Interesting facts about the AC72...

[JumbleDuck]

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?

From the road. The point, which is certainly not intuitive, is that at high speeds the power transfer is from the road wheels to the propeller. The propeller is pushing the car along, not the wheels. This works very well when the car speed is equal to the wind speed, not least because there is no aerodynamic drag force on the rest of the car at that speed.

 

[A.T.]

A very simple vector diagram where you only care about one vector.

Yes, a very simple vector diagram shows that you are wrong.

 

[toad_oftoadhall]

CMG is the way the boat is pointing,

When I used the term CMG I meant the 'net' course. The course from A to B where B is direct downwind. in this case the net course is the course the balloon takes. I accept the boat won't ever be on exactly that course but that's where it ends up.

If you prefer another term for that then fine - substitute any term you like.

You suggested doing the maths, the maths is done, please address it.

 

[JumbleDuck]

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

Two different things being discussed. Flaming is talking about the AC boats; others are talking about systems for going direct downwind.

 

[toad_oftoadhall]

Yes, a very simple vector diagram shows that you are wrong.

Top left diagram is the best one to visualize this - the grey arrow labelled downwind VMG can never equal the wind speed. When you see it drawn like it's obvious. When that grey arrow equals wind speed there is no useable wind anymore.

 

[toad_oftoadhall]

Two different things being discussed. Flaming is talking about the AC boats; others are talking about systems for going direct downwind.

No, Flaming is saying the AC boats go net dead downwind faster than the wind. He's saying they have to sail about at bit to do but they do beat the balloon to the point dead downwind.

 

[Tempus]

Top left diagram is the best one to visualize this - the grey arrow labelled downwind VMG can never equal the wind speed. When you see it drawn like it's obvious. When that grey arrow equals wind speed there is no useable wind anymore.

But on the diagram it's already bigger than the true wind speed!!!

 

[flaming]

Top left diagram is the best one to visualize this - the grey arrow labelled downwind VMG can never equal the wind speed. When you see it drawn like it's obvious. When that grey arrow equals wind speed there is no useable wind anymore.

Right... Well I disagree, mainly because the evidence of the AC72s achieving 1.8 times wind speed!

There's no reason for that to be a limit because the angle sailed always means there is plenty of wind over the sails.

 

[toad_oftoadhall]

Right... Well I disagree, mainly because the evidence of the AC72s achieving 1.8 times wind speed!

You suggested doing the maths. I've done the maths, and I'm satisfied.

You're seen a normal sail boat sail on a net course DDWFTTW so you're satisfied.

We can call it a day, fair winds to you.

 

[flaming]

You suggested doing the maths. I've done the maths, and I'm satisfied.

You're seen a normal sail boat sail on a net course DDWFTTW so you're satisfied.

We can call it a day, fair winds to you.

Your maths suggests it's not possible.
Observation clearly shows it is possible.

But you're satisfied with your maths.

Are you a politician?

 

[JumbleDuck]

Top left diagram is the best one to visualize this - the grey arrow labelled downwind VMG can never equal the wind speed. When you see it drawn like it's obvious. When that grey arrow equals wind speed there is no useable wind anymore.

No. The grey arrow is only one component of the boat's speed, in the direction of the wind. Even when that cancels out the wind speed there will still be the component of boat velocity at right angles to the true wind.

In the top right diagram that would give "true wind" as it is, "apparent wind" horizontally from left to right and "head wind" at forty five degrees to both of them.

 

[JumbleDuck]

Your maths suggests it's not possible.

Actually, the maths he quoted clearly shows that it is possible. As, of course, does reality.

 

[A.T.]

the grey arrow labelled downwind VMG can never equal the wind speed.

Sure it can. It can even be greater than true wind, and you still get a propulsive force from the sail, as shown in the diagram.

When that grey arrow equals wind speed there is no useable wind anymore.

Nope. If the downwind VMG = true wind, then the apparent wind comes directly from the left, at 45° to the boat axis. That is even easier to use for propulsion than what is shown in the diagram.

 

[flaming]

Actually, the maths he quoted clearly shows that it is possible. As, of course, does reality.

Ok, his interpretation of his maths shows...

 

[toad_oftoadhall]

Your maths suggests it's not possible.
Observation clearly shows it is possible.

But you're satisfied with your maths.

Are you a politician?

You suggested a mathematical solution, presumably because you realized that observations with so many variables can be misleading. I did the maths. As the grey arrow on the diagram increases the green arrow decreases until the magic point where it disappears.

My observation sailing windsurfers and cats that will beam/broad reach at incredible speeds with apparent wind miles forward matches that.

Perhaps you should follow your own suggestion and do the maths to double check your observation? I'd like to see how it differs to mine.

 

[Alan ashore]

Top left diagram is the best one to visualize this - the grey arrow labelled downwind VMG can never equal the wind speed. When you see it drawn like it's obvious. When that grey arrow equals wind speed there is no useable wind anymore.

I respectfully suggest that the expression "useable wind" gives a clue to a misunderstanding. I think what you are saying is that when VMG Downwind = VT (True wind speed), then the apparent wind will have no component in the same direction as the true wind. That is true, but the apparent wind is no less useable because of it!

Ice yachts have been beating balloons from the top to the bottom of the course for a very long time.

That is all.

 

[toad_oftoadhall]

Nope. If the downwind VMG = true wind, then the apparent wind comes directly from the left.

Remind me what energy source creates that apparent wind?

 

[Iain C]

Ice yachts have been beating balloons from the top to the bottom of the course for a very long time.

That is all.

I've done the sums, and actually ice yachts don't exist...

 

[Alan ashore]

I've done the sums, and actually ice yachts don't exist...

+1

 

[JumbleDuck]

You suggested a mathematical solution, presumably because you realized that observations with so many variables can be misleading. I did the maths. As the grey arrow on the diagram increases the green arrow decreases until the magic point where it disappears.

The green arrow is the true wind. It is independent of boat speed and is completely unaffected by any changes in the grey arrow.