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    <p>If you think your formulas are exact, you can comput an exact formula with trig, like so:</p> <pre><code>Rin = 2 f sin(w/2) -&gt; sin(w/2)= Rin/2f Rout= f tan(w) -&gt; tan(w)= Rout/f (Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -&gt; cos(w) = 1 - 2(Rin/2f)^2 (Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1 -&gt; (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1 </code></pre> <p>However, as @jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:</p> <pre><code>Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...) </code></pre> <p>By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.</p> <p>Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.</p> <p>Edit: as per your request, the equivalent scaling factor for the above formula:</p> <pre><code>(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1 -&gt; Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2) </code></pre> <p>If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.</p> <p>The inverse formula should be something like:</p> <pre><code>Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 ) </code></pre>
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