Aperture f-stop multiplier?

[Greg Lockrey]

Let's just round it to 1.15............

Yeah, that would be within the two stop error range for film.

[Emmanuel BIGLER]

Number of full f-stops = Log_2 (muliplying factor) = Log_2(10) x Log_10(muliplying factor) = 3.32 Log_10(muliplying factor)
muliplying factor for exposure time = 2^{number of full f-stops} = 10^{0.30 x number of full f-stops}

Correspondence between..
half f-stops -> multipliyng factor

-6 > 0.0156 = 1/64
-5.5 > 0.022
-5 > 0.031X = 1/32
-4.5 > 0.044X
-4 > 0.0625X= 1/16
-3.5 > 0.088X
-3 > 0.125X = 1/8
-2.5 > 0.177X
-2 > 0.25X = 1/4
-1.5 > 0.353X
-1 > 0.5X = 1/2
-0.5 > 0.707X
0 > 1X
0.5 > 1.41X
1 > 2X
1.5 > 2.83X
2 > 4X
2.5 > 5.66X
3 > 8X
3.5 > 11.3X
4 > 16X
4.5 > 22.6X
5 > 32X
5.5 > 45.2X
6 > 64X

----------------

Same by clicks of 1/3 of f-stop. for the uncompromising user of Compur and Prontor Professional shutters !

(Since Compur & Prontor view camera shutters have been discontinued, Copal shutters mounted by Schneider are supposed to have clicks by 1/3 f-stop)

clicks by 1/3 f-stop -> multiplying factor

-6 > 0.0156 = 1/64
-5-2/3 > 0.0197
-5-1/3 > 0.0248
-5 > 0.0313 = 1/32
-4-2/3 > 0.0394
-4-1/3 > 0.0496
-4 > 0.0625 = 1/16
-3-2/3 > 0.0787
-3-1/3 > 0.0992
-3 > 0.125 = 1/8
-2-2/3 > 0.158
-2-1/3 > 0.198
-2 > 0.25 = 1/4
-1-2/3 > 0.315
-1-1/3 > 0.397
-1 > 0.5 = 1/2
-2/3 > 0.63
-1/3 > 0.794
0 > 1
1/3 > 1.26
2/3 > 1.59
1 > 2
1+1/3 > 2.52
1+2/3 > 3.17
2 > 4
2+1/3 > 5.04
2+2/3 > 6.35
3 > 8
3+1/3 > 10.1
3+2/3 > 12.7
4 > 16
4+1/3 > 20.2
4+2/3 > 25.4
5 > 32
5+1/3 > 40.3
5+2/3 > 50.8
6 > 64

[Nigels]

OK, thanks guys. I think I understand it best in terms of the aera of a circle or simply the area of a square. Since we are dealing with comparative ratios I think we can ignore the constant of pie and just see it as calculating the diffence between areas of squares. Thats why we use sq.roots. Anyway I ran the cube root of the square root of 2 through a speadsheet and got a full set of 1/3 stops from f/1.0 right through to f/256 using the multiplier of 1.12246205. Right, I'm off to get a life now.

[Leonard Evens]

There seems to be a lot of confusion about this. The relation between the F-number mulitplier and the number of stops change is

F-number multiplier equals the square root of 2 raised to the number of stops which is the same as 2 raised to half the number of stops.

In this case, half the number of stops is 1/2 of 1/3 or 1/6, so the F-number multiplier corresponding to a change of 1/3 stop is 2^(1/6) = 1.122462048, which is as far as my calculator goes.

The corresponding formula in the other direction, which is more useful, is

Number of stops change = 2 x log(F-number multiplier)/log(2).

It doesn't matter what kind of logarithm you use.

I don't know where all the other numbers people came up with came from, but they are wrong.

I will try to explain just where all this comes from, but I don't know if you will like the explanation.

The rationale for this is that the F-number is defined to be the ratio of the focal length to the diameter of the aperture, so, for fixed focal length, it is inversely proportional to the diameter. On the other hand, exposure is proportional to the area of the aperture and hence to the square of the diameter. Stops refer to exposure, with one stop meaning a doubling of exposure. That is where the 2 comes from. To increase exposure by one stop, you need to multiply the diameter by the square root of 2, i.e., divide the F-number by the square root of 2. Similarly, if you stop down one stop, that multiplies the F-number by the square root of 2.

You might ask why the F-number is related to the diameter instead of the area. The reason is that the F-number is used for many optical calculations having nothing to do with the exposure, such as depth of field calculations, and for those, the diameter of the aperture, not its area, is the relevant parameter.

[Greg Lockrey]

I don't know where all the other numbers people came up with came from, but they are wrong.

Of course you are correct. The rationale I was using to derive at my number was that Nigel was using a square as a basis for his area instead of the more correct area of a circle formula using pi r squared. Since he "rounded off" with the square, I did too. This has become a very informative mathmatical thread.

The F number is in fact a rounded off number if you want to get real precise about the effect on area of the circle as it pertains to doubling/halving the area. It's close enough for photographic purposes.

[Helen Bach]

You might ask why the F-number is related to the diameter instead of the area. The reason is that the F-number is used for many optical calculations having nothing to do with the exposure, such as depth of field calculations, and for those, the diameter of the aperture, not its area, is the relevant parameter.

It also means that the f-number is dimensionless. That's a very convenient property for a number. If it wasn't dimensionless the f-number would be different for different unit systems.

best,
Helen

[Struan Gray]

That's some heavy siffering. Aren't we dealing with an area of a circle and how it multiplies itself by 2 from one diameter to the next? You're going to have to explain cube's and sixth roots to me when you are just dealing with square areas.

Others have given the formalism, but since you asked here's my take on things.

The real key is to ask yourself what is meant by "A third of a stop"? A whole stop is a doubling of the light reaching the film, and when we say a third of that most people intuitively think that we are looking for an increase that when made three times will double the light exposure.

Let's say we have an exposure time of 2 seconds. One stop less exposure is simply 1 second. Easy. It is tempting to think that one third of a stop less exposure must therefore be 1.666 seconds, i.e. that subtracting a third of a second gives the right answer because doing that three times gets us to 1 s, which is one stop less.

The problem is that the amount to subtract depends on your starting exposure. If your starting exposure is 4 s, you need to subtract two-thirds of a second. If your starting exposure is 8 s, you need to subtract 1.333 s.

The reason is that a stop is defined as a ratio, and that fractions of a stop need also to be defined as ratios. If you do that, and define a portion of a stop as an amount to multiply your base exposure by, you automatically adjust for the size of the base exposure.

In this case, you need to find a number that when multiplied by itself three times gives you a total factor of one half. That is (1/2)^(1/3), the cube root of one half, or 0.7937. Exposure times need to be multiplied by this number if you want to reduce the exposure by one third of a stop (or by its reciprocal, 1.260, if you want to increase exposure by one third of a stop). For two thirds of a stop, you just do the multiplication twice, and for three thirds, or a full stop, three times.

In the case of apertures the same thing applies. You need to reduce the area of the aperture by a factor 0.7937 if you want to reduce the exposure by one third of a stop. Thus the diameter of the aperture must be reduced by the square root of this ratio, which is 0.8909. This number is a square root of a cube root, which is a sixth root - in this case, the sixth root of 0.5. The f-number is proportional to the reciprocal of the aperture diameter, so the multiplier for the f-number comes out as the sixth root of two, or 1.225.

[Greg Lockrey]

Thanks Straun for taking the time to answer my question to your thought process. It is interesting that you took the approach of deciding that 1/3 of a stop was the equivalent of 1/3 of the increase or decrease of a ratio of exposure rather than a 1/3 of the distance of the radius of the aperture. In my mind it should work out the same, but since we are dealing with a area of a circle and not a square, you may have a point. I'm just too old, tired or lazy to figure this out to that extreme since for "photographic work" it doesn't matter anyway.

[Struan Gray]

Provided you keep the shape the same, it does not matter whether you consider squares or circles (or hexagons, or Siemens stars, or outlines of your dog). What does matter is whether you are considering linear quantities or area quantities.

If you double the length of the sides of a square you multiply its area by four.

If you double the diameter of a circle you multiply its area by four.

f-numbers are a combination of linear quantities, while the light allowed through your lens is determined by an area. You need to square the first to determine the second.


I have toyed with the idea of using fractal apertures in the Waterhouse slots of my process lenses, in which case I would actually lose the square relationship. A gimmick, but with the right artspeak it would probably sell well :-)

[Greg Lockrey]

Yes, you are correct. Where I screwed up was I took my 1/3 radius increases from 1 to 2 equating that of an f/stop ratio, I should have gone from 1.4 to 2.

Hey...I work 12 hours a day and take care of a wife with MS.