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    Re: Distance by Vertical Angle Adaptation
    From: Frank Reed
    Date: 2009 Jul 28, 22:36 -0700

    Hi Greg, you wrote:
    "Yes it does appear that small angles are not covered by this
    compromise formula (below 8' for 250ft , 10' for 500ft, 12' for
    And since it doesn't work well without iteration on the calculation, it's 
    probably better to go for the more complete solution rather than using this 
    correction for the obscured portion of the object *after* calculating a 
    preliminary distance. 
    Here's how we can get the "more complete solution":
    We have two formulas from your original post. First there's the simple 
    relationship that applies to any measured angle, 
      D = 0.566*H/a
    when H the height (or length along the arc perpendicular to the line of sight) 
    is in feet, a in minutes of arc, and the result D is in nautical miles. Next 
    there's the formula based on the maximum range of visibility for objects at 
    height h and H which tells us that the remaining height visible above the 
    horizon is 
      H = H0 - [D/1.17 - sqrt(h)]^2
    with D in n.m. as above and h, the height of the observer in feet. Now we take 
    that first formula, invert it to give H in terms of D and a as
      H = a*D/0.566
    and set that equal to the second:
      a*D/0.566 = H0 - [D/1.17 - sqrt(h)]^2.
    This can now be solved for D. But D is inside that squared term on the right 
    so we have to expand that and then solve for D. Skipping the algebra, the 
    result is
      D = 1.21*[sqrt[a1^2+0.94*(H0-h)] - a1]
    where a1=a-0.97*sqrt(h)) or a1=a-dip with the dip calculated for the observer 
    at height h (in feet). With only a very small difference, this is identical 
    to the equation used to generate Bowditch Table 15.
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