Elaine Harmon wrote: > > Hi guys, here's another one: with all these complicating factors of wind, > current, boat design... can we always say that the most energy-efficient > route from one point to another (assuming homogeneous conditions, of > course) is in a straight over-the-ground course between the 2 points? ... I have worked this out for a current (using some simplifying assumptions), but not for a side wind. The latter is more complicated, due to the slide slipping. (The current moves you sideways with the water, but not sideways relative to the water. The wind blows you sideways relative to the water, in addition to the former.) In the simpler case of a current, there are some circumstances where you are better off if you do not ferry across, but instead paddle with your boat pointed straight across and then paddle forward along the bank into the current to recover from your sideways drift. When would this be true? The following draws from my earlier posting on this subject: For the case of a current with uniform strength, my very rough model suggests that the break-even point is for a ferry angle of about 57 degrees, which occurs when the ratio of paddling speed to current speed is approximately 1.19. In other words, if you can paddle more than 19 percent faster than the current, your ferry angle will be less than 57 degrees, and it will take less time to ferry across than to paddle at a straight-across heading (with zero ferry angle, followed by a paddle up current). On the other hand, if your paddling speed is less than 19 percent faster than the current, your ferry angle would have to exceed 57 percent, in which case it will take you less time if you paddle a straight-across heading (zero angle), even though you have to paddle up current once you reach the other side. Now for the math! Using some fairly straightforward trigonometry, it can be shown (if I haven't made an error) that the ratio of the straight-angle time (including the up-current paddle) to the ferry-angle time, st/ft, equals (1+c/p)cos(a), where "c" is the speed of the current, "p" is the paddling speed, and "a" is the ferry angle (in degrees) which itself is a function of c/p. The necessary ferry angle "a" equals the inverse sine of c/p. In other words, the time ratio st/ft is a function only of c/p. This ratio equals one when c/p is approximately equal to .83867, which corresponds to a ferry angle of 57 degrees. If the ratio c/p exceeds .83867 (i.e., if p/c is less than 1.1924), then st/ft is less than one and ferrying will actually take more time than paddling at a straight-across heading, even though you have to paddle up current. (I have checked this against a few simulations, and this seems to work, but as always someone should check the math...) Next time you have to cross a current, ask yourself if you can paddle more than 19 percent faster than the current. If so, then you MIGHT save time by ferrying. I say that you might save time, because the above analysis assumes a current of constant strength. But I have never seen such a current. It is typically slower near shore, even in the absence of eddies. And with eddies it is, of course, a whole new ballgame. So actually, it is quite a bit more complicated than in the simple model above (it always is). Add your own fudge factor to the 19 percent rule. But perhaps the above "rule" provides a crude staring point. I would certainly welcome a more complex model, if someone knows of one or has the desire to develop one. Again, the case of a side wind would be more complicated. But given that ferrying is not always faster for a side current, the same may be true in the case of a side wind--but the parameters would certainly change. Dan Hagen *************************************************************************** PaddleWise Paddling Mailing List Submissions: paddlewise_at_lists.intelenet.net Subscriptions: paddlewise-request_at_lists.intelenet.net Website: http://www.paddlewise.net/ ***************************************************************************Received on Sat Oct 02 1999 - 14:56:36 PDT
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