 The Clarinet BBoard
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Author: Dave Beal
Date: 2006-02-11 03:47
Any physicists in the house? I'm trying to understand the acoustics of conical tubes, like saxophones and oboes. Cylindrical woodwinds make sense, with max pressure at the reed and a node at the bell. And I understand why the wave function in a conical tube must have a (1/r) factor, because within the cone the wave radiates like a sphere. But I don't get why this affects the wavelength of the wave. Can anyone explain why a closed conical tube has even harmonics?
BTW, the University of New South Wales has an excellent set of webpages on the acoustics of musical instruments: http://www.phys.unsw.edu.au/music/. But their discussion of this topic baffles me.
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Author: george
Date: 2006-02-11 04:30
The Physics of Musical Instruments, by Neville Fletcher & Thomas Rossing has it all.
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Author: Shorthand
Date: 2006-02-11 05:05
The solution is on pages 8 & 9 of these lecture notes:
http://www.physics.mcgill.ca/~guymoore/ph224/notes/lecture20.pdf
No, I didn't actually know, but I Googled "Standing wave cone".
The differential equation has to be solved in spherical coordinates (remember a cone is a section of a solid sphere.)
The solution is actually closely related to the solution of the radial portion of the hydrogen atom wave equation. (Think about it twice and it shouldn't surprise you - they're both 1/r^2 environments. The boundary conditions are a little different, but not so much that it knocks you into different solution territory.)
The whole discussion of Spherical Bessel functions (along with any other sort of mathematical monster of physical interest) is best found in Mathematical Methods for Physicists. In a pure math text, it probably won't make sense.
Isn't this a question for a sax or oboe forum? We aren't supposed to know this stuff.
Post Edited (2006-02-11 05:07)
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Author: Chris P
Date: 2006-02-11 07:19
I'd like an explanation in layman's terms - why does a closed expanding conical bore (oboe, bassoon, sax, etc.) follow the same harmonic series as an open cylindrical bore (flute) and not a closed cylindrical bore?
All too often the behaviour of a closed expanding conical bore isn't explained, and just mentioned that it shares the same behavioural characteristics as an open cylindical bore. Does this mean no-one has done any real research or what?
But please explain in layman's terms - no equations please.
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Author: BobD
Date: 2006-02-11 11:42
If I had a true understanding of waves perhaps I'd have a chance. Being able to see water waves doesn't help since there are no standing water waves.
Bob Draznik
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Author: Tony Pay ★2017
Date: 2006-02-11 11:53
Chris P wrote:
> Does this mean no-one has done any real research or what?
>
> But please explain in layman's terms - no equations please.
Reminds me of the story of the man who wanted to know the secret of life, and journeyed through great hardships and climbed a very high mountain to see a wise man.
"Can you tell me the secret of life?" he asked the sage.
"Sure," came the reply.
"Great! What is it?"
"Well, I can tell you; but the trouble is, you won't understand me. First, you have to make a deep study of mathematical logic and recursive function theory."
"You're kidding!"
"No, I'm sorry, but that's the way it is."
"Well....nah, skip it."
Tony
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Author: Dave Beal
Date: 2006-02-11 15:29
Well, I still don't know how to solve differential equations in spherical coordinates, but I think I understand the conical solution now. As stated in the reference cited by Shorthand, the solution is (1/r)sin(kr), where k = 2pi/lambda. The key is that as r approaches 0, (1/r)sin(kr) goes to 1, not 0. That's why the conical tube can have the same harmonics as an open cylindrical one, but still have a pressure peak at r=0 (the reed).
My thanks to everyone who responded.
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Author: Lelia Loban ★2017
Date: 2006-02-11 16:22
>>
"Well, I can tell you; but the trouble is, you won't understand me. First, you have to make a deep study of mathematical logic and recursive function theory."
>>
That fellow didn't read the late lamented Douglas Adams, evidently, or he'd have gone to the planet where one discovers that the secret to Life, the Universe and Everything is 42.
Lelia
http://www.scoreexchange.com/profiles/Lelia_Loban
To hear the audio, click on the "Scorch Plug-In" box above the score.
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Author: Tony Pay ★2017
Date: 2006-02-11 20:59
Lelia Loban wrote:
>> That fellow didn't read the late lamented Douglas Adams, evidently, or he'd have gone to the planet where one discovers that the secret to Life, the Universe and Everything is 42.>>
...but, but...he wouldn't have understood it:-)
Funnily enough, it turns out that the number 42 is implicated in some of the recent and esoteric speculations about the Riemann Hypothesis, probably the most famous undecided problem in mathematics. There is a series of which the first 3 terms are 1, 2, 42, and then something like 14420 -- sorry, I can't check.
And don't forget:
"Rule 42 -- all persons more than a mile high must leave the court!"
(Lewis Carroll)
Tony
Post Edited (2006-02-11 21:08)
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Author: Gordon (NZ)
Date: 2006-02-12 05:19
Don't waves in the sea speed up as they approach the shore (i.e. shallower water, or enter a narrowing entrance? (Maybe not) Could the reverse happen as a wave travels down tube of increasing diameter?
Could this help with a layman's explanation?
Over to you.... :-)
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Author: Don Berger
Date: 2006-02-12 12:56
I believe that Dr. Arthur Benade gave several explanations re: acoustics of musical instruments in his numerous writings, the most elementary I believe is in his "Horns, Strings and Harmony" which many of us may have available. He also discusses some of the effects of "departures" from true cones and cylinders which are common. Read and understand [perhaps, like me] . Don
Thanx, Mark, Don
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Author: Shorthand
Date: 2006-02-12 13:33
Chris,
Sorry I didn't see your post earlier. I was really trying to answer Dave's question. I'm afraid there is no way I know of to really explain this kind of stuff to someone without vector calculus except to show them pictures - just like your chemistry teacher did when discussing orbitals.
Dave,
The bit that happens at r=0 is not odd. The math is a little odd because the limit of sin(r)/r as r approaches 0 is one because for small r, sin(r) approaches r. The wierd part is what happens at the other boundry.
Basically think of a sax a sphere with the oscillator at the center. The open surface (think of light reflecting off the inside surface of glass) serves as a reflector (just like it does in the clarinet) to set up the standing spherical wave. The cone is just a slice of that sphere.
Its hard to explain wave dynamics to any "layman". Most people simply don't have experience working with them. You can usually talk about linear waves, or even surface ones, but standing waves in 3 dimensions is very very hard to discuss - chemistry teachers have been waving their arms at it (orbitals) for years with little success other than throwing pictures at the kids.
The mathematical solution is not something I truly understand, and I do have an undergrad in physics (and math, actually, but that's a BA). If you want to understand the math, you really need to look at "Mathematical Methods for Physiscists" - its one of the best written math texts you'll ever see.
I skimmed the spherical bessel function solution when I first posted, but didn't really delve into it once I remembered how simiar it was to the hydrogen atom, I was happy.
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Author: Phat Cat
Date: 2006-02-13 03:02
You don't actually have to solve the differential equations to derive the resulting wave forms. When a source oscillates in open space, it spreads out spherically in all directions. At a distance of r from the source, the sound wave covers a spherical surface whose area is proportional to r-squared, so the intensity (loudness) of the wave is proportional to 1 / r-squared. Since the intensity is the square of the pressure (amplitude or velocity), the pressure of the wave is proportional to 1/r.
In the case of the oboe (or saxophone), the bore of the instrument can be thought of as a cone, to a first approximation. As the sound wave travels from the reed to the bell it expands along the cone just as it would in open air, so the pressure of the wave is again proportional to 1/r.
The sound produced by a woodwind is actually the result of a standing wave in which the low pressure at the open bell causes a reflection back to the reed, which travels back to the bell, etc. This means that the wave function is a combination of sine and cosine functions. In the case of a conical bore, the sin and cos functions must be scaled by 1/r to make the intensity decrease along the bore.
Since the pressure is largest at the reed at smallest at the bell, the wave function must have a maximum at 0 and become 0 at the bell. As the previous poster pointed out, the function (1/r)sin(r) actually takes the value of 1 in the limit as r approaches 0. Thus the wave can take form
(1/r) sin(2Lr/n)
where L is the length of the pipe and n = 1, 2, ... correspond to the fundamental and the various harmonics.
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Author: Bassie
Date: 2006-02-13 07:02
Well, that makes sense to me, PhatCat. What you're saying is that the conical pipe essentially matches an open-ended waveform (like in a flute) to a closed end? Cunning.
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Author: Phat Cat
Date: 2006-02-13 12:47
Bassie:
Not exactly. A flute is a cylinder that is open at both ends and so the pressure is a planar wave having a minimum at both ends. This means the pressure wave function will be sin(2Lr/n).
In contrast, a conical bore essentially matches the section of a 3 dimensional sphere that is bounded by its wall. Since this pressure wave expands, its wave function is weighted by 1/r. Since it has a maximum at the reed and a minimum at the bell, it is of the form (1/r)sin(2Lr/n).
For completeness sake, the clarinet is a cylindrical bore that is closed at the mouthpiece and open at the bell. So its pressure wave is planar with a maximum at the closed end and a minimum at the open end. This means the wave function is cos(4Lr/m) where m is an odd integer.
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Author: Bassie
Date: 2006-02-13 13:26
I think I see what you mean, PhatCat.
But this 1/r weighting makes the wavelength look very odd towards the closed end of the pipe. What's the best way to interpret that?
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Author: Phat Cat
Date: 2006-02-13 14:05
Bassie:
Since the 1/r weighting damps the wave function towards the bell, the wave function gets squished compared to that of a clarinet or flute. The musical interpretation of this funky waveform is a sound richer in overtones..
To understand the latter, realize that our “ears” essentially resolve a musical sound into the fundamental, which we hear as pitch, plus overtones which we hear as timbre. The overtones correspond to the harmonics that make up the sound's Fourier series. In simple terms, every periodic waveform can be expressed as a weighted sum of *undamped* sin and cos functions. This sum is its Fourier series.
Since it takes a bunch of undamped wave forms to constitute the “funky” wave form, you’ll hear them as richer overtones.
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Author: Dave Beal
Date: 2006-02-13 18:10
It's interesting that the standing wave is caused by a mismatch of acoustic impedance at the bell. I first learned about wave reflections and impedance mismatches in electrical engineering classes 30 years ago. Isn't it cool how the same concepts show up in other places in nature?
Phat Cat, in your wave equations, I think the "L" should be in the denominator. The sine argument has to be dimensionless.
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Author: Chris P
Date: 2006-02-13 18:35
It's still Greek to me, but I'm on the way to understanding.
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Author: David Peacham
Date: 2006-02-13 19:05
Trumpets and trombones are cylindrical pipes and are closed (sort of) at one end. They sound both odd and even harmonics.
OK, you say, but they aren't really cylindrical, are they. They have a damn great bell flare.
I have in my hand a trombone with the bell section removed. Apart from the mouthpiece, it is very definitely cylindrical. (It wouldn't slide too well if it weren't.)
With the slide closed, I can easily make it sound three notes, which are (albeit out of tune) E, B, E. These are the second, third and fourth harmonics.
If I extend the slide, I can sound a series approximating to G, D#, G#, C#, E. I think this is a pretty funky variant on G, D, G, B, D: the 2nd to 6th harmonics.
Why is this possible?
Before you ask, if I remove the mouthpiece I can't get much of a note at all.
-----------
If there are so many people on this board unwilling or unable to have a civil and balanced discussion about important issues, then I shan't bother to post here any more.
To the great relief of many of you, no doubt.
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Author: Phat Cat
Date: 2006-02-13 19:09
Dave:
The functions are correct. I'm a math guy so the dimensionless thing doesn't resonate with me. L is just a scaling constant that normalizes the domain of r so that it takes values from 0 to 1 instead of 0 to L.
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Author: Phat Cat
Date: 2006-02-13 22:22
David:
Although the tubing of brass instruments is indeed cylindrical for a portion of its length, it is the flare of the bell which provides the characteristic "brassy" timbre (and not the metallic construction of the instrument body as the name implies and many assume). The bell on the clarinet or other woodwinds is relatively short with a mild flare and only interacts with the few notes in which all holes are covered. In contrast, the bell of brass instruments is longer with a more pronounced flare and affects all notes. It effectively makes the instrument conical as far as the overtones are concerned.
The reason the bell gives the characteristic brass sound is that the curvature of its flare is small relative to the overall length of the pipe. Higher harmonics are emphasized more because their shorter waves are better able to expand out the rapid flare. In fact it is the harmonics whose wavelengths are comparable to the curvature of the bell that resonate most.
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Author: Shorthand
Date: 2006-02-14 03:11
Coronets are mroe conical, Trumpets cylindrical - A Euphonium is the conical version of a Trombone.
Its not nearly as absloute as it is in the woodwinds, but that's the main difference between the two. (Ovbiously you can't make a slide on a conical-bore instrument.)
Phat, your reasoning is actually flawed. One issue with the reference I gave is that it didn't draw the cones.
See:
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/clocol2.html
(Which is the refrence I generally choose for these things when someone isn't asking for serious math - which you don't need for clarinets.)
Even though the velocity doesn't come back to a full node at the bell of a Sax, the sax is a 1/2 wavelength instrument just like the flute, but backwards in terms of pressure and velocity. (The clarinet is a 1/4 wavelength instrument.) Because of the 1/r^2 relationship, the harmonic frequency of the system is at a normal wavelength in linear propication of 2l, but the pressure and velocity profiles in the instrument don't look that way.
The fourier transform is nasty (and not very useful) as bessel functions are actually NOT of uniform period (though they converge towards periodicity as x -> inf).
From a physicist's perspective this is where we actually thank the Intelligent Designer for sending us mathematicians that love this stuff even though we generally dispise them and their attitude personally and then shamelessy use what Bessel did without spending useless time trying to "understand" it.
The important part of physics is setting up the probelm right - once you've done that, you can always whack on it with a computer if you really need to.
I don't understand the ins and outs (or even really the derivation) of Bessel functions and I don't want to. However, I've had enough quantum mechanics to understand that you can throw most of your linear wave propogation intuition out the window and you'd better stick to the math.
See also:
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/opecol.html#c1
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Author: Bassie
Date: 2006-02-14 07:59
So the question can be phrased as follows:
Why does the clarinet tube resonate at a quarter wavelength, while the saxophone tube resonates at a half wavelength?
Is it right, then, to say the following:
The tapered tube of the sax increases the pressure / velocity rato of the wave towards the mouthpiece, to such an extent that the only solution that matches the boundary conditions at the mouthpiece is a half-wave solution (actually, a Bessel function, but a half-wave in approximation)
?
(Shorthand - I'm becoming very suspicious of the pictures in your original web reference. Surely a half-wave-like solution, even expressed in Bessel terms, should have a velocity node somewhere in the pipe in the fundamental?)
Post Edited (2006-02-14 08:58)
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Author: David Peacham
Date: 2006-02-14 08:11
Phat Cat wrote:
"bell of brass instruments is longer with a more pronounced flare and affects all notes. It effectively makes the instrument conical as far as the overtones are concerned."
Please reread my posting. I was able to sound both odd and even harmonics on the trombone slide alone, having *removed* the bell section.
Further experimentation reveals that I can fit my clarinet mouthpiece over the trombone tubing in place of the bell section, thereby using the clarinet mouthpiece to blow air through the slide "backwards". With this setup, it is easy to produce a very deep note, which appears to be the "chalumeau register", an octave below the fundamental obtainable with the trombone mouthpiece, and two octaves below the second harmonic which is the lowest note readily obtainable with the trombone mouthpiece. I cannot overblow this chalumeau note, nor bend it easily; in contrast, the trombone mouthpiece produces a series of notes that are unstable in pitch.
It seems to me very odd that the choice of mouthpiece has such a radical effect on the overtone series. It does suggest that any attempt to explain the overtone series purely in terms of the shape of the tubing is flawed.
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If there are so many people on this board unwilling or unable to have a civil and balanced discussion about important issues, then I shan't bother to post here any more.
To the great relief of many of you, no doubt.
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Author: Phat Cat
Date: 2006-02-14 10:07
Bassie:
The short answers:
The fundamental of a clarinet is at 4L and the overtones are odd because its cylinder generates planar pressure waves and its boundary conditions are a maximum at one end and a minimum at the other end. The wave functions that meet these requirements are of the form cos(4Lr/n) where n = 1, 3, 5, …
The fundamental of a saxophone is at 2L and it has both even and odd overtones because its cone generates pressure waves that spread with spherical symmetry and its boundary conditions are also a maximum at one end and a minimum at the other. The wave functions that meet these requirements are damped sine waves of the form (1/r)sin(2Lr/n).
You can see these graphically on Pipes and Harmonics page at the excellent link of the University of New South Wales posted above.
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Author: Bassie
Date: 2006-02-14 10:30
My argument is that you can get the same answer by considering a plane sine wave in a narrow pipe with varying characteristic impedance.
What I need now is an obvious physical explanation for where the nodes and antinodes of velocity go!
(sorry for so many edits of this post)
*
David - sounds like a trombone is ordinarily open at both ends. Having said that, I've heard a trombonist (on the Royal Institution Xmas Lectures, no less) play sub-harmonics, so who knows what's possible.
Post Edited (2006-02-14 13:05)
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Author: Phat Cat
Date: 2006-02-14 13:42
Bassie:
I think I understand your question now. The wave functions I listed plot to the red lines in the graphs and measure the pressure of the air molecules. The blue lines in the graphs measure the vibration of the air molecules, or as you say, the velocity. The velocity functions are the (first) derivatives of the pressure functions. The derivatives for the clarinet and flute are straight-forward. That of the conical pipe is more complicated but you can compute it from Calc 101 using the formula for the derivative of a product.
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Author: Shorthand
Date: 2006-02-14 13:43
Bessie,
The fact that the velocity profile doesn't come back to a true node inside the length of the instrument yet the freqency is such that it should in a linear system is very confusing to me too. That's why I keep telling people to throw their intuition out the window in this system - the math is more similar to a hydrogen atom than the clarinet.
The diagrams are consistent across all the sources I've seen, though.
The 0 Bessel = sin(x)/x, which is what we are talking about. Sorry for the confusion - it was primarily mine. The harmonics are simply shorter wavelength versions of the zero-order bessel. (sin(kr)/r) (I've learned a lot over the course of this thread.)
The higher order bessel functions have the wrong boundry condition at r=0. The zero order converges to 1 as r -> 0, (equals 1 for all physical intents and purposes) but the others are 0 at r=0, so that would mean 0 pressure at the mouthpiece - ovbiously not a valid sax solution.
So P(r) = Po * sin(kr)/r.
Velocity is the derivative:
d/dr sin(kr)/r = -sin(kr)/r^2 + k*cos(kr)/r
That's why the velocity doesn't come back to a node while pressure does. The next mode is just P(r) = Po * sin(2kr)/r.
That changes the derivative somewhat as now you have 2 phases and now their proportion is different.
-sin(2kr)/r^2 + 2k*cos(2kr)/r
Post Edited (2006-02-14 13:50)
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Author: Chris P
Date: 2006-02-14 13:52
I've experimented woth clarinet and sax mouthpieces on brass instruments - but a soprano sax (or whichever fits best) mouthpiece on a BBb tuba (4 valve compensating) produces notes so low that they're only felt as pressure waves or a series of fast repeated popping sounds that brighten up with increased breath pressure.
Opening a water key will send them into a higher register.
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Author: Shorthand
Date: 2006-02-14 14:34
Attachment: Conic Tube Waves.pdf (14k)
I did a quick-and-dirty graph in Excel of the pressure and its derivative (there's clearly some constants missing) of the fundamental and 1st harmonic. See the attachment.
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Author: Bassie
Date: 2006-02-14 14:52
Thanx, shorthand. It's that velocity antinode where the flute would have a node that gets me!
I was taught that a standing wave can always be thought of as a superposition of incoming and outgoing waves - so I'm going to go away and ponder that for a while.
Having said that, I've seen weird things happen with spherical waves before. Near to a small antenna. Where light waves pass through a focus. All so Nature can avoid dealing with a singularity in the equations... deep. :-D
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Author: Chris P
Date: 2006-02-14 15:09
Not related, but what causes the water-like shimmering reflections in the road during summer when the roads are bone dry? And is this the same as a mirage (the optical illusion, and not the alcoholic beverage partnered with 'Taboo' in the '80s which probably induced optical illusions)?
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Author: Phat Cat
Date: 2006-02-14 15:15
Chris:
Both the non-alcoholic phenomena you mention are caused by refraction of light thru air of varying density. The density variation is caused by heat radiating off the surface.
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Author: Shorthand
Date: 2006-02-15 15:00
Chris, yes, that's a textbook mirage.
The road is hot (really hot) and so the air near it is also really hot, so much less dense. Therefore you get a lens effect that causes you to see sky where the road should be.
Bassie,
Yes, you CAN always think of a standing wave as a superposition of incoming and outgoing travelling waves. That is no less true in a spherical environment - its just that your intution has a hard time getting used to spherical waves as there are no real analogues in our everyday world. For pretty much everyone here, a Sax, oboe, or Bassoon is probably their most common encounter with speherical waves.
However, as the function isn't periodic, you can't do a Forier transform into a bunch of sine waves. Its a linear combination of 0-order spherical bessel functions instead - no big deal, just different (and harder to imagine). It is kind of cool.
I have to wonder how much math Dr. Sax knew, or did he just say "I will make a single-reed Basson!" and tweak until he got the Boehm system to work on a cone. The math we're talking about here was brand spanking new at the time. (OK maybe 40 years old.)
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Author: J B Lansing
Date: 2006-02-15 22:25
Just to muddy tthe waters. I am involved with early music and Phat Cat's explaination of the role of the bell in brass instruments is interesting in that Coronetos are conical "brass" instruments made of wood and have a conical bore with little flare at the bell. They deffinately have a brass sound. Sackbuts are early trombones with a different, less flared bell and are somewhat less "brassy". Now I think I undersatnd why.
Coronetos also have finger holes which may distrupt their conical nature on shorter notes. They did not benifit from modern acoustical science.
J B
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Author: Bassie
Date: 2006-02-21 09:34
Dave,
I'm going to have another crack at this.
(Thanks, PhatCat, for straightening out my rusty physics circuits!)
It's all because of what happens to a travelling spherical wave when it's brought to a focus (e.g. the point of a cone). The wave cannot behave like the 'sine' waves we all know and love, or odd things would happen (like the energy density in the wave rising to infinity). Instead, the pressure in a spherical wave behaves very differently near a focus. In particular, there is a 180 degree phase change - and thus (a) a closed end acts like an open end (b) the standing waves created in a conical pipe look really weird.
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Author: tictactux ★2017
Date: 2006-02-21 10:08
What's with the recorder?
I consider it being an inverted cone (wide near the mouth, narrow on the opposite end) but open(-ish) on both ends. (window/lip and foot joint). The whole setup looks - physically - even more weird than a sax or a flute.
Never considered its complex physics back then in those dreadful lessons...
--
Ben
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Author: David Peacham
Date: 2006-02-21 10:24
Recorders have varied through history between being contracting-conical to being much more nearly cylindrical. Most organ pipes are effectively cylindrical recorders without toneholes.
Flutes likewise; modern flutes are roughly cylindrical, but baroque and classical flutes, and even the first Boehm flutes, are contracting-conical.
-----
Still no answer to the conundrum I posed above: why does a trombone slide, which is perfectly cylindrical, behave entirely differently depending on whether it is "played" with a trombone mouthpiece or with a clarinet mouthpiece?
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If there are so many people on this board unwilling or unable to have a civil and balanced discussion about important issues, then I shan't bother to post here any more.
To the great relief of many of you, no doubt.
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Author: Shorthand
Date: 2006-02-21 20:59
A recorder is generally a form of flute and is an open open-ended instrument using the vortex street from a small air jet blown across an edge to create the tone:
http://hyperphysics.phy-astr.gsu.edu/hbase/music/edge.html
The transverse flute is a newer design (Boehm's magnum opus) and uses a parabolic reflector at the head joint to maximize volume - which requires the transverse design:
http://hyperphysics.phy-astr.gsu.edu/hbase/music/flute.html
I don't even want to think what would happen in a conical-bored flute, and I'm NOT pulling Mathematical Methods for Physicists off the shelf again for this!
[edited URL's]
Post Edited (2006-02-21 22:42)
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Author: Don Poulsen
Date: 2006-02-21 21:14
If the note produced by the clarinet mouthpiece on the trombone is an octave lower than the fundamental produced by buzzing lips, this would seem to indicate that, whereas the clarinet mouthpiece is acting as a closed end, the trombone mouthpiece/lip combination is somehow acting as an open end. I cannot think of why this would be so at the moment.
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The Clarinet Pages
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