Digression #3

A note about the digressions: The digressions are intended as "extra" material that is not really important to understand. They're here in case you're interested and are willing to slog through some technical talk on the way to grasping an esoteric concept. If you don't really care, or you just don't want to spend the time grappling with the math or whatever, then just skip the digressions. You won't be missing anything important.

1. Why does the guide number equation work?

Remember the guide number equation:  GN = aperture * distance.

Does it seem completely mysterious that this simple equation should work? And how is it that aperture and distance have any relationship, let alone so simple a relationship?

First off, if you're rusty on aperture numbers and what they represent, brush up on the topic with digression #1. If you're up to speed, then skip it.

Now to remind you of something you may have encountered in a high school physics class. Light intensity falls off like the square of the distance from the source. That is,

    I = I₀ / r²,

where I is the intensity measured at any point, I₀ is the intensity of the light measured at the source, and r² is the distance between the point in question and the source of the light. All this means is that if you double your distance from the strobe, the light is 1/4 as intense. Triple the distance, and you're down to 1/9 the brightness.  

Let's say you measure the distance between the strobe and the subject to be 2 feet and you get a proper exposure. Then if you move to 4 feet from the subject, you will need four times the light from the strobe in order to get proper exposure. How do you get four times the light?  The easiest way is to open up your aperture two stops.  Not one stop, but two. That means that if you were at f/8 and 2 feet, you'll need to open up to f/4 at 4 feet, right?   So let's look at those numbers.

2 feet * f/8 = 16
4 feet * f/4 = 16

Holy cow, we got 16 both times!  Is that a coincidence?  Let's try another combination, in reverse. What distance would make f/5.6 work for us? Hmmm, f/5.6 is half the light we had at f/4, so we need to go to a place that is twice as bright as it was at 4 feet. We know that's not 2 feet, right, because 2 feet is 4 times as bright as 4 feet. So how about something in the middle somewhere?  3 feet? Well, 3 feet will be (4/3)² times as bright, or 1.78 times as much light, so we're close but not quite there. How about 2.9 feet?  That's (4/2.9)², or 1.9 times as bright. OK, what about 2.8 feet?   That's (4/2.8)², or 2.04 times as bright. Close enough. OK, 2.8 feet must be the place. So now we have another number for the chart:

2.8 feet * f/5.6 = ....yeah, that's right: 15.7, or almost exactly 16!

What's going on here?  Is all this some weird numerology thing?  

Nope. Let's look at the numbers we just calculated. What's 4 divided by 2.8?  1.4. So you had to divide your distance by a factor of 1.4 in order to double the light. So answer me this: what is the square of 1.4?  I'll save you the trouble: 1.96. Sounds a lot like 2 to me. Coincidence?  Not at all: we wanted to double the light intensity, so we had to change our distance by a factor of the square root of two. Remember the equation that started all of this: I = I₀ / r². If you re-arrange that to get the distance all by itself, you get r² =  I₀ / I, or r = sqrt (I₀ / I). So if you want to halve your intensity, you have to multiply your distance by the square root of two. One third as bright? Square root of 3. One quarter as bright?  Square root of 4, namely 2.   You get the idea.

So how does this have anything to do with apertures?  Remember from digression 1 that the aperture numbers represent doubling (or halving) the light intensity of the next nearest number, but the numbers themselves are related to each other by factors of the square root of two. Awfully convenient, that, because that square root of two to double the light intensity by changing apertures is the same square root of two to halve the intensity by changing distance!  So if you multiply your aperture by root 2, thus halving the light reaching the lens, you can compensate by moving closer by a factor of root 2 to double the light intensity. The two factors of root 2 cancel out, and you're left with a constant, namely the guide number.

2. Can you make that a little simpler?

Here's another way to think of it:  as you get farther away from your strobe, its light covers a larger area. That area goes up like the square of the distance. That is, if you double your distance from the strobe, you quadruple the area the strobe light covers. Since the same amount of light covers a larger area, the intensity of the light has to decrease. The intensity decreases in proportion to the area, namely to the square of the distance from the strobe.

To compensate for the increase in area covered by the strobe, you have to increase the active area of the lens by the same proportion. That is, if you double the area lit up by the strobe, you have to double the area of the lens opening that admits light to the film.

Doubling your distance from the strobe means quadrupling the area it covers. Quadrupling the active area of the lens means doubling the diameter of the lens opening, which in turn means cutting the aperture number in half. So doubling the distance means cutting the aperture number in half...we doubled one and halved the other, so their product remains constant. There's the reason behind the guide number equation.

If you're not interested in math, stop reading now. If you're willing to follow a little algebra, here's the proof for what I just said.

3. Can you make that more complicated?  (or, Derivation of the Guide Number Equation)

We know that aperture numbers actually represent the diameter of the lens opening. Thus the formula for the active area of a lens is

A = k d_ap²

Where A is the active area of the lens, k is a constant, and d_ap is the diameter of the aperture. Furthermore, the light admitted to the film through the lens is directly proportional to the active area of the lens, so

I _f = I _i * A, or  I = I_i kd_ap²

Note that I_i represents the incident light falling on the lens, and I_f represents the light falling on the film. Now we're ready for the next step. Recall the distance equation we started with, namely I = I ₀ / r ². we need to rewrite that a little, because we want an expression that gives us the incident light falling on the lens as a function of the initial light emitted by the strobe.   Also note that we're ignoring the lens-to-subject distance. If you assume that the lens-to-subject distance is the same as the strobe-to-subject distance, or that they are related proportionally, all this does is introduce one more constant into the equation.  We're going to roll all of the constants together anyway, so ignoring one more won't change the results of the equation.

Anyway, here's the light intensity equation we want:

Insert that for I_i on the right hand side, and get

Let's divide both sides by I₀ so we collect all the intensity terms on one side:

Now let's invert both sides to get

But aperture numbers actually represent the focal length of the lens divided by d_ap, so what we want to do is replace d_ap with f/ N_a where f is the focal length of the lens, and N_a is the aperture number. Then

Since the focal length of the lens is a constant in this discussion (we're not changing lenses, right?), f² is a constant, thus so is kf². Let's roll all of them into one new constant, K². Then

Now multiply both sides by K² and take the square root of the whole equation, and you get

K(I ₀ / I _f)^1/2 = rN_a    (remember that an exponent of 1/2 means "square root")

This is the guide number equation, in disguise.  First off, forget the constant K. It's just a multiplicative number. If it makes you feel good, roll it into the guide number. Speaking of that, I₀ obviously has to be related to the guide number. Let's work on that in a second. Clearly, r is the strobe-to-subject distance, and N_a is the aperture setting.

But what the hell do we do with (I ₀ / I _f )^1/2?

It seems likely that I₀ is related to the strobe power, because all along, I₀ was the intensity of our light source. But what about I _f? Remember that we said that I _f represented the light falling on the film? Well, that's true, but if you want to make sense of the guide number equation we just derived, you have to re-interpret I _f to represent the amount of light required for proper exposure of the film. In other words, it's directly tied to the film speed. For our purposes, it might as well be the inverse of the film speed because the higher the film speed, the less light is required. So, 1/I _f = the film speed (times a constant, of course :) )

But what about the fraction, and that nettlesome square root?  Well, remember how we showed that doubling the strobe intensity only multiplied the guide number by root 2 = 1.41?  OK, then doubling I₀ does the same thing to I₀^1/2, so clearly I₀^1/2 is directly related to the guide number. Now we need to figure out what I_f really is. Recall how doubling film speed multiplied the guide number by root 2?  Well, doubling film speed here means halving I_f, which...um...yeah, multplies (I₀ / I_f)^1/2 by root 2 (dividing by 1/root2 is the same as multiplying by root 2, right?). So I_f^1/2 has to figure into the guide number, too.

I've got an idea: why don't we identify (I₀ / I_f)^1/2 as the guide number?  Does that makes sense?  Sure, because the guide numbers are always given "in feet, at ISO 100."  So if you double your film speed, the guide number goes up by a factor of root 2 for the reasons we just spelled out.   Similarly, if you turn down the strobe power to 1/2 max, I₀ just dropped by a factor of 2, so the guide number dropped by a factor of root 2, which is just what ought to happen.

So let's recast the equation in terms of the interpretations we just made. In other words, replace (I₀ / I_f)^1/2 with the guide number, r with the strobe-to-subject distance, and N_a with the aperture. Then we get a brand new equation:

GN = aperture * distance.

Ta-da!

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