# Flash guide number f/stop calculation

I have been reading the B&H Speedlight Buyer's Guide. I am confused specifically about the guide number calculation. I understand the principle and how the inverse square law is used. Here is the example from the article that I am basing my calculation on:

Simplifying this a bit, a realistic example would be if you have a flash with a guide number of 100—photographing a subject 25' away will require the use of f/4 for proper exposure. Likewise, a subject 50' away requires f/2 or a subject 5' away requires approximately f/22.

My confusion lies with the case when the subject is 5' away. The f/stop required here is in fact exactly f/20 (since using the guide number equation: 20*5=100). f/20 is 4 2/3 (4.66) stops away from f/4. I have no problem with this at all.

My confusion comes when I try to compare this value to what I get when I try to calculate using the amount of light needed. If the subject is 5' away then it is 5x closer than when it is 25' away. Using the inverse square law this means that we would need the flash to be 5^2=25 times less bright to get proper exposure (correct me if I'm wrong?).

I want to compare this value to how many stops of light this would be. Each time the lens is stopped down one f/stop, half the amount of light is allowed through. So in the case of this example if we stop down from f/4 to f/16, this is reducing the exposure by 4 stops (or 16x less light). If we stop down once more to f/22 we now reduce the exposure by 5 stops (or 32x less light). But in this case we want 25x less light to get a proper exposure at 5'. Here's the dealbreaker: f/20 is 2/3 of a stop past f/16. However, 25 is NOT 2/3 of the way between 16 and 32. It's actually 9/16 (0.5625). Meaning that 25x less light is actually 4.5625 stops less bright, rather than 4.66 stops. As far as I can tell these two values should be exactly the same. What am I doing wrong? I must me missing something here, but I can't work out what.

I hope you understand what I'm trying to say here. As far as I can tell there shouldn't be a discrepancy between the two values. Any help shedding some light on this would be greatly appreciated.

• f/22 is the nearest "full-stop" to f/20. All of the values used in the quote included in your question are full-stop numbers. You can only use f/20 if you can use 1/3-stop increments. May 10, 2016 at 18:46

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:

• 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.

Using the guide number 100, the f/number setting for 5 feet is 100 ÷ 5 = 20. Thus we would set the camera to aperture f/20. For a subject 25 feet distant, the math is 100 ÷ 25 = 4. Thus we set the aperture to f/4. Using the inverse square law, this is indeed a 25x reduction.

The problem is actually the complications of dealing with the focal ratio number set (f/numbers). The f-number set is based on the geometry of circles. The basic increment of the f-number is a 2x change in light energy. What we must do is to halve or double the surface area of the circular iris diaphragm.

Factorial: Multiply the diameter of any circle by the square root of 2 and you compute a revised circle with twice the surface area. This value is 1.414…. however we can round to 1.4 for the accuracy we need. Conversely, if we multiply the diameter of any circle by 0.707 (inverse of 1.4) we compute a revised circle with half the surface area. Using the square root of 2, a number set emerges. We call this number set the focal ratio of lenses or f/number for short.

This set is:

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

Note: Each number going right is its neighbor on the left multiplied by 1.4. Conversely each number going left is its neighbor on the right divided by 1.4.

The same scheme apples to create a number set in ½ stop increments. This time the multiplying factor is the fourth root of two which is 1.189….

This ½ f/number set is:

1 – 1.2 – 1.4 – 1.7 – 2 – 2.4 – 2.8 – 3.4 – 4 – 4.8 – 5.6– 6.7 – 8 – 9.5 – 11 – 13.5 – 16 – 19 – 22

Additionally we can compute this set in 1/3 stop increments, the multiplying factor is the sixth root of 2 which is 1.1224…….

This 1/3 f-stop number set is:

1 – 1.12 – 1.26 – 1.4 – 1.59 – 1.8 – 2 – 2.25 – 2.5 – 2.8 – 3.2 – 3.5 – 4 – 4.5 – 5 – 5.6 – 6.3 – 7 – 8 – 9 – 10 – 11 – 13 – 14.3 – 16 – 18 – 20 – 22 -25 – 29 – 40 – 45

f/4 to f-5.4 = 2x reduction

f/4 to f/8 = 4x reduction

f/ 4 to f/11 = 8x reduction

f/4 to f/16 = 16x reduction

f/4 to f/18 = 21x reduction

f/4 to f/20 = 25x reduction

f/4 to f/22 = 32x reduction