That method only works for two cabinets. Here's the one for more than two:
So basically you invert them all, add them & then invert again:
1/8 + 1/8 + 1/8 = 3/8 = 8/3 = 2.66666
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That method only works for two cabinets. Here's the one for more than two:
So basically you invert them all, add them & then invert again:
1/8 + 1/8 + 1/8 = 3/8 = 8/3 = 2.66666
If all cabs are the same impedance and all wired in parallel, you just divide the Impedance of one of the cabs by the number of cabs. Works no matter how many cabs you have.
One 8 Ohm cab = 8/1 = 8
Two 8 Ohm cabs = 8/2 = 4
Three 8 Ohm cabs = 8/3 = 2.67
Four... 2
Eight... 1
One 8 Ohm cab = 8/1 = 8
Two 8 Ohm cabs = 8/2 = 4
Three 8 Ohm cabs = 8/3 = 2.67
Four... 2
Eight... 1
No, that math is wrong.
Z(total) = 1 / (1/Z1 + 1/Z2 + 1/Z3)
= 1 / (1/8 + 1/8 + 1/8)
= 8/3
= 2.67
Report back after you try it.
I've thought about getting 2 more 210AVs so I could run 4 of them, but I don't have a 2ohm capable head.
I've thought about getting 2 more 210AVs so I could run 4 of them, but I don't have a 2ohm capable head.
No, it depends entirely on how much power it is getting. Lower overall impedance means the amp is pushing more power overall.
Now usually the power per cabinet does go down a little bit when you run multiple cabinets, but not by much.
If you run a number of cabinets of equal impedance, the cabinets split the power produced by the amp equally.
Man all this serious helpful talk in a thread that has so much potential for ridiculous wall of speakers memes.....so disappointed with TB, too much maturity tonight.
But for reals though, I was hoping to see a wall of speakers.
But for reals though, I was hoping to see a wall of speakers.
- Feb 12, 2006
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- Disclosures
- Development Engineer-Mesa Boogie, Development Engineer-Genzler (pedals), Product Support-Genz Benz
I would think that the pair of your 210's with the amp set to the 4/8 ohm position would be a reasonable way to go. I would be careful and use common sense if using the deep switch at high volumes however.
two 4 Ohm cabinets
four 8 Ohm cabinets
eight 16 Ohm cabinets
sixteen 32 Ohm cabinets
but only because the amp is capable of a 2 Ohm load
for a 4 Ohm load minimum amp (which are quite common)
you could only attach half as many
four 8 Ohm cabinets
eight 16 Ohm cabinets
sixteen 32 Ohm cabinets
but only because the amp is capable of a 2 Ohm load
for a 4 Ohm load minimum amp (which are quite common)
you could only attach half as many
Just so I'm following everyone:
I have an Acoustic Image micro amp with two speakon outputs. If I run two 8 ohm acme cabs , each with two speakon jacks- one into each output of the amp with separate cables, my total load is 4 ohms on the amp. If I run one cabinet speakon jack connected to the other cabinet and then a second speakon cable to only one output on the amp, is my load the same?
I have an Acoustic Image micro amp with two speakon outputs. If I run two 8 ohm acme cabs , each with two speakon jacks- one into each output of the amp with separate cables, my total load is 4 ohms on the amp. If I run one cabinet speakon jack connected to the other cabinet and then a second speakon cable to only one output on the amp, is my load the same?
Are all the Speakon connections wired in series?
Or are extra Speakon connections on the cabs wired for parallel?
Are both Speakon outputs from the amp series connections or is one series and the other parallel?
What do the Acoustic Image micro and Acme Cab manuals say?
Or are extra Speakon connections on the cabs wired for parallel?
Are both Speakon outputs from the amp series connections or is one series and the other parallel?
What do the Acoustic Image micro and Acme Cab manuals say?
That may be common in standalone power amps, but not bass heads. Yeah there are a few, but not many. GK 2001 comes to mind. The amp the OP has is certainly not designed like that.
It depends. Are your jacks in parallel to one channel, or is your amp dual channel like @blazebarlow mentioned above?
I was unable to find an exact match for 'Acoustic Image micro'.
The short answer is yes, the loads are identical, 4ohm, IF this is a single channel amp.
If it is designed for doublers, then each speakon output would only handle one channel and the loads would be 8ohm each.
Can you post the exact model of your AI for verification?
But with the same impedance cabs you can skip straight to 8/3 and avoid the math errors.
But mmbongo is correct.
The information he provides has to do only with the CABS.
The amp you hook them to doesn't matter.
Even if you don't connect them to an amp at all, his numbers are still correct.
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".
He said divide the impedance of the identical cabs by the number of cabs.
He did not say divide 8 by the number of cabs regardless of impedance
See post #22 above.
