wind-current equivalence factor

Does anyone recall the factor to apply to wind speed to get the equivalent current speed for use in calculating ferry angles on crossings?

For example, if you expect steady 20 kt beam winds, what speed do you figure that will drift your boat sideways, to be corrected by heading upwind (that is, ferrying)?

I seem to recall a factor of 1/6, that is, 20 kts of wind would be roughly equivalent to 3.3 kts of current. But I’m not sure, and that seems a bit high.

I realize that there are many other issues, like winds are rarely steady, constant and exactly abeam, and different boats and paddlers will react differently. I’ll adjust for those if I know the basic factor.

Thanks. --David.

nothing to back this up
but my brain is telling me that at some point someone told me it was every 10 knots is equivalent to a knot of current.

And that is worth exactly what you paid for it :stuck_out_tongue:

10 to 1 sounds reasonable…
… but still a bit high. That would mean a 20 kt wind will drive you roughly 2 kts sideways? Hmmm… maybe.

Of course, if you can see your destination, you can and should keep correcting, not only for initial estimates, but also your ferry angle based on current and for changing conditions in both wind and current.

If you can’t, however, see the destination or some intermediate point to make a correction from, then this initial guess becomes pretty important. Best then, perhaps, to overcorrect, hit the opposite shore (hopefully) on the upwind/upcurrent side, and slide toward the destination.

Make sense?


aiming off
Well, if aiming for a small target not easily seen (like a portage trail) at the end of an open water crossing, there’s no doubt that its best to aim off so as to deliberately miss to one side or the other. Then when you get to shore you know which way to go.

Nothing worse than getting there and guessing which direction to go, not finding it, going back the other way, then deciding you must have been right the first time, etc. etc.

Depends on the boat.
Some boats are more affected than others. Can vary by quite a bit.

Sea Kayak Navigation by
Franco Ferraro deals with this. Might want that. David Burch’s Kayak Navigation has an updated 4th edition. Might wannna look at that as well. Just my opinion, but don’t tidbit navigation. Stock up and know it. The one skillset sea kayakers ought to excel at should be down cold. Lotsa folks harp about Sea Kayaker Magazine not doing it for them, but they deal with nav/seamanship issues regularly. Some old articles I’ve retained because the person writing about one thing or another knows more than I know now. My 2 cents.


Burch book
Burch had the math on this. It has been quite awhile but just from memory I think 18 k wind equals 1.5 k current and equals approx the energy to paddle 3k per hour in calm air

But check the book !

Burch has tons…
… but mostly on head and tail winds (and a very detailed chapter on crossing current). I couldn’t find the crosswind issue. And that will be a lot different, since boats resist going sideways.

Of course, it’s very complex, and the wind is rarely directly abeam, boats differ, winds shift, etc. Sea state probably matters too.

I’m thinking the best way, when you can’t see the destination, is still to find some distant mark – a blip on the far shore or a distant, slow-moving boat – and do a 3-minute measurement of change in heading, even if you have to alter course a bit to do it.

Burch does show you how to do it
If memory serves, Burch does actually have a table and shows you just how to do this. The most important thing I learned from it is how much energy is saved by utilizing this information in regards to how to handle a crossing. I.e. do you cross by ferrying straight across a channel and then head straight down the other side, and so on. The other variable is to understand how in wind and current above a certain amount ferrying becomes impractical and one should wait out the conditions, like when crossing a tidal race or channel with high winds and current.


– Last Updated: May-07-09 1:20 PM EST –

Calculating Ferry Angle

The Amount of drift - referred to as Set - is 6 degrees for a current that is 1/10 of your paddling speed. This may start to sound difficult but it gets easier.

Use this ratio. It is clearer and very practical:

Current Speed ÷ Paddling Speed X 60° = Ferry Angle

an article on Ferry Angle calculation can be found here:

formula – thanks, but…
Thanks – knew that (and other methods). But the question I asked is – what if you have 15-20 kts of beam wind as well? That’s enough to throw you off a mile or two on a 3-4 mile crossing.

One way is to estimate the equivalent current for a given wind speed, and factor that into the calculation.

angle equation
Are you sure about that equation? It’s been a long day, but it would seem to me that at current speed = boat speed you should end up at 45 degrees (45, 45 90 right triangle for a 1x, 1x, squareroot 2 triangle).