Push-pull woofer alignment questions

[3-LockBox]

First off - I've read from several websites that seem to not be able to agree on two important issues.

1) Some state that PP configurations are like isobaric configurations, in that the Vas needed for cabinet volume is halved (assuming two woofers in parallel). But some say PPs are the same as two conventionally mounted woofers, in that the Vas requirement is doubled for a cabinet. Which is true?

2) Some state that even-ordered harmonics are cancelled, some say that odd-ordered harmonics are cancelled. Which is true? and Does it matter?

Also, if its possible to reduce the needed cabinet size by half using a PR, then how much bigger than the active driver would the PR need to be?

[john k...]

Well there is certainly a lot of misinformation in you post. First, if by PP you means two driver mounted in the same box but with the motor structure in the box and one flipped so the motor is out of the box and the driver wired out of place electrically (in phase acoustically) then the required box volume doubles as effectively Vas doubles.

The distortion issues are a little more complex. The common assumption is that because the drivers are electrically out of phase even order distortion generated in the motor will cancel. This seems true at first glance and is basically true for an infinite baffle or a dipole configuration. However, when a box enters the picture the loading of the box side of the driver changes the situation. When the box compliance is considered the forces acting on the driver include the pressure force on the box side of the cone and there is not guarantee that the even order distortion will cancel.

[Svante]

I can imagine that people could call all of these four ways of mounting two drivers "push-pull".



Box A is the standard isobaric way. It is push-pull because when one driver pushes from the magnet, the other one pulls towards the magnet. This causes even order harmonics caused by an asymetric motor to cancel.
This way of mounting the drivers requires half the box volume compared to a single driver.

Box B is also push-pull for the same reason. Even order harmonics are cancelled. Reaction forces on the box are cancelled. When one driver pushes left, the other one pulls right. Wee, another definition of push-pull...
It requires twice the box volume compared to a single driver.

Box C is push-pull in the sense that reaction forces on the box are cancelled, but not in the sense of motion relative to the magnets. Even order harmonics remain.
It requires twice the box volume compared to a single driver.

Box D is as A, but requires twice the volume of a single driver, ie 4 times that of A.

Odd order harmonics cannot be cancelled by the way the drivers are mounted.

[Svante]

Originally posted by john k...

The distortion issues are a little more complex. The common assumption is that because the drivers are electrically out of phase even order distortion generated in the motor will cancel. This seems true at first glance and is basically true for an infinite baffle or a dipole configuration. However, when a box enters the picture the loading of the box side of the driver changes the situation. When the box compliance is considered the forces acting on the driver include the pressure force on the box side of the cone and there is not guarantee that the even order distortion will cancel.

I left that out in my post, but since you bring it up...

In what sense do you mean that these effects would "cancel the cancellation"? Is it because the motion of the two drivers become different, or is it because the air inside the box is nonlinear?

The first case is true if the volume of air between the drivers is not neglectible. If it is small compared to the box volume, though, the motion of the two cones will become nearly identical and the harmonics will almost cancel.

The second case is more complicated. One could imagine a nonlinear suspension of a single driver that exactly compensates for the nonlinearity of the air inside the box. Using two such drivers would ruin this delicate balance and instead increase the amplitude of even harmonics.

But I think these are little things, it is better to first understand the basics, and deal with this after that.

[planet10]

In Svante's diagram, C is more accurately termed push-push. B & D are the same.

Attached is push-push push-pull pair (B139s).

dave

[planet10]

Originally posted by 3-LockBox
Also, if its possible to reduce the needed cabinet size by half using a PR, then how much bigger than the active driver would the PR need to be?]

A PR is sort of a replacement for a very large port, so if you had a box where half the internal volume is taken up by a porrt then pretty much.

dave

[navin]

Originally posted by planet10
In Svante's diagram, C is more accurately termed push-push. B & D are the same.
Attached is push-push push-pull pair (B139s).

Dave what would be the volume required in this box?

I would assume it would be the same as one woofer in a sealed box.

[navin]

Originally posted by planet10

A port is sort of a replacement for a very large port, so if you had a box where half the internal volume is taken up by a porrt then pretty much.
dave

I assme you meant a PR is sort of a replacement....:-)

[planet10]

Originally posted by planet10
B & D are the same.

topologically the same... with clever driver coupling B could also be push-push

dave

[planet10]

Originally posted by navin


I assme you meant a PR is sort of a replacement....:-)

Yes... i fixed my post... brain got ahead of mu 2 fingers

dave