What is the relationship between the focal length and field of view for an equal-area ("equisolid") projection fisheye lens? Alternatively, what is the relationship between the 180-degree image circle size and the focal length for such a lens?
Ultimately I want to find the field of view of a Tokina 10-17 mm fisheye zoom at focal lengths other than 10 mm on APS-C (at 10 mm Tokina claims 180 degrees diagonally).
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.
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.
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).
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.
2 arctan(r/f), where r = 15 mm is the half-diagonal of APS-C.
\$\endgroup\$
The angle of view calculation requires trigonometry: This is the formula I use in Excel.
=((ATAN((d/2/f)))x180/9.8596
d=the dimension of the frame 35mm full frame 24mm height by 36mm length diagonal 42.36mm f=focal length x=multiply
10mm = 100⁰ height – 121⁰ length 127⁰ diagonal
12mm = 90⁰ height – 113⁰ length 120⁰ diagonal
14mm = 81⁰ height – 104⁰ length 114⁰ diagonal
17mm = 70⁰ height – 93⁰ length 104⁰ diagonal
For the compact (APS-C) 16mm height by 24mm length by 30mm diagonal
10mm = 77⁰ height – 100⁰ length 110⁰ diagonal
12mm = 67⁰ height – 90⁰ length 100⁰ diagonal
14mm = 60⁰ height – 81⁰ length 82⁰ diagonal
17mm = 50⁰ height –70⁰ length 81⁰ diagonal
r for a point from the image centre is r = 2*f*sin(θ/2), where f is the focal length and θ is angle measures from the optical axis for that point. But it doesn't add up. Equal-area projection lenses that have a 180 degree diagonal FoV tend to have ~10 mm focal length (like this Tokina or the Nikon 10.5 mm). According to this formula they'd have a 20 mm radius 180 deg image circle. But the half-diagonal of APS-C is only 15 mm...
\$\endgroup\$
