 Quadratic-drag turning angle

For an ice mass in still water, approximate the wind stress by a fixed vector $a{\mathbf{e}}_x$ while retaining quadratic drag in the water. With mass per water-drag area $\mu =m/A_w$, water coefficient $b={\rho }_w{C}_w$ and $f=2\Omega \sin \varphi$, the steady balance is $a{\mathbf{e}}_x=bU\mathbf{u}+\mu f\hat{\mathbf{z}}\times \mathbf{u}$. Resolving along the drift and perpendicular to it gives $a\cos \theta =b{U}^2$, $a\sin \theta =\mu |f|U$. Thus $b^2{U}^4+(\mu f{)}^2{U}^2=a^2$ and $\tan \theta =\mu |f|/(bU)$. The direction is right of the wind in the northern hemisphere and left in the southern hemisphere. This angle neglects intrinsic boundary-layer turning, ocean currents and contact forces; it is not a universal observed sea-ice turning angle.