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Just wanted clarification on a few things. Woofers are constant acceleration devices within their passband. Which means that any woofer, regardless of its moving mass, will move at the same speed at a given frequency and SPL. SPL is acceleration and frequency is analogous to velocity (speed)  increase SPL you increase acceleration and if you reduce frequency you increase velocity.
I've known for sometime that mass has nothing to do with woofer speed  it only affects efficiency and driver Fs. I just want to brush up on the basics again. I think I'm a little rusty. Please correct me if I'm wrong on the above.
I've known for sometime that mass has nothing to do with woofer speed  it only affects efficiency and driver Fs. I just want to brush up on the basics again. I think I'm a little rusty. Please correct me if I'm wrong on the above.
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Different woofers will have more or less excursion for the same SPL depending upon the characteristics of the driver and enclosure. The frequency will not change, but the applied signal level (amplitude and thus slew rate) will change for different subs (or just drivers) for the same output SPL.
I have no idea what "if you reduce frequency you increase velocity" means. Something seems amiss with that statement...
I have no idea what "if you reduce frequency you increase velocity" means. Something seems amiss with that statement...
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Quote:
Originally Posted by DonH50
Different woofers will have more or less excursion for the same SPL depending upon the characteristics of the driver and enclosure. The frequency will not change, but the applied signal level (amplitude and thus slew rate) will change for different subs (or just drivers) for the same output SPL.
Different woofers will have more or less excursion for the same SPL depending upon the characteristics of the driver and enclosure. The frequency will not change, but the applied signal level (amplitude and thus slew rate) will change for different subs (or just drivers) for the same output SPL.
I'm not discussing excursion. Let me ask again, will two woofers move at the same speed assuming a given SPL and frequency, or are you suggesting that different woofers will move at different speeds to handle the same SPL and frequency?
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I have no idea what "if you reduce frequency you increase velocity" means. Something seems amiss with that statement...
Velocity is inversely proportional to frequency. Drop frequency, increase velocity. Increase frequency, drop velocity.
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No, because two woofers cannot be the same nor cannot exist in the same space at the same time.
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So a 12 inch woofer will move at a different speed assuming, let's say, 50 Hz at 90 dB, vs an 18 inch woofer handling 50 Hz at 90 dB?
Is that what you're saying?
Is that what you're saying?
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The physical speed at which a cone moves is determined by the combination of the excursion and the frequency being produced. If that excursion is, for instance, 10mm, and the frequency is 100Hz, the cone must move 10mm 100 times per second. You can do the remaining math to calculate the speed required to do so.The cones of two similar woofers with different sensitivities will move at different speeds to reach the same SPL, because as sensitivity goes up excursion goes down to reach the same SPL. The same applies to drivers with different cone areas, as when cone area goes up excursion goes down to reach the same SPL.
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I assumed all woofer cones moved at the same speed. In fact, I'm sure you mentioned something similar in another thread. When people say things like an 8 inch cone can move faster than the 12 inch, due to a lighter cone, I don't find it the least bit credible. I assumed that within their bandwidthlimited region, all woofers were constant acceleration devices.
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I'd suggest a course in physics and EE.
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I thought from the title the title it was about the old Wiggins (Adire) paper.
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I don't agree. Volume displacement determines SPL at any given frequency, so two drivers of similar cone area will have the same excursion requirement for a given SPL and frequency. Increased sensitivity only means one will use less power than the other to produce that SPL.
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@A9X308: What about woofer (piston) area? That is one thing I had in mind, as you have stated, that makes the excursion vary for different woofers at the same SPL. Yeah, if they are the same area, in the same enclosure, but that was not my assumption. I do not know if it was the OP's assumption.
@goneten: For a given frequency two different woofers will cycle at the same rate, but the magnitude of the cycle (woofer excursion) required to reach the same SPL depends on woofer and enclosure characteristics. If one woofer requires greater excursion to reach the same SPL, then it will exhibit higher peak slew rate than one that moves less distance.
As for velocity, for the same environmental conditions the propagation velocity of sound is the same and independent of frequency (to first order over the audio range). If you are talking about the velocity of the cone, that (peak slew rate) depends upon the amplitude (and where in the cycle, but I assume you mean the maximum slew rate of the cone?) Those are two very different things. I think I am not understanding your question or terminology or both.
I have texts that discuss all this in detail but they are from grad classes 30'ish years ago, I don't remember all the details, and they are far from a simple read.
@goneten: For a given frequency two different woofers will cycle at the same rate, but the magnitude of the cycle (woofer excursion) required to reach the same SPL depends on woofer and enclosure characteristics. If one woofer requires greater excursion to reach the same SPL, then it will exhibit higher peak slew rate than one that moves less distance.
As for velocity, for the same environmental conditions the propagation velocity of sound is the same and independent of frequency (to first order over the audio range). If you are talking about the velocity of the cone, that (peak slew rate) depends upon the amplitude (and where in the cycle, but I assume you mean the maximum slew rate of the cone?) Those are two very different things. I think I am not understanding your question or terminology or both.
I have texts that discuss all this in detail but they are from grad classes 30'ish years ago, I don't remember all the details, and they are far from a simple read.
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Quote:
Originally Posted by DonH50
As for velocity, for the same environmental conditions the propagation velocity of sound is the same and independent of frequency (to first order over the audio range). If you are talking about the velocity of the cone, that (peak slew rate) depends upon the amplitude (and where in the cycle, but I assume you mean the maximum slew rate of the cone?) Those are two very different things. I think I am not understanding your question or terminology or both.
As for velocity, for the same environmental conditions the propagation velocity of sound is the same and independent of frequency (to first order over the audio range). If you are talking about the velocity of the cone, that (peak slew rate) depends upon the amplitude (and where in the cycle, but I assume you mean the maximum slew rate of the cone?) Those are two very different things. I think I am not understanding your question or terminology or both.
Slew rate of the cone? The deeper a driver plays the same SPL, the faster it moves. If you keep SPL fixed but drop the frequency, you increase cone velocity. I'm not sure I agree with you that velocity depends on amplitude. If amplitude is fixed the acceleration of the cone is fixed  if you drop an octave in frequency from 80 Hz 40 Hz, the velocity must double as the duration in time for each cycle has doubled. Velocity is an integral of acceleration.
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Volume displacement is the area of the cone times excursion: bore x stroke if you will.
If you had a 12" sub and an high eff pro speaker with the same Sd both in an infinite baffle, to produce the same SPL at the same frequency within their passbands, they would be showing the same excursion. The difference is one would need 1015dB less power to do it.
The equation for sound pressure (in pascals) is
p=U*rho0*f / (2*r)
where U is the volume flow in m3/s
rho0=1.2 kg/m3
f is the frequency in Hz
r is the distance
U is also equal to V*2*pi*f, where V is the volume of air displaced in m3.
To get from sound pressure (p) to sound pressure level (SPL)
SPL = 20*log10(p/pref);
where pref=0.00002 Pa
If you had a 12" sub and an high eff pro speaker with the same Sd both in an infinite baffle, to produce the same SPL at the same frequency within their passbands, they would be showing the same excursion. The difference is one would need 1015dB less power to do it.
The equation for sound pressure (in pascals) is
p=U*rho0*f / (2*r)
where U is the volume flow in m3/s
rho0=1.2 kg/m3
f is the frequency in Hz
r is the distance
U is also equal to V*2*pi*f, where V is the volume of air displaced in m3.
To get from sound pressure (p) to sound pressure level (SPL)
SPL = 20*log10(p/pref);
where pref=0.00002 Pa
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I'm pretty sure I learned the velocity/acceleration of the cone as a function of frequency/SPL from Dan Wiggins. I'll see if I can dig up some posts he made on this.
I think we can all agree that mass has nothing to do with woofer speed. But if we take an 18" cone and an 8" cone, assuming identical excursion, the 8" cone will need to move further in the same amount of time to reproduce the same SPL. So an 8" cone can't possibly move faster than a 18" cone assuming identical SPL and frequency, otherwise you've changed the frequency.
If I remember correctly, Dan Wiggins used to say that the fastest sub in terms of physics, is the one that can play the lowest, the loudest.
I think we can all agree that mass has nothing to do with woofer speed. But if we take an 18" cone and an 8" cone, assuming identical excursion, the 8" cone will need to move further in the same amount of time to reproduce the same SPL. So an 8" cone can't possibly move faster than a 18" cone assuming identical SPL and frequency, otherwise you've changed the frequency.
If I remember correctly, Dan Wiggins used to say that the fastest sub in terms of physics, is the one that can play the lowest, the loudest.
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Try this http://www.linkwitzlab.com/SPL%20nomographs.htm
For a fixed SPL & fixed effective diameter the acceleration is constant. This means velocity and hence displacement drop with increasing frequency. Two different sized transducers will have different accelerations to produce the same SPL.
For a fixed SPL & fixed effective diameter the acceleration is constant. This means velocity and hence displacement drop with increasing frequency. Two different sized transducers will have different accelerations to produce the same SPL.
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Hi Goneten,
Now, I have certainly presented this in overly simple terms  the true behavior, especially the relationship to SPL, is more complex then I have described.
Quote:
Sorry, but I believe you've said that backwards, probably inadvertently. Ignoring SPL, if the excursion is a fixed value x, then the cone needs to travel a total distance of 2*x in one cycle. If you drop the frequency in half, the cone will need to travel that same 2*x distance in twice the time. For 80 Hz, the velocity would average (80*2*x), or 160*x per second. for 40 Hz, the velocity would average (40*2*x), or 80*x per second. If you double the frequency, you double the average velocity.Quote:
Again, if the duration in time has doubled, the velocity will be half.. . . if you drop an octave in frequency from 80 Hz 40 Hz, the velocity must double as the duration in time for each cycle has doubled.
Quote:
If you increase the amplitude, you are increasing the distance that the cone need to traverse. If the frequency does not change, then the velocity needs to increase to traverse the increased distance in the same amount of time.I'm not sure I agree with you that velocity depends on amplitude.
Quote:
Put another way, velocity is the derivative of position, and acceleration is the derivative of velocity. Discounting harmonic distortion in the speaker, if you are outputting a sine wave, then the cone's position, velocity and acceleration are all sine waves, each 90 degrees different in phase from the previous derivative. So if both amplitude and frequency are fixed, then the acceleration of the cone is fixed. As frequency increases, velocity and acceleration increase proportionately. If amplitude increases, velocity and acceleration increase proportionately as well.. . . If amplitude is fixed the acceleration of the cone is fixed . . .
. . . Velocity is an integral of acceleration.
. . . Velocity is an integral of acceleration.
Quote:
Yes, he would be correct. I think it is simply the direction of the relationship that you have wrong.I'm pretty sure I learned the velocity/acceleration of the cone as a function of frequency/SPL from Dan Wiggins.
Quote:
Yes, the mass only effects the power needed to reach that SPL. And the 8 inch cone does need to move further in order to match the SPL of the 18 inch cone.I think we can all agree that mass has nothing to do with woofer speed. But if we take an 18" cone and an 8" cone, assuming identical excursion, the 8" cone will need to move further in the same amount of time to reproduce the same SPL.
Quote:
Again, you have the sign inverted. The 8 inch cone must move faster since it needs to move further in the same amount of time.So an 8" cone can't possibly move faster than a 18" cone assuming identical SPL and frequency, otherwise you've changed the frequency.
Now, I have certainly presented this in overly simple terms  the true behavior, especially the relationship to SPL, is more complex then I have described.
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Quote:
Originally Posted by MarkHotchkiss
Hi Goneten,
Sorry, but I believe you've said that backwards, probably inadvertently. Ignoring SPL, if the excursion is a fixed value x, then the cone needs to travel a total distance of 2*x in one cycle. If you drop the frequency in half, the cone will need to travel that same 2*x distance in twice the time. For 80 Hz, the velocity would average (80*2*x), or 160*x per second. for 40 Hz, the velocity would average (40*2*x), or 80*x per second. If you double the frequency, you double the average velocity.
Hi Goneten,
Sorry, but I believe you've said that backwards, probably inadvertently. Ignoring SPL, if the excursion is a fixed value x, then the cone needs to travel a total distance of 2*x in one cycle. If you drop the frequency in half, the cone will need to travel that same 2*x distance in twice the time. For 80 Hz, the velocity would average (80*2*x), or 160*x per second. for 40 Hz, the velocity would average (40*2*x), or 80*x per second. If you double the frequency, you double the average velocity.
Hi Mark! Thanks for chiming in. With regards to the above, then I'm not sure what Dan is talking about, because I've read a couple of posts from him where he indeed claims that acceleration is tied to SPL and velocity is tied to frequency which is fine  however, he says that the lower and louder you go, from a physics standpoint, the faster the operation of said woofer. That, at a given SPL, velocity increases as frequency decreases. I'll be happy to send you links to where he said this.
Quote:
Again, if the duration in time has doubled, the velocity will be half. If you increase the amplitude, you are increasing the distance that the cone need to traverse. If the frequency does not change, then the velocity needs to increase to traverse the increased distance in the same amount of time.
Makes sense.
Quote:
Yes, he would be correct. I think it is simply the direction of the relationship that you have wrong.
The thing is, I've always known that the higher the frequency, the faster the cone has to move  velocity has to increase. Sounds basic. I've replied on this subject before many times, and I would like to think I had a handle on it then. What I'm saying above is based on what Dan has said on the subject of woofer operation, so I may be misunderstanding Dan then, but if so, then I have no idea what he's talking about.
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For a given frequency and SPL, the woofer with the smaller radiating area has to move further, and therefore faster.
Quote:
In fact, I'm sure you mentioned something similar in another thread. When people say things like an 8 inch cone can move faster than the 12 inch, due to a lighter cone, I don't find it the least bit credible.
I agree with you, because woofer cones follow that wellknown rule of Newtons, which says that F = MA. If you want more A (acceleration which implies higher velocity), just increase the F (force). Increasing the force is simply a matter of forcing more current through the voice coil.
There is no rule written in stone that says that an 8 inch cone is lighter cone than a 12 inch cone. In fact the 12 inch cone on a socalled full range speaker can easily be lighter than the cone on an 8 inch subwoofer.
The hidden agenda in all this is how displacement varies with frequency and constant SPL.
http://www.alpsadriaacoustics.org/archives/Full%20Papers/Uskokvic_The%20Use%20of%20Optocouplers%20in%20Measuring%20Loudspeaker%20Cone%20Displacement.pdf
Please notice that displacement is proportional to the inverse of frequency squared, and for a 16 cm diaphragm displacement required for 100 dB SPL becomes impossibly large around 75 Hz.
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I know. I can't disagree with that as it makes perfect sense. My earlier replies on this from 2006() basically echoed the above. Like I said, something that Dan posted is confusing the hell out of me, so perhaps you can figure out what Dan is saying here, as I respect Mr Wiggins on the subject of woofer operation :
http://www.audiogroupforum.com/csforum/showthread.php?t=2824
Quote:
Originally Posted by Dan Wiggins
Actually, from a physics standpoint, the LOWER you play, the faster you are... Drivers are a constant acceleration device; BLi=ma, and since a driver is a power transformer (electrical power in, acoustic power out), we are transforming current (power) in to audio (acceleration) out. Hence constant acceleration, with a sclar constant (BL/m) relating the two.
Now, drop your woofer down in frequency, say one octave, and keep the SPL the same. What happens? Acceleration stays constant (same SPL), but velocity increases (integral of acceleration). Why does velocity increase? You're integrating acceleration over twice the time (drop one octave means halving frequency, which means twice the duration in time for each cycle). So velocity doubles. And of course, if you look at excursion, you get the wellknown quadrupling of excursion for every octave you go down (integral of velocity is position, which is excursion, and that means it is quadrupled as you drop an octave).
So technically, to the physicist, the fastest woofer would be the one that can play the lowest the loudest. Speed  velocity  increases for all drivers as you drop in frequency and maintain the same SPL.
Now, if you're talking transient response, then it's an inductance issue you want to look at. But barring inductance (say the crossover for the woofer is well below the inductive rolloff of the driver), speed is really an issue of who plays lower louder.
Dan Wiggins
Adire Audio
Actually, from a physics standpoint, the LOWER you play, the faster you are... Drivers are a constant acceleration device; BLi=ma, and since a driver is a power transformer (electrical power in, acoustic power out), we are transforming current (power) in to audio (acceleration) out. Hence constant acceleration, with a sclar constant (BL/m) relating the two.
Now, drop your woofer down in frequency, say one octave, and keep the SPL the same. What happens? Acceleration stays constant (same SPL), but velocity increases (integral of acceleration). Why does velocity increase? You're integrating acceleration over twice the time (drop one octave means halving frequency, which means twice the duration in time for each cycle). So velocity doubles. And of course, if you look at excursion, you get the wellknown quadrupling of excursion for every octave you go down (integral of velocity is position, which is excursion, and that means it is quadrupled as you drop an octave).
So technically, to the physicist, the fastest woofer would be the one that can play the lowest the loudest. Speed  velocity  increases for all drivers as you drop in frequency and maintain the same SPL.
Now, if you're talking transient response, then it's an inductance issue you want to look at. But barring inductance (say the crossover for the woofer is well below the inductive rolloff of the driver), speed is really an issue of who plays lower louder.
Dan Wiggins
Adire Audio
And :
Quote:
Originally Posted by Dan Wiggins
Richard,
Run the math. A 1" diameter dome, moving 0.15mm one way at 2 kHz will generate ~100 dB SPL. In 500 uSec the dome travels 0.3mm total. Peak velocity will be 0.3 m/sec.
To generate 100 dB SPL at 31.25 Hz, a 10" woofer (typical Sd of 345 cm^2) will need to move around 10mm one way (20mm total). In ~32 msec, the cone travels 20mm total. Peak velocity will be 0.31 m/sec, actually HIGHER than the tweeter.
Note that as the frequency increases, and SPL stays the same, peak velocity drops. The speed drops as you increase frequency of reproduction (bolded)
Richard,
Run the math. A 1" diameter dome, moving 0.15mm one way at 2 kHz will generate ~100 dB SPL. In 500 uSec the dome travels 0.3mm total. Peak velocity will be 0.3 m/sec.
To generate 100 dB SPL at 31.25 Hz, a 10" woofer (typical Sd of 345 cm^2) will need to move around 10mm one way (20mm total). In ~32 msec, the cone travels 20mm total. Peak velocity will be 0.31 m/sec, actually HIGHER than the tweeter.
Note that as the frequency increases, and SPL stays the same, peak velocity drops. The speed drops as you increase frequency of reproduction (bolded)
There is more of this in that thread. Not only in that thread, but he has said similar things in other threads too.
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This chart shows two speakers, both driven to 100dB at 100Hz. The lighter trace is the excursion of the lower sensitivity speaker, the darker trace that of the higher sensitivity speaker. Both speakers are loaded with Eminence LAB 12 drivers.Originally Posted by A9X308
I don't agree. Volume displacement determines SPL at any given frequency, so two drivers of similar cone area will have the same excursion requirement for a given SPL and frequency. Increased sensitivity only means one will use less power than the other to produce that SPL.
I don't agree. Volume displacement determines SPL at any given frequency, so two drivers of similar cone area will have the same excursion requirement for a given SPL and frequency. Increased sensitivity only means one will use less power than the other to produce that SPL.
Bill Fitzmaurice Loudspeaker Design
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Thanks all, for the equations and explanations. A3X908  those ring a bell, might doublecheck to my old text this weekend (I kept it out where I could find it, fortunately).
Wiggins states excursion quadruples if you halve the frequency, which implies velocity must increase to cover the twice the distance in the same amount of time as before (assuming everything is linear). Arny's plot says essentially the same thing. I had forgotten that relationship.
From the first post: "Which means that any woofer, regardless of its moving mass, will move at the same speed at a given frequency and SPL." When the first post said "any woofer" my first thought was no, it depends upon area at least if not other parameters. Mass does not enter into it, except for how much power it might take to move it. Stiffness enters into the system when you start looking at distortion; one thing I remember clearly from my old text is the pictures of cone modes and the B&W laser interferometry system showing breakup in real cones that was touring about the same time (early/mid 1980's).
Wiggins states excursion quadruples if you halve the frequency, which implies velocity must increase to cover the twice the distance in the same amount of time as before (assuming everything is linear). Arny's plot says essentially the same thing. I had forgotten that relationship.
From the first post: "Which means that any woofer, regardless of its moving mass, will move at the same speed at a given frequency and SPL." When the first post said "any woofer" my first thought was no, it depends upon area at least if not other parameters. Mass does not enter into it, except for how much power it might take to move it. Stiffness enters into the system when you start looking at distortion; one thing I remember clearly from my old text is the pictures of cone modes and the B&W laser interferometry system showing breakup in real cones that was touring about the same time (early/mid 1980's).
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I don't believe a "smaller" diameter speaker will be any faster than a bigger one , surely it depends on the motor of the sub? A 12" with a humungous magnet (motor) will be "faster" than an 8" with a wimpy one...
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The mass of the cone and its "effective' mass due to the air it is moving are certainly greater with a larger cone.
This causes a time lag when a signal is being responded to. The time lag is greater when mass is greater, therefore a larger cone may accurately described as "slower" in most cases.
This is why many manufacturers of "highend' subwoofers (for example Vandersteen), use 3 or 4 smaller drivers rather than a single larger one, for faster response to the input signal.
There is also the issue of pistonic distortion.
Ideally, the cone acts as a piston. This means that the outer edge of the cone responds as quickly as the center of the cone. Unless the cone is made of solid metal, or something equally rigid, this will not happen.
The inner part of the cone normally starts to move, and the cone flexes before the outer part of the cone starts to move. This is another type of time lag and distortion. It "smears" the shape of the sound wave produced.
The solution is to make the cone very very light and very very rigid across its surface. That is nearly impossible to do, but all manufacturers have this as their goal.
Some very expensive ones use a solid thin metal cone machined from a solid piece of very light metal, which costs thousands of dollars per cone.
This causes a time lag when a signal is being responded to. The time lag is greater when mass is greater, therefore a larger cone may accurately described as "slower" in most cases.
This is why many manufacturers of "highend' subwoofers (for example Vandersteen), use 3 or 4 smaller drivers rather than a single larger one, for faster response to the input signal.
There is also the issue of pistonic distortion.
Ideally, the cone acts as a piston. This means that the outer edge of the cone responds as quickly as the center of the cone. Unless the cone is made of solid metal, or something equally rigid, this will not happen.
The inner part of the cone normally starts to move, and the cone flexes before the outer part of the cone starts to move. This is another type of time lag and distortion. It "smears" the shape of the sound wave produced.
The solution is to make the cone very very light and very very rigid across its surface. That is nearly impossible to do, but all manufacturers have this as their goal.
Some very expensive ones use a solid thin metal cone machined from a solid piece of very light metal, which costs thousands of dollars per cone.
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There is also the amp's impedance and the inductance of the coil. For woofers I agree and think the "smaller is faster" argument quite specious.
The only good documentation I recall offhand for supporting smaller cones had to do with FM distortion and was in an old Audio article, based upon an AES paper. Cone breakup/modes could also be a factor, but for decent woofers I think we're past that given their low frequency and generally good (rigid enough) cone material.
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Provide details of the rest of the WinISD simulation you posted.Also, care to show how my maths is incorrect? Or Arny's a few posts later?
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Just getting back to the physics again, did anyone read that thread I linked to? What is Dan saying?
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Originally Posted by commsysman
The mass of the cone and its "effective' mass due to the air it is moving are certainly greater with a larger cone.
This causes a time lag when a signal is being responded to. The time lag is greater when mass is greater, therefore a larger cone may accurately described as "slower" in most cases.
The mass of the cone and its "effective' mass due to the air it is moving are certainly greater with a larger cone.
This causes a time lag when a signal is being responded to. The time lag is greater when mass is greater, therefore a larger cone may accurately described as "slower" in most cases.
Bzzt  apparent violation of Newton's laws of motion which are valid as long as you stay away from substantial fractions of the speed of light.
I repeat without fear of any reasonable contradiction, subject to the constraints given above that F = MA.
If you want something to move fast, push on it harder. With modern high energy magnets, high temperature adhesives and low cost high powered amplifiers  far from mission impossible.
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Its a long thread. I can find no fault with Dan Wiggins' contributions. Please be more specific.
BTW Bill Fitzmaurice's post just shows that adding an acoustic transformer (AKA waveguide or horn) to a direct radiator changes the physics of how it operates.
The wiggly plot is probably from some kind of horn, transmission line or tapped horn enclosure, while the more gradually curved one is from a sealed box that is also quite a bit smaller, in all likelihood.
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The above is true if you demand that equal amplifier power be used.
Of course the days of SOTA 25 wpc tubed amps are long gone, so it is very practical to just find a more powerful amplifier if you want the cone to move faster.
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I'm surprised you are asking me to be specific. In that thread I even sent you quotes from Dan
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Originally Posted by Dan Wiggins
Actually, from a physics standpoint, the LOWER you play, the faster you are... Drivers are a constant acceleration device; BLi=ma, and since a driver is a power transformer (electrical power in, acoustic power out), we are transforming current (power) in to audio (acceleration) out. Hence constant acceleration, with a sclar constant (BL/m) relating the two.
Now, drop your woofer down in frequency, say one octave, and keep the SPL the same. What happens? Acceleration stays constant (same SPL), but velocity increases (integral of acceleration). Why does velocity increase? You're integrating acceleration over twice the time (drop one octave means halving frequency, which means twice the duration in time for each cycle). So velocity doubles. And of course, if you look at excursion, you get the wellknown quadrupling of excursion for every octave you go down (integral of velocity is position, which is excursion, and that means it is quadrupled as you drop an octave).
So technically, to the physicist, the fastest woofer would be the one that can play the lowest the loudest. Speed  velocity  increases for all drivers as you drop in frequency and maintain the same SPL.
Now, if you're talking transient response, then it's an inductance issue you want to look at. But barring inductance (say the crossover for the woofer is well below the inductive rolloff of the driver), speed is really an issue of who plays lower louder.
Dan Wiggins
Adire Audio
Actually, from a physics standpoint, the LOWER you play, the faster you are... Drivers are a constant acceleration device; BLi=ma, and since a driver is a power transformer (electrical power in, acoustic power out), we are transforming current (power) in to audio (acceleration) out. Hence constant acceleration, with a sclar constant (BL/m) relating the two.
Now, drop your woofer down in frequency, say one octave, and keep the SPL the same. What happens? Acceleration stays constant (same SPL), but velocity increases (integral of acceleration). Why does velocity increase? You're integrating acceleration over twice the time (drop one octave means halving frequency, which means twice the duration in time for each cycle). So velocity doubles. And of course, if you look at excursion, you get the wellknown quadrupling of excursion for every octave you go down (integral of velocity is position, which is excursion, and that means it is quadrupled as you drop an octave).
So technically, to the physicist, the fastest woofer would be the one that can play the lowest the loudest. Speed  velocity  increases for all drivers as you drop in frequency and maintain the same SPL.
Now, if you're talking transient response, then it's an inductance issue you want to look at. But barring inductance (say the crossover for the woofer is well below the inductive rolloff of the driver), speed is really an issue of who plays lower louder.
Dan Wiggins
Adire Audio
Now please read the above quote. I've bolded the parts that are relevant. You say you don't disagree with his contributions? So then please explain why people are telling me the exact opposite of what he is saying.
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