[Vo]:Personal:Little help with UFO sighting?

[Horace Heffner]

For background information, the attached is a graphic showing my  
derivation of the formula used below.  The umbra line, shown as  
horizontal, and the line segment x directly above the viewer is  
roughly the same as the altitude to an object on the umbra line,  
depending on the latitudinal distance from the viewer.

The altitude h to the directly overhead sun midline, with r factored  
out, is given by:

   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

Given time after sunset t we have:

   theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi)  
radians

Earth radius, r_earth, at Hawaii is about 3951 mi.  Here are some  
numbers:

t (min) theta (radians) h (miles)
                
1       0.00436331944   0.03760073165
5       0.02181659722   0.93976780755
10      0.04363319444   3.75594358
20      0.08726638889   14.973936498
30      0.13089958333   33.506081478
60      0.26179916667   130.1553394
90      0.39269875      279.3533269


Since the above is time after total sunset, you don't have to correct  
for the angular width of the sun.  However, even total sunset is not  
good enough to black out an object though, due to light  
diffraction.   Clearly not enough time, i.e. "shortly after sunset",  
passed to rule out an airplane.

<<inline: Umbra.jpg>>

Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/