On my trombone (a regular Bb 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 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 : Ab (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
More details here: http://en.wikipedia.org/wiki/Acoustic_resonance#Resonance_of_a_tube_of_air
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
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