Interesting facts about the AC72...

[toad_oftoadhall]

The losses are lumped together into the transmission. This is a valid simplification, if you just want to show that steady state is possible at v > v^W in principle, despite some losses. Then it doesn't matter where you put the losses.


That's not what I said. There is no contradiction between rolling at windspeed with 100% transmission loss and the formula. If you don't transmit any power, you can loose 100% of it, and it doesn't matter

The formula just gives no useful information for this case, and doesn't have to. You don't care what happens at v <= v_W, if you just want to show that steady state is possible at v > v_W.


The formula is correct. It follows directly from P = F * v_rel:
http://en.wikipedia.org/wiki/Power_(physics)#Mechanical_power
It's just not useful at v_rel = 0, but quite usefull for any v_rel > 0.

Just like Coulomb's force formula is not useful at r = 0, but quite usefull for any r > 0.

Ok, I say the Formula's self evidently wrong, and I say why. I've had a quick look through Stroud which is the only maths book I have to hand and (unsurprisingly) that doesn't help. I've had a quick look on t'Web with the same result. Throwing further words around isn't going to resolve anything so I'll save us both some time and drop out.

As an aside it seems to me it is useful at the v_rel = 0 point!!! If the formula is correct then v_rel = 0 looks pretty conclusive to me! But that's another argument, and I doubt either of us have time for it.

PS: Why is Coulomb's force formula not useful at r = 0?

 

[A.T.]

Ok, I say the Formula's self evidently wrong, and I say why.

Your "why" is based on misinterpreting the formula. Again: If you don't transmit any power, you can loose 100% of it, and it doesn't matter.

I've had a quick look through Stroud which is the only maths book I have to hand and (unsurprisingly) that doesn't help. I've had a quick look on t'Web with the same result.

What were you looking for in maths books? Try a physics book.

Why is Coulomb's force formula not useful at r = 0?

Because it gives you infinite force, which is not physical. Similarly applying P = F * v at v = 0 gives you an indeterminate force. Simple formulas often have limiting cases, where they become useless. But that doesn't make them useless in general, because they work well for all other cases.

 

[toad_oftoadhall]

What were you looking for in maths books? Try a physics book.

Great, cite a physics book that uses this Formula in this context and I'll do just that.

 

[A.T.]

Great, cite a physics book that uses this Formula in this context and I'll do just that.

Physics books are not collections of all possible applications of all formulas. This was a qualification exam for the International Physics Olympiad. Using a common application example, that is all over physics books, would be a bit too easy.

 

[mjcoon]

* The boats go directly downwind 1.8 times faster than the wind. So if you let a balloon go as you went around the top mark you would easily beat it to the bottom mark.

I think this has been debated elsewhere (!). I have just one question: as the boat accelerates downwind it presumably passes through the same speed as the wind on its way to 1.8 times. At that point there is zero apparent wind. So where does the extra energy come from?

Mike.

 

[flaming]

I think this has been debated elsewhere (!). I have just one question: as the boat accelerates downwind it presumably passes through the same speed as the wind on its way to 1.8 times. At that point there is zero apparent wind. So where does the extra energy come from?

Mike.

It's not going downwind, it's sailing at an angle.

The vector downwind is 1.8 times the speed of the wind, so by gybing back and forth it can effectively achieve 1.8 times true wind speed downwind.

 

[mjcoon]

I would install bucket seats and reclining peddling positions much like those low level recumbent peddle thingies..

Like French onion peddlers?

Mike.

 

[mjcoon]

It's not going downwind, it's sailing at an angle.

Oh, I misunderstood "directly downwiind"; it means at an angle?

Mike.

 

[JumbleDuck]

Yes, Hull aero drag and rolling resistance are completely neglected and therefore cannot explain the 10m/s speed despite 100pc loss. Looks like the formula is simply wrong.

No. You are wrong. It is precisely because the various drags are ignored that the prediction can be seen to be right. What speed would you expect a frictionless cart to make in a 10kt wind?

 

[JumbleDuck]

It's just not useful at v_rel = 0, but quite usefull for any v_rel > 0.

When the cart and wind are moving at the same speed, the propeller can still be providing thrust.

 

[flaming]

Oh, I misunderstood "directly downwiind"; it means at an angle?

Mike.

What it means is that the boat is capable of getting to the leeward mark 1.8 times faster than a balloon released at the windward mark at the same time. Hence it's VMG directly downwind. That doesn't mean it's going directly downwind, but that it's capable of getting to a destination directly downwind at 1.8 times the windspeed, despite sailing a larger distance.

 

[toad_oftoadhall]

Oh, I misunderstood "directly downwiind"; it means at an angle?

Mike.

My reading matches yours. Even if we assume downwind tacking the net effect of all this downwind tacking is that it HAS sailed directly downwind as evidenced by the balloon comment. Sounds unlikely to me.

 

[flaming]

My reading matches yours. Even if we assume downwind tacking the net effect of all this downwind tacking is that it HAS sailed directly downwind as evidenced by the balloon comment. Sounds unlikely to me.

Unlikely, yet true.

Do the maths, in 18-19 knots of true wind, both boats were sitting at over 37 knots boat speed, and hitting 40.

Gybing through about 45-50 degrees. So making about 155 - 160 TWA.

It's all there on the replays....

 

[Tempus]

Sounds unlikely to me.

It may sound unlikely but it was on the TV broadcast to the world!

If you go onto the America's cup site you can re-run the races, see the VMG and see the wind speed. Just because the physics is hard doesn't make it impossible. It's a very similar idea to being able to tack and make progress upwind.

Sorry, Flaming got there 1st

 

[toad_oftoadhall]

Unlikely, yet true.

Do the maths, in 18-19 knots of true wind, both boats were sitting at over 37 knots boat speed, and hitting 40.

Gybing through about 45-50 degrees. So making about 155 - 160 TWA.

It's all there on the replays....

I have done the Maths:

Wind speed = 10kts.
VMG = 10kts.
CMG = same as wind direction.

Wind available to the boat = 10-10 = 0.

You simply cannot be sailing away from the wind (ie the downwind component of your course the same as the wind) at the same speed as the wind and still use that wind to power you.

 

[toad_oftoadhall]

Just because the physics is hard doesn't make it impossible.

Physics for the boat speed may be hard. The maths to calculate the wind speed is a doddle. A very simple vector diagram where you only care about one vector.

 

[flaming]

I have done the Maths:

Wind speed = 10kts.
VMG = 10kts.
CMG = same as wind direction.

Wind available to the boat = 10-10 = 0.

You simply cannot be sailing away from the wind (ie the downwind component of your course the same as the wind) at the same speed as the wind and still use that wind to power you.

I respectfully submit that you're wrong... As evidenced by the fact that the AC72s VMG downwind at 1.8 times the wind speed.

But CMG is different from VMG. CMG is the way the boat is pointing, and it is NOT equal to TWD, so apparent wind is not 0.

Remember the AWA difference between upwind and downwind on these boats is 4 degrees. 4!

 

[Tempus]

I have done the Maths:

Wind speed = 10kts.
VMG = 10kts.
CMG = same as wind direction.

Wind available to the boat = 10-10 = 0.

You simply cannot be sailing away from the wind (ie the downwind component of your course the same as the wind) at the same speed as the wind and still use that wind to power you.

Your maths are too simple, that' the problem, and to expand:

if you are travelling at 45deg (for simplicity) off dead downwind, your direct downwind component is BOAT SPEED/Sqrt(2) - remember pythogoras, you are travelling down the hypotenuse of a triangle with both sides equal. Hence if your actual boat speed is 37 knots your VMG downwind is 25.5 knots. The boats were doing this in 16 knots of true wind speed. The thing to remember is that a sail or wing exerts force at 90deg to the wind, hence you can get these effects.

 

[Iain C]

I think you're missing the point. Imagine an old style Olympic course with a windward mark, a leeward mark and a gybe mark.

Boat starts sailing from the windward mark, trying to get to the leeward mark. At the exact same moment a balloon is released from the windward mark and drifts DDW towards the leeward mark.

The boat, however has other ideas. It aims for the gybe mark, going fast, much faster than the true wind, gybes at the mark, and hoons down the leeward mark, beating the balloon. Job done.

Many boats will do this, not just AC72s. 18s will do it, Moths will do it and I wouldn't be surprised if the 49er is coming pretty close...

 

[A.T.]

When the cart and wind are moving at the same speed, the propeller can still be providing thrust.

Of course. But it is so called static thrust, that cannot be related to shaft power and airspeed via the simple relationship: P * eta = F * v

And the AAPT solution doesn't analyze the situation at windspeed using this approach, they just use it to show v > v_w doesn't violate conservation of energy.