Speaker compliance

[shatteredsquare]

When we push it in instead there's no limit for the internal pressure (until we create a black hole at least) and the delta can be much larger than 1 atm.

Is this cabinet mathematically sealed or real-world sealed? If it's sealed sealed in science land it will pull a vacuum traveling outward with the same resistance force as is created in a pressure increase when traveling inward...

 

[DLC86]

Is this cabinet mathematically sealed or real-world sealed? If it's sealed sealed in science land it will pull a vacuum traveling outward with the same resistance force as is created in a pressure increase when traveling inward...

Nope, I explained why above

 

[shatteredsquare]

Yeah but I didn't see anything about pulling a vacuum on the out-stroke, just black hole density on the in-stroke. The force of air compression on the in-stroke would be equal to the vacuum pulled on the outstroke, the cab would add resistance both ways

 

[DLC86]

I'll give you an example that's pretty easy to picture.

Imagine you have a syringe which contains exactly 1mL of air (at normal pressure) and its tip is perfectly sealed.
Assume you want to pull the barrel out to reach the 2mL mark, that's easily possible with a bit of force and you'll basically halve the pressure inside since the volume doubles.

Now let's assume you want to push it in by the same "distance", that would mean you should make it reach the 0mL mark and if the air has no way to escape that means infinite pressure (or black hole) and infinite force needed.

By applying the same force you used when you were pulling the barrel out, it would only reach the 0.5mL mark (half volume, double pressure), so it'll basically move half way by applying the same force.

EDIT: errata corrige.
Since when pulling it out the difference between external and internal pressure is 1-0.5= 0.5 atm, when you apply the same force to push in you can just reach the point where the internal pressure is 1.5atm and the difference is 0.5 atm again, that won't be at 0.5mL but at the 0.66mL mark.

 

[200man]

Thanks for this thread...I haven’t played with this parameter yet!

 

[shatteredsquare]

That makes sense in that scenario to 0 mL, but the sealed cab scenario is like 5 mL of air in the syringe, and you're looking for the force on the plunger pulling from 5 mL to 6 mL, and pushing from 5 mL to 4 mL, it's not going to cap out on pressure in the cab, the speaker isn't going to move but a half inch both ways, max excursion.

 

[yeky83]

Well, that question came to my mind a few days ago actually, and I asked it to a dear friend of mine who has a PhD in theoretical physics (we call him Sheldon ).
He basically told me that when we pull the speaker out the delta between external and internal pressure can be 1 atm at most (assuming the speaker had no physical displacement limits, that means vacuum inside and normal atmospheric pressure outside).
When we push it in instead there's no limit for the internal pressure (until we create a black hole at least) and the delta can be much larger than 1 atm.
This also means that the force needed to push in the speaker by, let's say, 1 inch is more than the force needed to pull it out 1 inch. Pressure basically changes with a different trend so the delta will be larger in the first case.

The question though is if this difference of pressure delta is appreciable for typical speaker displacement range and cab internal volume, we didn't have the chance to do the math and discuss it in detail.

Maybe @FractalAudio can think about this, even though he probably already did it long time ago

I think @Rex already alluded to an answer here:

Things get interesting at the extremes of the physical movement of the speaker cone, where the speaker runs out of room to move, and compliance takes a nonlinear nosedive.

If you really want to get into it, this is a great paper:
https://www.klippel.de/fileadmin/_migrated/content_uploads/Loudspeaker_Nonlinearities–Causes_Parameters_Symptoms_01.pdf

 

[Rex]

Well, that question came to my mind a few days ago actually, and I asked it to a dear friend of mine who has a PhD in theoretical physics (we call him Sheldon ).
He basically told me that when we pull the speaker out the delta between external and internal pressure can be 1 atm at most (assuming the speaker had no physical displacement limits, that means vacuum inside and normal atmospheric pressure outside).
When we push it in instead there's no limit for the internal pressure (until we create a black hole at least) and the delta can be much larger than 1 atm.
This also means that the force needed to push in the speaker by, let's say, 1 inch is more than the force needed to pull it out 1 inch. Pressure basically changes with a different trend so the delta will be larger in the first case.

The question though is if this difference of pressure delta is appreciable for typical speaker displacement range and cab internal volume, we didn't have the chance to do the math and discuss it in detail.

Maybe @FractalAudio can think about this, even though he probably already did it long time ago

"Sheldon" has a solid grasp of the theory, but ask him to consider the practical side. Look at the surface of a 12" speaker cone and the fraction of an inch that it can move back and forth. Compare that tiny sliver of air to the volume of the cab. Heck, have him run the numbers. You'll see that it's effectively linear.

 

[shatteredsquare]

I think @Rex already alluded to an answer here:

​

If you really want to get into it, this is a great paper:
https://www.klippel.de/fileadmin/_migrated/content_uploads/Loudspeaker_Nonlinearities–Causes_Parameters_Symptoms_01.pdf

Why in the ever loving f u q do they not explain when you're in school that for every adolescent enthusiastic interest one could have, there is an equivalent scientific niche rabbit hole that one could spend 3 lifetimes studying? I would have paid way more attention given some context.

 

[DLC86]

"Sheldon" has a solid grasp of the theory, but all him to consider the practical side. Look at the surface of a 12" speaker cone and the fraction of an inch that it can move back and forth. Compare that tiny sliver of air to the volume of the cab. Heck, have him run the numbers. You'll see that it's effectively linear.

This is pretty basic stuff for him and probably finds it somewhat boring

Anyway.. Yeah, that's why I wrote this:

The question though is if this difference of pressure delta is appreciable for typical speaker displacement range and cab internal volume, we didn't have the chance to do the math and discuss it in detail.

We didn't do the math but it could be possible that with small cabs and high power the effect could be noticeable to some extent.
I'll investigate more

That makes sense in that scenario to 0 mL, but the sealed cab scenario is like 5 mL of air in the syringe, and you're looking for the force on the plunger pulling from 5 mL to 6 mL, and pushing from 5 mL to 4 mL, it's not going to cap out on pressure in the cab, the speaker isn't going to move but a half inch both ways, max excursion.

Obviously the effect wouldn't be as extreme as in the example, but it could still be noticeable in some circumstances .
As I said, I'll do the math next time I have the opportunity.

 

[Rex]

Why in the ever loving f u q do they not explain when you're in school that for every adolescent enthusiastic interest one could have, there is an equivalent scientific niche rabbit hole that one could spend 3 lifetimes studying? I would have paid way more attention given some context.

Good teachers spend their entire careers trying to explain that. Experienced teachers know that only a handful of their students will understand the message.

 

[DLC86]

So... today I've read a bit about the matter and tried to calculate the effect of internal pressure on the movement of the speaker.

I wasn't able to find all the necessary data about a particular guitar speaker to make a precise real-world example, these often have pretty poor datasheets, so I looked at a bunch of 12" PA mid-woofers and took an average of those values hoping a typical guitar speaker is not too far off from those.

If we take a medium sized 1x12" enclosure we can consider an internal volume of 30 liters a realistic value (taking into account the volume occupied by the speaker itself).

To calculate how the internal volume varies as the speaker moves we have to know the SD (aka diaphragm area, which seems to be tipically around 500 cm^2 for 12" speakers) and multiply it by the excursion.
I chose the Xmax value (maximum linear displacement) for that to have a real-world value and, in the datasheets I've seen, it ranges between 0.5 and 1.4cm, so I took 1cm as a medium value to simplify the math a bit (I'll call it X1).

(note: we guitar players probably often exceed the Xmax of our speakers since we love "break-up" so much)

So we have:

ΔV1= SD * X1 = 500cm^2 * 1cm = 500 cm^3 = 0.5 liters

So when, for a given voltage applied, the speaker moves forward by 1cm, the internal volume goes from 30 (let's call it Vr, volume at rest) to 30.5 liters (V1). This causes a variation in the internal pressure that can be calculated as follows:

P1 = (Pr * Vr) / V1 = (1 * 30) / 30.5 = ~0.9836 atm

(Pr is the pressure at rest which is the same as external pressure, IOW atmospheric pressure, 1 atm)

The difference between external and internal pressure is:

ΔP1= Pr - P1 = 1 - 0.9836 = 0.0164 atm

Now let's see what happens when the speaker moves backwards by the same 1cm excursion.

V2= Vr - ΔV1 = 30 - 0.5 = 29.5 liters

P2= (Pr * Vr) / V2 = (1 * 30) / 29.5 = 1.0169 atm

ΔP2= Pr - P2 = 1 - 1.0169 = 0.0169 atm

It seems ΔP is a tiny bit larger when the speaker moves backwards (0.0169 vs 0.0164) and this means it would need a bit more power to reach the same excursion it had when moving forward.

To have a clearer picture though, let's try to quantify by how much the cone excursion varies for the same voltage applied.
To do this we start from the first ΔP1 and use the inverse procedure to find out the what's the excursion happening when we have the same ΔP in the opposite direction.

P3= ΔP1 + Pr = 0.0164 + 1 = 1.0164 atm

V3= (Pr * Vr) / P3 = (1 * 30) / 1.0164 = ~29.5159 liters

This time around volume has varied by:

ΔV2= Vr - V3 = 30 - 29.5159 = 0.4841 lts = 484.1 cm^3

This means linear excursion is:

X2= (X1 * ΔV2) / ΔV1 = 1 * 484.1 / 500 = 0.968 cm


To summarise, a speaker in a sealed cab with the specs listed above will produce an asymmetrical sound wave of which the negative half will be smaller by ~3.2%.
This percentage will be higher for smaller enclosures and longer excursions (so it also depends on the frequency since low frequencies produce much more excursions).

Is 3.2% a perceivable difference? I don't know, probably not. But I'll try to investigate more anyway.

It's also more complex than that since we should take into account compliance and other speaker's characteristics to really understand how it can move.
But I think I can say that this internal pressure variation surely has at least some influence on the whole system.

If someone has more knowledge than me about this stuff, please let me know if there's any error in my calculations or if the values I chose are not realistic enough for guitar speakers.


EDIT: corrected a few typos and changed the nomenclature in formulas to make it a bit easier to understand.

 

[Rex]

So... today I've read a bit about the matter and tried to calculate the effect of internal pressure on the movement of the speaker.

I wasn't able to find all the necessary data about a particular guitar speaker to make a precise real-world example, these often have pretty poor datasheets, so I looked at a bunch of 12" PA mid-woofers and took an average of those values hoping a typical guitar speaker is not too far off from those.

If we take a medium sized 1x12" enclosure we can consider an internal volume of 30 liters a realistic value (taking into account the volume occupied by the speaker itself).

To calculate how the internal volume varies as the speaker moves we have to know the SD (aka diaphragm area, which seems to be tipically around 500 cm^2 for 12" speakers) and multiply it by the linear displacement.
I chose the Xmax value (maximum linear displacement) for that and, in the datasheets I've seen, it ranges between 0.5 and 1.3mm, so I took 1cm as a medium value to simplify the math a bit.
(note: we guitar players probably often exceed the Xmax of our speakers since we love "break-up" so much)

So we have:

SD * Xmax = 500cm^2 * 1cm = 500 cm^3 = 0.5 liters

So when, for a given voltage applied, the speaker moves forward by 1cm, the internal volume goes from 30 (let's call it V1) to 30.5 liters (V2). This causes a variation in the internal pressure that can be calculated as follows:

P2 = (P1 * V1)/V2 = (1 * 30)/30,5 = ~0.9836 atm

(P1 is the pressure at rest which is the same as external pressure, IOW atmospheric pressure, 1 atm)

The difference between external and internal pressure is:

ΔP= P1-P2 = 1 - 0.9836 = 0,0164 atm

Now let's see what happens when the speaker moves backwards by the same excursion.

P3= (1 * 30)/29.5 = 1.0169 atm

ΔP= 1 - 1.0169 = 0.0169 atm

It seems ΔP is a tiny bit larger when the speaker moves backwards (0.0169 vs 0.0164) and this means it would need a bit more power to reach the same excursion it had when moving forward.

To have a clearer picture though, let's try to quantify by how much the cone excursion varies for the same voltage applied.
To do this we start from the first ΔP and use the inverse procedure to find out the what's the linear excursion happening with that ΔP.

P= ΔP + P1 = 0.0164 + 1 = 1.0164 atm

V= (1 * 30)/1.0164 = ~29.5159 liters

This time around volume has varied by:

30 - 29.5159 = 0.4841 lts = 484.1 cm^3

This means linear excursion is:

484.1 / 500 = 0.968 cm


To summarise, a speaker in a sealed cab with the specs listed above will produce an asymmetrical sound wave of which the negative half will be smaller by ~3.2%.
This percentage will be higher for smaller enclosures and longer excursions (so it also depends on the frequency since low frequencies produce much more excursions).

Is 3.2% a perceivable difference? I don't know, probably not. But I'll try to investigate more anyway.

It's also more complex than that since we should take into account compliance and other speaker's characteristics to really understand how it can move.
But I think I can say that this internal pressure variation surely has at least some influence on the whole system.

If someone has more knowledge than me about these things, please let me know if there's any error in my calculations or if the values I chose are not realistic enough for guitar speakers.

Have another look at your numbers. In your xMax values, you made a leap from millimeters to centimeters.

 

[shatteredsquare]

what about the vacuum it pulls against on the outstroke though...it has to push against the sealed volume as well as pull against the sealed volume, unless the cab is pressurised with nitrogen or something.

*googles pressurised speaker cabinet patents*

 

[Rex]

So we have:

SD * Xmax = 500cm^2 * 1cm = 500 cm^3 = 0.5 liters

Actually,

SD * Xmax = 500cm^2 * 0.1cm = 500 cm^3 = 0.05 liters

If you think about it, there's no way that a 12" speaker displaces half a gallon of air. That's be pretty violent.

 

[iaresee]

If you think about it, there's no way that a 12" speaker displaces half a gallon of air. That's be pretty violent.

Guys in the room next to ours seem to loosen bowels with their volume, so maybe?

 

[Rex]

Guys in the room next to ours seem to loosen bowels with their volume, so maybe?

lol

 

[JJunkie]

I just drilled some holes into the back of my monitors. Problem solved

 

[DLC86]

Have another look at your numbers. In your xMax values, you made a leap from millimeters to centimeters.

Actually,

SD * Xmax = 500cm^2 * 0.1cm = 500 cm^3 = 0.05 liters

If you think about it, there's no way that a 12" speaker displaces half a gallon of air. That's be pretty violent.

Nope. In the datasheets I've seen xmax is expressed in mm and the values range between 5 and 14. I just converted it to cm cuz I had to multiply it by SD which was in cm^2.

Take a look at this page for example:
http://www.usspeaker.com/12.htm
At least for speakers having the Xmax listed in the specs, the lowest value I see is 4.8mm, the highest is 14mm. (there's also one speaker for which they say literally "Xmax of 8.0mm & 52mm Peak To Peak!")

Furthermore yesterday at rehearsal I looked at the speakers in my cab closely and they surely move more than 1mm at a reasonable volume.

PS: I'll also add that Xmax is not the physical limit of the speaker, it's just the point where the voice coil movement starts to get non-linear (an interesting read here: https://speakerwizard.co.uk/driver-ts-parameters-xmax/)

The physical limit is called Xlim or Xmech and is often more than double the Xmax, so maybe I've been conservative in my calculations actually.

EDIT: oh, I realized now that you were probably referring to this:

it ranges between 0.5 and 1.4mm, so I took 1cm as a medium value to simplify the math a bit.

That's just a typo, I had to write cm. Corrected in the original post 

 

[DLC86]

what about the vacuum it pulls against on the outstroke though...it has to push against the sealed volume as well as pull against the sealed volume, unless the cab is pressurised with nitrogen or something.

*googles pressurised speaker cabinet patents*

Did you read my post at all?

So when, for a given voltage applied, the speaker moves forward by 1cm, the internal volume goes from 30 (let's call it Vr, volume at rest) to 30.5 liters (V1). This causes a variation in the internal pressure that can be calculated as follows:

P1 = (Pr * Vr) / V1 = (1 * 30) / 30.5 = ~0.9836 atm

Now let's see what happens when the speaker moves backwards by the same 1cm excursion.

V2= Vr - ΔV1 = 30 - 0.5 = 29.5 liters

P2= (Pr * Vr) / V2 = (1 * 30) / 29.5 = 1.0169 atm