I saw a speaker design that employs 3 subwoofers/woofers/mids in isobaric fashion and was wondering how something like this would be modeled. One driver fired forward, the second fired into the back of the first, and the last woofer fired into the back of the second while the back of the third driver fired into the main enclosure. Would doing this decrease Vas by a factor of three?
Just me, but I think the middle driver does nothing. It just moves with the air in between the front and back drivers.
The speaker I saw was a very small cabinet.
But, does having the three woofers like this allow an even smaller box to be used than using two woofers?
But, does having the three woofers like this allow an even smaller box to be used than using two woofers?
Having three speakers means three times the mass plus the mass of the air in the isobarik connections. They would all need to see the same signal, which means wiring in series or parallel. Could you use 1/3 the cabinet volume? Sure, but you also would need three times the power to get the same SPL. The volume of the woofer connections becomes significant at some point.
If you instead put those three drivers in boxes which were too small, you could equalize and get much more SPL. So IMO, any flavor of Isobarik is a waste of cone area that could be used to make sound. If you think your woofer requires isobarik, get a different woofer.
If you instead put those three drivers in boxes which were too small, you could equalize and get much more SPL. So IMO, any flavor of Isobarik is a waste of cone area that could be used to make sound. If you think your woofer requires isobarik, get a different woofer.
Im not sure, aka im speculating, but if the middle driver does not interact with the outside air(in the box also) then its compliance wouldnt be factored into vas, so the net box size would be based on two drivers plus the space wasted for the non clamshell mounting of not just one internal driver as usual but in this case two internal drivers. If this is true, eliminating the middle driver would actually make the box requirement smaller.
Where did you see this speaker?
Where did you see this speaker?
Three drivers connected in "series" acoustically would have three times the Mms, and Vas would be a third of that of the original driver's value. Sd and Qts remain the same.
If the drivers are connected in series electrically, Bl and Re would be three times the original values. If they are connected in parallel electrically Bl would remain the same, and Re would become a third of the single driver.
All assuming that the volume of air between the drivers is negible.
So, the middle driver definitely has an impact.
If the drivers are connected in series electrically, Bl and Re would be three times the original values. If they are connected in parallel electrically Bl would remain the same, and Re would become a third of the single driver.
All assuming that the volume of air between the drivers is negible.
So, the middle driver definitely has an impact.
nunayafb said:
Think of the three drivers mounted together without the box. Also consider the volumes of air between the drivers as very small and incompressible. If you push the outermost driver's cone, then this wil force all three cones to move by the same amount. In order tom move them all you need three times the force compared to pushing just one of them. Thus, the suspension is three times as stiff, and Vas is one third.
Svante, I did some more research after your post made logical sense and the people at htguide forums got into this as well(not much else on the web about it) it seems that the concensus is that the third driver does indeed reduce Vas to one third. In addition, a company whos goal is too minimize cost decided that a 3rd woofer would benefit them by minimizing wood and its associated shipping costs, I doubt they just speculated.
Heres where I got hung up:
Vas=Rho*Cms*C^2*Sd^2 (Rho - air density, C - speed of sound)
So obviously there exists a linear relationship betweent Vas and Cms, however there also exists relationship between Vas and the surrounding air density and Sd. This suggests that Vas is dependent on rho, C and Sd which we know to be true. The inner and outer driver both interact with this air, but the middle one simply follows the other two with no change in pressure anywhere around it. So I can see how Vas could be reduced by the third driver but is it exactly 1/3? Does the lower Cms only partially reduce Vas due to the lack of air interaction?
Now Im pretty sure Im wrong but I cant explain why, can you poke some holes in my logic?
Thanks
Heres where I got hung up:
Vas=Rho*Cms*C^2*Sd^2 (Rho - air density, C - speed of sound)
So obviously there exists a linear relationship betweent Vas and Cms, however there also exists relationship between Vas and the surrounding air density and Sd. This suggests that Vas is dependent on rho, C and Sd which we know to be true. The inner and outer driver both interact with this air, but the middle one simply follows the other two with no change in pressure anywhere around it. So I can see how Vas could be reduced by the third driver but is it exactly 1/3? Does the lower Cms only partially reduce Vas due to the lack of air interaction?
Now Im pretty sure Im wrong but I cant explain why, can you poke some holes in my logic?
Thanks
nunayafb said:
🙂
Well, as you say Vas=Cms*rho0*Sd²*c² . The sound pressure in the air between the drivers and also inside the box is negible compared to the atmospheric pressure. Even if it reaches 160 dB, which it can inside a too small box, the absolute pressure and thus rho0 is only modulated by about 3%.
c is largely independent on the pressure.
So, the bottom line is that rho0 really is a constant, at least with the precision that is of interest for loudspeaker building.
In fact, I'd say that the effect of the mass of air stuck between the drivers (that affect Mms) has a far more important impact. Still the effect of it is very small.
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