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Re: Stellarium for sight reduction math practice?
From: Paul Hirose
Date: 2014 Sep 05, 21:27 -0700
From: Paul Hirose
Date: 2014 Sep 05, 21:27 -0700
> I've been playing with Stellarium and it altitude values to practice the math of a sight reduction. I'm assuming I don't have to do any sextant corrections if I use the apparent versus the geometric values for Stellarium's "Alt" generate my H_o? Is the only difference between the two that the apparent incorporates refraction (A2 Tables of the NA)? In astronomy, "geometric" refers to where a body actually is at a given moment. "Apparent" refers to where it *appears* to be. The difference is due to: 1) The time for light to travel to the observer. This is ignored for stars since light time is implicitly included in star catalog data. 2) Light deflection by the gravitation of other bodies. These bodies are often (usually?) ignored except the Sun, and that too is ignored unless great accuracy is necessary. 3) Aberration of light due to the observer's velocity through space. This causes an cyclic annual variation in the apparent coordinates of every body by a maximum of 20″ from the "un-aberrated" coordinates. The effect can be ignored when modest accuracy is adequate, as in a long term almanac. In addition to geometric and apparent, there are also astrometric coordinates, which are adjusted for light time only. (In the case of a star, the catalog coordinates are adjusted for parallax and proper motion only.) None of these includes refraction. In fact, The Astronomical Almanac definition of "topocentric place" specifically includes aberration and excludes refraction. I don't know if Stellarium uses these terms in the way I described.