= Quadratic-drag free-drift turning angle
{title2=$\tan\theta=mf/(\rho_w C_w q)$}
For stationary level water and wind stress approximated by $F=\rho_a C_a U^2$ in the wind direction, equilibrium with quadratic water drag gives $F^2=(\rho_w C_w)^2q^4+(mf)^2q^2$. The signed angle to the right of the wind is $\theta=\arctan[mf/(\rho_w C_wq)]$. This neglects the floe velocity relative to the wind in the air drag; the exact relative-wind equation adds $\rho_a C_a|U-v|$ to the angle denominator.
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