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    Re: Distance by Vertical Angle Adaptation
    From: Greg Rudzinski
    Date: 2009 Jul 31, 12:12 -0700

    Frank's modified Bowditch formula for distance by vertical angle is
    now inked into my piloting note book and useful for generating a table
    of distances for frequently used object heights.
    Table for Anacapa Light at 250 ft. from a height of eye of 6 ft.
    MOA      Distance (NM)
    1'       19.8
    2'       18.6
    3'       17.4
    4'       16.3
    5'       15.3
    6'       14.3
    7'       13.4
    8'       12.6
    9'       11.9
    10'      11.2
    15'       8.5
    20'       6.7
    25'       5.5
    30'       4.7
    40'       3.5
    142/MOA (NM for MOA above 40')
    On Jul 29, 6:51�pm, Greg Rudzinski  wrote:
    > Frank,
    > I've practiced a bit with your formula and find it satisfactory if I
    > use an index card to keep track of things. The head calculation is
    > quite a test and I have to admit that I couldn't get the same result
    > two times in a row ;-)
    > Greg
    > On Jul 29, 6:19�pm,  wrote:
    > > I wrote:
    > > "D = 1.21*[sqrt[a1^2+0.94*(H0-h)] - a1]
    > > � where a1=a-0.97*sqrt(h)) "
    > > And Greg, you wrote:
    > > "The over the horizon distance by vertical angle problem remains begging 
    for a calculator/slide rule friendly formula."
    > > Well, it is what it is, right? This certainly isn't hard to work on a 
    calculator, and in fact it's not that hard to do in your head. Compare it 
    against the equation in Bowditch which supposedly "explains" the origin of 
    Table 15. In what school of obfuscation do they teach people to put 0.000246 
    in the denominator of an equation?? And the difference between tan(x) and 
    x/3438 is much less than 1% for any angles that might ever be used for this 
    calculation, so why do they bother with explicit use of the tangent in the 
    equation? To make it look more "serious"?
    > > What should be said, and isn't said enough, is that these equations (and 
    tables) only provide an estimate of the distance since they depend on a mean 
    state of the terrestrial refraction. Terrestrial refraction (meaning 
    refraction of light traveling nearly horizontally close to the Earth's 
    surface) on average bends light rays downward by about one minute of arc for 
    every 6 nautical miles, but a range of 0.75 to 1.25 minutes in the same 
    distance should be expected as normal variation (on the order of one s.d.). 
    This matters for angles over the horizon because the we're measuring an angle 
    between things which may be at very differnt distances. If the horizon is 
    five miles away, and the vessel beyond the horizon is twenty miles away, then 
    a small increase in refraction lifts the image of the vessel much more than 
    the horizon (the distance to the horizon itself changes so the comparison is 
    not one-to-one).
    > > And it's possible to have even larger variations in terrestrial variation 
    than this. If the air gets cooler with altitude more rapidly than normal, the 
    terrestrial refraction can be zero (never less than zero for more than a few 
    seconds so that really is the lower limit). If, rather than cooling with 
    altitude, the atmosphere gets warmer with altitude at just the right rate, 
    the terrestrial refraction can be 6 minutes of arc in 6 nautical miles which 
    means that horizontal light rays remain parallel to the ground from all 
    distances --and that means that the Earth looks flat and objects at all 
    distances are in front of the sea horizon (the sea horizon would not actually 
    be visible in this case, and note that this is not an upper limit).
    > > -FER
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