环境光渲染(一）

vec3 irradiance = vec3(0.0);  

// tangent space calculation from origin point(x'=up, y'=N, z'=left)(left-hand)
vec3 up    = vec3(0.0, 1.0, 0.0);   //x'
vec3 left = cross(up, N);			//z'
up         = cross(N, left);		//x'

float sampleDelta = 0.025;
float nrSamples = 0.0; 
for(float phi = 0.0; phi < 2.0 * PI; phi += sampleDelta)
{
    for(float theta = 0.0; theta < 0.5 * PI; theta += sampleDelta)
    {
        float cosTheta = cos(theta);
        float sinTheta = sin(theta);
    
        // spherical to cartesian (in tangent space)
        vec3 tangentSample = vec3(cos(phi) * sinTheta, cosTheta, sin(phi) * sinTheta);
        // tangent space to world
        vec3 sampleVec = tangentSample.x * up + tangentSample.y * N + tangentSample.z * left; 

        irradiance += texture(environmentMap, sampleVec).rgb * cosTheta * sinTheta;
        nrSamples++;
    }
}
irradiance = PI * irradiance * float(nrSamples);

2.2.2 蒙特卡洛估算

使用蒙特卡洛采样估算积分

\int_{a}^{b} f (x) d x \approx \frac{1}{N} \sum_{i = 1}^{N} \frac{f (X_{i})}{pdf (X_{i})}

其中 $pdf (x)$ 函数是采样时使用的概率分布函数。方法1：对于要计算的这个二维积分,如果使用的随机采样是 $ϕ$ 在 $[0, 2 π]$ 之间均匀分布， $θ$ 在 $[0, π / 2]$ 之间均匀分布，那么

pdf (ϕ) = 1 / (2 π), pdf (θ) = 2 / π

\begin{aligned} \frac{1}{π} \int_{ϕ = 0}^{2 π} \int_{θ = 0}^{π / 2} L_{i} (ϕ, θ) \cos (θ) \sin (θ) d θ d ϕ \\ = & \frac{1}{π} \frac{2 π}{n_{1}} \sum_{j = 0}^{n_{1}} (\frac{π}{2 n_{2}} \sum_{i = 0}^{n_{2}} L_{i} (ϕ_{j}, θ_{i}) \cos (θ_{i}) \sin (θ_{i})) \\ = & \frac{π}{n_{1} n_{2}} \sum_{j = 0}^{n_{1}} \sum_{i = 0}^{n_{2}} L_{i} (ϕ_{j}, θ_{i}) \cos (θ_{i}) \sin (θ_{i}) \\ = & \frac{π}{N} \sum_{i = 0}^{N} L_{i} (ϕ_{i}, θ_{i}) \cos (θ_{i}) \sin (θ_{i}) \end{aligned}

其中

\begin{aligned} ϕ_{i} & = 2 π ξ \\ θ_{i} & = \frac{π}{2} ξ \end{aligned}

$ξ$ 表示均匀分布在[0,1]之间的随机变量

方法2：上面的采样方法在极点位置会比较密集，在赤道位置比较稀疏，由于漫反射是均匀分布的，如果采样点也是均匀分布在半球面上的，收敛速度会比较快，这种情况下

pdf (ϕ) = 1 / (2 π), pdf (θ) = \sin (θ)

\begin{aligned} \frac{1}{π} \int_{ϕ = 0}^{2 π} \int_{θ = 0}^{π / 2} L_{i} (ϕ, θ) \cos (θ) \sin (θ) d θ d ϕ \\ = & \frac{1}{π} \frac{2 π}{n_{1}} \sum_{j = 0}^{n_{1}} (\frac{1}{n_{2}} \sum_{i = 0}^{n_{2}} L_{i} (ϕ_{j}, θ_{i}) \cos (θ_{i})) \\ = & \frac{2}{n_{1} n_{2}} \sum_{j = 0}^{n_{1}} \sum_{i = 0}^{n_{2}} L_{i} (ϕ_{j}, θ_{i}) \cos (θ_{i}) \\ = & \frac{2}{N} \sum_{i = 0}^{N} L_{i} (ϕ_{i}, θ_{i}) \cos (θ_{i}) \end{aligned}

其中

\begin{aligned} ϕ_{i} & = 2 π ξ \\ θ_{i} & = \arccos (1 - ξ) \end{aligned}

vec3 irradiance = vec3(0.0);  

vec3 up = vec3(0.0, 1.0, 0.0); //x'
vec3 left = cross(up, N);	//z'
up = cross(N, left);	//x'

for(uint i=0; i<sampleCounts; i++)
{
    //http://holger.dammertz.org/stuff/notes_HammersleyOnHemisphere.html
    vec2 uv = Hammersley(i, totalCounts);

    // tangent space sample point
    float phi = uv.y * 2.0 * PI;
    float cosTheta = 1.0 - uv.x;
    float sinTheta = sqrt(1-cosTheta*cosTheta);
    vec3 tangentSample = vec3(cos(phi) * sinTheta, cosTheta, sin(phi) * sinTheta);

    // tangent space to world
    vec3 sampleVec = tangentSample.x * up + tangentSample.y * N + tangentSample.z * left; 

    irradiance += texture2D(s_texSkybox, sampleVec).rgb * cosTheta;
}

irradiance = 2*irradiance / float(sampleCounts);