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ported question

1K views 6 replies 4 participants last post by  Capt Steve 
#1 ·
If I wanted to port 2 12s and use one chamber for both, would I just have to double the recommended volume of the box AND the port size?
 
#6 ·
Just by going to a 6" [circle] you'd be more than doubling your area.

4" port has an area of about 12.57 square inches.

6" port has an area of about 28.27 square inches. That's 2.25 times the area.

As far as length, I'm not sure if it should be the same length or not. Maybe change the length to compensate for the larger area, but I don't know port calculations very well, just math...
Hope this helps figure port size.
 
#7 ·
Port size is a function of the volume of the box, port diameter, and the tuning freq. It has nothing to do with the number of drivers.

The formula is ((14630000*(D/2)*(D/2))/((Fb*Fb*Vb*1728)-(1463*(D/2)))) where D is diameter of the port, VB is box volume, and FB is tuning freq.

For example, if you have a box that is 1 cu/ft and you want to tune it to 35 hz using a 2" round port, that port would be 5.448 inches long.

Now if you doubled the size of the box, 2 cu/ft and you wanted the same 35hz, you would need a 2" round port of 1.992 inches long.

If you used a 4" round port for a 2 cu/ft volume, the length of the port would need to be 10.896 inches long and 24.71 inches for 1 cu/ft.

Slotted ports are a little different, but are still a function of port volume, box volume, and tuning freq.

Of course this does not take into account the volume loss due to the port itself which would also need to be calculated and subtracted from the box volume. It also does not take into account tapered ports.
 
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