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'pipe organ method - water level'
1999\01\19@232438 by Tjaart van der Walt

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John Payson wrote:
>
> An open flue pipe will produce a fundamental whose wavelength is equal
> to the twice length of the pipe plus a small extra amount related to
> its diameter.  The overtones produced will have wavelengths of
>    2L/n + k
> where 'L' is the pipe length, 'n' is a positive integer, and 'k' is
> about the same for all overtones (note that the overtones produced are
> not exact harmonics).
>
> A closed flue pipe will produce a fundamental whose wavelength is equal
> to four times the length of the pipe, plus a small extra amount related
> to its diameter.  The overtones produced will have wavelengths of
>   4L/(2n-1) + k
> where the parameters are as above.  Note that the fundamental pitch
> will be about an octave lower than an open pipe of the same length, but
> the pipe will have a much less "full" sound.

...now I would like to bring it closer to being on topic...

I have a private application where I want to measure water
level in a pipe. Bob Blick put me onto a nice way of doing
this with an LM3909. There is an application note describing
a self-resonating circuit where the speaker forms part of
the resonator.

I have just completed a telemetry device driven by a PIC
that can measure frequencies from 0-64kHz in 1Hz steps.

I now want to use this LM3909 circuit in a pipe to measure
the level, but I would prefer to know what the pipe sizes
should be before I start cutting...

I want to measure a maximum level difference of 2m with a
maximum frequency deviation of 1 octave. What should the
diameter be?


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1999\01\20@122137 by John Payson

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|I have just completed a telemetry device driven by a PIC
|that can measure frequencies from 0-64kHz in 1Hz steps.

|I now want to use this LM3909 circuit in a pipe to measure
|the level, but I would prefer to know what the pipe sizes
|should be before I start cutting...

|I want to measure a maximum level difference of 2m with a
|maximum frequency deviation of 1 octave. What should the
|diameter be?

If you're after a maximum deviation of an octave with a
length deviation of 2m, then your effective pipe length
will need to vary between 2m and 4m (i.e. 2m above the
water).  Diameter should not be at all critical, and the
pipe need not be straight (provided that the bends are
reasonably nice); the important thing to note is that any
sudden changes in pipe diameter or direction will cause
audio reflections.

Depending upon the phasing of your resonating circuit,
a 2-4m pipe will vary in pitch from 20-40Hz or from 40-80Hz.
Those frequencies are fairly low, so you might want to go
for the higher one (I think it may require using a seperate
speaker and mic, though).  Would be interesting to know how
it works out.


Attachment converted: wonderland:WINMAIL.DAT (????/----) (00027B30)

1999\01\20@174105 by leo.perretti

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Hi all,

I am a pipe organ restorer, with 20 years of experience, so I think I
can give some suggestions on the subject.
First of all, the explanations given by John Payson are generally right
with some inaccuracies. The frequencies generated by an open flue pipe
are given by F=(v*n)/(2L+k), where v is the speed of sound, which in
turn is equal to 20.055*sqr(T) m/s; T is the temperature in Kelvin
degrees (sqr is square root, of course); n is an integer giving the
order of the overtone, and k is a constant which takes into account the
diameter of the pipe; for open pipes it is equal to 2*sqr(S), where S is
the section of the pipe. For closed pipes the frequencies are
F=(v*(2*n+1))/(4L+k), with the same constants. If we set n=1 we obtain
the fundamental; for n>1 (integer) we have the overtone frequencies,
which are exact multiples of the fundamental, as can be verified in
actual fact. For closed pipes, only odd overtones are allowed.

After theory, let's go to the practical stuff

>I want to measure a maximum level difference of 2m with a
>maximum frequency deviation of 1 octave. What should the
>diameter be?

Before you think to the diameter you must determine the length of the
pipe. If you want to measure 2 m. of difference within an octave, you
need a pipe 4 m. long; in fact, the resonant length at maximum level is
half the length at minimum, so the frequency is double (one octave
range). A pipe partially immersed in a liquid can be regarded as a
closed pipe, if we assume that the liquid is dense enough with respect
to the air contained in the pipe. When designing organ pipes, a ratio of
1/3 or 1/4 the circumference with respect to the length is considered a
good  proportion for closed pipes, then for a pipe 4 m. long we should
use a circumference of about 1 m., hence a diameter of more or less 30
cm.. This is the right proportion for a pipe that should produce a
musically pleasant tone; for the purpose of measurement a smaller
diameter can be used, but take into account that the narrower the
diameter, the stronger the overtones, then the higher the risk that the
oscillator hooks with the overtones, giving false measurements.
Maybe 4 m. are too much for your application. I can suggest to bend the
tube one or more times, so it can be arranged as you want. Bending the
tube does not affect the resonant frequency, provided that the diameter
is constant in the curves.
Consider also that with this pipe dimension you work with very low
frequencies, as you can easily calculate.
What about enlarging the frequency range to more than an octave?

Best regards

Leonardo Perretti
leo.perrettispamKILLspamprojectpp.it

1999\01\20@234108 by Tjaart van der Walt

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Leonardo Perretti wrote:
>
> Hi all,
>
> I am a pipe organ restorer, with 20 years of experience, so I think I
> can give some suggestions on the subject.
> First of all, the explanations given by John Payson are generally right
> with some inaccuracies. The frequencies generated by an open flue pipe
> are given by F=(v*n)/(2L+k), where v is the speed of sound, which in
> turn is equal to 20.055*sqr(T) m/s; T is the temperature in Kelvin
> degrees (sqr is square root, of course); n is an integer giving the
> order of the overtone, and k is a constant which takes into account the
> diameter of the pipe; for open pipes it is equal to 2*sqr(S), where S is
> the section of the pipe. For closed pipes the frequencies are
> F=(v*(2*n+1))/(4L+k), with the same constants. If we set n=1 we obtain
> the fundamental; for n>1 (integer) we have the overtone frequencies,
> which are exact multiples of the fundamental, as can be verified in
> actual fact. For closed pipes, only odd overtones are allowed.
>
> After theory, let's go to the practical stuff

Thanks John and Leonardo. I was informed that production
on the LM3909 stopped last year. Pity. I have changed my
strategy now as follows :

I will use an external frequency source (62kHz or so) that
will be gated into the TMR/Cntr input of the PIC. Every
second I give a sync pulse (4us) that enables the counting.
When I receive the return pulse, I disable the counting again.
A simple blanking pulse to prevent the signal from the mic
from seeing the pulse from the speaker should prevent
zero distances. What do you think?

--
Friendly Regards          /"\
                         \ /
Tjaart van der Walt        X  ASCII RIBBON CAMPAIGN
.....tjaartKILLspamspam.....wasp.co.za  / \ AGAINST HTML MAIL
|--------------------------------------------------|
|                WASP International                |
|R&D Engineer : GSM peripheral services development|
|--------------------------------------------------|
| Mobile : EraseMEtjaartspam_OUTspamTakeThisOuTsms.wasp.co.za  (160 text chars) |
|     http://www.wasp.co.za/~tjaart/index.html     |
|Voice: +27-(0)11-622-8686  Fax: +27-(0)11-622-8973|
|          WGS-84 : 26¡10.52'S 28¡06.19'E          |
|--------------------------------------------------|

1999\01\24@130113 by leo.perretti
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Tjaart van der Walt wrote:

>I will use an external frequency source (62kHz or so) that
>will be gated into the TMR/Cntr input of the PIC. Every
>second I give a sync pulse (4us) that enables the counting.
>When I receive the return pulse, I disable the counting again.
>A simple blanking pulse to prevent the signal from the mic
>from seeing the pulse from the speaker should prevent
>zero distances. What do you think?

It should work; if you issue pulses every second, reflected pulses
should have time to disperse. Anyway, it is a good idea to stop the
upper edge of the tube with muffling material (foam or so), to prevent
resonance expecially when the length of the tube is near to the wave
length of the overtones of the 1Hz signal you generate. Note that the
open edge of a pipe still reflects the acoustic waves.
There is another tecnique I know, although I have never tested it. If
you feed the pipe with a continuous wave, you receive the reflected
signal after it has run along the pipe backwards and forwards, so its
phase is shifted as a function of the length of the pipe. You could then
detect the phase shift (with a PLL, for example) and determine the
level. In this case, however, you should absolutely eliminate the
reflected wave, and use a frequency the wavelength of which is
comparable with the length of the tube.
Just another idea.

Leonardo Perretti
leo.perrettispamspam_OUTprojectpp.it

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