finally solving dyadic error for chords

Sometimes just finding the right question is hard.

Suppose I have a just chord like 4:5:6:7. Let's not assume
octave equivalence. Now suppose I have an approximation to
this chord, a:b:c:d. Define the following errors for this
approximation:

| 5/4 - b/a |
| 3/2 - c/a |
| 7/4 - d/a |
| 6/5 - c/b |
| 7/5 - d/b |
| 7/6 - d/c |

Now let's make two different composite errors out of this
list: max-abs and RMS. We can also do the arithmetic mean
(RMS is a compromise between max-abs and the mean in terms
of imposing a penalty for a single relatively large error),
but for now let's just focus on max-abs and RMS.

Now, find a:b:c:d such that | max-abs - RMS | is maximal.

In the case of a triad, if we plot the sizes of the interior
intervals of the triad (a/b and b/c) on cartesian x & y axes,
then we can plot both max-abs and RMS error for the chord on
the z axis. Now replace these two plots on the z axis with
their difference. We are now interesting in triads located
under the local maxima on the z axis.

For tetrads we need a hypervolume. We only need one additional
dimension for each higher prime limit, since chords are always
uniquely defined by their interior intervals. And I think
the hill-climbing algorithm scales to higher dimensions and
can find local maxima there.

The idea is then to synthesize audio of these triads and
have listeners rank them. The rankings can then be compared
to the max-abs or RMS rankings and the best error averaging
method can be determined.

If the max-abs and RMS plots never cross on the z-axis (the
difference function is ever zero), then I think the rankings
will not differ. The last time Paul and I went around on
this, I think the consensus was that with triads, the rankings
will not differ, but with tetrads and higher-ads they will.

I will be happy to synthesize the chords if anyone can solve
this problem.

-Carl