Discussion
Which estimate of vorticity does a strongly curved trough with 20 metres per second flow give?
Answer: D
Dividing 20 metres per second by a radius of \( 5 \times 10^{5} \) metres gives \( 4 \times 10^{-5} \) per second from curvature alone, which is a substantial fraction of the Coriolis parameter and a clearly significant vorticity maximum. Shear would add to this but the curvature term alone gives a useful estimate, which is exactly why the natural coordinate decomposition is so handy for reading a chart by eye.
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