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saywhat?
12-13-2005, 11:18 AM
alright i know the whole Pi(r)^2 deal, but when i do that for a 6 inch port thats 33 inches long i get roughly .5 cubes. When I go to the12volt.com it tells me .44 cubes.

IamDeMan
12-13-2005, 11:49 AM
alright i know the whole Pi(r)^2 deal, but when i do that for a 6 inch port thats 33 inches long i get roughly .5 cubes. When I go to the12volt.com it tells me .44 cubes.

I usually just keep all measuremnts in CuIn when formulating and planning a box design, but I come up with .5396875 Cu.Ft for a 6" port 33 in long. This is also not inlcuding wall thickness. I am assuming the wall is 1/8" which then makes the displacment .5855983 Cu Ft. (EDIT: then again 3/4" of that port won't be taking up internal space, so subtract .013309 :p: )

saywhat?
12-13-2005, 12:04 PM
its actually an extra half inch of wall. and i now have it figured out, thank you.

saywhat?
12-13-2005, 12:10 PM
so 4.5 would be total diameter with wall thickness and all. 2.25 ^2=5x3.14=15.9

my port will actually be 27 inches so 15.9x27=429.3/1728=.24


.24x3=.72


does sub displacement really take much into effect in the final deal.

80INCHES
12-13-2005, 12:19 PM
depending on the sub ..some subs take up alot of air..some dont
4 example..a se10 takes up .14cubes of space in a box..while a 10w3 take up .05cubes

80

saywhat?
12-13-2005, 12:25 PM
yah, i mean on final tuning and all. cuz my ports are goin to take .72 cubes and i have 3.79 total. so i have room for one sub pretty much (.07 for a .09 sub displacement) but the other will take up .09. i think i should be perfectly fine though.