Re: coverage

Good answers have been given, but they have answered the question of why the
fall-off more than the question "why 4 stops of fall-off)". So here is my $0.02
worth of answer.

----- Original Message -----
From: "Stewart C. Russell" <scruss@sympatico.ca>
> >
> > Could anyone explain why "about 4 stops of falloff at ghe edges" ?
>
> that was from an off-list response. I was told that light falloff is
> proportional to cos^4 of the off-axis angle.
>
> So for 120 degrees symmetrical coverage, that would be cos^4 (120/2).
> As cos 60 = 1/2, that would be 1/(2^4), or four stops falloff.
>
> Or am I off somewhere?

It worked well in this case Stewart, but that explanation wouldn't work for
other angles, I think, but I may be wrong, I am known to be wrong from time to
time, actually my wife thinks I am wrong very often! but that is a topic for a
different NewsGroup!!

Here is my explanation:

Short answer: The intensity of light at OFF axis points on a film is equal to
the intensity of light at the center of the film (ON axis) multiplied by the
COSINE to the power of 4 of the angle of that off axis point (you may ask
why cosine^4, why not cosine^3, but you didn't ask that, so I will keep
going with this explanations). If we have a camera with a total
angle of view of 120°, it means the most OFF axis point on this piece of film
is 60° with respect to the center (60° on one side and 60° at the opposite makes
for 120° total). So, as Stewart said, Cosine of 60° is 0.5 and Cosine^4 would
be (0.5)^4=0.0625 which is nothing but multiplying 0.5 four times by itself
(0.5 x 0.5 x 0.5 x 0.5 = 0.625). Let's see: if we had 100 units of light
(whatever units) at the center of the film and the total angle of view is 120°,
the intensity of light at the corners of the film would be:

Intensity at corners = Intensity at center X [Cosine(60°)]^4
as we saw above [Cosine(60°)]^4 is 0.0625, so we have:
Intensity at corners = 100 X 0.0625
Intensity at corners = 6.25
That means that for the above camera the light intensity at the corners is 6.25%
of the light at the center, now your question: "why 4 stops":

As it turns out, 1 stop of difference means double (or half) the intensity of
light. For instance, if f16 gives you 50 units of light at the center of a
film, you can bet that f/11 will give you 100 units. To find out how many stops
of separation exist between 2 intensities of light, we just need to double the
lower intensity until it reaches the higher intensity, the number of times you
doubled the lower intensity is the number of stops. To double any number we
just multiply by 2. So lets see how many STOPS are there between 6.25 units of
light and 100 units of light:

6.25 x 2 = 12.5 (first doubling = 1 stop)
12.5 x 2 = 25 (second doubling = 2 stops)
25 x 2 = 50 (third doubling = 3 stops)
50 x 2 = 100 (fourth doubling = 4 STOPS)

There is your answer, that is why there are 4 stops of fall off from center to
corners on a 120 angle of view flat film plane camera.

Long answer: the short answer ended up being so long that I will omit a long
one. :-)

Guillermo
Received on Wed Jun 25 08:59:38 2003