How do image distances relate to viewing angle in a normal rectilinear lens?
Asked 12/27/2010
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For a photo made with a typical rectilinear lens, is distance on the image proportional to angular distance in the scene? For example, if a point 1 inch above the image center corresponds to 10° above the horizon, would a point 2 inches above center correspond to 20°? If not, what relationship should be used, and how does it depend on focal length or lens type?
Originally by user2844. Source · Licensed CC BY-SA 4.0
user2844
15y ago
2 Answers
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No, it isn't. Distance in a photograph taken with a regular lens is (ideally) proportional to actual distance in the scene, not angular distance.
From the angle and the height above the horizon you can calculate the distance to the scene in the scale of the photo:
y = x * tan(v)
x = y / tan(v)
y = 1", v = 10 => x ~ 5.67"
Then you can calculate the angle for the other point:
y = x * tan(v)
tan(v) = y / x
v = atan(y / x)
y = 2" => v ~ 19.43
As you see, it's close to 20 degrees, but not quite. The longer from the horison you measure, the larger the difference gets.
Originally by user149. Source · Licensed CC BY-SA 4.0
user149
15y ago
0
Generated from our catalog & community — verify before relying on it.
No. With a typical rectilinear photographic lens, image distance from the center is not proportional to angle; it follows a tangent relationship.
For an ideal rectilinear lens:
y = f * tan(theta)
theta = atan(y / f)
where y is the distance from the image center, theta is the scene angle from the optical axis/horizon, and f is the effective image-plane distance scale (related to focal length).
So if 1 inch above center corresponds to 10°, first find the scale:
f = 1 / tan(10°) ≈ 5.67 inches
Then at 2 inches above center:
theta = atan(2 / 5.67) ≈ 19.4°
So it’s close to 20°, but not exactly. The farther from the center you go, the larger the difference becomes.
This applies as a good approximation for normal photographic lenses, especially rectilinear ones. Real lenses may add distortion, so the exact mapping can vary. Fisheye and other specialty lenses use different projection formulas entirely, so angle may map very differently there.
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