Riddle Me This...

[Mario Marengo]

I just had one of those "wait a second!" moments...

You're standing in a pitch-black world and there's a single incandescent light bulb lit where you're standing -- very low power: its intensity dies off to nothing at 100 meters.

There's a gigantic mirror 200 meters in front of you, facing in your direction.

Do you see the light bulb reflected in the mirror?

(all our shaders say "yes", but I'm thinking "no")

[michael]

yeah...you'll see nothing.

even if the mirror was only 100 meters away -> 0 -> reflected 100% still = 0

what do your shaders say?

are they not takng into account the falloff?

[Mario Marengo]

yeah...you'll see nothing.

even if the mirror was only 100 meters away -> 0 -> reflected 100% still = 0

Good. Maybe I'm not going crazy yet.

Anyone have a good argument supporting the opposite?

(BTW, if we agree that falloff plays a role, then 50 meters would be the cutoff distance since it's a return trip there and back, but that's just nitpicking).

what do your shaders say?

are they not takng into account the falloff?

Yes, light shaders have falloff, but traced reflections (think of the light bulb as a constant-shaded object being reflected) typically don't (and by "typically" I mean all shaders I've ever looked at, including, sadly, my own).

OK. Now let's make the thought experiment a little tougher:

Now we'll assume the light is strong enough to get to the mirror and back, but we're going to put the light bulb in two different positions on a plane parallell to the mirror:

A: Place the light bulb directly on top of your head. (close to zero degrees between viewing and incident directions)

B: Now move the light bulb directly up by a few hundred meters -- imagine it's at the top of a very tall building and you're standing on the street. (say 60 degrees between viewing and incident directions).

Assuming the intensity of the light reaching the mirror is identical in both cases, will the reflected intensity also be the same for both?

(again, most shaders say "yes", but I'm now wondering whether irradiance should be taken into account, making the answer "no")

[michael]

yeah something is wrong with all this....

in the real world it's all about light energy....no matter what

the same thought experiment could be done with the /heat/ of the bulb...it should decay and be reflected accordingly

[crunch]

I just had one of those "wait a second!" moments...

You're standing in a pitch-black world and there's a single incandescent light bulb lit where you're standing -- very low power: its intensity dies off to nothing at 100 meters.

There's a gigantic mirror 200 meters in front of you, facing in your direction.

Do you see the light bulb reflected in the mirror?

(all our shaders say "yes", but I'm thinking "no")

Sorry Mario, I say "yes" too (just like the shaders). The reason that the light fades away after 100m is that the light is uniformly scattered on a spherical distribution. Which means that only a portion of the light hits the floor (for example).

However, with a mirror, the reflectance distribution is a much narrower beam, so it picks up more of the illumination from the light source.

Now, the one that gets me, is the mirror that's in the mountains reflecting the sun, over 20 miles away, you can still see a hand mirror. But, I think it's the same thing (i.e. that the mirror is reflecting the sun into your eyes pretty well).

Edit: Sorry, are you saying that if you're 100m away from the light, and you look at the light, you don't see the bulb shining there? If that's the case, then you have a participating media blocking the light. Otherwise, looking in a mirror gives you the same amount of illumination as looking directly at the bulb.

[Mario Marengo]

Sorry Mario, I say "yes" too (just like the shaders). The reason that the light fades away after 100m is that the light is uniformly scattered on a spherical distribution. Which means that only a portion of the light hits the floor (for example).

Right. I agree with that too (heck, I've been going "yes, no, yes, no..." for the past hour ). And "yes" was my answer until something I'm doing right now made me doubt it -- I'll get to that in a sec.

However, in your example, what's true for the floor should also be true for your eyes, shouldn't it? -- energy expands radially and so only a portion reaches your eye, just as it did the floor, and that portion diminishes in some proportion to the distance (inverse square or otherwise). And yet... I also know that I can still see the light source even when all surfaces around me have stopped reflecting any illumination from it (forgetting the mirror for a second).

There's also the issue of the source's projected area and how it increases in inverse proportion to the cosine of its angle to N, thereby decreasing the overall energy arriving at a given point... the second riddle.

OK. I'm in the middle of finishing something now, but I'll explain the actual problem that started me thinking about all this as soon as I'm finished. As you pointed out, it has to do with reflectance distributions -- ones that are very close to, but not exactly, mirror-like -- a narrow spike (but not a delta function) in some direction -- and how to weigh samples based on it.

[Allegro]

I'm not a mathematician... but I want to say yes to this as well...

At night, you can see a campfire from miles away... but would you still be able to notice that illumination from the fire on your face from said distance? I think no.

I think falloff is partially to do with the reflective properties of the objects... most objects absorb light, no? So in a real world situation, you don't normally get a return on the light that's as intense as what was originally sent out... highly reflective objects however wouldn't behave the same way though... I think a lot of it is also contrast. The light falls off to a point where we don't perceive it, but I don't think it entirely goes away... seeing a reflection though would be very focused, and thus high contrast.

Edited July 11, 2008 by Allegro

[Jason]

I say No - unless your "dying off" is due to atmospheric scatter as crunch says. If you're not modeling attenuation by atmosphere and instead are doing the "physically correct" 1/r^2 falloff - (which is due to the visible exposure area getting smaller) then no - it'll just be very faint and not extinct. In fact, if you're doing scatter too, it'll still be visible, just fainter.

The test, to me, is: imagine that light source is as bright as the sun (or brighter!). You'd probably see it unless a completely opaque object blocks all the light. If it's not 100% opaque, then *some* light will be visible.

[anamous]

Wow, I was just wondering about the exact same thing, which means that the cross-pollination of Mario's brain with the forum's hive-mind is starting to show some positive side effects And I came to the conclusion that the answer must be "yesno"... wait, what was the question again?

The idea is that what our eyes and virtual cameras see is only the portion of light reflected back into them. This amount increases as we go from diffuse reflection lobes over glossy ones and finally to specular (or retrospecular) ones - this means that the amount of light that we get from a specular reflection of a light source is higher (in CG, often 100%) than the amount a diffuse surface would reflect to us from the same light source, simply due to energy conservation laws (a diffuse surface scatters the light in a lot of directions, whereas the perfectly specular surface scatters all the light in one direction). So, yes, there definitely should be a visible reflection, but also no, since when dealing with the distances you describe, the reflection will be a very small dot, and it's energy will still fade on its way to our eyes. And it's size would be below the sampling threshold

At least that's how I validate my shaders... but I'm not too sure.

cheers,

Abdelkareem

[sibarrick]

Maybe think of it like this

Look up at the night sky and you see stars, now imagine there is only one, you'd still see it right. But hold up a piece of paper and it won't look like its illuminated by the star. However every so often a photon from the star will surely hit it. But look directly at it and you will see it twinkling away. So surely this must be some kind of perception trick that our photon detectors (or eyes) and our brain play on us, related perhaps to persistence of vision. Maybe that's why they twinkle, who says science takes the beauty out of nature.

[kensonuken]

I would say that the dynamic range in the perception of human eye is arround10,00,000: 1 so when we are able to distinguish the brighter and darker star then we can say that yes the photon still reaches the surface but for the medium that we use either film or video it has a darker level which is not zero and probably there is no use in looking for the value where the threshold is reached...