Surely a double of power gives only 3dB gain.
Perhaps I over-simplified earlier.
Doubling the radiation resistance (much like a horn) doubles the % efficiency, giving +3dB in the transition from full space to half space.
Furthermore, the half of the energy that was wrapping around the cabinet is now being radiated forwards. A double in energy forwards gives another +3dB.
I'm also fairly sure that, for a point source, sound does follow an inverse square law - you lose 6dB per doubling of distance. True line arrays AFAIK don't follow this.
Chris
Remember that you go back to 2PI space as soon as the sound hits the back wall, Olsons tests were made in an anechoic room or in the open air Also, as others have said, this is ignoring that the driver itself is directional.
The result is that 6dB is way too much
The result is that 6dB is way too much
Yes, yes, and yes.
Drivers are by their nature very low in efficiency. Another way of saying that is tht the acoustic load is a very small part of what they drive. In fact the environment in which they are placed has virtually no impact on cone motion, whether on a boundary, or near a boundary when considering Allison wall reflection issues.
In the end efficiency or power radiated is totally a function of the acoustic resistance presented to the driver by the mounting circumstances (1/2 space, free space, horn loaded, etc.)
I don't have a copy of Beranek but did find his curves in Martin Colloms "High Performance Loudspeaker" (2nd ed. page 17) where he shows the usual radiation resistance and reactance for 3 cases of a piston "on an infinite baffle", "on a cylinder face", and "in free air". On a cylinder face is physics shorthand for no baffle, but the cylinder, sometimes termed "on the end of an infinite tube" is how we explain that the back wave isn't interfering. It is similar to your "small enclosure". For us, these first two are what we have been calling 2 pi and 4 pi respectively.
In the range of interest the radiation resistance is double on the infinite baffle (2pi) case as it is in the 4pi case. That means that the same woofer will automatically be 3dB more efficient (watts out over watts in) in 2 pi than in 4 pi.
As a side note, a good number to remember is that 1 acoustic watt is 109dB at 1 m (omni source) and 112dB for a half space hemispheric source. Always 3 dB apart.
So the explanation for the 6dB difference is that 3 dB comes from efficiency doubling and the additional 3dB comes from rear radiation folding around to the front. Again the second "3" is dependent on the source being totally non directional, so we tend to see 6dB difference at lowest frequencies transitioning to somewhat less.
At higher frequencies, say once we get past the region of diffraction wiggles in the curve, the baffle is sufficiently large to constrain radiation to 2 pi and response doesn't change at all, whether the baffle is finite or infinite.
David S.
I've heard of SPL being analogised to Voltage. Would it be fair to analogise a speaker to a current source?
Dave, not to nit pick but this is where I have the problem with your semantics. You do not have 3dB due to radiation resistance and another 3dB because of non directional (or radiating into 4PI). What you are saying is that a) there is a 3dB decrease in SPL because the radiation resistance is lower by 3dB when the source radiates into 4PI space, and b) there is another 3dB decrease because the source is radiating into 4Pi space. Sounds weird because you are mixing cause and effect.
The way it works is the source is mounted in free space (4PI) or in an infinite baffle (2PI) and the pressure as a function of position is found by solving the wave equation to relate the strength of the source (volume velocity) to the radiated pressure. The result is that at low frequency a source of constant strength radiates a uniform pressure field which decreases as 1/r for which the pressure in 4PI is 1/2 the pressure in 2PI with the direct consequence that 4PI SPL is -6dB and the power is -3dB compared to 2PI. Radiation resistance comes from a similar argument. The radiation impedance is defined as the ratio of force applied to the medium (air) divided by the velocity of the source. It starts from the same point, the relationship between pressure and source velocity. The reduction is radiation resistance and efficiency is also a consequence of the change in environment. It is not a cause, it is an effect. Pick up Kinsler and Frey and it's all laid out. Beranek just quotes the K&F result.
Anyway, go on with the discussion. I not going to chime in further.
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Hi,
I don't think the cases are different and pressure change is the total
answer for why SPL for 2pi and 4pi must theoretically change by 6dB.
Total power radiated is a different matter, that is 3dB due to the two
different solid angles, and matches the 3dB increase in "efficiency".
Its really a loss though, as all drivers are specced assuming 2pi loading.
rgds, sreten.
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We are all dancing in circles.
The pressure change is due to two independent reasons: one of them is an efficiency doubling when going to 2 pi. We can argue which is cause and which is effect but it seems to me that the equivalent circuit would necessarily include the radiation resistance that changes (doubles in 2 pi). The radiated power curve can be calculated. The computed power output would raise by 3 dB alone. Finally, when converting that power response to a pressure response the radiation angle would need to be factored in and the second 3dB would be picked up.
The fundamental premise is that radiation resistance is different, by a factor of two, for the two cases. Can any of you find evidence that that is not the case?
David S.
The pressure change is due to two independent reasons: one of them is an efficiency doubling when going to 2 pi. We can argue which is cause and which is effect but it seems to me that the equivalent circuit would necessarily include the radiation resistance that changes (doubles in 2 pi). The radiated power curve can be calculated. The computed power output would raise by 3 dB alone. Finally, when converting that power response to a pressure response the radiation angle would need to be factored in and the second 3dB would be picked up.
The fundamental premise is that radiation resistance is different, by a factor of two, for the two cases. Can any of you find evidence that that is not the case?
David S.
Hi,
It is a case of cause and effect.
4pi to 2pi doubles pressure and quadruples SPL.
Simply from considering a fixed volume displacement.
The other subtleties are a necessary consequence of the above.
The above says radiation resistance must double, and so must efficiency.
Your trying to put the cart before the horse IMO, in terms of analysis.
It is a case of what is used to show what must be the actual case.
rgds, sreten.
(J.K. has said all of this already, in his way .....)
It is a case of cause and effect.
4pi to 2pi doubles pressure and quadruples SPL.
Simply from considering a fixed volume displacement.
The other subtleties are a necessary consequence of the above.
The above says radiation resistance must double, and so must efficiency.
Your trying to put the cart before the horse IMO, in terms of analysis.
It is a case of what is used to show what must be the actual case.
rgds, sreten.
(J.K. has said all of this already, in his way .....)
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I had a few hours on an airplane to mull this over, and I pulled out my Beranek text last night when I got home. The solution of the wave equation for a radiating source clearly yields the result that sound pressure decreases with 1/r. My earlier confusion was me being tired & trying to compare two unrelated effects (baffle/no baffle, double measurement distance).
With respect to the baffle step loss of 6dB, the short answer is: John K is correct. Dave is almost correct, but operating on the effects of the basic principles John outlined. Also, there are not actually two 3dB effects combining, which will be laid out in (3).
First, we should clearly define the building blocks.
- Specific acoustic impedance (zs): a constant property of air for a given bulk temperature and pressure; ratio of sound pressure to particle velocity (zs = p / u)
- Acoustic impedance (Za): not a constant; ratio of sound pressure to volume velocity (Za = p / U)
- Volume velocity (U): Rate of flow of acoustic medium through a given imaginary surface. Given a source that continually displaces the same amount of air, U is a constant for any continuous enclosing measurement surface. (U = u * S)
- Particle velocity (u): The velocity of an infinitesimal element of the acoustic medium.
- Enclosing surface area (S): The surface area of an imaginary boundary placed around a source radiating sound. In this case, hemispheres & spheres.
- Sound Pressure (p): The pressure of at an infinitesimal point in the acoustic medium relative to the bulk pressure in the acoustic environment.
(1)
So, we have (Za) = (p/U) = (p/(v*S)) = (zs/S). Dave, your acoustic impedance is shown to increase by 2x when S is cut in half since (zs) is a constant. Doubling of acoustic impedance is an effect in this case. Certainly, you can work with the effects to show the 6dB change, as you have been doing.
(2)
Using the same basic blocks, it is also demonstrable that sound pressure is halved when transitioning from 2pi space to 4pi space when measuring at a given distance from the source (wavelength of emitted sound is >> source dimensions).
Take: (p/U) = (zs/S) and rearrange, solving for (p).
Now: (p) = ((zs*U)/S), and remember that (zs) is a constant, as is (U). So, transitioning from 2pi to 4pi space doubles (S), the other two parameters are constant, and thus (p) is halved. Since sound pressure level (SPL) is 10*log10((p1/p2)^2), we get a 6dBSPL change in SPL. SPL and sound pressure are related, but different things. I guess we all get lazy and stop specifying the units associated with "dB".
(U) is a constant because the source continues to move the same amount of air, namely (U). In 2-pi space, (U) is measured as flowing through an enclosing hemisphere of any radius (ignoring the fact that real air is not adiabatic). In 4pi space, (U) is measured through an enclosing sphere of any radius (with the same "ideal" air).
(3)
Here are some further thoughts about the acoustic impedance angle. I assume that this is where Dave has been going.
The speaker, our acoustic source, presents a very high acoustic impedance as compared with air. If you build the equivalent analog circuit in your head, it boils down to a Thevenin equivalent source driving a resistive load (Za) (we are looking at a single frequency, so all the reactive components in the load get lumped into a single load element). The source resistance is very very high as compared to the load resistance. Volume velocity (U) is the analog of current, and sound pressure (p) is the analog of voltage.
Anyway, an electrodynamic speaker is actually a constant FORCE source, but we are wrapping up all the mechanical/acoustic circuit transformers & elements into the voltage (p) source in out Thevenin equivalent circuit. So we have a constant-pressure source with a very high impedance output, into a low impedance load. You could also describe it as a current (U) source, given the high output impedance.
OK, so doubling the load resistance (Za) will effectively double the voltage (p) across it since the output impedance is so high. Current (U) through the load will barely decrease. Shazam, here is our 6dB increase, all-in-one. So, we get a ~6dB increase in sound pressure level (SPL). At the same time, we only receive a 3dB increase in sound POWER. Remembering Ohm's law, Power = I*V and our analog (Power = U*p), we see a constant I (U) and a doubling in V (p), and hence a doubling of power, which is a 3dB change.
It all boils down to this:
- Doubling your free-field measurement distance yields a halving of pressure and a quartering of sound power...-6dB change in both.
- Transitioning from 2pi to 4pi space yields a halving of pressure (-6dB), and a halving of sound power (-3dB).
I HOPE that I was able to spell out the equivalent circuit stuff clearly enough. It is one of those things that draws itself just fine in my head, and I can only hope that it was typed clearly enough for others to "see" as well.
As far as I am concerned, CASE CLOSED!
EDIT:
I made a couple of clarifications in the last section and removed mention of "power" in one place since it just sounded confusing. Re-read it if you read this before my edit was made!
With respect to the baffle step loss of 6dB, the short answer is: John K is correct. Dave is almost correct, but operating on the effects of the basic principles John outlined. Also, there are not actually two 3dB effects combining, which will be laid out in (3).
First, we should clearly define the building blocks.
- Specific acoustic impedance (zs): a constant property of air for a given bulk temperature and pressure; ratio of sound pressure to particle velocity (zs = p / u)
- Acoustic impedance (Za): not a constant; ratio of sound pressure to volume velocity (Za = p / U)
- Volume velocity (U): Rate of flow of acoustic medium through a given imaginary surface. Given a source that continually displaces the same amount of air, U is a constant for any continuous enclosing measurement surface. (U = u * S)
- Particle velocity (u): The velocity of an infinitesimal element of the acoustic medium.
- Enclosing surface area (S): The surface area of an imaginary boundary placed around a source radiating sound. In this case, hemispheres & spheres.
- Sound Pressure (p): The pressure of at an infinitesimal point in the acoustic medium relative to the bulk pressure in the acoustic environment.
(1)
So, we have (Za) = (p/U) = (p/(v*S)) = (zs/S). Dave, your acoustic impedance is shown to increase by 2x when S is cut in half since (zs) is a constant. Doubling of acoustic impedance is an effect in this case. Certainly, you can work with the effects to show the 6dB change, as you have been doing.
(2)
Using the same basic blocks, it is also demonstrable that sound pressure is halved when transitioning from 2pi space to 4pi space when measuring at a given distance from the source (wavelength of emitted sound is >> source dimensions).
Take: (p/U) = (zs/S) and rearrange, solving for (p).
Now: (p) = ((zs*U)/S), and remember that (zs) is a constant, as is (U). So, transitioning from 2pi to 4pi space doubles (S), the other two parameters are constant, and thus (p) is halved. Since sound pressure level (SPL) is 10*log10((p1/p2)^2), we get a 6dBSPL change in SPL. SPL and sound pressure are related, but different things. I guess we all get lazy and stop specifying the units associated with "dB".
(U) is a constant because the source continues to move the same amount of air, namely (U). In 2-pi space, (U) is measured as flowing through an enclosing hemisphere of any radius (ignoring the fact that real air is not adiabatic). In 4pi space, (U) is measured through an enclosing sphere of any radius (with the same "ideal" air).
(3)
Here are some further thoughts about the acoustic impedance angle. I assume that this is where Dave has been going.
The speaker, our acoustic source, presents a very high acoustic impedance as compared with air. If you build the equivalent analog circuit in your head, it boils down to a Thevenin equivalent source driving a resistive load (Za) (we are looking at a single frequency, so all the reactive components in the load get lumped into a single load element). The source resistance is very very high as compared to the load resistance. Volume velocity (U) is the analog of current, and sound pressure (p) is the analog of voltage.
Anyway, an electrodynamic speaker is actually a constant FORCE source, but we are wrapping up all the mechanical/acoustic circuit transformers & elements into the voltage (p) source in out Thevenin equivalent circuit. So we have a constant-pressure source with a very high impedance output, into a low impedance load. You could also describe it as a current (U) source, given the high output impedance.
OK, so doubling the load resistance (Za) will effectively double the voltage (p) across it since the output impedance is so high. Current (U) through the load will barely decrease. Shazam, here is our 6dB increase, all-in-one. So, we get a ~6dB increase in sound pressure level (SPL). At the same time, we only receive a 3dB increase in sound POWER. Remembering Ohm's law, Power = I*V and our analog (Power = U*p), we see a constant I (U) and a doubling in V (p), and hence a doubling of power, which is a 3dB change.
It all boils down to this:
- Doubling your free-field measurement distance yields a halving of pressure and a quartering of sound power...-6dB change in both.
- Transitioning from 2pi to 4pi space yields a halving of pressure (-6dB), and a halving of sound power (-3dB).
I HOPE that I was able to spell out the equivalent circuit stuff clearly enough. It is one of those things that draws itself just fine in my head, and I can only hope that it was typed clearly enough for others to "see" as well.
As far as I am concerned, CASE CLOSED!
EDIT:
I made a couple of clarifications in the last section and removed mention of "power" in one place since it just sounded confusing. Re-read it if you read this before my edit was made!
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Sorry Dave, it pressure, period! The driver moves. A pressure field is generated. That is the physics. The radiation impedance, radiated power and efficiency can not be determined until the pressure the pressure field is known. They are all functions of the pressure field. Ask yourself how is radiation resistance determined? It is (defined as) the ratio of the force the source exerts on the air in contact with the source (= - the force the air exerts on the source) divided by the velocity of the source. The force in question is the pressure integrated over the surface of the source which obviously can not be determined until the pressure field is known.
This is not a chicken/egg thing. Radiation impedance is a quantity derived by engineers so that can construct a simple (or not so simple) equivalent circuit model of driver motion.
Oh and thank you John K for setting everything onto the right track. Now that I got the basic principles worked out & properly explaining observed phenomena, I can go to sleep at a reasonable hour! Sheesh, I thought I was "done" with Physics 101 almost 10 years ago!
Only one small error here. Doubling radial distance reduces pressure by 1/2 and SPL by 6dB. It also reduced intensity by 6dB. But sound power remains constant since power is the integral of the intensity integrated over the area of the surface through which the wave passes. For uniform radiation, this reduces to P = I x A where a goes like R^2. So I goes like 1/R^2, A goes like R^2, and P = constant, not dependent on R (unless there are other dissipative mechanisms present).
I swear I hear clucking.🙄
If you say the pressure has doubled and therefor the acoustic impedance must have gone up, and I say the acoustic Impedance goes up and therefor the power must have doubled, that sounds chicken and egg to me. Yes, I understand that acoustic impedance is the measurable ratio of pressure to velocity. It can be rigorously derived as well.
Let me think about what you and BMWman are saying and see if I can make sense of it.
Good discussion by the way. Have we lost the rest of our audience?
David
My god, to think that I was JUST explaining to a coworker that sound power is constant for a given source and its relationship to intensity! So yes, substitute INTENSITY for POWER in a few places in my (long) previous post. Why on earth doesn't this forum allow edits after one makes a new post? I get mixed up too easily when dealing with properties measured at a point and properties measured across a surface...gotta work on that.
Acoustic impedance is constructed of more basic physical elements. That is why it isn't a chicken & egg situation. We weren't around to see if the egg was laid or hatched first. However, air molecules were moving as a result of other objects pushing them around since forever, and humans certainly were "there" when acoustic impedance was born...we created it! Acoustic impedance is a human construct (well, so are pressure & velocity, but acoustic impedance was born of them). Nature just has a bunch of atoms / molecules that can't occupy the same space as other ones at the same time.
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Hi,
Not yet but your risking boring them to death 😉.
Everything is derived from the pressure equations, not the other way round.
You cannot rigorously derive what you are saying without them, and its
pointless to attempt to do so when it's already in black and white.
Try to, without using the pressure equations, or anything derived from them.
Its simply not physics, its like trying to prove the Pythagoras theorem
without using it, most people can't, they give an example that shows it
assuming it to be true, that is not a proof - and therein lies the pudding.
Without considering pressure you will be utterly lost.
rgds, sreten.
This has been a fun thread. It seems really easy to get lost in discussions of much higher-order effects on here, so it is sort of refreshing to kick things down into the basic physics from time to time. Getting down into it & seeing just how simple a phenomena like sound really is (particles knocking into each other) allows for an appreciation of just how complicated we humans are. The air is just doing what is has always done. We, on the other hand, get all worked up trying to make it do something specific and have to invent all sorts of descriptions & explanations for its behavior to try to better get it to do what we want!
Since my employer gave me the opportunity to sit through a multi-day lecture on sound measurement and human perception of sound, it is still pretty fresh in my mind how much more complicated we are than air molecules banging around. The physiology of the ear is nothing short of astounding. Then add in the most powerful non-linear, non-causal, time-variant signal processing units on the planet, and it is a wonder that we have done as well as we have in sound reproduction and control.
This isn't me trying to downplay anything about acoustic engineering. To get where we are with it, we have needed some damn talented individuals that with a really solid grasp on math and physical science. It just strikes me as a little funny (now), that we can get a seemingly unending font of technical challenges from such a simple natural behavior. I remember having the same feeling in a number of thermodynamics & strength-of-materials courses when it finally *clicked* in my head. That is probably why I went & spent so much time nagging the issue in this thread...that *click* is almost like a drug! I am really liking acoustics more and more because it pulls from so many fields; thermodynamics, materials, electrical engineering, psychology/cog-sci, and more. I guess most of them are derived from physics, which is such a huge field that it has to be broken up into large sub-fields to maintain some sense of what is involved.
Thanks guys!
Since my employer gave me the opportunity to sit through a multi-day lecture on sound measurement and human perception of sound, it is still pretty fresh in my mind how much more complicated we are than air molecules banging around. The physiology of the ear is nothing short of astounding. Then add in the most powerful non-linear, non-causal, time-variant signal processing units on the planet, and it is a wonder that we have done as well as we have in sound reproduction and control.
This isn't me trying to downplay anything about acoustic engineering. To get where we are with it, we have needed some damn talented individuals that with a really solid grasp on math and physical science. It just strikes me as a little funny (now), that we can get a seemingly unending font of technical challenges from such a simple natural behavior. I remember having the same feeling in a number of thermodynamics & strength-of-materials courses when it finally *clicked* in my head. That is probably why I went & spent so much time nagging the issue in this thread...that *click* is almost like a drug! I am really liking acoustics more and more because it pulls from so many fields; thermodynamics, materials, electrical engineering, psychology/cog-sci, and more. I guess most of them are derived from physics, which is such a huge field that it has to be broken up into large sub-fields to maintain some sense of what is involved.
Thanks guys!
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Yep, lot's of chuckling. 😀 But Dave, Please pay attention! 😱 You can not have acoustic impedance until you know what the pressure is. Yes rigorously derived by first finding the pressure field acting on the source. 🙂 That's why I pointed to K&F. They lay itall out. First they find the pressure field.....
But this is fun so let's take it to another level.
We can write the equation for driver velocity as
U = F /(Zm + Zr)
where Zm is the mechanical impedance of the driver and a function of the moving mass, Mms, mechanical resistance, Rms, and suspension compliance, Cms. Zr is the radiation impedance. F is the force applied to the driver by the motor. Now, for most drivers Zm is much greater than Zr and the driver motion is unaffected by the change in loading between 2 and 4 Pi. Therefore, as we have discussed, the volume velocity will remain essentially constant and the efficiency in 2PI will increase by 3dB. BUT, when we are dealing with very low mass devices, in the limit where Zm goes to zero, then for the same force applied to the driver doubling of Zr in 2Pi will result in making U lower by 1/2 and the radiated SPL in 4PI and 2Pi will then be constant when the same driving force is applied to the driver. Since the force is BL x I, constant force means constant current and the power dissipated in the voice coil will remain constant. Thus, with source of negligible Zm the efficiency in 2Pi will actually decrease by 3dB. So for a zero Zm source with constant power dissapated in the VC going from 4Pi to 2Pi results in no change in SPL, a 3dB (factor of 2) increase in Zr, and a 3dB decrease in efficiency.
I only throw this out because we have been saying that the volume velocity remains constant, but not addressed why that might be reasonable. (Actually I threw this out because it mucks things up so we can argue some more.
)I only throw this out because we have been saying that the volume velocity
remains constant, but not addressed why that might be reasonable. (Actually I
threw this out because it mucks things up so we can argue some more.
Hi,
Its a theoretical aside, for low efficiency drivers its very reasonable to
assume driver motion is dominated by Zm and Zr has no real effect.
Hence you assume identical driver motion for the two cases, which
cannot be true for an impossibly light impossibly stiff driver.
rgds, sreten.
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