Question for the matched cabs crowd

[walterw]

...I'm going to go ahead and admit that I can't figure out the formula.

Here's what I *do* understand: if I want to plug two cabs into my 4 ohm head, they better BOTH be 8 ohms.

what works in my head to get me in the ballpark is to think "half the average" of two cabs:

two 8Ω boxes? the "average" between them is still 8, half that is 4; 4Ω is the right answer.

a 4Ω and an 8Ω? the "average" between them is 6, half that is 3; the right answer is 2.7Ω, close enough for me to know if my amp is safe with that or not.

 

[walterw]

I got a chance to examine the other guy's 2x10 in person tonight. It's a 4 ohm cab. To recap: I had that, plus my 8 ohm 4x10, plugged into a GK400RB, which is rated for 4 ohms.

Sounds like what I had was bass-ackwards.

yes, but:

if that 2x10 had been internally re-wired series instead of parallel, so that the two 8Ω speakers in it totaled up to 16Ω instead of 4Ω, it would have been just right!

you would have had all 6 tens getting equal power from the amp, and they all would have added up to 5.3Ω, totally fine for any bass head.

(the average between 16 and 8 is what, 12? half that is 6, which is at least in the ballpark of what you would get, 5.3Ω.)

i keep wondering why no manufacturers have latched onto this, offering an 8Ω 4x10 and a matching switchable 4/16Ω 2x10; the 2x10 at 4Ω for little gigs, the 4x10 by itself for most gigs, and the full 6x10 stack for the big stuff.

 

[Downunderwonder]

The impedance rule is add the reciprocals of the cabs impedances and then take the reciprocal of that sum.

1/4 + 1/8 = 3/8

8/3 = 2.67

 

[Foz]

or... product [multiple the 2 numbers] over [make a fraction] the sum [add the two numbers] - thus:

a pair of 8 ohm cabs = 8x8 divided by 8+8 = 64/16ths = 4 ohms

or an 8 ohm and a 4 ohm = 8x4 over 8+4 = 32/12ths = 2.666666... [round to 2.7 ohms]

Pro tip: if the cabs are the same ohms then just cut the number in half - two 4's make 2 ohms.

 

[Foz]

Your friend is clueless

+1

 

[iualum]

The impedance rule is add the reciprocals of the cabs impedances and then take the reciprocal of that sum.

1/4 + 1/8 = 3/8

8/3 = 2.67

Yepper.

 

[iualum]

or... product [multiple the 2 numbers] over [make a fraction] the sum [add the two numbers] - thus:

a pair of 8 ohm cabs = 8x8 divided by 8+8 = 64/16ths = 4 ohms

or an 8 ohm and a 4 ohm = 8x4 over 8+4 = 32/12ths = 2.666666... [round to 2.7 ohms]

Pro tip: if the cabs are the same ohms then just cut the number in half - two 4's make 2 ohms.

Yepper again.

 

[lwknives]

Dude, my mixed cabs sound boss!

Yah, but science man!

 

[nojj]

She deafened me..........with Science.

 

[DigitalMan]

That is what I thought until I bought an svt 2. Then you need four ohm cabs! It will only do 2 or 4 ohm loads. Its crazy. I guess they mostly made it for the fridge.

I think your situation agrees with the corollary. A good fit but for a limited addressable market opportunity.

 

[Bassist4Eris]

The impedance rule is add the reciprocals of the cabs impedances and then take the reciprocal of that sum.

1/4 + 1/8 = 3/8

8/3 = 2.67

And the light in my brain finally comes on! Thank you Downunderwonder!

Foz and walterw, your methods also make sense to me, and are very helpful.

My thanks to you all.

 

[grenadilla]

I have an Ampeg 210 with ports on the back. The cardboard ducts have been lost for years. The cabinet is only 200 watts and it sounds ok. It is on a real heavy-duty Ampeg 410 (400 or more watts, about 122 lb. ) It is front ported. The 210 is 8ohms and the 410 is 4ohms. The Marshall MB450H is 450 watts @2 ohms. The rig is loud and sounds pretty good. My question is how much is the sound degraded by having a rear-port cabinet on a front-port cabinet? It must be a little muddy, right?

 

[Bassist4Eris]

yes, but:

if that 2x10 had been internally re-wired series instead of parallel, so that the two 8Ω speakers in it totaled up to 16Ω instead of 4Ω, it would have been just right!

you would have had all 6 tens getting equal power from the amp, and they all would have added up to 5.3Ω, totally fine for any bass head.

(the average between 16 and 8 is what, 12? half that is 6, which is at least in the ballpark of what you would get, 5.3Ω.)

i keep wondering why no manufacturers have latched onto this, offering an 8Ω 4x10 and a matching switchable 4/16Ω 2x10; the 2x10 at 4Ω for little gigs, the 4x10 by itself for most gigs, and the full 6x10 stack for the big stuff.

This is really cool too, especially since the other guy was probably hoping to be able to augment that 2x10 with another cab at some point. Knowing he can have it modded to work for him will probably be welcome info when I see him again.

Thanks again, everyone. This impedance business wasn't even what I originally posted about, but it was what I really NEEDED to learn. Some derails are good!