Example bin-centred 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:

The along-beam bin separation distance, $r_0=0.1/\cos 20^{\circ }=0.1064\text{m}$, so evaluating $\epsilon$ estimates on the basis of $r_{\text{max}}$ of $\sim 1.5\text{m}$ implies $n_{\text{rmax}}=1.5/0.1064\approx 14$ bins. The lowest number bin over which this separation distance can be centred is bin 8 $(n_{\text{rmax}}/2+1)$ and is evaluated using data from bins 1 and 15. The highest number bin for which the separation distance can be evaluated is 23, using data from bins 16 and 30. So the bin range for which $1⩽\delta ⩽14$ can all be evaluated is $8⩽n⩽23$.

Note that the structure function can be calculated for bins outside this range, but not all of the desired separation distances can be evaluated, limiting the range available for the regression. Care should therefore be exercised in comparing the resulting $\epsilon$ estimates.

For each profile $t$, the squared velocity difference ${\mathrm{\Delta }}^2(n,\delta ,t)$ is evaluated for each of the bins in the range $8⩽n⩽23$ and each of the separation distances $1⩽\delta ⩽14$.

1. If $\delta$ is even then the squared velocity difference is based on the bin velocities for the bins symmetrically spaced either side of the evaluation bin. So for the first profile in the segment $(t=1)$, bin $n=8$ and separation distance $\delta =2$:
   
   ${\mathrm{\Delta }}^2(8,2,1)={[v^{\prime }(7,1)-v^{\prime }(9,1)]}^2$
   
   similarly for $\delta =10$:
   
   ${\mathrm{\Delta }}^2(8,10,1)={[v^{\prime }(3,1)-v^{\prime }(13,1)]}^2$
   
   
2. If $\delta$ is odd the calculation is complicated by the fact that the separation distance cannot be exactly centred on the bin, so the mean of the squared velocity difference options centred on the bin edges is used. Again using the example of the first profile in the segment $(t=1)$ and bin $n=8$, but with separation distance $\delta =3$:
   
   ${\mathrm{\Delta }}_{\text{lo}}^2(8,3,1)={[v^{\prime }(6,1)-v^{\prime }(9,1)]}^2$
   
   ${\mathrm{\Delta }}_{\text{hi}}^2(8,3,1)={[v^{\prime }(7,1)-v^{\prime }(10,1)]}^2$
   
   and
   
   ${\mathrm{\Delta }}^2(8,3,1)=\frac{{\mathrm{\Delta }}_{\text{lo}}^2(8,3,1)+{\mathrm{\Delta }}_{\text{hi}}^2(8,3,1)}{2}$
   
   similarly for $\delta =13$:
   
   ${\mathrm{\Delta }}_{\text{lo}}^2(8,13,1)={[v^{\prime }(1,1)-v^{\prime }(14,1)]}^2$
   
   ${\mathrm{\Delta }}_{\text{hi}}^2(8,13,1)={[v^{\prime }(2,1)-v^{\prime }(15,1)]}^2$
   
   and
   
   ${\mathrm{\Delta }}^2(8,13,1)=\frac{{\mathrm{\Delta }}_{\text{lo}}^2(8,13,1)+{\mathrm{\Delta }}_{\text{hi}}^2(8,13,1)}{2}$

Having evaluated the squared velocity difference for the range of bins and separation distances for each of the profiles, the structure function is evaluated by taking the mean across the 300 profiles in the data segment i.e.

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

if all options are evaluated.

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