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Frank:

You are, as always, perfectly correct.  I was more trying to explain why
the radar horizon is beyond the visible horizon than to explain
refraction due to atmospheric density.  I fretted about simplifying
things too much.  Thanks for filling in the details way better than I
ever could have.

Lu

FrankReedCT@aol.com wrote:
> 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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