posted
Hay there I just started reading up on the abc box. Way cool. Two questions. 1. Dose a guy just use the anufacturs recamended box and port sizes. 2. Is the midle port all on one side or is it split 1/2 and 1/2 like in the photo below? I might just try this with my 8 Vega 10" if you think that would rock. I think I have the room with my new Mazda B2200 with the full walk through cab. Sorry 3rd Question are both ports on the same face as the woofer?
posted
You have it illustrated perfectly - it will perform best this way.
You can use the manufacturers recommended sizes if they give you a range to work in and you go right to the top of the range. If not, you need to use computer modelling to get it right. Also, don't forget that you need to figure ports using a pair, not just one.
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And that's the bottom line, cuz: LORD DUKK SAYS SO!
The Big Show The Dukk says: Know your bass: PORT your damn box!! Have HoleSaw, Will Travel!
posted
Ok I'm dum Length for a box with two 4" ports? How dose number of ports afect box lenght. He is what I fell I nead. Box 1.5cu.ft w/dispacment of diver and 1 and 1/2 ports displacment figured in. Then one box attached to that at .75cu. ft. w/ 1.5 port dispacement figured in. I don't have the vega book in front of me but when I do I will use real numbers.
Just incase you didn't know. When you want to use more than on port in a box you cut the volume of the box into the number of ports you want to use. If you have 2 cubes of box and want 2 ports figure the port length for 1 cube. This takes into account the frictional losses in the ports. You will notice the more ports you add the longer they get.
------------------ ******************* Founder: Street Sounds Car Audio Club ******************* If It Don't Read 150+ it Ain't Loud......
Posts: 289 | From: Howell, MI | Registered: May 1999
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Here are the number that came with the subs. Box net = 1.6Ft^3 port = 4" @ 13.25"long Port displacment = .1Ft^3 Woofer Displacment = .07Ft^3
(Or two 3" ports at 17.75"long)
For main box I did this... Vbox= 1.6Ft^3 + .07Ft^3[diver] + .1Ft^3 [port] +.05Ft^3 [1/2port] to get 1.82Ft^3
Then for second 1/2 I did this... Vboxa = 1/2(1.6Ft^3) + .1Ft^3 [port] + .05Ft^3 [1/2port] to get .95Ft^3
That is how I under stood the first few post.
How about if I use the books paramitors for 3" port with two and only used the one 3" in the main box, one in the secondary box, and one in the divider. All ports as the book says are 17.75" long. Is that the figuring you ment?
Thank for all the help This prodject realy cought my eye.
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