plane on a conveyor belt?

[rpm]

Reading this thread has made me dumberer.

[2bighorn]

Originally posted by hitech
Would you agree the rolling friction is generating a torque about the axle?

Definately

I looked up the bearing manufacturers and  the bearings for your 2ft wheel able to withstand dynamic radial loading of your mass combo (plane and wheels). I ignored all thermal properties, etc.
Torque per bearing is about 0.94 lb ft
(all numbers are rounded)
At 10,000 rpm that would be equivalent of 3.58 horsepower loss due to bearing friction.
or at 2ft wheel diameter that'd be about 714 mph of forward speed relative to the belt.

At 1,000,000 rpm we'd lose equivalent of 358 HP due to bearing friction,
at million rpm that'd be about 71,364 mph forward speed relative to the the belt.

So hypotetically speaking, on normal runway, if you'd speed fast enough you'd reach the point where bearing friction would be greater than engine thrust.

What we have to account for is, that in case of powered belt, the plane's thrust does not have to fight friction as much, since belt is contributing to wheel rotation whenever it's equal or faster than plane's air speed, so we actually have to ad resulting force (belt/wheel friction torque) to forward thrust.

[hitech]

2bighorn: I'm not interested in the values, what I am interested in , is you drawing a diagram showing how the torque friction of the bearing you are calculating turns into a directional force.

The reason I am asking this, it is the first step to understanding how the force accelerating the wheel acts exactly the same as the bearing friction.

[eskimo2]

Originally posted by 2bighorn
They have all the same total mass, different mass distribution, hence difference in energy for a particular part ie wheels.

The heavier wheels with more inertia (more angular momentum) are attached onto lighter airframes with less inertia (less kinetic E), or
the lighter wheels with less inertia (less angular momentum) are attached onto heavier airframes with more inertia (more kinetic E).

Total E needed to lift them up and fly is the same. Just because some of the E ain't as visible it doesn't mean it ain't there.

2Bighorn,

So your explanation is that Bob’s and Chuck’s planes have “invisible energy” stored somewhere?  How do you not see what is happening?

The heavy spinning wheels in Al’s plane have both 50 mph worth of kinetic energy AND rotational inertia.  Chuck’s plane’s original wheels (inside the plane) have only kinetic energy.   The spinning wheels in Bob’s plane have both 50 mph worth of kinetic energy and rotational inertia, but are half the mass of Al’s.  Bob’s spare set are in the plane and have only kinetic energy.  This represents three completely different energy states.

Al’s plane must have traded some forward acceleration/power of his entire plane for the 2X rotational energy now stored in his wheels.  Bob’s plane also must have traded some forward acceleration/power of his entire plane for the 1X rotational energy now stored in his wheels, and he took off before and in less distance than Al.  Chuck’s plane lost nothing to rotational inertia and took off first and in the least distance.

[2bighorn]

Originally posted by hitech
2bighorn: I'm not interested in the values, what I am interested in , is you drawing a diagram showing how the torque friction of the bearing you are calculating turns into a directional force.

The reason I am asking this, it is the first step to understanding how the force accelerating the wheel acts exactly the same as the bearing friction.

If belt is stationary than thrust has to oppose both bearing friction and belt/wheel friction. If we add another source of E like conveyor and if it contributes to wheel rotation, then both share the workload of fighting bearing friction.

Not sure if it's that what you meant,:

[hitech]

You drew exactly what I wanted you to draw.

Now comes the miss understanding.

Draw the same diagram except this time lock the breaks so there is no rotation occurring.

Here is a simplified version, note i'm not showing the torques but lets assume it is on a real plane on a runway and not sliding . The torques are actual implied in just the forces shown.



There is a few mistaken concept in your drawing, the bearing friction force and the belt wheel friction force , both should be labeled torques.  

There is a fundamental difference in the labels, torques can be draw and converted around the tire, I.E. the corresponding Bearing and belt torques could just as well be drawn at the front of the tire facing up and down.

So as you are trying to show the torques can cancel. But they are not forces, the do not move/translate the wheel they just rotate it. And hence why it is very important to label them that way.

And this is where you made your original mistake , when you transcribed the belt friction force to a torque, you were still thinking of it as a force , but you rotated it to the top of the tire, with a torque that is no problem, but not with a force.It is why you came to the conclusion that the rotational inertia helped the plane moved vs held it back.


HiTech

[2bighorn]

Well, you asked to be drawn as a simple force. Of course there's a difference between torque and a force.

But even so, it can be drawn as a pseudo vector to show the direction of a moment of force in relation to body's rotation.

[2bighorn]

Originally posted by hitech
And this is where you made your original mistake , when you transcribed the belt friction force to a torque, you were still thinking of it as a force , but you rotated it to the top of the tire, with a torque that is no problem, but not with a force.It is why you came to the conclusion that the rotational inertia helped the plane moved vs held it back.

Doesn't really matter, you can attach that (pseudo) vector anywhere on the wheel as long as it is pointing (tangentially) in direction of rotation.

But basically, if a torque works in direction of rotation it adds to rotation and oppose those which work in the opposing direction (like bearing friction torque).

Since none really acts radially to our wheel, none contributes to movement of a wheel, just rotation and as such the belt/wheel friction torque does not contribute nor oppose in any way to forward/backward motion of the plane except for the amount of bearing friction torque.

[hitech]

Your drawing is correct.
In so far as a description of how to calculate torque.

But note that original force vector should now remain the same length and direction and be move to the center of the wheel.

And the torque vectors ( refer to the physics pages I posted as vector couples) can be drawn easier like so. This way they cancel each others force, but leave the torque.

Notice I cut your  pseudo vectors in half.



Do you agree that my drawing is an equivalent to yours and the outcome would be the same?

Doesn't really matter, you can attach that (pseudo) vector anywhere on the wheel as long as it is pointing (tangentially) in direction of rotation.

Yes that is basically the definition of torque.  But not force.

But basically, if a torque works in direction of rotation it adds to rotation and oppose those which work in the opposing direction (like bearing friction torque).

No disagreement from me here, that is what I call summing torques.

Originally posted by 2bighorn

Since none really acts radially to our wheel, none contributes to movement of a wheel, just rotation and as such the belt/wheel friction torque does not contribute nor oppose in any way to forward/backward motion of the plane except for the amount of bearing friction torque.

Not sure if we are in agreement here. If you apply a force (not a torque) anywhere on a wheel and in any direction, it will move (translate/accelerate (F=MA) unless countered by another force.

[2bighorn]

Originally posted by hitech
Do you agree that my drawing is an equivalent to yours and the outcome would be the same?

Since it's a pseudo vector of a torque, you have to move that vector along the rotation until the both pivot points are in the same spot.

Bellow I moved them at R distance in the direction of the rotation until pivot points are covered. Then you can ad or subtract them as you'd do with normal force vector.

Red is opposing force, but I still have to rotate it first.



Since those pseudo vectors represent moment of force, if I move them closer (or further) from the axis of rotation, I have to adjust their magnitude.

Originally posted by hitech
Not sure if we are in agreement here. If you apply a force (not a torque) anywhere on a wheel and in any direction, it will move (translate/accelerate (F=MA) unless countered by another force.

Not quite. Whenever you break down torque into that pseudo vector as the force, your force will have component which acts radially, and a component which acts tangentially.

Radial component of the pseudo vector is f cos alpha. Your belt friction torque angle is 90 degrees (relative to the line from pivot to the center) which gives cos 90, hence the radial component of the moment of force is 0.

[hitech]

Not quite. Whenever you break down torque into that pseudo vector as the force, your force will have component which acts radially, and a component which acts tangentially.

Radial component of the pseudo vector is f cos alpha. Your belt friction torque angle is 90 degrees (relative to the line from pivot to the center) which gives cos 90, hence the radial component of the moment of force is 0.

I think you are trying to say that only the radial component translates the wheel, if you are saying that, than I disagree.

Please do so more reading on this , your torque and radial force calc is correct, but you are missing the point that the translation force/acceleration is exactly the same  no mater what the radial and tangential compoents are.


Read the figure 12-26. Straight from my physics book

It is really saying that a force anywhere on an object acts the same as if it went threw the center of mass as far as translation is concerned.

Just think of what they are calling a couple as torque. How I transcribed your original force vector is identical to the 12-26 diagram. I summed the forces, was only 1 so it was easy ,and moved it to the cg. I then used your calcs for the couple/torque vectors.

So in another words any force any where on  an object can be replaced by the sum of the forces placed at the center mass, along with the resulting torque.




Another way to see that radial force is irelavent as far as translation goes, picture 2 people sitting on a teetor totor, both have feet off ground and equally balanced and everything is level. Using your thought that only the radial force would translate,  (obviously 0 in this case) you are saying that neither would excert a force on the ground threw the teetor totor.

HiTech

[hitech]

Ding Ding Ding Ding: Looked at your drawing some more, think I know the disconnect in how you and I  are viewing things. I think you are missing how the slowing of the tire creates a ground friction in the opposite direction.

Take the wheel you drew sitting with the back of the car jacked.  Lets assume we spun the tires

Now the bearing friction would create a counter clock wise torque slowing the wheel.

As below, if the car is in the air, bearing friction would  create no translation force to the tire or the care.



I assume we are in agreement so far.


Now think what happens if the car is moving forward and this wheel that is rotating slower than our car is moving forward, it will slide slightly forward creating friction with ground exerting a force backwards.


Now the contact with the surface will create a torque and force opposite the torque of the breaks, since the wheel is slightly slower this torque will basically speed up the wheel to match the surface speed.

(in a real wheel the inside of the wheel is slightly slower than the out side, and the rubber of the tire is bending / stretching slightly instead of change speeds, but the effect is the same)



So now lets use the method describe in the physics book to translate the ground friction force to the CG and to 2 couples.



Do you disagree with any of this?

HiTech

[Holden McGroin]

So are we convinced yet that the original question is horribly flawed?

It is....

1st.

The conveyor belt can only overcome rolling friction.  The main drag on an airplane is aerodynamic, rolling friction max is probably less than 10% during take off.  The thrust is entirely aerodynamic, the only rolling resistance is bearing friction, and the heating of the tire as it is deformed under load.

2nd.

If the airplane begins to roll forward, the conveyor belt pulls it back.  The forward speed of the airplane is completely countered by the speed of the conveyor.  Since my accelerating force is aerodynamic, my speed will be measured in the air.  The only place where this magical conveyor can perfectly balance is at zero.  

Any other airspeed is out of the range of the original problem.  It does not matter what the RPM of the wheels, if the speed balance is maintained, the only thrust I can apply must be equal to the rolling friction for the balance to be maintained.

Either one must agree that the original problem is fundamentally flawed or the answer is that the airplane can never have any airspeed and therefore will never fly, unless it is a VTOL.

3rd.

As most airplanes have the thrust to overcome rolling friction the aircraft in this problem (which has thrust only equal to rolling friction) could not accellerate to takeoff speed regardless of a moving conveyor. Even if catapulted it would not have the thrust to maintain flight.

Please end this horrible thread.

Holden
Oregon Registered Professional Engineer

[eskimo2]

Originally posted by Holden McGroin
So are we convinced yet that the original question is horribly flawed?

It is....

1st.

The conveyor belt can only overcome rolling friction.  The main drag on an airplane is aerodynamic, rolling friction max is probably less than 10% during take off.  The thrust is entirely aerodynamic, the only rolling resistance is bearing friction, and the heating of the tire as it is deformed under load.

2nd.

If the airplane begins to roll forward, the conveyor belt pulls it back.  The forward speed of the airplane is completely countered by the speed of the conveyor.  Since my accelerating force is aerodynamic, my speed will be measured in the air.  The only place where this magical conveyor can perfectly balance is at zero.  

Any other airspeed is out of the range of the original problem.  It does not matter what the RPM of the wheels, if the speed balance is maintained, the only thrust I can apply must be equal to the rolling friction for the balance to be maintained.

Either one must agree that the original problem is fundamentally flawed or the answer is that the airplane can never have any airspeed and therefore will never fly, unless it is a VTOL.

3rd.

As most airplanes have the thrust to overcome rolling friction the aircraft in this problem (which has thrust only equal to rolling friction) could not accellerate to takeoff speed regardless of a moving conveyor. Even if catapulted it would not have the thrust to maintain flight.

Please end this horrible thread.

Holden
Oregon Registered Professional Engineer

Holden McGroin,

So far everyone who has dismissed the idea that a rapidly accelerating treadmill can keep a plane at full power from moving has not been able to answer the questions at the end of this story. If you don’t believe or understand what Hitech and I have been trying to say, please read on.

Here’s a story that simplifies the problem: (Note that the term wheels in this story refers to wheels and tires)

Identical triplets Al, Bob and Chuck buy three identical bush planes. Since they live in Alaska, all three brothers buy and install large balloon “tundra tires” and wheels. The wheels, planes and brothers are identical. All three planes will take off from a normal runway in exactly 100 feet and at exactly 50 mph. The brothers fly their planes to an air show in Wisconsin. At the air show Bob finds and buys a set of fantastic wheels. These wheels are exactly like the wheels he has on his plane in every way except they have half the mass. Their mass is distributed in the same proportion as the wheels that he plans on replacing. Al thinks Bob is silly and is content with his old wheels. Bob thinks that Al will eventually want a set, so he buys a second set to give to Al on their birthday.

Bob finds a buyer for his old heavy wheels and installs a set of his new lightweight ones. He loads the second set into his plane so that it is balanced just as it was before. Bob’s plane now weighs exactly the same as Al’s and Chuck’s, but its wheels have half the mass.

Meanwhile, Chuck runs into a magician who sells him a set of magic wheels. These wheels are exactly like the wheels he has on his plane in every way except they have no mass. Chuck installs his magic wheels. He loads the second set into his plane so that it is balanced just as it was before. Chuck’s plane now weighs exactly the same as Al’s and Bob’s, but its wheels have no mass.

When the brothers leave the air show they request a formation take off. They line up wing tip to wing tip and apply power at exactly the same time. All three planes weigh exactly the same and must hit 50 mph to lift off. When Chuck’s plane lifts off his wheels stop spinning instantly since they have no mass. Since they have no mass, they also have no rotational inertia. When Al’s plane lifts off his heavy wheels are spinning at 50 mph and have considerable rotational inertia. When Bob’s plane lifts off his half-weight wheels are spinning at 50 mph and have exactly half the rotational inertia as Al’s wheels.

Where did the rotational inertia and energy in Bob’s and Al’s wheels come from?
How did the rotational inertia and energy now stored in Bob’s and Al’s wheels affect the take off distance of their planes?
We know that Al’s plane will still take off in exactly 100 feet; where will Bob’s and Chuck’s planes take off?

[Holden McGroin]

Magic wheels?

Lightweight wheels allow for a quicker accelleration.  This is why TDF bicycles have $5000 carbon titanium rims.  The rotational equivalent of F=Ma is Torque = (WK^2) angular accelleration.  Yes, there is some energy you are storing in the flywheel effect of the spinning wheel / tire set. I can add a relatively small amount of energy I put into spinning my wheels (which is what this thread has been doing since the first post)  I could build a machine that probably could have a wheel set of a 747 spinning to an equivalent of 100 mph on 30 seconds or so with a 5 hp motor.

So, I accelerate, but I don't, because I can only apply enough power to overcome the rolling friction due to this rediculous conveyor on which I find myself.  I CANNOT APPLY FULL THROTTLE BECAUSE THRUST AND ROLLING FRICTION* MUST BE BALANCED IN ORDER TO STAY WITHIN THE CONTRAINTS OF THE PROBLEM regardless of the magical qualities of the wheels.

*+ 1/2*k*M*R^2*(ratational speed Rad/sec)^2  This is the flywheel energy stored in the spinning wheels... you would need to diffrerentiate with respect to time to get the force at any one specfic time.

Either the aircraft is horribly underpowered or the original question is flawed.