Actually, Josh, this is basic high school math.  Unless you actually
    are a high school student and haven't gotten to logarithms and
    exponentials yet, you have no business trying to be an engineer.

You can get through high school math, with good grades, without gaining
much of an understanding about how logarithms get applied to the real
world.  More people tend to forget their basic calculus, "required" in
college but never really used.  I know that the last time I looked at my
college physics text, there were an awful lot of symbols that I don't
remember how to work with (gradients and laplacians and ... stuff.)

One need not be "trying to be an engineer" to want to play with PICs and
pretty colored LEDs.  If an artist (for example) gets led into technical
issues in their search for an effect, that's OK.  If they end up
understanding that there's method and science behind the scenes, that's
good (IMO), and if they end up actually digging in and relearning their
"high-school-level math", that's quite excellent.

Historically, this has been more common in analog electronics and music,
I think.  But I don't think that us "REAL Engineers" ought to be too upset
at having to explain things every once in a while.  It can be better than
being asked to IMPLEMENT an effect at a level we don't understand either.

So...

We're trying to get preceived linear (smooth) brightness changes in a LED
by matching the (supposedly) logarithmic response of the human eye/etc.

A "logarithmic series" is one where the next value in the series is
calculated by multiplying the previous value by a constant.

S0 = A
S1 = S0 * C = A * C
S2 = S1 * C = A * C * C
S3 = S2 * C = A * C * C * C
  :
Sn = A * C^N

So if S0 = 1, and Sn = 255, for instance, we know that A = 1, and 255 =
C^n, where N is the number of steps in the series.  To solve for C, take
the Nth root of both sides C = root(255, N) An Nth root is equivilent to
raising to the power 1/N (this is derived from logarithm theory), and this
is the "m" factor that Olin's program calculates, more or less:
m = max_out^(1.0/max_in - 1)
m = 255^(1/254)

BillW

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