In doing some simple math comparing, I came up with a question.
Consider the following:
18" PartsExpress PA 18" woofer with Xmax of 6mm.
12" Peerless XLS 12" Woofer with Xmax of 12mm.
Now with some volume calculations it seems that the 18" will displace 2x what the XLS will even though it has half the Xmax. Plus it has a 4" voice coil VS a 2" on the XLS. So could the 18" driver be just as viable? It seems the load would be better spread out due to the larger VC, plus even though it has a much less Xmax, its displacing so much volume it takes less physical movement to make the same amount of output. They are also at a very close price point.
Other than an enourmous box (but were talking home audio application), would the 18" driver be a good (not asking better 😉 ) choice?
Thanks for any help even if this turns into a smaller drivers have faster bass debate.
Consider the following:
18" PartsExpress PA 18" woofer with Xmax of 6mm.
12" Peerless XLS 12" Woofer with Xmax of 12mm.
Now with some volume calculations it seems that the 18" will displace 2x what the XLS will even though it has half the Xmax. Plus it has a 4" voice coil VS a 2" on the XLS. So could the 18" driver be just as viable? It seems the load would be better spread out due to the larger VC, plus even though it has a much less Xmax, its displacing so much volume it takes less physical movement to make the same amount of output. They are also at a very close price point.
Other than an enourmous box (but were talking home audio application), would the 18" driver be a good (not asking better 😉 ) choice?
Thanks for any help even if this turns into a smaller drivers have faster bass debate.
Well, I don't know which Parts Express driver you are talking about. Their Dayton line does not have an 18 incher, so I suppose you are talking about another brand that Parts Express happens to carry.
The Peerless XLS has a cone area of 466 sq cm. The Selenium 18 incher carried by Parts Express has an Sd of 1190 sq cm. So the 18 incher moves 27% more air, or about 3 dB higher SPL.
The big thing is, though, that most 18 inchers are PA type speakers, not intended for deep bass. They are intended to give big SPL in the bass range of 50 Hz on up. If you can find an 18 incher with the Thiele-Small parameters that can go deep in the size box you don't mind building, then go for it.
Most of the time, though, if you want deep bass, you will probably end up with a 12 incher or, at most, a 15 incher.
The Peerless XLS has a cone area of 466 sq cm. The Selenium 18 incher carried by Parts Express has an Sd of 1190 sq cm. So the 18 incher moves 27% more air, or about 3 dB higher SPL.
The big thing is, though, that most 18 inchers are PA type speakers, not intended for deep bass. They are intended to give big SPL in the bass range of 50 Hz on up. If you can find an 18 incher with the Thiele-Small parameters that can go deep in the size box you don't mind building, then go for it.
Most of the time, though, if you want deep bass, you will probably end up with a 12 incher or, at most, a 15 incher.
18" Dayton Profesional Driver
Fs: 25Hz
Vas: 16.12 Cu Ft 😱
Qms: 6.74
Qes: 0.36
Qts: 0.34
Seems good, Fs is similar to most good 12" drivers, but still very low.
Fs: 25Hz
Vas: 16.12 Cu Ft 😱
Qms: 6.74
Qes: 0.36
Qts: 0.34
Seems good, Fs is similar to most good 12" drivers, but still very low.
An externally hosted image should be here but it was not working when we last tested it.
Some classic errors here.
SPL does not depend on total swept volume, so you
cannot use this to compare different sized drivers.
An 18" has ~ 2.25 the cone area of a 12", in the case
of the XLS this is nearer 2.5, and for the same swept
volume will produce 4dB more output due to more efficient
air loading.
(Consider 1 driver versus 4 in series parrallel, 4 drivers
produce 6dB more output with the same input power as 1,
though the swept volume is the same in both cases.)
The difference between the drivers is probably around 7dB.
🙂 sreten.
SPL does not depend on total swept volume, so you
cannot use this to compare different sized drivers.
An 18" has ~ 2.25 the cone area of a 12", in the case
of the XLS this is nearer 2.5, and for the same swept
volume will produce 4dB more output due to more efficient
air loading.
(Consider 1 driver versus 4 in series parrallel, 4 drivers
produce 6dB more output with the same input power as 1,
though the swept volume is the same in both cases.)
The difference between the drivers is probably around 7dB.
🙂 sreten.
sreten said:
No, Sreten. Max SPL is determined by the maximum volume flow for a given (low) frequency. So if a driver has twice the Sd and half Xmax, the max SPL will be the same.
Remember, the free field sound pressure from a point source is jwU*rho0/(4*pi*r), where U is the volume flow, r is the distance and rho0=1.2 kg/m3.
It is true that if you double the area the sound pressure will double and the power will quadruple, and the efficiency will double. But that is a different story.
David,
If the 18's Qts is appropriate for your desired alignment and you aren't worried about box size, I would certainly go for the 18 over the 12 (assuming motor linearity is good, distortion is low, etc.). The positive effects of the higher efficiency would not be subtle, IMO, paying off in reduced dynamic compression and less distortion in your bass amp's output.
Big cones rock. I listen to a pair of 21-inchers. 🙂
If the 18's Qts is appropriate for your desired alignment and you aren't worried about box size, I would certainly go for the 18 over the 12 (assuming motor linearity is good, distortion is low, etc.). The positive effects of the higher efficiency would not be subtle, IMO, paying off in reduced dynamic compression and less distortion in your bass amp's output.
Big cones rock. I listen to a pair of 21-inchers. 🙂
I think what Sretan figured I was talking about sensitivity, which is given as SPL @ 1W/1M, which indeed is not determined by swept volume. However, I was talking about max output.
Hybrid:
Sorry I told you that Parts Express did not have 18 inchers in their own brand.
I went to the Parts Express home page, selected "Dayton Loudspeaker", and when the screen came up, I clicked 18" under "Woofers". I got a screen that said my search had returned no results.
Well, you have something here that will give you an F3 of 27 Hz in a ported box of 14 cu ft or so, which is pretty good for a 18" woofer. If you have the place to put the box, go for it.
One question. If you are willing to build a huge box, have you considered using two Shivas? They cost only a little more money, the box size should be 10 cu ft, the sensitivity should be 90 dB, the F3 should be much lower and the displacement for the two Shivas should be much more than for the Dayton.
Just something to think about. 🙂
Sorry I told you that Parts Express did not have 18 inchers in their own brand.
I went to the Parts Express home page, selected "Dayton Loudspeaker", and when the screen came up, I clicked 18" under "Woofers". I got a screen that said my search had returned no results.
Well, you have something here that will give you an F3 of 27 Hz in a ported box of 14 cu ft or so, which is pretty good for a 18" woofer. If you have the place to put the box, go for it.
One question. If you are willing to build a huge box, have you considered using two Shivas? They cost only a little more money, the box size should be 10 cu ft, the sensitivity should be 90 dB, the F3 should be much lower and the displacement for the two Shivas should be much more than for the Dayton.
Just something to think about. 🙂
Well as far as I'm aware SPL is swept volume and the
efficiency of air coupling (depends on diaphrapm size)
- what is wrong with the following logic ? :
A single driver is moving 10mm to produce a given SPL.
4 of the same drivers are connected in series/parallel
so they each have 1/4 of the power and move 2.5mm.
The swept volume is the same in both cases but the
4 drivers produce 6dB more SPL than the single driver.
????????
🙂 sreten.
efficiency of air coupling (depends on diaphrapm size)
- what is wrong with the following logic ? :
A single driver is moving 10mm to produce a given SPL.
4 of the same drivers are connected in series/parallel
so they each have 1/4 of the power and move 2.5mm.
The swept volume is the same in both cases but the
4 drivers produce 6dB more SPL than the single driver.
????????
🙂 sreten.
Sretan:
If the single driver is moving 10mm vs four identical drivers moving 2.5 mm, the SPL output will be the same.
However, the four drivers will be producing this output on only 1/4 the power-not 1/4 the power sent to each individual driver, one fourth the power sent to the entire unit of four drivers.
If you take an individual driver, cut the power to it by 3/4, the output of the driver will drop by 6 dB.
So four times the power equals 6 dB, for an individual driver.
One quarter the power = minus 6 dB, for an individual driver.
If the single driver is moving 10mm vs four identical drivers moving 2.5 mm, the SPL output will be the same.
However, the four drivers will be producing this output on only 1/4 the power-not 1/4 the power sent to each individual driver, one fourth the power sent to the entire unit of four drivers.
If you take an individual driver, cut the power to it by 3/4, the output of the driver will drop by 6 dB.
So four times the power equals 6 dB, for an individual driver.
One quarter the power = minus 6 dB, for an individual driver.
surely as the load impedance is the same due to the series/parallel arrangement, the overall power will then be the same, but 1/4 power to each driver.
Quick question;
When using a linkwitz transform (as provided by True Audio excel spreadsheet), is F(o) the -3db point on the original design and Q(o) is the original Qtc of the box? And F(p) is the desired -3db point and Q(p) is the desired Qtc of the transformed driver?
More importantly my real concern is F(o) the Fs of the driver itself or is it the -3db point of the driver in the particular box?
When using a linkwitz transform (as provided by True Audio excel spreadsheet), is F(o) the -3db point on the original design and Q(o) is the original Qtc of the box? And F(p) is the desired -3db point and Q(p) is the desired Qtc of the transformed driver?
More importantly my real concern is F(o) the Fs of the driver itself or is it the -3db point of the driver in the particular box?
sreten said:
It takes 1/4 of the voltage and 1/16 of the power to move 2.5 mm. Displacement is proportional to voltage, not power.
hybrid,
Fo for a Linkwitz Transform is the resonant frequency of the box, Fc. Likewise Fp is the resonant frequency that is desired.
The only case for when resonant frequency (Fo or Fp) is equal to the F3 of the speaker-in-box is when a Butterworth response is achieved at Qtc = 0.7071.
Rule of thumb:
if Qtc <0.7071 then Fc < F3
if Qtc >0.7071 then Fc > F3
Fo for a Linkwitz Transform is the resonant frequency of the box, Fc. Likewise Fp is the resonant frequency that is desired.
The only case for when resonant frequency (Fo or Fp) is equal to the F3 of the speaker-in-box is when a Butterworth response is achieved at Qtc = 0.7071.
Rule of thumb:
if Qtc <0.7071 then Fc < F3
if Qtc >0.7071 then Fc > F3
Hands up ! I admit it ! I don't know what I'm talking about !
I've been led astray by other posts on acoustic coupling.
🙂 sreten.
I've been led astray by other posts on acoustic coupling.
🙂 sreten.
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