help me understand

[Lschevelle 10+ year member]

can someone explain to me how to subtract sub, port and bracing from the total amount of airspace

Click to read more...

 

[SpeedEuphoria 10+ year member]

subs normally list displacement which is what you need to subtract

the port depends on aeroport or square port

bracing depends on what type of bracing there are lots of styles, 45's in corner, round dowels, threaded rod, picture frame, ect...

Most of it is math. the total box is pretty simple, L x W x H. Then you have to figure the internal.

Here's my example:

Largest box that fits through my trunk opening is 29.5" x 17" x 16.5 external. So the internal is 1.5" less of each of those dimensions(2 sheets of 3/4" wood), so 28" x 15.5" x 15"= 6510 cubic inches.

Since we all use cubic feet which is 12x12x12=1728.

So I divide my 6510/1728=~3.76cubes

My SA12's are .14 each so .28

my 6" aero is pi x r^2(times the length gives volume), radius is r and equal to 1/2 the diameter. The external dimension of the port is what you need, so ~6.3" which/2=3.15". 3.15^2 x pi(3.14159)= 31.17 now times the length of 17" = ~530 cubic inches divide by 1728=.306

I also have bracing using 1" dowel with 15.5 length and another 15" length, so add together 30.5. Using the same formula as above, 1/2=.5, .5^2 x pi x 30.5=~24 cubic inches divide by 1728=.013 cubic feet

so if I take all the displacements, .013+.28+.306=.599 cubic feet

so 3.76-.599=3.16 cubes after displacements

For the Aero port I fudge it a little since I dont know how to get the true displacement of the internal flared end.

Thats the idea, my box will end up a little bigger with some tricks up my sleeve and maybe inverting the subs