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    Re: Equation for dip?
    From: Frank Reed CT
    Date: 2006 Oct 2, 21:28 EDT

    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,
    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.
    42.0N 87.7W, or 41.4N  72.1W.
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