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A Proposal For A Bi-Directional Concertina Reed


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Tom,

 

1. (alternative)

An alternative way of seeing that the pressure on top of the tongue is greater than the static pressure at any point where the flowing air collides with the tongue is to look at conservation of energy of a tiny unit volume of air as it moves from the region above the reed (pressure P1, speed V1) to just after it collides with the top of the tongue (pressure p1, speed v1). The equation, of course, is Bernoulli's equation, since it fundamentally is based on conservation of energy. Although i think that air suffers some significant compression very near the tongue due to the collision with the tongue, i'll take your assumption that it is effectively uncompressed and apply the simple form of Bernoulli's equation between the two points. The result is P1 - p1 = 1/2 rho*v1^2 - 1/2 rho*V1^2. This is equivalent to the Work-Kinetic Energy Theorem of classical mechanics, applied to the motion of this tiny volume of air (and is one way of deriving Bernoulli's equation for incompressible flow). When this packet of air collides with the tongue's upper surface, it gives up some of its kinetic energy to the tongue, so that the right side of this equation is negative. Thus, again the result is p1 > P1.

 

If there is any air compression upon collision, the work required to perform the compression comes at the expense of the kinetic energy of the air packet, so that p1 will be greater than P1 by a still further amount.

 

ron

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Lukasz,

 

I have ceased trying to describe the entire motion of a conventional reed, particularly the transients, as being way too complicated for my expertise. Rather, i have confined myself to the initial conditions where the air is flowing and the tongue has not yet begun to move significantly. Under these conditions the dynamic pressure on the top surface of the tongue is greater than the static pressure on top (from straightforward application of Newton's laws of motion or alternatively, conservation of energy); i.e., p1 > P1. I believe that the dynamic pressure at the bottom of the tongue is less than the static pressure there; i.e., p2 < P2, though i had to make some additional assumptions to derive that. Thus, p1-p2 > P1-P2; i.e., the force due to air pressure on the tongue is larger when air is flowing than when not, and larger than that due to the simple difference between the static pressures. Whether it is sufficiently larger to explain the motion is an experimental question, but the fact that the reed with tongue bent slightly upward works in a concertina whereas a flat reed does not is certainly consistent with the result.

 

Incidentally, i think the same condition will apply when the tongue is moving as well, so long as the speed of the free end is less that the speed of the air flow.

 

I didn't quite follow your description of the positions of the straw in the three cases of your experiment. Is it convenient for you to provide simple drawings for the three cases?

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That does seem clearer, and also blends with Lukasz's experiment. I really had no idea in which part of the swing the increased force occurred, only that the amplitude need to reach a certain amount. It makes more sense to be when the reed closes the window. That has the relatively abrupt change in reed conditions I seemed to see in the trace. It had to be at a point where the return force or its momentum generated was sufficient, which I still would guess is due to the initial bias the set provides. Simply entering the window isn't enough, since a low set reed that chokes enters the window but doesn't come back up. Likewise a high set reed needs lots of Lukasz's straw to begin to enter the window.

One thing I am not clear on is the relative velocity of the reed tip to the air stream. It seems for all but low reeds and or high playing pressures, the reed tip velocity likely exceeds the unimpeded air flow velocity through the gap. If this is the case, the reed doesn't block the downward air flow at all, but pulls it on through. I could figure out the reed tip velocity, but don't know how to work out the air stream velocity. If this is the case, what happens then? Or does the reed tip never exceed the air velocity. ( amplitude always increases to match the feed tip to the air speed)?

Dana

 

Hi Dana, you raise an interesting question, one that I’ve asked myself, so now we have an opportunity to think more on it. In my first post, I did a calculation that gave a tongue tip maximum velocity (139 ft/sec) that corresponds to an air pressure difference of about 4.3 inches water column, which would cause the same velocity in an air jet experiencing that pressure difference. I concluded that, since 4.3 in.W.C. can be a typical blowing pressure, these velocities are about the same order.

 

Yet, this comparison may not be that relevant, if we are focused on understanding only tongue vibration. It’s probably more relevant towards understanding of the sound field, but staying with the tongue vibration, the presence of a jet is more important than the velocity of the jet. With its presence, we know that the pressure below the tongue is very close to P2, and that’s all we really care about. The jet forms because of flow separation at the bottom corner of the tongue, and the “entrance flow” into any hole is generally quite small, except only very near the hole. This is because the area feeding this flow is much larger than the area of the hole (imagine over a hole in a wall a hemispherical region with a radius reaching to where air motion – which is radially directed – first starts). But this implies that the jet velocity is greater than the tip velocity. Is it possible to form a jet if the tip is moving downward faster than the maximum air velocity in the jet? I think so, with flow separation occurring only at the top corner of the slot, and only with a lot of turbulence formed within the jet flow by the faster moving tip.

 

But if we look only at the pressure acting on the top surface of the tongue, very near the tip, in quasi-steady flow, if the air flow separates from the tip (Vt < Vj) and causes a jet boundary, there will be airflow parallel to this small upper area and probably a stagnated flow away from this area. Thus, most the top area does not experience much air flow. (There is no jet flow above the tongue.) And we must also acknowledge that the tongue velocity varies nearly linearly along its length, from maximum at the tip to zero near the other end. With some repetition, with quasi-steady flow, I can visualize a generally downward air movement above that impinges on the tongue on its way into the jet, and most of this airflow will stagnate on the top surface, thus regaining the stagnation pressure it started with, P1. (I’m assuming here an idealized perfect tongue/slot fit.) Any air that moves parallel to the top surface will produce a slightly lower pressure than P1 on the top surface (from Bernoulli), but parallel flow. For unsteady flow, I can visualize a true suction, where air above the top surface is pulled by the downward moving tongue, which would cause a rarefaction pressure pulse to combine with the sound field. So again, we have to decide on the flow regime before we can come to a clearer understanding of the effect of air motion on tongue vibration.

 

It’s my guess that, whatever the nature of the flow field, the pressure on the top surface of the tongue is at least approximately the stagnation pressure, P1, for the purpose of determining the pressure forces acting on the tongue and for the established periodic motion. We’d need to be more accurate if we wanted to understand the sound field.

 

With that said, I will gladly supply the equation necessary to calculate the air jet velocity, and for other interested people, also the calculation for the maximum tip velocity. In this latter calculation, I will neglect the fact that, with increasing blowing pressure, the sinusoidal midpoint where this velocity is maximum shifts downward a bit into the slot.

For the air jet velocity, from Bernoulli,

P1 + 0 = P2 + ½*rho*V^2/gc

where gc is a dimensional constant that depends upon the system of units used. In the Lbm/Lbf system English system, this constant is 32.2 Lbm/Lbf*ft/sec^2. Rho is the density of air, or 0.075 Lbm/ft^3. The pressure difference, (P1 – P2) is expressed normally as Lbf/ft^2, but to convert it to in.W.C., we need to incorporate the factors 144 in^2/ft^2 and 0.036 psi/in.W.C. Thus,

V = sqrt(2*32.2*144*.036*delP/0.075)

or

V = 66.7*sqrt(delP),

V is expressed in ft/sec, and delP is expressed in in.W.C.

 

For the maximum tongue tip velocity, we reason that, with sinusoidal motion, the tip goes from zero velocity to its maximum in ¼*T, where T is the period of oscillation = 1/f, where f is the frequency of oscillation. For a sine wave, the (maximum) amplitude is equal to twice the average, and so, given the displacement amplitude, A, and noting that there are 12 inches in a foot, we get,

Vt = 4*f*2*A/12

or

Vt = 0.667*f*A

Vt is expressed in ft/sec and A in inches. A is the height the tongue tip reaches above the slot.

I also have some data on tongue vibration amplitudes and frequency, so I’ll present some of the results here. But now, I have to go out of town for a couple days and will resume this interesting thread when I get back.

 

Best regards,

Tom

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Tom, thanks it nice to have math that I don't have to purée my brain to understand. It really is useful to see different effects as not mutually exclusive, but rather added to the overall wave form. Typically free reed waveforms start out nearly ( but not quite) sinusoidal. As the pressure / amplitude increases, the sides get steeper and the top levels off and begins to develop an asymmetric dip. This is les the case as the reeds go up in pitch, not being noticeable on my concertina reeds above G5. II had only noted this and associated it with the increase in harmonics with increased amplitude. I hadn't considered how reed velocity might influence it. Since good concertina reeds begin to respond at. .1 to.2 inches WC and the dip is beginning to form by 1 inch WC, reeds in a good concertina are already at high amplitude by 2 inches WC. ( proper chambers make a huge difference in amplitude without increasing pressure.). This sounds like at least in some circumstances the reed tip may exceed the general air velocity through the window if not that of any jets. Bears interesting possibilities relative to choosing a longer or shorter reed scale.

Safe Journey

Best Wishes,

Dana

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Hi Ron, your “1.(alternative)” post contains no less than a colossal error! The air in the top and bottom reservoirs, having static pressures P1 and P2, respectively, is at rest. This is the meaning of a “reservoir,” and as I made clear in my first post: “…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.” In other words, we consider the air in the bellows to be at rest. Do you really think the air inside the bellows is moving? If so, with what velocity? In any event, your V1 is zero, and the conclusions you made are incorrect.

 

I hope it helps if I elaborate on a few concepts necessary to understand the physics of how a free reed operates. These concepts are readily applied by a fluid dynamicist, and they must provide a theoretical basis for understanding our little friend.

 

Steady, quasi-steady, and unsteady flow:

Picture an edge view of a vertical plate, where the reservoir on the left is at pressure P1 and the reservoir on the right is at P2, with P1 > P2, and there is a hole (orifice) drilled through this plate. There will be airflow from left to right through this hole. First consider the pressure difference (P1 – P2) constant in time. At any point there is airflow, there will be no change in its pressure and velocity. The only changes are spatial (convective), as air molecules move from left to right through the hole. This is called steady flow.

Now suppose P2 stays constant, but P1 changes, and consider an air molecule just as it enters the left-side of the hole. Suppose in the time that it takes P1 to drop only 1% of its value, the molecule zips through the hole and reaches a distance more than ten hole diameters downstream. As far as this molecule goes, it experiences essentially the same situation as a corresponding molecule in the above case of steady flow. Thus, if the stay time of a fluid particle in the region of interest is much much smaller than the time it takes for the pressure to change, we call the flow quasi-steady. Many times, we conduct quasi-steady analyses when the stay time is only much smaller than the times needed for changes, and still get good results, say, within about 10%.

 

Now suppose that P1 changes 50% in the time it takes for the molecule to reach only to the middle of the hole. This molecule is thus influence by a range of appreciable pressure differences during its flow, and thus, it will experience changes in velocity in time as it flows. Such accelerations are time derivatives, and these cannot be neglected, as they are in quasi-steady flow. This is unsteady flow, and it can result in flow regimes that are entirely different than those experienced in steady, or quasi-steady, flow.

 

Let’s again consider P1 > P2, with constant values, as in steady flow, or with changing values having a characteristic time (period) that is much larger than the stay time, as in quasi-steady flow. Airflow will be from left to right through the hole. The air molecules on the left will be essentially motionless, except those very near the hole. Imagine yourself exiting a very crowded room through a small door. You move very slowly, until you are very near the door, and once within the doorway, your motion greatly increases. Similarly, when the approach of the air molecules gets to a distance approximately one or two hole diameters away from the hole, they experience significant convective acceleration (not temporal acceleration), increasing their velocity up to a maximum as they are ejected from the hole.

 

Air emanating from the hole forms a jet. This is because the airflow separates from the downstream edge of the hole. Separation occurs because it would require (theoretically, with “potential flow,” which is frictionless flow) an infinite velocity and zero absolute pressure for the flow to turn around the sharp corner. But since air has viscosity and it sticks to surfaces, the actual air velocity goes to zero at the hole/wall surfaces. The air cannot have enough momentum to round the turn, so it separates, forming a jet, shooting into P2. The sides of this jet are called a “free surface.”

 

The jet persists for many jet diameters into the right reservoir, and the static pressure within the jet must equal the pressure immediately outside the jet, or P2. Thus, the exit pressure the airflow sees is P2. The reason for this is as follows. Assume the jet pressure to be less than P2, as soon as the jet forms. The jet diameter would then contract as it moved to the right, into a region of greater pressure, which compresses the free surface boundary. This contraction would cause an increase of jet velocity, because in satisfying Conservation of Mass (the Continuity Equation), the same airflow passing through a region of smaller cross section must speed up (as did the air molecules to the left of the plate). With an increase of velocity, the pressure in the jet must decrease, from Bernoulli. Since the jet pressure was already assumed to be less than P2, it’s thus not possible for the jet to adjust to P2. Yet P2 is the boundary condition of the problem; the flow must end up with this static pressure. Thus, the jet pressure just to the right of the hole cannot be less than P2. You can apply the same reasoning when assuming that the jet pressure just to the right of the hole is greater than P2, causing the jet diameter free surface to increase outward, causing smaller velocities and higher jet pressures, and you arrive at a similar contradiction. Thus, the jet “sees’ the static pressure P2 upon exiting the hole, and the exact spot where the pressure is P2, the jet velocity is maximum, since the pressure difference driving the orifice flow is the maximum, (P1 – P2). Incidentally, this reasoning applies only to subsonic flow. With supersonic flow, different mechanisms play out in adjusting the jet pressure to P2, and in particular, normal and oblique shock waves form in order to first bring the flow subsonic.

 

With steady flow, or more generally, with quasi-steady flow, there must be a jet, and in fact, if we know that jets form in a particular flow situation, we can assume steady or quasi-steady flow. For oscillating flow going through zero, it’s true that, in the moments near zero velocity, the stay time greatly increases. But if the oscillation amplitude is large enough, the fraction of the oscillation period in which there is quasi-steady flow becomes close to unity, resulting in fairly accurate results.

 

With unsteady flow, the flow regime looks much different. There are probably no jets, or there are only momentary jets, with momentary flow separation, eddies and vortices. In this case, with P1 increasing, imagine an air molecule first in the center of the hole, and by the time P1 doubles, the molecule just exits the hole on the right. The rapidly increasing P1 causes time acceleration of this molecule, causing it and its neighbors to push with increasing pressure on the air just downstream, which resists, because of its inertia. This crowding of the air causes increasing pressures that would move any free surface radially away from the hole axis. Thus, a steady free surface is not likely to form. The airflow thus hugs against the right side of the vertical wall, and the discharge flow near the hole looks something like a mirror image of the approaching radial flow on the left of the hole, except with flow arrows reversed in direction.

 

The frictionless, steady, incompressible Bernoulli equation is written for two points 1 and 2 within the same streamline of flow:

P1 + ½*rho*V1^2 = P2 + ½*rho*V2^2

where the “p’s” are static pressure and the “v’s” are corresponding flow velocities. Ron, you are correct in stating that this equation is the statement of conservation of potential and kinetic energy for the mechanical system all the flow assumptions require. But you seem to forget that it must be applied to flow in the same streamline. When you write this equation relating the jet velocity only to P2 and not to P1, you violate this condition, because you do not acknowledge that the airflow originates where the total pressure is P1.

 

Writing this equation for a streamline that exits the vibrating reed, with the presence of a steady free surface (jet):

P1 + 0 = P2 + ½*rho*Vj^2

because, as I proved, the static pressure in the jet must equal P2. Writing Bernoulli for a streamline that originates where the pressure is P1 and that impinges on the top surface of the tongue, where the flow stagnates (reversibly diffuses to zero velocity):

P1 + 0 = p1 + 0

or,

p1 = P1

The pressure on the top surface where the flow is stagnated is equal to P1. This fact illustrates the case for a frictionless system, where pressure measures potential energy and velocity measures kinetic energy. The system is conserved. Potential energy (P1) is used up to generate flow (kinetic energy), and when you reversibly recover that kinetic energy back into potential energy, you wind up with the same potential energy you started with (P1). Your statement that p1 > P1 thus violates the law of Conservation of Energy. Now writing it for the same streamline as above, but at an intermediate (i) position upstream of the tongue, where the velocity has reached v1:

P1 + 0 = pi + ½*rho*v1^2

Thus, pi is the minimum static pressure in the streamline, where there’s a combination of potential and kinetic energy. I don’t know how to state this more clearly, without repetition.

 

Best regards,

Tom

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We are apparently talking about different regions, because two of the regions i am talking about are very near the tongue, above and below, and are certainly not reservoirs. Reservoirs are useful for one point of each of the two hypothetical streamlines (one above, one below) to which i applied Bernoulli's equation. One streamline follows air as it moves from the upper reservoir toward the tongue, colliding with it before drifting onward to pass through the gap in the end. The other follows a streamline just below the tongue, from the fixed end, where the air is essentially static, toward the free end, where it joins the general flow of air that has passed through the gap.

 

But we can dispense with Bernoulli's equation altogether to show that the pressure at the top of the tongue is greater than P1, when the air is moving, as it is just before the tongue has begun to respond significantly. Anyone who has stuck their hand outside the window of a moving car can feel the extra force (beyond static air pressure) that the air exerts as it collides with the hand.

 

In the present context, when the valve is opened, air that flows into the chamber will relatively quickly cause the static pressure in the chamber to rise to the level of that in the bellows (P1 reservoir). But the air flow doesn't just stop, and a significant amount of that moving air will strike the tongue (i.e., all of it doesn't just make a beeline for the gap). Whatever strikes the top of the tongue exerts an extra force (beyond P1*A) on the tongue as it collides with it, and as you know, extra force means extra pressure.

Edited by rlgph
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This cause and effect are well understood here. The inefficiency of the reed with wider clearances brings its own problems ie. the need for more force to create the same volume and slower response. Retaining the close clearances and having a more mellow sound is where the game is.

I am not suggesting a reed with wider clearances in general, only at the top edges. The width over most of the slot depth is the same.

I was mystified by your comment until I have looked back through your posting history; I now realise you might never have come across a real concertina reed.

 

Briefly, there are three common types of reed assembly used in concertinas, traditional reeds, as used in almost every English built concertina, accordion reeds used in hybrid concertinas, and what I term semi-hybrid reeds which have the geometry of accordion reeds but the outward appearance of traditional reeds.

 

Concertina reeds are only tight at the top, the reed window is relieved underneath, an important part of producing the concertina sound. Accordion reeds have relatively vertical walls in the frame and it is easy to imagine the clearance will be small throughout the downward swing but the reed itself is usually tapered which means as it travels down through the frame window it effectively shortens and increases the clearance. Filing the top edge of either type will decrease the efficiency of the reed.

 

 

I too wondered if rlgpf was looking at the physics of accordion reeds without realising that the construction of concertina reed frames is very different!

 

We are apparently talking about different regions, because two of the regions i am talking about are very near the tongue, above and below, and are certainly not reservoirs. Reservoirs are useful for one point of each of the two hypothetical streamlines (one above, one below) to which i applied Bernoulli's equation. One streamline follows air as it moves from the upper reservoir toward the tongue, colliding with it before drifting onward to pass through the gap in the end. The other follows a streamline just below the tongue, from the fixed end, where the air is essentially static, toward the free end, where it joins the general flow of air that has passed through the gap.

 

But we can dispense with Bernoulli's equation altogether to show that the pressure at the top of the tongue is greater than P1, when the air is moving, as it is just before the tongue has begun to respond significantly. Anyone who has stuck their hand outside the window of a moving car can feel the extra force (beyond static air pressure) that the air exerts as it collides with the hand.

 

In the present context, when the valve is opened, air that flows into the chamber will relatively quickly cause the static pressure in the chamber to rise to the level of that in the bellows (P1 reservoir). But the air flow doesn't just stop, and a significant amount of that moving air will strike the tongue (i.e., all of it doesn't just make a beeline for the gap). Whatever strikes the top of the tongue exerts an extra force (beyond P1*A) on the tongue as it collides with it, and as you know, extra force means extra pressure.

 

On a theoretical physics level I cannot dispute anything you say because I'm pathetically unqualified to do so but I would ask you this:

 

Do you understand the fundamental difference between traditionally made concertina reeds/frames and accordion reeds/frames? You should be able to describe this in ten words or less because I can and I'm stupid. ;)

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Upon further reflection, i realize that my description of the dynamic pressure on the top of the tongue does not apply once steady state flow is reached, when p1 = P1, as Tom says. So the pressure on the top surface of the tongue starts at P2, when the valve is just opened, rises to a maximum of approximately 2*P1 - P2, and then settles to P1 as steady state is approached.

 

The equation for the streamline just below the reed tongue, however, is true at steady state. With the assumption of negligible air speed just under the fixed end, the pressure along the bottom surface of the tongue is approximately p2 = P2 - 1/2 rho*v2^2, where v2 is the air speed (toward the free end) at some arbitrary point very near the bottom of the tongue between the fixed end and the free end. v2 approaches the speed of the air through the gap as the chosen point gets closer to the free end.

 

[incorrect statement deleted.]

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Do you understand the fundamental difference between traditionally made concertina reeds/frames and accordion reeds/frames?

 

Nope. But i ceased trying to describe the operation of the reed quite a ways back. Lately, I've been considering the pressure difference between top and bottom of the reed tongue initially -- before the tongue has begun to move with any significant speed.

 

Besides, since this thread started out as a design proposal, i can make the frame and tongue to have the shape that i want. ;-)

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[This post is largely made irrelevant by the next one.]

 

Tom,

 

As i said, i'm now satisfied with my understanding of the temporal variation of pressure on the top surface of the tongue during the initial stages, and understand how we disagreed so fundamentally on the application of conservation of energy. Although perhaps it's important that that pressure gets significantly greater than P1 prior to settling into steady state flow, i'm thinking now that what happens on the underside is more important.

 

Are you REALLY saying that since there are two reservoirs, one above the tongue and one below, that the pressures on the top and bottom surfaces necessarily match the reservoir pressures? I agree that that becomes true on the top, where a quasistatic situation can be achiieved with steady state, but disagree that it happens on the bottom. The importance of local dynamic pressures is critical to understanding the motions of surfaces, even in steady state -- e.g., the differing pressures on the top and bottom surfaces of an airplane wing, despite the presence of a reservoir all around the plane.

 

When you write down what you call "the" Bernoulli equation, you are aware, are you not, that for any given streamline, there are an infinite number of Bernoulli equations -- one for each pair of points along that streamline?

 

I ask these questions, not to be insulting, but to try to ascertain our points of disagreements.

 

ron

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Okay Tom,

 

I've re-read your description of the jet from the P1 side to the P2 side, and now understand the significant disagreement between us -- what happens to the air (near the hole) surrounding the jet on the P2 side. First, let me remind you (and any others patient enough to read all this verbage we're producing), the average thermal velocities of air molecules are around 500 m/s, faster than the speed of sound (and the flow speed of the jet). And because of large size of the hole, there are significant velocity components of air molecules within the jet, perpendicular to the direction of the jet. This, together with the random directions of thermal velocities of molecules on the P2 side, means that the boundaries of the jet are not at all sharp. So collisions between air molecules that move through the hole into the P2 side with those on the P2 side immediately cause some air on the P2 side near the jet to be pushed along with the jet (though not necessarily at the same speed). Air on the P2 side near the hole and near the boundary between the regions will move into the space vacated by that being pushed in the direction of the jet. At least two relevant results of this are

 

(1) You can't unambiguously say what the pressure in the jet is because its boundary is too fuzzy and there is a gradient of flow velocities in the jet boundary area.

 

(2) A flow of air is set up on the P2 side, along the barrier, toward the hole. Far from the hole the speed of this flow is approximately zero; as the air moves closer to the hole its speed increases; as it gets very close to the hole, collisitions with air coming through the hole cause the momentum increase and change in direction as the air moves along with the jet. Apply Bernoulli's equation to this boundary air flow and you find p2 = P2 - 1/2 rho*v2^2; i.e. the dynamic pressure on the P2 side near the boundary is less than P2.

 

The air flow in (2) is analogous to the air flow that i have described moving below the tongue from its fixed end toward its free end.

 

In steady state the pressure on the bottom tongue surface thus varies between P2 at the fixed end to P2 - (P1 - P2) at the free end, resulting in top to bottom pressure difference of P1 - P2 at the fixed end to 2*(P1 - P2) at the free end.

 

ron

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  • 2 weeks later...

Rlgph, in a reply to Lukasz, you mention that some "same condition" will apply once a reed is in motion. It wasn't clear to me exactly what you were referring to. But just as a side note, when a reed is set too high to start, at least up to a point, setting it in motion will allow it to continue as long as the ordinary difference in air pressure is maintained. If a reed is set too low to start, the same does not apply and the mechanically applied starting oscillation will decay almost immediately as it is damped out. If this wasn't the implication of your post, ignore this.

Dana

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Rlgph, in a reply to Lukasz, you mention that some "same condition" will apply once a reed is in motion. It wasn't clear to me exactly what you were referring to. But just as a side note, when a reed is set too high to start, at least up to a point, setting it in motion will allow it to continue as long as the ordinary difference in air pressure is maintained. If a reed is set too low to start, the same does not apply and the mechanically applied starting oscillation will decay almost immediately as it is damped out. If this wasn't the implication of your post, ignore this.

Dana

I don't know what you're referring to, but i would accept what i said about the operation of a reed only in a qualitative sense. What i have said about the initial pressure conditions is on much firmer ground. OTOH, i personally have a difficulty accepting anyone's description of the operation if they don't get the initial conditions right.

 

PS What you say about starting reeds that are incorrectly set desn't surprise me, but there's probably nothing to be gained from my saying why -- it would only be intuitive speculation.

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Just to confuse this thread a bit more, my latest accidental finding (rather unsuprising, but still interesting). I was testing my new reedpan yesterday and have tried a couple of accordion reeds on it. They are still unvalved, so when I have applied pressure, BOTH tongues of the accordion reed have spoken simultanously. In such conditions both tongues have smaller amplitude, the combined pitch is much more pressure dependant (note bends very easily, it is quite difficult to achieve a steady note) and the "reverse direction tongue" oscilates with small amplitude only (so it does not sink into the shoe), but it vibrates nevertheless. This "double reed" setting is much more sensitive to overall pressure and behaves very nice around a bellows reverse point giving almost uninterrupted sound [it starts to speak with very low pressures and cannot be chocked with sudden, high pressure push on the bellows. When one of the tongues was valved (with a masking tape) the same reed was very prone to choking with sudden bellows movement]

 

This finding got me wondering if such effect could be usefull for some kind of "another type", new concertina… Probably somehow electric in nature as this setup has significantly lower volume (and is very air inefficient) and would require some sort of a "second step amplification", but almost continuous sound and note bending capabilities could prove musically usefull...

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OTOH, i personally have a difficulty accepting anyone's description of the operation if they don't get the initial conditions right.

 

 

This was originally linked by Tom in a "MIDI latency" thread, but I think it should be linked here as well. http://www.public.co...g/PMA035061.pdf There is quite solid proof in this paper, that initial conditions in transient stage are very different from those in a steady oscilation phase and therefore a steady oscilation phase can have a completely separate description (which is in fact quite well researched and used in practice of reed/instrument construction). The missing "initial movement" problem does not in any way invalidate all other findings on the free reed behaviour that have been published and I personally don't have any objections to Tom's description of physics after the first few cycles (more precisely after Dana's "abrupt amplification" phase).

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I personally don't have any objections to Tom's description of physics after the first few cycles (more precisely after Dana's "abrupt amplification" phase).

I thought you had returned to your opinion that the pressure difference between the two sides of the tongue is much greater than P1 - P2, at least initially.

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That's why I said "after first few cycles". From what I understood, Tom has explicitly stated, that he has no idea (at least he don't have a definitive description) on how the tongue start it's movement, and in my post above I have only pointed out, that initial movement and steady oscilation phase are different enough to be treated as separate problems, and that the latter is quite well researched and that Tom's description and math can be used in practical aplications of reed and instrument design.

My opinion on the very first cycles is that even slightest (both spatial and temporal) effects matter and that they accumulate over first few cycles, so no "single first cycle" description of the reed start can ever be complete nor true. And yes, I think that some of those effects cause the momentary effective pressure difference acting on the tongue to be greater than P1-P2 (or at least that the whole "macro scale" P1-P2 approach is somehow fundamentaly inadequate in those very first cycles, as it results only in a "pressure valve", single airflow cut predictions), but at this point I have no more ideas on what they might be. I also think, that the most important observation about first cycles is that the entire oscilation starts with amplitudes smaller than the gap width and that the first few cycles never even reach the shoe, so there is no complete airflow cut-off during this first stage. Then comes Dana's "abrupt amplification" phase (this is IMHO the point in which first tongue sinking into the shoe occurs) and then we have the steady oscilation phase, in which Tom's math works as described by him. (to be more precise, I think that Tom's approach is also at least partially true in the "abrupt amplification" phase, as this is the transitional phase in which both "initial" and "steady" parts of the description should overlap).

And I also think, that without high-speed, smoke flow footage of the first cycles we cannot give any true and definitive answers to how the reed starts. Unfortunately I don't have such capabilities.

[sidenote: I was wondering recently, if Schlieren photography could be of any use in resolving the first cycles movement question… https://www.youtube.com/watch?v=lSFwH0BVd3Q]

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