Why are full fstops the numbers they are?

I think the familiar whole stops (well, down to f/90, anyway) are truncated, rather than rounded, to a couple of significant figures. Otherwise, f/22.627 would be written as f/23.

Whole stops, maybe (I've not checked past f/22). It's certainly not that simple for fractional stops, though - f/3.174802 (1/3 stop above "f/2.8") shows as f/3.2, correctly (for nearest) rounding to the larger denominator. But then a third of a stop from that, f/3.563595 gets truncated to "f/3.5".

I'm just going to put it down to historical accident and probably cumulative rounding errors. Sometimes trying to look for a logical explanation doesn't work. :-)

Whole stops, maybe (I've not checked past f/22). It's certainly not that simple for fractional stops, though - f/3.174802 (1/3 stop above "f/2.8") shows as f/3.2, correctly (for nearest) rounding to the larger denominator. But then a third of a stop from that, f/3.563595 gets truncated to "f/3.5".

 

I'm just going to put it down to historical accident and probably cumulative rounding errors. Sometimes trying to look for a logical explanation doesn't work. :)

Yes the conventional way of rounding off doesn't follow any rules. Because if it's simply truncation then I can certainly program my calculator to do that. Whenever I do any programming related to f/stop or shutter speed I have to do the enumeration and look up table.

At the time that our favorite numbers were chosen, meters weren't all that accurate.

Certainly there wouldn't have been much reason for an f/23 over an f/22.

ISO (and ASA values before them) are rounded to 1/3 stop.

Oh, yes. The question to me isn't about telling 22 from 23, it's about 22 being a worse approximation to the correct power of two than 23 is. It's like using 4 as an approximation of pi - it's not so much that you didn't use more precision, it's that even at the precision you're using, you don't have the best answer.

I can see how you get there by successive approximation (sqrt(2) is 1.414... or roughly 1.4; 1.4x2 is 2.8, 2.8x2 is 5.6, 5.6x2 is 11.2 is roughly 11, 11x2 is 22), although arguably we should have ended up with f/7.8 (5.6x1.4=7.84) rather than f/8 if we were allowed to do that.

I do agree that f/22 is within a third of a stop of the "right" (f/22.627...) answer though - roughly 8.92 stops down from f/1.0. f/23 obviously gets closer to the desired 9 stops (9.047...)

Executive summary: things invented before the electronic calculator, especially when used by artists rather than engineers, sometimes had somewhat screwy maths. And a cellphone with access to the internet and a calculator app is a dangerous way to allow me to wake up on a Saturday morning.

Oh, yes. The question to me isn't about telling 22 from 23, it's about 22 being a worse approximation to the correct power of two than 23 is....

Hmm... but "22" is so much

easier to remember than "23"...

http://bayouline.com/o2.gif

Hmm... but "22" is so much

easier to remember than "23"...

Well, I did remember "two little ducks" and didn't remember "the Lord is my shepherd", but that might just mean my interests are more ornithological than ecclesiastical.

23 is the smallest prime number that's not two away from another prime number. It's the number of chromosomes we typically have. 23 is the number of people you need to have before it's likely that at least two share a birthday. It's the lane number that the protagonists used in the Big Lebowski.

22 is relatively boring - it's just twice a prime. The best way I've got to remember it is Catch 22.

Does that help?

Down with 22, up with 23. :-)

Hmm... but "22" is so much

easier to remember than "23"...

http://bayouline.com/o2.gif

I do like 22 better than 23. Once you get used to it, it is hard to change.

Though, not so obvious, as well as I know two stops down from f/22 is f/45. Why not f/44?

It may have been somewhat later that f/45 lenses came out. I think the ones I know that

have it are Nikon lenses designed to use with extension tubes. Maybe MF and LF also have f/45.

I do like 22 better than 23. Once you get used to it, it is hard to change.

 

Though, not so obvious, as well as I know two stops down from f/22 is f/45. Why not f/44?

 

It may have been somewhat later that f/45 lenses came out. I think the ones I know that

have it are Nikon lenses designed to use with extension tubes. Maybe MF and LF also have f/45.

Technically, f/22 is really f/22.623 (sqrt(2)**9). Two stops is sqrt(2)**11, which is 45.2548.

Technically, f/22 is really f/22.623 (sqrt(2)**9). Two stops is sqrt(2)**11, which is 45.2548.

We know but we are discussing if you round f/22.623 to 2 significant digit should it be 22 or 23?

Sorry, Stock-Photos, that's not how it works.

The f-stop describes the ratio of the entrance aperture (the area at the front of the lens which gathers light) to the focal length of the lens. I think of the effect on illuminating the image as standing at the bottom of a well - you get the same amount of light reaching you if the well is twice as deep but also twice as wide (the hole you can see at the top of the well looks the same size).

T-stop is similar, but allows for the amount of light loss passing through the lens due to coatings and internal reflections - so a lens's f-stop of 2.8 might be a T-stop of 3, for example.

Actual results aren't quite that simple - all lenses have a bit of illumination fall-off towards the edge of the image circle, for example.

What you're describing seems to be the ratio between the f-stop and T-stop, assuming your definition of "light going into the front of the lens" already allows for the f-stop aperture.

I hope that helps.

Sorry, Stock-Photos, that's not how it works.

 

(snip)

I think it isn't quite right, but does have some ideas that are often not considered.

First, I think there should be a cos(theta) for the angle that the light hits the film at,

which will be more and more important at larger apertures.

But also, how much light hits the film at a give f/ value compared to how much with no lens at all.

Consider a hemispherical uniform light source, which through a lens should uniformly illuminate

the film. Next illuminate film with a hemispherical source and no lens. At what f/ value are

they equal?

At small apertures, the solid angle of light collected is about pi(d/2)^2/f^2

The largest solid angle that you can collect is 2pi, but pi(d/2)^2/f^2 can, for large enough d,

be greater than 2pi.

Ignoring issues such as the sensor having reduced sensitivity to light arriving at an angle, there's no f-stop that corresponds to capturing all the light for a hemisphere. Or rather, there is, but it's f/0 with a focal length of 0, so the maths says 0/0 - something described in computers as "not a number". Of course you can capture more than 360 degrees of incoming light with a fish-eye lens, but that leads to a much more complicated (and non-standard) definition of "focal length" compared with standard rectilinear projections.

If we take field of view out of the equation, we could look at a point light source, and ask how we can capture a hemisphere of the light being emitted by it and funnel the result to a point on the focal plane. That concept retains a projection that could be considered to have a focal length, but it does lead to the front of the "lens" being bowl-shaped, with the light point at the centre. (If you wanted to capture all of the light even emitted by the back of the source, you could put an elliptical mirror around the subject and focal plane, put the light at one elliptical focus, and the focal plane at the other - but it would only work perfectly for one point.) I'm trying to imagine the optics to get this effect (for half the light) through refraction without wrapping mirrors around the back of the plane, but that's complicated geometry for what, for me, is early in the morning.

it seems to me the relationship between the input and output of a lens is simple, i.e. the amount of light exiting a lens at a given aperture is the same no matter what the focal length. is that correct?

To a good approximation, that's correct - which is why we use f-stops in photography. In reality, a combination of light fall-off across the frame (which tends to be affected by aperture but also the rear nodal point) and optical inefficiency in the lens means it's probably only an approximation. But it's a good enough one to be getting on with.

Thanks Andrew, Briefly, as I know it’s early, is the relationship between input and output

linear

nonlinear (hard)

nonlinear (easy via taylor series expansion)