Stacking Threes on the Harmonic Series

You start out with the harmonic series, let’s say using 110 Hz (A) as the base frequency along with all of its multiples.

110 Hz, 220 Hz, 330 Hz, 440 Hz, 550 Hz, 660 Hz, 770 Hz, 880 Hz, 990 Hz, 1100 Hz, and so on …

Each multiple can be said to be an “interval”.

We have the *1 interval (which keeps us at 110), the *2 interval (which takes us to 220), the *3 interval (330), and so on.

There are other types of intervals within the series, namely the ones that occur between each frequency with one another, but that’s not the approach I’m going to take here.

Keep in mind that intervals are multiplication based and not addition based. Think of the 220 as the frequency you get from multiplying the base frequency by 2, not from adding 110 to the base frequency. If I started on 440, adding 110 to it wouldn’t have the same effect as it does when 110 is the base. Instead I’d have to add 440, or multiply by 2. It’s a small nitpick in wording, but it prevents some minor confusion early on.

From here, the goal is to try to build up a set of unique notes to fit within an octave. The logical thing to do, it would seem, is to take one of these intervals and apply it over and over again.

If we try that with the *1 interval, we end up staying in one place. No matter how many times we multiply by 1, we’ll always stay on the same frequency. In this case, we have just one note.

The next thing to try is the *2 interval. Now we have some movement, but while we aren’t stuck on the same note anymore, multiplying by 2 over and over (which is the same as multiplying by some power of 2) means we’re stuck in the same “pitch class”. If 110 is the note ‘A’, multiplying by 2 takes us to 220, which is also an ‘A’, and multiplying by two again takes us to 440, which is also an ‘A’.

Multiplier as Power of 2Multiplier as ValueBase Freq. * Multiplier
* 2^0* 1110 (A)
* 2^1* 2220 (A)
* 2^2* 4440 (A)
* 2^3* 8880 (A)

The same is true for dividing by 2 over and over. 55 is ‘A’, 27.5 is ‘A’, and so on.

Multiplier as Power of 2Multiplier as ValueBase Freq. * Multiplier
/ 2^0/ 1110 (A)
/ 2^1/ 255 (A)
/ 2^2/ 427.5 (A)
/ 2^3/ 813.75 (A)

So that didn’t work.

The next thing then is to try the *3 interval. We start at 110 and multiply by 3. We get 330. That’s not the same frequency/note, nor is it the same pitch class, so that’s good. With 330 as a new starting point, we multiply by 3 again, and we get 990. That’s not the same pitch class as either of the previous two. (Seeing that it’s not the same pitch class is a matter of seeing that you can’t get to 990 by multiplying the other frequency (110 or 330) by any power of 2.)

It looks like we’ve got something. Let’s continue

Multiplier as Power of 3Multiplier as ValueBase Freq. * Multiplier
* 3^0* 1110
* 3^1* 3330
* 3^2* 9990
* 3^3* 272970
* 3^4* 818910

None of the added pitches are in the same pitch class as any of the previous ones, so this seems to be working great, but the question now is when do we stop?

Since the purpose of this is to construct a set of unique notes within an octave, it would make the most sense to stop when we get to some power of 3 interval that’s equivalent to some power of 2 (the base frequency * a power of 2 means landing on, and thus returning to, the base pitch class). But we know that that’s impossible. There is no power of 3 that’s equivalent to a power of 2. So the next best thing is to try to get as close as we can. At each interval, we’ll look for the nearest power of 2 that’s still less than (or equal to) the power of 3 we’re at, and stop when we get one that’s close enough.

And at this point, we can actually ignore the frequencies themselves (e.g., 110 Hz, 330 Hz, etc) and focus exclusively on the intervals. This works no matter what your original base is.

Power of 3Power of 3 ValueClosest Power of 2Power of 2 ValuePower of 3 / Power of 2
3^012^011.0000
3^132^121.5000
3^292^381.1250
3^3272^4161.6875
3^4812^6641.2656
3^52432^71281.8984
3^67292^95121.4238
3^721872^1120481.0679
3^865612^1240961.6018
3^9196832^14163841.2014
3^10590492^15327681.8020
3^111771472^171310721.3515
3^125314412^195242881.0136

With 2^19 being so close to 3^12, we can stop there, and mark that ratio as functionally equivalent to 1.0000.

So now our set of intervals looks like this

Power of 3Power of 2Power of 3 ValuePower of 2 ValuePower of 3 / Power of 2
3^02^0111.0000
3^12^1321.5000
3^22^3981.1250
3^32^427161.6875
3^42^681641.2656
3^52^72431281.8984
3^62^97295121.4238
3^72^11218720481.0679
3^82^12656140961.6018
3^92^1319683163841.2014
3^102^1459049327681.8020
3^112^171771471310721.3515

Now if we were to start with a base frequency of 110 Hz, our octave frequencies would be:

IntervalInterval * Base Frequency
1 / 1110.0000 Hz
3 / 2165.0000 Hz
9 / 8123.7500 Hz
27 / 16185.6250 Hz
81 / 64139.2188 Hz
243 / 128208.8281 Hz
729 / 512156.6211 Hz
2187 / 2048117.4658 Hz
6561 / 4096176.1987 Hz
19683 / 16384132.1490 Hz
59049 / 32768198.2236 Hz
177147 / 131072148.6677 Hz

Putting them in order, we get:

IntervalInterval * Base Frequency
1 / 1110.0000 Hz
2187 / 2048117.4658 Hz
9 / 8123.7500 Hz
19683 / 16384132.1490 Hz
81 / 64139.2188 Hz
177147 / 131072148.6677 Hz
729 / 512156.6211 Hz
3 / 2165.0000 Hz
6561 / 4096176.1987 Hz
27 / 16185.6250 Hz
59049 / 32768198.2236 Hz
243 / 128208.8281 Hz

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