leong Posted February 12, 2013 Share Posted February 12, 2013 As we all know, to move across current to a waypoint you need to angle your kayak upstream enough so that your strokes work to counterbalance the downstream drift. The proper heading angle is called the crab angle (or ferry angle). Holding your heading at the crab angle, you will travel a straight line to the waypoint and, obviously, this is the shortest distance to the waypoint. Under the simple assumptions of a right-angle current, the ferry angle is calculated as the arcsine of (Current Speed/ Paddling Speed). A fairly decent approximation is Current Speed ÷ Paddling Speed X 60°. With a GPS, all you have to do is keep the arrow (line or whatever) lined up to the top of the display.But, instead, suppose you want to cross the river in the minimum time and are indifferent to where you arrive on the other side of the river. This could be important; for example, if you are heading from an island to shore when a storm is coming soon you may want to get to any place on the mainland shore ASAP.Although the crab angle will minimize the distance traveled across the river, it won’t minimize the time to get to the other side of the river, etc. Can anyone guess the correct angle to travel at to minimize the time of travel. Although it’s not easy to derive the answer, it turns out that the minimum-time angle doesn’t require any calculations.PSI was kayak fishing yesterday and decided to head closer to shore where I saw birds diving at a fish-feeding blitz all along the shore. I just wanted to get somewhere near shore ASAP before the feeding blitz ended. There was a strong cross current and cross wind. I guessed right for the correct heading angle which I analytically validated at night because there was nothing interesting on TV. PPSAlthough I chose an optimum heading angle the feeding blitz ended before I got to it. But I did catch one small barracuda while trolling on the way home. Quote Link to comment Share on other sites More sharing options...

billvoss Posted February 13, 2013 Share Posted February 13, 2013 Assuming the river canal is perfectly straight with uniform width and current, and that wind effects are irrelevant, then I believe the fastest crossing angle if the current is N knots is the same as the fastest crossing angle if the current is 0 knots. Simply paddle straight across ignoring the current. That will provide the shortest path relative to the water around you, and thus the fastest crossing.Though I'm not bored enough to create a formal proof! Quote Link to comment Share on other sites More sharing options...

djlewis Posted February 13, 2013 Share Posted February 13, 2013 As we all know, to move across current to a waypoint you need to angle your kayak upstream enough so that your strokes work to counterbalance the downstream drift. The proper heading angle is called the crab angle (or ferry angle). Holding your heading at the crab angle, you will travel a straight line to the waypoint and, obviously, this is the shortest distance to the waypoint. Under the simple assumptions of a right-angle current, the ferry angle is calculated as the arcsine of (Current Speed/ Paddling Speed). A fairly decent approximation is Current Speed ÷ Paddling Speed X 60°. With a GPS, all you have to do is keep the arrow (line or whatever) lined up to the top of the display.But, instead, suppose you want to cross the river in the minimum time and are indifferent to where you arrive on the other side of the river. This could be important; for example, if you are heading from an island to shore when a storm is coming soon you may want to get to any place on the mainland shore ASAP.Although the crab angle will minimize the distance traveled across the river, it won’t minimize the time to get to the other side of the river, etc. Can anyone guess the correct angle to travel at to minimize the time of travel. Although it’s not easy to derive the answer, it turns out that the minimum-time angle doesn’t require any calculations.PSI was kayak fishing yesterday and decided to head closer to shore where I saw birds diving at a fish-feeding blitz all along the shore. I just wanted to get somewhere near shore ASAP before the feeding blitz ended. There was a strong cross current and cross wind. I guessed right for the correct heading angle which I analytically validated at night because there was nothing interesting on TV. PPSAlthough I chose an optimum heading angle the feeding blitz ended before I got to it. But I did catch one small barracuda while trolling on the way home.Same as Bill's answer. Assuming straight, parallel banks and parallel current (and neglecting turbulence effects), the time-optimum angle has to be 90 degrees to the banks, that is aim "straight across". Why? That will get you across in exactly the same time as if there were no current. It can't possibly be any faster -- the crossways component of the current is zero, so no boost is possible. There's pretty much your proof, without the formal math, which is just simple vector trig anyway.Too much time on yer hands, Leon? Quote Link to comment Share on other sites More sharing options...

leong Posted February 13, 2013 Author Share Posted February 13, 2013 Same as Bill's answer. Assuming straight, parallel banks and parallel current (and neglecting turbulence effects), the time-optimum angle has to be 90 degrees to the banks, that is aim "straight across". Why? That will get you across in exactly the same time as if there were no current. It can't possibly be any faster -- the crossways component of the current is zero, so no boost is possible. There's pretty much your proof, without the formal math, which is just simple vector trig anyway.Too much time on yer hands, Leon?Oh well, I was hoping for some wrong answers first. But, yes, you’re both correct. Simple logic says that the current at right angles to the direction you want to go doesn’t affect the time to get to the other side. This problem is analogous to firing a bullet perfectly horizontal vs. just dropping the bullet from the same height. Neglecting air resistance, both bullets will take the same amount of time to reach the ground.Having too much time on my hands as you say, I derived a formula for the time to reach the other bank as a function of the heading angle. The function is hard to differentiate with respect to the heading angle, but with a little help from a Maple symbolic processor, I set the derivative to 0 and indeed a heading angle of 0 minimized the function.Okay, I lied a little :-) … it wasn’t that I had too much time on my hands. It was that I was looking for any excuse to avoid preparing my income tax returns.A few other thoughts:If you think about distance over water (DOW) and distance over ground (DOG), the ferry angle course minimizes DOW at the expense of DOG. The solution to this posted problem does the opposite.Suppose the current in a river is much slower near the banks than it is in the middle and you want to cross the river to a waypoint directly on the other side from where you start. Then the ferry angle may not be the optimal heading to use. Instead, it might be faster to reduce the ferry angle somewhat so you get to the opposite bank sooner and then head back upstream. Quote Link to comment Share on other sites More sharing options...

jason Posted February 13, 2013 Share Posted February 13, 2013 Suppose the current in a river is much slower near the banks than it is in the middle and you want to cross the river to a waypoint directly on the other side from where you start. Then the ferry angle may not be the optimal heading to use. Instead, it might be faster to reduce the ferry angle somewhat so you get to the opposite bank sooner and then head back upstream. That assumes that you can paddle upstream at all. Many times you will ferry across a river/current and at best you can avoid dropping too far down with no hope of making up the ground that you have lost. Quote Link to comment Share on other sites More sharing options...

leong Posted February 13, 2013 Author Share Posted February 13, 2013 That assumes that you can paddle upstream at all. Many times you will ferry across a river/current and at best you can avoid dropping too far down with no hope of making up the ground that you have lost.Yes, you’re right. The specifics (values such as paddling speed, current speed and how it varies in the river) would determine the optimum course to the waypoint. That’s theory, but it may not work in practice. My mechanics professor once said something like, "In theory, practice is simple. But in practice, practice may be difficult". Quote Link to comment Share on other sites More sharing options...

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.