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Stokes Law is not going to apply in this situation because the water flow around the ball will be turbulent not laminar. The way to see this is to calculate the Reynold's number. For a sphere this is approximately given by: $$ Re \approx \frac{\rho_wdv}{\mu} $$ If we feed in $\rho_w = 1000$ kg/m$^3$, $d = 0.00317$ m, $v = 37$ m/s and $\mu = 0.001$ Pa.s ...


A single measurement like this has a lot of noise on it - and random signal is always going to have some random correlation. You should definitely not pay too much attention to the stuff that is in the tail of the correlation distribution - it's all noise. The fact that the built in function does not produce negative values is related to you only looking at ...

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