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A Proposal For A Bi-Directional Concertina Reed


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In the remote chance that someone is interested in testing this design, it occurred to me that there is an additional parameter that you can play with. In a conventional reed with the tongue bent upwards, the initial air flow is not simply around the end, but also around the sides of the tongue. In the design that i proposed air initially flows only around the end of the tongue, which as Lukasz suggested, would mean a lower volume of sound, depending on how big a notch is filed in the end of the frame.

 

The amount of initial air flow (and thus volume of sound) can be increased in my design by filing the inside edges of the frames, not just at the ends, but along the sides as well. Whether one can get acceptably loud volume (and pleasing tonal quality) is a question that can best (probably only) be answered by trying the design out.

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In the remote chance that someone is interested in testing this design...

 

So I guess you are not going to try it yourself.

Nope. I don't have the equipment or the skills to attempt such a thing.

 

And when i get my Wicki Peacock in a few weeks, i plan to spend my spare time learning to play it. I currently am a medocre player on the Scottish smallpipes, guitar, and chimera (a 5 string instrument that i designed), and am near medocrity on the bass guitar. My goal is to rise to mediocrity on the Hayden duet ;-) .

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I struggle to grasp any of the foregoing technical theory but I do like your avatar !

Thanks. It's a caricature drawn by a nephew of mine.

 

Curiously, it was ported over automatically from an acoustic guitar forum that i participate in.

Edited by rlgph
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By widening the gap at the sides of the reed you will at least - if getting a sound at all - lose the characteristics of quality concertina reeds as so many of us are loving them I'd guess...

Edited by blue eyed sailor
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Injecting a dose of reality into a creative conversation can end it prematurely and great opportunities for good ideas can be lost. Now the heat has gone from the conversation, a couple of things. Good thinking about the twin frame single reed, but it seems to be predicated on thinking reeds only need to be tight in clearance at the tip. In fact they need to be tight all around the reed in order to get the most power from the reed. So the reed with frames above and below would need to be tight around the reed at both points, as in the pic I have altered, (it should be at the bottom of this post.) The places where the frames need to be tight on the reed are marked in red.

 

Leaving aside the greater complexity in creating this shape in a reed frame, the reed when it travels will be making three distinct parcels of air. Two, roughly the same length, produced when the reed has passed the second tight area, and another after the reed has passed the first tight area and only lasting until the second. This parcel of air would be shorter that the other two and would presumably create a second much higher note. So you would have a note, a much higher note, and then the first note again every time the reed swings a full cycle.

post-74-0-89611100-1423308803_thumb.jpg

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By widening the gap at the sides of the reed you will at least - if getting a sound at all - lose the characteristics of quality concertina reeds as so many of us are loving them I'd guess...

 

I'm not suggesting that the frame slot be widened throughout, only that an inner edge be filed down slightly -- as on the end. The cross section of the side of the frame would look pretty much like the cross section at the end, so that the narrowest part of the slot would still be the same.

 

I suspect the added complexity (and hence increased cost) in manufacture would exceed the 2-reed established approach.

 

I don't think the manufacturing of the basic structure would be significantly different, but i would guess there would be considerable expense in the time required for tuning. I stated this right up front (2nd paragraph) of the original post.

 

 

Injecting a dose of reality into a creative conversation can end it prematurely and great opportunities for good ideas can be lost. Now the heat has gone from the conversation, a couple of things. Good thinking about the twin frame single reed, but it seems to be predicated on thinking reeds only need to be tight in clearance at the tip. In fact they need to be tight all around the reed in order to get the most power from the reed. So the reed with frames above and below would need to be tight around the reed at both points, as in the pic I have altered, (it should be at the bottom of this post.) The places where the frames need to be tight on the reed are marked in red.

 

Leaving aside the greater complexity in creating this shape in a reed frame, the reed when it travels will be making three distinct parcels of air. Two, roughly the same length, produced when the reed has passed the second tight area, and another after the reed has passed the first tight area and only lasting until the second. This parcel of air would be shorter that the other two and would presumably create a second much higher note. So you would have a note, a much higher note, and then the first note again every time the reed swings a full cycle.

 

The shape is not particularly complex, and in fact would be trivial for CNC manufacture (since the top and bottom frames are just mirror images of each other). There would be a good bit of one-time experimentation to determine optimum angles and depths. And you are correct that with two cutoffs the frequency would be higher -- about twice as high in fact, as i said earlier. A graph of the volume of air flow vs time for a conventional reed looks qualitatively like this:

Conventional.jpg

 

Note that this is basically a square wave. One of the properties of a square wave is a relatively large amount of energy going into higher harmonics, which is why to some people concertinas and accordions sound harsh.

 

The corresponding qualitative graph for the bidirectional reed i expect will be something like this:

BiDirectional.jpg

Two obvious characteristics that i have tried to show are the higher frequency and the lower amplitude -- the latter would be controlled by the size of the notch in the end and sides.

 

Based on the forces applicable, i believe that the cutoff time duration for the bottom cutoff may be somewhat larger than that for the top, which i have tried to illustrate. Depending on the magnitude, this asymmetry could well lead to some strange sounds.

 

The shape of the wave in this case is less square because the cutoff and reinstitution of the air flow is more gradual because of the filed edges. This has some possibly important implications for a conventional reed, which i talk about in the next post.

Edited by rlgph
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I suspect the added complexity (and hence increased cost) in manufacture would exceed the 2-reed established approach.

I do not expect, and never have expected, this proposed bidirectional reed to replace the conventional approach, although i do think it is worthwhile to investigate experimentally, because i may be wrong in my expectation, or because such experimentation may tell us something about how to improve the conventional approach.

 

For example, i think that filing the inner edges of a conventional reed frame may lead to less abrupt cutoff and recovery of the air flow, and thus a more trapezoidal shape for the wave form. This shape puts relatively less energy into higher harmonics, resulting in a more mellow sound, something that would be desireable to many.

 

With this i expect that it's time (many would say long past time) for this thread to end. I've enjoyed it because of the mental stimulation it has brought -- something i've missed since retiring as a physics professor about 10 years ago.

Edited by rlgph
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Greetings, thanks to all you posters for an interesting thread. I’d like to make a few comments that I hope can correct what I see as misconceptions.

 

“It will continue to move downward until the air flow is effectively cut off as the reed end comes very close to the inside end of the lower frame. (Assuming that the frames are such that this cutoff occurs before the spring force gets too large.) When the air stops flowing, the partial vacuum is no longer maintained and the force due to air pressure is greatly reduced.”

 

Incorrect. Rlgp, continuing with the quasi-steady assumption that has been assumed in this thread, let the pressure of the motionless air in the reservoir above the reed be P1 and the pressure of the motionless air in the reservoir below the reed be P2. When the tongue first enters the slot, the pressure force acting on the tongue becomes maximum, and is equal to (assuming a perfect seal), (P1 – P2)*A, where A is the flat area of the tongue.

 

Confusion arises over this point I think when people imagine a “suction” or “vacuum” under the reed caused by the air velocity there, and that this suction is somehow greater than what's physically possible. We know from the steady state Bernoulli principle (p + 0.5*rho*V^2 = constant, where p is static pressure and V is fluid velocity) that the pressure under the reed is less than P1. P1 is the stagnation pressure (some fluid dynamicists call it total pressure) of the air in the top reservoir. The pressure under the tongue can never be lower than P2, the exhaust pressure the air ultimately experiences after leaving the upper reservoir. The above Bernoulli equation would yield an imaginary velocity if the airflow is claimed to experience a pressure less than P2. During the start of motion, the pressure the tongue experiences varies over its surfaces, but is bounded by the expression P2 <= p <= P1, where <= is read, “less than or equal to.” To see this, just picture the tongue frozen inside the slot where it stops, with the air still. You only have two reservoirs at different, uniform pressures, separated by a still reed tongue.

 

“Note that there is an asymmetry in the operation. When the tongue is below its equilibrium position, the spring force is acting upward and that due to air pressure downward (with the latter decreasing as the air flow slows).”

 

Incorrect. Rlgp, the “latter” increases as the air flow slows. Before it slows, the net pressure under the tongue is at some pressure Pi, intermediate between P1 and P2, and when the air flow stops, that pressure decreases to P2, causing higher pressure difference and more force on the tongue.

 

Let me take this discussion one step further. If you maintain the quasi-steady assumption, you must consider the formation of jets in the airflow. The pressure in such a jet must equal the pressure in the ambient air immediately surrounding the jet (for subsonic flow). Thus, if a jet forms very soon after air starts flowing and continues into the lower reservoir, as the tongue moves into the slot, the total force on the tongue will be the maximum, (P1 – P2)*A once the jet forms.

 

In truth, the air motion through and around the tongue is at least somewhere and sometime not quasi-steady, but unsteady. Unsteady flow is flow in which time derivatives are important. Changes in velocity, or acceleration, cause inertial effects of the air to become important – important in our formulations and in our imaginations. It is the frequency of the reed vibration that mainly determines the flow regime, whether unsteady or quasi-steady. Quasi-steady means simply that there is motion, but with all time derivatives negligible compared to the other terms in the governing equations. From an order of magnitude study, you can show that, for the smallest reeds (those with the highest pitch), the steady, periodic airflow field is mostly unsteady, and for the largest reeds (lowest pitches), the periodic flow field is appreciably quasi-steady. The dividing line is somewhere in the vicinity of 1,000 Hz. It thus seems to me that jet formation will be minimal for reeds with pitch much above 1,000 Hz. With unsteady flow, the description of the flow field becomes too difficult for our imaginations to guide us very well. It’s my guess that the start of vibration, before periodic flow is established, is entirely unsteady.

 

“Note that if you want the air flow to be cut twice per cycle you need to adjust the thickness of the frames and the V-shaped cutout so that the reed end passes very close to the frames at both ends of the oscillation. In this case, though, the frequency would be twice what it would be if the cut is made only on the low pressure side.”

 

Incorrect. Rlgp, the human ear/brain system is not so easily fooled. The period of oscillation of the vibrating tongue determines the pitch we perceive, because our hearing system perceives the fundamental frequency as the pitch. There is no way the two pulses of air that you imagine can be precisely the same. One simple reason for this is that they occur in different places that have different geometries. Mathematically speaking, a Fourier transform will produce the same fundamental frequency that we perceive in the pitch. You can verify this for yourself with audio software, constructing a wave form consisting of two pulses, starting with two identical shapes, then altering one of them in the most minute way. You may be surprised at how good our ears are.

 

“In a conventional reed, as well as in this design, the maximum distance of the oscillation from the equilibrium position on the high pressure side is smaller than on the low pressure side because on the high pressure side both the pressure force and the restoring spring force act together to bring the motion to a stop, whereas on the low pressure side (after the air flow is cut off), only the spring force brings the motion to a stop.”

 

Partially correct. Rlgp, you are correct in saying that the amplitude of tongue motion above is smaller than that below. They differ by about 30% for average blowing pressures, and the reason is that the persistent downward net pressure forces on the vibrating tongue shifts the equilibrium position of (biases) its motion, which is very close to sinusoidal. But nowhere does pressure force serve to stop the reed motion (vibration is not stopped). Spring force is the only force that stops it. While speaking in periodic motion, when the tongue reaches its highest point above the shoe, it is bathed in P1, and my guess is that it experiences only very small unbalanced pressure forces.

 

“Fig.1 represents the typicall reed in it's starting position. The lift force is orders of magnitude higher than the force from pressure gradient.”

 

Incorrect. Lukasz, the “suction” caused by the air velocity illustrated by your arrows cannot possibly cause a force on the tongue that is larger than (P1 – P2)*A. If you haven’t already read my explanation above in explaining the role of the Bernoulli equation, I suggest you do so. Assuming a quasi-steady flow field, the airflow would not hug the underside of the tongue as you draw it, and it’s likely that it would form a jet, with the pressure inside the jet equal to P2. In the unsteady case, however, things can be much different, and perhaps there is some kind of airflow, generally as you draw it (in which case, the flow field will be Potential, or Irrotational). The unsteady case is too difficult to discuss here, but even in that case, I see no mechanism for a “lift force” to be “orders of magnitude” higher than pressure forces. Can you provide good physical reasoning for your statement?

 

There is another complication here, when the tongue has established its periodic motion, even with the quasi-steady assumption. In this case, when the reed is moving upward through the slot, its top surface collides with air, producing a dynamic pressure that is higher than P1 by the increment 0.5*rho*Vr^2, where Vr is approximately the maximum tip velocity of the reed. From my own measurements, an accordion tongue of pitch 696 Hz vibrates with amplitude of about 0.3 inches (forgive the English system of units). This means that the average velocity in one quarter cycle is 0.3/12*696*4 = 69.6 ft/sec, and the peak velocity will be twice that, or 139 ft/sec. The dynamic pressure produced when this tongue collides with air is 0.5*0.075*139^2/32.2/144 = 0.156 psi, or 0.156/0.036 = 4.3 inches water column above P1, which is roughly about the normal bellows pressure (P1 - P2). In other words, for a brief moment during the swing cycle, the tongue experiences on its top surface, only near its tip, a pressure force, while moving upward, approximately twice the bellows overpressure. Also, when the tongue is moving downward, when it first enters the slot, a similar event occurs on its lower surface. Both of these over-pressures are contrary to motion and will produce aerodynamic drag, and since they occur for both tongue directions, their effect on tongue motion may largely cancel. However, they should have a noticeable effect on the sound field, and for some reeds produce a feature of the tone we perceive. It’s also difficult to judge the force on the tongue that these over-pressures cause, since we don’t know the areas and the duration times involved. Since these over-pressures oppose tongue vibration, we do know, in spite of them, the tongue does indeed vibrate very well!

 

“Fig.2 represents your design. You can see how the lift is created on both sides of the tongue, with values much smaller that on the "classic" assymetric reed.”

 

Incorrect. Lukasz, keeping with the quasi-steady assumption, the pressure on the top of the tongue near the tip is approximately P1, simply because the airstream, in striking the tongue, stagnates there, approximately producing the airstream’s stagnation pressure over some undefined area near the tip, which is P1, and causing a downward force on the tongue. But over the vast majority of its upper surface, the tongue experiences pressure very near P1. In any event, I see no “lift force” pushing upward on the tongue in this instance during the start transient.

 

“Fig.3 represents an additional lift force created when tongue passes through the shoe. It is again working towards the resting point and propels the resonant oscilation slightly.”

 

Incorrect. Lukasz, I make the same comments here as I do in the above paragraph. You seem to think that, wherever there is a moving airstream on the tongue, it creates a “suction,” or “lift force.” This is not correct. The airflow effect on the tongue depends upon the angle that the flow makes with the tongue. When it’s perpendicular towards the tongue, the pressure on the tongue is the stagnation pressure of the flow. When it’s parallel to the tongue, the pressure on the tongue is equal to the pressure in the airstream. The steady state Bernoulli equation allows you to calculate the difference in pressure from the air velocity, or conversely, to calculate the velocity from the pressure difference causing the airflow.

 

“This is the moment when gap suction aerodynamic effect occur, sucking air from above the corner, creating dynamic underpresure on the closer surface of the paper, moving the paper and closing the gap.”

 

Incorrect. Lukasz, here, there is no “sucking air from above the corner,” since such an airflow would necessarily have to move counter to your breath as it loops around the paper’s edge. This would violate thermodynamic laws. I don’t understand the term, “dynamic underpressure,” but I fear it has something to do with a belief that moving air near flat surfaces causes “suction”. The only external forces that could act here are pressure forces, and we need to understand the magnitudes of the various pressures (and the areas they act on, along with their duration) in order to understand the effects of the forces they produce.

 

“From a physicist's point of view, there are only two forces involved -- the force due to the pressure gradient and the internal spring force.”

 

Incorrect. Rlgp, a real physicist would acknowledge a third force, the inertial force, due to acceleration of the masses involved. But I can understand why you would ignore this force, because you are reasoning from a quasi-steady assumption. Although the air flow, in some instances, can be considered quasi-steady, as with the lower pitched reeds, the tongue motion is indeed always unsteady, which means that its inertial forces are of the same magnitude as the forces you mentioned. In Newton’s second law, F = ma, the two forces you mentioned are contained in F, but you are neglecting the right side of the equation.

 

“Let me re-emphasize: The pressure difference between the top of the tongue and the bottom of the tongue is dynamic, and apparently gets a lot larger than the static pressure difference between the inside and outside of the bellows.”

 

Incorrect. Rlgp, you have given no physical line of reasoning that can justify your conviction that the pressure difference across the tongue is larger than the pressure difference between the top and bottom reservoirs, when the tongue is starting to vibrate. I have presented a calculation above that suggests there can be an approximate doubling of pressure difference on a small portion of the tongue when the tongue collides with air in both directions, as it moves with its maximum velocity, and during its established periodic motion. Other than the possibility that this collision can have a significant effect on the sound of the reed, it’s difficult to conclude much about its effect on tongue motion because we don’t really know the area and time duration over which these over-pressures act. In fluid dynamics, there is a term “dynamic pressure,” which is well defined as = p + 0.5*rho*v^2, where p is static pressure and v is flow velocity, but you have not invoked this definition.

 

“And yes, dynamic pressure (lift force) on the low pressure side of the tongue is what drives the oscilation, which I was telling you from the very start of this whole debate.”

 

Not clear. Lukasz, perhaps there’s a language problem. Your use of the term “dynamic pressure” is not conventional, as understood by fluid dynamicists. Dynamic pressure is not necessarily connected with a “lift force.”

 

“As the end of the tongue approaches the frame below, the air flow is blocked, so that the downward force due to pressure (or whatever terminology you want to use) is greatly reduced.”

 

Incorrect. Rlgp, this is a re-statement of the same error. When the airflow is blocked, the pressure force pushing down on the tongue increases to a maximum, as I explained above.

 

“The spring force will bring the motion to a stop and send the tongue back toward equilibrium. It will pass through equilibrium and move into the region above.”

 

Correct! Rlgp, how does that happen? This is one thing about the motion of a free reed that I do not understand: what mechanism causes the tongue to return to a point that is above its equilibrium position? We know it must, and it must return to higher and higher positions during the start transient, if it is to speak. But how does it do that, especially because, in travelling upward, it works against a maximum pressure difference? Anyone?

 

“Remember, however, if the oscillation amplitudes are such that the air flow is cut at both the top and bottom of the swing, the frequency of the reed with two frames will be twice that of the one with a single frame.”

 

Incorrect. This is a repeat of the same error I corrected above.

 

“This is false, I suspect that the diagram is just that, a diagram, showing an idealised view and not a scale drawing of an actual reed in operation.”

 

It’s both true and false. Theo, the very high pitched reed tongues do not travel below the reed plate, although the medium and lower pitched reed tongues do.

 

“This parcel of air would be shorter that the other two and would presumably create a second much higher note. So you would have a note, a much higher note, and then the first note again every time the reed swings a full cycle.”

 

Incorrect. Chris, with due respect, the pitch perceived will be a single note, equal to the frequency of vibration of the tongue, as I explained at the beginning. I’m amazed that this misconception has been so ingrained in people who busy themselves with the operation of the free reed. Our little friend holds many surprises for us. There is no such thing as a note produced during each “full cycle.” A musical tone is produced by regular, periodic vibrations. It’s not possible for the free reed to produce two different notes at the same time. All the pressure pulses that occur within one cycle are repeated periodically, and it’s the periodic occurrence of all pulses that produces the tone we perceive, once we have heard the tone for a sufficient number of cycles, and the sufficient number of cycles depends upon the cycle frequency.

 

“And you are correct that with two cutoffs the frequency would be higher -- about twice as high in fact, as i said earlier.”

 

Rlgp, as I said earlier, incorrect.

 

“For example, i think that filing the inner edges of a conventional reed frame may lead to less abrupt cutoff and recovery of the air flow, and thus a more trapezoidal shape for the wave form. This shape puts relatively less energy into higher harmonics, resulting in a more mellow sound, something that would be desireable to many.”

 

Rlgp, are you just guessing that any trapezoidal shape will put less energy into the higher harmonics than any rectangle, or have you studied Fourier transforms of both rectangles and trapezoids? But I think you are being distracted by such thinking. You may imagine the shape of a pressure pulse, but I can assure you that the pulses that actually occur will be far different. It’s one thing to assert what will happen, based upon imagination, and it’s another thing what Nature actually does, especially when your thinking does not invoke all of the physical mechanisms at work.

 

“With this i expect that it's time (many would say long past time) for this thread to end. I've enjoyed it because of the mental stimulation it has brought -- something i've missed since retiring as a physics professor about 10 years ago.”

 

Rlgp, may I ask, what grade level of Physics did you teach, and what level of formal education have you had in Physics? Thanks.

 

Let me make some comments of a general nature. The free reed vibrates primarily as a simple harmonic oscillator. With high blowing pressures, the second mode of vibration can be slightly excited, and the jagged pulses of air forcing the motion do cause a small amount of overtones in the lowest mode cantilever vibration, but these are confined to about 10% of the motion. This system also demonstrates what we call self excited oscillation, whereby the oscillation itself determines when the forcing function acts during the cycle. In order for there to be sustained vibration that overcomes friction and produces audible sound energy, a force must be applied at any time when the tongue motion is in the same direction. Much like a child on a swing, with a parent pushing at the right moment and in the direction of swing motion, the condition P1 > P2 ensures that, for some of its downward motion, the tongue experiences a downward (forward) kick once per cycle. Mathematically, this means that the product of complex pressure and complex velocity has a non-zero real part, and this translates in physical terms to mean that work is being done by the pressure difference (P1 – P2), and that work fuels all the air and tongue motion, establishing a sound field we perceive as a musical tone. Such a simple view might satisfy some, though I acknowledge that accurate understanding of the details of this description can prove challenging.

 

Best regards,

Tom

www.bluesbox.biz

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Truly fascinating reading, Tom! And, assuming you have it right, I must say that I always find it rewarding to follow certain - more or less common - misunderstandings in order to learn not just about some subject or matter but about the human mind and way of thinking as well... Hard to imagine how people without all this advanced knowledge are not just maintaining things but appear to have invented everything...!

 

Best wishes - Wolf

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Not to start any further debate, but just to clarify what I had in mind all this time, I have finally found a proper english name for what I have called a "suction effect" - I have previously used the direct translation from polish "zjawisko przyssawania" but it is called a Bank Effect in english: http://en.wikipedia.org/wiki/Bank_effect (as you can see now it is not a partial vacuum which makes drinking with straws work). This was my first physics debate in english and I can only apologise for not using proper english terminology (I have tried to find this name then, but have failed to…)

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Łukasz, according to the WP article it's after all also called "bank suction" or "stern suction" - however, the somewhat easily misleading phenomenon reminds me at sailing close to the wind (in fact another nautical theme...), as the heeling is not induced by the wind pushing the sail and mast aside and downwards but due to the aerofoil effect created by the wind skimming along the sails and thus reduced pressure on their lee side...

Best wishes - Wolf

Edited by blue eyed sailor
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This is one thing about the motion of a free reed that I do not understand: what mechanism causes the tongue to return to a point that is above its equilibrium position? We know it must, and it must return to higher and higher positions during the start transient, if it is to speak. But how does it do that, especially because, in travelling upward, it works against a maximum pressure difference? Anyone?

 

 

This is exactly why I think that (P1 – P2)*A is not a maximum force acting on the reed and an airflow through the gap must create force larger than this - you are seeking an additional mechanism that causes an amplified return in the first cycle, I seek the different mechanism which draws the tongue into the shoe in the first place, so the return mechanism is plain and logical.

 

This force created by the gap airflow and it's relative value to the pressure applied to the tongue is easiest to observe on the lowest/largest possible reeds as the scale and frequency of movement is observable with a naked eye. If I experiment with my lowest reed on my tuning rig, this is what happens:

 

1) if I act on the bellows with normal force (pulling on the bellows, so I can observe the tongue motion above the shoe) the tongue is pushed into the shoe (just the thickness of the tongue below the edge, so the whole reed is flat) by the (P1-P2)*A force and does not speak, it acts only as a pressure valve. (by "normal force" I mean the force that makes treble reeds speak)

2) in order to make it speak, I must start the airflow through the gap with a much lesser force on the bellows (so that (P1-P2)*A is to small to draw the tongue into the shoe and in the same time is too small to keep it inside on the "agaist the pressure" part of motion). The tongue start to vibrate (at the stage in which it does not even sink into the shoe with all it's thickness and almost all of the oscilation happens slightly above the shoe) and only then I can increase the force on the bellows to make it louder. (I can recreate this reed-starting behaviour with weights attached to the bellows, so the eventual unsteady motion of the hand is not a factor in this)

 

Another "paper experiment" for you to ilustrate that lift is generated: take a small piece of paper (I have used 5x10cm strip) and place it so it is horizontal (rest it on your finger) and has one (short) edge firmly held against your lower lip and then blow on it. As long as you blow there is an airflow speed difference between the upper and lower sides of the paper, so you can take your figer away and the paper will stay horizontal. The curvature of the paper is not important, it does not need to be an airfoil - an airfoil shape is necessary to divide the single pressure enviroment into two enviroments of different pressure but in this case we divide the enviroment by introducing the selective airflow by blowing just on the upper side(*). As soon as you stop blowing, the paper bends naturaly downwards, but you can lift it again by blowing (in this "freely hanging" starting position it does resemble an airfoil at the begining, but it becomes more flat the harder you blow and this shape is not essential to lift the paper as shown above in "resting on the finger" starting position).

 

If you look at the closeup of the reed tongue at the gap, it looks very similiar to this experiment: due to the tongue/gap/shoe geometry there is significant airflow only on the single (lower) side of the tongue. And my above example with the low reed shows that this effect occur at pressure lower than needed to push the tongue into the shoe and that this is essential for the reed to speak. So I must disagree with you Tom on your statements that my illustrations are incorrect.

 

My interpretation seems to be coherent with all different reed types: free, beating and double (and double oboe reed is the most close realisation of the "two sheets of paper" experiment, just with the hinge points on the oposite ends of sheets). And especially beating and double reeds cannot be explained by the approach in which (P1-P2)*A is the largest acting force, as this would just cut the airflow completely and hold the reeds closed as long as there is high pressure acting. And as you Tom admit, your interpretation fails to explain how the reed goes back against the pressure to higher position. Mine may of course be entirely wrong, but it does at least explain this and can be illustrated by all sort of different experiments.

 

(*) to be perfectly correct there is a secondary airflow on the lower side of the paper to ensure the continuity of the flow speed gradient across the enviroment (the flow is laminar, no turbulence is created or necessary), and is caused by the upper airflow, but it is much slower than the upper flow.

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It seems that i was incorrect in saying that it was time for the thread to end.

 

First let me deal with two specific points brought up by the quote below:

 


“From a physicist's point of view, there are only two forces involved -- the force due to the pressure gradient and the internal spring force.”

 

Incorrect. Rlgp, a real physicist would acknowledge a third force, the inertial force, due to acceleration of the masses involved. But I can understand why you would ignore this force, because you are reasoning from a quasi-steady assumption. Although the air flow, in some instances, can be considered quasi-steady, as with the lower pitched reeds, the tongue motion is indeed always unsteady, which means that its inertial forces are of the same magnitude as the forces you mentioned. In Newton’s second law, F = ma, the two forces you mentioned are contained in F, but you are neglecting the right side of the equation.

 

* * *

 

Rlgp, may I ask, what grade level of Physics did you teach, and what level of formal education have you had in Physics? Thanks.

 

With regard to the first quote, i assume that you accept Isaac Newton as a "real" physicist. In his second law, which you have written down, the forces go on the left side; the right side describes the effect of the net force. So far as i know, Newton never considered ma to be a force. The vast majority of physicists in the world do not consider ma to be a force, from the point of view of Newtonian classical mechanics. Sometimes a different point of view is taken by moving ma to the left side to get F - ma = 0; then by treating ma like a force, a dynamic situation can be treated formally like a static one. Similarly, to look at a dynamics problem in a non-inertial reference system, a portion of the ma term may be moved to the left side. This portion is usually referred to by "real" physicists as "ficticious forces" (e.g., centrifugal force, Coriollis force). Neither of these cases is the approach taken by Newton. I have been careful to state that my analysis is in terms of Newtonian classical mechanics, and so when i say force, i mean only force, not ma.

 

In answer to your question, i have a PhD in physics from the University of Colorado, and i taught physics from freshman level through Masters' level for about 25 years, mostly at the University of New Orleans. I have produced just over 50 refereed research papers and written a supplementary junior/senior-level textbook on classical mechanics. I consider myself (as did my doctoral commitee and the physics faculty at two universities at which i was employed and a theoretical physics division within the former National Bureau of Standards) a "real" physicist. (For those reading this, i apologize for "tooting my own horn"; i usually go out of my way to avoid doing so, but ttonon's suggestion that i am not a real physicist got under my skin.)

 

I have no significant expertise in fluid dynamics, which is why i have taken pains to describe my understanding of the motion of concertina/accordion (C/A) reeds using the language of Newtonian classical mechanics.

 

 

 

Now, with that out of the way, we need to agree, or at least explicitly address, an experimental question. Early in our discussion (on the original thread), and repeated several times subsequently, Lukasz stated that the force due to the static pressure difference could deflect the tongue only a very small amount, and thus was insufficient to explain the dynamics of a C/A reed. He cited as additional evidence of this the fact that a reed whose tongue is flat does not work. (Am i correct in my assessment of your statements, Lukasz?) I personally think that it must be true, and that the major contribution to the pressure difference that drives the tongue comes from the air motion. However, this is a question that should be settled based on experiment, because if it is not true, all the words that Lukasz and i expended were of no import.

 

Nevertheless, there are some important statements in ttoton's extensive discussion that are not correct. First, the Bernoulli's equation that he quotes is true only for incompressible fluids, a condition does not apply here, (at least) on the high pressure side of the tongue. There the air density increases as air "piles up" in the region very near the body of the tongue as the latter brings the vertical component of the air's drift velocity to a stop.

 

I shall use capital letters to indicate (essentially static) pressure within the bellows and above the reed (P1) and the ambient air pressure below the reed (P2). With the bellows compressed, P1 > P2. I shall use lower case letters to represent the dynamic air pressure at the tongue's upper surface (p1) and lower surface (p2). Although we cannot use the simple form of Bernoulli's equation, when the air is flowing we can say that p1 > P1, because in addition to the force due to the static air pressure, there is an additional downward force on the tongue as it brings the downward component of the air's drift velocity to (essentially) zero as the air collides with the tongue -- f1 = P1*A + fdrift and p1 = f1/A.

 

On the bottom side of the tongue, the simple form for Bernoulli's equation is probably approximately true. Thus if we consider two points -- one just below the fixed left end of the tongue where the air is nearly stationary (pressure P2), and one at an intermediate point just below the tongue (pressure p2) where the air is moving -- we have P2 = p2 + 1/2 rho*v^2, from which one finds p2 < P2.

 

Thus, in contrast to the inequality ttonon gives, the actual relationship is p2 < P2 < P1 < p1. So the force due to the pressure difference at the top and bottom of the tongue is (p1-p2)*A > (P1-P2)*A. I think that this dynamic force (while the air is flowing) is substantially greater than the static force on the right side of the inequality. As i understand what Lukasz has said, i think he also believes this, though he should of course speak for himself.

 

Because of the length of ttonon's post, i'm not going to address it point by point. I will simply say that i still think that my description of the motion of the tongue that i gave in my first post in this thread is essentially correct.

 

 

Now, with regard to the sound produced by a C/R reed. The sound produced by the oscillating tongue is nearly inaudible. What we hear is the sound produced by the puffs of air flow, alternating with periods of almost no air flow. This air wave is, to a pretty good approximation, a square wave, as can be seen by the recorded signal shown in the Physics Today paper that sqzbxr was so kind to provide in the previous thread -- http://www.public.coe.edu/~jcotting/Cottingham_pages.pdf. For a conventional reed, which cuts the airflow once every cycle of the oscillating tongue, the frequency is the same as the tongue frequency.

 

The very sharp jumps of the square wave come about because the cutoff and restoration of the air flow is very abrupt in the conventional reed due to the shape of the slot in the reed frame. With the top interior edges of the frame filed down somewhat as i suggested, the cutoff and restoration of the air flow should not be so abrupt. I qualitatively illustrated the resulting wave form as trapezoidal, though the actual shape will be more complicated. In the mode of operation of the bidirectional reed in which the tongue cuts the air flow at both the top and bottom of its swing, puffs of air occur twice per tongue cycle -- once when moving down through the V region, and once when moving up. These puffs will be separated by periods of essentially no air flow, when the tongue is in the body of top or bottom frames. The duration of the puffs will be less than the duration of the puffs in a conventional reed (and the rate of air flow may also be less), so the volume of sound will be less, but the frequency will be twice the frequency of the tongue oscillation.

 

As to my statement about less energy going into higher harmonics of a trapezoidal wave in comparison to a square wave, the amplitude of the Fourier component of the nth harmonic for a square wave drops off as 1/n. The amplitude of the Fourier component of the nth harmonic of a trapezoidal wave drops off as 1/n^2. The relative energies associated with these harmonics thus drop off as 1/n^2 and 1/n^4, respectively -- significantly less energy goes into the higher harmonics of the trapezoidal wave. Less energy in higher harmonics should mean a more mellow sound.

Edited by rlgph
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