How can an vehicle move faster than the wind that is powering it?

[Topher925]

As I'm sure we all know, that when you go sailing your fastest speed isn't directly downwind but when you are traveling at an angle to it. This allows sailboats and iceboats to go several times faster than the wind speed that is powering them.

But how is it that a vehicle such as http://www.boingboing.net/2007/02/06/video-can-a-vehicle-.html" , that is powered by the wind, can go DIRECTLY downwind faster than the wind that is powering it? Apparently, this is true of iceboats as well. Could someone please explain this so me.

 

[LURCH]

It's probably a hoax. I suppose the principle on which it is meant to operate is that the blades are at an angle to the wind, and this allows them to go many times faster than the wind (like a sailboat or iceboat going cross-wind). But of course, once the entire contraption gets going at the same speed as the wind, it will experience no wind at all. What powers it then? I couldn't guess, but whatever it is, it proves that this vehicle can travel when there is no wind at all.

 

[pixel01]

It's a hoax all right. But the idea can be: there's a flywheel which can store energy when the wind is strong, then when there's weak or no wind, the flywheel makes the cart move.

 

[FredGarvin]

I stopped reading when they put the car on a treadmill. I wonder if it took off?

 

[Topher925]

It's a hoax all right. But the idea can be: there's a flywheel which can store energy when the wind is strong, then when there's weak or no wind, the flywheel makes the cart move.

Pixel, consider this as an ideal situation not an realistic one. This "contraption" should still be able to constantly move faster than the wind assuming the wind has a constant velocity. Also consider the ground to be perfectly level and the cart's motion is only being influenced by the wind.

Also, does this also hold true for an ice boat? Consider the same phenomenon if the iceboat has a zig-zag type path and its magnitude velocity is greater than that of the powering wind. Could the ice boat's velocity component in the direction of the wind ever exceed the speed of this wind?

This isn't a trick question or a riddle, but a simple application of physics. There are a lot of people that believe this is not a hoax, but I consider most people on this forum to be experts in the area of physics so I would like your expert opinions.

 

[K.J.Healey]

Looks like a hoax with an RC car pulling it with some fishing line.

 

[mgb_phys]

Ironically it could go faster than the wind if the wind was from the side.
Then the power only depends on the wind speed and the size of the turbine - which you can make arbitrarily large.

 

[Topher925]

Ironically it could go faster than the wind if the wind was from the side.
Then the power only depends on the wind speed and the size of the turbine - which you can make arbitrarily large.

Yes, I agree. And is why sailboats typically do not move directly down wind in order to achieve fast speeds.

I found this topic on anther forum, and this is one of the forum members explanation/logic about how this thing might work.

Imagine you were standing on the stationary trolley with your back to the wind. When the trolley is traveling at the same speed with the wind, what to you feel? still air? Yes!... BUT what are the blades doing in this 'still air'...they are creating lift!?

You are traveling the same speed as the wind so you should feel nothing, but the blade disc is creating lift...in still apparent wind!. So the blades are creating lift over and above the wind speed.

So there are two components pushing the trolley along.
1. The thrust from the wind direct.
PLUS
2. The Thrust from the disc.

I've understood this just a bit more even whilst typing this post!

Is their something wrong with this persons logic? Does it disobey any physical laws?

 

[Trond]

You have to let go of the idea that it's the wind that is powering it by making the propeller turn. If that is the case it can never work.

Check out this:
http://www.ayrs.org/DWFTTW_from_Catalyst_N23_Jan_2006.pdf

http://www.flixxy.com/sailing-yacht-research.htm

Can a wind powered vehicle sail faster than the wind? Yes - it can!
"The key point is that the propeller is a propeller, not a wind vane, and when the cart is rolling, the wheels are powering the propeller, not the other way around. With the right gearing, the propeller will always push backwards against the air, whether or not the air is moving forwards or backwards relative to the cart. The tailwind and the propeller action combine to make the wheels spin fast enough to keep the whole system rolling faster than the wind". Definitely counter-intuitive."(Jack Goodman)

 

[spork]

Topher refuses to believe that ice-boats can maintain a downwind tack (at an angle to the wind) with speed sufficient to give them a downwind velocity component faster than the wind. I know it seems non-intuitive to some; but it's just a normal day out for the ice-boating community.

Here's just one document that talks a bit about it:

http://www.nalsa.org/Articles/Cetus/Iceboat Sailing Performance-Cetus.pdf

And a couple of diagrams from the article:

I'm well aware this doesn't relate *directly* to going STRAIGHT downwind faster than the wind, but once you realize your downwind component can be much faster than the wind, the steps to get to that nifty cart in the video are somewhat easier.

 

[rcgldr]

Topher refuses to believe that ice-boats can maintain a downwind tack (at an angle to the wind) with speed sufficient to give them a downwind velocity component faster than the wind.

and he would be correct. The diagrams appear to be wrong. In the second diagram, the boat is probably going 8 degrees from perpendicular to the wind, not 8 degrees from parallel to the wind.

Assuming Vb of 70 mph, and an angle of 8 degrees downwind of perpendicular, the boat would be going 9.74mph downwind (a bit over 1/2 the downwind speed), and 69.32mph cross wind. Relative to the air, the cross wind component remains the same at 69.32mph, the "up wind" component (boat is moving downwind slower than the air) is 18 mph - 9.74 mph = 8.26 mph. The boats speed relative to the air or it's "apparent wind" is sqrt(69.32^2 + 8.26^) = 69.81mph.

 

[spork]

The diagrams appear to be wrong. In the second diagram, the boat is probably going 8 degrees from perpendicular to the wind, not 8 degrees from parallel to the wind.

And you base this on what exactly? This will be news to the iceboat community where they do it all the time. And how would you explain the GPS data?

 

[DaveC426913]

In the second diagram, the boat is probably going 8 degrees from perpendicular to the wind, not 8 degrees from parallel to the wind.

Did you allow for "apparent wind" in determining this? The angle of the wind from the PoV of the vehicle will be quite different from the angle of the wind from an exterior PoV. (This is why sailboats seem to have a headwind when in fact, they're more like 45 degrees from head-to-wind).

 

[russ_watters]

And you base this on what exactly? This will be news to the iceboat community where they do it all the time. And how would you explain the GPS data?

I'm not sure about the underlying issue (I believe sailors actually do go downwind at an angle for the purpose of increasing their speed to their destination), but the GPS data does not support the contention. I don't know what the calculation is based on (there is no explanation of where the data came from), but it does not match the GPS output.

 

[rcgldr]

From that article: "Beta (β) is the angle between the apparent wind and the yacht vector." If Beta is small, then the apparent wind is mostly a headwind, with a small crosswind component:

If Va is 55mph, and Beta is 8 degrees as stated in the second diagram, then the apparent wind components, relative to the boat, break down into a head wind of cos(8) x 55 mph = 54.465 mph, and a cross wind of sin(8) x 55 mph = 7.655 mph. Still this doesn't give an overall picture of what's going on.

Wings divert the air, they don't accelerate air backwards in the opposite direction of travel. If the wind speed is 18mph, then the downwind component of the ice boat must be less than 18mph. If the speed of the boat is 70mph, the angle between wind direction and boat direction must be greater than 75.1 degrees (arccos(18/70)).

I'll research this and reply later.

 

[ThinAirDesign]

Topher, so much of debate is about definitions. I'd like to ask you a question that will help me (and perhaps others) determine exactly what your question and position is.

Many folks wonder ... “Can a ice boat (or other sailing craft) travel downwind faster than the wind itself?”. Most of us will concede that the ‘land speed’ of these craft greatly exceeds that of the wind, but there is much debate as to the downwind component of the path or the velocity made good (VMG) of the craft.

My scenario setup and question to you Topher, is this:


On a frozen lake, with a steady 15mph wind from north to south a start point is defined. Directly south, one mile away a finish point is defined. The ice boat takes a ~45degree track (say SW) and comes across the start point at full speed. At the moment the ice boat passes the start point, a neutral buoyancy balloon is released, floating to the south in the wind. At the ~halfway point, the ice boat tacks to the SW and takes aim at the finish point.

Can the ice boat win the race to the finish point, or will the balloon always win?

Thanks Topher (and anyone else who cares to answer).

JB

 

[Topher925]

Thinairdesign, you are right, it is always good to make sure you have the problem or question clearly defined. The scenario you mentioned is a good one and is an example where me and spork with many others from RR disagree.

Can the ice boat win the race to the finish point, or will the balloon always win?

Assuming the balloon always has the same velocity as the wind and the ice boat is only being powered by the wind, then yes the balloon will always win. The best thing that can happen is a tie but the ice boat will never beat the balloon.

If anyone doesn't clearly understand the problem, let me know and I can draw a simple diagram.

 

[rcgldr]

Most of us will concede that the ‘land speed’ of these craft greatly exceeds that of the wind ... Can the ice boat win the race to the finish point, or will the balloon always win?

Bad analogy, since the boat could build up a lot of speed perpendicular to the wind before crossing the start line and then covert this speed into downwind speed, essentially coasting against a relative headwind.

 

[spork]

I'm not sure about the underlying issue (I believe sailors actually do go downwind at an angle for the purpose of increasing their speed to their destination)

No question about it. That's exactly what sailors do, because their downwind VMG (velocity made good) is significantly better on a downwind tack than on a direct downwind run.

but the GPS data does not support the contention.

I can't imagine how you come to that conclusion. The true wind is 10 mph, the ice-boat speed is 40 mph, and the angle between them is 35 degrees. That gives the ice-boat a downwind velocity component of 28 mph. That's nearly 3 times the wind speed.

In the vector diagram they show an ice-boat going 70 mph at 25 deg from direct downwind in an 18 mph wind. This gives the ice-boat a downwind velocity component of more than 63 mph (in other words 3.5 times the wind speed).

Guys, there's no debate on this part of the question. Ice boats do this all the time. They chuckle at the notion that there are people that think it can't be done. Additionally, it's very easy to show *how* it can be done given the most basic performance parameters and a few vectors.

the boat could build up a lot of speed perpendicular to the wind before crossing the start line and then covert this speed into downwind speed, essentially coasting against a relative headwind.

The performance I'm describing is steady state.

 

[ThinAirDesign]

Assuming the balloon always has the same velocity as the wind and the ice boat is only being powered by the wind, then yes the balloon will always win. The best thing that can happen is a tie but the ice boat will never beat the balloon.

Ok, excellent ... it's always good to be on the same page with the questions.

Your answer is very common -- there's something about this "outrunning" our power source that just screams "over-unity", but in the end it turns out that most folks gut reactions are just plain wrong.

Given a reasonable range of wind (say 5mph to 35mph), the balloon just gets it's *** kicked three ways to Sunday.

I've been through this discussion many, many times. I've showed the charts, gps plots, vectors, etc, etc. Finally just very recently I got motivated and joined the Yahoo groups for Land Sailors just to get confirmation from the folks that go out on Sunday and actually do this.

I posted the exact question I posted for you. Here are a few of the responses and the link to the discussion: http://sports.groups.yahoo.com/group/2nalsa/

First, I got a response via email from the President of the NALSA (North American Land Sailing Association). It follows:

If the wind was less than 5 MPH you would need a special craft (which there are a couple of) to beat the balloon.

If the wind was between 10 to 30 MPH, I don't think the balloon could ever win.

In winds above 35 you would have to set up a boat to sail in those high winds, which could be done but you would have to survive the run without crashing.

Following is a sampling of the responses from the thread itself:

Welcome John,if you measure out 2 identical tacks at 45 degrees as you
proposed you travel 1.41 miles to reach the same 1 mile mark. If it
was blowing 15mph that means to have a dead heat you would only have
to sail an average speed of 21.15 mph.

That would be pretty easy to do in even the most basic low performance
dirt or ice boat.

.

If a sail was a parachute instead of a wing they would be right to have reservations...and if it wasn’t for drag, once you got it rolling you’d have a perpetual motion machine; which is how it feels to a landsailor who has stalled in light air and can’t get going but has someone sail past running on speed-generated lift, momentum and whatever tiny amount of wind that is still blowing. I think really efficient landsailers have been clocked at around 4 times the wind speed not the 1.4 times faster required to win the race with the balloon.

.

These boats have a lift to drag ratio of between 3 and 12 or more.
That means, since we fly sideways, racing that neutrally buoyant
balloon in your thought experiment that is blown by the wind, we can
sail 3 to 12 times further than the balloon travels in the same time.
At our optimum L/D speed (were that possible), we would just get to
leeward mark as the balloon got there, but we'd be flying fast and
tipping over! But we can sail deeper (slower) than best L/D speed and
beat the balloon to the leeward mark with ease.

.

Yes. The basic sailing performance relationship is:

Vb = Vt * sin(gamma - beta) /sin(beta)
Vb: yacht speed
Vt: true wind speed
gamma: course sailed relative to true wind (gamma = 0 is straight
into the wind)
beta: apparent wind angle, measured between course over the ground
and the apparent wind vector

The closer the yacht can sail to the apparent wind (small beta), the
faster it goes. A typical beta for an efficient landyacht is on the
order of 14 degrees, which yields a maximum boat-speed ratio of 4
times the true wind, and a maximum Vmg downwind of 2.5 times the true
wind.

Perfomance in these craft is all about L/D after all and it turns out it doesn't even take a high performance craft to beat the wind -- even the most basic can do it. Of course the higher performance machines leave the balloon cryin' for it's mommy.

I urge you to go to the forum yourself and quiz these guys if you are a doubter. The math supports it, the vectors support it, the gps plots support it, and the land sailors will get a good chuckle out of someone telling them they are violating the laws of physics as they rip across the lake bed with VMGs of 2-3x of the wind speed.

JB

PS It's not just land sailing craft that can do this ... the very highest performing racing sailboats also can achieve VMGs of greater than the winds, though just barely. A personal friend of mine (Google Stan Honey) is a world renoun racing navigator (and engineer) with countless world records to his name -- he tells me he's raced numerous sailboats with the performance to beat the balloon. He also chuckles at the notion it can't be done.

 

[ThinAirDesign]

Bad analogy, since the boat could build up a lot of speed perpendicular to the wind before crossing the start line and then covert this speed into downwind speed, essentially coasting against a relative headwind.

Jeff. No tricks involved here. Everyone is talking steady state.

JB

 

[OmCheeto]

As I'm sure we all know, that when you go sailing your fastest speed isn't directly downwind but when you are traveling at an angle to it. This allows sailboats and iceboats to go several times faster than the wind speed that is powering them.

But how is it that a vehicle such as http://www.boingboing.net/2007/02/06/video-can-a-vehicle-.html" , that is powered by the wind, can go DIRECTLY downwind faster than the wind that is powering it? Apparently, this is true of iceboats as well. Could someone please explain this so me.

It cannot go faster than the wind when going directly downwind.

The bowsprit was probably full of NiMH batteries.

 

[spork]

It cannot go faster than the wind when going directly downwind.

A traditional sailboat cannot go directly downwind faster than the wind. The cart in the video can, because it exploits the energy available at the wind/ground interface - not just the wind/vehicle.

In fact I know a guy that's offered a standing $100K bet against any taker to prove this can be done.

 

[OmCheeto]

I urge you to go to the forum yourself and quiz these guys if you are a doubter. The math supports it, the vectors support it, the gps plots support it, and the land sailors will get a good chuckle out of someone telling them they are violating the laws of physics as they rip across the lake bed with VMGs of 2-3x of the wind speed.

Forum? what forum?

The one thing I'm missing here is the human factor.

How much energy did the drivers of these vehicles impart to their vehicles?

There's a lot of energy involved in cranking a winch.

We've yet to discuss the wind-up factor...

 

[OmCheeto]

A traditional sailboat cannot go directly downwind faster than the wind. The cart in the video can, because it exploits the energy available at the wind/ground interface - not just the wind/vehicle.

In fact I know a guy that's offered a standing $100K bet against any taker to prove this can be done.

can or cannot.

this seems to be double negative wage based on a poor understanding of physics...

I'll wager that your friend works on wall-street and that this thread should be moved to GD in the next few moments, or else...

 

[spork]

can or cannot.

this seems to be double negative wage based on a poor understanding of physics...

I thought the wording seemed pretty clear. If you think it can't be done you can put up $100K against his $100K, and he'll prove it can be done.

"it" means building and demonstrating a vehicle that is powered solely by the wind and goes directly downwind faster than the wind - steady-state.

But you probably shouldn't let on what all the reasons are that it'll be impossible. Afterall - why tip him off. You should just take the money and run. You want me to put you in touch with him? His poor understanding of physics could be your gain - eh?

I'll wager that your friend works on wall-street

Cool - I'll take that bet!

...and that this thread should be moved to GD in the next few moments, or else...

What's "GD"? "or else" what?

 

[ThinAirDesign]

Forum? what forum?
...

If you can't find the forum when I provide you a direct link to it in the post, there's not much I can do for you.

How much energy did the drivers of these vehicles impart to their vehicles?

There's a lot of energy involved in cranking a winch.

There's also a lot of energy involved in you yanking our chain.

We've yet to discuss the wind-up factor...

We just did.

JB

PS: This is the Physics Forum right?

 

[rcgldr]

Vmg faster than downwind speed

OK, I'll bite, or at least I'm intrigued. In order for this to work, the sail is going to have to divert air from the apparent wind direction to true upwind direction, while at the same time diverting air opposite of the true direction of the ice boat to generate thrust (relative to the ice boat direction). I can't figure out how to draw a diagram where this combination occurs. I'm taking into account that there are two sets of coordinates involved. The apparent wind (lift and drag), the true wind, and the true direction of the ice boat.

I've seen the equations from that forum thread, and some generic statements, but more data and a diagram would help here. For the diagram, how about the example that shows a true speed of 18 mph, an actual boat speed of 70mph, and apparent wind speed of 55mph. The diagram doesn't show the effective angle of attack (equivalent of a flat board with little drag) for the sail, what would the sail angle be in this case?

Take the caes of an ice boat going down wind, can you show the expanded equation for Vb, gamma and beta in this case? You'll probably have to use the two parameter form of arctan(y, x) to deal with the zeroes.

Let's assume an ice boat with a Beta around 8 degrees with a true wind of 5 mph due north. The goal is to maximumize Vmg due north. Could you break this down into the boat heading and the sail angle (both the true angle relative to north, and angle relative to apparent wind), and the Vb and Vmg achieved?

 

[spork]

Jeff, I'm truly impressed. Rarely does anyone so quickly come at this with an open mind. I've drawn up a diagram that I think will explain how this works physically/aerodynamically. I'll post it in a few minutes. It will take a little explaining, but it's really reasonably straightforward.

Until then, please consider a related concept. Put a yo-yo on your table such that the string is wound under the axle and exits horizontally. Pull the string smoothly and horizontally. The yo-yo will move towards your fingers (in other words it will move faster than that which is pulling it) and will roll up the string. If you replaced your fingers with a drogue 'chute, the wind could pull the string and the yo-yo would go downwind faster than the wind. Of course it will reach the end of the string, so it doesn't satisfy the steady-state criteria. But it's a neat and surprising trick that gives a glimpse into what we're getting at here.

You can do the same thing by grabbing the middle of a bicycle spoke directly beneath the hub. Move the spoke horizontally, and notice that the bike moves twice the speed of your hand, in the same direction.

 

[spork]

O.K. I'm back. And here's my diagram. This diagram represents an ice-boat on a 45 degree downwind tack (i.e. 45 degrees downwind as measured from the true wind direction - the wind direction a stationary observer would feel).

Now, as I said, this diagram will take some explanation. So here goes...

Unfortunately, I freehanded this diagram in M.S. Word. So it's not to scale. I can easily produce one to scale in another program if need be.

The vertical arrows at the bottom obviously represent the true wind speed and direction.

The lowest of the diagonal vectors represents the speed and direction of travel of the ice-boat. I have it here going downwind at 45 degrees, and moving at twice the speed of the wind. If my final force vector (after considering lift and drag as it operates on the sail) has any component toward the front of the boat, this tells us the boat will still be accelerating in that direction. When the net force is perpendicular to the boat's velocity (after accounting for lift and drag on the sail) the boat is on a steady state course. I show the former case here.

Note: the two key parameters here are the L/D (lift to drag ratio) of the sail (including the mast, chassis, etc.) and the L/D of the skates on the ice. I have intentionally ignored the L/D of the skates on the ice for two reasons. First to keep the diagram simple and easier to follow, and secondly, because the L/D of the skates can just as easily be lumped in with the L/D of the sail. If this is a concern I'll be happy to include those vectors on my next diagram. In any event, I think we should be able to agree that the ice-boat could be running on rails with low-friction wheels such that the L/D of the blades (rails in this case) would be so ludicrously high that we can ignore it.

A word on L/D:

L/D is literally the lift-to-drag ratio. On a wing (which is just a type of sail) the L/D also describes the glide ratio. So a glider that had an overall L/D (wing, fuselage, and all) of 30, would descend 1 foot for every 30 feet it moves forward.

More on L/D:

L/D is almost always a measured quantity - not a theoretical one. How it is measured is important to understand in order to understand the diagram. Lift is always measured as the force perpendicular to the free-stream (the relative wind over the wing - but not the disturbed wind in the immediate vicinity of the wing). Drag is always parallel to that free-stream and operates in the same direction (i.e. it opposes the velocity of the craft).

Now we come to the apparent wind vector. This is the vector whose tail is at the head of the boat's velocity vector. The apparent wind vector is simply the true wind minus the boat's velocity. This should make sense as you would expect to feel zero apparent wind on a boat that's going the same direction as the wind at the same speed as the wind. THIS is the vector that describes the wind seen by the sail (and everything else on the boat).

This apparent wind vector represents the free-stream that the sail sees. So we know the lift will be perpendicular to it and the drag will be parallel to it. You can see the "lift", "drag", and "resultant force" vectors drawn relative to the free-stream. The length of the "lift" and "drag" vectors is given by the L/D of the craft.

I don't recall what L/D I used for this drawing. It appears to be maybe 6:1 (I'll measure it if anyone cares). High performance sailplanes get 50:1.

So in the end we have the resultant vector acting on the vessel - against the keel/skates. If that force has a forward component, the boat can still accelerate forward from it's current state. I think you can see that if you extend the resultant force vector to the boat's velocity vector, it does in fact have a forward component.

If you wish to consider the L/D of the skates (the one part I intentionally left out), you simply have to compare the cross-component to the forward component of the resultant force vector. As long as that ratio is less than the L/D of the skates, you're still accelerating.

I hope this makes sense, but I realize it's may be difficult to absorb though vectors and verbage. It's a lot easier if I can wave my hands around and point at things.

I'll be happy to expand on, or clarify, any aspect as necessary.