thegr8cody Posted July 8, 2007 Report Share Posted July 8, 2007 how do u measure the square inches of aero ports i need around 240 square inches of port Quote 1998 gmc sierra h/u:eclipse cd3200 sub:dd9518g d1.5 amp:jl e300/4 sub amp:dd m2a front stage:id ctx65cs dc power 270 alt lots of kicker 1/0 Link to comment Share on other sites More sharing options...
andym85 Posted July 8, 2007 Report Share Posted July 8, 2007 pi*r^2 pi=3.14 r=radius of port example: a 4 inch areo, radius of 2" 3.14*2^2=12.56sq" of area lol...so 19.1 4"aeros or 8.49 6" aeros...dont think that is gonna happen... Quote i likes me some audio stuff... Link to comment Share on other sites More sharing options...
emperorjj1 Posted July 8, 2007 Report Share Posted July 8, 2007 thats confusing Quote J. JMy CardomainFINISHED COBALT SS/SC DUAL ALTERNATOR PICS theres no such thing as too expensive when it comes to upgrades like that, because imo if you are gonna spend to upgrade then do it correctly rather then be a cheap ass ricer Link to comment Share on other sites More sharing options...
Derrick824 Posted July 8, 2007 Report Share Posted July 8, 2007 Aeroports don't work the same way a slot port does. You don't need as much port area with aeros. Quote Link to comment Share on other sites More sharing options...
thegr8cody Posted July 8, 2007 Author Report Share Posted July 8, 2007 so pi means? u do port radius x the depth of the flare? Quote 1998 gmc sierra h/u:eclipse cd3200 sub:dd9518g d1.5 amp:jl e300/4 sub amp:dd m2a front stage:id ctx65cs dc power 270 alt lots of kicker 1/0 Link to comment Share on other sites More sharing options...
Derrick824 Posted July 8, 2007 Report Share Posted July 8, 2007 so pi means? u do port radius x the depth of the flare? pi = 3.14. Think back to geometry & algebra in highschool. amatt85 did a good job explaining it already. Radius = half of the diameter. 4" port = 2" radius, 6" port = 3" radius. For a 6" port its 3.14*(3*3)=28.26 For a 4" port its 3.14*(2*2)=12.56 When you multiply that by the depth thats calculating for volume. Thats how you'll figure displacement of the ports. For a 6" port thats 10" long 28.26*10=282.6in or 0.16ft. I hope this helps cause I really don't know any easier way to explain it. Quote Link to comment Share on other sites More sharing options...
andym85 Posted July 8, 2007 Report Share Posted July 8, 2007 ^^yep..pi is just a constant number, kinda like a mile is a constant number...a mile is 5280ft. and pi is 3.14159...(i dont think scientist have actually found an ending decimal place yet, it just goes out to infinity, never repeating)... Quote i likes me some audio stuff... Link to comment Share on other sites More sharing options...
thegr8cody Posted July 8, 2007 Author Report Share Posted July 8, 2007 alright so say i have a 10 inch pipe,radius is 5",soi say 5 x5 = 25 x 3.14 =78.5 then my depth for 32 hz tuning is 5.35 so do i times 78.5 x the 5.35 that comes out to 419 square inches of port area? Quote 1998 gmc sierra h/u:eclipse cd3200 sub:dd9518g d1.5 amp:jl e300/4 sub amp:dd m2a front stage:id ctx65cs dc power 270 alt lots of kicker 1/0 Link to comment Share on other sites More sharing options...
andym85 Posted July 8, 2007 Report Share Posted July 8, 2007 no. your port area is 78.5sq" and your port displacement (or volume) would be 419cubic". area is a two dimentional measurement. volume is a three dimensional measurement (the third dimension being depth)...hope that clears it up! Quote i likes me some audio stuff... Link to comment Share on other sites More sharing options...
thegr8cody Posted July 8, 2007 Author Report Share Posted July 8, 2007 well if i need 220 square inches of port vented how many would i need in aero port if i used pvc and shaved the end would that be just as good Quote 1998 gmc sierra h/u:eclipse cd3200 sub:dd9518g d1.5 amp:jl e300/4 sub amp:dd m2a front stage:id ctx65cs dc power 270 alt lots of kicker 1/0 Link to comment Share on other sites More sharing options...
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