= Quadratic-drag turning angle
{title2=$\tan\theta=\mu|f|/(\rho_wC_wU)$}
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 $b=\rho_wC_w$ and $f=2\Omega\sin\phi$, the steady balance is $a\mathbf e_x=bU\mathbf u+\mu f\widehat{\mathbf z}\times\mathbf u$. Resolving along the drift and perpendicular to it gives $a\cos\theta=bU^2$, $a\sin\theta=\mu|f|U$. Thus $b^2U^4+(\mu f)^2U^2=a^2$ 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.
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