 Quadratic-drag free-drift turning angle

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 $F^2=({\rho }_w{C}_w{)}^2{q}^4+(mf{)}^2{q}^2$. The signed angle to the right of the wind is $\theta =\arctan [mf/({\rho }_w{C}_{w}q)]$. 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.