[Frost]

Ok, I've decided to make a new thred on this since we've (I've) been going on about in Liquid's nice 'Vader' thread. (Sorry Liquid! )

Anyway, my oppinion about this has become a lot clearer since my last post (self doubt)...

I first asked a very smart colleague of mine about this and he confirmed my analysis. Reflections on plastic, glass, water, etc (that is not a mirror or metal based) only reflects the lighter portion of the scale. He told me that the eye captures photons reflected on the surface where white has the most and black has the least. So in reflective objects, you will be adding the photons from the source scene and adding this photon value to the diffuse of the object, this making it whiter, and never darker (as that would include removing photons, which would require a filter). Metals and mirrors are a totally different, but didn't go into as I couldn't understand much more of his technical mumbo jumbo. Basically, metals are found in most things which are perfectly reflective such mirrors, CDs, polished chrome, etc., and they are one of the few materials that actually seem to absorb photons or represent darker values in reflections. This only became clearer to me on my way home and seeing endless examples of this as it was dark and raining...

Here are some more drawings I made to try to clarify this point more clearly.



(Ever notice this when looking outside at night with lights in the house? )

[Isric]

Wow Frost, that makes total sense.
I had never even thought of any of this before, but dang, it works.
Thank you very much

[[Shizo]]

Imagine how realistic would look a white Lamborgini when drawn with that nice reflections technique? Or sweet sweet bottle of vodka .,.. barely touched by morning frost ...,,.,.,...,,.,.. ahhh

[Sergenth]

I will have to "reflect" on this information. Please don't "pun"ish me for my puns :P

[Snake Grunger]

That's what happens when a 2D painter has done some coding in the past

[-- Transcendent --]

Hmmm, you missed out one, a dark (black) reflective sphere. Would the reverse (only dark objects refected) happen ? Anyway, it's making lots more sense now. Thank you so much.

Dhabih should compile a list of threads with technical art tips like this one ... it'll make this forum a golconda of artistic knowledge.

[Frost]

Well, this was meant more as a discussion thread more than a lesson... I know I'm 'generally' right, but I'm sure there are some flaky things here...

A pitch-black reflective sphere would reflect close to the same as a chrome sphere, except the light would be less intense especially in the mid tones (but no, it would not reflect only the dark environment).

I just suggest to people to look around and see for themselves -- I wish I knew the rules to this myself. =]

[WitchLord]

Let's think of the physics of this. I'm no physics master but I do know some theory of light . (I have a Master of Science degree after all )

When the light or photons hit an object they are either absorbed by the material or reflected. Thus a red object absorbs all photons that aren't red. A black object absorbs all photons. A perfect mirror reflects all photons thus the mirror itself doesn't have any color. (To be exact, the photons don't have any color either, it's the energy (wavelength) of the photons that determine the color.)

Let's consider Frost's scene above with the red ball. Light shines on the checker board, in the black squares the light is completely absorbed, but in the white squares all light is reflected. This is why only the white squares is visible in the reflection on the ball, the black squares don't reflect any light onto the ball.

I'm anticipating a question here: Since the white squares reflect all light, why aren't they mirrors? Well, that's because the light is reflected in all different directions, thus all colors from the scene is mixed into white color.

I know I haven't answered all questions and more are bound to be raised. But as a 3D graphics programmer this is an area that interests me a lot, so I'll keep visiting this thread for a while to see if I can answer some questions that may be stated. (Of course I'm not saying that I'm the only one that can answer the questions )

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- WitchLord

[Frost]

Hey, thanks a lot Witchlord!

[Kaiju]

Frost: Great explanation. Much clearer than your previous one. I understand what you are talking about now.

WitchLord: That is a nice run down on color theory, but I think it will just confuse those who are not familiar with it. No offense, but it really doesn't pertain to the subject of Frost's thread. The question isn't "why is the ball red?" it is "what does the colored ball reflect?".

For those who want to learn more about color theory here is a good site that Fred Flick Stone mentioned: http://www.were-here.com/forum/color_theory/index.html

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Kaiju
My Home

[This message has been edited by Kaiju (edited November 15, 2000).]

[WitchLord]

I thought I answered both those questions. But I can see how it can be a little confusing

- The ball is red because the red light is reflected whereas all other is absorbed.

- The ball reflects only the light part of the scene because the shadows contain no light to reflect.

Although the article you gave us is interesting I don't see how it helps in the discussion about reflections.

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- WitchLord

[Liquid!]

FROST - I can see where you're going with it now. But there's something that I'm not quite sure about. Read the next question.

WITCH - I think its reasonable to assume that your analysis is correct. However, if the "black" squares on the checkerboard were blue, what would happen then? Considering that blue does not absorb all light, it will reflect the photons, and thus be mirrored in the red ball in a redish (technically purply) hue. I guess the only thing that won't be reflected is black because of its absorbing property.

Correct?

FROST- If the above is true, then you're right that the "black wall" in the vader thread would not influence the green sphere, however, that is only true if it were perfectly matte [thus perfectly absorbant] in the "black" areas. But since it is somewhat reflective it would show up on the green sphere, if just as a shadow.

Now if the wall were to be a particular color I would venture to guess that it would influence the color of the green sphere. Again, this is taking Witch's analysis further.

I could be wrong here.

-c

[Kaiju]

quote:

---

Originally posted by WitchLord:
When the light or photons hit an object they are either absorbed by the material or reflected. Thus a red object absorbs all photons that aren't red. A black object absorbs all photons. A perfect mirror reflects all photons thus the mirror itself doesn't have any color. (To be exact, the photons don't have any color either, it's the energy (wavelength) of the photons that determine the color.)

---

That is color threory. The ball being red has already been established. My comment meant that you were going to deep with your explanation. The link is for those who are not familiar with color theory. So, those who are not familiar with it can understand what we are talking about. And yes.. I don't think it has anything to do with this thread just like I said previously.

By the way, I also disgree that the red ball will not reflect the black from the checkerboard. If the surface is as shiny as the one in the demo then it should reflect. Granted it will not be an extreme black, but it should be a darker red.

Neat discussion by the way Frost.

[This message has been edited by Kaiju (edited November 15, 2000).]

[WitchLord]

First of all, I'll have to admit that I'm not 100% sure that what I say is correct. Which makes me even more eager to find the answer

Let's consider the life span of a single photon. It hits a surface with the speed of light, at first it has two choices: Either it bounces right back or it tries to go into the surface. The more reflective the surface is the more likely the first alternative is. But if the photon chooses the second alternative, it once more has two choices: Either it decides to stay inside or it leaves again. This choice is based on what color the surface is. If the photon has the same color (wavelength) as the surface it leaves again. But in another direction, thus giving diffuse reflection.

With this in mind, a blue square in above scene would reflect of the red ball (giving a purplish tint). But if the red ball wasn't reflective then the blue color wouldn't be seen on the ball, because the blue photons are absorbed by the ball.

Take the same scene with a red, reflective ball and a blue and white checker board. Now shine the light so that it doesn't directly light the ball. Only the light that is reflected of the checker board hits the ball. What would this look like? I say that the white squares would light the ball, almost as if a faint light source lit it directly from below (due to diffuse reflection), thus giving a red color to the bottom. In addition the white squares are also directly reflected on the sphere giving pink if not white squares on the ball. The blue squares doesn't add to the diffuse reflection of the ball since the color is absorbed, but they are reflected on the ball adding to the red giving a purple tint.

Tell me if I'm wrong or not

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- WitchLord

[Kaiju]

quote:

---

Originally posted by WitchLord:
Take the same scene with a red, reflective ball and a blue and white checker board. Now shine the light so that it doesn't directly light the ball. Only the light that is reflected of the checker board hits the ball. What would this look like? I say that the white squares would light the ball, almost as if a faint light source lit it directly from below (due to diffuse reflection), thus giving a red color to the bottom. In addition the white squares are also directly reflected on the sphere giving pink if not white squares on the ball. The blue squares doesn't add to the diffuse reflection of the ball since the color is absorbed, but they are reflected on the ball adding to the red giving a purple tint.

---

Correct. I understand why you are breaking it down to the photon level now. Instead of just explaining what is reflected you are trying to explain why it is reflected. So, sorry for stepping on your toes.

Here is a question for you if blue is subtracted from the light and absorbed by the blue squares.. then how does the lightsource that reflects on the ball contain blue? or does it not contain blue to give it a purple tint? Could it be that the light is reflected back through the light containing blue.. therefor a smaller amount of photons will hit the red ball.. leaving just purple?

[DarkBlade]

Can anyone put all that in english, for those of us who don't speak science?

(Please? I"m kinda lost....)

~DarkBlade

[Frost]

Liquid & Kaiju: I see your point, but yes, I guess it all depends on the material under the gloss layer. But for the examples above, the objects are not shaded as to not confuse the point of reflections (I should have mentioned that). But from all examples I have gathered by looking around, I can't say that any of them really reflect darkness, but more like that darkness is from the base shading of the object itself (which doesn't involve reflection).

Indeed, if a colored light is present, it affects how the base color looks -- a red light on a green surface for example I beleive gives an ugly tone of sickly-brown -- but that once again deals with the diffuse or base color of the object, and not the reflection. Reflections override the diffuse color, so going back to the example of a reflective green surface lit by red, the base color of the surface will be lit by red causing it to be tinted to brown, but the highlights will be as red as the light is, and not brown -- whatever is black in the environment would be a brown color (the base diffuse surface color).

You can think of reflections as being a coating of glass on a surface - it is sepperate from the color of diffuse.


[This message has been edited by Frost (edited November 15, 2000).]

[WitchLord]

Kaiju:

You're not stepping on any toes. In any case, I usually wear hard boots so I wouldn't feel it anyway

You make one error in your last reasoning, the blue isn't subtracted from the light when it hits the blue squares. No, it's all the other colors that are absorbed by the squares, thus leaving only blue light to be reflected onto the sphere.

DarkBlade:
I'll give it a try. Think of all surfaces as containing two parts, one reflective part and one color filter. The reflective part just reflects all incoming light as it is, where as the filter part filters away all light colors that doesn't match the color of the surface. The combined light that is reflected from the surface is the direct reflection + the filtered light. This is why shadows aren't reflected, there is no light to reflect.

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- WitchLord

[Kaiju]

Witchlord: I asked about the blue being subtracted, because I thought in order for the blue squares to be blue... they absorb blue light. Which I thought would subtract it out. Kinda the opposite of what you said. If the "color filter" filters away the light that doesn't match wouldn't that be everything except blue? So when you say "The combined light that is reflected from the surface is the direct reflection + the filtered light." I would think the filtered light would contain no blue, therefor just leaving a purplish tint on the red ball. Blue is not as strong in the combined light. See what I'm saying or am I just babbling for the sake of babbling now?

[This message has been edited by Kaiju (edited November 16, 2000).]

[-- Transcendent --]

Kaiju, I can't grasp what you are getting at, but I detected a flaw in your first statement. to make an object blue, blue light has to be reflected off it, while other colors are absorbed. Black absorbs all colors while white reflects all colors, am I right ?

[Kaiju]

I said in order for the blue squares to be blue they would have to absorb blue from the light. Not reflecting the color blue, but everything else.

"Black absorbs all colors while white reflects all colors, am I right ?" That statement is correct.

[WitchLord]

The light that hits the retina of the eye is the color that you see. For you to see a blue object then blue light has to eminate from the object.

Black objects absorb all light, so no light at all hits the eye.

White objects reflect all light, so all colors of the light hits the eye.

Blue objects reflect only blue light.

But I think I know what you are really confused by. The red sphere absorbs blue light, yet there will still be purplish tint that the eyes see. The purplish tint comes from the blue light that has been reflected on the surface of the ball before the light entered the material and could be absorbed.

I could go into even more details about the properties of photons and atoms to describe why this is so but if I do that I will really confuse people But if anyone wishes to know then it's just to ask and I'll be happy to explain it.

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- WitchLord

[Frost]

Actually Witchlord, if you have the time, I'd like to get an idea of it if you don't mind spewing a few lines on it (although I'm sure I won't understand all there is to it).

(sort of dissapointed no one seems to be reading or participating besides us 3 or 4...)

Oh, and Witchlord, how would you go about doing this in RT using D3D or OGL? Additive blending would be too bright and clamp too white too soon... I guess using the DX8 pixel shaders you could scale the reflective values based on the previous stages? Anyway, I've had that in the back of my mind for a few days...

[This message has been edited by Frost (edited November 16, 2000).]

[Kaiju]

WitchLord is right. I was wrong.. partially.

Ok WitchLord I'm with you now. I think somewhere I confused myself and wasn't paying attention. I was thinking the object only absorbed the color that we see and reflected the rest. But now I realize it is the other way around. So, it makes sense why blue is reflected on to the red ball. Hehe... pretty funny. I had myself convince and I've probably totally confused some of the folks reading this.

[WitchLord]

I'll try to do a thorough explanation of the physics behind all this tomorrow. Right now it's late, too late in fact as I have a meeting early in the morning. I want to make some simple drawings too to illustrate what I'm writing so you'll have to wait until I get home tomorrow, which should be in about 20 hours from now

But remember, my master degree is in computer science and not physics so I may not be entirely right. But then again, if it had been in physics you probably wouldn't understand anything anyway. Not to be demeaning, it's just that they tend to complicate things even more than I do

Good night, and see you all tomorrow. Maybe I'll even post my latest picture

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- WitchLord

[WitchLord]

Ok, let me see...

When photons hits a surface some of them bounces right off it again with no alteration of the color. The ratio of photons that are reflected as opposed to entering the material depends on what the material is but also on angle between the photons direction and the surface normal. The more perpendicular to the surface the photons are traveling the less photons are reflected. The incidence angle is always equal to the reflection angle. This is the reason why when lighting an object from slighty behind you draw this kind of glow on the fringe of the siluette of the object. (I hope you understood that last sentence because I couldn't think of another way to say it)



The photons that are not reflected on the surface, enters the material and eventually collides with the atoms (at least for opaque objects). When the photons collide with the atoms the electrons are given the energy of the photon. If it's enough energy the electron will jump to a higher level of energy, i.e. further away from the core of the atom. But for some reason, something to do with quantum physics I believe, the electrons can only exist at discreet distances from the core, so extra energy from the photon dissapates. When the electron finally desides to get rid of its extra energy and jump back closer to the core it throws out a new photon with energy equal to the level the electron was at. Because of the discreet levels this photon usually doesn't have the same color as the one that collided with the atom in the first place. This is why some colors are absorbed and others reflected in a material. Also, the photon that is thrown from the atom is thrown in a random direction which is why we can see objects even though no light is directly reflected into our eyes.



I'm sure that this explanation is very confusing, but I can't do a better job without feedback so ask questions and I'll try to explain them. Eventually you should all understand it (if you are interested that is).

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- WitchLord

[This message has been edited by WitchLord (edited November 17, 2000).]

[pierre]

quote:

---

Originally posted by WitchLord:
Ok, let me see...

When photons hits a surface some of them bounces right off it again with no alteration of the color. The ratio of photons that are reflected as opposed to entering the material depends on what the material is but also on angle between the photons direction and the surface normal. The more perpendicular to the surface the photons are traveling the less photons are reflected. The incidence angle is always equal to the reflection angle. This is the reason why when lighting an object from slighty behind you draw this kind of glow on the fringe of the siluette of the object. (I hope you understood that last sentence because I couldn't think of another way to say it)

The photons that are not reflected on the surface, enters the material and eventually collides with the atoms (at least for opaque objects). When the photons collide with the atoms the electrons are given the energy of the photon. If it's enough energy the electron will jump to a higher level of energy, i.e. further away from the core of the atom. But for some reason, something to do with quantum physics I believe, the electrons can only exist at discreet distances from the core, so extra energy from the photon dissapates. When the electron finally desides to get rid of its extra energy and jump back closer to the core it throws out a new photon with energy equal to the level the electron was at. Because of the discreet levels this photon usually doesn't have the same color as the one that collided with the atom in the first place. This is why some colors are absorbed and others reflected in a material. Also, the photon that is thrown from the atom is thrown in a random direction which is why we can see objects even though no light is directly reflected into our eyes.

I'm sure that this explanation is very confusing, but I can't do a better job without feedback so ask questions and I'll try to explain them. Eventually you should all understand it (if you are interested that is).

(I'll add the illustrations soon)

---

Hi thread, and hi witchlord,

I have been following the threads dealing with this subject for a while and feel that this has gotten some miles out of hand.

Witchlord (where in sweden by the way? just wondering),

First of all, let me say that, yes, your explanations has taken on a very scientific view of looking at things, which in many cases, and especially in your field of working, is good and essential, but in here, most of it is quite hard to understand if you are not familiar with the theory behind the physics of light (photons/electro magnetical wavelengths). On this forum, you will find ARTISTS foremost SCIENTIST, secondhand. (I think)

The first quoted passage is correct, but it is not all the truth though. Another reason why surfaces that are not perpendicular to our eye, seem to be brighter than if we would have looked straight at them, has to do with the structure of the surface, and actually more so by that particular fact.
I will explain that later, but for the moment, just think of velvet, how it looks, and why it looks that way. (think as you read, and i will explain it later if it is not clear). This brings me to a way of explaining this to an ARTIST. I know that if I was not familiar with photon mumble jumble, i would much more appreciate to have this being explained to me as if I was a 5 year old.

First of all,

There is no such thing as a:

100 % reflective surface

100 % absorbing surface

100 % transmitting (transparent) surface

which then brings us to the fact that ther is no such thing as a e.g.:

"red" (and only red) surface since that surface, even though it would absorb the bulk of the wavelengths of the spectra that does not cause our cones of our retina to stimulate our brain to "see" red, it does not absorb all of them. A red surface, will indeed, reflect wavelenghts other than red, even though, compared to the reflected red, would not be much.

One can go how deep you want in explaining the physics behind light, and if one wants to go all the way, it would require some deep understanding of some areas of very advanced math, which I think is of no, or very little interest, to most guys on this forum anyways.

I believe most artists only bother themselves with knowing that this light is "intense", "reddish", "volumetric" etc, etc, not the deep physics that caused it.

I don't bother my brain with trying to get the source code of photoshop to know how it is coded, I simply just use photoshop.

This is how I would explain it, knowing that no surface is 100 % reflective, absorbing, and transparent, only have percentage degrees of that. Also, stating that no surface is 100 % reflective is the same as saying that no surface is 100 % smooth, which is exactly be the prerequisite for a surface to be 100 % smooth:

This is what cause object to get a mirror like appearance:

The light that hits them, has the same leaving angle as the incoming angle, thus the light that reach our eyes is not so much distorted that it would distort an image of say my face in the mirror (nevertheless it is distorted to a certain degree, remember, we have no 100 %) Also, it does not reflect and absorb all the wavelenghts perfectly, so we do have a slight distortion on the color aspect to but not so much that we couldn't call it a "mirror" or "chrome".

This "chrome" surface has a certain degree of a diffuse aspect but very little. Have you noticed that many people, when want to paint chrome, the instinctly use greyish tones for that? well, it is not strange, just think of this: What do you get if you blend all color equally? probably a greyish brown, but idealistically we would get a grey color, and that is exactly what is happeing, a metal surface that is somewhat or very rough, takes on a grey color, because it reflects most colors equally and thus blend them equally because if its rough surface. But only if the environment that it reflects contain somewhat a good balance between all colors and if lit by a fairly white light.

On this rough metal surface you would inded see a shadow (on the object itself, caused by the form of the object itself). Why? becuause it is rough, and this rougness is microscopical, very often, (microscopical bumps, crests, dents, call it whatever you want, on bigger bumps, crests, dents etc etc. These microscopical bumps, we call them bumps, are mirrors in and of themselves, but the joint of them is what we see and it is the joint of their reflective characteristics that gives us the major impression.)

It is also this roughness that cause us to see shadows on the metal surface (shadow caused by the surronding light, but almost as often from the light of the shadow itself) If you look closele, you will see a shadow cast on a mirror too, both on the glass and the silver emulusion. Even though it would be very subtle compared to say a shadow on the ground, nevertheless, it is there, more or less.

Lets go back to the velvet, most artists have noticed that velvet material seem very dark when its surface reflect the the light somewhat perpendicular to our eyes. And if the surface of the velvet is turned away from our eyes, it seems brighter, right? why? Mainly because the velvet material is contanied oug of relatively long straws, and the brightest areas of these straws is the top of the straws. Becuase the length of these straws and ther grouping, most of the areas on the velvet will be hidden in shadows, thus some areas receiving very little light. Now when looking perpendicular to the velvet surface we will see the bright areas of the straws, BUT, we will also see alot of the areas hidden in shadow, and these areas are for more than the bright areas, thus it will appear to us as darker as opposed to the velvet seen from another angle that is not perpendicular to our eye. Why? consider this, if you turn away the velvet from our point of view, so that we do not have the normal of it hitting our eyes, what will happen? you will see more of the bright (the top of the straws) areas of the velvet, whereas the areas that are in shadow will decrease as opposed to when you looked at the velvet perpendicular to your eye.

It is the same principle with ALL objects. The way too look at everything is as 100 % smooth 100 % reflective surfaces (even though there is no such thing) and go from there, and how do you go from there? By considering some major aspects:

Amount of rougness

Type of roughness

The balance of the reflective wavelengths (i.e. COLOR)

Amount of Transparancy (which always contain the refractive index, but we are not computers and we cannot perfectly calculate how light will refract, just as little as we can calculate any other physical behaviour perfectly)

Balance of the Absorbing Wavelengths

and ofcourse the Light hitting everything


Simple is good and I can bet that an artist (in this sense, most artist in here and out there that paint/draw with brushes/pencils with paint, let it be traditional or digital, very often, depending of course on how he/she works, does not calculate photons and the energy of them when in interaction with other other atoms, photons, electrons etc etc. In many senses, most senses, even though you are an artist in the above sense, it is good, very good to have a knowledge of the physical behaviour of our reality, but it is all a matter of how deep? and how will it help your specific area of work or goal. What is the best way to approach it?

Many artist, I belive work as "shaders" not "raytracers" (if i can make that analogy). A raytracer would calculate every ray of light from the pixel to our eye. A shader would not care so much of the perfect jorney of the lightray, it would break the rule, as long as the final result would look convincing.

So the artist remember how a backlight sphere look like, maybe not so much of the math and physics behind it, but he knows the visual appearance of the final result, and he simplifies the explanation of it so that it can be applied to any other object and ligthing environment.


Just my opinion.


[This message has been edited by pierre (edited November 17, 2000).]

[WitchLord]

quote:

---

Originally posted by Frost:
Oh, and Witchlord, how would you go about doing this in RT using D3D or OGL? Additive blending would be too bright and clamp too white too soon... I guess using the DX8 pixel shaders you could scale the reflective values based on the previous stages? Anyway, I've had that in the back of my mind for a few days...
[/B]

---

Sorry, I almost forgot about your question.

It's quite simple actually. What you want to do is to have the final image contain alpha parts of the reflection and (1-alpha) parts of the wall texture with light. If you have the reflection in a cube texture it would be simple to do it in a single pass. But if you wish to render the reflection the back buffer you need to do it like this:

1. Render the reflection
2. Render a transparent black polygon over it. This will subtract (1-alpha) from the colors
3. Render the wall texture modulated with the light. Use additive blending with (1-alpha)

And yes, it would probably be more simple with the DX8 pixelshaders. But I don't have a card that supports that so I can't use those yet.


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- WitchLord

[WitchLord]

quote:

---

Originally posted by pierre:
Witchlord,

I saw your pic of your girlfriend, it is nice and enough to call yourself an artist.

I have also some background of 3d math, but i have noticed that it does not help me that much when I am about to paint an idea. Sure it helps me, but in a total different way.

Just what I think.

---

Thanks

I meant that I have no education in art. I have learnt how to draw on my own. I agree, my knowledge of 3D doesn't help me much either when I want to paint a picture.

I don't regret my education though. My main interest is programming and 3D graphics. Painting is a secondary hobby for me. Although that may change as I get better at painting.

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- WitchLord

[Kaiju]

Ok.. so what your saying is Frost is wrong.

(Wow... that is the longest damn post I have ever seen. hehe)