[geocentrism] moonlight
- From: "philip madsen" <pma15027@xxxxxxxxxxxxxx>
- To: "geocentrism list" <geocentrism@xxxxxxxxxxxxx>
- Date: Sun, 4 Jan 2009 20:19:09 +1000
John asked about the math of the light from the moon. I thought I'd drum up a
geometrical diagram to show what I was on about. Note illuminating an object is
not quite like reflection from a mirror. If the moon was a mirror, most of its
surface would appear black with a bright sun reflecting from a part of it. The
moon makes its own light, as peculiar to its own color by reason of the full
spectrum white light of the sun, and then radiates this color back out to us
the observer, absorbing the rest as heat.
In this diagram A B is the same length as CD The light falling on AB is equal
to that falling on CD. But in the case of AB it falls on the curved hypotenuse
of the triangle, which is a much larger area than AB, and so this part of the
surface will be much darker and less illuminated than that curved surface
between CD which is almost equal to CD. I say it would be much darker to an
observer over to the left of this picture. But the total number of lumens
radiating back between AB and CD are equal in quantity. Thus the disc will
shine like a plate as represented by the line ABCD.
The incoming radiation is spread out as it hits the larger curved surface,
providing less watts per sq meter than that what is provided on the surface
between C and D. But all the outgoing radiation from the surface curved between
A and B is concentraded back between A and B , as observed, and is equal to all
the radiation radiated back from the surface between Cand D.
I hope that explains what I mean by the lumens radiated are equal over the full
face of the moon. Just look at any ball in the sunlight.
Philip.
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