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From: Frank Reed (no email)
Date: Wed Jan 11 2006 - 00:29:15 EST
Bill, you asked:
"Rather than pay sleuth working backwards from your estimate of
horizon distance of 4-5 miles (statute for Lake Michigan charts, or nautical
miles?), do you recall an approximation of your height of eye for the beach
shots? "
About fifteen feet.
As for the skyscraper heights, I'm not satisfied with them yet so I'm gonna
do a few more trials in order to put some error bounds on the results.
In another message you wrote:
"It would appear Bowditch significantly underestimates lift due to refraction
by today's standards."
Well, I wouldn't say it that way. The "problem", such as it is, is that the
tables in Bowditch are calculated for a specific value of the terrestrial
refraction --around 0.16 or so. Now of course, that's better than nothing. But
the point I want to make is that we can use different values for the
refraction because we can calculate the actual value of the terrestrial refraction
that exists under verious atmospheric conditions. Specifically the refraction
rate is given by
Q*[(1.79km)/s]*(1-h/s)
where h is the mean height of the ray, s is the "scale height", Q is the
usual temperature and pressure factor given by Q=(P/1010mb)/(T/283deg celsius).
[Note that the factor of (1-h/s) can be dropped in most cases of interest for
navigation.] The "scale height" applies just to the lower part of the
atmosphere, say, the lowest kilometer, where the refraction is taking place and it
can be calculated based on the temperature gradient or "lapse rate". The
scale height is typically around 10km, but it can be infinite if the lapse rate
close to the ground is -34.1 degress per km (which is rare but possible). If
the scale height is infinite, it means that the atmosphere has constant
density as a function of height in the layer near the ground, and of course an
atmosphere of constant density does not refract light rays. The lapse rate can
also be positive and the scale height can be 5km or 2km or lower. Notice that
the scale height governs most of the possible variaton in the terrestrial
refraction, though it also depends on variations in temperature and pressure to a
lesser extent.
"Remember good old 4134 from the first term? 3438 / (1 - 0.1864).
Coincidence? (Radius = 3438 nm)"
Yep. Good sleuthing on those constants! Now it's time to derive the equation
from scratch. There are several approaches to this. Let's start with the
refraction-free version and then apply the substitution on the radius of the
Earth.
Draw yourself an arc of a circle representing the curved surface of the
Earth. Mark a point A at height h above the surface for the observer's position,
mark point B at height H for the object's position at some distance from A (H
and h should be roughly comparable heights and both should be much less than
the radius of the Earth, and mark C for the center of the Earth at the
center of the arc. Now draw a triangle connecting all three points. The lengths of
the two long sides are R+h and R+H. The length of the other side doesn't
matter. We want the length of the arc along the Earth's surface connecting the
base of h to the base of H. This length is proportional to the angle in the
triangle at the Earth's center, C. Let's call that angle phi. With a sextant we
are measuring the altitude of point B from point A so that means that we
know the angle in the triangle at point A, let's call that gamma. Finally the
angle at the third corner, B, is related to the other two --it's just 180
degrees-(gamma+phi). OK so far? Now apply the law of sines to the big triangle.
And expand using the rules for sums of angles. You should get this equation:
(R+h)/(R+H) = sin(phi)*cot(gamma)+cos(phi).
Using the relationship between the measured altitude and gamma, you can
replace cot(gamma) with -tan(alt). And really, we're done at that point. The
resulting equation lets us solve for phi for any given values of R,h,H and
measured alt. Of course since phi is buried inside two trig functions, that used to
be considered a bit of a problem from a calculational standpoint. We can
eliminate that problem by assuming that phi is a "small angle". We can then
replace sin(phi) by phi and cos(phi) by 1-(1/2)*phi^2. When you do that, you get
a quadratic equation in phi as follows:
(1/2)*phi^2 + tan(alt)*phi - (H-h)/R = 0.
We can solve for phi using the quadratic formula (you know the one...
x=(-b+/-sqrt(b^2-4ac))/2a...). And finally, the distance is R*phi. Throw in a
factor of 3438 to convert to minutes of arc. Note that none of this involved
refraction. Finally, finally, substitute for the radius of the Earth using
R=R/(1-beta) to account for refraction. Voila. Table XV.
-FER
42.0N 87.7W, or 41.4N 72.1W.
www.HistoricalAtlas.com/lunars
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