Down wind faster than the wind
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Does everyone else agree with this?
ThinAirDesigns
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And yet we're having this discussion, even though it's the case.
You really should stop with the "typos" and starting adding more "nots". This would turn out a lot better for you in the end. Just one extra "not" per statement should do just fine.
humber some day you'll come to the realization that you don't just get to "rule out" the laws of physics the you haven't yet come to understand. Most important of all, the cart cares not which laws you understand, it just obeys the ones that are real and keeps on going.
We love to do wind test. As we've become fond of saying ... we can do wind test until they turn our power off. Been there, won that.
JB
Instead of having a propeller spun by wheels rolling along the ground why not have it spun by another propeller that is being spun directly by the wind? Both propellers are fixed to either end of a free spinning horizontal shaft running the length of the cart. The upwind propeller is spun directly by the wind and also acts as a sail to move the cart along. The upwind propeller also spins the downwind propeller which has an opposite pitch to thrust against the wind causing the cart go faster with the wind. The upwind propeller has more surface area than downwind one.
This would be best tested using a hovercraft.
Would it work? If not why not?
This would be best tested using a hovercraft.
Would it work? If not why not?
Michael C
Graduate Poster
Michael C
Graduate Poster
It would not work.
The cart uses the difference between the speed of the ground and the speed of the air to produce its motive force. If there is no difference (= no wind), it won't work: it has zero velocity with respect to both ground and air.
If both the propellers of your hovercraft are only in contact with the air, then there is no difference of speed to be exploited. The hovercraft will have zero velocity relative to the air, meaning that it will move at the same speed as the wind.
Brian-M
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True. Power out is never more than power in.
Also true. The gears only exchange speed for force, and vice-versa.
Yes, there will be.
Acceleration is determined by force, and gearing exchanges force for speed.
No, it won't be.
Acceleration is determined by force.
This device exchanges force for speed.
The final result is, it'll take longer to accelerate (less force to cover the same distance) but will reach a higher final velocity.
Remember, force and power are two different things.
That's why I brought it up. The wheel-driven object and the propeller-driven object are behaving in exactly the same manner, using exactly the same principles.
Only the medium being used to turn the wheel/propeller is changed.
No, no, no... You've clearly misunderstood this.
The geared cart moves in the same direction as your hand is pushing the plank of wood, only faster.
If the gearing was reversed, with big cogs on the bottom wheels, and small cog on the top wheel (or if the existing cart was simply turned upside-down), then the geared cart would move in the opposite direction of the hand, but only at the same speed as your hand is pushing the board.
The top wheel is geared to the bottom wheels with a ratio of 1/2, and so the cart moves in the same direction as it is being pushed. If it was geared with a ratio of 2/1, the opposite would be true.
Sorry, average difference of what to what will be zero?
I'm fairly sure I don't understand what you are trying to say here.
I'm equally sure you are completely misunderstanding how the geared cart works too.
From there, nowhere.
Because the top wheel is geared to turn backwards at half the speed of the cart, when the cart is moving twice the speed of the plank, the surface of the wheel in contact with the board is effectively stationary.
In other words, the force from the board is being applied over zero distance.
Power = Force times Distance
It doesn't matter how much force the plank has, if it's not applying it over any relative distance, it's not transferring any power to the cart, and so it can't accelerate any further.
The geared cart has nothing to do with the treadmill.
You've already admitted that a cart can travel faster than the medium pushing it.
What's wrong with using the wind as the medium pushing it?
I notice you still haven't answered my earlier skateboarder/car questions.
Any time you feel like explaining the differences in the two situations to me, I'm ready to listen.
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Brian-M
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Except for Humber? Probably.
You want to gear down, not up. The air should be coming out the back of the propeller at a lower speed than the surface of the treadmill. This way the cart can be pushing on the air with more force than the treadmill is pushing on the cart. (The cost of more force for the same power is always less speed/distance).
H'ethetheth
fishy rocket scientist
But the cart is motionless in still air on the treadmill. You really are losing me here.
First of all: What? The first sentence seems to mean that you think that things that move as fast as the air do not mave as fast as the air.
The second one has been addressed by me and presumably by many before me judging by the movie that shows the cart self-starting. The acceleration to windspeed is stable.
If that will make them relevant to the discussion, go for it.
But moving it around does not add any new information. Velocity is relative; the two situations are equivalent.
There are reasons for this: It has been found to work, and more importantly, it can be reasoned to be true, and observed to be true.
I'm sorry, I'm not going to say it again. You seem to have had a very odd physics teacher.
No! If you sit on a missile you can know your velocity, but only if you define a frame of reference! Velocity is meaningless without a frame of reference, that's why I asked you how fast the earth is travelling. You gave the correct answer, but for some reason you think that this is not true for things on earth.
No, it is not redundant. It can be enlightening, because it shows you that you don't have to measure flight in the air only; you can use a wind tunnel.
No. As an aerospace engineer I can assure you that this is a failure of your understanding.
The only force the cart needs to overcome when it is riding at wind speed is the friction in the mechanism. The energy to do this comes from the fact that, in this situation, the ground is passing by at wind speed.
OK. So far, so good.
Wait a minute. I think we have a problem now Houston. You could operate it in a wind shear with one prop in each airmass. But both props in a common airmass wouldn't get you anything. We have to exploit the energy available at the interface of the wind and road (or wind and wind - if we have two different winds).
ThinAirDesigns
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No, but I do.
This will work exactly as well as gearing the front tires of your car to the back tires of your car and testing to see if it will move on level ground without starting the motor.
It will also work exactly as well as removing your sail and replacing it with an extra keel.
A: there must be two mediums
B: they must be moving relative to each other
C: the device must have a tool to interface with both mediums.
JB
Not if they counter rotate, or otherwise oppose. You still need a differential force, or a second that is common to both.
They can be in series or parallel, but not both at the same time.
Holding a seance? You will need more than two to make that idea fly.
Then perhaps is was just a typo.
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ThinAirDesigns
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Sure. (Unlike torque, thrust will need some interface to be converted to/from torque.)
Many others have given correct and useful answers, so I'll only say what hasn't been said yet.
At zero relative speed to air, such a device would experience no force - none of its propellers would start to rotate. This is different from the cart, which cannot be at rest with respect to both ground and air (as long as there's wind).
If a device started with some relative speed wrt air, there would be a limited source of energy available for it (basically the kinetic energy of the device in the air's reference frame). This energy could, in theory, be utilized to temporarily propel a part of the device at a speed faster than the "wind speed" the device experiences. For example, let your air-floating device consist of the OP cart plus a large stone slab (with a lot of inertia). The cart part will be able to move over the stone slab with greater than "wind speed". But doing this will slow down the stone slab wrt air, and when its speed is equalized with that of air, there will be no more "wind" and it won't work anymore.
ThinAirDesigns
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And yet in spite of your empty assertions, energy is extracted from the relative motion between two mediums every moment of every day the world over.
JB
ThinAirDesigns
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The only harmonic wheel I'm familiar with is music related and to that the answer would be no.
JB
Hi!
I'm new to this forum, so a quick description of my status:
1. The cart works.
2. I understan how it works (at least rouhgly).
3. I understand how a sailboat works (pretty damn well).
About humber's f=ma issue. Rocket motors can provide constant force, right? I just wonder, if we would apply this constant force to the cart (the plank operated one was at question), I think the wall of the room would be a great frame of reference!
Anyway, to the point.
It has been said that the cart is like an sailboat/iceboat running with the wind. I'll use the sailing term running in the meaning that it is trying to reach something that is in the direction to where the wind is blowing, zigzaggin or not. It has also been said that the cart's transmission/wheels have the same fuction as the keel of the sailboat. This I think is not entirely true. Although the transmission/wheels are essential to the cart in the same manner that the keel is essential for a sailboat, the way they operate is quite different. I believe (and try to show) that the aerodynamic properties of the propellor are of utmost importance in the operation of the cart.
The picture below describes how a sailboat works when tacking upwind (those of you who already know can skip this part). In black: the apparent wind angle is about 45 degrees, the airflow on the sail is mainly laminar and the total force from the sail is directed aproximately perpendicular to the wind (in real life probably a bit more backwards). Without the keel the boat would only drift in the direction of the force and not go upwind. But the force (any force) can be divided to tho components: a large component sideways (in relation to the boats heading) and a small component forward.
Oops. The picture would be here but newbies are't allowd to post links...
Can we fool the system like this (add http and replace X->.):
forumsXrandiXorg/vbimghostXphp?do=displayimg&imgid=14370
In red: in a similar way, water flow past the keel, just as the wind flows past the sail. Althoug the keel is symmetrical, the boat is always drifting a bit sideways (about 4-5 degrees) and therefore in relation to the waterflow the keel is not symmetrical and provides lift. The simple version is just to say that the keel stops the boat from moving sideways. And ofcourse the keel also drags a bit. Anyway summing all the forces we are left with a small force forward (in green). (Bring the z-coordinate to the picture, the sideways forces also cause momentum that makes the boat heel, but thats not relevant here.) BTW the force needed is VERY small, I think about 50 kg for a "family cuiser" of 3000 kg displacement.
And now over to the cart. Greens are the aerodynamic forces on the blade, red and purple are the forces spinning/braking the prop/wheels through gearing.
Lets look at the propeller blade pointing directly up, we are looking from above the cart down. Another picture (I must rely that you can get to the pictures).
forumsXrandiXorg/vbimghostXphp?do=displayimg&imgid=14371
In picture 1 the cart is startin. The propellor/sail is too much perpendicular to the wind so the flow is turbulent. Anyway it results in forward force and a force trying to spin the prop CCW. But part of the forward force is transferred via gearing to rotate the prob CW. The result has been seen, the cart starts to move and propellor to spin CW.
As the speen increases (pic 2) the apparent wind to the sail comes from the right (because the blade is moving left). Now the angle is better and the flow probably only partly turbulent and the force from the sail increases A LOT. Again the force has a forward component and a CCW component. Some of the F-componet is transferred via gearing to form a CW component. This is ofcourse also seen as a braking force on the wheels. And then also the CCW component of the sail again produces a forward force on the wheel and makes the braking force smaller. The math must be really tricky because all the forces are connected together, I won't even try myself. I must note that these are totally "freehand" pictures just describing the idea, they are not at all intended to be in any scale or to reflect exact magnitudes or directions of forces involved.
In pic 3 the speed has increased so, that the apparent wind angle is optimum. The forces are coupled together like before ofcourse. The cart is stil accelerating.
Pic 4 is the interesting one. In sailing terms, we are pinching, trying to go too high. No more force the way we want, lots of force the way we don't want. Sailboats sails would be flapping like a flag, the carts propellor produces only drag and tries to slow down its spinning. But this results in forward torque in the wheels!?!
Conclusions:
There must be a balance somewhere between pic3 and pic4. Going at steady state, there is zero torque on the gearing. The wheels and the propellor are just freely spinning. (Assuming no friction in the system ofcourse). The forward force is equivalent to the drag of the cart frame. Am I correct? A fantastic little beast I would say!
In a sailboat, most of the force from the sail is unwanted and the keel resists this force perpendicular to the boats movement. The drag from both keel and sail are totally unwanted. On the cart the situation is different. Most of the force is directed the way we wan't it - forward. It is the propeller bearings that can be related to the keel. Only that they deliver the force we want, and do not oppose a force we don't want. On the cart the propellor blade's drag is interesting also, since a part of it is tranferred to a force in a direction we want, forward torque on wheel (to be precice, lowering the braking power).
My conlusion is, that the gearing is (not exactly but) more like the sheet of a sailboats sail. If we consider an iceboat capable of "tacking downwind", they start with their sails loosely sheeted, perpendicularly exposed to wind. As they accelerate, the apparent wind moves to forward and the sail is sheeted inwards accordingly and so on until we are "tacking downwind". The cart is like an iceboat with the sails sheeted tight all the time because of the gearing. It would be easy but very slow to accelerate a tightly sheeted iceboat from standstill towards downwind. But then again for an iceboat there is no transmission that could ever make it move backwards...
My interpretation of the fysics has been done from a point of view of a sailor. I'm not into propellor efficiencies. It might well be, an propably is, that doing all the calculations, these different explanations are in fact equal in a mathematical/physics sence. Just wanted to provide and alternate way of expaining our little hero.
To the guy trying to build one (ynok?). I think the shape of the propellor is of utmost importance here. Maybe even more important than the friction or exactly right gearing. If you think that the propellor must be able to freely "glide" through airflow "without any power", you understand that it must be precicely designed for this. Any prop designed just to transform torque into thrust won't necessarily do whaterver the efficiency. At least thats what I think.
That that. Huh. Back to work.
huh34
PS. I might have been able to put the pictures as attachments.
I'm new to this forum, so a quick description of my status:
1. The cart works.
2. I understan how it works (at least rouhgly).
3. I understand how a sailboat works (pretty damn well).
About humber's f=ma issue. Rocket motors can provide constant force, right? I just wonder, if we would apply this constant force to the cart (the plank operated one was at question), I think the wall of the room would be a great frame of reference!
Anyway, to the point.
It has been said that the cart is like an sailboat/iceboat running with the wind. I'll use the sailing term running in the meaning that it is trying to reach something that is in the direction to where the wind is blowing, zigzaggin or not. It has also been said that the cart's transmission/wheels have the same fuction as the keel of the sailboat. This I think is not entirely true. Although the transmission/wheels are essential to the cart in the same manner that the keel is essential for a sailboat, the way they operate is quite different. I believe (and try to show) that the aerodynamic properties of the propellor are of utmost importance in the operation of the cart.
The picture below describes how a sailboat works when tacking upwind (those of you who already know can skip this part). In black: the apparent wind angle is about 45 degrees, the airflow on the sail is mainly laminar and the total force from the sail is directed aproximately perpendicular to the wind (in real life probably a bit more backwards). Without the keel the boat would only drift in the direction of the force and not go upwind. But the force (any force) can be divided to tho components: a large component sideways (in relation to the boats heading) and a small component forward.
Oops. The picture would be here but newbies are't allowd to post links...
Can we fool the system like this (add http and replace X->.):
forumsXrandiXorg/vbimghostXphp?do=displayimg&imgid=14370
In red: in a similar way, water flow past the keel, just as the wind flows past the sail. Althoug the keel is symmetrical, the boat is always drifting a bit sideways (about 4-5 degrees) and therefore in relation to the waterflow the keel is not symmetrical and provides lift. The simple version is just to say that the keel stops the boat from moving sideways. And ofcourse the keel also drags a bit. Anyway summing all the forces we are left with a small force forward (in green). (Bring the z-coordinate to the picture, the sideways forces also cause momentum that makes the boat heel, but thats not relevant here.) BTW the force needed is VERY small, I think about 50 kg for a "family cuiser" of 3000 kg displacement.
And now over to the cart. Greens are the aerodynamic forces on the blade, red and purple are the forces spinning/braking the prop/wheels through gearing.
Lets look at the propeller blade pointing directly up, we are looking from above the cart down. Another picture (I must rely that you can get to the pictures).
forumsXrandiXorg/vbimghostXphp?do=displayimg&imgid=14371
In picture 1 the cart is startin. The propellor/sail is too much perpendicular to the wind so the flow is turbulent. Anyway it results in forward force and a force trying to spin the prop CCW. But part of the forward force is transferred via gearing to rotate the prob CW. The result has been seen, the cart starts to move and propellor to spin CW.
As the speen increases (pic 2) the apparent wind to the sail comes from the right (because the blade is moving left). Now the angle is better and the flow probably only partly turbulent and the force from the sail increases A LOT. Again the force has a forward component and a CCW component. Some of the F-componet is transferred via gearing to form a CW component. This is ofcourse also seen as a braking force on the wheels. And then also the CCW component of the sail again produces a forward force on the wheel and makes the braking force smaller. The math must be really tricky because all the forces are connected together, I won't even try myself. I must note that these are totally "freehand" pictures just describing the idea, they are not at all intended to be in any scale or to reflect exact magnitudes or directions of forces involved.
In pic 3 the speed has increased so, that the apparent wind angle is optimum. The forces are coupled together like before ofcourse. The cart is stil accelerating.
Pic 4 is the interesting one. In sailing terms, we are pinching, trying to go too high. No more force the way we want, lots of force the way we don't want. Sailboats sails would be flapping like a flag, the carts propellor produces only drag and tries to slow down its spinning. But this results in forward torque in the wheels!?!
Conclusions:
There must be a balance somewhere between pic3 and pic4. Going at steady state, there is zero torque on the gearing. The wheels and the propellor are just freely spinning. (Assuming no friction in the system ofcourse). The forward force is equivalent to the drag of the cart frame. Am I correct? A fantastic little beast I would say!
In a sailboat, most of the force from the sail is unwanted and the keel resists this force perpendicular to the boats movement. The drag from both keel and sail are totally unwanted. On the cart the situation is different. Most of the force is directed the way we wan't it - forward. It is the propeller bearings that can be related to the keel. Only that they deliver the force we want, and do not oppose a force we don't want. On the cart the propellor blade's drag is interesting also, since a part of it is tranferred to a force in a direction we want, forward torque on wheel (to be precice, lowering the braking power).
My conlusion is, that the gearing is (not exactly but) more like the sheet of a sailboats sail. If we consider an iceboat capable of "tacking downwind", they start with their sails loosely sheeted, perpendicularly exposed to wind. As they accelerate, the apparent wind moves to forward and the sail is sheeted inwards accordingly and so on until we are "tacking downwind". The cart is like an iceboat with the sails sheeted tight all the time because of the gearing. It would be easy but very slow to accelerate a tightly sheeted iceboat from standstill towards downwind. But then again for an iceboat there is no transmission that could ever make it move backwards...
My interpretation of the fysics has been done from a point of view of a sailor. I'm not into propellor efficiencies. It might well be, an propably is, that doing all the calculations, these different explanations are in fact equal in a mathematical/physics sence. Just wanted to provide and alternate way of expaining our little hero.
To the guy trying to build one (ynok?). I think the shape of the propellor is of utmost importance here. Maybe even more important than the friction or exactly right gearing. If you think that the propellor must be able to freely "glide" through airflow "without any power", you understand that it must be precicely designed for this. Any prop designed just to transform torque into thrust won't necessarily do whaterver the efficiency. At least thats what I think.
That that. Huh. Back to work.
huh34
PS. I might have been able to put the pictures as attachments.
Attachments
But if you consider the ice boat in its final configuration on a downwind tack with VMG faster than the wind, it will be sheeted in quite closely. If we now consider this same ice boat sitting motionless in the exact same configuration and orientation, it will feel the same two opposing forces as our prop. The wind will be hitting the sail from an angle closer to its trailing edge. If we look at the resulting force vector it will have the effect of wanting to make that sail go both left (assuming a starboard tack) and downwind. Moving left would cause the ice boat to go upwind based on the constraint of its blades. But the downwind force will still win out. All we'd have to do is change the angle of the sail in order to have the upwind force win out. This is equivalent to changing the advance ratio on our cart.