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From this post, I understand that the flash guide number (GN) is given by the following formula:

$$ \text{Guide Number} = \,\frac{\text{Shooting Distance}\times \text{f-number}}{\text{ISO factor}} $$

I'm assuming that the GN calculated via the equation above will be based on a flash that is firing at full power. Am I correct to assume that if I set the flash to fire at half power, that the guide number is effectively halved? i.e. the equation will now be:

$$ \text{Guide Number} = \,\frac{\text{Shooting Distance}\times \text{f-number}}{\text{ISO factor}} \times \text{Power Level}$$

where power level can be the following 1, 1/2, 1/4, 1/8, ... 1/64, etc?

scottbb
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CadentOrange
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  • It's useful to remember that this is a formula for determining what guide number should be in a certain situation, not for determining what it is from the other factors. For that reason, I think it's more useful to think of power as a way to adjust the guide number rather than a factor on the right side of the equation. – mattdm Jan 01 '12 at 15:24
  • And with that in mind, note that I actually covered this question at the end of my answer to the question you linked to. :) – mattdm Jan 01 '12 at 15:29

1 Answers1

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The guide number is inversely proportional to the power squared. This is due to the way that light intensity diminishes with distance, at twice the distance light is spread over four times the area, so each bit of that area receives 1/4 of the light.

So the actual formula needs to take into account the square root of the power level:

$$ \text{Guide Number} = \,\frac{\text{Shooting Distance}\times \text{f-number}\times \sqrt{\text{Power Level}}}{\text{ISO factor}} $$

The guide number has the same inverse square relationship to the sensitivity, as detailed by the ISO factor, defined as follows:

$$ \text{ISO factor} = \sqrt{\text{ISO}\over 100} $$

Substituting and bringing sensitivity onto the top of the fraction gives a formula which you can simply plug numbers into:

$$ \text{Guide Number} = \text{Shooting Distance}\times \text{f-number}\times \sqrt{\frac{\text{Power Level}\times 100}{\text{ISO}}} $$

scottbb
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Matt Grum
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  • I see, it's the square root of the power level. Thanks for that! – CadentOrange Jan 01 '12 at 15:14
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    It is not proportional to the power squared, but rather the energy squared. Generally the power of a flash has little relevance, its the energy that counts. A 500 microsecond flash at one power level and a 250 microsecond flash at twice the power are basically equivalent exposure-wise. – Olin Lathrop Jan 01 '12 at 16:21
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    @Olin: this might be a case of technical language vs. camera jargon vs. plain english. Flash manufacturers certainly use the term "power" in this sense, even when the adjustment is really done by cutting the impulse time. – mattdm Jan 01 '12 at 18:53
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    @mattdm: Thanks for clearing that up. As a engineer and student of physics, it bugs me when terms like this get used contrary to their formal definition. I guess this is just photography jargin I have to get used to. – Olin Lathrop Jan 01 '12 at 20:35
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    @Olin Lathrop not only that but flash manufacturers measure flash "power" output in watt seconds! – Matt Grum Jan 03 '12 at 12:47