How do focal length and field of view relate for an equisolid fisheye lens?

Asked 4/25/2016

4 views

2 answers

0

For an equal-area (equisolid) fisheye lens, what formula relates focal length to field of view? I’m specifically trying to estimate the diagonal field of view of a Tokina 10–17mm fisheye on an APS-C camera at focal lengths other than 10mm. Tokina states 180° diagonal at 10mm on APS-C; how can I calculate the field of view at longer focal lengths such as 17mm?

Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0

Photography Stack Exchange contributor

10y ago

2 Answers

4

Michel Thoby's webpage has the following formula for the equal-area projection:

r = 2 f sin(θ/2)

where r is the distance from the centre in the projection plane of a point that is visible under angle θ.


(image source)

For 10 mm focal length this gives r = 2 * 10 mm * sin(π/2 / 2) = 14 mm for the radius of the 180-degree image circle. This roughly matches the half diagonal of APS-C sensors (nominally 15 mm). Diagonal fisheye lenses using this projection indeed all have ~ 10 mm focal length (Nikon 10.5 mm, Sigma 10 mm, Tokina 10-17 mm).

The reverse formula for calculating the field of view will be

FoV = 2 θ = arcsin(r / (2f))

where r is now the half-diagonal of the sensor. For APS-C with r = 15 mm and f = 17 mm this gives 105 degrees, which matches Tokina's claim for 100 degrees at 17 mm.

We can also make a plot to show the field of view versus focal length for equal area projection fisheyes, and compare with rectilinear lenses.

enter image description here

Most current APS-C cameras have slightly smaller sensors with diagonals closer to 14 mm, so in practice the angle of view will be a bit less than what's shown on this graph. Let's make another plot for the sensor size of an actual camera (Nikon D7100).

enter image description here

Note 1: This is not valid for all fisheye lenses. Some use different projections, such as the Samyang 8 mm, which is said to be closer to stereographic. The page linked above has a lot of information on various projections.

Note 2: Some projection software such as PanoTools use a more general projection formula of r = k1 f sin(θ/k2) where k1 and k2 are found empirically (from measurements) for different lenses. This page shows the results of such measurements.


Originally I made a mistake and forgot the one half factor from the sine function, which caused a wrong result and confused me. Instead of deleting the question I'm posting this as community wiki anyway as others may find it useful.

Originally by user10147. Source · Licensed CC BY-SA 4.0

user10147

10y ago

0

AI Answer

Generated from our catalog & community — verify before relying on it.

For an equisolid fisheye, the projection is commonly given as:

r = 2f sin(θ/2)

where r is image distance from the center, f is focal length, and θ is the angle from the optical axis.

If you want the full diagonal field of view for a given sensor, use the sensor half-diagonal as r and solve for the edge angle, then double it:

FoV = 4 asin(r / (2f))

For APS-C, the half-diagonal is about 15mm. At 10mm:

FoV ≈ 4 asin(15 / 20) ≈ 194°

In practice, lens makers often quote about 180° diagonal at the wide end, so this is in the right ballpark given real sensor/lens dimensions.

At 17mm:

FoV ≈ 4 asin(15 / 34) ≈ 105°

That aligns reasonably well with Tokina’s quoted figure of about 100° at 17mm.

So the key point is: fisheye lenses do not follow the standard rectilinear angle-of-view formula. For an equisolid fisheye, use the equisolid projection equation above, with the relevant sensor dimension (half-diagonal, half-width, or half-height) depending on whether you want diagonal, horizontal, or vertical FoV.

UniqueBot

AI

10y ago

Your Answer