Downwind faster than the wind. Poll

[bbg]

You obviously sail your boat different to the way I do mine. When I have the apparent wind ahead of the beam, I tack to move the apparent wind from port to starboard of the bow (or reverse). When the apparent wind is behind the beam, I gybe to get the apparent wind to change sides across the stern. Somewhere else in the world the wind is blowing in a different direction, that doesn't change my apparent wind or my definition of tack and gybe.

By the way, are the blades of this contraption travelling dead downwind?

I suspect I do sail my boat differently, but that doesn't change the terminology. Yours is the first post I have ever seen in which the terminology of tacking and gybing is done by reference to the apparent wind, rather than the true wind. Probably because for most boats the direction of the apparent and true wind is (at the point of tacking and gybing) the same.

But, even if we will have to agree to disagree on that point, by your own terminology it is only these extreme boats or similar that are able to "tack" downwind (I assume that they don't slow to below TWS during the instant of a gybe, but don't know for sure).

You would have to accept that on your leadmine, or on mine - neither of which are able to exceed TWS on a broad reach - the correct terminology would be to gybe downwind, because the stern of the boat will pass through the eye of both the apparent and the true wind.

And it doesn't make much sense to me to have different terminology for the same maneouvre, depending solely on the speed of the boat at the time of the maneouvre. But if you want to make up terminology, I can't stop you.

 

[mikemonty]

So what do you call it when a foiling moth gybes - staying foiling the whole time - and the apparent moves around the bow. Just search Youtube for moths gybing to see.

Personally I'd call it "coasting"
The sails are not drawing during the gybe and start to pull again on the other side of the gybe. the boat is slowing down as this happens.

Take that from an old short board windsurfer who knows what happens if you fail to have enough speed going into the gybe.

 

[RAI]

That would work but:

It contravenes the original remit of using the wind to power a vehicle dead downwind at more than wind speed. You are adding in stored energy, albeit originally derived from the wind. You are in fact using the wind to extend the range of a battery driven vessel.

OK, I just saw all this stuff about conservation of energy and thought batteries do that too. But if its contravenes the laws of nature...

 

[nmeyrick]

Quick question, and this should be pretty easy because I gave up physics after school and there are obviously lots of people out there far smarter than me. Has anyone ever come across the equation:

KE = 1/2*M*V2

And can they tell me what it means?

 

[RAI]

But if you want to make up terminology, I can't stop you.

Thanks for not trying. I agree my yacht, and perhaps yours too, gybes down wind. But these modern machines actually tack down wind and they call it tacking too. See the BMW Oracle web site, for example. My yacht can't sail faster than the wind down wind either, whether by gybing or running with a Spi. But those hairy machines do have an average speed down wind much greater than the wind speed. So why can only very fast yachts do it and not a machine specifically designed to do it by using similar principles?

 

[RAI]

Quick question, and this should be pretty easy because I gave up physics after school and there are obviously lots of people out there far smarter than me. Has anyone ever come across the equation:

KE = 1/2*M*V2

And can they tell me what it means?

OoooH! That's Kinetic Energy! Where M is the mass of air in the wind and V is the wind speed. Lots of that about. Much more than in a little trolley, that would be 0.5mv2.

 

[halfway]

Keen_Ed "Except, that of course, this is exactly true. Reduce drag and your boat will go faster. Why do you think racing sailors spend so much of their time polishing the bottom of their boats? Why do they drysail 45' boats? To reduce drag. To go faster. "

You are taking about a system like a hill. Take away the air and the wheel friction and start rolling. The speed is limitless.

In the case of a yacht there is another issue. As you go faster the driving force diminishes as the apparent wind stalls the aerofoil. So the speed is limited by more than just friction. You have to improve the aerofoil.

At some point you have more drag on the foil than lift and the acceleration grinds to a halt.

 

[RAI]

At some point you have more drag on the foil than lift and the acceleration grinds to a halt.

I bet that happens with this "trolley with prolly" too. Only thing is, at what speed?

 

[nmeyrick]

OoooH! That's Kinetic Energy! Where M is the mass of air in the wind and V is the wind speed. Lots of that about. Much more than in a little trolley, that would be 0.5mv2.

Good spot RAI, thanks. Maybe you can help me with my problem then.

You're right the wind has kinetic energy and so does the cart. In the case of the cart v2 is the square of its velocity, so this little equation tells me that to increase the velocity of the cart in relation to the ground I need to increase its kinetic energy.

But as the cart is travelling at windspeed it feels no aparent wind, so the energy can't be coming from there - so where does the energy come from?

 

[xxyyzz]

You're right the wind has kinetic energy and so does the cart. In the case of the cart v2 is the square of its velocity, so this little equation tells me that to increase the velocity of the cart in relation to the ground I need to increase its kinetic energy.

But as the cart is travelling at windspeed it feels no aparent wind, so the energy can't be coming from there - so where does the energy come from?

When BMW Oracle is hammering it toward a downwind mark, it also must accelerate and so must also increase its kinetic energy (more so because it's tacking it is taking a longer path and so must go faster and experiences more drag), where does that energy come from?

 

[Keen_Ed]

But as the cart is travelling at windspeed it feels no aparent wind, so the energy can't be coming from there - so where does the energy come from?

You have to ignore the apparent wind felt by the cart. It's irrelevant.
You have to think about the apparent wind felt by the blades of the prop.

 

[Ubergeekian]

Well anyway, I was thinking if I get a big enough wind generator on my boat and an electric motor to drive the prop., plus a bit of electronics and some extra batteries, I should be able to power downwind, at least at hull speed. Does that count as DDWFTTW? On a calm day?

No, because you'd be doing it the wrong way round. You need to harvest energy from the passing water and use it to push against the passing air.

 

[nmeyrick]

From its apparent wind. Thats irrelevant as its a completely different scenario.

I did not question whether there is energy out there in the universe, just how the cart gets it in this particular case.

 

[RAI]

But as the cart is travelling at windspeed it feels no aparent wind, so the energy can't be coming from there - so where does the energy come from?

It comes from the airfoiled blade of the rotating propeller, which is not in still air when the cart is rolling at wind speed down wind. It, like the wing-sail on BMW Oracle, see the apparent wind. Just as that boat can sail faster through the water than the wind speed, so can the blade travel faster than the wind speed. It appears that boat can also travel 3 x faster down wind than the wind speed. So, just maybe, the propeller blade can do the same, before it runs out of thrust.

 

[snowleopard]

Quick question, and this should be pretty easy because I gave up physics after school and there are obviously lots of people out there far smarter than me. Has anyone ever come across the equation:

KE = 1/2*M*V2

And can they tell me what it means?

Yes. It's what we're all talking about and what I based a calculation of the available energy on it a couple of pages back.

Kinetic energy is the energy anything possesses by virtue of its velocity.

Kinetic energy - 1/2 x Mass x velocity squared.

On its own it isn't a terribly useful value. What really matters is how much of that energy you can extract. If you take a wider frame of reference, everything on Earth has a huge amount of kinetic energy due to the rotation of the Earth and an even bigger amount due to rotation around the sun. The only way we could get at that huge energy is by the Earth colliding with a stationery object in its orbit!

You get energy out of a wind powered system by changing the velocity of the air. If it is going past you at 10 metres per sec and you do something to slow it to 5 m/sec you have extracted half its usable kinetic energy.

Putting that into practice, let's see how much we can get out of a wind of 10 m/sec. Let's take a flat sail of area 1 sq m. square on to the wind. When it is stationary it stops 10 cubic metres of air every second. A cubic metre of air weighs 1.2 kg so in our equation, mass is 12 kg, velocity is 10 m/sec so the energy we extract is

1/2 x 12 x 10^2 = 600 Joules. (10^2 means 10 squared for non-IT types)

If our sail is now going downwind at 7 m/sec it can only reduce the velocity of the wind by 3 m/sec and only 3 cu m of air hits the sail in that time so the amount of energy extracted is

1/2 x 3.6 x 3^2 = 16.2 Joules.

And of course when the sail is moving at wind speed, the energy extracted is zero.

So where does the 'magic' trolley get energy to go faster than the wind?

Well it's all down to the change in velocity of the air. Instead of stopping the wind, the spinning prop sends it back the way it came. If the prop pushes the wind back from its blades imparting a backward thrust of 6 m/sec, let's see what has happened to the kinetic energy in the wind:

The mass of air that has gone through the prop is the 6 cu m that have gone through the fan, i.e. 7.2 kg .

To start with it was doing 10 m/sec, now having been given a 6 m/sec push it is only doing 4 m/sec so its change in velocity is 6 m/sec The amount of energy extracted from that air is therefore

1/2 x 7.2 x 6^2 = 129.6 Joules.

And that is the energy that goes to overcoming frictional losses in the machine.

 

[RAI]

No, because you'd be doing it the wrong way round. You need to harvest energy from the passing water and use it to push against the passing air.

Well, I did think of doing away with the underwater propeller and using the wind turbine as an air propeller. Then I could charge batteries on one day at anchor and drive the propeller with stored energy on the next. After all, I would be getting the energy from the wind over water speed on the first day and push my boat along in a dead calm the following day. How's that for conserving energy?

 

[halfway]

snowleopard: "There is a finite amount of energy in a column of air travelling at the speed of the wind. It is the mass of that column of air x 1/2 the square of the wind speed."

So double the prop size. Bigger column. You get double the energy and can go faster. Since the energy is in your view relative to the ground only, you can keep doing this.

As it is just a prop blowing, you can use pressure velocity shaping of the inlet and outlet ducts to keep the prop air speed reasonable.

Just keep going until just short of mach 1!


The issue is that there is a much lower fundamental limitation. That is to do with the aerofoil chosen. You are just using your ground speed to provide a gybe angle on a foil. It is a limited system for a given design. It is not anything to do with the energy being relative to the ground.

The available energy diminishes rapidly as the foil forward speed brings the apparent wind around so that the lift takes more energy out of the wheels on the floor than it creates by the forward component driving the vehicle along.

 

[Ubergeekian]

But as the cart is travelling at windspeed it feels no aparent wind, so the energy can't be coming from there - so where does the energy come from?

Why do you think that zero apparent wind means you can't take energy from the wind? All kinetic energy is relative to something. If you think o fit relative to the ground, both the cart and the wind have kinetic energy. Use a fan to push back against the wind: the cart speeds up, the wind slows down and conservation of energy works as well as it always does!

 

[Keen_Ed]

snowleopard:
So double the prop size. Bigger column. You get double the energy and can go faster. Since the energy is in your view relative to the ground only, you can keep doing this.

More lift also = more drag.

You can't generate lift without drag.

 

[xxyyzz]

No, your bigger prop increases the mass of the kart thus increasing wheel friction, air drag as it is big and the amount energy required to accelerate the cart.

Friction is also proportional to the square of velocity and once you are going faster than the wind you get a whole new set of friction from the air moving over the cart.

The energy available from the wind should be constant.