"Can you post the exact model of your AI for verification?"
Acoustic Image Focus, series III combined with a set of ACME B1s
Acoustic Image Focus, series III combined with a set of ACME B1s
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"Can you post the exact model of your AI for verification?"
Acoustic Image Focus, series III combined with a set of ACME B1s
thats if you connect all the speaker together and run from a mono bridged output...many bass amps have a seperate left and right or A and B speaker output...for running stereo output......each speaker connects to its own channel and the ohm rating doesnt change...engage the bridge mono switch..the outputs of the two channels are summed together..you only have one speaker cable coming out of the amp that goes to one speaker...and you run a speaker cable from speaker cab 1 to speaker cab 2...now you have affected the impedance of the cabinets..essentially cutting it in half...wired like this..two 8 ohm cabs become 4 ohms..
So you're saying 3 four ohm cabs in parallel is also 2.67 ohms? I don't really see how you "skip straight to 8/3".
whether you use seperate channels on an amp or bridge the channels together does matter...If I use channel A and B seperately on my ampeg SVT pro 4 and have a 4 ohm cabinet at each channel..its a 4 ohm system...if I bridge the two channels together..and jump the speakers together with a speaker cable..I am at a bridge mono load...it is now a 2 ohm system...the amp matters....all amps are different...you cant do this with all amps..but you can with amps that have stereo options...which essentially are two independent power sources in the amp for each channel...that can be used independently or together...running the channels independently...it is 4 ohms...bridging the channels is 2 ohms..
I know that is the simplification, I think it's worth using the actual formula I originally posted - the real formula - just in case you have a 610 (usually 5.3 ohms nominally) or want to compute the value directly for one 8 and one 4.No, for identical 4 ohm cabinets it would be 4/3
3x8 ohm cabs makes 2.6 ohms. I believe your d800 can be switched to work at 2 ohms so the answer is yes.Hi everybody,
I have a mesa boogie subway d800, and several cabs, eden 115, eden ex112, a pair of ampeg svt 210 av, and i was wondering if i could connect the pair of ampeg AND the eden 115 (all 8 ohm each) with the d800.
I know it could be a stupid question, and i beg your pardon, but i found nothing about it on the web and i haven't the skills to resolve it by myself, so i ask if someone can help me.
Thank in advance
I know that is the simplification, I think it's worth using the actual formula I originally posted - the real formula - just in case you have a 610 (usually 5.3 ohms nominally) or want to compute the value directly for one 8 and one 4.
I feel simplifications and shortcuts should only be used by those already well-versed in the formally correct way to compute it. Only they know the restrictions and special cases where the simplification actually works. The OP itself is a bit of a testament to that, isn't it?

Actually there is a simple formula for two cabs of different impedance as well.I know that is the simplification, I think it's worth using the actual formula I originally posted - the real formula - just in case you have a 610 (usually 5.3 ohms nominally) or want to compute the value directly for one 8 and one 4.
I feel simplifications and shortcuts should only be used by those already well-versed in the formally correct way to compute it. Only they know the restrictions and special cases where the simplification actually works. The OP itself is a bit of a testament to that, isn't it?