A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Frank Reed
Date: 2020 Mar 31, 15:29 -0700
Tony Oz, a week ago, you wrote:
"Weather permitting I'll try to check my impression that if I set my sextant to the exact value of the geocentric distance between the Sun's center and Venus - Venus slightly "overshoots" the Sun's center. I.e., the apparent distance is smaller than the geocentric one. By how much? Difficult to tell, I estimate it as 2~3 arcminutes."
That would be typical, yes. It's just the difference in refraction between the two bodies adjusted by the geometric fraction that falls along the angular arc between the Sun and Venus. It's easiest to see this in the vertically-aligned case. Imagine the Sun is 15° high in the west, and Venus is 40° above it (55° high on nearly the same azimuth). Work it out for yourself. How much would the angle between them be changed by refraction? Refraction lifts Venus by a fraction of a minute of arc. It lifts the Sun by a few minutes of arc. Thus the angle between them is shortened by the difference in the refraction. Easy, right? Now draw out a case where they're not aligned vertically. Suppose Venus is above the Sun, but a ray drawn from the Sun towards Venus leaves the Sun at an angle, let's say, 45° to the left of vertical. Refraction lifts the Sun vertically, but only about 71% (the cosine of 45°) of that refraction acts along the arc between the Sun and Venus. You can think through other simple cases, and yes, this sort of analysis can be directly extended to lunars.
Finally, suppose Venus and the Sun are, let's say, 32° apart, and they're both above 45° in altitude. How much would the distance between them be affected by refraction? ...Remember this one?? It's one part in 3000, very nearly.