Well I've got a 2ohm Nemesis NA-600 head, but i don't really feel like running it at that, so I curious; If i were to get a 4ohm cab (say and ampeg 610hlf) and a 8ohm cab (a 15), how many ohms would the amp be running at? 3? mmm?

I'm thinkin' (from your wording) that you may not fully understand Ohms, and power et.al. Eden gives one power rating for this amp: 600 watts @ 2 ohms. That doesn't mean that the amp will not function with a 4 or 8 ohm load.......the power rating will be much lower though, say probably on the order of 300 watts for a 4 ohm load and 150 watts for an 8 ohm load. A Loud speaker cab will present a varied level of resistance....that varies with frequency. The impediance # we get from the speaker/cab manufacturer is an average. PJR

i don't fully understand ohms, in the sense that i don't know the formulas to get (x ohmx + y ohms = z ohms) the overall impedance, i do realize that it runs lower at higher ohms but i also realize that i most definitely have a lot to learn. perhaps i'll take physics.

Bottom line: Yes your ohms with those two cabs (assuming they are wired in parallel) will be about 3ohms and if your amp is "safe" down to 2 ohms you should be OK. The general way you can figure your ohms is: Parallel: Average the cabs (Ex: 4, 6, & 8 ohms gives you a 6 ohm average) then divide the average by the number of cabs (Ex. continued: 6ohm average divided by the 3 cabs gives you a 2 ohm load). Series: Just add them up. Note: Even though you may daisy chain your cabs, most cabs are wired in parallel already so you won't be getting a series load.

2.67 ohms is correct. IMO I don't think its a good idea to mix different types of cabs on the same amp channel. I know a lot of people do it but I don't think its a good idea. And when each cabinet has a different nominal impedance like yours do I think even less of it. It may work for you for years and you may be happy with it, but I don't like it. YMMV.

Actually, 2.66666~(infinate) to be totally anal . I was being a little to general in my previous explanation of parallel ohmages (this is how we get general/ballpark ohmages on the fly). The actual, correct formula is 1/R = 1/R1 + 1/R2 + 1/R3 + (etc.,etc.) 1/R being your final load. 1/R1 being the resistance of one of your cabs, 1/R2 being another and so on.

Phats, you remind me of one of my electronics instructors) If your calculator had 10 digits, that's what you had to give for your answer! Anyhoo, don't forget about the product over the sum method which works for 2 resistances in parallel and is simpler: Rt=R1xR2/R1+R2 (Rt=total resistance)

Wow! A new formula I wasn't aware of. Maybe its because, as you can see (from my previous post) my calculator only has 6 digits (damn Radio Shack) .

I almost forgot: Any time you have resistances in parallel, the total resistance will always be lower than the lowest resistance in the circuit, regardless of how many resistances you have.