1
\$\begingroup\$

I'm looking to take a 2D texture (previously rendered from the user's perspective), and overlay it overtop of a cubemap.

Since a cubemap has 6 textures, I need to run a shader over top of all 6, and use some formula for calculating the UV from a given view.

I'm at a loss for how to calculate this.

Here's an illustration enter image description here

If the user were to spin around in 1 spot, they would build up full panoramic cubemap.

The intention is to reuse end-of-frame data to update a persistent cubemap that follows the user.

I know how to build manual cubemaps by rendering a scene 6 times, once on all axes, however in this case I wish to take an already rendered view and transplant it onto the correct spots on a cubemap.

\$\endgroup\$
2
\$\begingroup\$

I think I was going about this the wrong way. I didn't need to do some unruly math to convert spherical coordinates into UV coordinates for all 6 faces, I just needed to project the quad properly.

I made a fragment shader that simply passes through the texture it's provided, and the texture coordinates are just a quad's UV's (with no alterations).

#version 430 core
layout (location = 0) uniform sampler2D screenTexture;
layout (location = 0) out vec3 OutputColor;    
in vec2 TexCoord;

void main()
{       
    OutputColor = texture(screenTexture, UV).rgb;
}

Recall my goal: to project what the user sees into a cubemap.

In the vertex shader, I position the quad at the far plane of the user's frustum, and then render it from the perspective of each of the 6 viewing angles of the cubemap.

void main()
{       
    TexCoord = (vertex.xy + vec2(1.0)) / 2.0;

    // Strip the translation from the matrix
    mat4 InvRotVMatrix = vMatrix_Inverse;
    InvRotVMatrix[3][0] = 0;
    InvRotVMatrix[3][1] = 0;
    InvRotVMatrix[3][2] = 0;

    vec4 WorldPos = (InvRotVMatrix * pMatrix_Inverse * vec4(vertex.xyz, 1));
    gl_Position = Proj * View[View_Index] * WorldPos;
}
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.