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Re: Question about EEb bass

On 26th November Dave Taylor wrote:

> Using the formula c (speed of sound) = f (frequency of 2nd harmonic) *
> lambda (length of wavelength/tube), the total tubing would be engaged with
> all 4 valves down, giving a 2nd harmonic of E natural at approximately
> 41.2 Hz. The speed of sound is iirc about 321 metres/second. Rearranging
> to give lambda=c/f produces tube length = 321/41.2 = 7.79 metres = about
> 25 feet 6 inches. It will lose a bit because of the various bends in the
> tube and the bell flare. Call it 25 feet to the nearest foot.
>
> Dave Taylor
> p.s. if my value of the speed of sound is wrong, then this needs to be
> redone, replacing 321 with the appropriate value.
>

Seeing this elegant solution to the problem of the length of tube in a tuba
inspired me to re-read an article which was published in Scientific American
in July 1973 on the Physics of Brasses (re-published in the book "The
Physics of Music"). After wading through the delights of the Horn Equation
and comparisons with quantum mechanics I think David's analysis needs a
little modification. The standing wave set up in the instrument has a node
at the bell end and an antinode at the mouthpiece, therefore the tube length
will be equal to an odd multiple of half-wavelengths. In the case of the 2nd
harmonic the length of the tube will be 3 half wavelengths. David calculates
the wavelength of a low E natural to be 7.79m, that makes the
half-wavelength 3.895m and the length of tube 11.685m (or approximately 38ft
4inches). This seems a bit long to me, perhaps we should ask Boosey &
Hawkes.

I don't think it matters what length you use as long as you can justify it
to the people who loose!

Mike Stevenson


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