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Old     (mikeski)      Join Date: Aug 2003       03-31-2011, 10:21 PM Reply   
For Grant:

Calculating the sub box volume for the X-star:

1. Determine the desired box volume
a. JL specs a NET interior volume of 1.25 cuft.
b. JL specs the volume of the JLW6v2 at 0.068 cuft
c. So each sub should have 1.318 cuft or 1.318 x 1728 = 2,277.5 cu.in. interior volume or 4,555 cu. In. per box of a pair of subs
2. Box volume: W x L x H = 4,555 cu.in.
a. Z – 1.5” = H, 22-1.5” = 20.5 fixed
b. X – 1.5” = W, max dimension of 23”
c. Y – 1.5” = L, max dimension of 16” (for each box section housing a pair of subs)
d. (X – 1.5) x (Z – 1.5) x (Y – 1.5) = 4,555
e. Since Z is fixed then (X-1.5) x 20.5 x (Y – 1.5) = 4,555 but there is that little triangle cut out of the top so we need to remove that space from the volume calculation. The best way is to figure the area of a square less the area of the triangle then multiply the result times the length to determine the volume.
f. The sides of the triangle adjacent to the right angle are 9.625 and 8.5”
g. {[(X – 1.5) x (Z – 1.5)] – (9.625 – 8.5)/2} x (Y – 1.5) = 4,555
h. Or (((X – 1.5) x 20.5) -40.9) x (Y – 1.5) = 4,555
i. Solving for X = (((4555/(Y – 1.5))+40.9)/20.5)+1.5
j. Or solving for Y = (4,555/((X – 1.5) x 20.5) – 40.9) + 1.5
3. Pick an X or Y value and calculate the remaining variable
a. X = 14: Y = 22.65”
b. Want a longer but narrower box? OK Y = 25: X = 11.8”
c. Using the equations you can play with each dimension
4. So is there enough room to run a ported box?
a. ((23 – 1.5) x (22 – 1.5) – 40.9) x (16 – 1.5) = 5798 cu in or 2899 per sub or 1.68 cubic feet each, once you deduct the volume of a slot port it gets a little tight but you could easily run 2 big 3” ports and be within the spec…
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Old     (chucktronics)      Join Date: Nov 2007       04-01-2011, 11:15 AM Reply   
Is this like "guessing the number of Jelly Belly in the jar gets you the Jar?............JK


Double Jeapardy........"what is no" there is not enough space to port this properly

If you are questioning the system design "G" is working on this project concists of 4 drivers and two of the enclosures noted above

1/ you would have to run 2 3 inch ports per driver and that is still not enough . With a x section of a 3 inch port at approx 7 sq inches ,2 of them is around 14 and this driver needs around 18 sq inches of port . About 12 sq inches per cube of net enclosure space per speaker is required. Your port needs to be in the neighborhood of about 34 to 40 sq inches per enclosure

2/ What surface would you mount these ports to . with two enclosure and four speakers in the cabin ,are you gonna fire 2 ports forward and 2 ports back towards transom? Not very inefficient,and could lead to a handful of acoustical nightmares.And, How are you going to obtain the length required to tune correctly.
Old     (05mobiuslsv)      Join Date: Apr 2006       04-04-2011, 5:51 PM Reply   
I feel like I just watched a pros vs joes episode.

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