# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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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
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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