 The Clarinet BBoard
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Author: Joris
Date: 2001-10-23 12:07
In contrary to what I stated earlier on, a smaller bore-size results in a lower pitch.
Ron told me I was wrong, but of course I had to measure for myself. It's only a small effect (by far not as large as changing the length of the tube) solely caused by wall-losses (which become more importend for small diameter tubes).
The reason I thought it was the other way around is that everywhere in litterature (eq. Benade) they use an equivalenth length of the mouthpiece based on it's volume (they replace the mouthpiece with a cylindrical tube that has the same length).
This assumes that the smaller part at the beginning of the mouthpiece has the same effect as a larger diameter with a shorter length. (therefore a higher and not lower pitch). Until now the difference in pitch between the model and a real life mouthpiece was assumed to be caused by the reed movement... Maybe an interesting subject to study
P.S.: the measurements were done on an artificial mouth with constant lip tension and static input air pressure. The playing frequency was determined using a microphone linked to a fast fourrier transform device.
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Author: Gordon (NZ)
Date: 2001-10-23 12:28
Was the bore air temperature constant?
And I'm not sure that yhou can apply mouthpiece acoustic idiosycrasies to the rest of the bore.
As I understand it the efffective length of the standing wave is the length of the tube to the first tone hole, with various modifications, e.g. mouthpiece corrections already described, tone hole diameter and length. Another is an addition of a small fraction of the diameter of the open end. The exact fraction is pitch dependent. One of the reasons for the flare is to make this fraction more significant in one register than another, and possibly to attempt to correct inharmonicity (phenomenon of the overtones of a particular note being out of tune with the fundamental.)
My knowledge of this is very limited. But I suspect that the change in pitch from moisture content of timber or temperature expansion of plastic would be very small compared with the pitch changes resulting from changes in bore air temperature.
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Author: Joris
Date: 2001-10-23 13:33
The air temperature was constant (292.5K) (very easy to maintain constant, because the artificial player doesn't have `warm' breath like a human.
My original research was about the effect of the registerhole on the tuning of the seccond register. (It differs from note to note, try playing notes in the seccond register with and without register key: depending on the position of the key hole, some notes are get a higher pitch with a open register key-hole and some stay about the same.
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Author: Joris
Date: 2001-10-24 07:44
May I ask where that formula came from?
Because the 0.4D looks a lot like the 0.6r to 0.83r used to incorporate radiation losses. These only give the radiation losses at the open end of the tube, and say absolutely nothing about in tube disturbences (Neville H. Fletcher & Thomas D. Rossing ``The Physics of Musical Instruments'' Springer-Verlag, New York). In real life the radiation effect is smaller (Benade), more in the order of 0.7r (0.35D), and there is an effect on tuning due to wall losses. This is dependend on local in tube circumstances. The playing frequency drops with (alpha*c) / (2*pi). The wall-loss coefficient is (in simplified form) dependend on frequency and tube radius. According to Benade (1968), for a clarinet alpha is about 3*10^-5*SQRT(f)/r
For dayly use c/4L is by far accurate enough, but since we were talking about the effect of tube diameter, this one is needed.
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Author: Mark Charette
Date: 2001-10-24 13:35
As I said, it is the classical formula, and comes straight from the textbooks (Fundamentals of Musical Acoustics and others) showing the effect of D, which is significant. The calculation of <b>effective</b> D (in this formula's case D is assumed to be a on a cylinder, no tone holes, perfect reflector - just like any <b>other</b> classical formula would use) is non-trivial. I'm sure you have used the perfect gas law a few times, which as everyone knows is "imperfect" when applied to real-world mixtures of gases and radiation losses but nevertheless is quite useful.
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Author: Gordon (NZ)
Date: 2001-10-25 10:45
OK. Accept the 0.4 as approximate.
Let's look at an example using f=v/4(L + 0.4D)
where v is the local velocity of sound, L the length, and D the diameter.
In round figures take a mid-range note as L=200 (mm)
Take D as 15 mm and compare this the same instrument grossly expanded to 16 mm.
In the former (L + 0.4D) = 206 mm; In the latter it = 206.4 mm
The difference is 0.4 mm. This represents a 0.4/206, ie 0.2% change in frequency.
Surely 0.2% pitch change is insignificant by any musician's standards. This is about 10 times less than most players can even detect.
Or have I missed something?
Joris wrote "The air temperature was constant (292.5K) (very easy to maintain constant, because the artificial player doesn't have `warm' breath like a human." Did you measure the air temperature throughout the bore as 393.5, or was that the ambient air temperature?
I assume the air for the artificial player came form some form of compressor or blower, which is inefficient and heats during operation, affecting the temperature of the air inside the bore, which is the temperature that is significant. Or did your artificial player have some means of accurately controling its output air temperature to maintain constant bore temperature? Note that even its output air temperature would be higher than the temperature inside the bore because of the pressure change after passing the reed. To carry out this experiment with validity the temperature of the ENTIRE bore air would have to remain constant - not an easy trick!
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Author: Joris
Date: 2001-10-25 13:57
Since there is at least 20 meters of non insulated tube from the storage barrel of the compressor to my setup, I could very well assume the air to have taken room temperature. But to be sure I watch both in-tube and room temperature during the measurements, and they don't differ much (not measurable with the instruments I have).
Furthermore, you talk about temperature differences in the tube due to accoustic pressure. Just calculate how much the temperature will rise and decline from a pressure difference of 20mbar at 400Hz, you'll see that it is by far more neglectable than the 0.2% difference in tuning due to bore diameter.
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Author: Gordon (NZ)
Date: 2001-10-25 23:04
Fair enough.
I'm not sure of the term 'acoustic' pressure. I meant the drop in pressure from inside the 'mouth' - to atmospheric inside the bore, after going through the constricting orifice between the reed and mouthpiece. I'm not sure that acoustics comes into this drop.
If I blow air onto my hand through an OPEN mouth the air is far warmer than if I do so through a small lip orifice that causes a drop in pressure. Drop in pressure = drop in temperature ..... "Charles law".
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Author: Joris
Date: 2001-10-26 08:00
In fact the pressure doesn't just drop. The pressure in the mouthpiece oscillates between the mouth pressure and the atmospheric pressure. The pressure difference between the instrument pressure and the atmospheric is called the accoustic pressure. (and it changes with the playing frequency)
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