Mchart 130 Report post Posted June 25, 2014 Hi I'm new here I hope I'm not making mistakes with this post! if so please tell me! I'm rendering a scene on a render target, I've then to access its depth information in a post-processing stage. In this stage I need linear depth info. I've been browsing the internet quite a lot and found a lot of different approaches/alternatives. What do you think is the way to go? I'm a bit of a newbie with this stuff so can you give me an hint/pointer on the possible GLSL code? Thank you very much :) Micheal 0 Share this post Link to post Share on other sites
Samith 2460 Report post Posted June 25, 2014 (edited) Typically you'll read the value from your non-linear depth buffer and reconstruct the view-space z position at that point. // let s and t be two values that satisfy the following two equations: (s*f + t)/f = 1 // f is the distance from the eye to far plane (s*n + t)/n = -1 // n is the distance from the eye to near plane // the above is simply the mapping the occurs when you do a projection transform // s and t are found in your projection matrix at positions 3,3 and 3,4, respectively // let d be the value in your depth buffer // the (xxx + 1) * 0.5 is to map the [-1,1] clip-space coordinates to the [0,1] space in the depth buffer d = ((s*z + t)/z + 1) * 0.5 2d = (s*z + t)/z + 1 2d - 1 = (s*z + t)/z 2d*z - z = s*z + t 2d*z - z - s*z = t z*(2d - 1 - s) = t z = t / (2d - 1 - s) // reconstructed z value That math is just off the top of my head and might not be totally correct, but you should be able to get the picture from that. You basically just want to use your knowledge of the projection matrix you used when transforming the verts to do the inverse of that transformation and reconstruct the view-space z. Edited June 25, 2014 by Samith 0 Share this post Link to post Share on other sites
Mchart 130 Report post Posted June 25, 2014 Typically you'll read the value from your non-linear depth buffer and reconstruct the view-space z position at that point. // let s and t be two values that satisfy the following two equations: (s*f + t)/f = 1 // f is the distance from the eye to far plane (s*n + t)/n = -1 // n is the distance from the eye to near plane // the above is simply the mapping the occurs when you do a projection transform // s and t are found in your projection matrix at positions 3,3 and 3,4, respectively // let d be the value in your depth buffer // the (xxx + 1) * 0.5 is to map the [-1,1] clip-space coordinates to the [0,1] space in the depth buffer d = ((s*z + t)/z + 1) * 0.5 2d = (s*z + t)/z + 1 2d - 1 = (s*z + t)/z 2d*z - z = s*z + t 2d*z - z - s*z = t z*(2d - 1 - s) = t z = t / (2d - 1 - s) // reconstructed z value That math is just off the top of my head and might not be totally correct, but you should be able to get the picture from that. You basically just want to use your knowledge of the projection matrix you used when transforming the verts to do the inverse of that transformation and reconstruct the view-space z. Thank you Samith, just one question. In which space t and s are distances eye-near/eye-far? If I look in my projection matrix (built via glm::perspective) at position 3,3 and 3,4 what I see is something like -1.03 and -1 where my clipping planes are [0.1, 8] 0 Share this post Link to post Share on other sites
Samith 2460 Report post Posted June 25, 2014 (edited) Thank you Samith, just one question. In which space t and s are distances eye-near/eye-far? If I look in my projection matrix (built via glm::perspective) at position 3,3 and 3,4 what I see is something like -1.03 and -1 where my clipping planes are [0.1, 8] Yep. The s and t values are computed given the near and far planes. In my post above I said that s and t are two values that satisfy the equations (s*f + t)/f = 1 and (s*n + t)/n = -1, but I didn't provide any actually solution for s and t If you are interested, the solutions for s and t can be found like so: (sf + t)/f = 1 (sn + t)/n = -1 // here you have two linear equations and two unknowns. you can solve // for s and t with your basic high school algebra techniques eq1: sf + t = f eq2: sn + t = -n // subtract eq2 from eq1 eq3: s(f-n) = f+n s = (f+n)/(f-n) // substitute s into eq1 and solve for t eq4: sf + t = f f(f+n)/(f-n) + t = f f(f+n) + t(f-n) = f(f-n) f(f+n) - f(f-n) = -t(f-n) ff+fn - ff+fn = -t(f+n) -2fn/(f+n) = t // now you can check s and t given a near and far plane of 0.1 and 8.0 // s = (f+n)/(f-n) = (8-0.1)/(8+0.1) =~ 1.025 // t = 2fn/(f+n) =~ -0.19 Some caveats: these computations depend on some nitpicky details about your projection. For example, if you use your projection matrix to transform a right handed coordinate space into the left handed clip space, then the original equations will be slightly different (ie s*-f + t = f instead of sf + t = f, etc). Also, the s and t values will be found in your projection matrix in the 3,3 location and 3,4 location if your projection matrix is column major, but they'll be in 3,3 and 4,3 if your projection matrix is row major. I don't know enough about your exact environment to give you all the answers, but I hope the algebra I've posted at least gets you started down the right path. Edited June 26, 2014 by Samith 2 Share this post Link to post Share on other sites