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-   -   In push/pull woofer mounting, Why Vas will be half that of single driver? (http://www.diyaudio.com/forums/multi-way/1373-push-pull-woofer-mounting-why-vas-will-half-single-driver.html)

haggy 6th December 2001 06:25 AM

I can't understand why Vas will be half that of single driver in push/pull woofer mounting system.
If anyone knows or have some info to share, please advise.

Thank you very much.

AudioFreak 6th December 2001 11:02 AM

Vas will be half because the air trapped between the 2 cones is at constant pressure..... this means the 2 drivers act as 1 driver with a cone of twice the weight which therefore has 1/2 the Vas requirement.

AndrewJ 6th December 2001 03:28 PM

A correction to AudioFreak's post. He is correct that the mass is doubled over a single unit, but it isn't the mass that determines Vas (he may be confusing Vas with Vb, the box volume requirement). It is the stiffness of the spider plus surround. Since there are now two drivers intimitely coupled but with the same surface area as a single driver, their effective stiffness is doubled, so the Vas, which is defined as the volume of air having the same stiffness as the driver, will be halved.
Andrew

AudioFreak 6th December 2001 10:12 PM

I stand corrected. Thanks for that.

haggy 7th December 2001 01:42 AM

thak you for your information...

haggy 8th December 2001 03:46 AM


Vas(single driver)=1.4*10^5*Cms*Sd^2
and,
C=1/k(stiffness)
effective stiffness is doubled, so the Cms will be Cms/2.
so, Vas'(push/pull)=1.4*10^5*Cms/2*Sd^2=Vas/2
but, why is Sd same that of sinle driver. in entire system, two didrivers are driving. so will Sd be doubled?

and
I don't know why Qts will be same that of single driver too.
Qms=(1/D)*(Mms/Cms)^1/2
Qes=[Re/(Bl)^2]*(Mms/Cms)^1/2
Qts=Qms*Qes/(Qms+Qes)
So, for being the same single driver's Qts,I think Mms must be doubled and the other parameters are same that of single driver for being the same single driver's Qts.
but.. why?

I don't know the reason....

If anyone knows these reason.. Please advice

thanks

haggy 10th December 2001 02:10 AM

Sorry.. I misunderstand that problem!
 
sorry...
I misunderstanded it.
In push pull type, Vas will be twice.
In compunded woofer system, Vas will be same of half of singlee driver.
because, in push pull type, Cms will be half and Sd is twice of single driver.
so
Vas'=1.4*10^5*Cms/2*(2Sd)^2 =2Vas
and in compunded type, Cms will be half and Sd is same
of single driver.
so,
Vas'=1.4*10^5*Cms/2*Sd^2=Vas/2

but I can'nt understand that Qts will be same of single driver.

somebody help me... please

[Edited by haggy on 12-10-2001 at 01:29 AM]

AudioFreak 10th December 2001 02:29 AM

In any isobaric/compound configuration (clamshell, magnet to magnet, cone to magnet) Vas will be half that of a single driver. Sd will be that of a single driver, stiffness is doubled, Re is dependant on how the drivers are wired (series or parellel), Mms is doubled..... and there are others ........ basicly, almost every parameter changes.

If the drivers are simply mounted on opposite panels of the enclosure and operating in push/pull config then Vas will be double that of a single driver.

[Edited by AudioFreak on 12-10-2001 at 02:09 AM]


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