# Need an Equation for converting Vertical to horizontal FOV

## Main Question or Discussion Point

This is a representation of a square, and a triangle that creates an angle between that square. In this case, the horizontal and vertical components to the Angle of View are the same. Now, change this square into a rectangle, and the relationship to the horizontal and vertical components are changed by a proportion. Let's assume it's a 16:9 proportion. So this should have technically been correct...16/9 (1920/1080) should equal 160/90...but this turned out to give me a not so right answer.

I went and used this calculator, the results of a 90 Degree Vertical FOV yields "121" degree Horizontal FoV. for 1920/1080 pixels.

This calculator is nice and all, but
a) it's not exact enough for the calculation for what i'm doing, and...
b) What am i missing that it yields 90x121 rather than 90x160? I understand that as the AOV's approach 180 degrees, the FOV approaches infinity, so i know the two arn't the same, but it also means that the equation isn't just defined by a proportion.

Before anyone jumps to the comments, i want to point out that no, this situation is strictly mathematical in nature and has no analog to the physical world, or measuring distances in terms of feet or inches.

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What am i missing that it yields 90x121 rather than 90x160?
You are missing the tangent function.

You are missing the tangent function.
hmm okay. But what if the distances are not available to me? The only distance i have available would be the size in pixels, of the box (in our example, 1920/1080).

mfb
Mentor
Then all you can get is a ratio of the tangent function of the angle, but not a ratio of angles.

• hutchphd
Then all you can get is a ratio of the tangent function of the angle, but not a ratio of angles.
But how can that be true? How is it then, that this website has a functioning FOV calculator? There must be some relationship that is based solely on the angles, given a certain aspect ratio. Not saying that a trig function has nothing to do with that, but there must be more to this problem then what your saying.

mfb
Mentor
The "relationship based on the angles" is the fixed ratio of their tangent function values.

##\tan(\beta) = \frac{16}{9}\tan(\alpha)## for suitable ##\alpha##, ##\beta##. Or, solved for one angle, ##\beta = \arctan(\frac{16}{9}\tan(\alpha))##

The "relationship based on the angles" is the fixed ratio of their tangent function values.

##\tan(\beta) = \frac{16}{9}\tan(\alpha)## for suitable ##\alpha##, ##\beta##. Or, solved for one angle, ##\beta = \arctan(\frac{16}{9}\tan(\alpha))##
Okay I see. This helped me out a a lot, it’s essentially a proportion, just a proportion within a proportion (since the Opposite and adjacent lengths can be arbitrary so long as you have a single angle)

Thanks