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    Re: Refraction
    From: Marcel Tschudin
    Date: 2005 Aug 8, 00:52 +0300

    George, here a qick reply before going to bed; we are here three ours later.
    Please remember I am not an expert, but one who just tries to get
    familiarised with the subject.
    > I don' t really understand the distinction that Marcel makes between
    > "terrestrial" and "astronomical" refraction. It's all refraction,
    > integrated along different parts of the light path. But I agree, this is
    > the first, and perhaps most difficult, step in the procedure.
    I understand your comment, it's all refaction. But from what I notice this
    distinction between terrestrial and astronmical refraction is actually made.
    (Have look with Google.) I have not found a definition for one or the other,
    but what I noticed is that in navigation or geodesy the approach is
    different than in astronomy. This probably for reason of the dominant
    factors. I guess for the astronomical refraction the change of density of
    the different air layers is dominant factor and for the terrestrial
    refraction the (change of) temperatures and air circulation over the surface
    of the earth (sea, forest, meadow, rocks, desert etc.). May be the
    difference could also be expressed this way: Astronomical refraction has one
    value for 0? altitude, knowing this may vary a lot; terrestrial refraction
    tries to handle the disturbences around 0? altitude.
    Refering to the approach in trying to calculate refraction for negative
    altitudes, it needs a further clarification:
    The reason for calculating the tangent point, is because I assume there an
    astronomical refraction for 0? altitude (ca. 34') corrected with pressure
    and temperature at this point. From the tangent point to the observer I
    intend to calculate the additional refraction with roules for terrestrial
    Isn't what you propose here:
    > A ray of light passing horizontally through Marcel's tangent point
    > undergoes a known refraction, on its way from outer space. If the tangent
    > was at sea level this would be 29 arc-minutes if Saemundsson's formula is
    > correct. At a higher altitude for the tangent point, it would need
    > adjusting by the ratio of air density at that altitude to air density at
    > sea level (call this factor D1). That adjustment factor should be roughly
    > correct if the tangent point is somewhere in the lower atmosphere, but
    > that
    > may not be so at higher levels.
    > If that ray of light then carries on to disappear into outer space again,
    > then (from symmetry) it would undergo a further deflection of 29 D1
    > minutes, so in all 58 D1 minutes caused by the Earth. This is the total
    > refraction the Earth would cause to that ray, as viewed by an observer far
    > from the Earth.
    > But Marcel's observer is not far from the Earth. He is at some altitude
    > within the atmosphere. So the refraction he observes will be 58 D1
    > minutes,
    > less the further refraction that would have occurred if that ray, instead
    > of meeting the his eye, had carried on into space.
    > If the observer is at a height where the density relative to sea level is
    > D2, and he was looking at a negative angle of A degrees, what is needed is
    > the refraction  he would see if looking the other way, at a star at a
    > POSITIVE apparent angle of 4 degrees, from a layer where the air density
    > (relative to that at sea level) is D2. This is a familiar problem, with a
    > well-known answer. Then that refraction needs to be subtracted from 58 D1
    > minutes.
    what already was mentioned?
    R(-4?)   approx.=  R(0?)  +  (  R(0?)  -  R(+4?)  )   ???
    > To get the answer to step 1, it may help if I quote from Young's "Sunset
    > Science" article ...
    Does someone of you have a pdf-file of this article? I would be very
    interested in it. (It's the only one I couldn't download.)
    >.... ray, the product nRsinz is invariant, where n is the local refractive
    > index, R is the radius from the centre of the Earth, and z is the local
    > zenith distance of the ray. I only quote, please don't ask me to justify
    > or
    > interpret that rule.
    You can find here a very good explanation for this
    This web page is a very special on refraction; here is the main entry to the
    So, good night for now

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