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Marcel Tschudin posed an interesting question--

>While searching with Google, I came across this mail list. May be some one
>out here may be able to help me answering the following question:
>
>
>
>How do refraction values for negative elevations have to be calculated,
>such as e.g. the horizon from a plane? (I am interested in the range of 0?
>to approx. ?5?.)
>
>
>
>Is Bennett?s approximation also valid for negative elevations? If not,
>what other approximation formulae should be used, or, where can one find
>some benchmark values?
>
>
>
>I am interested in formulae for both, refraction from apparent position
>and from physical position.

===============

Following that, there have been several resposes from the Nav-l community.

By the way, several of those responses from Roibert Ene were misdated to
have a September date rather than an August one. I hope he will correct his
calender, because my email reader, which puts correspondence into strict
date order, keeps mis-sorting Robert's contributions.

==============

Marcel's question puzzled me considerably, until it emerged that was
referring to bubble-sextant observations, with respect to the true
horizontal, not altitudes measured up from the observed horizon with an
ordinary sextant.

It's not a question I am familiar with, nor is the table HO249, so I have
little to offer in the way of a positive contribution. I've been reading
the correspondence with interest, and have some comments to make about
suggestions made earlier.

Marcel wrote-

>I also was wandering whether the approximate formulae could be used by
>calculating the Refraction R for e.g. -2? the follwing way:
>
>R(-2?)  =  R(0?)  +  (  R(0?) - R(+2?)  )
>
>If this would be correct then one would not need separate formula for
>negative elevations.

I am sure that suggestion would not work. It would be true only if the
refraction was linear with altitude, which is VERY far from being the case.

==================

Fred Hebard wrote-

>I assume you have Meeus' formula for normal observations.
>Meeus' formula, as transcribed by G. Huxtable, is:
>tan(90-0.99914*S-(7.31/(S+4.4)), where S is the sextant altitude in
>degrees.  I don't know whether this formula blows up at 0, but you
>could try it and see whether it gives the same results as HO249's Table
>6.

No, it doesn't blow up at 0 degrees, but it does at -4.4 degrees: This
empirical formula doesn't pretend to give even an approximate answer for
negative altitudes.

By the way, that formula wasn't "transcribed" by me from Meeus, but
tinkered-with slightly. This was to avoid an infinity arising during the
calculation of tan, for an angle of 89.92 degrees (which is in the 'useful'
range). The coefficient of S is adjusted to 0.99914 rather than 1.0, just
to ensure that refraction goes to zero at an altitude of 90 degrees, as
symmetry says it must.

=================

Robert Eno wrote-

>If I had a scanner, I could scan the table for you.  Here are a few examples:
>
>Height of observer: 0 feet
>Sextant Altitude:  minus 1 degree
>Correction: minus 35 minutes
>
>Observed Altitude = minus 1 degree 35 minutes.


That seems a bit odd, and unphysical. From a height of 0 feet, how can a
sextant altitude possibly be -1degree? Also, the quoted value for the
correction, at -35 minutes, is rather a surprise, considering that the
adopted value for refraction at 0 degrees altitude is -34 minutes, and
refraction increases very quickly as the altitude decreases towards zero.

==================

I can suggest a useful analysis of such low-level refractions by Andrew T
Yoiung and George W Kattawar, titled "Sunset Science  2. A useful diagram",
in Applied Optics , vol 37 no 18 (20 June 1998), pages 3785 to 3792. Andy
Young is an acknowledged authority on such atmospheric optics, including
mirages.

George.






===============================================================
Contact George at george@huxtable.u-net.com ,or by phone +44 1865 820222,
or from within UK 01865 820222.
Or by post- George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13
5HX, UK.

   

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