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perrogrande007 
Member - Posts: 9
Member spacespace
Joined: October 17, 2004
Location: United States
Posted: October 17, 2004 at 4:43 PM / IP Logged  
I'm putting 4 12WO's in the back of my Jeep Wrangler.  If I can, I might get in 6 of them.  My question has to do with polyfill for the box.  JL says that the optimum box is 1.5ft^3 per speaker....I have about 10ft^3 to play with, maybe a little more, but I would like to use as little space as possible.  If I make a sealed box, what is the smallest box I can make per speaker (going to make separate compartments) and how much polyfill can I use to make each speaker "see" 1.5ft^3???  Can I do something as small as .75 per sub and make up the difference in polyfill?  I have heard you can also use fiberglass in place of the fill, is this correct?  Am I going to have a change in SQ?  I want it to hit hard, but I am a stickler for SQ.  I have a 12" RF Power  in there right now and it hits really hard, but sounds like crap, so it's got to go.
boxmaker85 
Silver - Posts: 433
Silver spacespace
Joined: September 19, 2004
Location: United States
Posted: October 17, 2004 at 6:06 PM / IP Logged  

I would personally use the smallest box that JL recomends (1 cu. ft.) and stuff that w/ polyfill (don't use fg it's too messy and polyfill does a better and easier job).  That will let the sub "see" about 1.3 (ish give or take .1 cu ft) cu ft which is close to the 1.5 optimal while still remaining in the advised range.  That will still keep sq and hit hard.

Also, you have a wrangler?  And you have how much room to play with?  10 cu ft?  Try measuring that again.  My dad just bought one and cargo space is horible even w/ the back seat out.  Good luck putting 6 12" (or even 4 for that matter) in the back.  But if you can I think that you'll be the first.

perrogrande007 
Member - Posts: 9
Member spacespace
Joined: October 17, 2004
Location: United States
Posted: October 17, 2004 at 11:51 PM / IP Logged  

Well, the back seat is out, and I have more than 30X30X18 usable space...I can actually get 35X35X15 or so...so that makes 18375 in^3, which is 18375/1728=>10ft^3....or is my math a little fuzzy?


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