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Field of View

Fisheye and Rectilinear Lenses by Bob Atkins, 2004

With the advent of digital cameras having non-standard sensor sizes there seems to be quite a lot of confusion concerning focal length, field of view and digital multipliers and how they relate. This article is intended to try to clear up some of that confusion.

First let's define a few terms:

Focal length: The focal length of a lens is defined as the distance from the optical center of a lens (or, the secondary principal point for a complex lens like a camera lens) to the focal point (sensor) when the lens is focused on an object at infinity. It's a primary physical characteristic of a lens which can be measured in an optical lab. It remains the same no matter what camera the lens is mounted on. A 7mm focal length lens is always a 7mm focal length lens and a 300mm focal length lens is always a 300mm focal length lens

Field of View: The field of view of a lens (sometimes called the angle of coverage or angle of view) is defined as the angle (in object space) over which objects are recorded on the film or sensor in a camera. It depends on two factors, the focal length of the lens (see above) and the physical size of the film or sensor. Since it depends on the film/sensor size it's not a fixed characteristic of a lens and it can only be stated if the size of the film or sensor it will be used with is known. For a lens used to form a rectangular frame, three fields of view are often given; the horizontal FOV, the vertical FOV and the diagonal FOV

Digital multiplier: Digital multiplier is a term that's come into use with the increased use of digital cameras with a sensor smaller than the frame size of a 35mm camera. Since the angle of view of a lens depends on both the focal length of the lens and the size of the image you can define a "digital multiplier", which is the factor by which a lens's focal length would have to be increased to give the same angle of view was the lens has on a digital sensor. For example, a 100mm focal length lens mounted on a digital camera with a "1.6x" multiplier sensor has the same field of view on that camera as a 160mm lens would have when mounted on a full frame 35mm camera. It's still a 100mm focal length lens, but it acts like a 160mm lens would on a full frame camera.

What we're really most interested in from a photographic viewpoint is the Field of View. If we want a wideangle shot we want a wide field of view (say, 84 degrees horizontal). If we want a "normal" shot we want a "normal" field of view (say, 40 degrees horizontal)  and if we want a telephoto shot, we want a narrow field of view (say 6.5 degrees horizontal). For those used to thinking in terms of 35mm cameras these would correspond to lenses with focal lengths of 20mm, 50mm and 300mm respectively. However for 4x5 camera users, they'd think in terms of  a wideangle 80mm lens, a 200mm normal lens and a 1200mm telephoto lens. So again, FOV isn't determined by focal length, it's defined by focal length AND format size. That's why when we come to APS-C format digital SLRs (with a sensor approximately 15mm x 22mm) the wideangle lens is now 12.5mm, the normal lens is now 32mm and the telephoto lens is now 188mm. Note that these number are the same as the 35mm numbers divided by a "1.6x digital multiplier" (or in this case, a "1.6x digital divider").

Rectilinear and Fisheye Lenses

There are two types of lens you'll find in photographic use.

The first is the rectilinear lens, This is the typical lens which renders all straight lines in the subject as straight lines in the image (see diagram below). It's pretty much the way our eyes see things and it's exactly the way a pinhole cameras sees things. For normal and telephoto use, a rectilinear lens is ideal, however for extreme wideangle use it isn't. Objects near the edges of the frame in very wideangle shots are "stretched". It's also impossible to make a rectilinear lens with 180 degree (hemispheric) coverage. In fact it's very difficult to make a rectilinear lens with more than about 100 degrees of horizontal coverage

The second type of lens is the fisheye lens. A fisheye lens renders straight lines which don't run through the center of the frame as curved (though lines running through the center remain straight). Objects at the edges of the frame are not stretched, but they are distorted. It's easy to make a lens with a diagonal coverage of 180 degrees ("full frame fisheye") or even with a horizontal, vertical and diagonal FOV of 180 degrees ("circular frame fisheye") - though this results in a circular image with the rest of the frame dark. Fisheye lenses were first made for scientific use, since with hemispherical coverage they can image the entire sky on a single frame and so were useful for astronomical and meteorological studies. The first "fisheye" camera was a pinhole camera that was filled with water, but luckily technology has come up with more convenient ways to make fisheye images!

lenses1a.jpg (14698 bytes)

The illustrations above show the pinhole model of rectilinear and fisheye lenses. In a fisheye lens wide angle rays are bent in more towards the center of the frame. To do this with real lenses a very large, very strongly curved negative front element must be used as shown in the lens diagrams below:

8mmf28optic.jpg (11776 bytes)     raytrace.jpg (15565 bytes)


Calculating Field of View

Rectilinear Lenses

The field of view of a rectilinear lens is very easy to calculate using simple trigonometry. It's given by:

FOV (rectilinear) =  2 * arctan (frame size/(focal length * 2))

Here "frame size" refers to the dimension of the frame in the direction of the FOV, so for 35mm (which is 24mm x 36mm), frame size is 36mm for the horizontal FOV, 24mm for the vertical FOV and 43.25mm for the diagonal FOV.

Fisheye Lenses

The situation is rather more complex for fisheye lenses because there is no such thing as a "Fisheye" equation. Instead there are several different "mapping equations" or "projections" which different fisheye lens manufacturers have used.

Probably the most common is the equisolid angle projection, and the FOV is given by:

FOV (equisolid fisheye) = 4 * arcsin (frame size/(focal length * 4))

Also popular is the equidistance projection, and for the the field of view is given by:

FOV (equidistance fisheye) = 2 * (frame size/focal length)*57.3
(the 57.3 is to convert from radians to degrees).

Less common are the orthogonal projection which gives the following field of view:

FOV  (orthogonal fisheye) = 2 * arcsin (frame size/(focal length *2)

and the stereographic projection which gives:

FOV (stereographic fisheye) = 4 * arctan (frame size/(focal length * 4))

Of course just like the fact that rectilinear lenses are rarely ever truly rectilinear (they suffer from barrel and pincushion distortion), so fisheye lenses usually don't follow the exact mapping suggested by these equations. This is generally of no consequence unless you are trying to do scientific studies involving precise conversion of points in a fisheye image to  "real world" coordinates.

You can think of the various rectilinear and fisheye projections as being somewhat analogous to map projections. We all know the earth is a sphere, but we can represent it on a rectangular map with horizontal and vertical straight lines representing latitude and longitude using a Mercator projection. This could be looked on as an analogy to a rectilinear lens mapping. However just like a rectilinear lens tends to stretch out objects at the edge, such a map projection stretches out areas near the poles. Fisheye lens projection would then correspond to various map projections where latitude and longitude lines are no longer straight, but where, say, areas are proportional, such as azimuthal equal area. Each mapping scheme distorts "reality" in some way.   We're more used to seeing one than the other, so with think of one as "normal" and one as "distorted", but that's not strictly true.

maps.jpg (18342 bytes)

The plot below shows how the field of view relates to frame size for a given focal length lens  for a rectilinear lens and four types of fisheye lens. As you can see, the rectilinear lens can never get to a 180 degree FOV,  no matter how large the frame size, but all the fisheye lenses can. You can also see that for all the lenses, the field of view increases with frame size.

map3.gif (9587 bytes)

C and D are equidistance and equisolid angle fisheyes respectively (most common)
B and E are stereographic and orthogonal fisheyes respectively (little used)

Note that you can't just take any lens and use a very large frame to get a wide field of view. Lenses have an image circle which is the diameter of the largest image that the lens can form. Outside that diameter the lens vignettes, cutting off the image due to the limited size of the optical elements or other characteristics of the design.. Lenses designed for use on full frame 35mm cameras must be designed to have an image circle of at least 43.5 mm, since the diagonal dimension of the 35mm frame is 43.25mm. It's very difficult to make short focal length lenses with large image circles.


Using the above information we can calculate, for example, the field of view of a full frame fisheye lens designed for 35mm use when used on an APS-C camera. Lets take the example of a 15mm  fisheye lens. Let's assume it uses equisolid angle projection, so the FOV is given by 4 * arcsin (frame size/(focal length * 4)).

For a 24 x36mm frame this gives a horizontal FOV of  147.5degrees, a vertical FOV of 94.3 degrees and a diagonal FOV of 185 degrees. Canon give numbers of 142, 92 and 180 for their 15/2.8 fisheye lens, so the mapping isn't exactly equisolid angle, but it's a typical full frame fisheye with approximately 180 degrees diagonal coverage

For a 22.7 x 15.1mm sensor (APS-C) the numbers become: Horizontal FOV = 88.9 degrees,  Vertical FOV = 58.3 degrees, diagonal FOV = 108.1degrees. If you "defish" a fisheye image, i.e. convert the image to rectilinear mapping, you keep the horizontal and vertical FOV, stretch the edges of the image and reduce the diagonal FOV. So if you "defished" the image you'd get an image with approximately and 88 degree horizontal FOV and a 58 degree vertical FOV. This corresponds to the horizontal FOV of a 19mm lens and the vertical FOV of a 22mm lens. How is this possible? Well the 1:1.5 ratio of vertical to horizontal if the APS-C sensor is changed when the image is "defished" and becomes closer to 1:1.7


All original text is (©) Copyright 2004   Robert M. Atkins [www.bobatkins.com]   All Rights Reserved

Article created 2004

Readers' Comments

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William Nicholls , November 15, 2004; 12:16 P.M.

Correct me if I'm wrong, but "field of view" relates the angle of view to subject distance. Angle of view is the combination of focal length and image frame size. Field of view would have to be calculated on the spot for a given shot, angle of view is the measurement that's useful for format plus focal length comparisions.

Bob Atkins , November 15, 2004; 01:25 P.M.

It depends what you're talking about. For binoculars "Field of View" is usually given as "abc ft at xyz yards". For camera lenses (and astronomical telescopes) the term "Field of View" is often used in place of the possibly more technically correct "Angle of view". It should more properly be called "Angular Field of View", but it usually isn't. By similar reasoning, binocular Field of View might more properly be called "Linear Field of View".

Kjeld Olesen , November 16, 2004; 10:47 A.M.

Clearly, field of view and angle of view are NOT synonyms. Angle of view is quite obvious ... the angle taken in by the lens. The field of view is the dimension of the subject relative to the distance to it. I.e. if the subject is 50 meters across and 100 meters away, then the FOV is 50%.


Kjeld Olesen

Bob Atkins , November 16, 2004; 03:06 P.M.

"...that which we call a rose By any other name would smell as sweet..." - Shakespeare (Romeo and Juliet).

..."The amount of sky that you can view through a telescope is called the real (true) field of view and is measured in degrees of arc (angular field). The larger the field of view, the larger the area of the sky you can see..." - Celestron website

"...The field of view is approximately 170 degrees from right to left..." - NASA website describing the cameras on the Mars Rover

"...The Field of View (FOV) is measured in degrees of the horizon. It determines how much of the scene is visible and is directly related to the Focal Length of the lens..." - Webreference.com

"...(Field of View is) the horizontal or vertical angular expanse visible through a camera..." - Apple.com quicktime reference

Don't be so pedantic.

"Field" of View can be linear of angular. Take your pick. "Field" in this context doesn't mean the width of a piece of earth covered in grass...

Alan Bower , November 17, 2004; 10:42 A.M.

Regarding the Digital multiplier you say "For example, a 100mm focal length lens mounted on a digital camera with a "1.6x" multiplier sensor has the same field of view on that camera as a 160mm lens would have when mounted on a full frame 35mm camera. It's still a 100mm focal length lens, but it acts like a 160mm lens would on a full frame camera."

This statement is not absolutely correct. The 100mm lens on the digital camera acts like a 100mm lens shot which has been propotionately cropped.

An important point often foregotten, is that the relative magnification of foreground to background objects changes when you use a different focal length the get the identical HFOV. With a 1.5x camera you can take a full frame rectiliear 16mm shot and it will look very different from a 24mm shot on a full frame film camera when the subject comprises foreground and backgrond objects which are spatially separated. The 16mm shot will cover the same horizontal field but the camera to subject distance must be much shorter and the foreground objects will be much larger than the 24mm shot.

Bob Atkins , November 17, 2004; 12:41 P.M.

My comments in this article relate only to field (or angle!) of view.

The subject of differences between "equivalent" lenses on different size formats with respect to depth of field for example, is covered in articles like "Depth of Field and Digital".

Perspective is determined only by distance from the subject. Cropping doesn't change perspective.

To get the same shot (FOVwise) with a 16mm lens on a "1.5x mutiplier" camera and a 24mm lens on a full frame camera, you need to be at the same distance from the subject.

William Nicholls , November 18, 2004; 11:08 P.M.

It's not really pedantic to be precise when discussing camera optics in a technical article rather than one discussing telescopes focused at infinity or optics like binoculars where "field of view" is based on a specific reference distance.

Most photo calculator software offers separate field of view and angle of view calculations and they provide different information. Field of view is probably most useful for technical large format work (figuring out the lens you need to photograph a large painting in the limited space of a museum, for example). Field of view calculations provide a linear dimension, angle of view is expressed in degrees.

There are surely thousands of misuses you can cite, but that doesn't make another any more correct.

Kjeld Olesen , November 19, 2004; 08:46 A.M.

Digital multiplier:

"... you can define a "digital multiplier", which is the factor by which a lens's focal length would have to be increased to give the same angle of view was the lens has on a digital sensor..."

While correct, this illustrates why we should distinguish clearly between AOV and FOV. A common way to use the cropping factor is to say that when using the x mm lens on a cropping camera, the FOV gets cropped by exactly the cropping factor. The same can not be said for AOV.

I.e., with rectilinear lenses it is the FOV (according to my description above) and NOT the AOV that get cropped by the cropping factor. There is no practical difference here with regard to telephoto lenses, but a significant difference with respect to wide angles. I.e. while the hAOV of a 12 mm lens on full frame is 111 degrees it is 87 on a 1.6x cropping camera, yet the ration 111/87 is only 1.27!!!

However, with fish eye lenses, specifically the C model in the figure, it is the AOV that get cropped.

As to Alan Bower's comment I would like to add another issue about the misuse of the term "acting like", i.e. a 100 mm lens "acting like" a 160 mm lens.

A 100 mm lens requires 100 mm of extension to give 1:1 magnification. A 160 mm lens requires 160 mm of extension to give 1:1 magnification.

So, no matter how you crop the image, the 100 mm lens will still act like a 100 mm lens.

And I must stress that pedantism is in place when discussing technical issues on reputable websites. If the facts are not presented correct, someone is going to refer to it and use it to impose misconseptions on other people, and at the end there will be no common way to express anything.

Alan Bower , November 22, 2004; 11:02 P.M.

I wanted to clarify my recent comments on this topic. The non-full-frame digital sensors provide an image "crop" at the film plane - no more, no less. It doesn't matter for theoretical subjects at infinity, 2-dimensional subjects or distant landscapes. I have done some testing and am convinced that no real problems are created within the range of 35mm to about 100mm. By selecting the approprate focal length and possibly changing the camera to subject distance you can take photographs full-frame or "cropped" which are identical.

However, when you have 3-dimensional compositions with near and far objects and you are using long telephoto lenses or very wide angle lenses the crop factor may be significant. I am particularly concerned about wide angle shots. My favorite lens is a 24mm Nikkor f/2.8 and I like to get close to my primary subject to emphasize the spatial relationship between the principal subject and the background objects. The Nikon 1.5x crop factor equivalent is 16mm.

I have uploaded two images. Both were taken from the same shooting location with the same camera pitch angle. One is a full-frame shot taken with a 28mm f/2.8 Nikkor. The second was taken on film with a 17mm f/3.5 Vivitar which was cropped to 22.5mm x 15.0mm (Canon EOS 20D) giving a 1.6x crop factor.

When you view the two images two things should be evident.

Firstly the 17mm shot is significantly lower quality. If we are to adopt the smaller digital sensors stuffed in a body designed for 35mm full frame then we need lenses to match the format - for wide angle that would be 1.5x or 1.6x primes which match the quality, size and price of the 28mm, 24mm and 20mm lense available for full frame. To-date I don't think there are any 16mm primes which match my 24mm Nikkor.

Secondly, the use of a wider angle lens changes the FOV and, when the camera is pitched, the converging line distortion is more pronounced. In the 17mm sample, the nearby supporting columns of the stucture are flared more at the bottom and the background buildings lean more. All as a result of needing to use a wider angle lens due to the frame crop of the digital sensor.

Let's put more effort towards full frame sensor development - not more slow and expensive super wide angle zoom lenses.

Alan Bower , November 23, 2004; 11:58 P.M.

Here are the images which were left out of my last post.

Image Attachment: sample.jpg

Kevin Yong , November 24, 2004; 07:21 A.M.

I have read Bob's article on DOF and done similar experiments to Alan's and some of his comments and samples reflect what I have found.

Can someone confirm or correct me on these ideas...

It seems it would be IMPOSSIBLE to get 2 IDENTICAL pictures taken with a full frame SLR and a 1.5x DLSR. The only way to get identical images is to use the same focal length and then crop the full frame image.

If one was to use 2 different lenses to get "equivalent" FOV, the DOF of both images would be different.

If the same focal lenth was used but the DSLR image was taken by moving the shooting position back to get the same FOV, then the Perspective would differ. I.e the position of background objects in relation to the subject would differ.

This is not really a problem for most people, including myself but may pose problems for some specific situations. I have a 10D. I don't mind the crop factor, but like Alan, I would like a full frame to regain the capabilities of quality fast wide primes. (I defish a sigma fisheye currently for this purpose.) If there was a good fast 12mm and 15mm primes (even if they were EF-S) then I would not see a need to go full frame.

Ilkka Nissila , November 24, 2004; 10:57 A.M.

Nikon and Canon both make 14 mm f/2.8 lenses which cover 24x36mm. I suspect that they do fine on digital cameras as well. Has anyone compared the quality from e.g. 14 mm Nikkor + D70 with 20 mm Nikkor and a film SLR?

Jeroen Wenting , November 26, 2004; 07:10 A.M.

The important point is that someone who knows what he's talking about (even if he's oversimplifying at places) is finally attempting to debunk in public the marketing myth (which by now is believed by millions of people all over the world and taken as gospel) that a digital camera somehow magically increases the focal length of a lens.

Many digicams are sold with incorrect focal lengths marked on them by even major brands and those that don't usually have a mention of the "equivalent focal length". This is what started the myth probably.

Mark Ci , November 28, 2004; 12:49 A.M.

Alan, I have no idea what you think you see in those shots, but whatever it is is either in your imagination or due to inaccuracies in your setup. Either way, your conclusions are completely incorrect. The apparent size relationships between objects at different distances are fixed by geometry -- specifically trigonometry. Optics are simply not involved. Perspective is therefore entirely determined by shooting position.

Mark Michaelson , January 02, 2005; 12:58 A.M.

Too bad Kodak didn't have the daring to market a full-body "110 instamatic film" SLR 30 yrs ago, or a "disk" SLR 20 yrs ago. If they had, then this whole digital SLR FOV business would have been resolved by now - or perhaps, nobody would have the idea of marketing a 2/3 frame digital SLR today.

christophe kiki , June 05, 2010; 06:37 A.M.


I 've a question regarding the fov process :

you said fov = 2 * arctan ( frame size / (focal lenght*2))

that mean for 50mm : 2 * arctan(43.266 / (2* 50)) and for me the result of this is 0,81668160956 .... this result have no sense, regarding the info I found on the web I should find : 46.7930.

I tcould be kind of you to explain me how do you use your mathematical formula.

thanks a lot.


Sandy Preaux , July 11, 2013; 03:15 P.M.

Although this is an old post here is the answer to the above question:

the value inside the arctan is in degrees not radians.

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