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From: Marcel Tschudin (no email)
Date: Fri Jun 16 2006 - 19:20:42 EDT
On 6/17/06, Bill <> wrote:
>
> Now that I have the basic concept, what happens on a larger scale? We
> could
> use Frank's beach shots as an example (Indiana to Chicago, approx. 22 nm).
> Assume a thermal inversion as Frank stated. What will that do to the
> horizon relative to a horizon with well-mixed air (shift it up or down)
> and
> what will it do to building tops around 1100-1500 feet above water level
> (raise or lower them)?
>
> The problem I was having using the current Bowditch Table 15 formula was
> that the distance kept falling short, meaning if the T15 constants are
> correct, the angle measured was too large. Therefore if the horizon was
> shifted up, the building tops must have been shifted up to a greater
> extent.
>
> I understand the problem as stated may be too complex for a simple answer,
> but try a if-then.
>
> I also strongly suspect another problem area is the Bowditch constants.
Just an example of what a strong inversion (sea is cold air is increasing
warmer) may cause.
Lets assume an observer
at 10m above sea, at his place the temperature is 10°C and the pressure 1010
mb.
Lets assume a simplified standard atmosphere
where the temperature decreases by 6.5°C per 1 km height, this up to the top
of the troposphere at about 11 km height and having constant temperature
further up.
For this observer the dip of the horizon is as calculated with the Bowditch
formula, i.e. 5.6 arc min.
Lets assume now an inversion layer
in the lowest 50m of the standard atmosphere (as described above) with a
temperature increase of 6°C over 50m height (+0.12°C/m).
For the above observer the dip would now be reduced, it would be now only
0.92 arc min.
May be someone else can confirm this. How about you, Frank?
Marcel
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