Analytically, the focal length is described by the lens maker's equation: 1/f = (n - 1)(1/R1 + 1/R2), where R1 and R2 are the radii of curvature, f is the focal ...

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The SilverFast. Resolution Target is equally well suited for film and flatbed scanners with transmissive unit. 2. Resolution of a Scanner. 2.1 dpi, what exactly ...

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I'm looking to understand where my calculations may be going wrong and what a standard and accepted trig. equation would be to allow me to properly calculate the vertical and horizontal FOVs (in mm).

Solved: My problem in Excel was that when I was dividing by 2 in my =TAN(HFOV/2), the 2 was in radians and not degrees. Changing the formula to =TAN(HFOV/degrees(2)) solved the problem.

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I am having some trouble finding a standard way to calculate the horizontal and vertical field of view of an image based on a set of initial input parameters.

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where \$h\$ is the horizontal sensor dimension, \$f\$ is the focal length of the lens, and \$\mathrm{WD}\$ is the working distance. The horizontal FOV is also in millimeters, for which I calculate.

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The first Edmund is the easiest. Just similar triangles. It converts to: \$ f\,/\,h = \text{WD }/\text{ HFoV}\$. So the mm units on the left cancel out, and the feet units on the right cancel out, and no units, and no trig, no need for half angle. Works just as is, rearrange at will.