Deriving a Dipole Surround Speaker from a Standard Monitor Speaker

[hddummy]

Sooner or later, I’m going to start building the speakers for my future home theater. I’m pretty certain that I want dipoles for the surrounds, but I can’t find any projects that people have made. I could try to design one from scratch, but I’m apprehensive. I can figure out two ways around this.

First, I can select an appropriate 4ohm monitor, build two of them into one cabinet, and wire them in series to create an 8ohm load to the receiver. I’m worried about the cost of duplicate crossovers.

Second, I think I know how to modify the XO from an 8ohm monitor to work with double the drivers and I want your help to know if my theory is sound. If I take two tweeters and wire them in parallel, the combined inductance and resistance will be half of a single tweeter. Now, if I wire a small inductor and resistor in series with the two tweeters to bring the effective values of the entire system back to the same as it was with just one tweeter, the rest of the XO shouldn’t react any differently. The overall sound output should be the same because I now have two tweeters, but each is outputting half as much power as the original one. I don’t know if this is good or bad since I can imagine I might want higher output from the tweeters in a dipole since their pointing in different directions. Regardless, I can play around with the padding resistors to get the levels that sound good.

Am I completely off base here, or do you think this will work? I’m going to find a good DIY monitor project to start with, recreate the project in Speaker Workshop, and then modify it as I’ve described. Hopefully, it will prove my method correct

[hddummy]

This plan doesn't really work. The XO network is largely dependant on the impedance curve of the driver. Just substituting an inductor and resistor to make up for the difference of having two drivers will not replicate the impedance curve. I suppose theoretically, this plan would work if you are using the driver only in the frequency range where it's impedance is more or less constant. Back to the drawing board i suppose.