The significant wave height is defined as [math]H_s = 4 \sqrt{\sigma^2}[/math], where [math]\sigma^2[/math] is the variance of the surfave elevation.

 

The variance of a sine wave is [math]\dfrac{a^2}{2}[/math], where [math]a[/math] is the amplitude of the wave. A single wave [math]sin(t)[/math] thus has a variance of [math]\sigma^2 = \dfrac{1}{2}[/math], and a significant wave height of [math]H_s = 4 \sqrt{\sigma^2} = 4 \sqrt{\dfrac{1}{2}} \approx 2.83 m[/math], while the wave height of the sine wave is only [math]2 \cdot a = 2 m[/math].

 

How is this possible?

From the definition, it sounds like this is applied to wave systems observed in nature, which have a certain amplitude distribution. Wikipedia says that they typically follow a Rayleigh distribution. Does a sine wave follow a Rayleigh distribution? http://en.wikipedia.org/wiki/Significant_wave_height

Ah, thank you. It's obvious now.

You should also take note that as waves progress to shallow(er) water the wave height increases and the profile departs markedly from sinusoidal, in particular the crest amplitude and trough depth are no longer equal or equal to half the sinusoidal amplitude, a. See the Reid and Bretschneider 'Breaking Index Curve'.