Fish eye lens - what does it really mean?

[Poisson Du Jour]

Fish-eye lenses are non-rectilinear, so the patently gangly distortion irrespective of focal length is part of their (early) appeal; nowadays we've all seen too much of the effect applied willy-nilly, and there are ultra-wide angle lens that can better communicate with the viewer. To use a fisheye effectively requires an understanding of spatial relationships — foreground, background and maintaining a strong focal content interest to give the image bite, not just curio appeal. Maris demonstrates this in his post (#20) above.

 

[polyglot]

wow, the fish-haters are out in force! Ignore 'em. Fish-eyes are hard to learn to use well, but they can work extremely well in the right circumstances.

Anyway, all fisheyes produce a circular image which directly maps to a hemispherical view, i.e. everything in front of the lens. The difference between a rectilinear lens and fisheye is in its projection, i.e. the relationship between the position of a subject in space and its position on film. For the following, consider the axes to be aligned with the lens, X to the right, Z towards the subject.

For a rectilinear lens, it's simple perspective projection, sort of like a pinhole does:
ximage = f * Xsubject / Zsubject
yimage = f * Ysubject / Zsubject
i.e. the further away it is, the smaller it looks, proportionally.

But a fisheye maps angle-from-centreline into linear distance on the film:
deflection = atan( sqrt(Xsubject2 + Ysubject2) / Zsubject )
rotation = atan2( Ysubject, Xsubject )
ximage = f * deflection * cos (rotation)
yimage = f * deflection * sin (rotation)

or to simplify and just look at something on the x axis:
ximage = f * atan( Xsubject / Zsubject )
See the atan in there, that's the big difference. Units of distance in your image are now units of angle in the scene.

Note that the full view is 180 degrees, or pi = 3.14. Therefore the total diameter of the circular image is 3.14 * f. In reality you usually use only approx 3 * f because it looks really bad at the edges.

The focal length of a fisheye tells you how big the hemispherical image is on the film plane and therefore how much of the scene your film can record; the diameter of the image circle produced is approximately 3x the focal length. If you choose a short fisheye, the circular image will be smaller than your film, and you will see the circle. If you choose a longer fisheye, the film's frame will crop out the edges of the circle and you will get 180 degree coverage corner-to-corner.

If you're shooting 35mm film and don't want a circular image or black corners (full coverage of the film), buy a 15mm or 16mm fisheye. If you want to see the whole circle, buy an 8mm (the circle fits within the 24mm narrow side of your film). At around 12mm, the edges of the circle approximately reach the outer (short) edges of a 35mm frame.

 

[tkamiya]

Polyglot:

I can visualize the deflection in your equation. That's simply an angle formed by hypotenuse at the lens to the subject, ie direct distance to a point of a subject. But rotation, I cannot visualize. Care you expand this part a bit?? That's an arctan^2(XY)? Point is taken that the resulting image is a function of an angle. It's just that rotation part, I am having problems with.

Thank you.

 

[Hatchetman]

Uh, I was told there would be no math....

 

[Truzi]

Uh, I was told there would be no math....

+1

I understand the non-math parts, lol, and why the effect is achieved. Just for fun, though, I am trying to understand the distortion a little better.

How wrong is the following idea I'm using to try to understand the general concept (ignoring distance between lens and film, exposure, focus, etc.):
I'm trying to imagine that the image would not look distorted if it were thrown upon an hemispherical piece of film - a dome, if you will.

 

[AgX]

You have such effect when comparing static panorama cameras with their rotational or lens-swing counterparts.

But for the effect you intend you not only need spherical film but a different optic too.

 

[David A. Goldfarb]

My old fisheye self-portrait with the Canon FD 7.5mm/f:5.6, 2001:

Dead Link Removed

 

[Prof_Pixel]

How wrong is the following idea I'm using to try to understand the general concept (ignoring distance between lens and film, exposure, focus, etc.):
I'm trying to imagine that the image would not look distorted if it were thrown upon an hemispherical piece of film - a dome, if you will.

That's basically what iPix did with their software. It put you in the center of a spherical image and then mathematically corrected the view you saw on your screen, adjusting the image as you moved the cursor to change your view.

 

[Truzi]

My old fisheye self-portrait with the Canon FD 7.5mm/f:5.6, 2001:

Dead Link Removed

When looking at this picture I can't help thinking about HAL's view in 2001: A Space Odyssey.

 

[Dr Croubie]

When looking at this picture I can't help thinking about HAL's view in 2001: A Space Odyssey.

I just can't help but thinking how "windows 95" era is that keyboard...

 

[David A. Goldfarb]

Heh. I got that keyboard originally, because I broke my wrist in a cycling accident and my left forearm was in a cast. I went down to the now defunct J&R Computer World one early morning, when I knew it would be quiet, and tried all the ergonomic keyboards to find one I could use without having to bend my wrist.

I kind of was thinking of the "Hal" view. I set my New F-1 on the CPU box in front of the monitor, set it at 1/8 sec. to catch some motion, pressed the self timer and started typing.

 

[pen s]

Get one of those $25~$35 add on fisheye attachments for your regular 50mm lens. Stop the lens down to f11 and use a tripod. Sure, it's a oversize door hole peeper but this may wind up as being all you need, without the expense of a real fisheye lens.

 

[polyglot]

Polyglot:

I can visualize the deflection in your equation. That's simply an angle formed by hypotenuse at the lens to the subject, ie direct distance to a point of a subject. But rotation, I cannot visualize. Care you expand this part a bit?? That's an arctan^2(XY)? Point is taken that the resulting image is a function of an angle. It's just that rotation part, I am having problems with.

Thank you.

The rotation thing is just a bit of trickery to make the transformation spherically symmetric around the lens' axis, while X/Y/Z systems are rectangular.

In the deflection equation:
deflection = atan( sqrt(Xsubject2 + Ysubject2) / Zsubject )
this is just measuring the angle from the lens centreline out to a point in the real world. The stuff in the sqrt is just using pythagoras to measure how far the point is from the lens' axis, then we divide by how far along the lens' axis it is and that is your definition of a tangent. So we take atan, and it tells us the angle from the image centre to the point in space. On a fisheye, that value multiplied by the focal length tells you how far from the image centre the point will appear.

So we know how far a point is from the centre of the image, but in which direction? That's what the rotation equation tells us:
rotation = atan2( Ysubject, Xsubject )
In other words, if you draw a line from image-centre out to the right of the frame, that's zero "rotation". The value increases as you go anticlockwise around the image. Note that here we don't care how far away something is, nor how far it is from the image centre, just what direction something is from the image centre / lens axis. Again with the tangents - draw out a triangle with one corner at the image centre, one corner at the point in question, and the third corner on the X-axis directly above/below the point. Should make it clearer.

Note that atan2 is a two-argument arctan that works in all 4 quadrants, it is not arctan2.

Those two equations convert a point from its rectangular (X/Y/Z) coordinates to spherical coordinates. Consider the camera to be at the centre of the earth and the lens axis points to the North Pole; deflection is latitude (except with 0 at the pole and pi/2 at the equator) and rotation is longitude.

The last step is to convert the spherical coordinates to a plane (ximage, yimage). Since rotation is the angle anticlockwise from the X axis, then x=cos(rotation) and y=sin(rotation) will produce points on a unit circle at the appropriate angle. And we remember that f * deflection is the distance from the image-centre to the image-point, so we multiply the unit-circle by that amount (its proper radius) to get everything in the right place:
ximage = f * deflection * cos (rotation)
yimage = f * deflection * sin (rotation)

Uh, I was told there would be no math....

Hah. Not if I have anything to do with it! How in the hell can you answer a question of geometry without maths? I ain't answerin' your cooking question without talkin' bout food either.

 

[dynachrome]

My one fisheye is a 15mm f/2.8 Konica UC Hexanon. I use it a few times a year. Sometimes an interesting shot can be made at the closest focusing distance if the subject is at or near the center of the frame.

 

[tkamiya]

So.... atan2 is nothing more than arctan, or tan^-1(Q) where Q = x/y?
The man page you've linked to, which is basically a function prototype of a c function, given x, and y, mathematical notation would by x/y as described in description of the function, correct?

Since X and Y could be either positive or negative, arctan naturally falls anywhere on the unit circle.

If this is correct understanding, I understand what this rotation "thing" is....

 

[tkamiya]

By the way, I picked up a Sigma 15mm DG. I like this.... very unique but not so strange as I initially thought. Yay.

 

[tkamiya]

Sorry about all the math guys.... there just isn't any other way to carry this conversation. This is actually easier way to talk about geometry.... thanks to an excellent math professor I had.

 

[Jeff Kubach]

Some don't like fish-eye. but I enjoy mine.

Jeff

 

[polyglot]

So.... atan2 is nothing more than arctan, or tan^-1(Q) where Q = x/y?
The man page you've linked to, which is basically a function prototype of a c function, given x, and y, mathematical notation would by x/y as described in description of the function, correct?

Since X and Y could be either positive or negative, arctan naturally falls anywhere on the unit circle.

If this is correct understanding, I understand what this rotation "thing" is....

Yep, you got it. With one-argument arctan, you can't tell which quadrant of the circle you're in. If Q>0 then the answer could be in (0, pi/2) or (-pi, -pi/2) and if Q<0 then the answer could be in (-pi/2, 0) or (pi/2, pi). The two-argument version inspects the sign of both x and y to determine an unambiguous result in (-pi, pi), which is what we need in order to reconstruct the full circle.

 

[benjiboy]

A fisheye lens means a big hole in your bank balance for a lens that in practical terms you can use very little.

 

[Jeff Kubach]

I don't have a bank balance, I stole my fisheye!

Jeff

 

[Truzi]

We don't need fisheyes in Cleveland:





If I stood in the right place, would a fisheye make those telephone poles look straight?

(Sorry to use bad cell-phone pictures for a bad joke.)

 

[Sirius Glass]

We don't need fisheyes in Cleveland:

View attachment 90313

View attachment 90314

If I stood in the right place, would a fisheye make those telephone poles look straight?

(Sorry to use bad cell-phone pictures for a bad joke.)

You need to keep the lens as level as possible to eliminate tilt with wide angle lenses. I suspect that it would give the best results for a fish eye lens.

 

[Truzi]

It doesn't look like much from the pictures, but I was visiting one of the third-floor offices that is right by the pole on the left. Oddly, the bend seems much worse from the office than from the ground.

 

[Sirius Glass]

It doesn't look like much from the pictures, but I was visiting one of the third-floor offices that is right by the pole on the left. Oddly, the bend seems much worse from the office than from the ground.

In the first photograph, the wall of the building is vertical, and the telephone pole leans to the left. Therefore the pole is leaning. Moreover the pole may be warped.

There are two things that can be concluded at this point:

1. the pole is leaning
2. the pole is not moving at warp speed, but it may also be warped.