Some musings on measuring bellows activity in electronic Anglo concertinas

[ttonon]

3 hours ago, Richard Mellish said:

It is not divided between the reeds, though the resulting air flow is the sum of the flows past all the reeds.

Richard, thanks for asking the same question I asked, though in more detail. When Weary didn't answer after I asked twice, I gave up. 

 

Weary, in the conventional concertina, the lower frequency notes consume much more airflow than do the higher frequency notes. I'm not sure you understand the basic physics here. If you can devise the right questions you need to gain that understanding, please ask me, and I will gladly explain. 

 

Regards,

Tom

[rgammage]

Deleted

Edited May 16 by rgammage

[wearyhacker]

3 hours ago, ttonon said:

Weary, in the conventional concertina, the lower frequency notes consume much more airflow than do the higher frequency notes. I'm not sure you understand the basic physics here. If you can devise the right questions you need to gain that understanding, please ask me, and I will gladly explain.

 

I think we are talking at cross purposes here. I am talking about the abstract mathematical model of a low cost electronic concertina. This concertina in the real does not have reeds in it, does not have valves (pads)  in it and does not have any midi interfaces. It is in electronic instrument terms a "rompler". In other words it uses soundfounts that contain looped samples of real concertina reeds to play notes. It has buttons and leaky bellows to go some way towards emulation of the feel of a real concertina.

 

Modelling the pnuematic impedance of a pad and reed is way out if scope.

 

If you can provide me with such a pad/reed model I would be very interested  in looking at it.

[ttonon]

Weary, do you use two identities in posting to this group, rgammage and wearyhacker?

[rgammage]

2 minutes ago, ttonon said:

Weary, do you use two identities in posting to this group, rgammage and wearyhacker?

Not on purpose. I have helped rgammage recover his account, and accidently posted a reply whilst logged into his account.

 

I have posted again under my own login and I am trying to delete the first post.

 

I cannot find a way to completely remove the post. Can you help there?

 

[ttonon]

1 minute ago, rgammage said:

Can you help there?

I see no obvious way.

 

Tom

[RatFace]

One problem with a single load cell is that it's only going to work with a fairly fixed "bellows" assembly, and that won't feel very real. It won't allow for "bending" as well as any real in/out movement.

 

The advantage of a pressure sensor is that it should allow for all the normal movement... but perhaps air pressure sensors aren't great (I need to build some bellows to test mine out!), especially as the pressure changes reported here are quite low.

 

How about using three load cells in series with a bellows - arranged like this (this only shows two, because it's a 2D concertina - just imagine three!):

 

The bellows would anchor directly to one end, and to a plate at the other. The plate is connected to the other end by three load cells. By averaging the three force measurements, that will be a good proxy for the air pressure. If the bellows are bent, or hit against the knee (tango style!) it should still give you the pressure, even if the forces are wildly different.

 

The standard HX711 strain gauge amplifies will run at 80Hz - that's been enough for me, though I don't know if it would have too much latency for an Anglo. It might be possible to run them with an external clock at a higher rate? The force sensitivity itself is really good.

 

There's also the question of what to do about the air - I'm not sure how constantly leaky bellows would feel. It should be possible to have some servo actuated valves at one end that would be connected to the microprocessor (plus maybe a conventional air button).

 

[Don Taylor]

Has anybody measured the differential air pressure  inside the bellows at full volume?

 

There are air pressure sensors available but knowing the range of pressures to be measured would be useful.

[RatFace]

Pressure is force divided by area.

Force is about 15N (1.5 kg max)

Area of end is 0.015m^2 or so

 

So I estimate pressure change is about 1000Pa

 

Atmospheric pressure is 10^5 Pa so this is a variation of 1% or so 

[ttonon]

4 hours ago, RatFace said:

but perhaps air pressure sensors aren't great (I need to build some bellows to test mine out!), especially as the pressure changes reported here are quite low.

I don't know why you would have such suspicions. As an engineer who worked with many different pressure transducers, I know there are many available with much more accuracy than you'd need with response times in the ms. For instance, here's one that's less than $100:

https://www.omega.com/en-us/pressure-measurement/pressure-transducers/series-616kd/p/616KD-53

This is a differential transducer with a range -10 to +10 inches of water, ideally suited for typical concertina operation. 

 

Pressure transducers have been available for many decades and there are certainly some that can handle this application. 

 

Regards,

Tom

[wearyhacker]

5 hours ago, RatFace said:

How about using three load cells in series with a bellows - arranged like this (this only shows two, because it's a 2D concertina - just imagine three!):

 

The bellows would anchor directly to one end, and to a plate at the other. The plate is connected to the other end by three load cells. By averaging the three force measurements, that will be a good proxy for the air pressure. If the bellows are bent, or hit against the knee (tango style!) it should still give you the pressure, even if the forces are wildly different.

 

There's also the question of what to do about the air - I'm not sure how constantly leaky bellows would feel. It should be possible to have some servo actuated valves at one end that would be connected to the microprocessor (plus maybe a conventional air button).

 

 That is a really interesting idea. I think the load cells are actually in parallel, the more the merrier. Maybe the bellows could be made airtight and the servo could incrementally open the valve depending on the number of buttons that are currently pressed. Hmm.

[RatFace]

2 minutes ago, wearyhacker said:

 That is a really interesting idea. I think the load cells are actually in parallel, the more the merrier. Maybe the bellows could be made airtight and the servo could incrementally open the valve depending on the number of buttons that are currently pressed. Hmm.

Yes, load cells in parallel with each other, but in series with the bellows!

 

[Chris Ghent]

On 5/13/2024 at 4:38 AM, ttonon said:

Hi Chris, good to hear from you. 

 

In my experience, a typical bellows differential pressure for free reeds is from about something less than an inch of water to up around six inches or a little larger for loud players. An inch of water is equal to 0.249 kilopascal. 

 

Since pressure difference is the driver, that's approximately true for all free reed instruments. The arm force on the bellows required for that depends on the bellows cross sectional area. Since concertinas have a relatively small cross-sectional area, less force is required than for full size accordions to produce the same volume. 

Tom,

 

I don’t understand the meaning of pressure “differential”.
 

Some time ago I spent time gauging the starting vacuum of reeds as a way to producing more efficient reeds.  In the tuning rig (which has a  variable voltage fan and two magnehelic gauges, one a 0 to .5”) I found reeds needed to start at less than .1” WC to be efficient.  The full power of the rig peaks at 1.6”WC which I use as a tuning -pressure.  I no longer use this starting assessment method because I know what I need to do to achieve that result and see no need to check it every time.  
 

Consequently when the operating pressure in the bellows (I drilled a hole in the bellows frame and sealed a tube into it) turned out to be less than .1 I was a little surprised but accepted it.  It was easy to create -pressure of many inches in the bellows but as soon as I pressed a button it dropped away to almost nothing.  
 

I’m not saying you are wrong, more that our experiences must be complimentary in some way and I don’t know what that way is.  
 

Cheers

 

Chris

[Chris Ghent]

On 5/13/2024 at 4:38 AM, ttonon said:

Hi Chris, good to hear from you. 

 

In my experience, a typical bellows differential pressure for free reeds is from about something less than an inch of water to up around six inches or a little larger for loud players. An inch of water is equal to 0.249 kilopascal. 

 

Since pressure difference is the driver, that's approximately true for all free reed instruments. The arm force on the bellows required for that depends on the bellows cross sectional area. Since concertinas have a relatively small cross-sectional area, less force is required than for full size accordions to produce the same volume. 

Tom,

 

I don’t understand the meaning of pressure “differential”.
 

Some time ago I spent time gauging the starting vacuum of reeds as a way to producing more efficient reeds.  In the tuning rig (which has a  variable voltage fan and two magnehelic gauges, one a 0 to .5”) I found reeds needed to start at less than .1” WC to be efficient.  The full power of the rig peaks at 1.6”WC which I use as a tuning -pressure.  I no longer use this starting assessment method because I know what I need to do to achieve that result and see no need to check it every time.  
 

Consequently when the operating pressure in the bellows (I drilled a hole in the bellows frame and sealed a tube into it) turned out to be less than .1 I was a little surprised but accepted it.  It was easy to create -pressure of many inches in the bellows but as soon as I pressed a button it dropped away to almost nothing.  
 

I’m not saying you are wrong, more that our experiences must be complimentary in some way and I don’t know what that way is.  Happy to be wrong myself. 
 

Cheers

 

Chris

[ttonon]

49 minutes ago, Chris Ghent said:

I don’t understand the meaning of pressure “differential”.

Hi Chris, I think you'd agree that for a reed in an instrument, pressure differential refers to the difference between static pressure inside the bellows where there's little air motion, and that in the outside air, where there's little air motion. In a test rig, it's basically that static pressure difference upstream and downstream of the reed. 

 

I'm not sure our findings are contradictory and there may be some differences because I work with the bellows sizes in accordions, and you work with concertinas. (Is your Magnetohelic gauge a Dwyer?) I don't doubt your starting pressures at a couple of tenths of an inch WC and tuning at about 2 inches or above seems okay, though from my experience, your tuning pressure of 1.6-inch WC may be on the low side. That corresponds to a force on the concertina bellows of only 2 pounds (see below). But if concertina players usually play at bellows pressures less than full-size accordion players, it may be representative. 

 

The pitch of a free reed is a function of driving pressure, but since most people cannot perceive a change of several cents in actual playing, all the refinements here may not mean much. 

 

The statement in my post above refers to the maximum pressures encountered during playing. I'm sure different players use different pressure maximums, but in order to utilize the full dynamic range of a free reed, there can't be too much difference. That assumes that concertina reeds and accordion reeds have similar dynamic properties. Also, my consideration of what is maximum is for when I do calculations on my theoretical fluid dynamical model of free reed vibration. So it may be a bit too large compared to the vast majority of actual performance. I've been away from that project for a couple of years, and when I get back to it, I'll review all this. 

 

The typical full-size accordion has internal bellows dimensions of about 6.5 x 17.5 inches^2 and let me guess that for a concertina they are more like 6 x 6, for a cross-sectional area ratio of about three. For a 6-inch WC bellows pressure, the accordion player exerts about 25 pounds force on the bellows, and this is certainly a maximum, but I don't doubt that some beefy players do it momentarily. And that pressure is easier to achieve with the smaller melodians. 

 

The area difference means the concertina player needs exert about 1/3 the force or only about 8 pounds for the 6 inch WC. 

 

 

1 hour ago, Chris Ghent said:

It was easy to create -pressure of many inches in the bellows but as soon as I pressed a button it dropped away to almost nothing.  

 

I don't understand. Were you powering the bellows with your arms, or were you supplying the air with a fan? If the latter, I'd guess that you need a stronger fan. If the former, I'm surprised that you couldn't easily maintain a constant pressure when pushing the button.  

 

Best regards,

Tom

 

 

 

 

 

 

 

[ttonon]

Chris,

 

After all that verbiage, I decided to measure the force I use to play my full size accordion but found that the batteries to my small digital scale are dead.

 

Maybe a reader here could do the measurement for a concertina. It shouldn't be difficult to place a small digital scale between one's palm and push to sound a note to record the force. Press a button the opposite side from the scale, so that the button force doesn't complicate the measurement. Take a number of measurements, with the note barely sounding, to "normal" volume, to maximum volume/force expected in a performance.

 

When I get new batteries, I'll do the same with the accordion.

[RAc]

Apologies if this should be a stupid remark (it probably is), but shouldn't it also be possible to measure the *time* it takes for some mechanically determineable point in a fake bellows set to another such point - the idea being that the stronger the force exercised, the faster the travel, so the force can be deducted from the time?

 

After all, if there is one thing that even less powerful computers are doing fairly well, then it is dealing with time in at least the microsecond range (which should be sufficient for this use case)...

[ttonon]

15 minutes ago, RAc said:

the idea being that the stronger the force exercised, the faster the travel, so the force can be deducted from the time?

What "travel" do you mean? It would help if you explain what is travelling and what are the points of the two measurements.

 

Regards,

Tom