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From: george huxtable (no email)
Date: Mon Aug 08 2005 - 03:52:52 EDT
Marcel wrote-
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
>refraction.
>
>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°) ) ???
=================================
Response from George.
No, not the same at all, unless we are talking completely at cross-purposes.
In my argument above I have taken a ray of light that has a particular
height at the tangent-point, and then attempted to show how to calculate
refraction for observers at different heights (each such height
corresponding to a different negative angle) along that particular ray.
On the other hand, Marcel's expression-
>R(-4°) approx.= R(0°) + ( R(0°) - R(+4°) )
was, I took it, intended to apply to an observer at a particular height,
looking at different negative angles. In which case, each of those
different rays would correspond to a different height at its tangent-point.
In which case, the bending of each such ray would be different. In that
situation, Marcel's expression would be incorrect.
If I have misunderstood, I hope Marcel will put me right.
======================
>>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.)
I don't know if a digital version is available. The journal, Applied
Optics, comes from the Optical Society of America.
The article was No. 2 of a series on "Sunset Science", and It's likely they
would all be of interest to someone like Marcel (or me). Part 1 "The mock
mirage" was in Appl. Opt. 36, 2689-2700, (1977), and presumably further
parts have appeared since part 2.
I have an email address for Andy Young, who I have always found to be a
helpful correspondent, at-
In an earlier posting, Marcel wrote-
>The web page here http://www.iol.ie/~geniet/eng/refract.htm#Terrestrial
>(this time in English!) mentions the following values for terrestrial
>refraction, this time as K=1/k
>
>K=4.91 at noon,
>K=10.64 during sun set/rise and night (equinox, wind speed [4 m/sec] (at 10
>m height) and latitude 53°)
I've had a look at that website, which comes from Victor Reijs, a name
well-known to many nav-l listmembers. There, he gives an edited version of
Alexander Thom's refraction corretion methods.
Thom's name may ring a bell. He became well-known (notorious might be a
better word) for constructing extravagant theories about the technological
and astronomical knowledge of the builders of megalithic monuments, based
on flimsy statistics.
He made many surveys of those megalithic monuments, which were highly
competent and have not been questioned. His surveying techniques should be
treated with respect, even if his archaeoligical conclusions have been derided.
I have many papers of Thom's, but not that of 1973 which Reijs' website
draws from. Working backwards from the expressions for dip in that website,
it appears to me that the values of K quoted by Marcel correspond closely
to the radius of curvature (in units of Earth radii) of light near the
surface of the Earth, influenced as it is by dendity and thermal gradients.
So tne lower the K value, the more the light is curved.
The usual value adopted for the radius of curvature of light near the
Earth's surface is abour 6 or 7 earth radii, somewhere between Thom's
values, at different times of day, of 4.91 and 10.64. Giving those
quantities to two decimal places provides a spurious illusion of precision.
Those values would alter the predicted dip by about 4%, either way, from
its normally adopted value.
Reijs quotes a formula given for "terrestrial refraction", used to
calculate "the apparent altitude, seen from local point, of a distant
object, such as the top of a mountain". It seems to me extremely unliky
that the thermal gradients near the sea surface, which would give rise to
these local changes to K and thus to dip, would be maintained at the same
value right up to altitudes such as the top of a distant mountain. But
clearly it's not possible to assess the worth of Thom's work without
reading his original book. All I can suggest is that his suggested values
for K , and thus for curvature due to refraction, be taken with a grain or
two oif salt, and not accepted as god-given.
George.
===============================================================
Contact George at ,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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