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#1 |
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diyAudio Member
Join Date: Dec 2001
Location: Seoul Korea
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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. |
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#2 |
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Warp Engineer
On Holiday
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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.
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#3 |
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diyAudio Member
Join Date: Sep 2001
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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 |
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#4 |
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Warp Engineer
On Holiday
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I stand corrected. Thanks for that.
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#5 |
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diyAudio Member
Join Date: Dec 2001
Location: Seoul Korea
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thak you for your information...
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#6 |
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diyAudio Member
Join Date: Dec 2001
Location: Seoul Korea
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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 |
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#7 |
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diyAudio Member
Join Date: Dec 2001
Location: Seoul Korea
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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] |
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#8 |
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Warp Engineer
On Holiday
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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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