Does anyone know what the displacement is roughly of a 2.5 port *10" and a 6.5" speaker? It would be in a box approx 14*10*9.5"??

THanks

[ December 19, 2004, 11:07 PM: Message edited by: ChizzerZ24 ]

 

Pi*R^2*L = Port Displacement in in^3

The speaker displacement will be negligible (probably less than 0.02 ft^3) and the port displacment is around 0.034 ft^3 assuming 1/8" thick PVC ...

0.05 ft^3 isn't going to be audible, so don't even worry about it ...

When you start getting into larger ports (esp. slot ports) and larger subwoofers, then it's a lot more necessary to take the displacements into account ...

 

Sounds good! Thank!

 

Yeah, wasn't so much it was an issue didn't know if it needed to be taken into consideration. [img]smile.gif[/img]