Example forward-difference: Difference between revisions

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The square of the velocity difference between bins separated by <math>\delta</math> bins is then evaluated for each <math>t</math>.  So for <math>t_1</math>, bin 1 and <math>\delta=1</math>, we get:
The square of the velocity difference between bins separated by <math>\delta</math> bins is then evaluated for each <math>t</math>.  So for <math>t_1</math>, bin 1 and <math>\delta=1</math>, we get:


:<math>\Delta{v}(1,1,1) = \left[v^\prime(1,1) - v^\prime(2,1)\right]^2</math>
:<math>\Delta{v^\prime}(1,1,1) = \left[v^\prime(1,1) - v^\prime(2,1)\right]^2</math>


For bin 1 the squared velocity difference can be evaluated for <math>1\leqslant\delta\leqslant29</math>, whilst for bin 2 it is restricted to <math>1\leqslant\delta\leqslant28</math>, reducing by 1 with each bin, so that for bin 29, it can only be evaluated for <math>\delta=1</math> and there are no options for bin 30.  This is summarised as follows:
For bin 1 the squared velocity difference can be evaluated for <math>1\leqslant\delta\leqslant29</math>, whilst for bin 2 it is restricted to <math>1\leqslant\delta\leqslant28</math>, reducing by 1 with each bin, so that for bin 29, it can only be evaluated for <math>\delta=1</math> and there are no options for bin 30.  This is summarised as follows:

Revision as of 17:35, 12 November 2021

Consider the example of an ADCP with a beam angle of 20, 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 square of the velocity difference between bins separated by δ bins is then evaluated for each t. So for t1, bin 1 and δ=1, we get:

Δv(1,1,1)=[v(1,1)v(2,1)]2

For bin 1 the squared velocity difference can be evaluated for 1δ29, whilst for bin 2 it is restricted to 1δ28, reducing by 1 with each bin, so that for bin 29, it can only be evaluated for δ=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 e.g.

D(1,1)=t=1300D(1,δ,t)

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