# [geocentrism] Re: moonlight

• From: Neville Jones <njones@xxxxxxxxx>
• To: geocentrism@xxxxxxxxxxxxx
• Date: Sun, 4 Jan 2009 11:55:56 -0800

-----Original Message-----
From: pma15027@xxxxxxxxxxxxxx
Sent: 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. No, it isn't. There are images of illuminated spheres to be found on the Internet and these show the Lambertian intensity profile that would be expected. Neville.

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.