User:Arseniiv/Timbres

From Xenharmonic Wiki
Jump to: navigation, search

Here are some approaches to picking harmonics for timbres for this and that purpose, aside of just taking out entire sequences of multiples of, say, 5 from a harmonic timbre.

Golden-harmonic timbres

When you want the golden ratio interval (≈833.1 ¢) to sound nice, you can take a timbre with harmonics 1 : φ : φ² : φ³ : ..., but this set of harmonics looks pretty scarce. What can you populate it with to still handle φ interval nicely but also to be more interesting and to make the timbre more adjustable?

Note that to construct a harmonic timbre from a “bare octave-allowing timbre” 1 : 2 : 4 : 8 : ..., one can just take sums of various subsets of {1, 2, 4, 8, ...} and take all of them as the new timbre. One then recovers all the natural numbers: 3 = 2 + 1, 5 = 4 + 1, 6 = 4 + 2, 7 = 4 + 2 + 1 and so on (of course you know your binary). We can apply the same sums-of-subsets construction here, but with a caveat: as φn = φn − 1 + φn − 2, we probably should disallow subsets like {φ², φ³, φ⁴}: in this one, φ⁴ effectively contained twice, and its sum is “incorrect”. (That’s easy to do: just disallow subsets which contain {φn, φn + 1} for some n.) Proceeding this way from powers of φ, we get intervals

1, φ, φ + 1 ≡ φ², φ + 2, 2φ + 1 ≡ φ³, 2φ + 2, 3φ + 1, 3φ + 2 ≡ φ⁴, 3φ + 3, 4φ + 2, 4φ + 3, 4φ + 4, 5φ + 3 ≡ φ⁵, 5φ + 4, 6φ + 3, 6φ + 4, 6φ + 5, 7φ + 4, 7φ + 5, 8φ + 4, 8φ + 5 ≡ φ⁶, ...  (G1)

We can note that neighboring intervals in this list differ either by 1 or φ − 1 ≈ 0.68, so they are spaced quite nicely to not be immediately a dissonant mess. (As in harmonic timbres they are all spaced by 1 and that sounds nice, given the greater harmonics are very quiet in regard to the small ones. And 0.68 is pretty close to 1 and is rarer encountered.)

Now multiply an interval r from this list by φ. As it’s a sum of powers of φ with no exponents differing by just 1, so is r φ. We can place other rules on powers in these sums, given these rules behave well under multiplication by φ.

We can slightly depart from a sums-of-subsets approach, filtering all possible m φ + n intervals in another way: as earlier, include each power of φ, and also as earlier make differences between adjacent intervals 1 or φ − 1, but no other constraints. Though I feel the intervals picked, considered as points (m, n) in the plane, should be close to the polygonal chain with vertices φk.

The following ASCII art illustrates such a planar representation for the interval list (G1) constructed above. It’s easily seen we can change an angle here and there, e. g. add 2φ + 3 while leaving out 3φ + 1.

  | 0 1 2 3 4 5  n
--+--------------→
0 | o-@   .   .
  |  /    .   .
1 | @-@-o .   .
  |    /  .   .      @ — powers of φ
2 |   @-o .   .      o — other intervals
  |    /  .   .      - — adding 1
3 |   o-@-o   .      / — adding (φ − 1)
  |      /    .
4 |     o-o-o .
  |        /  .
5 |       @-o .
  |        /  .
6 |       o-o-o
  |          / 
7 |         o-o
  |          / 
8 |         o-@-...
  |
m ↓

Initially I came to this scheme by taking base-Fibonacci numeral system but treating each Fibonacci number as a power of φ. I tried to compact the description but it might have gone hard to understand, so feel free to comment.

And I think something in this vein may be possible for any other interval which is a root x of a low-degree polynomial equation xn = ... with integer coefficients (or even rational ones?). And I hope very much such a timbre sounds well — hadn’t tested that yet.

Another timbre

Now I think (G1) has its harmonics too close. We can fix this without remorse if we treat 1 as somewhat distinct from all others and start really adding two chosen differences only from φ. In that case we can choose 1 and φ (we may just scale all of (G1) by φ, effectively skipping some harmonics that are too close to their neighbors):

1, φ, φ + 1 ≡ φ², 2φ + 1 ≡ φ³, 3φ + 1, 3φ + 2 ≡ φ⁴, 4φ + 2, 4φ + 3, 5φ + 3 ≡ φ⁵, 6φ + 3, 6φ + 4, 7φ + 4, 8φ + 4, 8φ + 5 ≡ φ⁶, ...  (G2)

Other findings without structuring

We can use a similar approach to build a simple “√2-enduring” timbre:

1, √2, 2, (√2 + 1), 2√2, √2 + 2, 4, (√2 + 3), 2√2 + 2, (3√2 + 1), 4√2, 3√2 + 2, 2√2 + 4, √2 + 6, 8, ...  (S1 and S2)

Here we also can either use differences √2 − 1 ≈ 0.4 and 2 − √2 ≈ 0.6 right from the start, or we can start adding 2 − √2 and 2√2 − 2 ≈ 0.8 just after reaching 2, effectively skipping half of the harmonics of the first timbre each time we go from an even power of √2 to the next odd power.