Example forward-difference: Difference between revisions
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The velocity data from a single beam for a single data segment can therefore be visualised as: | The velocity data from a single beam for a single data segment can therefore be visualised as: | ||
[[File:Velocity data.png|frameless|center| | [[File:Velocity data.png|frameless|center|400px]] | ||
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: |
Revision as of 10:14, 14 November 2021
Consider the example of an ADCP with a beam angle of [math]\displaystyle{ 20^{\circ} }[/math], 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 [math]\displaystyle{ \delta }[/math] bins is then evaluated for each [math]\displaystyle{ t }[/math]. So for [math]\displaystyle{ t_1 }[/math], bin 1 and [math]\displaystyle{ \delta=1 }[/math], we get:
- [math]\displaystyle{ \Delta^2(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]\displaystyle{ 1\leqslant\delta\leqslant29 }[/math], whilst for bin 2 it is restricted to [math]\displaystyle{ 1\leqslant\delta\leqslant28 }[/math], reducing by 1 with each bin, so that for bin 29, it can only be evaluated for [math]\displaystyle{ \delta=1 }[/math] 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.
- [math]\displaystyle{ D(1,\delta) = \sum_{t=1}^{300}\Delta^2(1,\delta,t) }[/math]
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