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

THE SAND-RECKONER 83
Archimedes has, lastly, to compare the diameter of the sun
with the circumference of the circle described by its centre.
Aristarchus had made the apparent diameter of the sun ylo^h
of the said circumference ;
Archimedes will prove that the
said circumference cannot contain as many as 1,000 sun's
diameters, or that the diameter of the sun is greater than the
side of a regular chiliagon inscribed in the circle. First he
made an experiment of his own to determine the apparent
diameter of the sun. With a small cylinder or disc in a plane
at right angles to a long straight stick and moveable along it,
he observed the sun at the moment when it cleared the
horizon in rising, moving the disc till it just covered and just
failed
to cover the sun as he looked along the straight stick.
He thus found the angular diameter to lie between T^^R and
^oR, where R is a right angle. But as, under his assumptions,
the size of the earth is not negligible in comparison with
the sun's circle, he had to allow for parallax and find limits
for the angle subtended by the sun at the centre of the earth.
This he does by a geometrical argument very much in the
manner of Aristarchus.
84 ARCHIMEDES
the sun when it has just cleared the horizon. Draw from E
two tangents EP, EQ to the circle with centre 0, and from
C let CF, CG be drawn touching the same circle. With centre
G and radius GO describe a circle : this will represent the path
of the centre of the sun round the earth. Let this circle meet
the tangents from G in A, B, and join AB meeting GO in M.
Archimedes's observation has shown that
?bfB> I PEQ >^ B;
and he proceeds to prove that AB is less than the side of a
regular polygon of 656 sides inscribed in the circle AOB,
but greater than the side of an inscribed regular polygon of
1,000 sides, in other words, that
T^R >IFCG > aio^-
The first relation is obvious, for, since GO > E0,
Z PEQ > Z FGG.
Next, the perimeter of any polygon inscribed in the circle
AOB is less than -4 T 4 - GO (i.e. -~- times the diameter)
Therefore AB < ^•-\ 4 - CO or T^g CO,
and, a fortiori, AB < tJq CO.
Now, the triangles GAM, COF being equal in all
respects,
AM= OF, so that AB = 20F= (diameter of sun) > CH+ OK,
since the diameter of the sun is greater than that of the earth
;
therefore
for
CH + OK < ^CO, and HK > ^CO.
And CO > CF, while HK < EQ, so that EQ > T %%CF.
We can now compare the angles OCF, OEQ
;
Z OCF
TOEQ
> tan OCFtan
0EQ_
EQ
> CF
Let the circles with centres 0, G represent sections of the sun
and earth respectively, E the position of the observer observing
g2
Doubling the angles, we have
IFGG >^%.IPEQ
> 2omo^
^ 203
Z1 '*
> T
9
o
9
o, a fortiori.
since I PEQ > ^i?,