View Full Version : Can someone help me calculate internal volume?
12-07-2011, 12:25 PM
Wood thickness 0.75
Width at all sections is 13"
All dimensions are outside dimensions
I need vol before displacement
The box is for a SA10, which im assuming is about 0.12 displacement?
12-07-2011, 12:29 PM
When i split it into 2 boxes, i get 1.38ft^3
12-07-2011, 12:58 PM
Actually its 1.5685221 cu ft before sub displacement subtract .12 for sub 1.4485221 cu ft after displacement ir .12 is correct for sub
12-07-2011, 01:02 PM
LxWxH then divide by 1728 to convert cu in to cubic ft dont forget to subtract wood thickness on all sides
12-07-2011, 01:04 PM
I finall different recommendations for this sub:
Displacement 0.10 ft^3
Sealed box 0.5 cu. ft.
Ported box 0.75-1.0 cu. ft.
1.25 - 1.5 cubic feet tuned to 30 - 35 Hz
Yields F3 in the mid to high 20s
0.55 cubes for 0.707 Qtc alignment
Yields F3 of 50 Hz
12-07-2011, 01:11 PM
Ive used multiple calculators and get ~1.4ft^3
12-07-2011, 01:44 PM
I get 1.65cu ft before displacement :(
12-07-2011, 01:47 PM
why is this so hard lol...
im taking 2 boxes:
12.75 x 12 x 13
8.25 x 15 x 13
and putting each into 12v calc:
add them up
subtract .1 for displacement
12-07-2011, 02:47 PM
Are u subtracting 3/4 inch .for wall of box on each side. 1.56 before displacement 1.4 after not that .1 of a cube is a big deal but I figured using cu in and converted to cu ft about as accurate as youll get.
12-07-2011, 03:03 PM
Tune it to 30 hz and call it a day.
12-07-2011, 03:05 PM
pro-rabbit confirmed it was 1.39ft^3 before displacement, and going by the SA10 .10 displacement spec that would be 1.29ft^3...
Going to try a 3" x 15" aero port for a 30hz tuning will let you know how it sounds.
12-07-2011, 03:27 PM
Sweet not sure how my numbers got that fat but both figures show your right where you need to b either way. Let us jnow how it turns out.
12-07-2011, 03:31 PM
I would like to know said formula used by pro rabbit. The above mentioned is the only one I know for cu ft or maybe I hit a key wrong putting it in if he used same formula.
12-07-2011, 08:40 PM
12-08-2011, 11:29 PM
[(10.5*11.25*11.5)+(14.25*6.75*11.5)+(0.75*6.75*11. 5)]/(12*12*12) = 1.46 ft^3 before displacement.
Far left rectangle, far right rectangle, airspace you subtracted from the wood thickness in between the two main rectangles that actually exists, divided all by 12^3 to change in^3 to ft^3.
12-08-2011, 11:49 PM
maTh iz harD