Partials

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

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

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

Currently my range is from the B^b 2nd partial (I can get the pedal B^b not-very-reliably after I've been practising for a while and my lips are loosened up) up to about the 6^th partial, sometimes 7^th, 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 B^b 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 B^b (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 B^b(58.27Hz) to 1st partial B^b(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 : B^b
5 * 58.27Hz = 291.25Hz : D
6 * 58.27Hz = 349.65Hz : F
7 * 58.27Hz = 407.89Hz : A^b (this one's a bit flat, A^b 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.

More details here: http://en.wikipedia.org/wiki/Acoustic_resonance#Resonance_of_a_tube_of_air