math majors - how to compute guide number ratios?

<p>What is the formula for the power ratio between two flash guide numbers?<br /> <br /> For example, two guide numbers 160 and 80, which is a 2x ratio of GN.<br /> <br /> If this is considered as f/16 and f/8, then clearly this is two stops difference.<br /> If this is considered as 16 feet and 8 feet, then clearly this is two stops difference.<br>

Two stops seems the only rational possibility.<br /> <br /> Same with 160 and 40, a 4x ratio.. which is clearly four stops difference.<br /> <br /> But it goes like this:<br /> <br /> 4x 4 stops<br /> 3x 3.2 stops empirical<br /> 2.5x 2.6 stops empirical<br /> 2x 2 stops<br /> 1.4x 1 stop<br /> 1x 0 stop<br /> 0.7x -1 stop<br /> 0.5x -2 stops<br /> 0.25x -4 stops<br /> <br /> As opposed to looking it up in a f-stop chart, is there no formula similar to (GN A / GN B) = X stops power difference?<br /> <br /></p>

<p>try this:<br>

stops difference = 2 * log(GNa / GNb) / log(2)<br>

so, given GNa=100 and GNb=33.33 ...<br>

GNa/GNb=100/33.333=3<br>

2*log(3)/log(2)=3.17</p>

<p>All right! That seems to work great, thank you very much for the help.<br>

I cannot imagine that derivation, and would have never come up with it myself.</p>

<p>An alternative formulation of Rainer T's formula that is perhaps more intuitive is</p>

<p>(GNa / GNb)^2 = 2^(stop difference).</p>

<p>Here's the explanation for the formula's validity. The ratio of the guide numbers indicates the relative light intensity output as a linear distance, so in order to derive the ratio in light output per unit area, one must square this ratio. For example, if the ratio of guide numbers is 2:1, that means the stronger flash must output 2^2 = 4 times the light of the weaker flash to be able to reach twice the distance, all other settings being equal.</p>

<p>Since f-stops are base 2 logarithms of the amount of luminous intensity per unit area reaching the focal plane (that is to say, each full stop increase corresponds to an increase in exposure by a factor of 2), it follows that the number of stops change in exposure between the two guide numbers satisfies the above formula.</p>

<p>The equivalence to Rainer T's formula is demonstrated by simply taking the logarithm (to any base) of both sides:</p>

<p>log((GNa / GNb)^2) = log(2^(stops))</p>

<p>Then using the fact that log(x^y) = y log x, it follows that</p>

<p>2 log(GNa / GNb) = stops log 2,</p>

<p>where upon dividing both sides by log 2, we obtain the desired result.</p>

<p>Thanks Peter. Ranier's seems more practical to me, in that it has Stops = ...., by itself on one side of the equal sign. :) But yours does seem slightly less like magic... In that I could imagine that I needed the stops to be in an exponent, but could not get there myself. Been too many years for me... You guys are great, thank you for the help.</p>

<p>You're most welcome. A somewhat less mathematical way to look at it is to think about the effect of guide number and f-stop on exposure. If your guide number ratio is 4:1, then the stronger flash is 16 times more intense than the weaker. Since each additional stop of light corresponds to a factor of 2 increase in light intensity, then you need 4 stops to get 16x more light.</p>

<p>When the GN ratio is 8:1, then your stronger flash is 64x more intense than the weaker. How many powers of 2 equals 64? You need 2*2*2*2*2*2 = 2^6 = 64, so 8:1 GN = 6 stops EV.</p>

<p>Of course, this method only works if the GN ratio is an integer power of 2. But it helps in terms of understanding the underlying relationship. I've included a graph for convenience.</p><div></div>

<p>Yes, the graph helps, thanks. Interesting too, because those even integer alignments were really confusing me in the beginning.</p>

<p>Try to understand the "inverse square root" law of light distrubution, and do not have to remember graphs, logarithms, etc.</p>

<p><a href="http://www.scantips.com/lights/flashbasics.html">http://www.scantips.com/lights/flashbasics.html</a><a href="http://www.lbl.gov/abc/experiments/Experiment1.html"></a></p>

<p>Some basic principles to remember are:<br>

- twice the guide number requires four times amunt of energy<br>

- twice the distance - requires twice the guide number.<br>

- etc. as in the link.</p>

<p>The link has some good examples for FP ...</p>