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    Re: Equation for dip?
    From: Lu Abel
    Date: 2006 Oct 02, 19:50 -0700

    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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