Let's see how much power that amp puts out when driven into clip, which is (unfortunately) very common for subwoofer amps.
I see what you meant now. To answer your question, an amplifier will output a maximum of double the output power when driven into clipping, which will be 80W per channel in this case. This is because the RMS voltage of a squarewave is equal to its peak value and the RMS voltage of a sine wave is equal to V/root(2). R = V
2/R and V
RMS = V
P/root(2).
For example:
Suppose the maximum output voltage of the amplifier is 8V and the speaker impedance is 4Ohms
If the output is a sinewave, without any clipping, the peak voltage of the sinewave will be 8V. Calculate the RMS voltage:
V = 8/(root(2)) = 8/1.414 = 5.657V
R = 4
Calculate the power
P = V
2/R
P = 5.657
2/4 = 32/4 = 8W
Suppose it's driven into clipping, so the output is a squarewave. The RMS voltage of the squarewave is simply 8V.
Calculate the power
V = 8
P = V
2/R = 8
2/4 = 64/4 = 16W
See it's double!
In reality the user will not drive their amplifier so far into clipping to the point of it being so distorted, it's a squarewave, giving 80W per channel, a total of 160W. They most likely reduce the volume long before that point and the music won't be playing at a full duty cycle, so it's a non-issue