In a total eclipse of the Sun, the Moon just barely covers the Sun. The Sun itself is so distant that when viewed from anywhere on Earth, it covers practically the same patch of the sky, with a width of about 0.5°. Hipparchus concentrated on point E at the edge of the Moon (drawing), which during totality, when viewed from the Hellespont (point B) just overlapped point D on the edge of the Sun.
Viewed from Alexandria (point A), at that same moment, the point E only overlapped point C on the Sun, about 1/5 solar diameter short of the edge--which was why the eclipse there was not total. One-fifth of the Sun's diameter covers about 0.1° in the sky, so the small angle α (alpha, Greek A) between the two directions measured about 0.1 degrees. That angle is the parallax of the edge of the Moon, viewed from the above two locations.
In the times of Hipparchus, ships would sail from Alexandria to Greece, and he knew the Hellespont was north of the city. He might not have known the distance, but knew the latitudes of the Hellespont and Alexandria.
The local latitude can be shown to be equal to the elevation of the celestial pole above the horizon and today can be readily deduced by observing the height of Polaris, the pole star above the horizon. In the time of Hipparchus the pole of the heavens wasn't near Polaris (because of the precession of the equinoxes), but Hipparchus, who had mapped the positions of about 850 stars, must have known its position quite well.
The latitude of the Hellespont (from a modern atlas) is about 40° 20' (40 degrees and 20 minutes, 60 minutes per degree), while that of Alexandria is about 31° 20', a difference of 9°.
Assume also Alexandria is exactly due south. If furthermore r is the radius of the Earth, then the circumference of the Earth 2πr, where π= 3.1415926... ("pi", Greek lower-case P) is the ratio between the circumference and diameter of any circle. Since the entire circumference equals 360°, we get
AB = (2πr) · (9/360)
where the dot marks multiplication (algebra's equivalent to the x symbol).
The distance R to the Moon
The points AB are also located on another circle, centered on the Moon. The radius in that case is the distance R to the Moon, and because the arc AB covers 0.1°, we get
AB = (2πR) · (0.1/360)
Strictly speaking, each of the two arcs AB expressed in the above equations is measured along a different circle, with a different radius (and the two circles curve in opposite ways). However, in both cases AB covers only a small part of the circle, so that as an approximation we may regard each of the arcs as equal to the straight-line distance AB. That assumption allows us to regard the two expressions as equal and to write
(2πR/360)·0.1 = (2πr/360)·9
Multiplying both sides by 360 and dividing by 2π
0.1 R = 9 r
Dividing by (0.1 r)
R/r = 90
suggesting the Moon's distance is 90 Earth radii, an overestimate of about 50%.
A more accurate calculation
One reason an excessive value was obtained is that the Moon was assumed to be overhead at A or B. Actually, it is likely to be at some significant angle to the overhead direction, the "zenith" (see drawing).
Then the section cut by the angle α from the circle of radius R around E is not AB but AF (second drawing), which is smaller. Taking this into account reduces the distance.
We don't know where the Sun was during the 129 BC eclipse, but it must have been on the ecliptic (the words are obviously related!), which places it within 23.5° of the celestial equator, on either side. Assuming it was on the equator (that is, it passed overhead on the Earth's equator) and south of the reported observations (i.e. the eclipse occurred near noon) one can make a crude estimate of the correction, using simple trigonometry (see section M-8).
The Hellespont is around latitude of 40 degrees, and as the drawing shows, that is also the angle between the Moon's direction and the zenith. From the drawing (x marks multiplication)
AF = AB cos 40° = 0.766 AB
Repeating the preceding calculation for AF
AF = (2πR/ 360)·0.1
AF = 0.766 AB = 0.766·(2πr/ 360)·9
and in the end
R/r = 90·0.766 = 69
Final comments (optional)
According to "A History of Astronomy" by A. Pannekoek, the result obtained by Hipparchus was between 62 and 73 Earth radii. Today we know the average distance is about 60 radii, varying by a few Earth radii either way because of the ellipticity of the Moon's orbit.
In the absence of accurate timing, the method is almost guaranteed to produce an overestimate. The Earth rotates beneath the shadow spot cast by the Moon, which makes that spot sweep over a long strip, hitting many different locations at different times. The Hellespont was just one of many places where the eclipse was total. Similarly, Alexandria was just one of many locations where 4/5 of the Sun was covered. Randomly selecting point B from the first group and point A from the second may give a much longer baseline AB and a much larger (and incorrect) distance of the Moon. The fact Alexandria is almost exactly south of the Hellespont does not guarantee their peak eclipse times are the same, just that they are not too different.