# Harmonic lengths of a set of tubular drums?

I have a 200 cm long piece of a clear plastic tube with 20 cm outside diameter that I would like to cut into three pieces to make a bigger diameter version of the tubular drums that exists today in smaller diameters (Octobans etc).

Is there any mathematical theory that could help me to predict what lengths I should cut the tube into?

I can of course tune the drum heads to different notes but I'm looking for a way of knowing how to cut the tube into three pieces that will resonate harmonically together by themselves and also produce three clearly distinguishable pitches.

• I'd have thought a simple division by 6 would do it. One 1/6th, next 2/6 (1/3), last 3/6 (1/2), so allowing for mimimal kerf, 333mm, 666mm and 999mm. – Tim Apr 7 at 15:06
• @Tim: wouldn't 1/6 be the octave of 1/3? – Albrecht Hügli Apr 7 at 16:08
• how to cut the tube into three pieces that will resonate harmonically: what do you mean by this? which intervals? root, fifth, third? – Albrecht Hügli Apr 7 at 16:09
• Google 'octobans' and dimensions are stated there on Wiki. The maths could be worked out from those dimensions. – Tim Apr 7 at 16:26
• @CarlWitthoft - it's possible to play most instruments that way! Even cellos! Wonder if you've tried it..? Seriously, tubular bells et al, work best because the material they're made from - metal - has a better vibratory factor (maybe there's a better word?) than the plasic envisaged here.Miced up it could be interesting, though. – Tim Apr 8 at 14:50

Another solution might be:

Root = 1/2

3rd = 2/5

5th = 1/3

(Common quotient of 2, 5 and 3 is 30)

that is 15/30 (root) 12/30 (3rd) 10/30 (5th)

(15+12+10)= 37 parts

200cm:37=5.4cm (2mm=rest)

Root=15*5.4cm=81cm

3rd=12*5.4cm=64.8cm

5th=10*5.4cm=54cm

Sum=199.8cm (2mm = rest for cutting)

• This is very interesting Albrecht, thanks a lot! – Per Alenfelt Apr 8 at 7:41
• if I would only cut the tube into two pieces and go for the root and 5th, can I then simply use the same mathematics but start with: 3/6 + 2/6 => 5 parts? – Per Alenfelt Apr 8 at 7:51
• If my assumptions are correct - in this case 1/5*200=40 -> one part would be 40 and the 2 lengths 1/2:1/3 = 3/6:2/6 were 3*40 : 2*40= 120:80 you wouldn't have any rest for cutting but I think you could ignore it. Try this out with a cheap plastic tube ;) – Albrecht Hügli Apr 8 at 8:02

As the pitch is produced by the air in the tubes depending of the length you can choose the natural physical ratios of the length of the air wave known since Pythagoras.

If you want to have a triad (fifth, octave, third) 1/3 + 1/4 + 1/5 = 200cm = 20/60+15/60+12/60

that means the ratio has to be 20:15:12 (20+15+12=47)

You divide the 200cm:47=42.5cm (2.5mm rest)

5th = 20*42.5cm=85cm

8ve = 15*42.5cm=63.75cm (root)

3rd = 12*42.5cm=51cm

Sum = 199.75cm Rest difference = 2.5mm

(the rest 2.5mm you can use as fall out from cutting)