Merged DDWFTTW in a “Real” Wind

[spork]

Because the propeller is moving faster downwind than the driving wind and therefore hits the air as a headwind not a driving wind. Where the propeller hits the air therefore can't possibly be the transfer point.

Completely wrong.

Ynot,
...There is not a mysterious "cushion of air of prop blast and wind".

He's been told that a dozen times. He simply ignores it and insists on repeating his mistake as if it's fact.

It's a nit. Is this a drag statement, a rolling friction statement, KE statement, or something else? I don't see how the statement in blue can be true any way I've sliced it.

Would you please explain?

Happy to explain. When pushing a car by hand, I believe rolling resistance is by far the dominant reaction force. Rolling resistance is normally considered to be a constant percentage of vehicle weight. So pushing a car at 10 mph will require 10 times as much energy (per second) as it would to push that same car at 1 mph.

 

[Jack by the hedge]

An above WS cart accelerates that slower air before it, so the force in reaction to that, opposes the cart.
The blades of the rotor will experience the same force, and it makes no difference if it is turning or not.

That's not correct. It matters a very great deal that the prop is turning.

Consider a small volume of air immediately in front of the propeller blade (small compared to the width of the propeller blade).

The air is moving forward at windspeed, but the cart (and prop) are moving forward faster. So if the prop were not turning, the blade would contact that small volume of air and push it out of the way, transferring momentum from the cart to the air and creating a reaction force which acts to slow the cart.

But the propeller is turning. As the propeller advances towards the small volume of air, the propeller blade turns such that the spot on the blade which is immediately behind the volume of air is further across the face of the blade and, oweing to the pitch of the blade, it is further back. Although the cart has advanced further than the air, the part of the blade behind the volume of air is actually slightly further away. You can see, I hope, that this would tend to produce a lower air pressure in front of the blade and consequently divert the air in front, transferring momentum from the air to the blade and pushing the blade forwards.

The opposite happens on the rear face of the propeller blade: Each small portion of air, moving forward at windspeed, which comes into contact with the propeller blade finds, that although the propeller as a whole is moving forwards faster than the air, the point in contact with that volume of air is a spot on the blade which is constantly moving across the face of the blade and is advancing more slowly than the wind. So the wind behind the blade transfers momentum to the blade and speeds the cart up.

 

[mender]

Opposing winds aren't like opposing light beams that don't interact with each other. There is a region behind the propeller where the opposing winds interact with each other and create an area of pressure. I believe this area is quite close to the propeller blades. For the energy of the wind to transfer to the cart there has to be a point of transfer. I believe this point of transfer is this area of pressure. If anyone thinks the point of transfer is in a different place please explain exactly where it is.

Ynot, this is what I also initially thought was happening but through patient conversations with spork and JB and several others, I realized that the air wasn't stacking up behind the cart.

What is happening is that the propeller is accelerating the air molecules. The place where that is happening is in the immediate vicinity of the prop blades, and the energy transfer is in the same place. The air mass is accelerated to the rear, resulting in an equal and opposite force acting on the prop blades (F=MA). After the cart passes by, the air that has been slowed wrt the cart by the prop interacts with the surrounding air and is gradually brought back up to the speed of that surrounding air, but that happens well after the cart is gone and doesn't contribute to the thrust.

I'm a pilot and have experienced ground effect, and was originally thinking that the cart was being pushed by a similar cushion of air, just like you are. The problem with that is that there isn't "ground" right behind the prop, close enough to generate that ground effect, like the ground immediately beneath a wing does. It's a good thing that a wing can generate lift without ground effect otherwise flying would be very exciting! In the same way, the cart prop doesn't need or use that stacking effect but instead operates just like a wing, changing the momentum of the air instead, and remember that more lift (momentum change of the air) is generated by the top of the wing than the bottom. That means that there is more happening in front of the cart than behind.

I'm still not 100% sure that there isn't a very small "ground effect" due to the surrounding air but I am sure that it's much more about changing the speed of the air than it is stacking it up against "the following wind". Maybe someone has a comment about the actual amounts of each?

If the cart was operating in a vacuum and had a little cylinder of air instead of a prop, it would work just as well by changing the momentum of the air in that cylinder by shooting it to the rear. The KE change of that air would be very close to the same KE change that the prop causes on the air that passes through it for the same amount of force acting on the cart.

 

[humber]

Completely wrong.



He's been told that a dozen times. He simply ignores it and insists on repeating his mistake as if it's fact.




Happy to explain. When pushing a car by hand, I believe rolling resistance is by far the dominant reaction force. Rolling resistance is normally considered to be a constant percentage of vehicle weight. So pushing a car at 10 mph will require 10 times as much energy (per second) as it would to push that same car at 1 mph.

Give it up Spork, your silly ideas have been exposed, and you don't have the skill that a hoaxer would in defense. Nothing will save your TM, nor the trial result.

 

[windgrins]

Spork-Rolling resistance of a real device!

Happy to explain. When pushing a car by hand, I believe rolling resistance is by far the dominant reaction force. Rolling resistance is normally considered to be a constant percentage of vehicle weight. So pushing a car at 10 mph will require 10 times as much energy (per second) as it would to push that same car at 1 mph.

The reason I questioned this is because yesterday my riding lawnmower lost its drive belt and I had to push it by hand about 200 yards home! The rolling resistance was very high initially but as I moved it faster, (walking speed only, of course) the resistance either seemed to decrease or stay roughly the same.

The tires and axles seemed to actually have more resistance at the slow speed.

So, for an average car at these slow speeds, I'm not sure if rolling resistance is all that linear but it certainly may be.

Thanks for your explanation.

Regards,

WindGrins

 

[windgrins]

Well Stated JBTH!

That's not correct. It matters a very great deal that the prop is turning.

Consider a small volume of air immediately in front of the propeller blade (small compared to the width of the propeller blade).

The air is moving forward at windspeed, but the cart (and prop) are moving forward faster. So if the prop were not turning, the blade would contact that small volume of air and push it out of the way, transferring momentum from the cart to the air and creating a reaction force which acts to slow the cart.

But the propeller is turning. As the propeller advances towards the small volume of air, the propeller blade turns such that the spot on the blade which is immediately behind the volume of air is further across the face of the blade and, oweing to the pitch of the blade, it is further back. Although the cart has advanced further than the air, the part of the blade behind the volume of air is actually slightly further away. You can see, I hope, that this would tend to produce a lower air pressure in front of the blade and consequently divert the air in front, transferring momentum from the air to the blade and pushing the blade forwards.

The opposite happens on the rear face of the propeller blade: Each small portion of air, moving forward at windspeed, which comes into contact with the propeller blade finds, that although the propeller as a whole is moving forwards faster than the air, the point in contact with that volume of air is a spot on the blade which is constantly moving across the face of the blade and is advancing more slowly than the wind. So the wind behind the blade transfers momentum to the blade and speeds the cart up.

Well Stated JBTH! Glad to see you back!

WindGrins

 

[humber]

Ynot, this is what I also initially thought was happening but through patient conversations with spork and JB and several others, I realized that the air wasn't stacking up behind the cart.

What is happening is that the propeller is accelerating the air molecules. The place where that is happening is in the immediate vicinity of the prop blades, and the energy transfer is in the same place.

For a prop, the air leaving the stream tube has been accelerated by the rotor. To maintain the same mass/time flow at the input to that stream tube, as the output, ( thrust side) the cross sectional area behind the blade decreases.
That is in complete contrast with a turbine which extracts energy from the air. The air that is input to the stream tube, is decelerated, so slowed, and the stream tube to the rear increases in area. The two cases are mutually exclusive, is only; because tow different pressures can not exist at the same time.

The air mass is accelerated to the rear, resulting in an equal and opposite force acting on the prop blades (F=MA). After the cart passes by, the air that has been slowed wrt the cart by the prop interacts with the surrounding air and is gradually brought back up to the speed of that surrounding air, but that happens well after the cart is gone and doesn't contribute to the thrust.

When above WS, the cart is faster than the wind ahead of it. A prop would accelerate that air, and it would leave the stream tube, faster than the free stream, and then slow back to the free-stream velocity.
To do that, power is needed at the prop shaft. An engine. One would expect that a pilot sees that.

I'm a pilot and have experienced ground effect, and was originally thinking that the cart was being pushed by a similar cushion of air, just like you are. The problem with that is that there isn't "ground" right behind the prop, close enough to generate that ground effect, like the ground immediately beneath a wing does.

Ground effect is a matter of change of pressure between the object and the ground. No power comes from it.

It's a good thing that a wing can generate lift without ground effect otherwise flying would be very exciting! In the same way, the cart prop doesn't need or use that stacking effect but instead operates just like a wing, changing the momentum of the air instead, and remember that more lift (momentum change of the air) is generated by the top of the wing than the bottom. That means that there is more happening in front of the cart than behind.

F1 cars once used the ground effect to allow air pressure to increase wheel traction. Ground effect has nothing to do with the cart. That differs from the wind gradient, that that prop exploits as it turns. The wind was monitored at hun height, so there really is no record, no matter how one looks at it.
However, the cart exploits wind direction changes, the prevent it from the embarrassingly slow crawl that is seen when ddw, in a steady wind,

If the cart was operating in a vacuum and had a little cylinder of air instead of a prop, it would work just as well by changing the momentum of the air in that cylinder by shooting it to the rear. The KE change of that air would be very close to the same KE change that the prop causes on the air that passes through it for the same amount of force acting on the cart.

That thinking is why most pilots fly planes, but don't design them.

 

[humber]

The reason I questioned this is because yesterday my riding lawnmower lost its drive belt and I had to push it by hand about 200 yards home! The rolling resistance was very high initially but as I moved it faster, (walking speed only, of course) the resistance either seemed to decrease or stay roughly the same.

The tires and axles seemed to actually have more resistance at the slow speed.

So, for an average car at these slow speeds, I'm not sure if rolling resistance is all that linear but it certainly may be.

Thanks for your explanation.

Regards,

WindGrins

Tyres may require more work to initiate rolling, but that is not directly related to rolling resistance.
However, the explanation for your lawn mover ( and pushed car) is that from a standstill, you need to do work to accelerate the mass of the mower to add KE, as well as against the rolling resistance. When at constant speed, there is no KE demand.
Also, when accelerating, the reaction force at the wheel is larger, so the friction increases for that period. If there are pneumatic tyres, they will distort to accommodate that. If not, then the ground does, or sliding will take place if the applied force is larger than the former rolling resistance.

 

[69dodge]

Regarding a "cushion of air" behind the propeller: I'm not sure exactly what that means, but I'd imagine that the pressure right behind the propeller is greater than the pressure right in front of it, in the same way that the pressure below an airplane wing is greater than the pressure above it. My understanding is that, in the case of an airplane wing, the "pressure difference" theory and the "accelerating air downward" theory are alternative ways of looking at the same thing, and not two different effects, whose relative magnitude we could compare, for example. I assume the same applies to a propeller. Comments?

 

[mender]

For a prop, the air leaving the stream tube has been accelerated by the rotor. To maintain the same mass/time flow at the input to that stream tube, as the output, ( thrust side) the cross sectional area behind the blade decreases.
That is in complete contrast with a turbine which extracts energy from the air. The air that is input to the stream tube, is decelerated, so slowed, and the stream tube to the rear increases in area. The two cases are mutually exclusive, is only; because two different pressures can not exist at the same time.

True; that shows that the prop is definitely not "stacking" air behind it when acting as a prop. Hopefully, ynot will see how that proves the air cushion theory wrong. Thanks for adding that.

Since it is quite obvious that the cart prop is being driven by the wheels and the power flow is from the wheels to the prop as witnessed by the tension of the chain and the presence of the ratchet on BB, there will be a stream tube behind the cart that decreases in cross sectional area rather than increases.

An experiment that shows the shape of the stream tube might help by proving a visual confirmation that the prop is acting as a prop should.

 

[windgrins]

Ground Effect on airfoils

Regarding a "cushion of air" behind the propeller: I'm not sure exactly what that means, but I'd imagine that the pressure right behind the propeller is greater than the pressure right in front of it, in the same way that the pressure below an airplane wing is greater than the pressure above it. My understanding is that, in the case of an airplane wing, the "pressure difference" theory and the "accelerating air downward" theory are alternative ways of looking at the same thing, and not two different effects, whose relative magnitude we could compare, for example. I assume the same applies to a propeller. Comments?

You are correct. A propeller does take air from a lower to a higher pressure area as the flow through it, which at first may seem counterintuitive.

The comment about "ground effect" :

From Wikipedia:

The wing in ground effect, often described as a 'cushion', is thought to be an increase in air pressure which occurs below a wing when it gets close to the ground. The effect begins to be noticeable when the aircraft's altitude is within 1–1.5 times the length of its own wingspan and, when the altitude is within about half the wing chord, the effect can increase lift by as much as 10%. Due to the effect of spoilers and high wing loading, this effect is only dramatically noticed in smaller, less complex aircraft, usually weighing less than 12,500 lbs (5,670 kg).[citation needed] Ground effect is a major factor in aircraft "floating" down the runway, and is the reason that low-wing aircraft have a tendency to float more than the high-wing varieties.

Since the BB prop is working in relatively clean air, there is probably no ground effect operating. There is a reasonable expectation of a gradient of wind with the wind higher at altitude.

This is why the streamer at the top of the BUFC cart was so illustrative. It clearly shows that the cart was exceeding WS that any portion of the cart was experiencing (other than the streamer and pole, itself). It exceeded this also as illustrated by the streamer blowing almost directly back.

Regards,

WindGrins

 

[humber]

That's not correct. It matters a very great deal that the prop is turning.

Consider a small volume of air immediately in front of the propeller blade (small compared to the width of the propeller blade).

The air is moving forward at windspeed, but the cart (and prop) are moving forward faster. So if the prop were not turning, the blade would contact that small volume of air and push it out of the way, transferring momentum from the cart to the air and creating a reaction force which acts to slow the cart.

But the propeller is turning. As the propeller advances towards the small volume of air, the propeller blade turns such that the spot on the blade which is immediately behind the volume of air is further across the face of the blade and, oweing to the pitch of the blade, it is further back. Although the cart has advanced further than the air, the part of the blade behind the volume of air is actually slightly further away. You can see, I hope, that this would tend to produce a lower air pressure in front of the blade and consequently divert the air in front, transferring momentum from the air to the blade and pushing the blade forwards.

If the slower WS air is struck by the faster prop, the transfer of energy takes place over a very, very, short time period. Only when considered as a flow - an aggregate over time - does it make any sense to talk of the blade's motion. However, if that is considered as you suggest, the blade would "move away" from the molecule, so the force would be less than if it did not. It would be self-defeating when hoping to gain energy. Fortunately, rotors do not work like that.

The opposite happens on the rear face of the propeller blade: Each small portion of air, moving forward at windspeed, which comes into contact with the propeller blade finds, that although the propeller as a whole is moving forwards faster than the air, the point in contact with that volume of air is a spot on the blade which is constantly moving across the face of the blade and is advancing more slowly than the wind. So the wind behind the blade transfers momentum to the blade and speeds the cart up.

Same. The acceleration would fall, as the blade moves away. Rather like hitting a ball with a bat, and moving it backwards as you try to hit it the other way....but...

The shape of the airfoil causes the air to follow its profile, such that efficient transfer can take place. That is a useful by-product of what is known as the no-slip condition, http://en.wikipedia.org/wiki/No-slip_condition, and rotors are designed to exploit it.
Drag may be seen as a loss, but without it, the rotor efficiency would be very low. That is what so many "blue skies" thinkers and their imaginary props, forget. When losses are wished away, the mechanism that creates what they seek, is wished away too. It is for that reason, that all such analysis are "static", and use only algebra, otherwise the game stops there.

Impact time affects the transfer to or from the blade, so the longer the air stays in contact, the better. Ask a golfer. Keep the head in contact with the ball for as long as it is possible to apply force. Or study impact theory, JBTH.
The same effect accounts for the large static thrust, that falls when the prop is in motion.

 

[spork]

I am sure that it's much more about changing the speed of the air than it is stacking it up against "the following wind". Maybe someone has a comment about the actual amounts of each?

The flow through the prop disk and in the vicinity of the prop and craft is exactly the same as for an aircraft in flight. Nothing about the flow in the region could tell you that you're operating in a true tailwind.

The reason I questioned this is because yesterday my riding lawnmower lost its drive belt and I had to push it by hand about 200 yards home! The rolling resistance was very high initially but as I moved it faster, (walking speed only, of course) the resistance either seemed to decrease or stay roughly the same.

The typical model for rolling resistance is that the force is directly proportional to the weight on the tires. Therefore it should stay the same. As with all models, it's not perfect.

So, for an average car at these slow speeds, I'm not sure if rolling resistance is all that linear but it certainly may be.

It's not linear, but constant.

 

[mender]

The flow through the prop disk and in the vicinity of the prop and craft is exactly the same as for an aircraft in flight. Nothing about the flow in the region could tell you that you're operating in a true tailwind.

Will there be a difference in engine rpm when doing a static run-up in a private plane with no-wind conditions vs a tailwind? Not usually done, but I suspect there will be a difference, right?

ETA: it's been a while but I seem to remember the engine rpm being pretty static during the actual take-off run, so that would indicate that I'm wrong and you're right - again.

 

[windgrins]

You got me again????

It's not linear, but constant.

Spork,

If the rolling resistance is constant, ignoring aerodynamic drag and only considering rolling resistance, wouldn't a car need the same force to keep it at 1mph as 10mph, once achieved?

Since energy =f X d, I would assume that the energy would be the same but the power expended would be higher since the car is covering the same distance 10 times faster at 10mph.

Perhaps I'm being daft here. Please help me out!

Regards,

WindGrins

 

[Jack by the hedge]

If the slower WS air, is struck by the faster prop, the transfer of energy takes place over a very, very short time period.

The whole point of what I was explaining about the air immediately in front of the blade is that it is not struck by the blade.

Only when considered as a flow - an aggregate over time - does it make any sense to talk of the blade's motion.

Why only then? If you insist, you can analyse the blade's motion aggregated over time in terms of its angle of attack against the apparent wind. You'll get the same effect, though: the prop blade will accelerate the air backwards and the reaction force pushes the prop forwards.

However, if that is considered, the blade would "move away" from the molecule, so the force would be less than if it did not. So would be self-defeating when hoping to gain energy.

Please explain which force you are talking about here. It is far from clear what you mean.

Same. The acceleration would fall, as the blade moves away. Rather like hitting a ball with a bat, and moving it backwards as you try to hit it the other way....

I can't picture whatever you mean by that. Can you explain it any more clearly?

The shape of the airfoil causes the air to follow its profile, such that efficient transfer can take place. That is a useful by-product of what is known as the no-slip condition. http://en.wikipedia.org/wiki/No-slip_condition

Which explains why the blade behaves like a wing. Or a sail. Agreed.

Impact time affects the transfer to or from the blade, so the longer the air stays in contact, the better.

The propeller blades are always in contact with the air. Larger blades or extra blades would transfer energy from more of the air, agreed.

 

[spork]

Spork,

If the rolling resistance is constant, ignoring aerodynamic drag and only considering rolling resistance, wouldn't a car need the same force to keep it at 1mph as 10mph, once achieved?

Yes.

Since energy =f X d, I would assume that the energy would be the same but the power expended would be higher since the car is covering the same distance 10 times faster at 10mph.

Once again, I'm describing the amount of energy that would be expended in any given period of time. Thus pushing the car at 10 mph for 10 seconds would take 10 times the energy as pushing it at 1 mph for 10 seconds. So yes, by definition, the power required would be 10 times as great as well.

 

[spork]

Will there be a difference in engine rpm when doing a static run-up in a private plane with no-wind conditions vs a tailwind? Not usually done, but I suspect there will be a difference, right?

There could be. Keep in mind that much of the prop is stalled when running in static conditions. When running in reversed flow (something our cart never does after it passes wind speed), even more of the prop will be stalled.

We're talking here about a prop that's operating in a local headwind - just like the prop on a plane in flight - whether that plane is flying into a headwind or with a tailwind.

 

[humber]

The flow through the prop disk and in the vicinity of the prop and craft is exactly the same as for an aircraft in flight. Nothing about the flow in the region could tell you that you're operating in a true tailwind.

You claim the prop "harvests" energy from the wind. That is a forced position in avoidance of the stigma of perpetual motion, implied by your wheel-driving-the-prop-driving-the-wheel Harrypothesis.

If the cart is moving to the right above WS, then as the wind approaches the rotor from the right, the air pressure increases gradually, since the rotor acts as a barrier to the wind. There is a pressure zone to the front of the rotor, that is highest at the rotor, and falls gradually away to the free-stream pressure ahead.
The air pressure will drop immediately behind the rotor plane (to the left). It then gradually increases to the normal air pressure level of the free stream.

The typical model for rolling resistance is that the force is directly proportional to the weight on the tires. Therefore it should stay the same. As with all models, it's not perfect.

It is approximated as Frr= m*g*Crr. Crr remains essentially constant with velocity ,and that accounts for the effect, and not some inferred folk physics about the "percentage of the weight".

The reason that the Crr remains (near) constant under the same load, is that there is only a small amount of motion between the tyre and the road, because rotation, causes the tyre to peel on and off the surface. An extreme case is the track of a tank. For most of the track, there is not motion between the track and the road, so the friction is static, but without the expense of dragging. Wheels are good, huh?

It's not linear, but constant.

It is neither.

 

[windgrins]

Thought I was losing it-Thanks,

Yes.



Once again, I'm describing the amount of energy that would be expended in any given period of time. Thus pushing the car at 10 mph for 10 seconds would take 10 times the energy as pushing it at 1 mph for 10 seconds. So yes, by definition, the power required would be 10 times as great as well.

OK,

We are in violent agreement! Thanks for clearing this up. I thought I was losing it!

Regards,

WindGrins