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# Mathematical formulae and wind instrument help.?

Is there a mathematical formula for the arrangement of hole on a flute or better a recorder.

In addition has anyone every tried to make a saxophone mouthpiece ?

I am interested in trying to make a DIY version of a Xaphoon http://www.youtube.com/watch?v=j3Wnx2VYZq8

## Discussions

For a tube closed at one end (by a mouthpiece), the frequency (

f) of a note produced isf=^{v}/_{2L}where v = speed of sound in air (340m/s), and L is the length of the tube. For you, that is the distance between the mouthpiece and the first un-covered hole.Rearranging to solve for L, you're looking at L =

^{v}/_{2f}For instance...Google reveals that "middle C" is 523.28Hz

L =

^{340}/_{(2x523.28)}= 0.324873872mWith rounding, the length of a tube that gives a middle C = 0.325m = 32.5cm = 12.8 inches.

If the first hole uncovered by your finger is 32.5cm from the top, you will get a middle C.

Doesn't the diameter of the tube have an effect?

As long as the tube is significantly longer than wide, no.

The Wikipedia article agrees and disagrees:

http://en.wikipedia.org/wiki/Acoustic_resonance#Open

Oh, hang on...

I quoted maths above assuming that the tube resonated as a closed tube.

However, the maths for a closed tube traditionally places the sound generator at the open end. These Xaphoons have the sound source at the closed end.

Damn. I've over-thought myself.

They do indeed - it is a Saxaphone reed /mouth piece.

Is the measurement made from the tube end or from the sound source? in my case there may be some slight difference

The U-Tube video I linked to shows the original "inventor" making them by hand from bamboo BUT he is a musician and sax player and I am not but I would like to try to make one from PVC pipe.

My reason for wanting the maths is I have a profile of a Xaphoon and They say it is 12.5 inches long, I have made an "accurate" scale drawing f the picture and want to check the measurements I get from it.

Even flutes and recorders, in fact all musical wind instruments AFAIK, produce the sound at the closed end.

Thanks for all your efforts guys - any more input is welcome before I start dismantling the house plumbing!

I thought it works only for a kind of flute where you move a stick in and out of it, then it's a closed tube of a variable length and it is easy to calculate the length of a standing wave inside (even I can do it). But what happens if the flute is open at the end and in lots of little holes except one?

1) Air flow though the open holes is obviously significant, so we can't consider it a tube...

2) I have no idea of how to calculate a standing wave inside such a thing, but I can't believe it continues to be the same formula.

PS I think that a tube closed at one end is a thing where the wave meets a solid wall at the end.

This link might help? It has a link to an online calculator too :)

will follow your link Thanks

searching for 'woodwind recorder math' in google got me a pdf about how to make them...

and a dude says 'the math is difficult, but I wrote software to calculate it"...

http://www.logarithmic.net/pfh/design

I have not tested this software nor can I presume it's remotely safe.