TwinTurbo is correct. Take a simple sine wave. At the point of clipping (Vc), the voltage output is a sine wave of P-P voltage of 2Vc. That has an RMS voltage value of 0.7Vc. and the power into load R is (0.7Vc)²/R or 0.5Vc²/R

When the clipping is severe, the waveform is a square wave of P-P voltage of 2Vc. That has an RMS voltage value of Vc. and the power into load R is Vc²/R or twice that of a sine wave.

For other more complicated waveforms, I believe that ratio increases.

edit: a bit more detail. The RMS value of a sine wave is 1/√2 times the peak amplitude. or 0.707Vc

TT, I am not confusing audible power with electrical power. I said louder. The power being dissipated is increasing, but not that much. One you reach clipping, the limit is going from a sine wave to a square wave. If I recall correctly, a sine wave of the same amplitude as a square wave has about 88% of the power under the curve compared to the square wave. So the maximum gain in power would be about 13.6%.

In my original statement that started all this, I said something like that a low powered amp will not blow speakers rated for higher power. Obviously I did not mean 60w rated speaker and a 59 w amp.

hmm … well, let me put my math hat on. As the Beatles say “We Can Work it Out” … lol … Power calculations for AC signals is usually done via their rms value. Root mean square. That’s sort of the average voltage of the signal you use to calculate the signal power. You can’t use just the average voltage of the signal b/c the average of a sin wave for example is 0, but a sin wave has more than 0 power.

So let’s do rms technique. The rms value of a sin wave is 0.707 of the peak value. Isn’t that correct? And the rms value of a clipped sin wave (i.e. a square wave) is the same as its peak value, isn’t that right?

So comparing the two powers, given that power is proportional to voltage squared, so we have to use the square of the rms value – that’s correct right? — then the power comparison, sin wave vs clipped sin wave would be

Power of sin wave = .707^2 = ~0.5
Power of square wave = 1

Or do I need to join Ed Bill in the remedial math classes? … lol …

Your math makes since, but when I was studying this in school, it wasn’t done that way. School was over 40 years ago. It was a calculus function and I haven’t used calculus ever since graduating.

But as I recall, a square wave was the fundamental frequency and all the odd harmonics added in phase. The maximum power in the third harmonic was 1/9th of the fundamental (1/3 voltage x 1/3 current). The 5th harmonic had 1/25 th of the power, 7th has 1/49th ect. The total power of all the harmonics comes nowhere near the power in the fundamental so a square wave could not have twice the power of the sine wave.

Now I can simply turn your numbers around. The peak is .414 higher than the RMS and .414 squares is .172 so the power would go up 17.2% from a sine wave to a square wave. But I don’t think that is right either.

The reason square wave isn’t half is because a sinusoidal wave spends more time at extremes than in the center.

The first derivative of sin is cos…so at extreme displacement from the mean (sin(n)=1)…acceleration is 0…(cos(n)=0). Thus, “on average,” a sine wave spends more time at extremes than at zero.

If we were talking about a “zig-zag” wave (straight line from maximum to minimum,) then you’d be right.

When you use the RMS value of a square wave and compare it to the RMS value of a sine wave of the same amplitude, that cannot be done, they are completely different signals and in no way connected.

One small error, well a really big one at that in my earlier post. The square wave is the fundamental and all the odd harmonics algebraically added in phase. That one word makes a huge difference because at the peak of the fundamental, all the harmonics are out of phase algebraically from the fundamental. They are all in phase at the beginning and end of the fundamental waveform.

That lowers the amplitude of the resulting square wave compared to the peak amplitude of the applied sine wave. This not only works out mathematically but in the real world as well. If you overdrive an amp, the amplitude of the output waveform goes down as the input is increased and more clipping occurs.

A square wave is the sum of the instantaneous values of a sine wave and its odd harmonics at strengths inversely proportional to their order. Fundamental at 100%, third at 1/3, 5th at 1/5, 7th at 1/7 and on and on.
If you include the even harmonics in there along with the odd, you get a sawtooth wave.

A square wave definitely has a different frequency spectrum than a sin wave. It seems like if the load to which the power is delivered is not dependent on frequency … a resistor for example … it wouldn’t matter, the calculations above would be correct. But if the load changed vs frequency, then my calculations wouldn’t apply. I was just addressing the statement you cited above as being inaccurate

"

Actually, the power doubles from sine to square. See previous post. "

Now I can simply turn your numbers around. The peak is .414 higher than the RMS and .414 squares is .172 so the power would go up 17.2% from a sine wave to a square wave. But I don't think that is right either.

Don’t the two prototype signals we’re talking about for comparison share the same peak voltage of 1? I don’t see what you mean.

Since rock and blues musicians purposely overdrive their amps for distortion, does it harm your stereo to listen to it? How about a clarinet? Being a stopped tube, it only plays odd harmonics and so a square wave kind of has a clarinet tone.
The “warm” sound of overdriven tube amps comes mostly from the asymetrical clipping which puts even harmonics in the sound. A class A tube amp can be driven into cutoff yet not into saturation or vice versa depending on the bias voltage.

If the volume was loud enough it could damage the speakers, but the distorted signals heard in rock and blues music — like in Jimi Hendrix’s tunes — don’t seem any louder than the rest of the piece, just distorted. I expect part of the record making process is to limit how loud a segment of the track can be.

" I expect part of the record making process is to limit how loud a segment of the track can be. "

I think this brings us back to where it all started. I said a low power amp will not harm speakers with a higher power handling capability. Most of the time (not always) the higher power speakers are not as efficient, so the low power amp has to be turned up to its max volume. But at max volume, even if the amp is driving the final stage to saturation and clipping the signal, distorting it, creating high frequencies etc, it is still not loud enough to hurt the speakers, because it doesn’t have the power to begin with.

"Don’t the two prototype signals we’re talking about for comparison share the same peak voltage of 1? I don’t see what you mean. "

Yes and no. The two theoretical signals are different signals. The theoretical sine wave has a lower RMS than the theoretical square wave and the RMS of the square wave would have the same RMS of a much higher sine wave. Its apples and oranges.

Most of the time (not always) the higher power speakers are not as efficient, so the low power amp has to be turned up to its max volume.

I would think that if speakers were less efficient at low volume, they would distort the dynamics of a musical piece. Although the dynamics of most recorded music is already distorted by the use of compression, in fact compression is a necessary evil in music listened to in a noisy environment such as a car going down the highway, especially in classical music where there is much use of crescendos and decrescendos.
Without compression, you would find yourself turning up the volume during the second movement of “The William Tell Overture” so you could hear the solo woodwinds play that pastoral tune over the road and wind noise, and then turn the volume down for the third movement where the brass and percussion come in so it doesn’t blast your eardrums.
Really, the best possible speakers for a moving car listening environment would be a set of noise cancelling headphones, but I’m not sure if it’s legal to drive that way.
Remember when Ed Sullivan introduced the Beatles on his show? Am I the only one who watched it wishing all those screaming girls would just shut up? It really drives home the importance of a good signal to noise ratio.

Really, the best possible speakers for a moving car listening environment would be a set of noise cancelling headphones, but I'm not sure if it's legal to drive that way.