Why is wind against current a problem? - Page 4 - SailNet Community
Old 12-07-2018
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Re: Why is wind against current a problem?

Yes, you will get this effect at the mouth of the Potomac. We got the snot beat out of us on the way home from Annapolis one time (we live off the lower Potomac). The waves were not huge, maybe 4 or 5 feet or so, but the period was so short that they were square. We rattled that boat so much that an electrical wiring harness in the engine compartment came apart. This was on a 41' sailboat. What was interesting is that there was a traditional Chesapeake deadrise work boat that went past us and with that sharp bow just cut right through the chop. It was a miserable couple of hours, especially since it was our first day with the boat and we were just trying to bring it home!

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Re: Why is wind against current a problem?

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Originally Posted by capnJudd View Post

Furthermore, I suspect that the current has to be quite narrow compared to the wave train for this phenomenon to show up. If a local wind creates a north-bound wave train atop a 300-mile-wide area of ocean already marching south, I will claim that the steep wave action does not occur. So the question may change again, this time into: “Why is a (wide) wave train against a (narrow) current a problem?”
I am not sure that width has much to do with it. I cant speak to the mathematics but I can speak my own observations.

I do a lot of my sailing on a large local river. There are sections down river of me, which I do have the opportunity to sail on from time to time that get quite wide. The estuary in places is maybe 15 miles wide or more, but the current is still substantial because in addition to the substantial fresh water discharge, there is also a tidal range of about 17-18 feet. When the wind blows contrary to the current the whole river gets rough, steep cappy waves pretty much bank to bank. The potential fetch in this section of river is about 180 miles, so it can get pretty rough. There is a pic below illustrating the width of the river.

The Gulf stream, is pretty wide, about 60 miles on average apparently. Here is a link with a good visual and a thumbnail below. https://www.windy.com/-Show-add-more....419,-65.127,5

I am not sure if there is a specific 300 mile wide current you are thinking of, but of the examples I am aware of, the width of the current relative to the width of the wave train doesn't seem to have too much of an impact. You are less likely to have standing waves in very deep water, but that is more a depth thing than a width thing.
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Old 12-07-2018
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Re: Why is wind against current a problem?

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Originally Posted by MastUndSchotbruch View Post
There is lots of discussion (e.g. in the current thread of the Beneteau delivery crew being rescued in the Gulf Stream) about the dangers of a situation with the wind going against the current, like North wind in the Gulf Stream.

I do not doubt that this is true but my question is WHY? What is the underlying physics? I am actually trained as a physicist and one of the dogmas is that 'everything is relative.' I understand that the (velocity-) vectors of wind and current are added if they are contrary but that cannot be all. How much is the current in the Gulf stream at its highest, maybe 3 knots? So, if there is a 20knot wind against it this would mean there is a relative wind of 23 knots. From what I am reading here, the effect must be MUCH larger.

Again, I am not doubting that the effect is real, but can someone explain what is going on?

Thank you!
The way it was explained to me...

The wavelength (L in feet) in deep water is related to the period (P in seconds) by L=5.12P^2.

The wave speed (S in knots) in deep water is related to the period by S=3.03P.

So, a wave with a period of 4 seconds has a length of 80 feet and a speed of 12 knots.

If it moves into an area where there is a contrary current of 3 knots, the wave's speed drops to 9 knots.

A wave with a speed of 9 knots has a period of 3 seconds and a length of 45 feet.

The waves got steeper and they are coming with a greater frequency. In addition because the energy that was previously spread over 80 feet of sea surface is now concentrated in 45 feet and because the height of a wave increases as the square root of the energy, the wave height also increased making the waves steeper still. At a ratio of 1:7 the waves break.

It is a bad day in the Gulf Stream even if the wind is blowing in a place far away.

(The formula are from Bowditch (1984), vol 1, p826.)

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Re: Why is wind against current a problem?

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Originally Posted by Arcb View Post
Wind essentially causes a surface current on the water. The surface current is resisted by slower moving sub surface water. The friction of the subsurface water combined with the flow of the surface water creates a cyclical flow of energy. More wind on top results in faster surface current while sub surface current continues to create friction. The cyclicing energy causes waves to build in height.

A counter acting subsurface current results in greater subsurface friction counteracting the surface flow, resulting in steeper waves. The subsurface current reduces the frequency of the waves resulting in shorter frequency waves. As the frequency of the waves decreases the waves, instead of building in speed, they build in height. Higher shorter waves have less stability in their foundation resulting in relatively earlier breaking.

Shallow water works more or less the same way. Surface current on top, friction on the bottom forces the energy up causing steeper and eventually breaking waves.

Take a sheet of paper and push it across a flat surface like your kitchen table and the paper might stay flat. Push it faster and ripples may start to appear in the paper. Repeat experiment with a moderate friction surface like a carpet, the paper will show more folds. Repeat experiment again with a high friction surface like your driveway and paper will shpw even more folding and creasing.
I see that I started a really interesting discussion. I am not going to double the length of the thread by replying to every contribution so far but let me address some of them.

As for your arguments here, Arcb, I agree with everything you say. Except that I don't think this addresses the main question that I have which is relativity. Why is ANYTHING (wave height, shape, subsurface friction, whatever) about a 20 knot wind against a 2 knot current different than a 22 knot wind in the absence of a current, or a 24 knot wind with the current? This could potentially make a difference in shallow water, as you say, but the Gulf Stream (for instance) is a kilometer deep https://en.wikipedia.org/wiki/Gulf_Stream I doubt very much that a wave, of a height of at a max a few tens of feet can 'feel' the bottom that is 4000' below.
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Re: Why is wind against current a problem?

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Originally Posted by MikeOReilly View Post
As others have said, it’s not that the two force vectors are added. It is that water is encountering a counter force (wind) causing it to create standing waves. If there’s enough force the waves can be large and steep. This is what makes crossing the Gulf Stream challenging at times (not that I’ve ever done it — yet).
Well, I pretend that it actually is precisely the case that two vectors are added. That is simply the mathematical expression of "water is encountering a counter force (wind)".
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Re: Why is wind against current a problem?

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Originally Posted by MarkofSeaLife View Post
I ripped the following off the net...

The Scripps Institution of Oceanography (IIRC) has developed this simple graph describing the modifications of wave characteristics when under the effect of a current of given speed.
The two curves called H (height) and L(length) show how these two wave characteristics are modified by a favourable current (to the right of the vertical axis), or more interestingly by an opposing current (to the left of the vertical axis.

The horizontal axis shows the ratio between current speed and wave period, the vertical axis shows: 1. Hc/Hs, ratio between wave height with current and wave height without current; 2. Lc/Ls, wave length with current and wave length without current.

"R" is a measure of wave steepness, the ratio between height and length. It is usually accepted that beyond a 1:7 steepness the wave breaks. Suppose one is sailing in a 1:10 steep wave, say 2m high, 20m long, and all of a sudden an opposing current brings us along the dashed line in the graph.

Along line L one can find Lc (point A): it's 70% of Ls. The current shortens the wave, which is now only 14m long. Along line H one can find Hc: 1.4 times Hs, the opposing current brings wave height to 2.8m. The theoretical steepness of the resulting wave would be 1:5, in practice the wave would have broken already...

Of course, real world waves are not of constant length nor height, but rather a group of relatively smaller ones, followed by a group of higher waves and so on... in practice the boat finds herself like periodically jumping in a hole in the water, usually with the loudest crash. When water depth is sufficiently low, the effect is magnified.
Thanks for finding this. But I don't understand it. Do you have a reference where this is explained? The link where you found it? Thanks!
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Re: Why is wind against current a problem?

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Originally Posted by SanderO View Post
It is wind that creates waves. No wind... no waves. Wind interacts with the water at the boundary layer the surface to create waves. Wave characteristics are driven by depth of water, and "fetch". So the wave height and period is constrained by the parameters of the body of water. You won't find "large" waves in small harbor. If it is open to the sea... large waves could enter but will like break up.

You can see / learn the behavior if waves in say Long Island Sound. It has current going in and out 2x a day. And often has a persistent SW breeze. If you've sailed all over LIS you can see how the wind and waves (current) interact and what the results are,

The Race, Plum Gut and Hellgate are know for strong currents and confused seas and short tall waves when the wind direction is opposing the current direction (more or less). When they more or less align... the sea is calmer waves are lower and fewer white caps.

Shallower waters near the lee shore have steeper waves... in the lee of the shore the sea is calmer.

I suspect wave and wind interaction has been physically studied and measured in "towing tank" simulations and the behavior described in formulas.

Rapidly flowing rivers especially the narrow rapids with variegated bottoms will show confused seas and waves just as when two currents moving in different directions encounter one another.
Yes, that would make sense. I don't pretend that I understand the hydrodynamical factors that explain the effect of the bottom on wave shape etc but at least that is possible IN PRINCIPLE. I doubt that the effect of the bottom can be 'felt' by a wave in an ocean that is a mile deep so in that case, I cannot see how anything other than relative motion (of wind and current) can play a role. And yet, that is what is claimed to be the case.
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Re: Why is wind against current a problem?

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Originally Posted by Don0190 View Post
Think about the mass of the water moving along the stream at 2 knots, then hitting a wall. All that flow energy has to go somewhere, so it goes up into potential energy. ..................... Now replace the wall with a mass of air blowing into the stream having opposing energy of its own.

Add to this the thermal energy differences.
Why is water moving at 2 knots encountering a mass of air moving against it at (say) 10 knots different from the same water resting and encountering a wind of 12 knots? How would the water know the difference?
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Re: Why is wind against current a problem?

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Originally Posted by barquito View Post
i know just enough about physics to know when i am an idiot. I still don't get it. It seems like a frame-of-reference problem. The water in the gulf stream does not know it is going north. Actually if i am not mistaken, it is going east at about 24,000 mph due to the spin of the earth. So why wouldn't the surface of the gulf stream just react as if the wind is 4 kt faster than if the wind was from the south?

exactly!
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Re: Why is wind against current a problem?

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Originally Posted by ReefMagnet View Post
Well I'm no physicist, but here's my take.

If the wind is against a 4 knot current, the difference in apparent wind speed to the water isn't 4 knots, it's 8 knots. Because the wind running with the tide has it's influence on the water subtracted by 4 knots, too.

Now if we have, say, 20 knots of wind and there is an apparent 8 knot difference in the speed, the different apparent wind on the water is 16 knots vs 24 knots. Because the effective speed of the wind has increased by 50%, the energy delivered by the wind to drive the wave has increased by over a factor of two.

Then throw in the frictional effects of the wave meeting the apparent resistance of a 16 kn current, vs a 24 kn current and then include the shaping of the wave by the friction of the air that moves past it as well and BINGO! There's your difference with wind opposing current.

As for the spinning of the Earth. That's the Coriolis effect and it does influence wind flows but it's effect on the water is akin to the effect on a tennis ball when thrown up in the air inside a vehicle moving at constant speed and direction which is 2/10's of almost nothing.
I don't think the Coriolis effect plays a role here.

But maybe it is the first part of what you are saying, addition vs. subtraction of the speed of current to/from that of the wind: a 20 knot wind against a 4 knot current is equivalent to an 28 knot wind with a 4 knot current. If that is the case, my physicist's mind is fully satisfied. But my impression is that people claim that the wind-against-current effect is more than that. Am I wrong?
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