# how much polyfill does it take to simulate .5 Cubic feet?

Discussion in 'Audio' started by RingMaster4x4, Oct 29, 2003.

1. ### RingMaster4x41/2 ton statusPremium Member

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The title says it all. /forums/images/graemlins/thinking.gif

2. ### Greg72"Might As Well..."Staff MemberSuper Moderator

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3. ### RingMaster4x41/2 ton statusPremium Member

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Ok thanks for the link Greg. I will check it out.

4. ### RingMaster4x41/2 ton statusPremium Member

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I looked at it and I am confused. I have a box that is 2.45 CF per chamber. The driver has a Displacement - 432 cu in. There needs to be atleast 2.50 CF per chamber.. is there a formula or some thing I can use? Math is not one of my strong points. /forums/images/graemlins/thinking.gif

5. ### Greg72"Might As Well..."Staff MemberSuper Moderator

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[ QUOTE ]
I looked at it and I am confused. I have a box that is 2.45 CF per chamber. The driver has a Displacement - 432 cu in. There needs to be atleast 2.50 CF per chamber.. is there a formula or some thing I can use? Math is not one of my strong points. /forums/images/graemlins/thinking.gif

[/ QUOTE ]

Here are the basics:

2.45 CuFt = 4233 cu in
Driver Displacment = 432 cu in
Total Volume = 4233 - 432 (3801 cu in) which equal 2.2 Cu ft

Using the rule of thumb suggested, you can gain 30% more effective volume by stuffing with 1.5 Lbs/cu ft of enclosure volume....That would give you:

2.2 Cu Ft * .30 = .66 cu ft extra volume = 2.86 Cu Ft per side.

SO....you need to stuff 3.3 Lbs of polyfill into each side to get to that 2.86 Cu Ft volume.

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