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我正在按照本教程在 OpenGL 中实现 SSAO：Jhon Chapman SSAO

基本上，所描述的技术使用沿片段法线定向的半球内核。然后将样本的视图空间 z 位置与其屏幕空间深度缓冲区值进行比较。如果深度缓冲区中的值较高，则意味着样本最终处于几何图形中，因此应该遮挡该片段。

该技术的目标是摆脱对象平面变灰的经典实现伪影。

我有相同的实现，但有两个小区别

  • 我没有使用噪声纹理来旋转我的内核，所以我有条带伪影，现在很好
  • 我无法访问具有每像素法线的缓冲区，因此我必须仅使用深度缓冲区来计算我的法线和 TBN 矩阵。

该算法似乎运行良好，我可以看到片段被遮挡，但我的脸仍然变灰...... IMO它来自我计算我的TBN矩阵的方式。法线看起来不错，但一定有问题，因为我的内核似乎没有正确对齐，导致样本最终出现在脸上。

屏幕截图具有 8 个样本的内核和 0.1 的半径。第一个只是 SSAO 传递的结果，第二个是生成的法线的调试渲染。

这是计算 Normal 和 TBN 矩阵的函数的代码



    mat3 computeTBNMatrixFromDepth(in sampler2D depthTex, in vec2 uv)
    {
        // Compute the normal and TBN matrix
        float ld = -getLinearDepth(depthTex, uv);
        vec3 x = vec3(uv.x, 0., ld);
        vec3 y = vec3(0., uv.y, ld);
        x = dFdx(x);
        y = dFdy(y);
        x = normalize(x);
        y = normalize(y);
        vec3 normal = normalize(cross(x, y));
        return mat3(x, y, normal);
    }

和 SSAO 着色器

#include "helper.glsl"

in vec2 vertTexcoord;
uniform sampler2D depthTex;

const int MAX_KERNEL_SIZE = 8;
uniform vec4 gKernel[MAX_KERNEL_SIZE];

// Kernel Radius in view space (meters)
const float KERNEL_RADIUS = .1; 

uniform mat4 cameraProjectionMatrix;
uniform mat4 cameraProjectionMatrixInverse;

out vec4 FragColor;


void main()
{   
    // Get the current depth of the current pixel from the depth buffer (stored in the red channel)
    float originDepth = texture(depthTex, vertTexcoord).r;

    // Debug linear depth. Depth buffer is in the range [1.0];
    float oLinearDepth = getLinearDepth(depthTex, vertTexcoord);

    // Compute the view space position of this point from its depth value
    vec4 viewport = vec4(0,0,1,1);    
    vec3 originPosition = getViewSpaceFromWindow(cameraProjectionMatrix, cameraProjectionMatrixInverse, viewport, vertTexcoord, originDepth);

    mat3 lookAt = computeTBNMatrixFromDepth(depthTex, vertTexcoord);
    vec3 normal = lookAt[2];

    float occlusion = 0.;

    for (int i=0; i<MAX_KERNEL_SIZE; i++) 
    {
        // We align the Kernel Hemisphere on the fragment normal by multiplying all samples by the TBN        
        vec3 samplePosition = lookAt * gKernel[i].xyz;

        // We want the sample position in View Space and we scale it with the kernel radius
        samplePosition = originPosition + samplePosition * KERNEL_RADIUS;

        // Now we need to get sample position in screen space
        vec4 sampleOffset = vec4(samplePosition.xyz, 1.0);
        sampleOffset = cameraProjectionMatrix * sampleOffset;
        sampleOffset.xyz /= sampleOffset.w;

        // Now to get the depth buffer value at the projected sample position
        sampleOffset.xyz = sampleOffset.xyz * 0.5 + 0.5;

        // Now can get the linear depth of the sample
        float sampleOffsetLinearDepth = -getLinearDepth(depthTex, sampleOffset.xy);

        // Now we need to do a range check to make sure that object 
        // outside of the kernel radius are not taken into account
        float rangeCheck = abs(originPosition.z - sampleOffsetLinearDepth) < KERNEL_RADIUS ? 1.0 : 0.0;

        // If the fragment depth is in front so it's occluding
        occlusion += (sampleOffsetLinearDepth >= samplePosition.z ? 1.0 : 0.0) * rangeCheck;
    }  

    occlusion = 1.0 - (occlusion / MAX_KERNEL_SIZE);
    FragColor = vec4(vec3(occlusion), 1.0);
}

从深度缓冲区计算的法线 SSAO 通行证

更新 1

TBN 计算函数的这种变化给出了相同的结果

mat3 computeTBNMatrixFromDepth(in sampler2D depthTex, in vec2 uv)
{
    // Compute the normal and TBN matrix
    float ld = -getLinearDepth(depthTex, uv);
    vec3 a = vec3(uv, ld);
    vec3 x = vec3(uv.x + dFdx(uv.x), uv.y, ld + dFdx(ld));
    vec3 y = vec3(uv.x, uv.y + dFdy(uv.y), ld + dFdy(ld));
    //x = dFdx(x);
    //y = dFdy(y);
    //x = normalize(x);
    //y = normalize(y);
    vec3 normal = normalize(cross(x - a, y - a));
    vec3 first_axis = cross(normal, vec3(1.0f, 0.0f, 0.0f));
    vec3 second_axis = cross(first_axis, normal);
    return mat3(normalize(first_axis), normalize(second_axis), normal);
}
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1 回答 1

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我认为问题可能是您正在混合坐标系。您将纹理坐标与线性深度结合使用。您可以想象两个垂直表面稍微面向屏幕左侧。两者与垂直平面的角度相同，因此应该具有相同的法线对吗？

但是让我们想象一下，这些表面之一离相机更远。由于 fFdx/fFdy 函数基本上可以告诉您与相邻像素的差异，因此远离相机的表面将比靠近相机的表面在一个像素上具有更大的线性深度差。但是 uv.x / uv.y 导数将具有相同的值。这意味着您将根据与相机的距离获得不同的法线。

解决方案是计算视图坐标并使用它的导数来计算法线。

vec3 viewFromDepth(in sampler2D depthTex, in vec2 uv, in vec3 view)
{
    float ld = -getLinearDepth(depthTex, uv);

    /// I assume ld is negative for fragments in front of the camera
    /// not sure how getLinearDepth is implemented

    vec3 z_scaled_view = (view / view.z) * ld;

    return z_scaled_view;
}

mat3 computeTBNMatrixFromDepth(in sampler2D depthTex, in vec2 uv, in vec3 view)
{
    vec3 view = viewFromDepth(depthTex, uv);

    vec3 view_normal = normalize(cross(dFdx(view), dFdy(view)));
    vec3 first_axis = cross(view_normal, vec3(1.0f, 0.0f, 0.0f));
    vec3 second_axis = cross(first_axis, view_normal);

    return mat3(view_normal, normalize(first_axis), normalize(second_axis));
}
于 2016-10-28T00:07:04.320 回答