[geocentrism] Re: moonlight
- From: "philip madsen" <pma15027@xxxxxxxxxxxxxx>
- To: "geocentrism list" <geocentrism@xxxxxxxxxxxxx>
- Date: Mon, 5 Jan 2009 08:49:49 +1000
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.
- [geocentrism] Re: moonlight
- From: Neville Jones
- [geocentrism] Re: moonlight