evil84.github.io/content.json at master · evil84/evil84.github.io · GitHub

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{"pages":[],"posts":[{"title":"Precomputed Atmospheric Scattering","text":"","link":"/2022/12/11/2022-12-11-Precomputed-Atmospheric-Scattering/"},{"title":"Sampling BRDF","text":"[TOC] 基本概念 $Pdf(\\omega)$ : 立体角上的概率密度函数 （平常我们采样函数返回的pdf是这个） $Pdf(\\theta, \\phi)$ : 球面坐标系上的概率密度函数 $Pdf(\\theta) = \\int_0^\\pi Pdf(\\theta, \\phi) d\\phi$: $Pdf(\\theta, \\phi)$关于$\\theta$的边缘概率密度函数（marginal probability density function） $Pdf(\\phi | \\theta) = \\frac{Pdf(\\theta, \\phi)}{Pdf(\\theta)}$: $\\phi$关于$\\theta$的条件概率密度函数（conditional probability density function） 我们有在(0, 1)范围内均匀分布概率的随机函数，生成的随机数我们定义为$\\epsilon$,想计算出满足指定概率分布的随机数，这个时候需要用到逆变换采样(Inverse transform sampling)，首先对pdf求cdf：$cdf(x) = \\int_{-\\infty}^x{f(t)dt}$，然后令$\\epsilon = cdf(x)$, 然后我们只需要求出$cdf^{-1}(x)$即是满足该概率分布的随机数。 Uniform Sample Hemisphere我们需要对整个半球均匀采样，假设我们不知道半球的概率密度函数是什么，我们设它为: Pdf(\\omega) = c,由概率定义可知，其在半球上的立体角的积分为1： \\int_{\\Omega^2}{c}d\\omega = 1转换为球面积分： \\int_{\\Omega^2}{c}sin(\\theta){d\\theta}{d\\phi} = 1求解： \\int_0^{\\frac{\\pi}{2}}{sin(\\theta)} \\int_0^{2\\pi}d\\phi d\\theta = 2\\pi c要令$2\\pi c = 1$,那么可知$c = \\frac{1}{2\\pi}$，可知，$Pdf(\\omega) = \\frac{1}{2\\pi}$, $Pdf(\\theta,\\phi) = \\frac{sin(\\theta)}{2\\pi}$ 知道了$Pdf$,现在来计算采样向量，假设我们有2D随机变量E，由分层采样或者低偏差序列而来，x,y分别都在0到1之间随机分布，先计算关于$\\theta$的边缘概率密度函数： Pdf(\\theta) = \\int_0^{2\\pi} \\frac{1}{2\\pi} sin(\\theta) d\\phi = sin(\\theta)然后计算累积分布函数： CDf(\\theta) = \\int_0^{\\theta} sin(t) dt = 1 - cos(\\theta)令$\\epsilon_1 = 1 - cos(\\theta)$，可知$cos(\\theta) = 1 - \\epsilon_1$，又因$\\epsilon_1, \\epsilon_2$都在$(0..1)$范围内均匀分布，故可以用$\\epsilon_1$替换$1-\\epsilon_1$,这样可知$\\theta = cos^{-1}(\\epsilon_1)$。 继续求$Cdf(\\phi | \\theta)$： Pdf(\\phi | \\theta) = \\frac{Pdf(\\theta, \\phi)}{Pdf(\\theta)} = \\frac{\\frac{sin(\\theta)}{2\\pi}}{sin(\\theta)} = \\frac{1}{2\\pi}Cdf(\\phi | \\theta) = \\int_0^{\\phi} \\frac{1}{2\\pi} dt = \\frac{\\phi}{2\\pi}令$\\epsilon_2 = \\frac{\\phi}{2\\pi}$, 可知$\\phi = 2 \\pi \\epsilon_2$。 转换到球面坐标系，我们这里用UE的坐标系，z轴向上，球的半径为1，那么笛卡尔坐标系和球面坐标系的关系为： x = sin(\\theta) cos(\\phi)y = sin(\\theta) sin(\\phi)z = cos(\\theta)令$E.x = \\epsilon_2, E.y = \\epsilon_1$, 可知： cos(\\theta) = E.ysin(\\theta) = \\sqrt[2]{1 - E.y * E.y}cos(\\phi) = cos(2 *\\pi * E.x)sin(\\phi) = sin(2 *\\pi * E.x)这样就可以求出$x,y,z$分别的值了。 UE5代码参考： 123456789101112131415float4 UniformSampleHemisphere( float2 E ){ float Phi = 2 * PI * E.x; float CosTheta = E.y; float SinTheta = sqrt( 1 - CosTheta * CosTheta ); float3 H; H.x = SinTheta * cos( Phi ); H.y = SinTheta * sin( Phi ); H.z = CosTheta; float PDF = 1.0 / (2 * PI); return float4( H, PDF );} Cosine Weight Sample Hemisphere通常我们采样时，会根据brdf的形状生成想要的概率分布，常用的漫反射模型如lambert，oren-nayar等都可以用这个分布去采样，他们的特点是平行于法线的贡献高，所以需要在这附近多生成射线，越靠近垂直于法线的贡献低，生成的射线越少，也算是重要性采样(importance sampling)的一种。 同样，我们假设不知道$pdf$是什么，考虑到$cos(\\theta)$的加权，我们假设为： Pdf(\\omega) = c cos(\\theta)立体角上的半球积分，： \\int_{\\Omega^2} c cos(\\theta) d{\\omega} = 1转换到球面积分： \\int_{\\Omega^2}c cos(\\theta) sin(\\theta){d\\theta}{d\\phi} = 1求解： \\int_0^{\\frac{\\pi}{2}}{c cos(\\theta) sin(\\theta)} \\int_0^{2\\pi}d\\phi d\\theta = \\pi c要令$\\pi c = 1$，那么可知$c = \\frac{1}{\\pi}$, 考虑到$cos(\\theta)$加权，可知$Pdf(\\omega) = \\frac{cos(\\theta)}{\\pi}$， $Pdf(\\theta, \\phi) = \\frac{\\cos(\\theta) \\sin(\\theta)}{\\pi}$。 仿照上例，先计算关于$\\theta$的边缘概率密度函数： Pdf(\\theta) = \\int_0^{2\\pi} \\frac{cos(\\theta)}{\\pi} sin(\\theta) d\\phi = 2 cos(\\theta)sin(\\theta)然后计算累计分布函数： Cdf(\\theta) = \\int_0^{\\theta}2cos(\\theta)sin(\\theta) = 1-cos^2(\\theta)令$\\epsilon_1 = 1 - cos^2(\\theta)$，可知$\\theta = cos^{ -1 }\\sqrt{ 1 - \\epsilon_1 }$，同理用$\\epsilon_1$替换$1-\\epsilon_1$，可得$\\theta = cos^{-1} \\sqrt{\\epsilon_1}$ 继续求$Cdf(\\phi | \\theta)$： Pdf(\\phi | \\theta) = \\frac{Pdf(\\theta, \\phi)}{Pdf(\\theta)} = \\frac{\\frac{\\cos(\\theta) \\sin(\\theta)}{\\pi}}{2 cos(\\theta)sin(\\theta)} = \\frac{1}{2\\pi}Cdf(\\phi | \\theta) = \\int_0^{\\phi} \\frac{1}{2\\pi} dt = \\frac{\\phi}{2\\pi}令$\\epsilon_2 = \\frac{\\phi}{2\\pi}$, 可知$\\phi = 2 \\pi \\epsilon_2$。 知道$\\theta$和$\\phi$的值，其他计算同上，这里就省略了。 UE5代码参考：123456789101112131415float4 CosineSampleHemisphere( float2 E ){ float Phi = 2 * PI * E.x; float CosTheta = sqrt(E.y); float SinTheta = sqrt(1 - CosTheta * CosTheta); float3 H; H.x = SinTheta * cos(Phi); H.y = SinTheta * sin(Phi); H.z = CosTheta; float PDF = CosTheta * (1.0 / PI); return float4(H, PDF);} Microfacet BRDF Sampling接下来考虑微表面的情况，按Cook-Torrance公式来说，我们知道，对Specular Lobe形状影响最大的主要来自于法线分布函数（Normal Distribute Function)，所以我们需要生成满足法线分布函数分布概率的采样向量，注意，NDF并不完全等于概率密度函数，需要考虑到微观表面到宏观表面的投影，这里我们就不展开了，直接列公式（这里$D(m)$代表法线分布函数NDF）： \\int_{\\Omega^2}D(m)cos(\\theta_m)d\\omega = 1下面我们把常见的NDF都推导一遍，同时需要考虑各向同性和各项异性两种情况。 GGX isotropicD_{GGX}(m,\\alpha) = \\frac{a^2}{\\pi ( cos^2(\\theta)(a^2 - 1) + 1)^2}Pdf(\\omega) = \\frac{a^2 cos(\\theta)}{\\pi ( cos^2(\\theta)(a^2 - 1) + 1)^2}Pdf(\\theta, \\phi) = \\frac{a^2 cos(\\theta) sin(\\theta)}{\\pi ( cos^2(\\theta)(a^2 - 1) + 1)^2}Pdf(\\theta) = \\int_0^{2\\pi}\\frac{a^2 cos(\\theta) sin(\\theta)}{\\pi ( cos^2(\\theta)(a^2 - 1) + 1)^2}d\\phi = \\frac{2a^2cos(\\theta)sin(\\theta)}{(cos^2(\\theta)(a^2-1)+1)^2}Pdf(\\phi|\\theta)= \\frac{\\frac{a^2 cos(\\theta) sin(\\theta)}{\\pi ( cos^2(\\theta)(a^2 - 1) + 1)^2}}{\\frac{2a^2cos(\\theta)sin(\\theta)}{(cos^2(\\theta)(a^2-1)+1)^2}} = \\frac{1}{2\\pi}Cdf(\\theta) = \\int_0^{\\theta}\\frac{2a^2cos(t)sin(t)}{(cos^2(t)(a^2-1)+1)^2}dt = -\\frac{cos^2(\\theta)-1}{(a^2-1) cos^2(\\theta)+1}令： \\epsilon = -\\frac{cos^2(\\theta)-1}{(a^2-1) cos^2(\\theta)+1}可知： \\theta = cos^{-1}(\\sqrt{\\frac{1-\\epsilon}{(a^2-1)\\epsilon+1}})求解$\\phi$同上，不再赘述。 UE5代码参考： 1234567891011121314151617float4 ImportanceSampleGGX( float2 E, float a2 ){ float Phi = 2 * PI * E.x; float CosTheta = sqrt( (1 - E.y) / ( 1 + (a2 - 1) * E.y ) ); float SinTheta = sqrt( 1 - CosTheta * CosTheta ); float3 H; H.x = SinTheta * cos( Phi ); H.y = SinTheta * sin( Phi ); H.z = CosTheta; float d = ( CosTheta * a2 - CosTheta ) * CosTheta + 1; float D = a2 / ( PI*d*d ); float PDF = D * CosTheta; return float4( H, PDF );} beckmann isotropicD_Beckmann(m,\\alpha) = \\frac{e^{\\frac{-tan^2(\\theta)}{\\alpha^2}}}{\\pi\\alpha^2cos^4(\\theta)}Pdf(\\omega) = \\frac{e^{\\frac{-tan^2(\\theta)}{\\alpha^2}}}{\\pi\\alpha^2cos^4(\\theta)} cos(\\theta)Pdf(\\theta,\\phi) = \\frac{e^{\\frac{-tan^2(\\theta)}{\\alpha^2}}}{\\pi\\alpha^2cos^4(\\theta)} cos(\\theta) sin(\\theta)Pdf(\\theta) = \\int_0^{2\\pi} \\frac {e^{\\frac{-tan^2(\\theta)} {\\alpha^2} } } {\\pi\\alpha^2cos^4(\\theta)} cos(\\theta) sin(\\theta)d\\phi = \\frac{2{e^{-\\frac{tan^2(\\theta)} {\\alpha^2}}}}{ {\\alpha^2} {cos^3(\\theta)}}sin(\\theta)Pdf(\\phi|\\theta) = \\frac{Pdf(\\theta,\\phi)}{Pdf(\\theta)} = \\frac{1}{2\\pi}Cdf(\\theta) = \\int_0^{\\theta}\\frac{2{e^{-\\frac{tan^2(t)}{\\alpha^2}}}}{ {\\alpha^2} {cos^3(t)}} sin(t) dt = 1 - e^{\\frac{1}{\\alpha^2} - \\frac{1}{\\alpha^2 cos^2(\\theta) }} = 1 - e^{ \\frac{1} {\\alpha^2} (1-\\frac{1}{cos^2(\\theta)})}令： \\epsilon = 1 - e^{\\frac{1}{\\alpha^2}(1 - \\frac{1}{cos^2(\\theta)})}可知： \\theta = cos^{-1}(\\sqrt{\\frac{1}{1 - \\alpha^2ln(1-\\epsilon)}})求解$\\phi$同上，不再赘述。 Blinn PhongD_{Blinn}(m, \\alpha) = \\frac{\\alpha+2}{2\\pi}cos^\\alpha(\\theta)Pdf(\\omega) = \\frac{\\alpha+2}{2\\pi}cos^{\\alpha+1}(\\theta)Pdf(\\theta,\\phi) = \\frac{\\alpha+2}{2\\pi}cos^{\\alpha+1}(\\theta)sin(\\theta)Pdf(\\theta) = \\int_0^{2\\pi}\\frac{\\alpha+2}{2\\pi}cos^{\\alpha+1}(\\theta)sin(\\theta)d\\phi = (\\alpha+2)cos^{\\alpha+1}(\\theta)sin(\\theta)Pdf(\\phi,\\theta) = \\frac{Pdf(\\theta,\\phi)}{Pdf(\\theta)} = \\frac{1}{2\\pi}Cdf(\\theta) = \\int_0^{\\theta}(\\alpha+2)cos^{\\alpha+1}(t)sin(t)dt = 1 - cos^{\\alpha+2}(\\theta)令： \\epsilon = 1 - cos^{\\alpha+2}(\\theta)可知： \\theta = cos^{-1}((1-\\epsilon)^{(\\frac{1}{\\alpha+2})})求解$\\phi$同上，不再赘述。 Visible GGX isotropicGGX anisotropic从Physically-Based Shading at Disney可知： D_{GGX\\_aniso}(m, \\alpha) = \\frac{1}{\\pi\\alpha_t\\alpha_b} \\frac{1}{((\\frac{t \\cdot h}{\\alpha_t})^2 + (\\frac{b \\cdot h}{\\alpha_b})^2 + (m \\cdot n)^2)^2} = \\frac{1}{\\pi\\alpha_t\\alpha_b} \\frac{1}{((\\frac{t \\cdot h}{\\alpha_t})^2 + (\\frac{b \\cdot h}{\\alpha_b})^2 + (cos^2(\\theta))^2}其中： t \\cdot h = sin(\\theta) cos(\\phi)b \\cdot h = sin(\\theta)sin(\\phi)化简： ((\\frac{t \\cdot h}{\\alpha_t})^2 + (\\frac{b \\cdot h}{\\alpha_b})^2 + cos^2(\\theta))^2 = (\\frac{sin^2(\\theta)cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\theta)sin^2(\\phi)}{\\alpha_b^2} + cos^2(\\theta))^2 \\\\ = cos^4(\\theta) \\frac{1}{cos^4(\\theta)} (\\frac{sin^2(\\theta)cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\theta)sin^2(\\phi)}{\\alpha_b^2} + cos^2(\\theta))^2 \\\\ = cos^4(\\theta)(tan^2(\\theta) (\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})+1)^2可得： D_{GGX\\_aniso}(m, \\alpha) = \\frac{1}{\\pi\\alpha_t\\alpha_b cos^4(\\theta)(tan^2(\\theta) (\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})+1)^2}从而可知： Pdf(\\omega) = \\frac{1}{\\pi\\alpha_t\\alpha_b cos^3(\\theta)(tan^2(\\theta) (\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})+1)^2}Pdf(\\theta, \\phi) = \\frac{1}{\\pi\\alpha_t\\alpha_b cos^3(\\theta)(tan^2(\\theta) (\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})+1)^2}sin(\\theta)Pdf(\\phi) = \\int_0^{\\frac{\\pi}{2}} \\frac{1}{\\pi\\alpha_t\\alpha_b cos^3(\\theta)(tan^2(\\theta) (\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})+1)^2}sin(\\theta) d\\theta \\\\ = \\frac{1}{2 \\pi \\alpha_t \\alpha_b(\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})}Cdf(\\phi) = \\int_0^\\phi \\frac{1}{2 \\pi \\alpha_t \\alpha_b(\\frac{cos^2(t)}{\\alpha_t^2} + \\frac{sin^2(t)}{\\alpha_b^2})} dt \\\\ = \\frac{1}{2\\pi}tan^{-1}(\\frac{\\alpha_t tan(\\phi)}{\\alpha_b})令： \\epsilon_1 = \\frac{1}{2\\pi}tan^{-1}(\\frac{\\alpha_t tan(\\phi)}{\\alpha_b})可知： \\phi = tan^{-1}(\\frac{\\alpha_b}{\\alpha_t} tan(2\\pi\\epsilon_1))这里要注意，因为$tan^{-1}$的值域为$-\\frac{\\pi}{2}$到$\\frac{\\pi}{2}$，我们需要转换到$0$到$2\\pi$， 我们令$K = (\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})$，可知： Pdf(\\theta|\\phi) = \\frac{Pdf(\\theta,\\phi)}{Pdf(\\phi)} = \\frac{\\frac{1}{\\pi\\alpha_t\\alpha_b cos^3(\\theta)(tan^2(\\theta) (\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})+1)^2}sin(\\theta)}{\\frac{1}{2 \\pi \\alpha_t \\alpha_b(\\frac{cos^2(\\phi)}{\\alpha_t^2} + \\frac{sin^2(\\phi)}{\\alpha_b^2})}} \\\\ = \\frac{\\frac{sin(\\theta)}{\\pi \\alpha_t \\alpha_b cos^3(\\theta)(tan^2(\\theta)K + 1)^2}}{\\frac{1}{2 \\pi \\alpha_t \\alpha_bK}} \\\\ = \\frac{2sin(\\theta)K}{cos^3(\\theta)(tan^2(\\theta)K+1)^2}Cdf(\\theta|\\phi) = \\int_0^{\\theta}\\frac{2sin(t)K}{cos^3(t)(tan^2(t)K+1)^2}dt = \\frac{Ksin^2(\\theta)}{(K-1)sin^2(\\theta)+1}令$\\epsilon_1 = \\frac{Ksin^2(\\theta)}{(K-1)sin^2(\\theta)+1}$，可得： \\theta = sin^{-1}(\\sqrt\\frac{-\\epsilon_1}{K\\epsilon_1 - \\epsilon_1 - K})sin(\\theta) = \\sqrt\\frac{-\\epsilon_1}{K\\epsilon_1 - \\epsilon_1 - K}cos^2(\\theta) = 1 - \\frac{-\\epsilon_1}{K\\epsilon_1 - \\epsilon_1 - K} = \\frac{K\\epsilon_1 - K}{K\\epsilon_1 - \\epsilon_1 - K} = \\frac{K(1-\\epsilon_1)}{K + (1-K)\\epsilon_1}\\theta = cos^{-1} \\sqrt{\\frac{K(1-\\epsilon_1)}{K + (1-K)\\epsilon_1}}PBRT-V3代码参考： 123456789101112131415161718192021222324252627282930Vector3f TrowbridgeReitzDistribution::Sample_wh(const Vector3f &amp;wo, const Point2f &amp;u) const { Vector3f wh; if (!sampleVisibleArea) { Float cosTheta = 0, phi = (2 * Pi) * u[1]; if (alphax == alphay) { Float tanTheta2 = alphax * alphax * u[0] / (1.0f - u[0]); cosTheta = 1 / std::sqrt(1 + tanTheta2); } else { phi = std::atan(alphay / alphax * std::tan(2 * Pi * u[1] + .5f * Pi)); if (u[1] &gt; .5f) phi += Pi; Float sinPhi = std::sin(phi), cosPhi = std::cos(phi); const Float alphax2 = alphax * alphax, alphay2 = alphay * alphay; const Float alpha2 = 1 / (cosPhi * cosPhi / alphax2 + sinPhi * sinPhi / alphay2); Float tanTheta2 = alpha2 * u[0] / (1 - u[0]); cosTheta = 1 / std::sqrt(1 + tanTheta2); } Float sinTheta = std::sqrt(std::max((Float)0., (Float)1. - cosTheta * cosTheta)); wh = SphericalDirection(sinTheta, cosTheta, phi); if (!SameHemisphere(wo, wh)) wh = -wh; } else { bool flip = wo.z &lt; 0; wh = TrowbridgeReitzSample(flip ? -wo : wo, alphax, alphay, u[0], u[1]); if (flip) wh = -wh; } return wh;} beckmann anisotropic Burley 2012, “Physically-Based Shading at Disney”","link":"/2021/06/16/Sampling%20BRDF/"}],"tags":[{"name":"atmosphere scattering","slug":"atmosphere-scattering","link":"/tags/atmosphere-scattering/"},{"name":"Ray Tracing from the Ground Up","slug":"Ray-Tracing-from-the-Ground-Up","link":"/tags/Ray-Tracing-from-the-Ground-Up/"}],"categories":[{"name":"Realtime Rendering","slug":"Realtime-Rendering","link":"/categories/Realtime-Rendering/"},{"name":"Ray Tracing from the Ground Up","slug":"Ray-Tracing-from-the-Ground-Up","link":"/categories/Ray-Tracing-from-the-Ground-Up/"}]}