A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Frank Reed
Date: 2018 Dec 27, 10:51 -0800
There's a short, efficient calculation that yields a position for "Venus Center of Light" accurate enough for celestial navigation.
First, work in ecliptic coordinates if possible. In most applications for planetary coordinates, the process starts with ecliptic coordinates so this should be no problem (if not, the work is only slightly longer). In ecliptic coordinates the elongation, E, of Venus from the Sun is given by
cos(E) = cos(L) · cos(dλ),
where L is the true ecliptic latitude of the center of Venus, and dλ is the difference in ecliptic longitude between the Sun and Venus. In most cases, the elongation will be approximately equal to dλ.
Next we need the orientation of the illuminated edge of Venus relative to the ecliptic pole. The illuminated limb, of course, points toward the Sun. We draw the PVS triangle with corners at the ecliptic pole, Venus, and the Sun and solve for A, the position angle at Venus. That's given by
cos(A) = -tan(L) / tan(E).
The ecliptic latitude and longitude, L' and λ', of Venus Center of Light are then given by
L' = L + cos(A) · S,
λ' = λ +/- sin(A) · S.
Pick the +/- sign in the last depending on whether the celestial longitude of the Sun is greater or less than that of Venus.
Here S is the offset from the proper center of Venus to the Center of Light. That distance is given to good approximation by a simple expression which is 90% of the semi-diameter of Venus, and since Venus (at the cloudtops) is about 96% of the diameter of the Earth, the SD of Venus is about 0.96·p (p for parallax of Venus), so we have
S = 0.86 · p.
In summary, given the ecliptic lat, lon, and parallax of Venus (L, λ, p) and the ecliptic longitude of the Sun, Λ, calculate:
dλ = λ - Λ
S = 0.86 · p
cos(E) = cos(L) · cos(dλ)
cos(A) = -tan(L) / tan(E)
L' = L + cos(A) · S
λ' = λ +/- sin(A) · S.
The result is the ecliptic latitude and longitude of "Venus Center of Light". Any and all further computations, such as getting the GHA and Dec, should be worked treating this as a distinct celestial body.
The offsets should not be calculated or believed beyond a tenth of a minute of arc accuracy (by this quick method or any other method) since this is a subjective adjustment to the position of Venus which only applies when low magnification is used. Without magnification, the normal resolution of human vision is around one minute of arc or slightly better. The geometric center and the center of light are essentially indistinguishable with zero magnification. Using a 3x sextant scope that resolution of vision increases to one-third of a minute of arc or slightly better, and it's in these circumstances that the center of light has some small significance. An observer can detect the offset position of the center of light (just barely) but cannot resolve the disk of the planet. At higher resolution, the disk of Venus can be resolved as a small disk or crescent, and the true center should be used for sights. At no point is there any reason to calculate the offset beyond a tenth of a minute of arc accuracy, which is why the simple expression for S, above, works. In any case, this is a very small difference in coordinates that is dependent on the vision of the observer as much as astronomical details.