZ\
Hence all four arcs SV, VC, QC, QS are represented in the
auxiliary figure in one plane.
So far the procedure amounts to a method of grajMcally
constructing the arcs
required as parts of an auxiliary circle
in one plane. But Ptolemy makes it clear that practical
calculation followed on the basis of the figure. 1 The lines
used in the construction are SF= sint (where the radius =1),
FT=0Fsin(f), FG = OF sin (90° -0), and this was fully
realized by Ptolemy. Thus he shows how to calculate the
arc SZ, the zenith distance (= d, say) or its complement 8V,
the height of the sun (= h, say), in the following way. He
says in effect: Since G is known, and Z F / 0'G / = 90° — 0, the
ratios O'F' : F'T and O'F' : O'T' are known.
0'F f T)
[In fact 7^7777-, =
—, .
L n
O'T' crd. (180° -20)'
- E— —rr, where D is the diameter
*
of the sphere.]
Next, since the arc MS or M'S' is known [
= £],
and therefore
the arc P'S' [= 90°-t], the ratio of
[in fact 0'F'/D= {crd. (lS0-2t)}/2D.
It follows from these two results that
O'F' to D is known
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
/= crd.(l 80°-2Q erd
x
2D Yn
QO _
Lastly, the arc SV (= h) being equal to B'I\ the angle h is
equal to the angle 0'1'T in the triangle FQ'T'.
'
And in this
triangle O'F, the radius, is known, while O'T' has been found ;
and we have therefore
O'T' crd.(2/ (arc BV — arc B'X'), the figure gives
tan a) _ G
,
1
xg' s'F' _ s'F' err 1
Q , -jTy,- ,F rr - tan t. ffi ^
See Zeuthen in Bibh'otheca mathematics, h, 1900, pp. 23-7.
^ ,