Why Is The Full-Wave Voltage Not Twice As Large As That Of The Half-Wave Rectification?

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Why is the full-wave voltage not twice as large as that of the half-wave rectification?

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The answer to this question is very complex due to the ac voltage given by the rms value (which is the root of the means squared), effectively integrating the sine function over time.

But the full wave voltage not being double the half wave voltage can be demonstrated by simple circuit analysis (avoiding complex integration).

Consider a 1Vac rms sine wave fed into a 1 ohm resistance; by ohms law the current will be 1A and the power dissipated by the resistor 1W.
Now consider a modified circuit where a 1Vac rms sine wave has the positive cycle half wave rectified fed into a 1 ohm resistance and the negative cycle half wave rectified also fed into a separate 1 ohm resistance.
Ignoring any diode voltage drop, the power dissipation in each resistor will be half a Watt since the 1Vac source will be loaded by 1 ohm over the full cycle.

The voltage across each 1 ohm resistor will be a half wave rectified voltage; from the equation V = square root of (WxR), we get V=square root (0.5 x 1) = 0.707V rms, and not 0.5V rms that might otherwise be expected from visual inspection of the waveforms.