EXAMPLE PROBLEM III-1-1

FIND:

The statistics of the sediment size distribution shown in Figure III-1-2 and their qualitative

descriptions.

GIVEN:

The needed phi values are: φ05 = 0.56, φ16 = 0.80, φ25 = 0.93, φ50 = 1.37, φ75 = 1.87, φ84 = 2.08,

and φ95 = 2.48.

SOLUTION:

In phi units, the median grain size is given as:

φ50 = *M*dφ = 1.37φ

From Equation 1-1b, the median grain size in millimeters is found as:

From Equation 1-2, the mean grain size is found in phi units as:

From Equation 1-1b, this is converted to millimeters as:

From Equation 1-3, the standard deviation is found as:

σφ = (2.08 - 0.80)/4 + (2.48 -0.56)/6.0 = 0.32 + 0.32 = 0.64 φ

From Equation 1-4, the coefficient of skewness is found as:

αφ = 0.055 + 0.078 = 0.13

From Equation 1-5, the coefficient of kurtosis is found as:

βφ = 1.92 / 2.29 = 0.84

Thus, using Tables III-1-2 and III-1-3, the sediment is a medium sand (Wentworth classification),

it is moderately well sorted, and the distribution is fine-skewed and platykurtic.

Coastal Sediment Properties

III-1-13

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