NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: lunars from Mt. Everest?
From: Frank Reed CT
Date: 2003 Dec 17, 21:09 EST
From: Frank Reed CT
Date: 2003 Dec 17, 21:09 EST
Looking through the list archives (which I've been doing for the past few days so that I don't re-cover tired ground), I see that this business about refraction corrections based on significant altitude of the observer's location in the atmosphere has come up before. I didn't see a message posting the correction to the refraction, so here it is:
R = R0 * exp(-height/(32000 feet))
where R0 is the sea level refraction found in the Nautical Almanac. The refraction is directly proportional to the atmospheric density, and that decreases exponentially with height. In Denver or some other place that's 5000 feet above sea level, the sea level refraction, R0, is reduced by about 15%. If you observe an object in the sky at an altitude of 15 degrees, the sea level refraction is 3.6 minutes at sea level, but in Denver it would be 3.1 minutes. That's not a big deal for standard line of position sights, but it would be a major source of error for land-based lunars practice.
How about doing celestial on Lake Titicaca in the Andes? Out in the middle, you can be a dozens miles from land. And it's waaaay up there. The refraction difference due to altitude on Lake Titicaca would be greater than 0.1 minutes for any altitude below 70 degrees and bigger than 1 minute of arc for altitudes below 18 degrees.
By the way, the "scale height" in the equation above might be 33000 feet or 35000, but using 32000 will get you close enough for "practical" work.
Frank E. Reed
[X] Mystic, Connecticut
[ ] Chicago, Illinois
R = R0 * exp(-height/(32000 feet))
where R0 is the sea level refraction found in the Nautical Almanac. The refraction is directly proportional to the atmospheric density, and that decreases exponentially with height. In Denver or some other place that's 5000 feet above sea level, the sea level refraction, R0, is reduced by about 15%. If you observe an object in the sky at an altitude of 15 degrees, the sea level refraction is 3.6 minutes at sea level, but in Denver it would be 3.1 minutes. That's not a big deal for standard line of position sights, but it would be a major source of error for land-based lunars practice.
How about doing celestial on Lake Titicaca in the Andes? Out in the middle, you can be a dozens miles from land. And it's waaaay up there. The refraction difference due to altitude on Lake Titicaca would be greater than 0.1 minutes for any altitude below 70 degrees and bigger than 1 minute of arc for altitudes below 18 degrees.
By the way, the "scale height" in the equation above might be 33000 feet or 35000, but using 32000 will get you close enough for "practical" work.
Frank E. Reed
[X] Mystic, Connecticut
[ ] Chicago, Illinois
