This is a technical curiosity of mine. Almost everything I have read pertaining to baffle step compensation (and baffle diffraction simulators) indicates a 6dB loss across the frequency response for a baffled loudspeaker in free space. It is common knowledge that it happens because higher frequencies see a large reflecting surface relative to their wavelength and radiate into 2*pi space. Lower frequencies pass right around it since its size is relatively small compared to their wavelength, and radiate into 4*pi space. OK, so nothing new here.
Why is the loss not 3dB though? When a constant sound power source starts radiating into full space, which is twice the volume of half-space, the energy density is halved. This would indicate that sound intensity is also halved, and intensity is proportional to the square of pressure. So, wouldn't this be a 3dB drop?
The equation for calculating the sound pressure level in the room includes a directivity factor (1 for free-field, 2 for single-plane baffle). A 10*log(2) is 3dB. A 6dB drop would indicate a halving of sound pressure, rather than intensity (which is quartered). An acoustics consultant at the seminar I am at also says that the drop should be 3dB, based upon the same equation. So, where am I getting mixed up (since it seems unlikely that a simple mistake like this would be so widespread)? Thanks.
Aforementioned SPL equation on page 5.
http://www.eurovent-certification.com/fic_bdd/pdf_fr_fichier/1137149375_Review_67-Bill_Cory.pdf
Why is the loss not 3dB though? When a constant sound power source starts radiating into full space, which is twice the volume of half-space, the energy density is halved. This would indicate that sound intensity is also halved, and intensity is proportional to the square of pressure. So, wouldn't this be a 3dB drop?
The equation for calculating the sound pressure level in the room includes a directivity factor (1 for free-field, 2 for single-plane baffle). A 10*log(2) is 3dB. A 6dB drop would indicate a halving of sound pressure, rather than intensity (which is quartered). An acoustics consultant at the seminar I am at also says that the drop should be 3dB, based upon the same equation. So, where am I getting mixed up (since it seems unlikely that a simple mistake like this would be so widespread)? Thanks.
Aforementioned SPL equation on page 5.
http://www.eurovent-certification.com/fic_bdd/pdf_fr_fichier/1137149375_Review_67-Bill_Cory.pdf
Thinking about it in terms of distance from the source also seems to indicate a drop of 3dB.
A doubling of distance halves sound pressure and quarters intensity (-6dB). Doubling the radius of a sphere gives 8*volume and 4*surface area. Since intensity is power/area, it makes perfect sense. Increasing the radius by sqrt(2) gives 2*surface area (and 4*volume), with a halving of intensity and sqrt(half)'ing sound pressure (-3dB).
Now, this differs from the 2*pi-to-4*pi transition since, in that case, both volume and surface area of the virtual sphere double (whereas with a sphere of increasing radius has volume proportional to the square of surface area). Still, the SPL equation, taking directivity factor into account, also indicates -3dB. I wish I had my Acoustics (L.L. Beranek) textbook with me...I bet the answer is in there.
A doubling of distance halves sound pressure and quarters intensity (-6dB). Doubling the radius of a sphere gives 8*volume and 4*surface area. Since intensity is power/area, it makes perfect sense. Increasing the radius by sqrt(2) gives 2*surface area (and 4*volume), with a halving of intensity and sqrt(half)'ing sound pressure (-3dB).
Now, this differs from the 2*pi-to-4*pi transition since, in that case, both volume and surface area of the virtual sphere double (whereas with a sphere of increasing radius has volume proportional to the square of surface area). Still, the SPL equation, taking directivity factor into account, also indicates -3dB. I wish I had my Acoustics (L.L. Beranek) textbook with me...I bet the answer is in there.
Going from 4pi to 2pi there are always 2 things going on. First, the energy that was radiating behind the driver is now constrained to the front. Secondly the resistive air load on the driver is doubling. This is the difference between the airload of a "piston on the end of a pipe" and a piston on an infinite plane.
The second factor gives double the efficiency or +3dB in level. The first item gives a little less, typically 1.5 to 2 dB as the woofer would have some directivity, even at low frequncies.
The end result is 4.5 to 5 dB and a slight uphill slope for the 4 pi case.
David S.
p.s. I designed speakers professionally for over 30 years and never once added baffle step compensation. It was always incorporated into the woofer low pass network.
The second factor gives double the efficiency or +3dB in level. The first item gives a little less, typically 1.5 to 2 dB as the woofer would have some directivity, even at low frequncies.
The end result is 4.5 to 5 dB and a slight uphill slope for the 4 pi case.
David S.
p.s. I designed speakers professionally for over 30 years and never once added baffle step compensation. It was always incorporated into the woofer low pass network.
Thank you, David. I am not worrying about the topic for speaker design purposes at the moment, but rather out of intellectual curiosity. I get bothered when my grasp on a technical subject is fuzzy!
Perhaps some of my confusion lies in how you break the effect into two parts. Isn't the resistive load doubling BECAUSE the energy is being constrained to radiation through half the amount of space? I have seen a number of calculations/diagrams showing a frequency plot for a spherical enclosure baffle and a 6dB transition from low to high frequencies. That indicates a doubling (or halving) of sound pressure level and quartering of intensity...but it would seem that doubling the radiation space should double the intensity level, which is a 3dB change.
Sorry if I am being dense. Hopefully there is just some silly oversight that is confusing me here.
Perhaps some of my confusion lies in how you break the effect into two parts. Isn't the resistive load doubling BECAUSE the energy is being constrained to radiation through half the amount of space? I have seen a number of calculations/diagrams showing a frequency plot for a spherical enclosure baffle and a 6dB transition from low to high frequencies. That indicates a doubling (or halving) of sound pressure level and quartering of intensity...but it would seem that doubling the radiation space should double the intensity level, which is a 3dB change.
Sorry if I am being dense. Hopefully there is just some silly oversight that is confusing me here.
Now you've got me thiking about it and I am also confused.😱
I'm certain that the radiation resistance doubles in 2pi, as I've seen the curves many times. That guarantees double the power radiated, double the efficiency. but wouldn't that be +6, since it is double the power and in half the solid angle?
I don't know if the resistive load is doubling because of the space halving, just that it is.
Let me think about it some and see if I can come up with a better explanation.
David S.
Yeah, it is sort of a brain teaser. The acoustics consultant, who has been doing this since well before I was born, certainly sees no reason for a 6dB change...under ideal conditions. Certainly, real-world effects change things, although reflections and such will make the loss smaller rather than larger. Does John K (Zaph) post in here? This seems like something that he could probably elucidate quickly.
Hi,
Its the old chestnut, pressure doubles (as it must), but SPL quadruples, SPL is not pressure,
its pressure squared. Would better if SPL was called Sound Power Level, but it unfortunately isn't.
20log(P1/P2) = [10log(P1/P2)]squared is what you use for SPL. 10log(P1/P2) for pressure.
rgds, sreten.
Its the old chestnut, pressure doubles (as it must), but SPL quadruples, SPL is not pressure,
its pressure squared. Would better if SPL was called Sound Power Level, but it unfortunately isn't.
20log(P1/P2) = [10log(P1/P2)]squared is what you use for SPL. 10log(P1/P2) for pressure.
rgds, sreten.
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We know that there is a 6dB, (or "approaching" 6dB) difference for all the many curves, such as the orriginal Olson diffraction curves, that we have seen.
Sreten, PWL is the standard term for power level. I think that "pressure doubles (as it must)" isn't the total answer. Although I use that for two woofers in parallel, the woofer in 2pi vs 4pi is a little different case.
Maybe the best way to state it is, when going from 2pi to 4pi the acoustic load cuts in half for a 3dB loss across the board and another 3dB as frontal sound wraps to the back. At upper frequencies where the sound doesn't "wrap" as well you lose an intermediate amount for the observed 4 to 5dB.
Any takers?
David S.
Sreten, PWL is the standard term for power level. I think that "pressure doubles (as it must)" isn't the total answer. Although I use that for two woofers in parallel, the woofer in 2pi vs 4pi is a little different case.
Maybe the best way to state it is, when going from 2pi to 4pi the acoustic load cuts in half for a 3dB loss across the board and another 3dB as frontal sound wraps to the back. At upper frequencies where the sound doesn't "wrap" as well you lose an intermediate amount for the observed 4 to 5dB.
Any takers?
David S.
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Again, it boils down to simple physics. When the cone moved it displaces a given volume of air giving rise to a wave propagating into space. When the wave is constrained to expand into the front hemisphere the mass displaced by the wave will pass through an area equal to the surface of a hemisphere or some arbitrary radius, R. The mass displaced will be density times surface area times velocity. When the wave expands into free space the area the wave passes through is a sphere which will have twice the surface area of a hemisphere of the same R. Since the mass displaced by the driver is the same and the surface area doubles the velocity at that surface will be 1/2 that of the hemispherical case. The specific acoustic resistance of air in the free field is a constant thus, P = Z x U and since U is 1/2 in full space pressure is also 1/2 that of the hemispherical case. SPL goes like 10 x Log(P^2) or 20 Log(P). Since Pfullspace / P halfspace = 1/2 the full space SPL is -6dB compared to that of 1/2 space (20xLog(1/2) = -6)
Right, so the driver is better coupled to the air, and the sound isn't wrapping around.
3dB for each gives 6dB.
Think that makes sense.
Chris
Lets not keep it too simple. Pull out your text books. There will be two graphs given, one for the acoustic resistance of a piston in an infinite baffle (2 pi) and one for a piston at the end of an infinite tube (4 pi). Different acoustic resistances equate to different efficiencies for the two cases.
David S.
Thank you guys for all the posts. Substituting (volume velocity / radiation surface area) for particle velocity:
sound pressure = (volume velocity * spec. acoustic impedance) / radiation surface area (p=v*z/S); S varies, v & z are constant for the given frequency we conduct the imaginary experiment at. This jives with the -6dB seen at frequencies where the baffle is relatively small in a free field. Double area = half pressure = -6dB with other parameters equal.
Now, that would imply that pressure is inversely proportional to radius-squared (area term in denominator). It is widely known that pressure is inversely proportional to radius, not radius-squared. So a doubling of sphere radius yields 4x the area, which would quarter pressure and give -12dB. Yet, this is obviously wrong! Where is the disconnect?
How is it that pressure can be derived to be a function of radius-squared, which explains the baffle step loss, but conflicts with basic knowledge that it is a function of radius (double measurement distance, -6dB)?
Again, my apologies for being dense. I am sure that this is me making a stupid oversight somewhere (which is why this is bugging the hell out of me).
sound pressure = (volume velocity * spec. acoustic impedance) / radiation surface area (p=v*z/S); S varies, v & z are constant for the given frequency we conduct the imaginary experiment at. This jives with the -6dB seen at frequencies where the baffle is relatively small in a free field. Double area = half pressure = -6dB with other parameters equal.
Now, that would imply that pressure is inversely proportional to radius-squared (area term in denominator). It is widely known that pressure is inversely proportional to radius, not radius-squared. So a doubling of sphere radius yields 4x the area, which would quarter pressure and give -12dB. Yet, this is obviously wrong! Where is the disconnect?
How is it that pressure can be derived to be a function of radius-squared, which explains the baffle step loss, but conflicts with basic knowledge that it is a function of radius (double measurement distance, -6dB)?
Again, my apologies for being dense. I am sure that this is me making a stupid oversight somewhere (which is why this is bugging the hell out of me).
Hi Dave,
That is a consequence of the physics I described. I'll give you that for free space P <> ZoQ/4Pi and for the IB P<>ZoQ/2Pi, where Q is the source strength, and thus the radiation impedance seen by the source, Zs, doubles. I should have said "characteristic" instead of "specific" impedance. Zo is the characteristic impedance of air and is constant.
My explanation addresses why Zs doubles for 1/2 space. But it is still that the pressure doubles. That is the end of it. The pressure doubles thus the SPL is 6dB greater. There is no 3dB because Zs doubles and another 3dB because of wrapping around. In 2Pi there is a 3dB increase in pressure (factor of 2) compared to 4Pi, which is a 6dB increase in SPL (factor of 4).
Efficiency, on the other hand relates to power out divided by power in. Radiated power requires integrating the intensity over the radiating area; a sphere in 4Pi space or a hemisphere in 2Pi space. In 2 Pi space you have 4 times the intensity (I<>P^2) over 1/2 the area or twice radiated power compare to 4Pi space for the same power in, thus a 3dB increase in efficiency. It's all just dobling the pressure over 1/2 the area.
The thing about basic text books is that they tell you how P relates to Q in 2Pi and 4Pi space without bothering to tell you why one goes like 1/2PI and the other goes like 1/4PI.
It seems clear now why pressure changes by a factor of two when changing between half- and free-space. Area doubles, pressure halves (-6dB), and the math is easy to demonstrate.
Using the same rules, how do we get a halving of sound pressure with a doubling of measurement distance, when doubling the distance quadruples the area? How can that be mathematically demonstrated without conflicting with the previously demonstrated math?
Thanks. That seems to be the only lingering question.
Using the same rules, how do we get a halving of sound pressure with a doubling of measurement distance, when doubling the distance quadruples the area? How can that be mathematically demonstrated without conflicting with the previously demonstrated math?
Thanks. That seems to be the only lingering question.
john k said:
I think that we are agreeing more than disagreeing. Yes the level is 6dB greater, the curves all show that, and so the pressure must have doubled. If you also agree that the efficiency doubled, that explains (and must account for) 3dB of it. My explanation of efficiency doubling is the woofer imparting the same volume velocity into twice the radiation resistance. If you explanation makes sense to you, that is fine as long as we agree that radiated power has doubled.
All we need is an explanation of the other 3 and it seems obvious that constraining rear radiation to the front nicely deals with that.
David S.
POWER is constant in the discussion. WHERE the energy goes (or its distribution, rather) is varying when we switch from half- to free-space, but the source is constrained to radiating constant power. Maybe you meant, "...that is fine as long as we agree that radiated energy is constant and energy density has doubled."
Let me explain the imaginary test setup. I am imagining a piston radiator with a small enclosure preventing it from behaving as a dipole to the surrounding environment, floating in a big open air space. It is operated at constant power input at some low frequency that gives wavelengths >> enclosure baffle. Being imaginary, I can instantly add an infinitely large, infinitely rigid baffle that is flush with the enclosure baffle.
Are you saying that the addition of the baffle doubles the acoustic resistance, meaning that twice the energy is delivered to the air (for the same input since efficiency is doubled), and hence twice the power? Does 2x acoustic resistance = 2x efficiency?
Are you saying that the addition of the baffle doubles the acoustic resistance, meaning that twice the energy is delivered to the air (for the same input since efficiency is doubled), and hence twice the power? Does 2x acoustic resistance = 2x efficiency?
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