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Frank,

Sometimes shortcuts become long splices ;-(  The over the horizon
distance by vertical angle problem remains begging for a calculator/
slide rule friendly formula.

Greg

On Jul 28, 10:36�pm,  wrote:
> 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
> 1000ft)."
>
> 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.
>
> -FER
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