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Ok I think I found a solution by myself. I'll try to explain it here, let me know what you people think about it. INPUT: 3 grayscale 8 bit images OUTPUT: 1 HDR radiance map (single precision) So, to compute the HDR image I use the Devebec algorithm This algorithm outputs the radiance map computed from the 3 (or more) input photos. Every pixel value ...


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Firstly, an 8bit jpeg is 24bit color; 8bits per color. Your 32bit HDR is also 8bit per color plus an additional 8bit shared channel (Alpha). Secondly, not every correctly exposed image will have values from min (black/0) to max (white/255). HDR is rewriting values to fit w/in the display/reproduction capability... if in that process it eliminates any true ...


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The end result of High Dynamic Range Imaging (HDRI) is not to produce an image with dynamic range as high as the scene it attempts to reproduce. We do not have display technology available that can do that. At least not practical ones. The aim of HDRI is to take a very high dynamic range scene and reproduce it in a way that we can see the very bright and ...


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First, regardless of the particular dynamic range definition, calculating "anything vs zero" luminance is not useful, because then the ratio becomes infinite. Your: log2(max)-log2(0) is another way of saying: log(max/0) and we can't divide by zero (and log2(0) is undefined) So let's forget the 0 for now (zero means the image is not well exposed anyway ...


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