## Monday, March 14, 2011

### Partials

Aha - so this is what the different notes you can play without moving the slide are called !

On my trombone (a regular Bb bone) there's a harmonic series of notes I can (notionally !) play purely by changing my embouchure:

 Partial Note Fundamental [1st partial] 2nd 3rd 4th 5th 6th 7th 8th 9th 10th Bb Bb F Bb D F Ab- Bb C D

Currently my range is from the Bb 2nd partial (I can get the pedal Bb not-very-reliably after I've been practising for a while and my lips are loosened up) up to about the 6th partial, sometimes 7th, again once my lips have warmed up a bit, but not very reliably.

Interestingly the intervals of the harmonic series (of partials) form a pretty curve when you write them up on a stave, here's the Bb series:

It turns out that the reason for this curve is rooted in physics (as you'd expect), but, curiously, also has to do with our perception of sound;  the fundamental pitch of my trombone is the pedal Bb (at 58.27Hz), it transpires that each partial up is a multiple of the fundamental frequency, so the next partial up from the fundamental is 2 * 58.27Hz.  More details below for the curious.

That's the physics bit, now the perception bit:  what we hear as an octave interval is really the doubling of a note's frequency.

So from 58.27Hz to 2*58.27Hz explains the octave interval from Fundamental Bb(58.27Hz) to 1st partial Bb(116Hz).

The next partial is 3*58.27Hz = 174.81Hz which is an F (there's a table of frequencies here), and so on:

4 * 58.27Hz = 232.08Hz : Bb
5 * 58.27Hz = 291.25Hz : D
6 * 58.27Hz = 349.65Hz : F
7 * 58.27Hz = 407.89Hz : A(this one's a bit flat, Ab is more like 415Hz - another weird perceptional thing?)
8 * 58.27Hz = 466.16Hz : Bb
9 * 58.27Hz = 524.43Hz : C
...

Now you know :)

More on the physics

Open cylindrical tubes resonate at the approximate frequencies:
f  = nv / 2L
where n is a positive integer, L is the length of the tube, and v is the speed of sound in air (~343m/s).

A trombone with slide closed is about 2.8m, and has a bore of about 1.4cm so the fundamental resonant frequency is:
f = (1 * 343) / (2 * 2.8)
= 343 / 5.6
= 61Hz
Near enough, the formula is an approximation since the anti-node reflection point is a little past the end of the tube.

Solving for the other partials gives:
f=2: 122
f=3: 183
f=4: 245
f=5: 306
etc.