How do you calculate vertical and diagonal angle of view for an equidistant lens from horizontal AoV and aspect ratio?
Asked 7/27/2018
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For an equidistant fisheye lens, the image mapping is linear in angle: r = f·θ, where r is image height from the center, f is focal length, and θ is the off-axis angle.
Given the horizontal angle of view and the sensor aspect ratio A = width/height, how can you calculate the vertical and diagonal angle of view?
I’ve seen claims that, for equidistant lenses, horizontal and vertical angle of view scale linearly with aspect ratio. I’d also like to know the exact relationship, not just an approximation.
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
7y ago
2 Answers
2
For lenses with an equidistant mapping function, the mapping function is given by
r = ƒ∙θ
where
- ƒ is the focal length of the lens;
- θ is the angle of an object from the lens's optical axis; and
- r is the linear distance of the image of that object from the center of the camera's sensor.
You state that you know or are given the horizontal angle of view of the camera+lens system (we'll call it θh), and the aspect ratio (horizontal:vertical) of the imaging system, A = h/v. So, the horizontal measurement of the sensor is θh∙ƒ units (probably millimeters).
The vertical dimension of the sensor is just the horizontal dimension divided by the aspect ratio. So the vertical angle of view is just the horizontal angle of view divided by the aspect ratio:
θv = θw / (A)
The length of the diagonal of the sensor is found from the Pythagorean theorem:
d = √(h² + v²)
= √(h² + (h/A)²)
= √(h²(1 + 1/A²))
= h ∙ √(1 + 1/A²)
Because of the equidistant mapping function, the diagonal angle of view is just √(1 + 1/A²) times the given horizontal angle of view.
I saw here that the relation between horizontal and vertical angle of view is linear with the aspect ratio. By the way, what is the exact relation (that website specifies is approximated)?
The relation is exact, to the extent that a particular lens is described by the equidistant mapping function.
Originally by user11924. Source · Licensed CC BY-SA 4.0
user11924
7y ago
0
Generated from our catalog & community — verify before relying on it.
For an equidistant lens, image radius is proportional to angle: r = f·θ. That makes the sensor dimensions directly proportional to the corresponding half-angles of view.
If the aspect ratio is A = width/height and you know the horizontal angle of view θh, then:
- vertical angle of view: θv = θh / A
because the sensor height is width/A, and angle scales linearly with image distance for an equidistant projection.
For the diagonal, first get the sensor diagonal from width and height:
- diagonal = width · √(1 + 1/A²)
Since angle is also linear with image distance, the diagonal angle of view is:
- θd = θh · √(1 + 1/A²)
Equivalently, using vertical angle instead:
- θd = θv · √(A² + 1)
So unlike rectilinear lenses, where tangents are involved, equidistant lenses use a simple linear scaling between image size and angle. That means horizontal, vertical, and diagonal AoV relate by the same geometry as the sensor dimensions themselves.
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