Quote:
Originally Posted by iceman
But what's the answer, Dennis!?  |
Okay Mike, here’s how I did it, although I am open to review and any corrections for the procedure and maths, as it has been such a loooong time since I did this stuff at school!
Using Photoshop CS5 I measured the diameter of the Moon and the length of the airplane fuselage and came up with the following:
PART A: Relative angular size of Moon/Airplane.
Moon Diameter = 1370 pixels
Body Length = 760 pixels
Therefore Body Length/Moon Dia ratio = 0.5547. If the Moon subtends an angle of 30.3 arcmins, then the Airplane Body subtends an angle of 0.5547 x 30.3 arcmin = 16.8 arcmin.
16.8 arcmin / 60 = 0.28 degrees.
PART B: Line of sight distance from Chris.
As a good approximation, 107.75 feet / Tan 0.28 degrees should give the distance of the airplane from Chris; that is, some 21,500 feet which is approx 6568 metres.
As we know the altitude of the Moon above the horizon (ignoring Chris’ height above sea level) and have just calculated the line-of-sight distance from Chris, we can calculate the ground distance using Cosines.
PART C: Ground distance from Chris.
That is, Cosine 73°46’ x 6568m = 1.840 kms approx, giving the ground distance some 1.84kms from Chris’ observatory.
PART D: Height above ground of aircraft.
The height of the aircraft above the ground can be obtained using Sines as follows:
Sin 73° 46’ x 6568 m = 6306m approx. (ignoring Chris' elevation)
LUNAR LIBRATION CHECK:
As amateur astronomers, we are aware of the effects of Lunar Libration; attached is a Lunar Map Pro screen capture of the Moon as simulated for Blackbutt at 5:24pm on 16th Sept 2010.
Well, that was a pleasant stroll down memory lane - please feel to correct any bad assumptions and poor memory recall!
Cheers
Dennis