Home › Forums › Games202-高质量实时渲染 › 作业四疑问 This topic has 4 replies, 5 voices, and was last updated 4 years ago by Neon. Viewing 2 reply threads Author Posts 2021年7月5日 at 下午6:37 #8785 Score: 0 MoonParticipant 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。 2021年7月6日 at 下午5:00 #8808 Score: 0 YunHsiaoParticipant Karma: 19 pts 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 4 years ago by YunHsiao. This reply was modified 4 years ago by YunHsiao. 2021年7月8日 at 上午12:42 #8843 Score: 0 RyushinnParticipant Karma: 2 pts 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 4 years ago by Ryushinn. This reply was modified 4 years ago by Ryushinn. 2021年7月8日 at 上午8:52 #8846 Score: 0 MigoParticipant 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 2021年7月8日 at 下午8:49 #8854 Score: 0 NeonParticipant 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。 Author Posts Viewing 2 reply threads You must be logged in to reply to this topic. Log In Username: Password: Keep me signed in Log In