I was thinking... What would the impedence be for the M80's when your using them in a bi-amp configuration. I dont imagine it would stay at 4 ohms because youve either got an amp driving the two woofers, or the two tweets/mids. Maybe alan can give some insight.
Just out of general intuition, it would be safe to assume that the impedence would rise since you are distributing the load on two amplifiers (and thus halving the number of drivers that run in parallel). When you run two identical drivers in parallel, impedence is halved, as you most likely already know.
Now, I'd be sure that the woofers still are closer to four to six ohms, and that the tweeters and mids would be at eight or higher. I would also think it is safe to assume that the drivers are wired in parallel internally because since it's not a dipole speaker.
A dipole speaker is one designed such that the drivers fire in phase but in different directions (ie. QS8). A bipole speaker is designed so that the drivers (usually opposed outward to each other) fire out of phase. A few tidbits I've picked up on this board.
So, if the drivers in the M80 are wired in parallel it would be neither a bipole nor a dipole.
Sean
SeanF,
You got your terms mixed up. Bipoles are in-phase, di-poles are out-of-phase.
Peter
I sit corrected. Di-pole is OUT of phase, bipole is IN Phase.
Sean