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How do you calculate the flash power fraction, assuming you want to keep constant the other variables of ISO, aperture, and distance from subject to flash?

The Guide Number formula for a flash (strobe) is \$\text{GN} = \text{distance}\times \text{f-stop}\$

Let's say the Guide Number is 174 at ISO 200. You want to shoot at f/8. This gives a distance of

$$\begin{align} d &= {174\over 8} \\ &= 21.75\,\text{feet} \end{align}$$

Now, let's say you're shooting in a room that doesn't have that much room to play with.

But you do know you can move the flash 10 feet away.

What fraction of power do you use?

Is it linear, e.g. 21.75/10=2.175, so use 1/2? Or something else?

The numbers in this example are for the Nikon SB-800 and a Nikon D90 camera, but the principle is likely the same.

scottbb
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jfklein13
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  • GN doesn't double as ISO doubles, it increases by a factor of √2 (inverse-square law again), so your GN would be 272 in this situation. – ex-ms Aug 20 '10 at 23:17
  • Thanks Matt. Edit: Corrected Guide Number to what's published for the SB-800 at ISO 200, changed numbers in the example to make it easier to follow. – jfklein13 Aug 21 '10 at 03:28

3 Answers3

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Its not linear, its a square, so if you double the distance, you need 4 times the light. So in your example you would need 1/4 the power, or adjust the aperture and or ISO.

ex-ms
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BillN
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  • Aha! That might explain some surprises in recent shooting sessions.

    So in closed form the fraction of power is (d/(gn/f))^2?

     – jfklein13 Aug 20 '10 at 22:21
  • Yes, that's correct. Square of the ratio of the real distance and guide distance. (Though see my note about GN attached to the question.) – ex-ms Aug 20 '10 at 23:15
  • Going from 20' to 10' flash distance would increase exposure by x4 (2 stops). 2 stops less than full flash power would be 1/4th (2x2=4). 1/16th is 4 stops less (2x2x2x2=16), or what would be needed in the above example if the flash was moved to 5' from the subject. – Henry Peach Aug 23 '10 at 13:36
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    @Henry - Just to clarify the apparent error here; the example changed from an initial 40ish feet to the current 20ish after this answer was written. It was correct at the time. (And I'll correct it now, since BillN might not have noticed the change) – ex-ms Aug 25 '10 at 18:18
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Light falls off according to the inverse-square law. Basically it boils down to this equation:

$$ I \propto {1\over r^2} $$

where \$I\$ is the intensity and \$r\$ is the radius (which is subject distance for us) and the '\$\propto\$' means 'proportional to'. Anyways, a couple of good articles on the subject can be found at Cambridge in Colour and at Portrait Lighting.

scottbb
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Joanne C
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Personally, I just guess, take a shot and look at my histogram and tweak as needed. I just find it quicker and less complicated on my poor brain. It also affords me more brain power to keep talking to my subject.

matt burns
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  • Absolutely! I don't plan to go computing this all the time. My original reason for asking was for shopping for lights. I wanted to see some tables that show what power setting on a given light would be needed for a given aperture and subject distance. This discussion helped tremendously. – jfklein13 Aug 25 '10 at 21:25