Convert the shear probe data

copy&paste from V1 docx

Convert the shear-probe data samples into physical units using the standard equation,

$s=\frac{N_s}{2\sqrt{2}S{U}^{2}G\gamma }$

where $s$ is the shear signal in physical units of ${\mathrm{s}}^{-1}$ , $N_s$ are the raw numeric samples (the output of an analog-to-digital converter), $S$ is the calibrated sensitivity of the shear-probe in units of $\mathrm{V}/\mathrm{(}\mathrm{m}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}{\mathrm{)}}^{\mathrm{2}}$, $U$ is the speed of profiling in $\mathrm{m}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}$, $G$ is the gain of the electronics of the shear probe in units of $\mathrm{s}$, and $\gamma$ is the gain of the analog-to-digital converter used to create the samples in units of $\mathrm{c}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{t}\mathrm{s}{\mathrm{V}}^{\mathrm{-}\mathrm{1}}$.

The sensitivity of the shear probe to shear is proportional to the square of the speed of profiling. Thus, one should set a minimum speed for the conversion of the shear-probe data into physical units. Otherwise, the conversion may produce enormously large and quite unrealistic values. Realistic minimum speeds for the conversion into physical units are $0.05$ to $0.1\mathrm{m}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}$ because the shear-probe signal is likely to be dominated by electronic noise and the angle of attack will be large ($>20^{\circ }$) even for low levels of dissipation.