Why doesn’t the guide number calculation seem to match the inverse-square law in flash exposure?
Asked 5/10/2016
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I’m trying to reconcile flash guide numbers with the inverse-square law. For a flash with guide number 100, the usual formula gives:
- 25 ft → f/4
- 5 ft → f/20 (since 100 ÷ 5 = 20)
That part makes sense. But if the subject moves from 25 ft to 5 ft, it is 5× closer, so the light intensity should be 25× greater by the inverse-square law. So I’d expect the aperture to reduce the exposure by 25×.
My confusion is in converting that 25× change into stops. Going from f/4 to f/16 is 4 stops (16× less light), and f/22 is 5 stops (32× less light). Since 25× lies between 16× and 32×, I tried estimating the fraction of a stop and got a value that doesn’t seem to match f/20.
What is the correct way to compare the guide-number result (f/20) with the inverse-square-law result (25× change in light)?
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
10y ago
2 Answers
5
However, 25 is NOT 2/3 of the way between 16 and 32.
You just have to remember that the scale is exponential, not linear. 4 2/3 stops allows 24.66 more light, and 24.66 = 25.28132..., or roughly 25.
It's actually 9/16 (0.5625). Meaning that 25x less light is actually 4.5625 stops less bright, rather than 4.66 stops.
Again, you're mixing linear and exponential scales. If you want to know the actual value, you need the power of 2, i.e. the solution for x of 2x = 25, or equivalently, log2(25).
That's actually easy to calculate even if your calculator doesn't have a button for finding powers of arbitrary bases. Just use a base that you do have, like 10 or e, and convert to base 2, like this:
log2(25) = log10(25) / log10(2) = 1.39794 / 0.30103 = 4.64385
What am I doing wrong? I must me missing something here, but I can't work out what.
There are two problems:
- your math was wrong
- photographers play fast and loose with numbers
When we say that f/20 is 2/3 stop past f/16, we're using round numbers. Nobody is going to bother remembering that it's actually 0.64385 stops past f/16.
Originally by user4262. Source · Licensed CC BY-SA 4.0
user4262
10y ago
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The mismatch comes from treating stops as a linear scale. They’re logarithmic.
From 25 ft to 5 ft, the subject is 5× closer, so the flash illumination is indeed 25× stronger. To compensate, you need 25× less light through the lens.
A stop is a factor of 2 in light, so the number of stops is:
stops = log2(25) ≈ 4.64 stops
That agrees with the aperture change from f/4 to f/20, because f-numbers are based on the square root of 2 per stop:
f/4 → 5.6 → 8 → 11 → 16 → 20
That’s about 4.64 stops.
Why not 4.56 or “2/3 of the way” between f/16 and f/22? Because you were interpolating linearly between 16× and 32×. Exposure changes by powers of 2, not by equal arithmetic steps.
Also, aperture light transmission depends on the area of the opening, so exposure scales with the square of the f-number ratio:
(20/4)^2 = 25
So both methods are fully consistent:
- inverse-square law says you need 25× less exposure
- guide number says f/20 instead of f/4
- and f/20 gives exactly 25× less light than f/4
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