What hurts a subwoofer?

*COUGH* hurtED *COUGH*
hurts=curent

hurtED=past.

Dont worry I fixed it for you //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif
hurts=current

hurt=past

Don't worry I fixed it for you //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

(Don't worry I fixed it for you //content.invisioncic.com/y282845/emoticons/tongue.gif.6130eb82179565f6db8d26d6001dcd24.gif )

Thanks for the rep.

 
I have a question about how a amplifier running into full clipping will be pushing ~double power.
I wish I had an oscilloscope scope on me but, once an amplifier clips doesn't the amplitude of that wave stay the same if you keep on turning the volume up. For example using ramdom figures, if an amplifier has 1/8 of the sine wave clipped and the amplitude voltage (amplitude being peak to reference) is 40V, once you turn the gain up on the amplifier would it still be at the 40V I would guess that the more you turn the gain up the voltage should still be at 40V, if the voltage did not stay the same and it increased the amp will still have the same 1/8 clipped.

In order for the amplifier for the amplifier to put out more power, the amplitude must be increased such as your 2000w example, the only way else that 1000w amp can put out more power if it puts out more current. I hope you understand my question. Basically the question I'm asking, if an amplifier runs into clipping how does it increase its power, I sorta considered when the amp is clipping it is sorta at it's zenith meaning thats how much it can put out and that's it.
I *think* I understand your question. Lets take your example, and say an amplifier has a peak voltage of 40V.

For a sinewave, continuous power is found by the following formula;

Vrms = Vpeak * .707

So, for your example of 40V, Vrms = 40V *.707 = 28.28V

And, to make things simple, lets say we are powering a 1ohm load. So, to find power, we use ohms law of P = V^2/R

28.28^2/1 = 799.76w RMS

So, that amplifier would have a continuous power output (RMS) of right around 800w.

However, for a square wave (what would be "full clipping"), the volts RMS becomes the same as volts peak, or Vrms = Vpeak, or 40V

So, to find power again using ohms law;

40^2/1 = 1,600w

So, there's the math to show how and why clipping creates double the power.

 
Don't most SPL competitors drive their power to clipping???

yes and no

I know some that just crank all the knobs gain/boost and go burp and i know some that look for just slight clipping and then some that use pure clean power. I use clean power found out by an oscope.

 
Canoes... Yes, canoes.
Funny story... So I was fishing with my friend in a nearby lake in his canoe. We caught all we cared to catch, so we decided to head out. So we load the canoe into the back of my Explorer, and I knocked my sub box over... oops. So we get back to my friend's house, and I discover that the canoe didn't hit the box, but the sub itself, and went straight through.

So I can say I have killed a sub with a canoe. Beat that.
pffft, I killed a sub by calling it names //content.invisioncic.com/y282845/emoticons/imdaman.gif.bc6c552e19aa4ad8c47461144f40eb1a.gif

 
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