With AH and \theta, we only need \mu to get D'. Because the eclipse was total at H, and 4/5 total at A, it follows that \mu is 1/5 of the apparent diameter of the Sun. This quantity was well known by Hipparchus—he took it to be 1/650 of a full circle. The distance from the center of the Earth to the Moon then follows from D \approx D' + t.
