Yeah, too bad propellers don't have any angles.
Maybe someone could invent a way of putting angles into a propeller, so it would be able to do the cool faster-than-the-wind trick that an angled sail can do.
Respectfully,
Myriad
Down wind faster than the wind
- Thread starter my_wan
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Yeah, too bad propellers don't have any angles.
Maybe someone could invent a way of putting angles into a propeller, so it would be able to do the cool faster-than-the-wind trick that an angled sail can do.
Respectfully,
Myriad
Brian-M
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You're right (a very pretty picture
). I struggled with the wording for that post. What's the word for regarding velocity as a purely scalar quantity?I thought Humbler might be thinking of wind that way, but from his later posts, I no longer believe I have any idea what he's talking about.
Brian-M
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Aha! Now we come to the crux of your blatant misunderstanding of this device.
You seem to be under the delusion that the force on each side of a gearbox will always be the same. That's not true. The whole point of a gearbox is to exchange force for velocity, and vice-versa.
When you gear-down, you get less speed, but more force.
When you gear-up, you get more speed, but less force.
The total power remains the same but the force is different.
By gearing down, the cart is pushing itself up the belt by applying a greater force to the air than the belt is applying to the wheels. Since forwards force is larger than the backwards force, the cart moves forward.
The cost of this extra force (in order to maintain conservation of energy) is speed. This is why the cart can't work if the air is moving at the same speed as the ground/belt, the propeller wouldn't be able to push the air back, because the air would already be moving back faster than it could push it.
When the stationary person is hit by the 60 mph car, that person is accelerated backwards to 60 mph.
When the person moving at 60 mph hits a stationary car, that person is decelerated by 60 mph.
Decelleration is exactly the same thing as accelerating backwards. In both cases, the person is accelerated to 60 mph, in both cases they are both subjected to exactly the same change in kinetic energy.
There is no difference in the forces. They both receive the same injuries.
On a different note, you keep saying it requires over-unity just to reach wind-speed. Then how come hot air balloons, and bubbles travel at exactly wind speed, with no external power source?
Seymour Butz
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Because there's more than just wind "pushing" on the sail that provides the lift.
The sail uses Bournelli's principle to generate lift from the laminar flow across it. A teardrop wing shape is not necessary to achieve lift.
Plus, if the wind is blowing N to S (0 degrees ) at 10 mph and you sail at, say 45 degrees, you lose a small fraction of the 10 mph tailwind, and see maybe..... 8 mph tailwind. But you ALSO pick up extra energy by moving sideways to the wind direction, creating a self generated wind of another say..... 8mph. Then add in Bournelli's lift effect and there's now plenty of energy available.
However, sailing directly downwind eliminates Bournelli's since there will be no laminar flow across the sail.
Brian-M
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Easily done.
If you look closely at the gearing on this cart (or read the description)...
You will see that the top wheel is geared so that it always moves at half the speed of the bottom wheels, in the opposite direction.
Logically, this means that the bottom wheels always move at twice the speed of the top wheel, in the opposite direction.
Because gearing exchanges speed for force, half the speed equates to twice the force, and vice versa.
So...
Top wheel is geared to turn in the opposite direction of the bottom wheels, with a ratio of 1/2, giving it twice the force, and half the speed.
Bottom wheel(s) are geared to turn in the opposite direction of the top wheel with a ratio of 2/1, giving them half the force, and twice the speed.
Note: The fact that both the bottom wheels are connected by the drive chain/belt doesn't affect anything. I could have had the chain connected at one end to a free-standing cog with the same result, It was just easier to draw this way. Any force applied to the wheels through the belt is shared equally between them.
To do the math, we'll use signage to indicate direction. Positive to the right, negative to the left (just like on a number line).
OK, let's look at speed first.
Sa = Speed of top wheel
Sb = Speed of bottom wheel(s)
Sc = Speed of conveyor belt
St = Total speed of cart.
Since the bottom wheels are connected to the ground, and the ground is immobile, St and Sb will always be the same.
St = Sb
Because of the gearing, the bottom wheels will turn at twice the speed of the top wheel, and as it turns in the opposite direction, we use the negative sign, so...
St = Sb = -2Sa
If the cart were immobile, then the top wheel would turn at the same speed as the conveyor belt. But as the cart can move, we have to subtract the motion of the cart.
Sa = Sc-St
So if the belt was moving 2mph to the right and the cart was moving 1mph to the right, the top wheel would be turning at 1mph (that's 2-1). If the belt was moving 2mph and the cart was moving 1mph to the left, the top wheel would be moving at 3mph (that's 2-(-1)).
So we get...
St = Sb = -2Sa = -2(Sc-St)
Apply some algebra...
St = -2(Sc-St)
-St = 2(Sc-St) = 2Sc-2St
2St-St = 2Sc
St = 2Sc
In other words, total speed is twice the conveyor speed. Notice, the signage is the same, so it's travelling in the same direction.
So looking at speed alone, it must travel at twice the speed as the conveyor, in the same direction.
Now let's look at force.
Fa = Force being applied to the top wheel
Fb = Force applied by the bottom wheel(s)
Fc = Force applied by the conveyor belt
Ft = Total force on the cart
Obviously, the total force is equal to the sum of the forces applied to (or applied by) the top and bottom wheels, so...
Ft = Fa+Fb
Because of the gearing, the bottom wheel(s) have half the force of the top wheel, applied in the opposite direction, so...
Fb = Fa/-2
Fa = -2Fb
Since all the force applied by the conveyor belt is being applied to the top wheel alone...
Fa = Fc
Which means...
Fb = Fa/-2 = Fc/-2
So let's put all that together...
Ft = Fa+Fb = Fc + (Fc/-2) = Fc - Fc/2 = Fc/2
So...
Ft = Fc/2
Which means that the total of the forces on the cart is half the force being applied by the conveyor belt, in the same direction as the belt.
Put simply, if the belt is pushing the top wheel with a force of 2N, then the bottom wheels will push in the opposite direction with a force of 1N, and the cart will move along in the direction of the belt with a force of 1N.
From the math, it's clear that the cart will travel in the same direction of the belt at twice the speed of the belt, and half of the force of the belt.
Can you find anything wrong with my working-out?
Can you give me any valid reason why replacing the top wheel and the conveyor belt with a backward directed propeller and a wind would produce a different result?
ThinAirDesigns
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That's true on a traditional sailing rig where the sail and the hull are constrained to the same path. The sails on the cart have no such constraint as they continue on their angled path even though the hull (chassis) tracks DDW.
It's amazing to me how people who understand how a sailing rig can get VMGs above 1.0 can't grasp that our propsail is taking the exact same angled path as the sailing rig and sees the same wind as the sailing rig on an angled path and generates the same forces as the sailing on an angled path *because it is on an angled path*.
Ignore and/or explain away the above bolded point at your own peril. It's the truth and is the key to understanding the connection between sailing rigs VMG greater than 1.0 and the carts VMG greater than 1.0
JB
ThinAirDesigns
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Another physical demonstration of the simple principle of DDWFTTW.
http://www.youtube.com/watch?v=ZVSjA7Rhccs
Notice that the hand doesn't move as fast as the vehicle.
JB
http://www.youtube.com/watch?v=ZVSjA7Rhccs
Notice that the hand doesn't move as fast as the vehicle.
JB
Variable pitch propellers.
Your gear track employs the difference in motion of two concentric disks. OK.
If you have two devices, one geared and one not, but each weighs 1kg.
If you apply the same force to the same mass for the same time, you get the same velocity.
Seymour Butz
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The difference between a sail and a prop is that even while a sail is using Bournelli's to generate lift, it is not spinning, therefore no energy is needed to provide its lift. It's a pure energy extractor. It's like a wing on a plane. There is a drag component - on a plane, it "holds" the plane back. In a boat, it "pushes" the boat sideways.
The prop will consume energy to generate lift. Like, well, a prop on a plane. They also have a drag component, only this time, they are constrained by their bearings, and so to create their lift, they need a torque source to maintain steady state lift.
The wheels do not supply torque to the prop. The wheels CANNOT supply torque to the prop, since they are unpowered as dictated by the prop pitch on the carts. Nor can the prop supply torque to the wheels, they are designed to blow air backwards, not generate torque by the air blowing through them. I'll admit I had this concept wrong - I was assuming that the wind blowing through the prop generated the torque that then powered the wheels. But the cart on the street had the prop angled to blow, something that I hadn't noticed before.
So where is the torque coming from?
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Conservation of energy does not care how you use it. Just make sure you pay the bill.
Is the torque available at the tip of the prop, different from near the center?
The work input is the same. Call it what you will, but the net effect upon the cart is equalised. You are not taking into account that this condition is forced by feedback. The effect is to minimise the force available to drive the cart forward, which is why it stays still. The progress up the belt I have already explained. If you don't agree, then the cart is incredibly inefficient. All that force, speed and no progress.
A control could be added, so that the cart could be sent back and forth as desired. Just alter the balance a bit, either way.
When you are on the belt, you may say that you have a new frame of reference, but the all are the same, so I see what you see, literally.
Car on belt as usual, 10mph, 10 foot long belt. Get out of the car, you go backwards at Vbelt, until the end. Walk back to car.
How gar have you travelled, and what is your displacement?
20foot, 0foot. Car (same transit time) 20foot.
Infinite belt? No good, one way ticket.
Longer? Lower repetition rate.
The belt is not a road, its a treadmill belt. If you stick to your reference, then you must ride all the way around.
Net velocity = 0
Net Displacement = 0
Net Windspeed = 0
All zero, for all viewers.
To be equivalent the treadmill must be a road.
Both viewers are indeed equivalent, for all common components.
The treadmill is a force balance device, in still air, with a velocity ever so slightly more than zero.
I keep saying it because of the energy issue. just like others, I see a big hole in the energy budget.
Don't know about air balloons. Do they reach windspeed in direct down wind? They use lift as well I think. Traveling horizontally may be powered by the potential gained from extra height.
It makes no difference to use other examples, unless they can be directly compared, like another prop device, perhaps. Sails are not propellers. I have argued this from all angles, but sails remain for someone else to rebut, the next time. Just like the over-energy thing.
The yo_yo. The input is not V_wind, it's F_chute.
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How can you be able to do that, and yet reach the wrong conclusion about the cart??? ( The prop and wheels are in series, you know)
I will take a look, but my immediate reaction is if you already have enough force, then you will be able to accelerate the same mass, to the same velocity but without gears.
I am going to make my own model of this vehicle and want to make it as much as possible the same as the one in the video demonstration. I would like to know what the gear ratio is between the wheels to the propeller. I would appreciate any other design advice as well. I will test it in an outside wind and will use something like soap bubbles floating in the wind to establish what the actual air speed is at any given time. I don’t have much spare time at present so it may take a few weeks to get this done.
EXCELLENT. I'm still trying to get together a couple of parts to make this really cheap and easy for anyone to try. I'll post all the parts and build instructions. For now, it should actually be pretty easy to do with the following. I expect within a week or so I can have something much better to post.
Build instructions:
The gear ratio between the axle and prop shaft is 1:1.
JB's cart uses a tailrotor gearbox and torque-tube from a T-Rex 600 R/C
heli (H60133 "600 Metal Tail Torque Tube Unit" $61.99)
Mine will use a gear set from the T-Rex 500 R/C heli (H50096 Torque Tube
Front Drive Gear Set $10.99) and I will make the gearbox that holds those
gears using a mill.
The wheels are: GWS 4" Ultra-Light Wheel 102mm 6.3g ($2/set)
The prop is: GWS 14"x10" slow flyer plastic prop ($3 each)
The wheel size is relatively important as is the size and pitch of the prop.
These govern the advance ratio of the vehicle. That's what enables it to go
downwind faster than the wind. To make it go upwind, I'd simply buy a set
of the same style GWS wheels in a smaller diameter.
The axle is a 5mm carbon tube
As I say, I plan to put together something more comprehensive. But this
thing is pretty darn simple, and the above info along with a picture should
be enough for most folks.
Notes:
- Make sure the prop spins clockwise from behind when the cart is pushed
forward.
- We've used rubber bands on our wheels for better traction - when needed.
It works fine on our treadmill without
I forgot one tip on the build instructions... make sure you're sitting down when you put this thing on your treadmill. I can't be responsible if you hit your head on something sharp when you collapse out of complete surprise.
Believe it or not, I've measured a whole lot of bat tip speeds, and some of them are a bit over 100 mph. But your point is a valid one.
What!? They both develop lift from the air moving over them. One translates, the other rotates. No big difference.
You're mixing up force and energy as well as lift and drag. Lift is the force that pushes sideways on a sailboat - and the force that holds a plane up. Drag tends to slow them both down.
Yup - just like a wing or a sail. You can't make a lifting surface with an infinite L/D.
There is no torque on the main rotors of a helicopter in autorotation, but they have no trouble developing plenty of steady state lift.
Indeed they do. All you have to do is check out the drive belt on our larger cart. It's plenty easy to see that the wheels are turning the prop and not the other way around.
Note: This post is not aimed at those who are still struggling with understanding the basic concept of the device.
While the operation of the idealized device has already been explained in several ways in this thread, more realistic models including unwanted drag and friction have not received that much attention. In this post, I will try to focus on these practical considerations, and what effect they have on the vehicle's operation.
The ideal device (with drag only on the propelling surfaces) could be geared to accelerate up to an arbitrary multiple of windspeed. In the real world, this doesn't happen, because the force driving the vehicle decreases as f approaches 0, until it is overcome by friction and drag forces.
The next picture shows the effect of unwanted drag (modelled by Rayleigh's equation) on the vehicle's operation. CD*A for non-propelling surfaces is chosen to be 1/10 of CD*A for propelling surfaces. The graph shows how changing the leverage factor f affects the equilibrium speed of the vehicle, with windspeed being constant. Blue line shows the ideal model, red line the model with unwanted drag:
(At f=0, the ideal vehicle is in equilibrium at any speed.)
A few things to note:
- There is a maximum equilibrium speed to be reached.
- Equilibrium speeds are lower for backward movement (it is harder to move against the wind).
- For f>1, the unwanted-drag-model moves faster than the ideal model. This is because the ideal equilibrium speed is below windspeed, and so the drag on non-propelling surfaces actually accelerates the vehicle.
The next step is to include kinetic friction (modelled as Coulomb friction), the magnitude of which is arbitrarily chosen to equal 1/2 the magnitude of drag on non-propelling surfaces at reference windspeed (the windspeed in the preceding graph); static friction is assumed to be up to 1.5 times kinetic friction.
Compare with the preceding graph for effects caused by including friction in the model.
For another graph, we will fix the leverage factor f at 0.5 and examine how changing the windspeed affects the equilibrium speed of the vehicle; the units on the x-axis correspond to multiples of reference windspeed (the one used in preceding graphs):
(If you think you see the red line split near zero cart speed, you see correctly. Because static friction is greater than kinetic friction, the system actually has two equilibria for certain range of forces: one at zero speed, and one at non-zero speed.)
The final graph fixes both the windspeed (at reference windspeed) and the gearing (at f=0.5) and examines various forces on the vehicle in relation to its speed:
The windspeed is marked on the x-axis. The ideal device would reach equilibrium at the point where the blue line intersects x-axis (2x windspeed). The model with unwanted drag and friction reaches equilibrium at the point where the red line intersects x-axis.
While the operation of the idealized device has already been explained in several ways in this thread, more realistic models including unwanted drag and friction have not received that much attention. In this post, I will try to focus on these practical considerations, and what effect they have on the vehicle's operation.
The ideal device (with drag only on the propelling surfaces) could be geared to accelerate up to an arbitrary multiple of windspeed. In the real world, this doesn't happen, because the force driving the vehicle decreases as f approaches 0, until it is overcome by friction and drag forces.
The next picture shows the effect of unwanted drag (modelled by Rayleigh's equation) on the vehicle's operation. CD*A for non-propelling surfaces is chosen to be 1/10 of CD*A for propelling surfaces. The graph shows how changing the leverage factor f affects the equilibrium speed of the vehicle, with windspeed being constant. Blue line shows the ideal model, red line the model with unwanted drag:
(At f=0, the ideal vehicle is in equilibrium at any speed.)
A few things to note:
- There is a maximum equilibrium speed to be reached.
- Equilibrium speeds are lower for backward movement (it is harder to move against the wind).
- For f>1, the unwanted-drag-model moves faster than the ideal model. This is because the ideal equilibrium speed is below windspeed, and so the drag on non-propelling surfaces actually accelerates the vehicle.
The next step is to include kinetic friction (modelled as Coulomb friction), the magnitude of which is arbitrarily chosen to equal 1/2 the magnitude of drag on non-propelling surfaces at reference windspeed (the windspeed in the preceding graph); static friction is assumed to be up to 1.5 times kinetic friction.
Compare with the preceding graph for effects caused by including friction in the model.
For another graph, we will fix the leverage factor f at 0.5 and examine how changing the windspeed affects the equilibrium speed of the vehicle; the units on the x-axis correspond to multiples of reference windspeed (the one used in preceding graphs):
(If you think you see the red line split near zero cart speed, you see correctly. Because static friction is greater than kinetic friction, the system actually has two equilibria for certain range of forces: one at zero speed, and one at non-zero speed.)
The final graph fixes both the windspeed (at reference windspeed) and the gearing (at f=0.5) and examines various forces on the vehicle in relation to its speed:
The windspeed is marked on the x-axis. The ideal device would reach equilibrium at the point where the blue line intersects x-axis (2x windspeed). The model with unwanted drag and friction reaches equilibrium at the point where the red line intersects x-axis.
sol invictus
Philosopher
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- Oct 21, 2007
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No, not really.
Again - go back a few pages and look at Myriad's chain-driven wheel thingie. Then think about your question again, specifically whether that device gets its power from the upper chain or the lower one, and which one is providing the torque when it's moving at chain speed.
Thanks Spork good info and advice. As I said I will be testing mine in an outside wind not on a treadmill (to start with anyway). Do you think size is much of a factor for an outside wind test? Would yours work just as well in an outside wind and have you tested it this way? I realise that it’s virtually impossible to get an outside wind with a stable speed and direction but I don’t see this as being too much of a problem as the current wind speed and direction will be shown using bubbles. The vehicle has to beat the bubbles.
Great post! What we really need are three different threads. One for believers that want to discuss performance and theory, one for skeptics that are open to hearing how we think it works (and for whom reality counts), and the third is reserved for those that want to call us fools - and are absolutely 100% sure it's B.S. (but aren't willing to bet a dollar on their position).