Example forward-difference

Consider the example of an ADCP with a beam angle of $20^{\circ }$, configured with a vertical bin size of 10 cm, recording profiles at 1 second intervals with a data segment length of 300 seconds. The Level 1 QC of the data identified that good data was typically returned from bins 1 to 30.

The velocity data from a single beam for a single data segment can therefore be visualised as:

Note: $b^{\prime }$ is synonymous with $v^{\prime }$ in this figure and the one below.

The square of the velocity difference between bins separated by $\delta$ bins is then evaluated for each $t$. So for bin 4, $\delta =3$ and $t=2$, we get:

${\mathrm{\Delta }}^2(4,3,2)={[b^{\prime }(4,2)-b^{\prime }(7,2)]}^2$

For bin 1 the squared velocity difference can be evaluated for $1⩽\delta ⩽29$, whilst for bin 2 it is restricted to $1⩽\delta ⩽28$, reducing by 1 with each bin, so that for bin 29, it can only be evaluated for $\delta =1$ and there are no options for bin 30. This is summarised as follows:

The mean is then taken across the 300 profiles in the data segment i.e.

$D(1,\delta )=\sum _{t=1}^{300}{\mathrm{\Delta }}^2(1,\delta ,t)$

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