Jump to content

Why Do Reeds Need To Occur In Pairs?


Recommended Posts

Well, the short answer is, that the reed would not 'start' in one of the directions.

 

The reed tongue starts to play by being sucked into the slot of the shoe by the flow of air, then bounces back, and thus vibrates producing the note until the airflow is cut off again. This is why the setting of the reed is crucial for having a good responsive instrument - too low and there will not be enough draught under the tip (between the tongue and the edge of the slot in the shoe) to start the swinging of the reed, and if too high there will be too much air being sucked under the clearance before the reed starts vibrating.

 

Thus, the reed only functions in one direction - air flowing the wrong direction simply would not start the reed - and it's therefore not possible to have the reed operating for both bellows direction.

Link to comment
Share on other sites

As i understand it, the slight bend in the reed (toward the lower pressure side) allows air to flow around the end. The resulting turbulence on that side produces the time-varying force that causes the reed to vibrate. Without the bend the pressure difference would first have to open the reed up slightly to allow the air to flow -- hence the slowness to get the reed vibrating.

 

Now, consider a totally flat reed (with a close fitting, but non-binding frame of the same thickness as the reed). Without doing anything else to the reed, when the bellows is pushed or pulled, the reed will experience the delay while the pressure difference causes it to bend slightly and start the vibration. So what can you do to the reed to lesson the delay? Two possibilites come to mind. One, you can slightly file the end of the reed so air can begin to flow and set up the turbulence as soon as the pressure gradient is set up. Two, you can thin the reed at the fixed end by filing carefully on both sides, allowing a given pressure difference to bend the reed more easily and more quickly to set up the air flow.

 

Obviously a good bit of experimentation would be necessary to see if the scheme will work, and work satisfactorily, but the potential benefit in terms of cost and compactness is significant if a single reed can be set vibrating from either a push or pull of the bellows.

Edited by rlgph
Link to comment
Share on other sites

The sound of a concertina reed doesn't come from the reed itself vibrating. It comes from the airflow through the vent being totally cut off at so-many times per second (440 times for the "central A", etc). If you make the reed smaller than the vent, either by filing the tip or the sides, then it will never totally cut off the airflow through the vent, and you won't get a note.

Link to comment
Share on other sites

I am aware that the sound of a concertina comes only indirectly from the vibration of the reed. However, In the discussion of reeds on the Concertina Connection site, the picture of the "third position" shows the reed passing into the slot in the frame. That diagram (and the discussion) indicates that the reed is smaller than the slot in the frame, and thus never completely cuts off the air. How much smaller the reed can be than the slot and still produce an acceptable sound is an empirical question.

Link to comment
Share on other sites

After thinking more about it, although there will be turbulence, it's probably not necassary to produce the time-varying force necessary to start and maintain the reed vibration. The pressure difference between the two sides can start the reed moving, and the variation of that pressure difference with the position of the reed in its cycle (as air flows from one side to the other) can produce the time-variance of the external force.

Link to comment
Share on other sites

As i understand it, the slight bend in the reed (toward the lower pressure side) allows air to flow around the end.

 

 

You think it backwards… Reed tongues are bent to the higher pressure side: pressure is working towards closing the reed (pushes the tongue inside the shoe, not away from) and is released when tongue reaches the other side of the shoe. Then the energy acumulated in the tongue swings tongue back against the airflow and opens the reed for half a cycle, when another portion of air draws the tongue back into the shoe with more force. And this initial gap works as a miniature version of a bladeless fan - it is essential to draw the tongue into the slot with initial pressure lower than needed to hold the tongue on the other side permanently. It is dynamic not static process and takes a few cycles to achieve resonant and stable vibration. Without this gap (and in the oposite direction) a reed tongue works as a pressure valve which won't vibrate - it will just open in a static way when pressure gradient is high enough to bend it and let the air out from the higher pressure reservoire.

 

There are asian type free reeds, which are flat as the one you describe, but they work on a different principle and need a pipe resonator to sound and are mounted parallel to the airflow direction, not perpendicular as in concertinas or accordions.

Link to comment
Share on other sites

You think it backwards Reed tongues are bent to the higher pressure side: pressure is working towards closing the reed (pushes the tongue inside the shoe, not away from) and is released when tongue reaches the other side of the shoe. Then the energy acumulated in the tongue swings tongue back against the airflow and opens the reed for half a cycle, when another portion of air draws the tongue back into the shoe with more force. And this initial gap works as a miniature version of a bladeless fan - it is essential to draw the tongue into the slot with initial pressure lower than needed to hold the tongue on the other side permanently. It is dynamic not static process and takes a few cycles to achieve resonant and stable vibration.

I understand this; i am not thinking backwards here, though i apparently have not been clear. The main thing is that a time-varying external force is necessary to cause the oscillation, and the air leakage around the end of the reed is the primary way that the pressure difference is lessened, so that the external driving force due to the pressure difference is not constant.

 

Without this gap (and in the oposite direction) a reed tongue works as a pressure valve which won't vibrate - it will just open in a static way when pressure gradient is high enough to bend it and let the air out from the higher pressure reservoire.

 

Not true. Imagine a horizontal reed with a pressure difference such that the reed is initially forced downward. As the reed opens on the downward side, air flows through, reducing the pressure gradient. The more it moves down, the lesser the pressure difference and the greater the (internal) restoring force of the spring, causing the motion of the reed to stop, then accelerate back toward the equilibrium position. Its momentum will carry it through to the top side, where the restoring force and the newly increased pressure difference will stop its upward motion and drive it back downward, starting the cycle again. Each time the reed passes through equilibrium it cuts off most of the air flow. As mentioned earlier, it is this periodic cutting of the air flow that is responsible for the audible sound.

 

My description of the physics is right (though turbulence, which i've ignored here, may play some role) -- a flat reed will oscillate when subjected to a pressure gradient set up perpendicular to the reed by a push or pull of the bellows, and the oscillation will cause periodic cutting of the air flow. What i don't understand is why this cannot be used to make an acceptable concertina sound.

Edited by rlgph
Link to comment
Share on other sites

Sory, you have mislead me into believieng you have it backwards by this sentence: "As i understand it, the slight bend in the reed (toward the lower pressure side) allows air to flow around the end." as it is exactly oposite to reality.

 

 

My description of the physics is right (though turbulence, which i've ignored here, may play some role) -- a flat reed will oscillate when subjected to a pressure gradient set up perpendicular to the reed by a push or pull of the bellows, and the oscillation will cause periodic cutting of the air flow. What i don't understand is why this cannot be used to make an acceptable concertina sound.

 

 

 

Because it does not work like you describe… A reed without a gap does not speak, flat reed mounted perpendicular does not speak.

 

Perpendicular reed needs to be assymetric and require a gap and nothing can be done about it. It is just how reed physics works. Please make a simple experiment - take one of your bass reeds, bend it back flat into the shoe so there is no gap and try to make a sound. It won't speak at all and will be just poor pressure valve. [bass one is best for this purpose, because high tone reeds require only a slight gap to start, so even when flatted out, the tolerance around the tongue is a gap sufficient to start the movement at low pressure levels]

 

 

The more it moves down, the lesser the pressure difference and the greater the (internal) restoring force of the spring, causing the motion of the reed to stop, then accelerate back toward the equilibrium position.

 

 

This is the point where you get this wrong - as long as you have pressure it won't accelerate back, it will just stop bending more and will find stable bent position depending on the pressure gradient.

 

In more elaborate form: in the flat, gap-less reed the pressure is just released as through valve - spring just bends proportionally to the pressure pushing on it [and because tongue thickness is so small it happens with very minute pressure gradient levels]. As long as you squeeze your bellows with same force, the pressure inside is constant, the reed/valve opens and air just flow outside with a rate depending on reed stifness. This is exactly what happens when you blow air through reed in oposite direction - it just leaks air, it does not oscilate and in this case the initial gap seting is irrelevant (it can even be "negative", i.e you could set the tongue to be initially inside a shoe, this will just slightly increase "releasing pressure"). Of course there is some oscilation when you release the force on the bellows, but this is not resonant oscilation, just a depleting momentum of an elastic valve returning to equilibrium, making no audible sound at all.

 

One other thing: pressure gradient changes are not instantaneous. When you press your bellows or open the pad on the airhole it takes some small amount of time to start the airflow. This is the moment when reed oscilation "jump starts" through gap mechanism on pressure levels lower than needed for stable oscilation, so when that pressure is achieved the reed is already in it's swinging motion (the gap is a leak which makes effective pressure gradient lower at the start of the motion than when the tongue is inside the shoe and blocks the airflow). With fast enough, high pressure pump instantly and constantly feeding high pressure air through a reed you can choke even the properly set assimetric concertina reed and effectively turn it into a mute pressure valve. This is one of the reasons why different free-reed instruments require different reeds depending on operating pressure levels, as this can even occure on my Elise lowest bass note, when I squeeze a bellows the hardest I can, as it has accordion reeds, designed for lower pressure levels.

Link to comment
Share on other sites

Sory, you have mislead me into believieng you have it backwards by this sentence: "As i understand it, the slight bend in the reed (toward the lower pressure side) allows air to flow around the end." as it is exactly oposite to reality.

Oh, sorry. I should have said "toward the higher pressure side". That is what i was thinking. I should read what i write more carefully.

 

 

The more it moves down, the lesser the pressure difference and the greater the (internal) restoring force of the spring, causing the motion of the reed to stop, then accelerate back toward the equilibrium position.

 

This is the point where you get this wrong - as long as you have pressure it won't accelerate back, it will just stop bending more and will find stable bent position depending on the pressure gradient.

My statement you have quoted here is correct. It will only find a stable position if the pressure difference does not change, so that the force due to the air pressure exactly balances the restoring force internal to the spring. In this case, however, air is flowing around the end of the reed causing a decrease in the pressure difference. At the maximum downward swing, the restoring force is greatest, and the force due to air pressure difference is getting less due to the air flow. The net force is toward the equilibrium position and the reed acelerates in that direction. And that is true regardless of whether in equilibrium the reed is bent upward or not.

 

One other thing: pressure gradient changes are not instantaneous. When you press your bellows or open the pad on the airhole it takes some small amount of time to start the airflow.

True, but the air flow will start before the reed starts to move because the mass of the air molecules is much much less than the mass of the reed.

 

As i said before, the physics that i have described is correct. However, my description of the operation of a reed under the influence of a perpendicular pressure gradient does not depend on whether there is an initial gap or not. If you have done the experiment with a flat reed and found no oscillation, it means that i have left out some aspect of the physics contributed by the bent reed. Perhaps turbulence caused by the initial air flow around the bent reed is the explanation, producing a force that initially is significantly larger than that due to the pressure gradient.

 

Thanks for your thought provoking comments.

Link to comment
Share on other sites

 

As i said before, the physics that i have described is correct. However, my description of the operation of a reed under the influence of a perpendicular pressure gradient does not depend on whether there is an initial gap or not.

 

 

 

Then it is not correct, as the gap and asymetry are crucial in free-reed, accordion/concertina like instruments which is proven by almost 200 years of experimentation and you can replicate this in a 1 minute experiment involving a single reed and pair of lungs…

 

 

 

It will only find a stable position if the pressure difference does not change, so that the force due to the air pressure exactly balances the restoring force internal to the spring. In this case, however, air is flowing around the end of the reed causing a decrease in the pressure difference. At the maximum downward swing, the restoring force is greatest, and the force due to air pressure difference is getting less due to the air flow. The net force is toward the equilibrium position and the reed acelerates in that direction. And that is true regardless of whether in equilibrium the reed is bent upward or not.

 

 

 

You are describing a spring pressure valve here, not a free reed oscilator. What you are (correctly) describing here is the physical process involved in a static tongue bending caused by pressure gradient and air leaking through an unvalved reed in it's silent direction, until all bellows is compressed and there is no more higher pressure reservoire. At which point (and just then) the net force you describe is in fact accelerating the tongue back to it's resting position with some tiny amount of oscilation at the end of it's movement.

 

The process involved in producing sound by a free reed is entirely different, involves a gap and asymetry of the reed, and I have described it earlier. If you don't see your mistake at this point I have no more ideas on how to point it out to you...

 

 

Thanks for your thought provoking comments.

 

 

You're welcome :)

Link to comment
Share on other sites

 

As i said before, the physics that i have described is correct. However, my description of the operation of a reed under the influence of a perpendicular pressure gradient does not depend on whether there is an initial gap or not.

 

Then it is not correct, as the gap and asymetry are crucial in free-reed, accordion/concertina like instruments which is proven by almost 200 years of experimentation and you can replicate this in a 1 minute experiment involving a single reed and pair of lungs…

As i said in the post above, the physics is correct, but incomplete, since it does not agree with experiment. There must be other forces acting. My guess is turbulence, but that's only a guess.

 

 

It will only find a stable position if the pressure difference does not change, so that the force due to the air pressure exactly balances the restoring force internal to the spring. In this case, however, air is flowing around the end of the reed causing a decrease in the pressure difference. At the maximum downward swing, the restoring force is greatest, and the force due to air pressure difference is getting less due to the air flow. The net force is toward the equilibrium position and the reed acelerates in that direction. And that is true regardless of whether in equilibrium the reed is bent upward or not.

 

You are describing a spring pressure valve here, not a free reed oscillator.

What i am describing is the motion of a physical pendulum (the reed clamped at one end) subject to a perpendicular external force caused by a pressure difference between the two sides -- a pressure difference that changes as the reed moves due to air flow from one side to the other. It is certainly not a static system. Since my model does not agree with experiment, there is another significant force missing, and this force is related to the bent reed equilibrium position. Unless you can explain what this force is, neither one of us has satisfactorily explained the physics of the reed's motion.
Link to comment
Share on other sites

By "static" I don't mean stationary, but "equalized at all times" - in this case, in your approach, high pressure is always compensated by internal spring force. It does not mean that the tongue does not bend, only that forces are equalised in any given "time slice". I may use the wrong english word to describe it. "Dynamic" in my comments mean that the process does not equalize in such manner, and there is no such "missing force" in any given "time slice", but you have to consider the whole temporal evolution of a system consisting of high pressure reservoire, tongue spring, gap suction, airflow etc...

 

The reed is not a physical pendulum, it is a spring oscilator. The "missing forces" (if we must name it as such) that you choose to neglect are dynamic effects of gap suction, initial airflow increase, pressure gradient "momentum" and resonant oscilation effects in the first cycles of the reed movement and I have given you a very extensive and detailed description of that process.

And I'm not debating on how the reed work, I'm refering to you a well established knowledge, which for some reason you refuse to acknowledge… And you have even mentioned earier, that you have read an extensive essay on the matter on concertina-connection site, but you still refuse to agree, that symmetrical or flat reed simply won't work...

 

So I have one question for you: Your description does not differentiate sides and you assume that flat reed would work. In your model, what is the difference between the physics of sounding and silent direction of airflow through traditional reed? If there is none, and gap and asymetric thickness is irrelevant and flat reed should work, then why the reed sounds only in one direction and flat reed does not speak at all?

Link to comment
Share on other sites

By "static" I don't mean stationary, but "equalized at all times" - in this case, in your approach, high pressure is always compensated by internal spring force. It does not mean that the tongue does not bend, only that forces are equalised in any given "time slice". I may use the wrong english word to describe it. "Dynamic" in my comments mean that the process does not equalize in such manner, and there is no such "missing force" in any given "time slice", but you have to consider the whole temporal evolution of a system consisting of high pressure reservoire, tongue spring, gap suction, airflow etc...

I have taken great pains to point out that the forces are not equalized at all times. If they were there would be no acceleration, and therefore no oscillation.

 

The reed is not a physical pendulum, it is a spring oscilator. The "missing forces" (if we must name it as such) that you choose to neglect are dynamic effects of gap suction, initial airflow increase, pressure gradient "momentum" and resonant oscilation effects in the first cycles of the reed movement and I have given you a very extensive and detailed description of that process.

 

And I'm not debating on how the reed work, I'm refering to you a well established knowledge, which for some reason you refuse to acknowledge… And you have even mentioned earier, that you have read an extensive essay on the matter on concertina-connection site, but you still refuse to agree, that symmetrical or flat reed simply won't work...

Have you been reading my posts? Immediately after you told me the results of your experiment i accepted that the flat reed does not work. I have simply been trying to understand why not, from the point of view of physics. In order for oscillations to occur, the reed must be subject to time-varying forces. The only things that can exert forces on the reed are air molecules (external force) and the molecules that make up the reed (internal force). (Gravity is too small to have any significant effect.) My current guess as to why the bent reed works is that the air flow around the end of the read while the reed is still open on the high pressure side causes the air on that side to exert a significantly larger force than in the flat reed case, probably due to turbulence on the underside created as the air flows around the asymmetric reed configuration.

 

Now i have to go read the reference that sqzbxr provided (thanks sqzbxr) to see what it can tell me.

 

Link to comment
Share on other sites

Thanks very much for this reference. Although it doesn't satisfactorily answer my more recent puzzling about the underlying physics, it does provide references that should help. And it does answer the question that i posed in my original post before i became embroiled in the physics.

Link to comment
Share on other sites

@ rlgph: Yes, I have been reading all your posts carefully, and I'm trying as much as I can to point out to you where you make some false assumptions on how physics of free reed works...

 

I have simply been trying to understand why not, from the point of view of physics. In order for oscillations to occur, the reed must be subject to time-varying forces. The only things that can exert forces on the reed are air molecules (external force) and the molecules that make up the reed (internal force). (Gravity is too small to have any significant effect.) My current guess as to why the bent reed works is that the air flow around the end of the read while the reed is still open on the high pressure side causes the air on that side to exert a significantly larger force than in the flat reed case, probably due to turbulence on the underside created as the air flows around the asymmetric reed configuration.

 

 

One other thing, which is relevant to both this thread and our second physics discussion, which may lead you to understand free-reed physics and wasn't stressed enough previously: air pressure force acting on the reed is much lower than needed to bend tongue to it's maximum position - internal spring force is much greater than maximum bellows pressure force. It is the effect of resonant amplification of small tongue movements that makes this amplitude so large. The pressure in the bellows is high enough only to bend a plastic/leather valve to the roughly same order of magnitude of displacement as an oscilating metal tongue.

 

If you would make an experiment with a typical reed and measure tongue displacement, then your physics is a good description of what can be measured with a pressure acting in the silent direction of the reed. The tongue would bend a little, releasing high pressure - the higher pressure the more tongue displacement - but this displacement would be measured in fractions of a milimeter. And all forces would be equalised at all times, there is no need for aditional "missing time varying force" here.

 

With sounding airflow direction, what happens is this (in as elaborate form as I can think of): at first, when you squeeze the bellows, the air starts to flow through the gap. The suction effect of the gap is signifficant enough to move the tongue a little towards the shoe. Then when there is no more airflow (when reed sinks a bit into the shoe) and suction effect stops, the internal spring force (which is greater than pressure force acting on the tongue) is accelerating the tongue back towards it's resting position, but due to momentum it swings a bit higher than resting position. At this point the suction effect starts again, and acts towards drawing a tongue into the shoe, but this time there is a bit of added momentum from the spring force going back from it's highest position. A cycle closes, but with a bit of added energy. With each cycle the swinging motion increase a bit and after a few cycles the tongue goes through to the other side of the shoe, at which point airflow increases, making each cycle even more influential and within a few cycles the tongue achieves a stable oscilation, proportional to air pressure in the bellows, which now translates to volume (an oscilation amplitude) and frequency.

 

If you want to learn more about the strenght of suction effect in fluid mechanics, then read (and watch) about suction effect of ships passing each other in harbours as this is the most visual illustration of this effect. Or you can take two sheets of paper, hold them down close to eachother and blow gently between them and observe how they will be drawn to each other by a dynamic airflow.

 

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...