plane on a conveyor belt?

[lukster]

Originally posted by Casca
The last possible interpretation is the conveyor moves equal and opposite to wheel speed irrespective of the speed of the airplane.  If you read it like that you wind up in your basement filming a belt sander documentary.
 

I laughed twice at that.  

[Terror]

I see some confusion as to which direction the conveyor is turning....

Lets get out my inner ascii artist:

---->  direction of airplane
<----  direction of conveyor

OR

----> direction of airplane
----> direction of conveyor

In my opinion.  Doesnt matter.  The airplane will take of in either circumstance.  The rpm of the wheels would differ in each case, but in the end, the airplane itself will still accellerate and take off.

Terror

PS.  I didn't my inner self had such artistic capabilities!!!

[Mini D]

Originally posted by hitech
Mini D: In agreement so far.

And I also agree the tire static surface friction (I.E. Brake's locked how much force can the tire hold back) has to be as great as the total thrust, or the wheel would start to slide on the belt.

This is not what I said.

The wheel would not start to slide on the belt. The wheel turns. This is why your scenario just doesn't make sense. The friction would have to increase between the wheel and the bearings in order to stop the plane from moving forward. Brakes are just a device that press a much less effective bearing against the wheel. The friction between the belt and the tire is irrelevant as you are assuming they are always spinning at the same speed. This scenario does not present the posibility of friction between the tire and belt as a slowing mechanism... it eliminates it as a factor.

But once again, that is not what is normally considered as the rolling friction force transmitted to the plane. The rolling friction force is a constant at any speed the wheel turns.

Whether it is a constant or not is irrelevant, once again. There needs to be a counter force acting on the airframe that has the engine mounted to it. Basically, there is no rope in this situation.

As you accelerate the wheel a force is needed for this acceleration (energy is stored in the rotation of the wheel). That force can not be larger than the STATIC friction of the tire to the belt or the wheel would begin to slide. If it is not greater the force is transmitted to the belt and the plane. But note, that force is not what eskimo and I are considering friction.

The static friction, once again, is irrelevant. In order to move the wheel.. that is to get the energy from the engine to the wheel, you have to move your moment on the wheel relieving a bit of the downward force on the bearings and moving it to a portion of the bearings that is elevated causing the wheel to turn. Please understand that this is friction that is causing this rotation. You aren't turning an axle, you are moving it forward against the wheel. In order to prevent the plane from moving, an equal force would have to be pushing the wheel back against the axle which would actually cause even more forward acceleration on the wheel and be infinately absorbed. The act of trying to keep the wheel from moving forward would require infinite rotation, but it would not be enough because it would not be able to transfer that energy to oppose the thrust of the engine.

[lukster]

Eskimo's video is a good illustration of inertia and acceleration. If the belt were to continue accelerating the wheel would continue to exert force on the band stretching ever further. You can replace the band with the plane and see that a normal engine would never be able to overcome the force of a belt capable of extreme speeds. Of course there is friction at every point in this model where force is being applied.

[Casca]

Okeedokee.  Forget the conveyer for a moment.  Jack the wheel up and turn it with a torque wrench.  There will be a momentary reading to overcome the inertia of the wheel which will disappear when the wheel starts to spin.  The force vector created by this torque wrench applied at this position will be substantially in opposition to the forward travel of the airplane.  This is where the force is coming from.  It does not depend on friction from the axle bearings.



If this horse isn't dead it's starting to look a little peckish.

[lukster]

I'm not disagreeing with you Casca but for the plane to move forward it must take it's wheels with it. We could define the connecting point between the plane and it's wheels as the wheels axle. Force is therefore transferred through the axle and friction must result at that point.

[Casca]

Help.  I'm stuck in the conveyor belt thread and cant get up.

[lukster]

Originally posted by Casca
Help.  I'm stuck in the conveyor belt thread and cant get up.

hehe

Skuzzy, stop this crazy thing!

[lukster]

Even if you allow for a belt capable of infinite speed but limit acceleration the problem is calcuable. As seen on the belt sander, force is applied strectching the band so long as the belt is accelerating (take static friction out of the equation for now). When the belt reaches a constant speed the wheel returns to it's original position.

If the belt is allowed an accleration which induces a force on the wheel greater than the thrust the plane is capable of producing then the plane will never move forward. Assuming you allow an acceleration somewhat less than what the plane is capable of countering then you would calaculate the resulting difference over time to determine if the plane could reach flying velocity before the wheels were spinning so fast that wheel bearing heat became a factor.


I know all of this has been said. I'm really just summarizing.

[hitech]

lukster: Can you help either me or Mini D understand what we are missing from each others posts?

[eskimo2]

MiniD,

Read the story and answer the questions!  This story eliminates the belt, and illustrates the same problem.

Originally posted by eskimo2
Here’s a story that illustrates my idea:  (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 no 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 half the 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?

[eskimo2]

Originally posted by hitech
lukster: Can you help either me or Mini D understand what we are missing from each others posts?

MiniD and anyone else who does not understand the forces involved in accelerating the rotational velocity of a wheel needs to explain where the energy in Al’s and Bob’s planes’ wheels came from.  This really simplifies the problem.

[lukster]

Originally posted by hitech
lukster: Can you help either me or Mini D understand what we are missing from each others posts?

I think each of you are focusing on different aspects of the model. I think MiniD is looking at the friction induced by an overheating bearing and considers this to be the predominant factor.

I think you and Eskimo are looking at the acceleration of the belt and consider the force applied to the plane through the wheel simply as a result of overcoming the wheel's inertia as the larger force.



You can't throw in massless wheels Eskimo, that's like dividing by zero.

[Terror]

Originally posted by lukster

If the belt is allowed an accleration which induces a force on the wheel greater than the thrust the plane is capable of producing then the plane will never move forward. Assuming you allow an acceleration somewhat less than what the plane is capable of countering then you would calaculate the resulting difference over time to determine if the plane could reach flying velocity before the wheels were spinning so fast that wheel bearing heat became a factor.


I know all of this has been said. I'm really just summarizing.

An airplane can overcome the inertial forces without the conveyor belt, right?  It still has to accellerate the wheel.  Why would it not be able to overcome the force applied by the conveyor that is also accellerating the wheel?  I would estimate the inertial force applied by the conveyor to be double the normal forces needed to be able to take off.  Why would it be anymore than that?  

Wouldn't the inertial forces be calculable based upon the weight of the wheel, the radius from the center of rotation to the point of friction?

Something like:  (wikipedia ftw!)

I=k*M*R^2

where

I is the moment of inertia
k is the inertial constant (1/2 for a solid disk .. closest approximate to a wheel),
M is the mass, and
R is the radius of the object from the center of mass.

I just dont see the wheels (a small percentage of the plane's overall mass) generating enough inertial forces to keep the plane reaching take off speed.  The engine and thrust were designed to get the plane+load+20% to flight speed.

Terror

[lukster]

Originally posted by Terror
An airplane can overcome the inertial forces without the conveyor belt, right?  It still has to accellerate the wheel.  Why would it not be able to overcome the force applied by the conveyor that is also accellerating the wheel?  I would estimate the inertial force applied by the conveyor to be double the normal forces needed to be able to take off.  Why would it be anymore than that?  

Wouldn't the inertial forces be calculable based upon the weight of the wheel, the radius from the center of rotation to the point of friction?

Something like:  (wikipedia ftw!)

I=k*M*R^2

where

I is the moment of inertia
k is the inertial constant (1/2 for a solid disk .. closest approximate to a wheel),
M is the mass, and
R is the radius of the object from the center of mass.

I just dont see the wheels (a small percentage of the plane's overall mass) generating enough inertial forces to keep the plane reaching take off speed.  The engine and thrust were designed to get the plane+load+20% to flight speed.

Terror

We're talking rotation speeds much much great than what the plane's thrust is capable of generating. I'm not an engineer so I can't give you some realistic esitmates off hand but I will dig out my old physics book and get back to you with some numbers.