It's a totally naive laymans approach
I hope the formatting stays in place.

The feedback delay in the loop folds the signal back
so we have periods of a comb filter.
|          |          |          |
|__________|__________|__________|___

Now we want to fill the period densly with impulses:

First bad idea is to place a first impulse exactly in the middle

that would be a ratio for the allpass delay of 0.5 in respect to the comb filter.
It means that the second next impulse falls on the period.

|         |
|____|____|___


The next idea is to place the impulse so that after the second cycle
it exactly fills the free space between the first pulse and the period like this,
exactly in the middle between the first impulse and the period:

|           |           |
|     |     |  |        |
|_____|_____|__|__|_____|___

this means we need a ratio "a" for the allpass delay in respect to the combfilter loop that fulfills:

2a - 1 = a/2

Where 1 is the period of the combfilter.
Alternativly, to place it on the other side, we need

2a - 1 = 1 - a/2;


|       |       |
|   |   |     | |
|___|___|___|_|_|___

This gives ratios of 0.5. 0.66667 and 0.8

These are bad ratios since they have very small common multiples with the loop period. So we detune them slightly so they are never in synch with the loop period or each other.
That was my very naive approach, and surprisingly it worked.


The next idea is to place the second impulse not in the middle of the free space
but in a golden ratio in respect to the first impulse

|            |            |
|       |    |    |       |
|_______|____|____|__|____|

2a - 1 = a*0.618...

or

N*a mod 1 = a*0.618..

or if you prefer the exact solution:

a = (1 + SQRT(5)) / ( SQRT(5)*N + N - 2)

wich is ~ 0.723607  and the same as 1/ (1+ 0.382...) or 1/ (N + 0.382)

where N is the number of impulses, that means instead of placing the 2nd impulse on a*0.618
we can also place the 3rd, 4th etc for shorter AP diffusors.

(And again we can also fill the other side of the first impulse with 0.839643 And the solution for N = 1 is 2.618.. and we can use the reciprocal 0.381 to place a first impusle)

The pattern this gives for 0.72.. is both regular but evenly distributed so that each pulse falls an a free space, just like on a Fibonaccy flower pattern each petal falls an a free space,
forever.
(I have only estimated the first few periods manually, and it appeared like that Its hard to identify in the impulse response since I test a loop with 3 APs )

The regularity is a bad thing, but the even distribution seems like a good thing (?). I assume it doesn't even make a huge difference to using 0.618.. for a ratio though it seemed to sound better.
(And if you use 0.618, what do you use for the other APs?)

So it's not the solution I am looking for but interesting never the less.

I believe that instant and well distributed echo density is a desired property
and I assume that the more noise like the response is as a time series
the better it works also in the frequency/phase domain.

For instance you can make noise loops with randomizing all phases by FFT in circular convolution
that sound very reverberated.




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