作业四疑问

[Moon]

课程中提到u是sinTheta，为何作业中注释显示乘cosTheta。
计算Eavg的时候不是只需要Eu*u*2就行么，得出来的结果相差很大啊。

[YunHsiao]

μ 的定义感觉这里其实还是没那么严谨的，Kulla Conty 原作里面注释说 μ 就是 cos(θ)：

Notice that I’m using the Greek letter µ for the cos θ term here.
This is just to avoid too much trigonometry in the slides and
it’s closer to how the formulas are actually implemented.

E_avg 我理解也应该是不需要做 sample 的，因为已经有了加权平均的过程……

• This reply was modified 2 years, 9 months ago by YunHsiao.
• This reply was modified 2 years, 9 months ago by YunHsiao.

[Ryushinn]

1. 无论u是cosTheta还是sinTheta，Eavg的式子都可以写成2 * \integrate{ E(u) * u du }，所以采用课程的定义或者作业框架的定义应该都可以

2. 我也认为框架给的求Eavg这部分代码有点奇怪，我是直接Eu*u*2了，最终得到的结果比参考图要亮，不知道LZ是不是也偏亮。
而且个人感觉作业文档的Eavg参考图好像是错的：Eavg的图整体要比Emu的图暗一些(把两张图并排对比，roughness大的部分尤为明显)。

// 编辑：要在主楼里回复的..搞错辽….

• This reply was modified 2 years, 9 months ago by Ryushinn.
• This reply was modified 2 years, 9 months ago by Ryushinn.

[Migo]

μ是cos(θ)或者μ是sin(θ)，最后推导出来的积分式都是一样的，所以问题不大

[Neon]

mu是sinTheta,所以mu*dmu*dPhi=sinTheta*cosTheta*dTheta*dPhi,采样是对w的，dw=sinTheta*dTheta*dPhi,所以最后只剩一个cosTheta，个人理解是这样，不知道对不对。

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