I think I understand what you are saying here and can see how it applies to
something that covers the entire field of vision such as a wall, but can you
explain how it would apply to a small object against a dark background?  To
me the area that the pixel is gathering light from is increasing by r^2, but
since the object is fully in the field of vision and the background is dark,
there is no additional light to be gathered than what is reflected by the
object.  If you look at small object, say a red sphere (I just love those
things, gotta get me some real ones) under a dim light in a dark room, woul
you expect the apparent intensity of the ball's colour to be the same
whether you are 1m away or 10m away?

Please dispell any myths I have here as I need to know now.

-tgq


"Kari Kivisalo" <ray### [at] engineercom> wrote in message
news:3BD58A14.D268FC06@engineer.com...
> Trevor Quayle wrote:
> >
> > over distance (1/r^2), the energy stays the same but it is spread over a
> > larger area:
>
> Yes, but the observed area increases by r^2 so they cancel out.
> How the energy got on or leaves the surface (direct, radiosity,
transmission,
> reflection) doesn't matter, the emitted energy to observer's fixed
> space angle is not a function of distance.
>
> Do your own experiment. Look at a wall from 1 m distance. Note the
> brightness. Go to 3 m distance and compare the brightness. It's
> the same!
>
> A pixel gathers light from fixed space angle so the area it gathers
> the energy from increases by r^2. The energy arriving at the pixel
> from a fixed area on the surface decreases by 1/r^2. You don't have
> to be a rocket scientist to figure this one out.
>
> _____________
> Kari Kivisalo

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