#1
Did some searching and found a thread that explained all the complicated physics and whatnot behind this but I'm looking for a way to make it easier for myself and everyone.

Is there a formula or way to calculate if one head will work with a certain cab that is rated at less watts or different ohms (Ω. Or how to calculate how much power I'll get running so many watts through either 4Ω or 8Ω .

Say I ran a 600 watt head through a 300 Watt cab that is 4Ω its gonna blow up right? How do you figure that out.

If I ran the same head through a 300 watt cab that is 8Ω its gonna be fine though right?

Basically I'm looking for a general formula so I can walk into a music store, pick a head and cab and figure out how many watts i'll actually be able to get out of that cab with different heads, or how many watts i'll get out of a cab with different heads...

Thanks,
Confused
#2
I in Ampère/U in volt=Ω in Ohm (Is wat I thought it was)
It depends on how may volt and how many ampère your amp can handle.

And that 600 to 300 watt. The 600 watt has 4Ω?

Then it`s P=UxI
600=230x2,61.
2,61 ampère is alot
If you have 8Ω it will be 1,31 wich he probably can handle.
I don`t think it can handle that.

I=power in A(mpere)
U=...(don`t know the name) in V(olt)
Ω=Resistance in Ohm.

But you can adjust your volume can`t you? So there`s a bigger resistance in it then 4Ω I think.


By the way: I`m 15 so I don`t know if this is entirely right. I only know the formulas and did the math so...
#3
Quote by MaXiMuse
I in Ampère/U in volt=Ω in Ohm (Is wat I thought it was)
It depends on how may volt and how many ampère your amp can handle.

And that 600 to 300 watt. The 600 watt has 4Ω?

Then it`s P=UxI
600=230x2,61.
2,61 ampère is alot
If you have 8Ω it will be 1,31 wich he probably can handle.
I don`t think it can handle that.

I=power in A(mpere)
U=...(don`t know the name) in V(olt)
Ω=Resistance in Ohm.

But you can adjust your volume can`t you? So there`s a bigger resistance in it then 4Ω I think.


By the way: I`m 15 so I don`t know if this is entirely right. I only know the formulas and did the math so...

ok. this needs to be dumbed down.
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#5
You're overthinking everything.

Look, heads put out a certain amount of wattage at a certain resistance. Pretty much all heads tell you exactly the power they run at with any given resistance. For example, the Carvin BX600 pushes out 225W when connected to an 8 ohm load, 400W when connected to a 4 ohm load, and 600W when connected to a 2 ohm load. That's all.

NOW, if you have an amp that PUSHES 600W AT 4 OHMS and connect it to a 300W 4 OHM CAB, then you have a good chance of blowing it if you turn the volume past ~halfway. If the amp pushes 600W at another impedence, you may or may not have to worry.
Quote by Cody_Grey102
I was looking at a used Warwick Vampyre LTD 5'er for about $200. I went home to grab my wallet and came back and some jerk with an epic beard got it already..
#6
I was looking for something more like what thefitz posted. Except that when I've gone into guitar centers I've never been able to figure all of that out by just looking at the stuff and there are never enough staff to help me figure it out.

Maximuse I don't think i'm gonna be taking scratch paper and a calculator into a guitar center and working all that out. thanks though. I was hoping there was a simpler formula but I cant remember much of what I learned in physics last year.

Anyone else have suggestions or whatnot. I'm basically just trying to keep it so that I don't have to post a will this head work with this cab threads for every stack i look into.
#9
Did you read my post? Get cab that handles more power than the head puts out. Choose the impedence that'll give you the power you want. Done.
Quote by Cody_Grey102
I was looking at a used Warwick Vampyre LTD 5'er for about $200. I went home to grab my wallet and came back and some jerk with an epic beard got it already..