Quote Originally Posted by PugnacAutMortem View Post
But I can run one 2x12 for the left and one 2x12 for the right and have a stereo rig that way correct?
Disclaimer: I'm trying to not get into the gory details of AC signals, and how they actually change with frequency in this discussion.
So rather than having a "techy" debate in this thread, I'd ask that you start a new one, so we don't confuse the OP.

For a stereo sound system each channel (Left or Right) is considered independent of the other. Each channel is using it's own amplifier.
Although both amps may be located in one box, the "stereo amp" performs pretty much like 2 separate mono amps (one for L and one for R).
(Let's not get into "Bridged mono" configurations in this discussion...if you want to know how they work, ask about that as a separate question)

So your load is (2) 12"@8 ohm speakers per channel (left or right).
You have 2 identical "boxes'; one for L and the other for R

If wired in series, the load will be about 16 ohms.
If wired in parallel, the load will be about 4 ohms.
( I say about, because the impedance and reactance is frequency dependent, and the speaker responds differently at different frequencies. Thus an 8 ohm speaker is not exacty 8 ohms at all frequencies, but is close enough not to worry about it for this purpose).

Many amps are capable of driving 4, 8, or 16 ohm loads. (Check the specs)
Given the same voltage, a lower impedance load will offer more power delivered to the load.

Amps typically refer to Volts or Watts in terms of "RMS (Root Mean Square)" .
This is "sort of" a DC approximation of an AC signal (i.e. an audio signal) and useful for quick calculations.

For DC (or AC rms) and for general audio calcs, the DC equations are "close enough"
( P=IV or P= I^2R or P= V^2/R where P=power(Watts), I=current(Amps), V=Voltage (Volts) and R=resistance (Ohms))

EX: Say Voltage is 10V rms

Power P=V^2/R
for 16 ohm load : (10V rms)^2 / 16 = 6.25 W rms
for 8 ohm load : (10V rms)^2 / 8 = 12.5 W rms
for 4 ohm load: (10V rms)^2 / 4 = 25 W rms

So if your amp can handle a lower impedance load, wire them as 4 ohms, and you can get more power across the load.

Finally, YES, you should have
A "left" box : 2x 12" @8ohm (parallel) = 4ohms, as seen by the amp's LEFT channel
A "right" box: 2x 12" @8ohm (parallel) = 4ohms, as seen by the amp's RIGHT channel

Also, if this is for a not for guitar or bass, you would benefit by filtering (i.e a crossover network or low pass filter or eq. ) out the high end at or before the signal gets to the speaker. 12" speakers are not typically capable of producings "highs". Guitars and basses are not capable of producing highs, such as those produced by cymbals etc. and sort of "self filter" their sound.

There is still considerable energy in the high end that gets to the speaker, even if the speaker cannot reporduce it. Since the speaker is not optimized to reproduce highs, the high freq energy is often converted to heat and can fry the voice coil. (lots of other reasons this can happen as well- that's yet another discussion).

Great place for parts : www.parts-express.com

I hope this helps...please ask about things that I may have caused confusion with.

PS If you need any amps, I'm getting rid of my QSC 1450 and 850 amps, and other audio stuff (mixers, speakers, cables, etc.)
Let me know.