wildfire99 said:
You'll likely find four 15" drivers a better choice.
There's more of them to choose from (unless you're building your own drivers)
18's are really not slower of faster than smaller drivers if they are designed correctly. The problem is, its really easy to but a big motor on 10" driver relative to its size, the same ratio required for an 18" is ungodly expensive, and that’s why good 18's are expensive.
From what I've gathered, mass doesn't affect speed, SPL does. So SPL is inversely proportional to acceleration.
If I understand it correctly now, even scaling the magnet assembly to compensate for bigger and heavier drivers aren't required to achieve the same accelerations as smaller drivers.
An 18" driver just has to move as quick as it needs to. A 12" driver needs far more excursion to achieve similar levels of acceleration.
Apparently. . .
--Sincerely,
If I understand it correctly now, even scaling the magnet assembly to compensate for bigger and heavier drivers aren't required to achieve the same accelerations as smaller drivers.
An 18" driver just has to move as quick as it needs to. A 12" driver needs far more excursion to achieve similar levels of acceleration.
Apparently. . .
--Sincerely,
A 12" driver and an 18" driver will NOT have the same cone acceleration and speed at the same SPL. SPL and acceleration may be inversely proportional, but only with a CONSTANT cone area. If you change cone area, you change the equation.
Obtained SPL and required acceleration are proportional given the same radiating area.
Required acceleration is inversely proportional to radiating area given the same SPL.
Required acceleration is inversely proportional to radiating area given the same SPL.
Eva, so then if a driver can reproduce a 20 hz tone it is moving as quick as it has to to reproduce a 20 hz tone ?
So if a 18 " driver is reproducing, say, 20hz, it is moving as quick as it needs to because if it wasn't quick enough, it wouldn't be able to reproduce the signal ?
Is that correct ?
So an 8" driver will not be able to move quicker than a 18" driver at very low frequencies because it cannot move as much air, is that correct ?
And the ability for the driver to "react" to the signal, is based on what again ?
Thank you.
--Sincerely,
So if a 18 " driver is reproducing, say, 20hz, it is moving as quick as it needs to because if it wasn't quick enough, it wouldn't be able to reproduce the signal ?
Is that correct ?
So an 8" driver will not be able to move quicker than a 18" driver at very low frequencies because it cannot move as much air, is that correct ?
And the ability for the driver to "react" to the signal, is based on what again ?
Thank you.
--Sincerely,
Farther not faster for SPL. There is a liitle known property of multiple drivers called acoustic inductance. It can, in may cases, cause two systems (subs?) with equal cone area, but with different size drivers to behave alike; have similar FR. If one has smaller drivers that one is likley to have better transient response because total cone weight is less.
Why do we keep getting this confused....
For a 8" driver to have EQUAL SPL as a 18" driver playing the SAME frequency it must move both farther AND faster. This is simple trig/calculus.
I guess you werent paying attention earlier when this was declared false. Inductance is the critical factor in transient response, not moving mass.
Moving mass is however a major player in step response(basically how fast the driver goes from Xmax to a stop). Good step response may well be the percieved "fastness".
For a 8" driver to have EQUAL SPL as a 18" driver playing the SAME frequency it must move both farther AND faster. This is simple trig/calculus.
I guess you werent paying attention earlier when this was declared false. Inductance is the critical factor in transient response, not moving mass.
Moving mass is however a major player in step response(basically how fast the driver goes from Xmax to a stop). Good step response may well be the percieved "fastness".
Is a larger woofer better?
I would personally (and do) use 18" drivers in a system, although there are arguments for and against.
As I see it, a large diameter driver is better coupled to the surrounding space - the impedance has a slightly better match - than a smaller one. This fact alone should favour the larger driver for use at the lowest frequencies, although in horn loaded applications this might be less important as the horn structure matches the differing 'impedances'.
However, from a hi-fi point of view, there is a possibility of resonances occurring due to the mass of air contained within the deep cone of a large driver, and this could colour the sound somewhat, not that I've ever noticed this phonomenon, and presumably by crossing over below this resonance it could be avoided anyhow.
I'll always favour a large driver in a large cabinet - "Ye cannae chenge the laws of phezziks captain..."
As for why there aren't many 18" (hi-fi) drivers available, it's probably because big is no longer beautiful, as it was in the late 70s. Fashions change.
I would personally (and do) use 18" drivers in a system, although there are arguments for and against.
As I see it, a large diameter driver is better coupled to the surrounding space - the impedance has a slightly better match - than a smaller one. This fact alone should favour the larger driver for use at the lowest frequencies, although in horn loaded applications this might be less important as the horn structure matches the differing 'impedances'.
However, from a hi-fi point of view, there is a possibility of resonances occurring due to the mass of air contained within the deep cone of a large driver, and this could colour the sound somewhat, not that I've ever noticed this phonomenon, and presumably by crossing over below this resonance it could be avoided anyhow.
I'll always favour a large driver in a large cabinet - "Ye cannae chenge the laws of phezziks captain..."
As for why there aren't many 18" (hi-fi) drivers available, it's probably because big is no longer beautiful, as it was in the late 70s. Fashions change.
Until a single 12" can provide this cleanly:
An externally hosted image should be here but it was not working when we last tested it.
then yes, a larger driver is better for subwoofer use 😀
Hi,
it's not even as difficult as that
For a given SPL and frequency a big cone will need to move a certain distance to displace enough air.
A half size cone of quarter the area will need to move 4 times as far to displace the same volume of air.
If the cone is moving 4 times as far it is effectively moving 4 times as fast and will need 4 times the accelaration to keep up with the SPL of the twice as big driver.
Fortunately the half size cone will be about a quarter of the weight (or maybe a bit/lot less) so at 4 times the acceleration and a quarter of the weight the forces involved turn out to be the same. If the small cone is lighter than a quarter then the motor power could be less since it will have to generate less force.
All that and not a calculator in sight nor even much mental arithmetic. Sometimes the clever dicks don't need much brain power.
it's not even as difficult as that
simple arithmetic with a little bit of algebra thrown in will solve it.
For a given SPL and frequency a big cone will need to move a certain distance to displace enough air.
A half size cone of quarter the area will need to move 4 times as far to displace the same volume of air.
If the cone is moving 4 times as far it is effectively moving 4 times as fast and will need 4 times the accelaration to keep up with the SPL of the twice as big driver.
Fortunately the half size cone will be about a quarter of the weight (or maybe a bit/lot less) so at 4 times the acceleration and a quarter of the weight the forces involved turn out to be the same. If the small cone is lighter than a quarter then the motor power could be less since it will have to generate less force.
All that and not a calculator in sight nor even much mental arithmetic. Sometimes the clever dicks don't need much brain power.
well, you need to have weight to get low Fs, and in a big cone you use the weight as an active element, extra weight in a small cone is just dead weight.
Bass from a big cone simply sound different than bass from a smaller cone, doesnt it 😉
Bass from a big cone simply sound different than bass from a smaller cone, doesnt it 😉
I'll take that as a compliment.... thank you
You really must use calculus to PROVE how much greater the acceleration and speed will be for a smaller driver. How else are you going to derive velocity from position and acceleration from velocity???? Of course the answer is easy to menatlly guess, but as reasonable as the guess may sound the correct math is a much more accurate way to do things.
Hi Bassawdy,
no.
The cone/motor velocity, whether averaged or peak or rms, will be directly proportional to displacement and to frequency. There is no calculus nor complicated maths.
Double the displacement and you have double the velocity.
Quadruple the displacement and you have quadruple the velocity.
Half the piston size and you have quarter the piston area. Quadruple the displacement and quarter the area gives the same volumetric displacement and requires quadruple the velocity. No guessing required.
no.
The cone/motor velocity, whether averaged or peak or rms, will be directly proportional to displacement and to frequency. There is no calculus nor complicated maths.
Double the displacement and you have double the velocity.
Quadruple the displacement and you have quadruple the velocity.
Half the piston size and you have quarter the piston area. Quadruple the displacement and quarter the area gives the same volumetric displacement and requires quadruple the velocity. No guessing required.
So was I correct in my assessment on page 15 (last post) ? I would like for someone to correct me if I'm wrong.
A driver with a high mass will give you less efficiency but a lower Fs. Which means lower LF extension in a given box. But moving mass has nothing to do with acceleration (in terms of reproducing deep bass frequencies) because bass is slow, by definition, and does not require instantaneous acceleration.
And if a driver can reproduce 40hz then it is moving as quick as it needs to (meaning that it does not need to move any quicker) to reproduce that 40 hz. That goes for any deep bass frequency.
And we don't even have to go as far as using big car/small car analogies. An 18" driver can move as quick as an 8" driver in it's respective bandwidth.
SPL is acceleration. And acceleration is velocity ? And inductance is a measure of the current changes which translate to how the cone will react to movement ?
And current is a function of transient response too, because as the Adire paper said, if you had infinite current flowing through the coils, there would be infinite transients.
Do I have this right ? Please, those more knowledgable than me, correct me if I'm wrong on these points.
And finally, an 8" driver may move quickly, but that's only because it NEEDS to move quickly to move a sufficient amount of air to reproduce a given bass frequency. An 18" driver does not require high cone excursions to reproduce the same bass frequency.
So an 18" driver will have a better chance at reproducing those frequencies.
Am I correct ? I think I'm correct but I could be wrong. 🙂
--Sincerely,
A driver with a high mass will give you less efficiency but a lower Fs. Which means lower LF extension in a given box. But moving mass has nothing to do with acceleration (in terms of reproducing deep bass frequencies) because bass is slow, by definition, and does not require instantaneous acceleration.
And if a driver can reproduce 40hz then it is moving as quick as it needs to (meaning that it does not need to move any quicker) to reproduce that 40 hz. That goes for any deep bass frequency.
And we don't even have to go as far as using big car/small car analogies. An 18" driver can move as quick as an 8" driver in it's respective bandwidth.
SPL is acceleration. And acceleration is velocity ? And inductance is a measure of the current changes which translate to how the cone will react to movement ?
And current is a function of transient response too, because as the Adire paper said, if you had infinite current flowing through the coils, there would be infinite transients.
Do I have this right ? Please, those more knowledgable than me, correct me if I'm wrong on these points.
And finally, an 8" driver may move quickly, but that's only because it NEEDS to move quickly to move a sufficient amount of air to reproduce a given bass frequency. An 18" driver does not require high cone excursions to reproduce the same bass frequency.
So an 18" driver will have a better chance at reproducing those frequencies.
Am I correct ? I think I'm correct but I could be wrong. 🙂
--Sincerely,
Hi,
I'm not sure what you mean by this;
This
I'm not sure what you mean by this;
This
is saying that at the SAME frequency then the small speaker MUST travel faster than the big speaker to produce the same SPL.
How about the "Magnat Aggressor 5000 Death Match" 20" subwoofer
DIN Durchmesser 500
Ulower cut-off frequency 16 Hz
upper cut-off frequency 1500 Hz
Nominal Power handling 750 Watt
Max. Power handling 2000 Watt
Impedance 3 Ohm
soundpressure level 97 dB
Fs 24
Vas 324
Qts 0,43
Qes 0,59
Qms 1,6
Xmax 21
Link
DIN Durchmesser 500
Ulower cut-off frequency 16 Hz
upper cut-off frequency 1500 Hz
Nominal Power handling 750 Watt
Max. Power handling 2000 Watt
Impedance 3 Ohm
soundpressure level 97 dB
Fs 24
Vas 324
Qts 0,43
Qes 0,59
Qms 1,6
Xmax 21
Link
An externally hosted image should be here but it was not working when we last tested it.
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