Relative effect of wind and current on a boat with sails down

[RichardS]

Following on from a thread a few months ago about a strange phenomenon I noticed whilst anchored in a tidal stream in the Grado Lagoon, I tried to think mathematically about the effects of wind and current on a boat. However, I just can't figure out the physics.

Let's assume that we are talking about a long keeler with no sails and let's assume that the cross sectional area below the water line is exactly the same as the cross section above the water line. If the boat is positioned across the current with the wind directly blowing onto the opposite side of the boat, there must be a current speed and a wind speed where the two forces are in balance and the boat stands still, ignoring any differences resulting from drag coefficient differences between the below waterline and above waterline hull shapes.

As I understand it, the force exerted by water and wind moving at the same velocity is proportional to their density. However, seawater at 16C is about 850 times more dense than air at 20C.

It is clear that there is something I am failing to grasp here as, all other things being equal in terms of surface area acted upon etc, a boat sitting in a tidal stream of 5 knots with a 5 knot wind acting on the opposite beam is surely not going to be experiencing 850 times more force from the sea than the wind as it seems impossible to visualise an equilibrium ever being achieveable, even allowing for the fact that it is the square of the velocity that needs to be considered.

What am I missing, apart from half a brain?

Richard

 

[prv]

all other things being equal in terms of surface area acted upon etc, a boat sitting in a tidal stream of 5 knots with a 5 knot wind acting on the opposite beam is surely not going to be experiencing 850 times more force from the sea than the wind as it seems impossible to visualise an equilibrium ever being achieveable, even allowing for the fact that it is the square of the velocity that needs to be considered.

A 5-knot current is a rather strong tide. Intuitively, it's going to take an awful lot of wind to hold back a long-keeler beam-on to that tide with no sails up. I've no idea of the maths, but it feels like we might be into XKCD territory with the wind strength required.

(In practice, of course, the boat isn't going to sit there neatly beam-on and upright, but we'll ignore that.)

Pete

 

[RichardS]

A 5-knot current is a rather strong tide. Intuitively, it's going to take an awful lot of wind to hold back a long-keeler beam-on to that tide with no sails up. I've no idea of the maths, but it feels like we might be into XKCD territory with the wind strength required.

(In practice, of course, the boat isn't going to sit there neatly beam-on and upright, but we'll ignore that.)

Pete

But even if we use a 1 knot current ..... the square of that is 1, so we have to have a wind speed of around 30 knots to give a square of 850.

Can this really be true .... a 30 knot wind = a 1 knot current?

Richard

 

[Lon nan Gruagach]

Following on from a thread a few months ago about a strange phenomenon I noticed whilst anchored in a tidal stream in the Grado Lagoon, I tried to think mathematically about the effects of wind and current on a boat. However, I just can't figure out the physics.

Let's assume that we are talking about a long keeler with no sails and let's assume that the cross sectional area below the water line is exactly the same as the cross section above the water line. If the boat is positioned across the current with the wind directly blowing onto the opposite side of the boat, there must be a current speed and a wind speed where the two forces are in balance and the boat stands still, ignoring any differences resulting from drag coefficient differences between the below waterline and above waterline hull shapes.

As I understand it, the force exerted by water and wind moving at the same velocity is proportional to their density. However, seawater at 16C is about 850 times more dense than air at 20C.

It is clear that there is something I am failing to grasp here as, all other things being equal in terms of surface area acted upon etc, a boat sitting in a tidal stream of 5 knots with a 5 knot wind acting on the opposite beam is surely not going to be experiencing 850 times more force from the sea than the wind as it seems impossible to visualise an equilibrium ever being achieveable, even allowing for the fact that it is the square of the velocity that needs to be considered.

What am I missing, apart from half a brain?

Richard

Funny you should mention that, but I was on a 15m saily thing that was side on to the wind and stable.... so I assumed the current was going the other way.
Now, a few things to add. You are stationary, tied to a mooring? If not, how else do you know?
If you are on a mooring, then its likely to be shallow water and the current is strongest at the top, fading pretty evenly as you go deeper to zero(ish) at the bottom. There is much less of is effect upstairs with the wind.
5 knot tide is getting to be on the rather fast side, whereas 5 knot wind is less than a gnats, so that also is a rare case, more like a 15 knot wind and 2 knot tide.
Ignoring drag coefficient puts you at the back of the bus, you just must not do it, its as important as cross sectional area. And side on saily things are wind drag farms, everything is pointy front to back and square side on. Underwater though can be a very different matter with potentially only the keel being flat on. Certainly the boat I was on (a high latitude explorer) is virtually semi circle cross section, giving the hull very little drag sideyways.
And another thing. saily boats just love to go forwards, they are even designed to do so. So well it only takes a sail the size of the side of the boom to work. So, just a slight angle off the wind and off you go, cutting through the tide....

Think of it like this... on a mooring with wind and tide the same way the boat will swim, change the tide and the swim can easily become biassed to one side and just sit there with the wind blowing across ways.

 

[pvb]

Can this really be true .... a 30 knot wind = a 1 knot current?

That's probably nearer to the truth.

 

[prv]

Can this really be true .... a 30 knot wind = a 1 knot current?

On a boat with as much area below the water as above it, no rig, beam on? Yes, I could believe that.

Ever hove-to in a deep-bodied old gaffer?

Pete

 

[pvb]

But don't even ask about a catamaran in the same situation! The maths are brain-damaging.

 

[RichardS]

On a boat with as much area below the water as above it, no rig, beam on? Yes, I could believe that.

Ever hove-to in a deep-bodied old gaffer?

Pete

But don't even ask about a catamaran in the same situation! The maths are brain-damaging.

That's very interesting. I've never sailed anywhere with a proper tidal stream so 1 kt = 30 kts just seemed so far fetched. It just shows how much work the sails have to do.

With my cat's windage the formula will look more sensible so perhaps 1 kt water = 20 kts wind but it's still about double what I was expecting.

Richard

 

[davidej]

Equal cross-sectional area above and below the waterline seems intrinsically implausible especially when you take mast, boom and rigging into consideration.

ditto drag co-efficient.

 

[RichardS]

Equal cross-sectional area above and below the waterline seems intrinsically implausible especially when you take mast, boom and rigging into consideration.

ditto drag co-efficient.

The 50:50 was only a working hypothesis to make the calculation easy and get a handle on the figures ..... but if you look at diagrams of a traditional long keeler and look where the water line is, I doubt whether it's too far from reality.

The above and below waterline drag co-efficients will be different but it's mostly quite "rounded" so I doubt whether the difference in profile will affect the overall figures that much.

Richard

 

[Lon nan Gruagach]

The 50:50 was only a working hypothesis to make the calculation easy and get a handle on the figures ..... but if you look at diagrams of a traditional long keeler and look where the water line is, I doubt whether it's too far from reality.

The above and below waterline drag co-efficients will be different but it's mostly quite "rounded" so I doubt whether the difference in profile will affect the overall figures that much.

Richard

Youre avin a giraffe!

 

[westhinder]

Youre avin a giraffe!

Not exactly the typical deep-bodied long-keeler mentioned in the OP

 

[dom]

30:1 feels intuitively high to me too, then again the process will be nowhere near linear.

The math aspect of this question has got to be right up Mr Duck's neck of the river and it would be interesting to hear his take on it.

 

[Poignard]

Following on from a thread a few months ago about a strange phenomenon I noticed whilst anchored in a tidal stream in the Grado Lagoon, I tried to think mathematically about the effects of wind and current on a boat.
(...)

Can you provide a link to that thread, please?

 

[bbg]

30:1 feels intuitively high to me too,

Really? Trying running through air. Then try running through water fully submerged. Do you still think 30:1 resistance ratio is high?

 

[prv]

Youre avin a giraffe!

Not exactly the typical deep-bodied long-keeler mentioned in the OP

Exactly:

I did specify "with no rig" because it's very hard to work out a total area for all the gear, plus the drag on it probably behaves quite differently to a solid block of hull or superstructure.

Pete

 

[RichardS]

Can you provide a link to that thread, please?

http://www.ybw.com/forums/showthrea...stion-about-how-tides-affect-a-boat-at-anchor

It was actually the Marano lagoon I was anchored in rather than the Grado lagoon but I guess it's the same effect.

Richard

 

[LadyInBed]

Really? Trying running through air. Then try running through water fully submerged. Do you still think 30:1 resistance ratio is high?

I like that analogy
I'm sure I once saw a comparison chart of wind to water strength.

 

[RichardS]

Really? Trying running through air. Then try running through water fully submerged. Do you still think 30:1 resistance ratio is high?

I think that's the wrong comparison though. The relative resistance of air and seawater is 850:1 and that is what you would "feel" if you ran through them.

The 30:1 that we find high is to imagine that you were holding this theoretical 50:50 long keeler (Pete's picture above is very relevant) alongside a pontoon by hand and there was no wind blowing but a 1 knot tidal stream pulling the boat away from you, and then imagine that there was no tidal stream but a 30 knot wind blowing from behind you.

Would you have the same difficulty holding the boat alongside the pontoon in both these situations?

I would have thought that you could hold the boat against a 1 knot current but holding it against a 30 knot wind would be totally impossible .... but this is only based on gut feeling rather than any quantifiable experience.

Richard

 

[dom]

Really? Trying running through air. Then try running through water fully submerged. Do you still think 30:1 resistance ratio is high?

Yes okay, but even there it gets tricky. I'd imagine the simple approximation that the boat collides with air/water molecules at a rate proportional to 'v' (basis of the v^2 relationship) only holds over a specific range of Reynold numbers. At very low Reynolds numbers one will get laminar flow where drag is proportional to v (?), whereas at the extreme upper range it will roughly constant(?). Then there is the problem that the keel will act as a blunt object in certain aspects and a laminar aerofoil in others. Finally there is the problem that an anchored yacht balanced sideways in an opposing current/wind will be dynamically unstable. Happy to be corrected on any and all of this BTW.


Still it’s a long time ago since I did any of this and I must admit to finding it almost impossible to visualise all of this