On 9/14/2015 5:48 PM, Wayne.B wrote:
On Mon, 14 Sep 2015 05:37:09 -0600, Paul Cassel
wrote:


===

It's basically simple vector algebra. Assume a speed through the
water of 5 kts on a course of due east. Also assume a north flowing
current of 5 kts. The resultant course made good would be north east
at 7 kts.



I see where we disagree. My logic was course to the destination and
that's not enhanced by a current which is 90 degrees to the desired course.

I guess until Skip checks in, we'll not know exactly what he was about.

-paul

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