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Anacapa Light is actually 277ft but I find it a better visual to use
the peak of land under the light(250ft) for the vertical angle from
long distance. Short of the horizon the top of the light house is a
good visual so that .567(277ft)/MOA or 157/MOA can be used to get
distance.

On Jul 31, 12:12�pm, Greg Rudzinski  wrote:
> 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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