The Rendering Equation

Game Mechanics
#1

Here’s the rendering equation:

And here’s me trying to understand it:

It’s the integration part I don’t get. I know what integration is, and I know the basics of using it, but what does it have to do with anything? I know the omega subscript indicates that the integration has to be done over a hemisphere, but what does it mean for integration to be done over a hemisphere? Where am I? What is life? I don’t understand what’s going on :frowning:

I get the rest of it. But why is it an integral equation? Thanks :slight_smile:

#2
#3

Since fr is a bidirectional reflectance distribution function, the equation doesn’t treat transmission or scattering phenomena. So what has to be described for every surface point x is the amount of light that is reflected in some/eye direction w…the amount of light reflected in a given direction could come from any direction over the hemisphere around a given point - depending on how the surface material reflects light. But how many directions could you take over a hemisphere to really treat ALL incoming light? The answer is: infinte. Thats why you treat the hemisphere as some kind of projected surface. Depending on the opening angle, a direction from point x projects a small “patch” onto the unit sphere. Even in this small patch, you would have to take an infinte count of directions where light comes from for your calculation…so instead of describing the incoming light this way, you describe it with light coming through this patch alltogether.

The total amount of incoming light for a point x could further be described as a function over a horizontal and a vertical angle. the horizontal angle makes a whole 360 degree and spans a circle and the vertical one spans 180 degree, because they span a HEMIsphere around a surface point. But if you are not interested in function values from one direction only but a range of directions (of two angles!), you somehow have to build a sum, speaking in discrete terms. Or an integral, because an integral can be used to calculate a sum, if properly chosen.

I’m not an expert either, and hope I’m not telling you bullshit - but here’s something to read: https://en.wikipedia.org/wiki/Lebesgue_integration

Maybe this information does help you: The integral over Omega truly means a double integral with horizontal and azimuth angle that span a hemisphere.

#4

That’s an interesting diagram, but I still don’t really understand it. Thanks, though :slight_smile:

What do you mean exactly? Describe it with light coming through the patch altogether? :clue: not sure if I understand…

This is the part I don’t really get. An integral can be used to calculate a sum? Err…? Or maybe that’s explained in the lebesgue integration link? I’m checking it out right now. Thanks for taking the time to explain :slight_smile: Hopefully I’ll be able to understand a little more.

#5

The super short answer is conservation of energy. Now relook at the linked equation.

#6

By conservation of energy, are you referring to the law the a surface cannot reflect more light than it absorbs?

#7

I think I get what you meant by “an integral can be used to calculate a sum”. Basically, since a Riemann integral (where the x-axis is partitioned) can be represented as the limit of a Riemann sum. So technically, it’s a sum of the areas of the rectangles under the line on the graph.

And for Lebesgue integration, the lebesgue integral (where the y-axis is partitioned instead) can represented as the limit of a lebesgue sum.

Well, that’s it for now. Still trying to understand :slight_smile:

#8

1st term: Light being emitted from the surface that reaches the eye.
2nd term: Sum up all the light hitting the surface that reflects to the eye. It’s a product of terms to get a how much of that part.

If the light is hitting a surface significantly larger than it’s wavelength, then the surface can be modeled as a plane. The hemisphere is everything outside of the material.

#9

Yeah, but why is the integration considered a sum? I mean, I get the Riemann sum part, but I wanna understand the math behind the whole thing. And what does it actually mean to “integrate over a hemisphere”?

#10

Adding up an infinite number of small pieces.

#11

But what are these small pieces, then? Is it what’s shown between the integral symbol and the dw’? The light is calculated for all of the different possible directions in the hemisphere and then added up to get the final product?

#12

Yes, that’s the rough idea. Since we’re talking about a unit sphere, the projected area (the patch size) is determined only by dw. Or to make it clearer: the opening angle for the current dirrection. So if you take 100 “samples” in both directions, you have 1/100 * 2 pi (for your 360°) and 1/100 pi for your 180°-> every patch covers an area of 2/100 pi times 1/100 pi…but since the patch is not just a plane, but a bent plane on a sphere surface, and one side is stretched by a cosine (the top of the sphere is tighter…), you can not simply use these calculations. That’s what you need a proper integral and stuff for.