Calculating Port Length for Dual Woofers

[johnlinstromberg]

How can port length, for a bass reflex enclosure, be calculated when using two woofers within a single enclosure?

[damkor]

> How can port length, for a bass reflex
> enclosure, be calculated when using two
> woofers within a single enclosure?

The same way you calculated it in the first place, but this time for Vas, use twice the Vas of one driver.

Not sure if you mean you are trying to keep the same box that was appropriate for one driver, and now use that volume for two drivers. If so, then you aren't going to get the same low F extension. Fb will be higher. Basically, the tube should probably be shorter. Recalculate and remodel everything.

[Chris Roemer]

> The same way you calculated it in the first
> place, but this time for Vas, use twice the
> Vas of one driver.

> Not sure if you mean you are trying to keep
> the same box that was appropriate for one
> driver, and now use that volume for two
> drivers. If so, then you aren't going to get
> the same low F extension. Fb will be higher.
> Basically, the tube should probably be
> shorter. Recalculate and remodel everything.

As a wise man once said, build the box twice the size (as for a single woofer) and use TWO of the originally recommended ports.

[damkor]

> As a wise man once said, build the box twice
> the size (as for a single woofer) and use
> TWO of the originally recommended ports.

That doesn't work for me. The ports are too short. It's not too far off though. Could be I'm not calculating right.

Anyway, for two woofers, to get the same F3, the box SHOULD be twice as big for sure, and you should use a wider port. If John wants to give the spec.s, we can work something out.

[Jeff B.]

> That doesn't work for me. The ports are too
> short. It's not too far off though. Could be
> I'm not calculating right.

> Anyway, for two woofers, to get the same F3,
> the box SHOULD be twice as big for sure, and
> you should use a wider port. If John wants
> to give the spec.s, we can work something
> out.

Well, that instruction should work correctly. Keep in mind though, that the port tunes the box independently of the drivers used so if someone wants 3 cubic feet tuned to 30Hz the port is the same no matter what woofer(s) are used.

[damkor]

..since if you use two separate volumes, the total volume is twice what is was for one woofer, and you have two ports the saem size. What difference can it make to simply remove the barrier..? But the conventional port calculation says it does not work.

The reason is that root 2 (the increase in effective radius of two ports) squared divided by 2 (twice the Volume)is indeed one, so that left side fraction part of the equation equals out (the part with the Fb and Vb in it), but then you have the negative (1.463 times R). And that is variable. The bigger the radius you are dealing with, the more that estimate veers from true.

[Jeff B.]

> ..since if you use two separate volumes, the
> total volume is twice what is was for one
> woofer, and you have two ports the saem
> size. What difference can it make to simply
> remove the barrier..? But the conventional
> port calculation says it does not work.

> The reason is that root 2 (the increase in
> effective radius of two ports) squared
> divided by 2 (twice the Volume)is indeed
> one, so that left side fraction part of the
> equation equals out (the part with the Fb
> and Vb in it), but then you have the
> negative (1.463 times R). And that is
> variable. The bigger the radius you are
> dealing with, the more that estimate veers
> from true.

Well, theory and math both say it will work fine. Using the same program, that uses the standard port formula (I know because I wrote the program, so I know what formula was used) I figured the following: Using a 50 liter box, 1 3" diameter port, and an Fb of 30Hz it calculates a length of 9.74 in. Uisng a 100 liter box, 2 3" ports, and an Fb of 30 Hz it calculates a length of 9.74 in. This looks pretty close to me.

The end correction that you say is off, is adjusted by the added cross-sectional area to result in the same length in the end.