Nowadays, there are so many different image size formats in cameras that it is difficult to know what a particular focal length corresponds to. Many camera manufacturers will give you actual focal length, along with a “35 mm equivalent” focal length, and that helps, but it is mighty unwieldy. Has anyone ever come up with a unitless way of describing focal length, one that automatically corrects for differences in image size? f-stops take care of this for apertures; why can’t we have one for focal length?
I suspect you could use image magnification, but that’s not very intuitive.
You could also use Field of View in degrees, but once again, that’s pretty hard to visualize. I use FOV a lot for astrophotography.
It’s still necessary to calculate it for all lens/sensor pairs - there’s never going to be a way to describe what you want independent of the camera.
Well, Field Of View is pretty much what the OP is asking for, as it accounts for both the focal length and the sensor size and produces a single metric, where:
FOV = 2 arctan({SensorSize}/2{FocalLength})
It’s just not used very much in photography AFAIK because as you say it’s not very intuitive. But focal length is a very specific thing, an actual distance representing where the lens focuses parallel rays from a theoretical infinite distance. In my experience digital photographers with cameras that have less than full-frame sensors just get used to a different understanding of what focal length means to them than users of full-frame digital cameras or 35mm film cameras.
Why do you find it unwieldy? It’s a pretty easy conversion, and nearly all photographers can visualize the 35mm equivalent instantaneously.
A normal lens for 35mm film is considered to be 50mm focal length. The diagonal measurement of the image is 43mm, so the normal lens is 50/43 = 1.162. Substitute the focal length dimension for telephoto, wide angle, etc. Don’t these ratios hold true for all formats?
Dennis
FOV is clearly the right answer, as it copes with extremes better.
However the only reason f numbers are constant with the lens is that, well they are only a function of the lens. FOV is impossible to separate from the sensor size, so it is intrinsically impossible to give a number that conveys anything more useful than the focal length of the lens. Any lens design has a maximum field of view (which directly translates to the size of sensor it can cover without vignetting), that comes from physical limitations of the optics and how much you want to spend, but there isn’t a minimum. You can use as small a sensor as you please, and the FOV will drop accordingly.
I use a lot of medium format, and here the effective field of view is again different. You get used to mentally translating the FOV of focal lengths in your head all the time. If you know the diagonal size of the film/sensor you can cope with pretty much anything.
Are we using diagonal or image width? (technically, maximum picture width if portrait mode) Of course, some cameras have assorted aspect ratio options (as does the iPhone for rectangular or square) so even then, the width or diagonal is not fixed except in the largest (full sensor) format. There’s always a gotcha, which is why the 35mm equivalent is a better option.
I find it unwieldy because that’s a lot of numbers to etch onto the lens. You generally have to dig into the background material to find the 35 mm equivalent.
That’s what I envisioned; some sort of “normalization” function where larger numbers indicated telephoto and smaller numbers indicated wider angle. I was curious if anyone had ever done that, or if there was a name for it.
Thanks for all the responses, gang!
Eh? It works with binoculars. And that’s generally what you’re using when you talk about effective focal length. Most people divide by 50 (the “normal” magnification) to get a “times” zoom factor.
For wide angle lenses, I think it is more common to think of field of view, but it’s not a very useful metric for telephoto lenses. No one cares if the owl in the tree a mile away subsumes 0.5 degrees (in fact, that could be an owl the size of an airliner for all I know). You mostly care about how much closer the lens seems to bring the owl.
Actually, “35mm equivalent” is just as ambiguous when we are dealing with different aspect ratios. Is a “35mm equivalent focal length” one that provides the same image width, or the same diagonal FOV?
Magnification and focal length are not equivalent. Focal length is a conversion factor between angular size and physical size of the image. Magnification is a ratio between two angular sizes (e.g. true object size vs. object size produced by a telescope) or two physical sizes (e.g. original document and a photocopy).
Just as a side thread, FYI–for a non-photographer to begin with, even–FOV on first principles (“we’re going to need a bigger camera…or a bigger pair of eyes”) is gently introduced in Could the “visible light” images of some galaxies look the same to us as the Hubble?, where, as first replier here beowulff noted, the concept works most efficiently.
A standard* lens is one which gives the same angle of view as a human eye. It should be possible to hold up a print, at the most natural viewing distance, in front of the scene it depicts and have the print match up with the real scene. It has been ascertained by observing large numbers of people that the natural viewing distance for a print of reasonable size tends to be very close to the print’s diagonal. By simple geometry it can be shown that the desired field of view therefore requires the distance between lens and film/sensor to equal the format’s diagonal. Focused on a distant subject the focal length should therefore equal the diagonal.
For the mass market however, the normal print size is rather small and viewing at the “optimum” distance would be uncomfortably close for most people’s eyesight. As a result the somewhat longer 50mm focal length became the standard.
*“standard lens” is the usual terminology, not “normal”
For a camera with a built in lens, you just need the 35mm equivalent. For a camera with a detachable lens, you just need the actual focal length and you should know the multiplier for your particular camera.
Astrophotography? Cool!
Do you have any of your photos online anywhere?
At least for all the old photographers that learned how on film. In another few years that’ll be the minority though. I would not be surprised to see the 35mm format DSLR start being a different size and shape. Yes, there’s some backwards compatibility lock-in. But it’s not totally insurmountable.
The 60 to 1 rule is a quick and dirty tool we use in aviation for these kinds of calculations.
To a couple decimal places, 1 degree subtends 1 part in 60. As in a one degree change in angle will accumulate a 1 mile displacement over 60 miles. Up/down or left/right doesn’t matter.
The other thing is we use nautical miles that are about 6000 feet. So 1 degree in 1 nautical mile = 1 degree in 6000 feet = 100 feet.
For ballpark work a 6000 foot mile is close enough to a 5280 foot mile. Or apply a 10% fudge factor to get a tighter estimate.
So …
An owl subtending 0.5 degrees at 1 nautical mile is about 50 feet tall. A 0.5 degree owl at one statute mile is about 45 feet tall. On his scale you are the size of a smallish mouse vs. a typical owl. Be afraid. Be very, very afraid. :eek:
I’m not very accomplished, but I do like taking “astro" landscapes, like these:
In practical terms, 35mm-equivalent seems to be common, and as close to a universal standard as we’ll get for consumers (as opposed to engineering and hardcore photography geeks).
I’m currently lusting after a Fujifilm X100F, which has a fixed prime lens. It is widely described as being a ‘35mm’ camera and is often discussed in those terms, although the real focal length is 23mm (with a 1.5x crop factor).
Very cool photos.
In the last one, what comet is that?
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That was C/2011 L4 (PANSTARRS). The photo was taken west of Lake Pleasant on AZ 74.