Watts per square meter I can deal with.
However:
I need to find a good textbook on audio physics. I just don't have a sufficiently intuitive grasp of how some of the factors interrelate, to see when things are going sideways.
The reason this bugs me, is that Sound Intensity appears to be defined as the product of sound pressure, and particle velocity,
We've already determined that SPL shouldn't change due to the simple "slot halves the area, and doubles the velocity" slot-loading feature. The Volume Velocity remains the same, so the SPL should (to within a handwaving approximation) remain the same.
If the SPL doesn't change, and the particle velocity doubles, then the Sound Intensity should double. It doubles over a half-area slot, so if we do the math this way, Watts/meter^2 remains identical.
On the other hand, Sound Intensity also is written as proportional to the square of the particle velocity. The particle velocity doubled, so the square is 4x, over half the area, leaves us with this path through the math saying the Sound Intensity doubled.
It's got to be the impedance term that's mucking with this, but the electrical analogies for the sound properties just haven't clicked well enough for me to see how.
you guys are focusing too much on what happens at the exit of the slot rather than what happens in the far field. Of course you will see higher velocity and SPL at the slot exit just as you see higher SPL at the exit of a port in a vented box. But to see how the port and driver sum in the far field you have to scale the port down by ratio of Sd/(port area). You are fixated on what the velocity at the port exit is. That does not define the source strength. The source strength, and therefore the SPL in the far field, is given by the volume velocity = VV = source area x velocity. On the driver side it would be the cone velocity time SD. On the slot side VV would be VV = slot area x air velocity in the plane of the slot. The air velocity at the slot exit, at frequencies well below the slot resonance, will be cone velocity x Sd/(slot area). This then becomes VV = cone velocity x Sd/ (slot area) x (slot area) = cone velocity x Sd, same as the driver side. This applies at low frequency, below the slot resonance. So at low frequency the slotted OB woofer behaves as a dipole, just as I showed with my experiment/measurements. There may be some minor differences due to the way the baffle blocks the front or rear radiation and because of path length differences, but those are secondary issues.
Ah, no - I'm fixated on understanding where the (assuming a half-driver-area slot) 4x multiplier that appears due to the v^2 term, disappears based on a 1/2 term from the slot-area to driver-area ratio.
I don't have a problem with accepting your measurements or results, but I'm trying to understand how and why the terms come out the way they do. There's a v^2 term that distributes /something/ asymmetrically, with a ratio of 4:1, front to rear (again, assuming a 1/2-driver-area slot). There's a 1/2-area slot. Something else must be eating the extra factor of 2 on the front, or, there must still be 2x /something/ asymmetrically distributed front-to-rear.
I'm just trying to understand what ate the spare "2", or, what the "something" is, that's doubled. I don't have a dog in the fight regarding the answer, I just don't like not being able to make the math make sense.
whether a slot the area of the driver cone,
or a slot 1/2 of that,
as the slot increases the velocity of the air in the mouth of the speaker,
it also decreases the velocity of the cone?
and whatever results is dissipating into the same volume of air?
not an answer, just a thought.
or a slot 1/2 of that,
as the slot increases the velocity of the air in the mouth of the speaker,
it also decreases the velocity of the cone?
and whatever results is dissipating into the same volume of air?
not an answer, just a thought.
Guys, the Volume velocity (and so radiated pressure) below the high cutoff is unchanged by the slot but the slot along with the trapped volume between the radiator and the exit form an acoustic low pass filter In the form of a Helmholtz resonance.
If this is sized appropriately then one can get some gain at the high end of the band in exchange for a sharper roll off above that.
The horns we make all use this approach because the acoustic low pass filter reduces the harmonics all drivers produce, that “free sound” not being part of the input signal.
The difference between a horn at the exit side and a slot is what the acoustic resistance looks like to the radiation aperture, a horn presenting a much larger acoustic load.
Bottom line, like all things in acoustics, proper measurements trump a computer model. Regardless of it being essentially the same as the direct radiation case at /near the low cutoff, there is a reason it might measure higher up close.
Approaching a source with a large enough velocity with a microphone, one finds that up close, one may find levels that are higher than predicted. At least some of the time, it can be psudo-sound. Here the air is moving fast enough that it carries momentum as it impacts the microphone diaphragm and that error can cause a high reading (which can be greatly reduced by turning the mic 90 degrees)..
Best,
Tom Danley
Danley Sound Labs
If this is sized appropriately then one can get some gain at the high end of the band in exchange for a sharper roll off above that.
The horns we make all use this approach because the acoustic low pass filter reduces the harmonics all drivers produce, that “free sound” not being part of the input signal.
The difference between a horn at the exit side and a slot is what the acoustic resistance looks like to the radiation aperture, a horn presenting a much larger acoustic load.
Bottom line, like all things in acoustics, proper measurements trump a computer model. Regardless of it being essentially the same as the direct radiation case at /near the low cutoff, there is a reason it might measure higher up close.
Approaching a source with a large enough velocity with a microphone, one finds that up close, one may find levels that are higher than predicted. At least some of the time, it can be psudo-sound. Here the air is moving fast enough that it carries momentum as it impacts the microphone diaphragm and that error can cause a high reading (which can be greatly reduced by turning the mic 90 degrees)..
Best,
Tom Danley
Danley Sound Labs
if the slot "is sized appropriately", (maybe resonance F due to depth/width of slot?),
could "gain at the high end of the band in exchange for a sharper roll off above that"
be used to partially tame woofers with a peaky response at upper frequencies?
for general example, Peerless 830669 peak at 2 kHz,
appropriate depth/width of slot deliver rising FR slope (from 500-1000, say?) with steeper rolloff (6 dB/octave more/less?) above that?
http://www.parts-express.com/pedocs/specs/264-1118-peerless-830669-specifications.pdf
and a big thank you to Mr. Danley for what was already generously given!
could "gain at the high end of the band in exchange for a sharper roll off above that"
be used to partially tame woofers with a peaky response at upper frequencies?
for general example, Peerless 830669 peak at 2 kHz,
appropriate depth/width of slot deliver rising FR slope (from 500-1000, say?) with steeper rolloff (6 dB/octave more/less?) above that?
http://www.parts-express.com/pedocs/specs/264-1118-peerless-830669-specifications.pdf
and a big thank you to Mr. Danley for what was already generously given!
I am new to the debate and it is not clear to me quite what you are trying to equate with what but it looks like you might be trying to get all the energy in the near field to end up in the far field? If you halve the area, conserve the momentum then the local kinetic energy will double but this does not mean all of it will make it into the far field. Clearly something substantially different and more complex has to be going on compared to a purely outwardly propagating wave. However, for what an acoustician would call a "mass source" this complexity does not matter if the wavelength of the sound is substantially greater than the size of the source because the strength of the spherical sound waves in the far field will simply follow the amount of fluid displaced by the source. Whether the local rises in kinetic energy are lost or reversibly recovered by whatever is driving the mass source will depend on the local details but is irrelevant to the sound in the far field.
Me? At this point I'm just trying to get a grip on what the math says is happening. I'll get back to trying to figure out how I design the workshop audio once I convince myself that I understand enough of the basics to make a reasonable first pass (still have about 40 yards of concrete to pour for the last part of the pad, and the shell to get up, before any of this makes any difference).
I have to assume that Nelson Pass did something like look at the Sound Intensity is proportional to the square of particle velocity equation, and said "cool, with a half-driver-area slot, I double the particle velocity, and halve the area, so I get 4x the Intensity (watts per square meter) over half the area, which gives me an asymmetric double the energy out the front".
Likewise I have to assume that John K looked at the SPL is proportional to Volume Velocity equation, and said "Umm, no - Volume Velocity is the product of particle velocity and area, so (2*Velocity)*((1/2)*Area) gets you right back to the same SPL you started with -- and I measured it so I know that's working".
Me, I'm a cheap (and mostly broke) b*st*rd, so I can't afford umpteen gadgillion-dollar speakers to create perfect sound, but, I am building a building and it doesn't really cost me anything to incorporate architectural features that themselves function as components of the audio system - but I've got to get it right the first time, because once it's built, it's not coming back out without the building coming down.
In the face of this, I've got Nelson's math, that agrees with everything I can find about acoustic physics, and John's math that also agrees with everything I can find, and the additional equations : Sound Intensity is proportional to the square of SPL over Impedance, and, Sound Intensity is proportional to the square of particle velocity times Impedance.
On the face of it, these can't all be true at the same time, but, unless audio physics math is completely wonky, they _must_ all be true at the same time.
I grok electrical physics, kinetics, heck, if I close my left eye I can deal with quantum mechanics with reasonable fluidity. I just cannot yet grasp the workings of audio physics well enough to intuit my way out of a wet paper bag.
Unless I'm completely misreading the math, John K's analysis _and_ Nelson Pass' analysis must simultaneously be true. Likewise, the Sound Intensity must simultaneously be proportional to the square of the SPL (which doesn't change) and proportional to the square of the particle velocity (which does change).
Near field, far field, doesn't make any difference to me. Apples to apples, I need to understand well enough to make the math work out /somewhere/ before I start thinking about cutting lumber 🙂
The only thing that currently seems like a sense-making proposition to my addled brain, is if that mostly-ignored impedance term isn't "the same Z" in each of the equations. Clearly a half-width slot is going to present a different impedance than the full driver area presented to free air, so "they're different" seems reasonable, but, I don't grasp the fundamentals well enough yet to be comfortable saying that it's OK to stuff the disagreement between all of the above propositions into that term.
Anyways - I'm really just trying to develop enough understanding of how to use the math, that I don't do something stupid when building my building.
Well it looks like it may need to if you want to understand the answer to your question.
The difference in energy between your two slots is irrelevant to the sound in the far field if the wavelength of the sound is substantially greater than the size of the slots. What matters is the amount of fluid displaced by the air moving in and out of the slot (the volume velocity). Since you want to set this quantity to be the same for the two slots then the two slots will have the same SPL in the far field. But they will not have the same SPL in the near field. As you have noted the smaller slot has larger velocities/pressures/intensities and so will have more sound energy moving around in the near field and not being radiated to the far field.
I think the circulation of sound energy in the near field is probably they key bit of physics you do not understand but it is hard to be sure because it is not fully clear to me what you equating with what and why.
This is entirely possible - which is why I think I need to find a good teaching-text on this. Right now, I'm stuck with a bunch of equations relating different variables about sound, and my assumption is that they must all be true _simultaneously_ at any point in the field, near or far. If some of the equations need sticky-notes that say "this is only relevant in the far field" attached to them (which I have a growing suspicion is true for the SPL /approx volume velocity equation), then I'm afraid I haven't yet found an online resource that's sufficiently clear about those details for someone as simpleminded as myself.
I'm really not trying to be obtuse! At the end of it all, I'm trying to understand how John and Nelson can /both/ have correct analyses, which I believe that they both do. What I will call Nelson's "particle velocity squared" analysis says that a slot puts, not just a greater _density_ of energy, but _more_ energy out the front than an open-air driver. John's "volume velocity" analysis says that the additional energy doesn't matter, because a slot produces the same SPL as the bare drivers.
If I can't understand how more energy out, doesn't matter to the SPL, then I can't hope to make intelligent choices about system design. (and yes, I realize that this actually involves how the sound gets from near-field to far-field, but if I can't even make the math work at _a_ point, it hardly seems likely that I'm going to understand the more complex transfer process).
The only thing that seems to make sense to me, is if Nelson's additional energy, is eaten by the difference in impedance faced by the slot, versus bare drivers.
Unfortunately I only see arguments against his analysis that say "nope, you're wrong, slot-loading isn't allocating energy asymmetrically, it's just condensing it". If that counterargument is true, then I need to understand why the particle-velocity-squared equation doesn't hold in that case. If the counter-argument is /false/, then I can start trying to understand why that additional energy is irrelevant to the SPL.
Nelson is correct. The driver has had to do more work to squeeze the air through the narrower slot and double the kinetic energy.
If it is a compact monopole source then John is correct.
A spherical wave in the far field with the same volume velocity as the monopole source has the same amount of air flowing back and forth across the large sphere as the air flowing back and forth across the slot. This is what needs to be equated for a compact monopole source.
The energy flowing across the sphere is going to be significantly less than that flowing in and out of the two different slots. Equating energy is wrong. I will leave it to you as an exercise to calculate the kinetic energy for the two slots and a sphere with equal volume velocity.
What happens after the air flows out of the slot is that it spreads out and the velocity reduces in a complicated manner. The high velocity air moving in one direction from the slots will spread out in all 3 directions and reduce velocity. This will reduce the kinetic energy in the same way as your bigger slot has less kinetic energy than the smaller slot. As the kinetic energy is reduced the the internal energy will be raised so that total energy is not lost.
Thank you thank you thank you!
This finally gets me to where I can start to make sense of things. There are still some rough edges I need to think about, but the inability for KE to be conserved as KE, and a way to see what's going to be conserved and what's going to be converted to non-SPL-producing energy, definitely gets me closer to understanding.
Many thanks for taking the time to pull teeth!
Sorry, conservation of energy applies. You can not just throw away the 1st law of thermodynamics because you do not understand it.
Perhaps I abbreviated my reply too drastically - of course conservation still applies. However, it does not necessarily follow that the energy is conserved in a form that is accessible as contributing to SPL or Sound Intensity.
Hi,
You can handwave the numbers in physics but you can't
handwave the the physics, conjecture is pretty pointless.
A slot does not put out more energy, except at resonance.
The nature of that energy is different nearfield to a plain
driver and identical (near enough) farfield, its that simple.
However to be fair to a plain baffle it also exhibits proximity
effects nearfield that are no different to the slot loading case,
except for the extreme nearfield, and its all just basic physics.
Slot loading does nothing useful at the bass end of an open baffle farfield.
I'd also argue it does nothing really useful fullstop, it is wishful thinking.
(Other than allow a form of force cancellation for the bass drivers).
rgds, sreten.
You can handwave the numbers in physics but you can't
handwave the the physics, conjecture is pretty pointless.
A slot does not put out more energy, except at resonance.
The nature of that energy is different nearfield to a plain
driver and identical (near enough) farfield, its that simple.
However to be fair to a plain baffle it also exhibits proximity
effects nearfield that are no different to the slot loading case,
except for the extreme nearfield, and its all just basic physics.
Slot loading does nothing useful at the bass end of an open baffle farfield.
I'd also argue it does nothing really useful fullstop, it is wishful thinking.
(Other than allow a form of force cancellation for the bass drivers).
rgds, sreten.
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There would be a tiny bit of viscous loss with the slot however, the higher velocity is over a smaller area and so one moves down on the radiation resistance curve, meaning it is a less efficient source.
The net picture is there is no acoustic gain from the slot vs direct radiation except for that gained when making a band pass source (and then only when the upper and lower corners are close enough).
A horn is a different case as it presents a different radiation resistance than the direct radiating driver but that loading has a "high pass" corner set by the expansion rate.
Hi,
No he isn't. At least your interprepretation of his meaning is wrong.
rgds, sreten.
Oh Tom, now you have gone and done it. It was hard enough with out considering losses and the effect of source area on radiation resistance. 😀
Ok, I thought I was getting a handle on this, but, apparently, pfft!
I agree completely, you can't handwave physics.
So - and please keep in mind that I'm not intending to argue, but rather am trying to figure out where the handwaving _isn't_ in this:
Let's just look at the plane of the driver, or the plane of a half-driver-area slot.
Ignoring back-pressure or fluid effects that may decrease the particle velocity passing through the half-area slot, the driver pushes the same volume of particles out the half-area slot, so they're moving, to within an approximation, twice as fast.
Sound Intensity, watts per square meter, is defined in terms of the square of particle velocity, times the impedance.
Unless the impedance changes for the slot case by a factor of 2 (and I don't understand acoustical impedance well enough to know whether this is true) this leaves us, for the slot case, with 4x the watts per square meter, distributed over 1/2 the area, resulting in 2x the total energy.
That's the simple physics, based on the definitions as I can find them.
So - how do we get to:
without passing through handwaving?
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