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Back from a few days afloat, to find Marcel Tshudin's message of 8th August
awaiting a reply. We have been discussing how to tackle the calcularion of
refraction as observed from high altitudes, such as a mountain or an
aircraft, when the apparent altitude is negative (below the true horizontal.

To many listmembers, this may be an arcane and irrelevant matter, but
Marcel and I both seem to be getting a bit of fun out of it, and I ask them
to bear with us while we thrash the matter out between us, if we can. If
nothing else, it's providing us both with a lesson in clear thinking and
clear expression.

Marcel starts by wishing-

How would it be helpful now to discuss
>this together in front of a blackboard!

I expressed that same sentiment recently in a nav-l message, shortly before
Marcel joined the group. In the government lab where I used to work, every
office and every lab was provided with a blackboard, even the tea-room.
Especially the tea-room, I should say. One day, a new and enthusiastic
cleaning-lady started work, and before anyone arrived, she had carefully
erased every trace of what had been recorded on each blackboard along a row
of offices. What howls of anguish arose! Not only had the previous day's
arguments and discussions been deleted from the central area of every board
(akin to the RAM) but also the permanent stuff in each corner (effectively,
the hard-disc), phone numbers and the like, had been wiped clean as well.
It never happened again!
=============================

>The only difference between our
>views is to my oppinion how one would have to adjust for air pressure and
>temperature, but let us forget this detail for the moment.
>
>What George indicates with ...
>
>>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.
>
>... corresponds to twice the refraction for 0? altitude, this regardless
>where the tangent point is above the earth's surface (air pressure and
>temperature neglected), therefore:
>
>R(0?)  +  R(0?)

We don't really need to consider air temperature and pressure separately,
but just their combined effect on air density, which is what affects the
refractive index of air. This density-difference at different tangent-point
heights is what Marcel decides to neglect. In doing so, he is discarding
the baby along with the bathwater.

Observed from a fixed height (the top of Everest, say), the light from a
setting Sun, when its light arrives truly horizontally (at 90 degrees to
the zenith), has travelled no nearer the earth's surface than the height of
Everest itself. At the other extreme, a few minutes later, when the light
arrives at its most negative possible altitude (perhaps that's near -4
degrees to the observer, I'm not certain of the exact value) then it has
just skimmed the Earth's surface at low height in the surrounding plain, so
we know that the refraction between outer space and that tangent point will
be about 29 arc-minutes (or it would be if that plain was at sea-level).
Clearly, the refraction from outer space to the tangent-point, in that
first instance, for horizontal arrival, must be much less than 29 minutes.
because of the lower air density along the path. My assumption, without
thinking too seriously about the matter, is that if the air density varies
exponentially with height, the total refraction, seen by an observer
looking horizontally from the tangent point, along the path all the way
down to from space to the tangent point, is proportional to the air density
at that tangent point. Whether that is a good assumption or not (and the
details may bear a bit of arguing about) there's no doubt, surely, that
there's a large effect of density at that tangent point in reducing that
horizontal-refraction with increasing height of the tangent point.

So when our observer at the top of Everest observes the Sunlight arriving
from different negative directions, these rays have passed through very
different amounts of air. The quantity he refers to as R(0 degrees) , for
truly horizontal light, is not a constant but varies greatly with
tangent-point height and so with negative angle of altitude as seen by the
observer.


>>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.
>
>If the correction for air pressure is neglected, this corresponds exactly to
>
>-  R(+4?)
>
>And summarised now the refraction for an observer looking at an astronomical
>object at -4?:
>
>R(-4?)   approx. ? =  R(0?)  +  (  R(0?)  -  R(+4?)  )

But that will NOT be the case when the major effect of density changes is
taken into account.

>To my oppinion this resulting refraction can in this case be corrected for
>air pressure and temperature as indicated in the textbooks. To my
>understanding it is the temperature and air pressure at the eye of the
>observer which corrects/adjusts the complete line of sight.

No, you have to consider BOTH the density at the observer (which I labelled
D2) AND ALSO the density at the tangent point (which I labelled D1). And D1
is different for each different value of negative altitude seen by the
observer's eye.

>Sorry, George, if I still should have not yet understood what you ment. In
>this case I propose that the conference call with blackborad is soon
>invented for the internet.

I think that with that internet blackboard we would have thrashed the
matter out between us ages ago.

It may be worth commenting on another matter. For large angles of observed
altitude of an astronomical body, the path through the atmosphere is
relatively short, compared to the radius of the Earth. In which case the
geometry is very nearly plane-parallel. In which case, Snell's law applies,
and the sine of the angle of incidence to the vertical, above the top of
the atmosphere multiplied by the refractive index above the top of the
atmosphere (which is vacuum, so equals exactly 1) is exactly the same as
the sine of the corresponding angle at a lower point in the atmosphere,
multiplied by the refractive index at that point. So for such angles of
incidence (greater than +10degrees, say) the refraction depends only on the
incident direction and the air-pressure at the observer, and not at all on
any intervening layers along the way. No integration along the path is
called-for.  For shallower angles of incidence, as we are consiudering in
the discussion above, such simple approximations are no longer valid, the
curvature of the density contours has to be taken into account, and the
matter becomes much more complex.

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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