Why are shutter-speed calculations not linear?

[Michael Althaus]

Maybe I'm overcomplicating this and there is a really simple answer. This has been bothering me for awhile but I never really tried to find out:

Example:

Lets say my base exposure is 1/1000 = 0.001 seconds

If I want to underexpose by a full stop, I would expose for:  1/2000 = 0.0005 seconds
Therefore 1 stop underexposed means that I expose for 0.0005 seconds less

So far everything seams logical. 

However, let's just say I want to underexpose by 1/3 stop: So... if 1 stop under exposure is 0.0005 seconds less, I would expect that 1/3 stop less equals 0.000167 seconds less (0.0005/3)
Therefore, the exposure time would be 0.000833 seconds (0.001-0.000167) which is 1/1200 second. 

However, every shutter speed chart lists 1/1250 seconds for 1/3 underexposure (base is 1/1000). 

How come?

 

 

 

[Michael Althaus]

One more example: A shutter speed chart lists that 1/1600 (0.000167) is 2/3 stops less than 1/1000 (0.001).

A full stop less would be 1/2000 (0.0005). I would therefore expect that the steps would be as follows: 

Full stop under: 0.0005 s (1/2000)

2/3 stops under: 0.00067 s (1/1500) (Charts say 1/1600)

1/3 stops under: 0.00083 s (1/1200) (Charts say 1/1250)

No correction (base exposure): 0.001 se (1/1000)

Why is the scale on the chart not linear...

 

Edited March 21 by Michael Althaus

[Brian Drysdale]

I'd be surprised if anyone shooting video or film would use shutter speed for the fine exposure adjustments, they'd tend to use the iris for that. 

Some of these fast shutter speeds are carry over from stills photography, so you're less likely to use them on video or film unless you're shooting for a visual effect. You'll come across them when shooting slow motion. Film and Video shutter speeds have tended to be a combination of frame rate and shutter angle working from a 180 degree shutter, although this can vary from camera to camera. Plus you may be able to vary the shutter on individual camera models.

On video cameras, you tend to have shutter speed settings based around the frame rates used in PAL or NTSC counties, e.g. 1/50. 1/100 etc or 1/60, 1/120 etc

You may be able to work in 1/3 and 1/2 stop increments using shutter speeds on individual stills cameras.

Here's Wikipedia on the subject, there appears to have been an element of rounding up of the numbers for convenience ; https://en.wikipedia.org/wiki/Shutter_speed

 

[Mark Dunn]

Note also the fact that, although it's used more loosely in photography and film, the term "stop" refers strictly to the size of the aperture, which varies logarithmically with the diameter. This is why f-numbers have that 2, 2.8 4, 5.6 sequence- the multiple is the square root of 2. This applies equally to the intermediate f-numbers. They arose when mechanical shutters were nothing like as accurate as they are now and a tiny difference in shutter speed was an irrelevance.

[Nicolas POISSON]

Stops are a logarithmic scale. 1/3rd and 2/3rd of a stops are not linearly spaced. Otherwise it would mean full stops follow a logarithmic scale but intermediate values follow a linear scale, which would make a complete mess.

In Excel (enclosed file), the formula is:

Shutter speed -1/3rd = 2^(log(reference shutter speed; 2) + 1/3)

Shutter speed -2/3rd = 2^(log(reference shutter speed; 2) + 2/3)

Starting from 1/1000th of a second, this gives:

1/3 stop below: 2^(log(1000; 2) + 1/3) = 1/1260s

2/3 stop below: 2^(log(1000; 2) + 2/3) = 1/1587s

 

The series repeat with a factor of 2. Starting from 1/50th of a second:

50 - 63 - 80

100 - 125  - 160

200 - 250 - 320

400 - 500 - 640

800 - 1000 - 1250

1600 - 2000 - 2500

3200 - 4000 - 5000

6400 - 8000 - 10000

12500 - 16000 - 20000

Some values are rounded to help the memorization. For example, 2x640 should give 1280, not 1250. But then the following figures would be 2560 instead of 2500, 5120 instead of 5000, and so on. Or the other way around: 1250/2 should give 625 instead of 640, then 312.5 instead of 320, then 156.25 instead of 160. I am not sure, but I guess the actual value used by the device is very precise. It is just the way it is labelled that is simplified.

By the way, the ISO follows the same scale, for the same reason. As do frequencies on an audio graphics equalizer (octave is logarithmic scale like stops). The rounded value sometimes differs a bit. In audio, it is common to find 315Hz and 630Hz instead of 320 and 640.

For historical reason, there might be some additional shutter speed values that do not follow the logarithmic scale, such as 1/48 and 1/96 (multiples of motion picture 24fps) ,1/60, 1/120, 1/180 and 1/240 (60Hz countries). Using a photo lens with a clicked aperture, you will not be able to respect the triangle of exposure, as you will not have the required ISO or diaphragm intermediate values to compensate. But that is hardly a problem since in videography, one rarely use shutter speed as an exposure parameter. And even if we do, The difference between 1/48 and 1/50, or between 1/60 and 1/63, is so small that it is nothing you cannot manage in post. With a de-clicked aperture, you could compensate exactly, but I very much doubt you could be that precise.

stops_calculation.xls

Edited March 21 by Nicolas POISSON

[Michael Althaus]

Thanks for the replies

@Nicolas POISSON

Thanks for all the details and math, I really appreciate it.
I'm aware of how it works in practical terms (I've been shooting video and some film for over 20 years). I just never actually thought about the math behind "the numbers" and didn't realize the obvious issue you get with mixing a linear and log scale. I guess my question actually comes down to why the logarithmic scale applies and not a linear scale. Does it have something to do with how film reacts to light? 

Looking at your formula, I just realized that the change is always by a factor of the cube root of 2 (3√2 = 1.2599). In other words, one step up, is 26% more than the previous value (Example: 100 to 125 to 160 to 200... numbers are rounded). This applies to the shutter-speed, ISO, and the calculated area of the entrance pupil (using the unrounded numbers like f/1.41421 for the calculation). It is quite obvious if you think about it... just never really thought about it in that way (just knew the numbers by heart but never really why they are what they are).

 

[Nicolas POISSON]

My understanding is that it has more to do with the way human perception works. It just happened that film follows more or less human sensitivity to light. Audio levels are expressed in decibels, which is also a log scale (base 10 instead of 2), although recording medium are not inherently logarithmic. But human hearing is.

For the maths: it is just the property of x^(a+b) = x^a * x^b. The simplest example could be x^2 = x^(1+1) = x * x.

In our case: 2^(log(1000; 2) + 1/3)  =  2^(log(1000; 2) * 2^(1/3),  and 2^(1/3) is the cube root of 2.

Since the scale is usually split in 1/3rd of a stop, there are only three values to memorize, then you can guess all the other ones multiplying by 2 (one stop).

If you were to use a scale in half stops, then the multiplier would not be the cube root but the square root of two, and there would be only two base values to memorize. If you want to use 1/10th of a stop (some light meters use that), the multiplier is 2^(1/10) = 1.072.

 

By the way, the aperture series follows a square root of two, but for a complete different reason: there is a full stop between f/4 and f/5.6, not half a stop. It is just that photographers chose to express the series using the radius of the aperture, instead of its surface. Things would have been much simpler if the aperture scale was:

1 (instead of f/1),

2 (instead of f/1.4),

4 (instead of f/2)

8 (instead of f/2.8)

16 (instead of f/4)

32 (instead of f/5.6)

64,

...and so on.

Edited March 23 by Nicolas POISSON

[Nicolas POISSON]

If you want to calculate intermediate 1/3rd stops values of aperture, you have to cumulate the cube root of 2 (1/3rd stops) and the square root (radius vs. surface). This time, the basic maths are: (X^a)^b = X^(a*b)

The multiplier becomes :

sqrt(2^(1/3)) = ( 2^(1/3) ) ^ (1/2) = 2^(1/3 * 1/2) = 2^(1/6) = 1.122

For example, starting from f/2

f/2

f/2+1/3 = f/2.2

f/2+2/3 = f/2.5

f/2.8

f/2.8+1/3 = f/3.2

f/2.8+2/3 = f/3.6

f/4

...

Edited March 23 by Nicolas POISSON