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I have been trying to create a xylophone out of copper pipes for a school project and have been running into a few problems with calculating the length of each bar. I have been using this paper as reference, specifically equation (1). I have cut two pipes so far with lengths of 0.884m and 0.585m which played the notes B3 and A♭6 respectivly, which is far from the expected E2 and G♭3. For reference, my pipes have an inner radius of 0.25 in and an outer radius of 0.3125 in.

Even when I tried "reverse calculating" the speed of sound in the bar using (1) with both sets of data, I got answers both different from the expected 3700 m/s and from each other. Does anybody have any idea why this could be, either an error on my part or in my calculations? Is there another equation I should be using or is it more likly I am just making some silly error?

For reference, here are my calculations for the 0.884m bar:

K = 1/2 sqrt(0.25^2 + 0.3125^2) = 0.2 in = 0.0048 m

f = (pi * v * K * 3.011^2) / (8 * L^2) = (pi * 3700 m/s * .0048 m * 3.011^2) / (8 * (0.884 m)^2) = 80.9 Hz

This is around (a bit lower than) the frequency of E2, but when tested the tuner registers B3.

After some more tinkering around and reading through some comments, I am more lost than ever. I tried mounting the bars this time using rubber foam weather seal but the pitch was being registered even more spastic from the tuner. I have attached a video here, which hopefully can be of use.

COsborne
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    Are you using the pipes for resonators or as the vibrating bodies themselves? – phoog Jan 20 '20 at 04:19
  • @phoog I am using the pipes as vibrating bodies, not as resonators – COsborne Jan 20 '20 at 05:06
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    I suspect that you're getting an overtone. B3, for example, is the third harmonic of E2. A♭6 is the 9th harmonic of G♭3. I don't quite see how those pipe lengths would give fundamental pitches so far apart, though, so I'm not sure you've calculated the expected pitches correctly. – phoog Jan 20 '20 at 05:20
  • You're mixing systems of units in the question. Did you take this into account in your calculations? (Did you convert the inner and outer radius of the pipe to millimeters, the unit used in the paper?) – phoog Jan 20 '20 at 05:28
  • @phoog the paper uses millimeters initially but then converts those into meters when doing the calculations for L (at least that’s what I can gather). That’s a good point about the overtones though—I’ll try messing around with tuners/my stand/mallets to test that tomorrow afternoon and will update with the results. – COsborne Jan 20 '20 at 05:44
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    Can you record the sound it is making and post it here? Just to rule out harmonics being read by the tuner, which happens frequently. Or are you really sure that's not the case? Can you do FFT images? – NPN328 Jan 20 '20 at 06:02
  • That’s a good idea Von Huffman, when I return home this afternoon I will try uploading some audio files. – COsborne Jan 20 '20 at 13:10
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    How are you mounting these sections? If there's even a little damping or restraint you may find that the fundamental is being destroyed in favor of an overtone. In the meantime, if you can get some literature on vibraphone design (where the bar length and the resonant pipe lengths have to be matched up), that might help. – Carl Witthoft Jan 20 '20 at 14:35
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    For your analysis, I **strongly** recommend cutting one clean length, making sure you can hear a fundamental, and use that to estimate a single constant "Alpha" in the reduced equation frequency = Alpha/(length^2) . If a second bar 1/4 or 4X as long doesn't produce an octave, then you have to figure out how better to mount or strike your pipes. – Carl Witthoft Jan 20 '20 at 14:39
  • After looking through these comments I think the problem definitely is with my mounting. I'm going now to get some rubber weather seal as was used in the paper, but before I was just getting a buddy to stretch out two rubber bands at the nodes (0.224L and 0.776L) which I placed the pipe on and then struck at its center. Going to try the weather seal later tonight, but if anybody has any other mounting suggestions I would love to hear them. – COsborne Jan 20 '20 at 22:54
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    @VonHuffman just added a video reference to the thread if that will help. – COsborne Jan 21 '20 at 01:04
  • Place the supports at the ends of each pipe length? – Carl Witthoft Jan 21 '20 at 16:12
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    A minor point, the "xylo" in xylophone means wood. Since you are using metal, glockenspiel might be better. See this previous question: https://music.stackexchange.com/questions/24243/what-is-the-difference-between-a-xylophone-a-glockenspiel-a-marimba-a-xylorim/24245#24245 – badjohn Jan 22 '20 at 12:40

2 Answers2

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Try using the formula: A=L2f. f is frequency, L is length, and "A" is the number you get from doing all the measurements described below.

I tried to do it the way you described first, but I found a bunch of different speeds for the speed of sound in copper and I wasn't sure where I was going wrong. Then I found that formula.

If you know your pipe length and frequency, then you can calculate "A". Then, use that to find the length of pipe for the frequency that you're shooting for. I would cut each pipe a little long, then figure out my "A" again, just because my tools were not super accurate. Get it real close, and then file the ends to fine tune. It takes a while, but I was very pleased with what I was able to finish.

Edward
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Mike gerard
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Your radius is 0.25" but you use 0.5" in the calculations. I fixed that ran through your numbers and got 237Hz for n=1, which is not that far from B3=247Hz (actually it's about half way between B and Bb).

I was only working to two significant figures though. (Tip - it's less error prone to just put numbers in in the right units and THEN calculate ...)

EDIT : Here is my working at OP's request. Actually it was 236Hz, not 237Hz. But close enough for jazz (xylophone) ;-)

0.25" = 6.35mm

0.3125" = 7.94mm

K = sqrt(63.00 + 40.32) / 2 = 5.08mm = 0.00508m

f = [(pi * v * K) / (8 * L^2)] * m^2

If L = 0.884m

= [59.05 / (8 * 0.781)] * m^2

= (59.05 / 6.25) * m^2

= 9.44 * m^2

If n = 2, m = 5

f = 25 * 9.44 = 236Hz

(The frequency your tuner is picking up seems to be a harmonic.)

danmcb
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    Oops I think I accidentally typed 0.5in instead of 0.25in, thank you for catching that error. However, that 0.2in was still correct (I think) with 0.25in. How did you calculate 237 Hz? Is it possible that you could include your calculations? – COsborne Jan 20 '20 at 13:04
  • The video OP posted is a dead link now, but the pipe glockenspiels I found on Youtube produce very pure tones- It's very unlikely that the tuner is picking up any harmonic. – Edward Mar 16 '22 at 02:30
  • how do you know they are pure tones? the only instrument I know of that produces anything close to a sine wave is a flute. – danmcb Mar 16 '22 at 12:50