i have a 4 ohm cab. i am buying a 8 ohm cab to go along with the other cab. the 4 ohms is a single 15 the 8 ohm is a 410 i have a hartke 2000 only handles 4 ohms i was told on tabcrawler(which is down now)that i can splice a cable together to get 3.** ohms just enough to save the head. how would you splice the wire? and could i change the wiring in the 410 to 16 ohms? would this make 4 ohms? any suggestions

You probably won't want to hear this, but my suggestion is to do one of the following: (1) use your current cab by itself, or (2) sell your current cab and either (a) don't buy the 8 ohm you're considering and just get one good 4 ohm cab or (b) get two new 8 ohm cabs I don't think doing the wiring trick, whatever it is, makes sense. Anything that would increase the resistance up to the safe point would also reduce power delivered. You could also get a different amp, one that would take loads down to 2 ohms.

Actually, if you put a 4ohm cab and an 8ohm cab together it comes out to 2.67 ohms, unless I'm mistaken. There's simply no practical way to put those two cabs together and get good results. Even if you do rewire the 4x10 cab, the impedance of the 15" cab will be the lower of the two and because it's lower it will get most of the power leaving the 4x10 cab a little weak. I think your best options are to either sell the 15" cab and get one rated at 8ohms, or pull the driver out and replace it with an 8ohm driver.

i was told if you get a y cable. the split part to the cabs and then the one part where they come together it will be 3.98 or somehting close to that. i was told that even bass player mag had an article about it. if not what is a good cab for $400. i want a lot of bottom and i want highs. i do everything in my band i mean everything so what do you all suggest.

Ok, i just pulled out the calculator and figured it up. 4ohms + 8ohms = 3.75 ohms. Even if the amp will tolerate the slightly lower than allowed impedance, it still doesn't solve the problem of the 4ohm cab being way louder than the 8ohm cab. This is why it's not a good idea to mix cabs of different impedance ratings on a mono amplifier. You can get away with it on stereo power amps because the two channels operate independently of one another and don't care what impedance is on the other side.

well the 410 is louder than the 15 410= 40 inches of speaker 15= 15 inches of speaker so won't it even out i played the 410 alone and it was loud even though it was only running 120 watts and the 15 runs at 200 alone so won't this even out about?

It might and it might not because here is where speaker efficiency comes into play. If the 15" has an efficiency rating (hypothetically) of 103db and the 4x10 has an efficiency rating of 98db, the 4x10 box will never keep up. If it's the other way around then it might even out. But all things being equal, if you have enough power available it will be the 15" cab that will cause you to have to back off on the volume. It's the one receiving the most power and will reach it's peak long before the 4x10 cab does.

I have the same problem as you. I have a 4 ohm 1-15, an 8 ohm 4-10 and an 8 ohm 1-18. I bought a gk 2000rb... wonderful tone and very reliable if you get a later production serial number. 500 watts per channel at 4 ohms. When I need to move air I hook the two 8ohm cab in parallel (4 ohm) into the left channel and the 1-15 in the right channel then balance the volume to taste. If I do not need all that then either any one cab will work or I can combine in any configuration I need. 8ohm on one side 4 ohm on the other. The two amps have separate volumes so it is easy to get a nice even sound. I can even bridge and get 1000 watts @ 4 ohm. jam

No. You're trying to relate diameter instead of area. Swept area of the drivers is more important. A = pi * r^2 So a 15" driver would give a area of 15^2 = 255 * pi A 10" driver would have 10^2 = 100 * pi So you're comparing A15 vs. A410. So the equation would be like... 255 * pi / 100*4*pi = Speaker area ratio. So pi drops out, and the ratio comes out to 2.55/4. So the area of a single 15" is a little more than half of a 410. But this does NOT equate to volume, frequency response...etc. My 2 2x10's don't have nearly twice the volume of my 1x15, nor do they have nearly twice the bass. That's because I built them to complent each other. And resistance in a RLC circut is pretty easy. For parallel, 1/Ztotal = 1/Z1 + 1/Z2 +....... For series Ztotal = Z1 + Z2 +....... The only way you can increase the impedance total, is to rewire the 4x10. If it starts as 8ohm, then you can pretty much only have either 2ohm, or 32ohm. This because if this 4x10 cab comes as 8ohm, then it's either 4 32ohm drivers in parallel (8 ohm) or 4 8ohm drivers in parallel/series (two sets of 8 ohm seriesed together, 16ohm, then paralleled, 8 ohms). I'm willing to bet it's the 8 ohm situation. So you can only rewire to a 32 ohm (4*8ohm drivers in series) or 2 ohm (4 8ohm drivers in parallel). So now, the only real option is the 32 ohm option in order to try to shoot for a 4ohm load minimum. Sooooooo 1/Ztotal = 1/32 + 1/4 1/Ztotal = 1/32 + 8/32 1/Ztotal = 9/32 32 = Ztotal *9 Ztotal = 32/9 Ztotal = 3.56 (not good). Bottom line, Get two 8 ohm cabs. Then get a power amp that can run at 2ohms/side. Then you can bridge to 4ohms. Don't buy something just to modify it, and then it still won't work. HTH's

I did not want to say that without knowing what assumptions he made... But I can't see any possible way to get this result either. Matthias

My result? The formula for calculating ohms for a parallel circuit is 1/(1/a + 1/b), where a and b are the ohm ratings of the cabs (or whatever it is in question).

I think he was using 32 ohms and 4 ohms, as per his post. To show the 4-10 ohms after rewiring it in series.

Ah. I just glanced at the quoted material, which said "4 ohms + 8 ohms = 3.75 ohms," which is incorrect.

Sorry for being not clear enough , I referred to Mudbass' post. jamminji's comment makes sense. BTW I don't think that anyone wants a 32Ohm bass cab...

OK it's official, I'm a moron. The formula is... 1/R = 1/R1 + 1/R2 + 1/R3 + etc. ("R" = ohms) The way this comes out to 375 is by being in a hurry and forgetting to invert the bloody fraction in the final computation. Done correctly, 4ohms + 8ohms = 2.67ohms. 1/4 + 1/8 = 3/8 Therefore R/(1) = 8/3 or 2.67