# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Calculating Horizontal Parallax**

**From:**Frank Reed

**Date:**2014 Apr 2, 16:14 -0700

Sean,

You can also get HP's for planets by simple inverse proportion in distance, knowing that the Sun's HP is 9 seconds of arc. For example, we see Jupiter and Mars high in the night sky tonight. Jupiter is about 90° away from the Sun and crosses the meridian close to sunset. If you draw a little picture, and you know Jupiter's mean distance from the Sun is about 5.2 AUs (5.2 x the Earth's mean distance from the Sun), then you can see that the HP of Jupiter would be less than 2" of arc (from 9/5 --one significant figure is good enough here). This small angle is irrelevant for celestial navigation. Meanwhile Mars transits right around midnight so it's about 180° from the Sun. Given that the mean distance of Mars is about 1.5 AUs, it must be roughly 0.5 AU from the Earth right now yielding a parallaz of 18" or 0.3' of arc (from 9"/0.5) which is minor for navigation by common celestial altitudes but would be significant for something like lunar distances. So if you know the basic sequence of the planets' distances from the Sun in AUs --0.4, 0.7, 1.0, 1.5, 5.2, 9.5 from Mercury through Saturn-- and if you can picture their positions on an "orrery" or a flat "overhead" view of the Solar System, then you can usually estimate the parallaxes of the planets with a very short calculation: HP=9"/d. And of course, once you have that, then the parallax in the altitude is HP*cos(h) as usual.

Here's a fun puzzle: you look through a small telescope, and you see that the phase of Venus is very nearly half full. Knowing only that the Sun's HP is 9" and the mean distance of Venus from the Sun is 0.7 AU, what is the HP of Venus on that date? :)

-FER

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