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Under geostrophic balance, one can write

$$\frac{V^2}{R}+fV-fV_g=0$$ where $V:$wind speed, $V_g:$ geostrophic wind speed, $f:$ Coriolis parameter, and $R:$ radius of curvature.

Solving for $V$, we can get a relationship between $V$ and $V_g$ as $$V=-\frac{fR}{2}\pm\frac{\sqrt{f^2R^2+4fRV_g}}{2}$$

How does one then determine when to use $\pm$ for cyclonic and anti-cyclonic flows?


I was able to find a solution from these slides 40-41 and I am also aware that the $R$ can be positive and negative, and plays a part in being physically meaningful as seen here. However, I fail to understand why and how the $\pm$ signs come into play. To be clear, I am referring to the $\pm$ between the two terms on the RHS.

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It's a general problem that quadratic equations have two roots when only one is expected. One will then be unphysical and we have to decide which is the one we want.

In this case we can proceed as follows: When $|R|$ becomes very large (either positive or negative) then the centripetal acceleration becomes negligible and $V\to V_g$. The sign has to be chosen to make this happen. So (assuming that $f$ is positive) for $R$ positive we must take the postive square root and for $R$ negative we must take the negative root. To see that this is so it helps to write your quadrtic solution as $$ V=-\frac{fR}{2} \pm |fR| \frac {\sqrt{1+4V_g/fR}}{2}\\\approx -\frac{fR}{2} \pm |fR|(1+V_g/fR+\ldots) $$ The last expansion is valid when $|R|$ is large.

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  • $\begingroup$ Is it possible to extend your answer to include how cyclonic and anti-cyclonic flows affect this decision, or will that fall under a different kind of consideration? $\endgroup$ – fromzero Oct 29 '20 at 22:29
  • $\begingroup$ Ah, since anti-cyclonic flows have $R<0$, thus we will use the $-\frac{fR}{2}-...$ version as explained. Vice-versa, cyclonic flows have $R>0$ and $-\frac{fR}{2}+...$ version. $\endgroup$ – fromzero Oct 29 '20 at 22:32

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