5.3: Frequency of Upcrossings

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Page ID: 47246

Franz S. Hover & Michael S. Triantafyllou
Massachusetts Institute of Technology via MIT OpenCourseWare

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The first statistic we discuss is the frequency with which the process exceeds a given level; we consider upcrossings only of the positive value \(A\). Now let \( \bar{f}(A) \) be the average frequency of upcrossings past \(A\), in upcrossings per second. Then \( \bar{f}(0) \) is the average frequency of zero upcrossing, or \(1 / \bar{T}\), the inverse of the average period, \(E(T)\). The formulas are

\begin{align} \bar{f}(0) \, &= \, \dfrac{1}{2 \pi} \sqrt{ \dfrac{M_2}{M_0} } \\[4pt] \bar{f}(A) \, &= \, \dfrac{1}{2 \pi} \sqrt { \dfrac{M_2}{M_0} } e^{- A^2 / 2 M_0} . \end{align}

With \(M_0\) equal to the variance, the exponential here clearly is of the Gaussian form. Here is an example of the use of these equations in design. An fixed ocean platform is exposed to storm waves of standard deviation two meters and average period eight seconds. How high must the deck be to only be flooded every ten minutes, on average?

This problem does not involve any transfer function since the platform is fixed. If it were floating, there would be some motion and we would have to transform the wave spectrum into the motion spectrum. All we have to do here is invert the equation to solve for \(A\), given that \( \bar{f}(A) = 1/(60 \times 10), \, M_0 = 4, \) and \( \bar{T} = 8 \) or \( \bar{f}(0) = 1/8\):

\[ A \, = \, \sqrt{-2 M_0 \ln (\bar{T f}(A))} \, = \, 5.87 m. \]

This result gives a flavor of how valuable these statistics will be - even though the standard deviation of wave elevation is only two meters, every ten minutes we should expect a six-meter amplitude!