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Lu, you wrote:
"The surface of the earth,  especially seawater, is a conductive medium
and hence will bend  electromagnetic waves."

This is significant for long wavelength radio  waves, but totally irrelevant
to the behavior of visible light.

"This  effect is inversely proportional to the frequency of the
electromagnetic  wave.  Thus light bends very little and the visual
horizon is an almost  negligible amount beyond the geometric horizon.*

The ground wave effect  for visible light is a tad more than "almost
negligible". It's nil, zilch, nada,  certainly below one part in a trillion and I
wouldn't be surprised if it's below  one part in a trillion trillion (ok, even
that's not zilch, but it's certainly  irrelevant ).

The distance to the visible horizon is different  from the Euclidean
geometric distance because of the variation in the density of  the atmosphere. It gets
thinner as you go up, so rays of light necessarily bend  downwards allowing
us to see beyond the distance to the geometric horizon. This  variation in
density depends somewhat on the weather. If the temperature in the  air rises with
height above the water, and rises steeply enough, the bending due  to the
density gradient can easily be much greater than normal and allow us to  see many
miles beyond the normal horizon. It's worth nothing that this sort of
temperature gradient also promotes the formation of sea fog. So if you're in an
area noted for sea fog, you can also expect unusual refraction including
anomalous dip and the ability to see far beyond the normal horizon.

And  you wrote in a footnote:
" * Before one of the more erudite people on the  list jumps on me, I'm
fully aware that light IS refracted near the horizon  and we DO apply
refraction corrections when taking low-angle sights;  distance to the
horizon is a much cruder calculation.  In fact, I  believe the last place
in the 1.17 multiplier has jumped around as people  have argued about the
effect of atmospheric refraction."

Distance to  the horizon can be calculated in the same way as dip, range of
visibility,  distance by angle to the horizon, etc. The last place in the
multiplier does  indeed jump around in the literature because different sources
have made  different assumptions about the rate of change in atmospheric density
(which  itself depends primarily on the lapse rate in temperature). They're
trying to  give a value for a mean state of the weather.

Some mathy  details:
First: the curvature of the surface of the Earth is 1 minute of arc  per
nautical mile (this is a direct consequence of the definition of a nautical
mile). What this means is if I place a vector on the ground at the equator
perfectly horizontal at some point P pointing east and then travel 1 n.m. east  and
place a vector horizontally on the ground at some point Q, again pointing
east, then the angle between those two vectors is 1 minute of arc. The second
vector is tilted relative to the first vector by 1 minute of arc. Makes sense,
right?

Now suppose I shoot a light ray (aim a laser) horizontally at sea  level from
point P towards point Q. Under average conditions, the curvature of  this
light ray will be about 0.15 minutes of arc per nautical mile. Because it  is
proportional to horizontal distance traveled, this curvature can be accounted
for by pretending that the light ray does not bend but the Earth is slightly
less curved (bigger Earth radius). More generally, when you work out the
refraction, you'll find that the curvature of the trajectory of a light ray is
c = Q*(alpha0*Re/scaleHt)*(1-h/scaleHt)
where alpha0 is the  refraction constant for air, approximately 0.000281 at
standard Temp/Pressure,  Re is the actual radius of the Earth, h is height
above sea level, scaleHt is  the "scale height" of the atmosphere, and Q is the
usual temperature pressure  factor: Q=(P/P0)/(T/T0). The scale height is the
distance over which the  atmospheric density "e-folds" or decreases by a factor
of e (base of the natural  logs).

The scale height is usually about 11 kilometers. A very  reasonable range of
values to expect for the low level scale height is from 7 to  13 km
(corresponding to low-level lapse rates of +5 and -12 degrees C per km  respectively).
If the lapse rate should happen to be as low as -34.1 degrees C  per km, the
scale height is infinite and the curvature is zero. That's the case  I mentioned
in another message where the lowest layer of the atmosphere has  constant
density, not varying with height. In that case, the refraction is zero  and it is
just as if there is no air at all. If the lapse rate is positive and  rather
high, 129.6 degrees C per km, the scale height is about 1.7 km which  implies
a curvature of 1.0 minutes of arc per nautical mile --equal to the  curvature
of the Earth's surface. In that case, light rays remain parallel to  the
Earth's surface and it is "as if" the curvature of the Earth is zero.
Flat-earthers rejoice! The dip is zero at all heights (assuming that lapse rate  is
maintained over the range of heights under consideration).

Please  note: I'm working from my notes from last January and some of the
exact numbers  here may be based on slightly different assumptions. The general
conclusions are  correct though, I think.

-FER
42.0N 87.7W, or 41.4N  72.1W.
www.HistoricalAtlas.com/lunars


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