為什麼BRDF和照明不相關?


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關於this question,在接受的答案中,我可能有一個瑣碎的問題。

W.r.t。以下積分

$$L_o(x \ rightarrow \ Theta)= \ int _ {\ Omega_x} f_r(x,\ Psi \ rightarrow \ Theta)L_i(x \ leftarrow \ Psi)cos(N_x,\ Psi)d \ omega _ {\ Psi}$$

在接受的答案中,其中一項評論指出$ f_r(l,v)= f_r(x,\ Psi \ rightarrow \ Theta)$和$ L_i(l)= L(x \ leftarrow \ Psi)$不相關。像往常一樣...我想了解原因...我在這個主題上的參考是this book。在這本書中,brdf $ f_r(l,v)$被 定義 定義為

$$f_r(x,\ Psi \ rightarrow \ Theta)= \ frac {dL(x \ rightarrow \ Theta)} {dE(x \ leftarrow \ Psi)} = \ frac {dL(x \ rightarrow \ Theta)} {L(x \ leftarrow \ Psi)cos(N_x,\ Psi)d \ omega_ \ Psi},$$

從最後一個方程式中,我可以看到兩者實際上是相關的,除非我缺少某些東西。為什麼它們不相關?

關於所使用符號的註釋。

$ L(x \ rightarrow \ Theta)$:這是從$ x $向$ \ Theta $方向的輻射$ f_r(x,\ Psi \ rightarrow \ Theta)$是BRDF從$ x $中的方向$ \ Psi $進入並向$ \ Theta $方向離開$ L(x \ leftarrow \ Psi)$:方向$ \ Psi $向$ x $的輻射度

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When rendering an image you are trying to find the outgoing light from a point into the camera $L(x \rightarrow \Theta$).

To do this you solve "The rendering equation", which means you integrate the product of the brdf and the incoming light for every incoming direction.

$$L(x\rightarrow\Theta )=\int_\Omega f_r(x, \omega\rightarrow\Theta)\cdot L(x\leftarrow\omega)d\omega$$

Since you don't have the outgoing light you cannot use the definition of the brdf to calculate it.

$$f_r(x,\omega\rightarrow\Theta )=\frac{L(x\rightarrow\Theta)}{E(x\leftarrow\omega)}$$

Instead, you use a formula to compute $f_r(x,\omega\rightarrow\Theta )$ and some operation to find the incoming light $L(x\leftarrow\omega)$ (a raytrace, a env map sample, etc.), and use those to compute the outgoing light $L(x\rightarrow\Theta)$.

These two calculations are usually done in a completely separate way. So there is no way the incoming light can effect the brdf in any way.