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For helping with judging NLC candidates (are they NLC or not) I have a set of formulas to calculate the minimum altitude (in km) of the candidate given an observed altitude (in degrees) of the candidate (using known stars, like Auriga, as reference points) and the date/time (giving the Sun's altitude).

A typical observed altitude is 10 degrees, largely unaffected by refraction and a typical result is 70 km (ruling out any other type of cloud). The formulas are limited to only work in the horizontal direction (azimuth) of the Sun (that is under the horizon), but this is not a problem in practice.

Reasonable assumptions are used, except one: no atmospheric effects like refraction and scattering are accounted for.

  1. Are NLCs only visible to the naked eye if directly exposed to sunlight (the Sun directly visible by the NLC)? Or is scattered light enough to make them visible?

  2. What about refraction? Light coming to the NLC goes through near vacuum when it is close to the NLC so the refraction must be less than for an observer on the ground, but what is the exact figure (when the Sun is at the horizon as seen by the NLC)?

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Unfortunately, I live too far south (44° 44' N) to see noctilucent clouds, or so I thought, but they seem to be moving south...they were reported this week from the northern USA. –  Geoff Gaherty Jul 3 '11 at 21:21
    
@Geoff Gaherty: the season started late this year. I saw some magnificent ones around 2011-07-10 and weak, but distinct ones, yesterday evening (2011-07-20T23:00Z+1) through thin clouds (only the Moon and the brightest stars were visible.) –  Peter Mortensen Jul 21 '11 at 14:17
    
Two NLC movies from 2011-07-11. –  Peter Mortensen Jul 21 '11 at 21:28

1 Answer 1

I'm only "answering" by default of more competent others, and because I've just today become interested in the noctilucent cloud phenomenon.

  • 1) I don't know, but strong yellow to red illumination should indicate direct Sun light

What you're trying to do is find the NLCs vertical altitude, which might be 50 miles, and 100 miles or more slant distance from you. Perhaps you could team with another person to get two azimuth bearings, then use trigonometry against your altitude bearing to find the NLCs "flat Earth" vertical altitude directly.

  • 2) You can use a table of corrections for celestial navigation, check out this link

But, you'd have to apply corrections for "dip" and refraction (unless that's already in your formulas). Again: I suggest a "range and bearing" solution with a friend.

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