Factory Subwoofer Enclosure

Try this formula:

A. Account for real external dimensions of box, excluding the clearance from the bottom.
real external dimensions of box width = 29.25", height = 9.25", depth = 12"

B. Account for the 2 sub compartments
external width of box width = 29.25"
external width for each compartment width = 14.625"

C. Subtract the thickness of the wood (.75" per each side), including the middle divider piece (.375" per each side):
internal width 14.625"-.75"-.375" = 13.5"
internal height 9.25"-.75"-.75" = 7.75"
internal depth 12"-.75"-.75"= 10.5"

D. Multiply dimensions to find cubic inches.
width*height*depth= cubic inches
13.5 * 7.75 * 10.5 = 1098.6 in^3

E. Convert cubic inches to cubic feet.
find cubic inches in 1 cubic foot 1 ft^3 = 12 in^3; 12*12*12 = (1728 inches)
compute to cubic feet: 1098.6 / 1728 = 0.636 appx.

F. Account for speaker displacement.
subtact 0.1 ft^3 0.636 - 0.10 = 0.536 ft^3

:jawdrop:

Or I could just say that yes, you can put an 8" in there with a couple modifications. I've seen it done before and although I have not been overly impressed with the output of the setup, it is definitely possible.