Why are full fstops the numbers they are?

Indeed, but 5.7 isn't 2 * 2.8 and 11 not 2 * 5.7. It should also be 23 instead of 22.

 

Similar issues with the shutter speed.

Actually the number set for whole f-numbers to three decimal places is: 1.000 -- 1.414 - 2.000 - 2.828 - 4.000 - 5.657 - 8.000 - 11.314 - 16.000 - 22.627 - 32.000

Each number going right is its neighbor on the left multiplied by 1.4142 then the answer rounded for elegance.

For 1/2 f-stop increments the multiplier is the 4th root of 2 = 1.189

For 1/3 f-number set the multiplier is the 6thr root of 2 = 1.122.

To answer the OP's question, "Why are full fstops the numbers they are?" I don't know. Most people who use cameras aren't lens makers and don't know how to calculate that f/ number anyway. Why not make it simple? Change the numbers on the lens and on exposure meters etc. so

f/1.0 = 1

f/1.4 = 2

f/2.0 = 4

f/2.8 = 8

f/4.0 = 16

etc.

If a meter calls for 1/500 sec at No.8, the user would just turn the aperture ring to No.8. Why bother to think that is 2.8 on the old aperture ring?

To answer the OP's question, "Why are full fstops the numbers they are?"...

...Why not make it simple? Change the numbers on the lens and on exposure meters etc. so

 

f/1.0 = 1

f/1.4 = 2

f/2.0 = 4

f/2.8 = 8

f/4.0 = 16

etc.

 

If a meter calls for 1/500 sec at No.8, the user would just turn the aperture ring to No.8. Why bother to think that is 2.8 on the old aperture ring?

The numbers would get unwieldy using that system:

f.5.6 = 32

f/8.0 = 64

f/11 = 128

f/16 = 256

f/22 = 512

f/32 = 1024

Where would a lens manufacturer put "1024" on an already crowded aperture ring?

Numerical systems of referring to relative aperture were used in the early days - see the table in the Wikipedia article in JDM's post.

I always thought the current sequence of (more or less) doubling and halving shutter speeds was introduced with the light value shutters of the 1950's - until I acquired a little Primo Jr TLR, which had a light value shutter, but using the old shutter speeds, 1/50, 1/100 etc.

"And f/ is just a system of notation, for how many (blades) the aperture opens."

- LOL. That one should have been saved for posting on April 1st.

"And f/ is just a system of notation, for how many (blades) the aperture opens."

 

- LOL. That one should have been saved for posting on April 1st.

Yes. I haven't checked, but I really hope someone hasn't edited Wikipedia to say that.

Most people who use cameras aren't lens makers and don't know how to calculate that f/ number anyway.

That's actually quite frustrating, because the number is a useful one. A 100mm f/2 lens has an entrance aperture 50mm (100mm/2) across. A 50mm f/2 lens has an entrance aperture 25mm (50mm/2) across. Look in the front of the lens and you'll see it. The "f" stands for the focal length of the lens. Does it matter for calculating exposure? No. Is it useful when you then start thinking about how the aperture affects the depth of field in your scene? Yes. Which is why I go nuts when people say "f2" instead of "f/2" and start talking about bigger numbers for smaller apertures.

Why not make it simple?

Well, you could define "f/1" as 0, and then use the log base (sqrt 0.5) for the fraction. (For those with maths as rusty as mine, that means ln(f-stop as a fraction, so f/2 = 0.5)/ln(sqrt(0.5)).) Which, with a few common non-whole-stop values in parentheses:

(f/0.95 = -0.15)

f/1.0 = 0 (f/1.2 = 0.5)

f/1.4 = 1 (f/1.8 = 1 2/3)

f/2.0 = 2 (f/2.5 = 2 2/3)

f/2.8 = 3 (f/3.3 = 3.5)

f/4.0 = 4

f/5.6 = 5 (f/6.3 = 5 1/3)

f/8.0 = 6

f/11 = 7

f/16 = 8

f/22 = 9

f/32 = 10

...which is how people have to calculate the exposure difference between stops anyway. But then you'd really need a calculator to work out depth of field, whereas multiplying by 1.4 isn't so tricky, especially if you're allowed to round it off.

Now, we could also have my pet argument about how ISO is an irritating measurement if you're comparing sensor sizes and relative apertures and focal lengths, and redefine the whole scheme. But we probably shouldn't.

That's actually quite frustrating, because the number is a useful one. A 100mm f/2 lens has an entrance aperture 50mm (100mm/2) across. A 50mm f/2 lens has an entrance aperture 25mm (50mm/2) across. Look in the front of the lens and you'll see it. The "f" stands for the focal length of the lens. Does it matter for calculating exposure? No. Is it useful when you then start thinking about how the aperture affects the depth of field in your scene? Yes. Which is why I go nuts when people say "f2" instead of "f/2" and start talking about bigger numbers.

Huh! It's been over 40years since I took my first photography class in high scool and during that time I have never stopped to think why the f-stop designation is engraved f/ on the lens even though I would occasionally wonder why... Learn something new everyday... Thanks

Yes. I haven't checked, but I really hope someone hasn't edited Wikipedia to say that.

 

 

 

That's actually quite frustrating, because the number is a useful one. A 100mm f/2 lens has an entrance aperture 50mm (100mm/2) across. A 50mm f/2 lens has an entrance aperture 25mm (50mm/2) across. Look in the front of the lens and you'll see it. The "f" stands for the focal length of the lens. Does it matter for calculating exposure? No. Is it useful when you then start thinking about how the aperture affects the depth of field in your scene? Yes. Which is why I go nuts when people say "f2" instead of "f/2" and start talking about bigger numbers for smaller apertures.

 

 

 

Well, you could define "f/1" as 0, and then use the log base (sqrt 0.5) for the fraction. (For those with maths as rusty as mine, that means ln(f-stop as a fraction, so f/2 = 0.5)/ln(sqrt(0.5)).) Which, with a few common non-whole-stop values in parentheses:

 

(f/0.95 = -0.15)

f/1.0 = 0 (f/1.2 = 0.5)

f/1.4 = 1 (f/1.8 = 1 2/3)

f/2.0 = 2 (f/2.5 = 2 2/3)

f/2.8 = 3 (f/3.3 = 3.5)

f/4.0 = 4

f/5.6 = 5 (f/6.3 = 5 1/3)

f/8.0 = 6

f/11 = 7

f/16 = 8

f/22 = 9

f/32 = 10

 

...which is how people have to calculate the exposure difference between stops anyway. But then you'd really need a calculator to work out depth of field, whereas multiplying by 1.4 isn't so tricky, especially if you're allowed to round it off.

 

Now, we could also have my pet argument about how ISO is an irritating measurement if you're comparing sensor sizes and relative apertures and focal lengths, and redefine the whole scheme. But we probably shouldn't.

Andrew do you know what numbers Nikon use for aperture from f/1 to f/2 in 1/3 stop increments? I ask because the may round them off in some what different fashion than mathematically correct. I don't know because I don't have fast lenses like you do.

Nikon use for aperture from f/1 to f/2 in 1/3 stop increments?

Since Nikon has nothing faster than f/1.2, any guess about the 1/3 stop increments between f/1 and f/1.4 would be entirely academic. As for the f/1.2 lens, it only has stops at f/1.2 and f/1.4 and next f/2.

The 1/3rd stops between f/1.4 and f/2 on a Nikon, as far as I know, are f/1.6 and f/1.8.

Thanks, Wouter - I don't have a 1.2 and couldn't check whether it had a stop at 1.4. The 50 f/1.8 (both E and AF-D) does have a stop at f/2.

As Wouter says, the my D850 reports 1.6 and 1.8 when doing 1/3 stops with the Sigma 85 f/1.4. Set to half stop steps, it reports f/1.7, 2, 2.4, 2.8, 3.3...

Power aperture in movie live view seems to report 1/3 stops, although it's quite hard to be precise.

I don't believe anything faster than f/1.4 has electronics, and Nikon aperture rings (that I've seen) only have clicks at whole stops, except where wide open isn't a whole stop.

I can't vouch for the 1-series f/1.2 or the faster rangefinder lenses, though!

Andrew, as Alan already explained, the 1/3rd stop multiplier is the 6th root of 2 = 1.1225.

Therefore the 1/3rd stop increments between f/1 and f/2 are:

f/1.1

f/1.3

f/1.4

f/1.6

f/1.8

Rounded to 1 decimal place, or the nearest integer, as is the convention with stop numbers. Regardless of anything Nikon might think.

The half-stops are 1.2 and 1.7.

It should be remembered that these are just labels and the real stop diameter might be closer to its mathematically correct value.... or not!

Though how you figure that out with a pentagonal, heptagonal or nonagonal aperture I can't be bothered to work out. And it doubtless gets worse with semi-rounded iris leaves.

BTW, the reason (or so I've read) that Nikon only provides whole stop detents is to give enough distance between 'clicks' to guesstimate fractional stops and set the aperture accordingly. Not necessary with modern cameras of course.

ZZZZzzzz

Martin was last seen March 29th....

Joe: agreed about the theory (as in my "proposal" of a log2 system above) - BeBu just asked how Nikon indicate them, so I reported what the camera shows - which happens to be "correct", at least in the 1.4-2 range.

That said, the D850 reports the following "1/3" stops with the Sigma 85mm:

1.4, 1.6, 1.8, 2, 2.2, 2.5, 2.8, 3.2, 3.5, 4, 4.5, 5, 5.6, 6.3, 7.1, 8, 9, 10, 11, 13, 14, 16. Strictly speaking the two values in bold should be 3.6 and 5.7, though I imagine the convention of saying "5.6" avoids novice questions about its relation to 2.8.

The half-stop sequence is:

1.4, 1.7, 2, 2.4, 2.8, 3.3, 4, 4.8, 5.6, 6.7, 8, 9.5, 11, 13, 16. The bold values should be 3.4 and 5.7.

Nikon have several lenses with fastest apertures aligned to half a stop, as I showed in my list, including the f/1.2 lenses. With the f/1.2 lenses, you can only set the aperture with the ring on the lens (on an SLR - an FA could do it), so the granularity of partial stops is irrelevant, though I certainly concur that the f/2 stop on my f/1.8 lenses is a very small movement of the aperture ring. If the Nikkors had partial stops throughout, no doubt someone would complain (and the oldest cameras were a bit iffy with metering). Control from the camera means there's a choice in step size that the user can configure, making partial stops practical. I believe some Fuji cameras with aperture on the lens resort to a "shift" control on the body to support partial aperture sizes - but their aperture rings won't be implemented with a mechanical stop like Nikkors.

You can (and must) set the aperture from the camera with the 28-80 f/3.3-5.6G, however. At 28mm (and therefore an f/3.3 maximum), the "1/3" stop smaller than f/3.3 is f/3.5, as if the lens were actually f/3.2. With half stops, the camera jumps straight to f/4.

I no longer commonly use variable-aperture lenses, but the D850 reports the following maximum aperture sequence with the 28-80 as it zooms:

3.3, 3.5, 3.8, 4, 4.2, 4.5, 4.8, 5, 5.3, 5.6. If those are supposed to be 1/6 of a stop (which is how I'd implement a system that had to handle 1/3 and 1/2), the bold values should actually be 3.4, 3.6 and 5.7. I suspect that some odd rounding is traditional, however. I've not checked with a macro to see whether distance changes follow the same sequence.

Forgive me if I don't try to get one of my Canon bodies working to see whether it uses the same numbering scheme.

On a related note, Nikon's non-CPU lens data menu seems to include only maximum apertures that Nikon has actually used (or you could make with a teleconverter): 1.2, 1.4, 1.8, 2, 2.5, 2.8, 3.3, 3.5, 4, 4.5, 5, 5.6, 6.3, 7.1, 8, 9.5, 11, 13, 15, 16, 19, 22. I'm not sure it's exhaustive even then.

Ken: Yes. I think we answered the original question a while ago (if we'd gone on a tangent without doing so I'd worry). While the tangent may make more sense on the Nikon forum, I think it's harmless here. And I certainly didn't know this without checking, so I'm learning!

ZZZZzzzz

 

Martin was last seen March 29th....

- Ken, sometimes these threads, like Topsy, just grow.

Andrew: what's the point in worrying about f-numbers to such accuracy?

Unless you're about to drill a box of Waterhouse stops from sheet brass.

As I hinted, multi-leaf irises can't possibly be assembled, linearised or controlled to an accuracy that makes it worth bothering about more than one decimal place of calculation. And even then....

A theoretical f/1.4(1416) lens actually transmits more like a perfect f/1.8 lens, and so on. We work in values that double or halve the amount of light, for goodness sake! That's a sloppy tolerance by anyone's standard; so there's absolutely no point getting hung up on details of nomenclature that make two-tenths of bugger all difference to the end result!

And don't even get me started on studio strobes that pretend to control the flash output to 1/10th of a stop.

No argument, Joe. Tolerances are certainly limited, and T-stops aren't f-stops (though the latter is still relevant for depth of field).

Still, Nikon report the alleged f-stop numbers in their user interface, and do so at the accuracy I quoted (irrespective of the lens, which suggests basing it on theory). And if they're going to do that, it does appear that they're rounding incorrectly.

I'm certainly not suggesting that instead of f/5.6 Nikon should report f/5.6568542494923801952067548968388... - but it doesn't take many maths lessons to realise the actual denominator is closer to 5.7 than 5.6. So it's a curio that "5.6" is what we always say. Similarly we say "f/22", not "f/23" when talking about f/22.6274... although the lens I was testing didn't allow me to select that.

I'm assuming this is just convention, and people start with exactly 1.4 rather than sqrt(2) as a number to multiply. This doesn't quite give an excuse for why Nikon apply odd rounding to some intermediate numbers like 3.5 - to get to that number you already need to be doing some exponentiation, and may as well start with exactly 2.

But the topic of this thread does give us some leeway to discuss both the mathematical origin of the numbers, and then why the numbers don't quite line up with the maths. We're already in the land of theory with any of this.

In the real world, factors like vignetting come into play, which is why I assume Nikon have a slight discontinuity in the way the AI ring reports aperture once you've got fast lenses (or it might relate to how the viewfinder screen transmits brightness to the TTL meter, if your meter is in the prism). It's complicated - but that makes it interesting.

Thanks for clarifying, BeBu.

I suspect "5.6" is just a historical approximation caused by multiplying the approximation "2.8" by 2 (and never working with more precision). If someone once said "you multiply the number two stops before by 2" and everything was to two significant figures, you'd get here without having to explain fractional powers to people - although high school maths should cover it easily.

The 3.3/3.4 and 3.5/3.6 errors are harder for me to justify. I guess you can round either way, and especially for maximum apertures the marketing team would rather you rounded to the larger aperture - although there are rumours that some 200mm/2 lenses are actually a bit faster than specified.

Actually, I think the theoretical series (logically) starts with f/1, and meanders on from there.

Oddball apertures like f/3.3, f/3.5, f/4.8 and f/6.3 are 'traditional' from before the series was standardised.

Whether the geometric aperture of a nominal f/3.5 lens is closer to f/3.5 or to f/3.6 is totally irrelevant in practise. It is what it is. The approximate 5.8% difference in brightness is of no practical consequence.

I mean, why did many early shutters use the crazy sequence of 1, 2, 5, 10, 25, 50, 100, 200, 400 fractional seconds? Not everything in photography follows maths or logic... to state the bleedin' obvious.

IMO we've now gone too far in the opposite direction, by trying to use needless accuracy. Sorry to bang on about them, but decimal stops?

A ridiculous complication and irrelevance that nobody needs or can even implement on any camera or lens.

True, Joe, "1" is the start of the system (and arguably "1.4" then defines the step size) - although obviously you can go negative. Not that the "f/0.95" lenses are terribly meaningful (or, arguably, useful) - even the Barry Lyndon f/0.7 lenses were only a stop ahead of f/1.

And yes, it's all down to rounding of mostly irrelevant numbers. But it's still odd that they don't obviously align to sensible fractions of an f-stop. Maybe the f/3.3 lenses really are f/3.3, but as an electronic partial stop on a faster lens, it's weird. Not important, but weird.

As for fractions for shutters, I have to assume that's because we're writing the numbers in decimal rather than binary, and the fractions would confuse people otherwise - though I've not checked how close to accurate the stated numbers there are. Presumably, from other discussions on this thread, the actual timings are closer to binary fractions. Again, it would all be simpler talking in powers of two, but that may not help people wanting to do, for example, the 1/focal length calculation for shutter speeds. Someone would no doubt complain that 1/256s is needlessly complicated and can be "simplified" to 1/200s (and I'd guess not 1/300s).

But you've lost me with the last argument. With decimal stops, are you complaining about having control at the fraction-of-a-stop level, or that apertures are written to two significant figures (ish), and therefore the larger apertures get a decimal point? If the former, I think 1/3 stops are still relevant. If the latter, I suspect any other way of distinguishing f/1.0, f/1.2, f/1.4, f/1.8 and f/2 would be more verbose, possibly excluding the "powers of the square root of a half" representation.

"But you've lost me with the last argument. With decimal stops, are you complaining about having control at the fraction-of-a-stop level,..."

- Basically, yes.

I have an otherwise excellent Minolta meter that displays its aperture readings in 1/10th of a stop increments. However the display for, say, 2/10th stops below f/8 is displayed as f/5.6 subscript 8; an absolutely ridiculous way to show an aperture!

The Minolta meter isn't alone in using this infuriating way of showing apertures. It's used by Sekonic and the 'power' display on the back of many studio strobes as well.

I worked out that 1/10th stop represents only a 3.5% change in image brightness. So not only is the display format unfathomable and impossible to implement, it's also totally irrelevant in terms of a practical exposure.

Pretending to be able to control a flash output to 0.1 stop is even more misleading. The full-output pulse rises with a fast exponential curve, and decays at a much slower exponential rate. Therefore attempting to control the integrated amount of light by controlling the pulse duration requires an extremely complex calculation and precision of pulse-width. I very much doubt the designers of such strobes have the expertise to tailor the pulse-width to the required precision, nor the knowledge of the pulse shape to begin to compute such.

You only have to measure how imprecise the so-called 1/3rd stop decrements on speedlights are, to realise that accurate 0.1 stop control of a flash output is pure fantasy.

1/10 of a stop is about 7% rather than 3.5%. I would want my light meter to be more accurate than that whether or not I can control the light to that accuracy. Knowing accurate light level is important.That's why I wish I could afford the Minolta light meter and not the exposure meter. But put that aside for now.

Regarding the f/3.5 or f/3.6 the difference is small but I want to know which number Nikon uses so that I won't tell someone to set the aperture to f/3.6 which they can't do on their camera.

According to the service manual for the Nikon F3 the aim point for the shutter speeds are exact doubling (I.E. 1/256 and not 1/250) however if the speeds match the marked speed they are well within their tolerance.

Hmm. I might argue about the merits of accuracy when we're talking 1/10 of a stop, but 1/3 of a stop is absolutely visible - and it put me off camera interfaces (hello Fuji) where the aperture is constrained to a whole stop. It's one reason I'm a bit annoyed that AI-S lenses can't let the camera control the aperture, even though I'm aware of no reason why not. It may be less relevant for older cameras with less precision - but the shutter has been electronically calibrated since, I believe, at least the F5, and I'd hope we have at least a bit more relative accuracy these days. Specifying accurate decimal places is pushing the accuracy a bit far (except below f/1.2), but thirds of a stop are going to have to be described efficiently somehow unless we go with the logarithmic system and start having custom LCD elements for +/- 1/3 and 1/2.

We seem to be stuck with it, anyway - as with using ISO as a sensitivity measure and focal length for field of view (both of which have some disadvantages, although also advantages) the cost of changing would outweigh any benefit. But as BeBu says, it's helpful if we're all using the same terminology, even if it's wrong. If I start describing a lens as f/5.7, people will give me very funny looks.

The job of the lens is to project an image of the outside world onto the film or digital chip. Thus the camera system acts like a slide projector backwards, the film/chip being the screen. Now, how bright the image on the screen will be is a function of several factors. For this discussion we are only interested in the lens’s working diameter which is defined as the lens’s aperture. We need the ability to change the working diameter to make the screen image brighter or dimmer. Years ago it was determined that the best way to do this was to use an increment that either doubles or halves the screen brightness.

Now changing the subject (maybe):

You are the Captain of Cavalry “A” Troop, one hundred men with horses marching through the desert. Water is a problem. You bivouac for the night and you expect rain. You order the men to dig a circular pit 8 feet in diameter and line it with their canvas tent material. It rains as expected and the pit begins to collect rainwater. By your experience, you know an 8 foot diameter pit is adequate to collect rain water for your needs. Unexpectedly a lookout spots “B” Troop approaching --another 100 men with horses. You order your men to expand the diameter of the circular pit to accumulate water for 200 men and horses.

How big must the revised pit be to double the amount of collected rainwater? Answer: You multiply the pit diameter (8 feet) by 1.4142. This value is the square root of 2. The answer is 11.3 (rounded it’s 11 feet). You order the pit expanded to 11 feet diameter. Surprise, this new value causes the pit to accumulate twice as much water as before. Why? The surface area (catch basin) now has double the surface area; thus it can capture twice the amount of rain.

The lens opening or aperture is also a circular geometric figure. The area of any circle (thus its ability to collect rain or light) is doubled if you multiply its diameter by 1.4 (1.4142 rounded). Using this factor a number set emerges:

1 – 1.4 – 2 – 2.8 – 4 - 5.6 – 8 – 11 – 16 – 22 – 32 – 45 – 64

Note each number to the right is its neighbor on the left multiplied by 1.4 and then rounded. Each number to the left is its neighbor on the right divided by 1.4 and then rounded.

These are the mysterious values engraved on the lens barrel. The f-numbers are ratios. A ratio is dimensionless. Using a ratio allows us to compare a lens with all other lenses, as to its light transmission, without regard to the actual dimensions of the lens. f-number is short for “focal ratio