[geocentrism] Re: moonlight

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 did 
find one animation, but it seemed to be artificially lighted, shades of faked 
moon shots with divergent shadows, and or simulated, not like what a sphere 
lighted in sunlight would present in a darkened black room)

 
Thanks Neville, I tried to follow that up, the angles of incidence to the 
reflection did worry me when I first put up my attempt. I will need to have 
someone explain Lambertian reflection as applied to spheres in some simple form 
to understand what its about. I paste in a couple of confusing inputs below 
which seems to support my offering..  Wouldn't the surface profile have a lot 
to do with the re-radiation ? A smooth sphere would have a different effect to 
a cratered and pottered mountainous surface. I try to modify my diagram to 
explain this below . 

I found this, 
If a surface exhibits Lambertian reflectance, light falling on it is scattered 
such that the apparent brightness of the surface to an observer is the same 
regardless of the observer's angle of view. More technically, the surface 
luminance is isotropic. For example, unfinished wood exhibits roughly 
Lambertian reflectance, but wood finished with a glossy coat of polyurethane 
does not .....Isotropic radiation has the same intensity regardless of the 
direction of measurement,....... Of course a sphere complicates this.
Lambertian reflection is typically accompanied by specular reflection, where 
the surface luminance is highest when the observer's angle is the same as the 
angle of the light 
source....http://en.wikipedia.org/wiki/Lambertian_reflectance   
May we assume that with the full moon we as observers are close enough to being 
at the same angle as the light source, namely the sun. The problem of 
explaining that the moons sphere does appear evenly lighted , seems to resolve 
around these factors , where perhaps the "specular reflection" to lambertian 
reflection ratio is higher. This following work referred to spheres. 

"For rough and smooth surfaces, there are departures from Lambertian 
reflectance at the occluding limb. This is due to backscattering from 
microfacets which protrude above the surface. These departures from Lambertian 
reflectance limit the applicability of shape-from-shading to the surface.    
Ragheb, H.; Hancock, E.R.
Image Processing. 2002. Proceedings. 2002 International Conference on
Volume 2, Issue , 2002 Page(s): II-553 - II-556 vol.2 
http://ieeexplore.ieee.org/Xplore/login.jsp?url=/iel5/8052/22295/01040010.pdf?temp=x

My rough drawing:

From this exagerated image can it be seen that the reflection from the edge of 
the moon off the walls of the orange craters could be roughly the same as from 
the middle, directly back to the source of the light, and we the observer. 
Actually the light from the floors of the craters seen in the centre of the 
luna surface is slightly darker, due to different composition perhaps. 


I am sort of at a loss as to what this has to do with accusing the moon of 
having its own internal lumenosity. If it does , it is still under the 
surface..  I prefer to see this as an exercise in learning about Lambertian 
reflection. Having lots of knowledge without comprehension is of no value to me 
or anyone else really . 


Philip. 
  ----- Original Message ----- 
  From: Neville Jones 
  To: geocentrism@xxxxxxxxxxxxx 
  Sent: Monday, January 05, 2009 5:55 AM
  Subject: [geocentrism] Re: moonlight



   
    -----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.

JPEG image

JPEG image

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