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

[PhysicsAddict]

In order to understand this you need not understand sailing, relative wind, apparent wind or ANYTHING other than Newtons 1st law.

To get your head around this, imagine the cart facing west sitting on a long treadmill in a long windless hallway. Start the treadmill which runs towards the east and slowly increase the treadmill speed until the cart is at the perfect “break even” point. In other words, to an observer standing still in the hallway, the cart appears to also be standing perfectly still even though it is on the treadmill with its wheels spinning and propeller turning. This is the point where the cart goes EXACTLY as fast as the wind downwind. We don’t need it to go faster than the wind downwind yet. Right here at the break even point is the best place to get your head around it.

Now you don't need to understand anything more that two simple ideas:

1. You must understand Newton’s 1st law of motion - specifically pertaining to balanced and unbalanced forces. In order for the cart to appear to stand perfectly still on the treadmill, the forces pushing on the cart from the east must equal the forces pushing on the cart from the west.

In other words, let’s say that the treadmill is expending 10 Newtons of energy driving the cart east. Since the forces are balanced, the propeller on the cart must be expending 10 Newtons of energy driving the cart west in order to hold it stationary. Since the cart is not experiencing ANY wind pushing it at this point, all its energy driving it west must come from thrust generated by the propeller.

2. You must understand that mankind has yet to invent a machine simple or complex that outputs 100% of the energy it consumes. A propeller is at best 85% efficient. Add in the other components of friction and well it all goes downhill from there. In other words it would be impossible for the cart even get to this break even point. It will never generate thrust equal to the energy it consumes. Now to go even faster than the wind it will have to generate thrust IN EXCESS of the energy it consumes which of course is never going to happen.

Since you now understand these two simple ideas, you can now conclude that the video can only be either:

A. A hoax.
B. Some other "artifact" captured on film.

So here it is reduced to it's minimum components. Nothing to obfuscate here. Very simple.

If you contend that it is possible then all you need to answer is this simple non-obfuscated problem:

Let's suppose that the treadmill imparts a continuous 10 Newtons of force where the treadmill belt strikes the wheels. Please lay out the equations for me assuming your propeller is 90% efficient (that would be an awesome propeller BTW) and there is no friction in the inner gearing of the device. Show me where the device is able to generate continuous thrust in excess of 10 Newtons in order to break even and stand still.

It's a simple equation I assure you. If you need the equation I can point you toward it.

So if you would please lay it out for us where 10N into the propeller results in >= 10N of thrust out.

 

[uart]

I just wanted to address the problem of whether a wind power craft (sailboat or ice-boat etc) could zig-zag downwin faster than the speed of the wind. This I believe might be possible.

Starting only with the assumption that a craft could travel into the relative headwind (apparent wind) at an acute angle phi I calculated the downwind component of the craft velocity and determined when it was a maximum.

Note that this analysis doesn't address the case where the craft is traveling directly downwind because it doesn't properly consider the fact the relative wind speed will appraoch zero.

See attached figure where,

W = wind velocity.
V = craft velocity.
A = relative (or apparent) wind velocity.Applying the sine rule to the velocity magnitudes gives,

\frac{V}{\sin( \phi + \theta)} = \frac{W}{\sin \phi}.

BTW. Recall that \sin( \phi + \theta) = \sin(\pi - \phi - \theta)

Rearranging and simplifying gives,

\frac{V}{W} = \frac{\sin( \phi + \theta)}{\sin \phi} = \cos \theta + \cot \phi \,\, \sin \theta.

Define alpha as the ratio of downwind velocity component to wind-speed,

\alpha = \frac{V \cos \theta}{W} = \cos^2 \theta + \frac{1}{2} \, \cot \phi \, \sin(2 \theta).

Differentiating wrt theta (and simplifying) gives,

\frac{d \alpha}{d \theta} = \cot \phi \, \cos(2 \theta) - \sin(2 \theta)

Stationary point is at,

\tan(2 \theta) = \cot \phi,

\theta = \pi/4 - \phi/2. (assuming phi is acute).

You can show that this corresponds to a maximum. Substituting this back into the expression for alpha gives (after some simplification),

\alpha_{\max} = \frac{1}{2} \, ( 1 + \cosec \phi )

Since cosec(phi) is greater than one it follows that the maximum value of alpha is also greater than one.

 

[uart]

Sorry but latex seems to be hopelessly messed up today. I think it must be a bug in the forums hopfully they'll get it sorted out soon.

 

[ThinAirDesign]

So the propeller turns solely by itself as soon as it starts turning, is that what you are saying? (Y/N?))

N

Let's keep with the cart on the treadmill here and let's assume it's held back perhaps by a tensiometer, no wind. What's making the propeller spin?

Your question can't be answered as asked as there is a problem with it: If the cart is bing "held back", then the sails (prop) are obviously creating thrust. If there is thrust, then then the propeller is spinning. If the propeller is spinning, then there is wind over the prop. See, your "held back perhaps by a tensionmeter, no wind" is a question that just doesn't make any sense.

All I can do for you at this moment is the following ...

Here's the 'force path':

1: The relative wind powers the tacking sails (spins the prop).

2: The sails (prop) pull the cart forward. If you know the purpose of a "thrust bearing", it would be this bearing on the prop shaft transfering the load from the sails to the cart.

3: The cart's movement relative to the ground spins the wheels, axle and lower pulley.

4: The lower pulley *pulls down* on the belt. (this is key to remember, it's the side of the belt going DOWN that's under tension. If one used two spools of hi-strength line rather than a belt, the line would move from the upper to the lower spool)

5: The upper pulley rotates in response to the belt/line tension, constraining the prop to a fixed tacking angle. The gearing ratio between the wheels, these pulleys and the prop pitch determine the angle of the sail tack.

(Notice in the above, I didn't say "powering the prop", I said "constraining the prop" -- the line merely provides a 'keel or skate constraint', and there's a BIG difference between "constrain" and "power" -- it only takes a small force (less than the thrust bearing is absorbing) to provide the constraint. If the belt/line were *powering* the prop, we'd have a perpetual motion device on our hands, and we all know how those work out)

The apparent wind created by it's own rotation, is that what you are claiming? (Y/N)

See above.

JB

 

[DaveC426913]

If it can go downwind faster than the wind, then it can sail when there is no wind. After all, that would just be going "faster" than the stationary air around it.

In fact, it must do exactly that at some point during the test run in the video. Somewhere along that run, the craft is stationary relative to the air around it, yet contiues to accelerate. This means that, as stated above, the craft can sail during a dead-calm. This amounts to a sailing craft that needs no energy input to provide propulsion.

this is not true. You are dropping a piece of the puzzle.

The craft [in motion wrt ground but with no relative wind] is NOT the same as the craft [stationary wrt ground but with no relative wind]. They are not equivalent scenarios.

(I'm not for or against yet, I'm just proceeding through the arguments.)

 

[Trond]

See, your "held back perhaps by a tensionmeter, no wind" is a question that just doesn't make any sense.

Ok, how come Jack Goodman has done just that and gotten a result?

 

[ThinAirDesign]

If it can go downwind faster than the wind, then it can sail when there is no wind. After all, that would just be going "faster" than the stationary air around it.

In fact, it must do exactly that at some point during the test run in the video. Somewhere along that run, the craft is stationary relative to the air around it, yet contiues to accelerate. This means that, as stated above, the craft can sail during a dead-calm. This amounts to a sailing craft that needs no energy input to provide propulsion.

In the above, you're confusing "craft" with "sails". Unlike a traditional sailing vessel, the wind seen by the chassis of the craft and the sails of the craft are not the same.

At the moment the *chassis* reaches the speed of the wind, the sails are moving quite nicely through the air.

The above is no different than taking two ice-boats on mirror zig-zag downwind tacks and placing a sliding beam between them -- give me two seats and a windsock right in the middle of the beam between the boats. You and I sit in those seats and watch the windsock. Yep, sure enough ... at the moment we reach the real wind speed, the sock hangs limp. Do the twin ice-boats that we are riding on care? ... of course not -- they are zigging and zagging and see plenty of wind as they accelerate us both to a VMG of greater that 1.0 and the sock turns into our face.

See, the above is nothing at all like stating that our ice-boats can sail in dead-calm. If the wind over the ice stops, the boats stop and we stop. Same with the cart.

JB

 

[DaveC426913]

In order to understand this you need not understand sailing, relative wind, apparent wind or ANYTHING other than Newtons 1st law.

To get your head around this, imagine the cart facing west sitting on a long treadmill in a long windless hallway. Start the treadmill which runs towards the east and slowly increase the treadmill speed until the cart is at the perfect “break even” point. In other words, to an observer standing still in the hallway, the cart appears to also be standing perfectly still even though it is on the treadmill with its wheels spinning and propeller turning. This is the point where the cart goes EXACTLY as fast as the wind downwind. We don’t need it to go faster than the wind downwind yet. Right here at the break even point is the best place to get your head around it.

Now you don't need to understand anything more that two simple ideas:

1. You must understand Newton’s 1st law of motion - specifically pertaining to balanced and unbalanced forces. In order for the cart to appear to stand perfectly still on the treadmill, the forces pushing on the cart from the east must equal the forces pushing on the cart from the west.

In other words, let’s say that the treadmill is expending 10 Newtons of energy driving the cart east. Since the forces are balanced, the propeller on the cart must be expending 10 Newtons of energy driving the cart west in order to hold it stationary. Since the cart is not experiencing ANY wind pushing it at this point, all its energy driving it west must come from thrust generated by the propeller.

2. You must understand that mankind has yet to invent a machine simple or complex that outputs 100% of the energy it consumes. A propeller is at best 85% efficient. Add in the other components of friction and well it all goes downhill from there. In other words it would be impossible for the cart even get to this break even point. It will never generate thrust equal to the energy it consumes. Now to go even faster than the wind it will have to generate thrust IN EXCESS of the energy it consumes which of course is never going to happen.

Since you now understand these two simple ideas, you can now conclude that the video can only be either:

A. A hoax.
B. Some other "artifact" captured on film.

So here it is reduced to it's minimum components. Nothing to obfuscate here. Very simple.

If you contend that it is possible then all you need to answer is this simple non-obfuscated problem:

Let's suppose that the treadmill imparts a continuous 10 Newtons of force where the treadmill belt strikes the wheels. Please lay out the equations for me assuming your propeller is 90% efficient (that would be an awesome propeller BTW) and there is no friction in the inner gearing of the device. Show me where the device is able to generate continuous thrust in excess of 10 Newtons in order to break even and stand still.

It's a simple equation I assure you. If you need the equation I can point you toward it.

So if you would please lay it out for us where 10N into the propeller results in >= 10N of thrust out.

Hm. Can't find a flaw in this argument. It does seem to eliminate all the confusing components, leaving nothing but an over-unity paradox, wherein the wheels must drive the propellor to generate more thrust than the wheels are getting from the treadmill.

 

[ThinAirDesign]

Ok, how come Jack Goodman has done just that and gotten a result?

He hasn't -- you're simply continue to misunderstand what he's done. I can say this with certainty as to me he's merely a phone call away.

This device WILL NOT work when there is no wind relative to the ground.

JB

 

[Trond]

It will never generate thrust equal to the energy it consumes.

It doesn't have to, going down wind it needs less energy at same speed, just like an airplane going down wind, or you can obtain a higher speed then upwind. All relative the ground of course. Same speed relative the air. And I'm never talking relative the rotating blades. Always cart.

 

[ThinAirDesign]

Hm. Can't find a flaw in this argument. It does seem to eliminate all the confusing components, leaving nothing but an over-unity paradox, wherein the wheels must drive the propellor to generate more thrust than the wheels are getting from the treadmill.

Yes, it's all nice and it's all simple and it "eliminates all the confusing components" etc. Mission accomplished.

Problem is, it's just wrong as the wheels don't provide the force to turn the propellor to generate the thrust to drive the wheels.

If his argument were correct, ice-boats would not be able to achieve VMGs of over 1.0 and they do this in multiples.

JB

 

[PhysicsAddict]

He hasn't -- you're simply continue to misunderstand what he's done. I can say this with certainty as to me he's merely a phone call away.

JB

Indeed he has. Go read right here. He claims to certainly have tested it on a tensiometer.

http://www.ayrs.org/DWFTTW_from_Catalyst_N23_Jan_2006.pdf

 

[Trond]

Ok, please explain to me what part of the following I do not understand...

http://www.ayrs.org/DWFTTW_from_Catalyst_N23_Jan_2006.pdf

Line 32 and the 17 following lines, left column, starting with Fortunately...

Imho he holds the cart back and measures pull, line 40 "and tying the car to a tension gauge"

Doesn't this mean he held it back and measured tension?

 

[PhysicsAddict]

Yes, it's all nice and it's all simple and it "eliminates all the confusing components" etc. Mission accomplished.

Problem is, it's just wrong as the wheels don't provide the force to turn the propeller to generate the thrust to drive the wheels.

JB

Excellent... So let's stop right there and there on that point and work it out.

I'm saying that I am an observer in the same inertial frame as the cart on the treadmill in the hallway at the break-even point. I cannot feel any wind. Therefore the cart does not feel any wind.

So if the wheels are not powering the propeller and there is no wind that is powering the propeller then please explain what is it at this point that spins the propeller.

We do agree that the propeller is spinning correct? And that something does indeed have to power it correct? Please explain what that force is that is powering the propeller.

 

[Trond]

I'm too slow

And welcome aboard...finally someone that seems to be thinking in the same box as I do

 

[ThinAirDesign]

Indeed he has. Go read right here. He claims to certainly have tested it on a tensiometer.

http://www.ayrs.org/DWFTTW_from_Catalyst_N23_Jan_2006.pdf

Indeed he has not. You're confusing the argument between Trond and I.

Trond says the tensiometer test is done with "no wind" (that's his quote). This is simply not true.

The test was done on a treadmill and as any PhysicsAddict will know, there is the *same, exact* wind available to this device on a treadmill moving 10mph in a room as there is in a 10mph wind out on the street.

ALL of Jack's tests and video have been performed WITH WIND.

JB

 

[ThinAirDesign]

I'm too slow

And welcome aboard...finally someone that seems to be thinking in the same box as I do

LOL Trond -- it's a bit speedy back and forth, that's for sure. :-)

JB

 

[Trond]

Uh, the report says "in still air" did he misunderstand?

 

[Trond]

LOL Trond -- it's a bit speedy back and forth, that's for sure. :-)

Sure is

 

[PhysicsAddict]

Indeed he has not. You're confusing the argument between Trond and I.

Trond says the tensiometer test is done with "no wind" (that's his quote). This is simply not true.

The test was done on a treadmill and as any PhysicsAddict will know, there is the *same, exact* wind available to this device on a treadmill moving 10mph in a room as there is in a 10mph wind out on the street.

ALL of Jack's tests and video have been performed WITH WIND.

JB

Ok, call me dumb but directly from the link I posted:

For those who missed that July issue, No. 21, a vehicle on a treadmill in still air, with the wheels going eight miles per hour is the same as a vehicle going eight miles per hour down wind, in an eight mph following wind. If a car moves forward on a treadmill with no assistance, it is going faster than the wind

After leveling the track, putting a backstop on to get the car up to speed, and tying the car to a tension gauge, we started the treadmill and increased the speed in one mile per hour increments. ...

That really looks like a treadmill test with a tensiometer with no wind.

 

[Trond]

Sure does to me 2

 

[ThinAirDesign]

That really looks like a treadmill test with a tensiometer with no wind.

Are you really trying to tell me that you have the handle "PhysicsAddict" and don't understand simple physics frames of reference? -- I mean, no offense if you don't ... we all have to learn sometime, but this is very basic stuff and I assume a physics addict would have covered this a long time ago.

Wind is relative as is all motion.

To a treadmill moving 10mph in a room -- there is 10mph of wind. If I put you in a large enough room on a large enough treadmill would simply could not tell which was moving -- the room or the air. In fact, it matters not which -- all tests return the same results.

10mph wind in the street -- 10mph treadmill in the bedroom. Same exact "wind".

With the treadmill test, if you want the "wind" to stop -- you must turn off the treadmill.

JB

 

[PhysicsAddict]

Are you really trying to tell me that you have the handle "PhysicsAddict" and don't understand simple physics frames of reference? -- I mean, no offense if you don't ...
JB

So now I take it that we are arguing that Jack Goodman's definition of "still air" means that he actually had a room with 10mph air blowing across his treadmill.

Is this what you are now claiming? Please explain. If I am standing stationary next to the treadmill and the cart is on the treadmill at the break even point stationary with refrence to me) and I do not feel any wind since we are in "still air" please explain how the cart which is in the exact same of inertial reference as me (the observer) experiences a 10 mph wind.

To further calrify the question:

If I run on the treadmill which is set to 10mph and I am in a room with still air do you think I feel a 10mph breeze on my face?

 

[rcgldr]

red x

It's a .gif image.

It's working now, as stated, it takes a while before an attached image is approved, so I normally post a link as well when I do attachments.

fan blowing on sail

This is the same principle as a thrust reverser on a jet engine.

Not quite, the jet engine uses a compressor and heat to create a huge pressure jump. You'd need a strong power source and a prop with a pressure jump to higher than ambient to pull this off. A high efficiency prop wouldn't work, because the pressure jump occurs below ambient, and air is decelerating once it passes through the prop disk.

If it can go downwind faster than the wind, then it can sail when there is no wind.

The device is relying on the difference between wind speed and ground speed as a source of power, which is independent of the device speed. In the case of the treadmill, the treadmill is moving backwards while the air is still. In the case of the outdoor run, there's a tail wind and the ground isn't moving.

Although wind versus ground speed is independent of device speed, which allows for the concept of apparent wind to work, there is a limit to downwind speed (Vmg = Velocity made good). The sail has to divert the apparent wind so that it generates a force in the true direction of travel of the device.

For Vmg/Vt to be greater than 1, it would seem that this same diversion of apparent wind has to include an upwind component > (Vmg - Vt), otherwise there would be a net drag in the direction of the wind. I haven't seen this aspect of the issue explained.

Assuming the formula stated for a device with a Beta of 14 degrees, I made a graph of Vmg/Vt versus heading offset from true down wind. It peaks at about 2.56 at 38 degrees.

http://jeffareid.net/misc/dwvhdg.gif

Getting back to the original post, the issue is propeller efficiency, and the ability for the propeller situation to turn into a situation similar to the apparent wind situation of a ice or land sail boat. Even with no drag, there's an induced efficiency factor that limits the efficiency of propellers (I think this is related to the fact that a propeller operates in it's own induced wash). Propellers can have overall efficiency factors greater than 90%. I'm not sure what is required for the device shown.

 

[nixy2]

OK...Question 1.

Trolley with prop at zero degrees. (could even be just a rotating disc) Assuming no mechanical drag. What might be the highest speed the trolley could achieve?

and are the wheels producing thrust or drag at this condition?

 

[Trond]

Not quite, the jet engine uses a compressor and heat to create a huge pressure jump. You'd need a strong power source and a prop with a pressure jump to higher than ambient to pull this off. A high efficiency prop wouldn't work, because the pressure jump occurs below ambient, and air is decelerating once it passes through the prop disk.

Doesn't a thrust reverser just divert the airflow?

Just like on a water jet?

 

[rcgldr]

... cart is on the treadmill ... explain how the cart ... experiences a 10 mph wind.

It doesn't, it experiences 0 wind speed and -10mph speed at the wheels.

If I run on the treadmill which is set to 10mph and I am in a room with still air do you think I feel a 10mph breeze on my face?

If you stand on the treadmill while it's moving at -10mph you will experience the equivalent of a +10 mph tail wind.

fan on a sail similar to jet engine reverse thrust

Not quite, the jet engine uses a compressor and heat to create a huge pressure jump. You'd need a strong power source and a prop with a pressure jump to higher than ambient to pull this off. A high efficiency prop wouldn't work, because the pressure jump occurs below ambient, and air is decelerating once it passes through the prop disk.

Doesn't a thrust reverser just divert the airflow?

Yes, but note that the engine is also sucking in air. The intake accelerates the air backwards, the diverted nozzles accelerate the air forwards, with the result of opposing forces. There has to be an increase in kinetic energy of the air being accelerated forwards by the diverters before there's a net braking effect. In the case of a jet engine, there's a huge pressure jump well above ambient because of combustion. Note that a propeller operates in it's induced wash, the pressure of the air just before it crosses the prop disk is below ambient. If the pressure jump doesn't cause the pressure to exceed ambient, then the air starts deceleration once it's past the aft side of the propeller disk, reducing it's kinetic energy over time and distance, and in this case, the diverters would just reduce thrust, not reverse it.

 

[nixy2]

OK...Question 1.

Trolley with prop at zero degrees. (could even be just a rotating disc) Assuming no mechanical drag. What might be the highest speed the trolley could achieve?

and are the wheels producing thrust or drag at this condition?

Oh alright, I'll answer it myself.

In a 10 kt wind I'd suggest the highest speed of the trolley to be say 10kt... but no higher...and at this condition the wheels are producing neither drag nor thrust?

and as a matter of interest if I were sitting on this trolley, would I feel any wind at all?

 

[Trond]

If you stand on the treadmill while it's moving at -10mph you will experience the equivalent of a +10 mph tail wind.

I've done my share of running on a treadmill, but I've never seen one that moves so that I've experienced any tailwind and I have had the belt going at more than 10 mph

 

[PhysicsAddict]

It doesn't it experiences 0 wind speed and -10mph speed at the wheels.

If you stand on the treadmill while it's moving at -10mph you will experience the equivalent of a +10 mph tail wind.

Yes and yes. But we are not standing still on the treadmill. We are running at 10mph. Just like the cart.

So, if it experiences 0 wind speed and 10mph belt speed then what is it that is powering the propeller. Since ThinAirDesigns hasn't answered, I pose the question to you Jeff.