Is this enough port area

[UCF52 10+ year member]

Hey I was curious if ~ 50 in^3 is enough port area for a single 12 in a 2.25 ft^3 (net) enclosure.

Sub: JL 12" W6.v1 ~ seeing 360w

Net box volume: 2.25 ft^3

Port Diameter: Single 3"

Port Length: 7"

Port Area: 49.5 in^3

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[jellyfish420 10+ year member]

yes

 

[UCF52 10+ year member]

yes

Great! Thanks a lot man.

 

[Immacomputer 10+ year member]

A single 3" port is NOT enough for that sub.

It is also NOT 50in^2 of port area -- it's about 7 in^2 of port area. What you calculated there is volume, not area. Area is 2 dimensional, not 3 dimensional.

You did area times the length of the port. That is three different dimensions. Length times width is the cross sectional area. In the case of a round port, it's pi times the radius squared, or pi*r^2.

I would do two 4" round ports at about 30" each but you would have to keep the enclosure around 2.25 cubic feet.

 

[UCF52 10+ year member]

A single 3" port is NOT enough for that sub.
It is also NOT 50in^2 of port area -- it's about 7 in^2 of port area. What you calculated there is volume, not area. Area is 2 dimensional, not 3 dimensional.

You did area times the length of the port. That is three different dimensions. Length times width is the cross sectional area. In the case of a round port, it's pi times the radius squared, or pi*r^2.

I would do two 4" round ports at about 30" each but you would have to keep the enclosure around 2.25 cubic feet.

haha, yea you're right. it has been a long week //content.invisioncic.com/y282845/emoticons/furious.gif.fc81ca146dbff91fede3ed290dbc4f4c.gif

thanks for the input!

 

[jellyfish420 10+ year member]

A single 3" port is NOT enough for that sub.
It is also NOT 50in^2 of port area -- it's about 7 in^2 of port area. What you calculated there is volume, not area. Area is 2 dimensional, not 3 dimensional.

You did area times the length of the port. That is three different dimensions. Length times width is the cross sectional area. In the case of a round port, it's pi times the radius squared, or pi*r^2.

I would do two 4" round ports at about 30" each but you would have to keep the enclosure around 2.25 cubic feet.

oopps. maybe i should read the whole post. all i read was 50in^2 for 2ft box.