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.
Table of Contents:
 Why Does the Guide Number Equation work?
 Can you make that a little simpler?
 Can you make that more complicated?

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 = I_{0} / r^{2},
where I is the intensity measured at any point, I_{0}
is the intensity of the light measured at the source, and r^{2}_{
}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
= I_{0} / r^{2}. If you rearrange that to get
the distance all by itself, you get r^{2} = I_{0}
/ I, or r = sqrt (I_{0} / 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_{ap}^{2}
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_{ap}^{2}
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 lenstosubject distance. If you assume that the lenstosubject
distance is the same as the strobetosubject 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

I_{ 0}kd_{ap}^{2} 
I_{ f} = 


r^{ 2} 
Let's divide both sides by I_{0} so we collect all the
intensity terms on one side:
I_{ f} 

kd_{ap}^{2} 

= 

I_{ 0} 

r^{ 2} 
Now let's invert both sides to get
I_{ 0} 

r^{ 2} 

= 

I_{ f} 

kd_{ap}^{2} 
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
I_{ 0} 

r^{ 2} 

r^{ 2} N_{a}^{2} 

= 

= 

I_{ f} 

k(f/N_{a})^{ 2} 

k f^{ 2} 
Since the focal length of the lens is a constant in this discussion (we're not changing
lenses, right?), f^{2 }is a constant, thus so is kf^{2}. Let's roll
all of them into one new constant, K^{2}. Then
I_{ 0} 

r^{ 2} N_{a}^{2} 

= 

I_{ f} 

K ^{2} 
Now multiply both sides by K^{2 }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, I_{0} obviously
has to be related to the guide number. Let's work on that in a second. Clearly,
r is the strobetosubject 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 I_{0} is related to the strobe power, because
all along, I_{0} 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 reinterpret 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 I_{0 }does the same thing to I_{0}^{1/2},
so clearly I_{0}^{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 (I_{0} / 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 (I_{0} / 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, I_{0}
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 (I_{0} / I_{f})^{1/2}
with the guide number, r with the strobetosubject distance, and N_{a} with the
aperture. Then we get a brand new equation:
GN = aperture * distance.
Tada!
