NavList:

   

Reply

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

   

Reply