plane on a conveyor belt?

[hitech]

2bighorn: You are confusing the term force and torque.

In that setup you would translate 2 vectors and points. Ill do 2 dimensional standard x,y plane

Top of center wheel Vector A -1,0 at point 0,1
Bottom of center wheel Vector B 1,0 at point 0,-1

Translating those vectors to the center of Wheel leaves you with
Force Vector
VA + VB = 0,0

And using cross product of the vector and the point to figure torques
TA = -1
TB = -1

TA + TB = -2

So we have a net force vector of 0,0 I.E. the wheel dosn't move
And we have a net torque of -2 , therefore the wheel rotates.

Note I just choose values, the quantities can change , but as long as VA = - VB, the wheel does not move(translate) but rotates.

Also I found the pages I was looking for on the concepts



HiTech

[hitech]

And for those interested here are the code fragments from AH doing those calcs.

#define maCALC_TORQUE(DEST,FORCE,PNT) \
{ \
(DEST).Roll  = ((FORCE).x * (PNT).y - (FORCE).y * (PNT).x); \
(DEST).Pitch = ((FORCE).y * (PNT).z - (FORCE).z * (PNT).y); \
(DEST).Yaw   = ((FORCE).z * (PNT).x - (FORCE).x * (PNT).z); \
}


void simforceAddWorldForce(simFORCE_MODEL * ForceModel,
const madPOINT * Force,
const madPOINT * Pnt)
{
madATTITUDE Torque;

maINC_POINT(ForceModel->WldForce,*Force)
if(Pnt != NULL)
{
maCALC_TORQUE(Torque ,*Force,*Pnt);
maINC_ATT(ForceModel->Torque,Torque);
}
}

[mietla]

Originally posted by 2bighorn
Here, two identical conveyors, same power, same speed. Two equal and opposing forces work on the wheel.
According to you, wheel in the middle shouldn't rotate since NET FORCE SUM = 0.

Originally posted by hitech
No it will be translated to a torque and a force. Depending on all other torques and forces sums the object will move in the direction of the force sum. And rotate in the direction of the torque sum.

The original Hitech's statement is absolutely correct. If the vector sum of all forces is 0, there is no translation (linear movement). If the sum of all torques is 0, there is no rotation.

Rotation and translation are independent (the superposition rule). You can have none of them, one of them (either), or both happening at the same time.

In your diagram, the vector sum of forces is 0, therefore the wheel between the belt will not move (its CG will remain stationary, but the total torque acting on a wheel is not 0, therefore the wheel will rotate.

Reverse a direction of one of the belts, and you'll get a total torque equal to 0, and the sum of forces that is not 0. In this case the wheel will not rotate but it will move in a straight line.

[2bighorn]

Originally posted by hitech
2bighorn: You are confusing the term force and torque.

No I'm not. It's just another of your straw arguments.

I did not show the last drawing to argue about the detail, but to show where you made logical error since you apply different sets to your drawing where you cancel T  but for my drawing you ad them up.

In my conveyor case if I use simplified torque definition:  T = r*F*sin(angle) I can work with F vector alone since r and angle are always the same and final value really doesn't matter. The point is that T on wheel exercised by both conveyors is identical and by your original claim wheel shouldn't rotate.

Similarly, it proves that it matters where on the rotating object you apply force vector, which you initially claimed it does not.


If I use your logic from the first case (your drawing) the T cancels out and wheel stops, but not in mine, why? What is different?


It is really ridiculous to argue about it since you don't wanna admit the logical error of calculating angular moment of the wheel as opposing force to the plane's motion.

And whenever I point out that error you switch to ridiculous statements such as:
I don't have a clue about physics,
I don't know what force is,
I don't know what torque is,
there is no rotational motion, that's just a torque acting up,
or you bring up AH as holly grail of physics, etc


PS
I wouldn't use AH as supportive evidence in your case. AH is simulated world where you can change any physical law you wish. You can have 0 gravity, objects with no mass, etc. They can all have effect on final behavior, or they don't. You can often afford logical errors without visible consequences. It is just a model nothing more.

[2bighorn]

Originally posted by mietla
In your diagram, the vector sum of forces is 0, therefore the wheel between the belt will not move (its CG will remain stationary, but the total torque acting on a wheel is not 0, therefore the wheel will rotate.

That's the whole point. In his first drawing he used force vector, calculated the size and cancel them out and proclaim angular moment cancels plane's forward motion.

I commit to the same semantic error so he can point it out and with that understands error in his original claim where he cancels the torque instead of adding them up.

It is not as much exercise in calculating the physical forces or splitting the hair about definitions but logic.

[hitech]

The point is that T on wheel exercised by both conveyors is identical and by your original claim wheel shouldn't rotate.

I never claimed what you state. Possibly miss comunication.

What I do claim
Identical = do not cancel

opposit = cancel.

Either way it is as simple as  sum up the forces (note forces can be + & -) if resultant force sum is 0, the object will not move. Read the pages on the ladder, it assumes the sum of the forces must be 0 or the ladder would be moving.

And I do not do straw man arguments, analize my conveyor 910 fps^2 acceleration drawing using the method we described.

I did some simplification on that drawing wrather than using a full coradinate system. But the result is exatly as I describe.

And once again any 2 forces in opposit directions of the same magnitude ,just like your conveyors no mater where they are aplied to an object will create no movement(transaltional) to the object.  Than can or can not create torque, and hence rotation depending on where they are applied to the object.

If you believe my drawing is incorrect analize it again with any numbers you wish.

HiTech

[hitech]

Bighorn: Please take my diagram and analyze all forces involved .
I assume you still belive that the friction force on the belt can not be transfered to the frame?

Based on my sum of forces being 0 argument? Which you belive to be incorect?

[2bighorn]

Hitech I never argued your calculation is wrong.

I just claim that conveyor acceleration you calculated as to create friction necessary to stop wheel from rotation and hence oppose the plane's thrust in sufficient amount to prevent it from take off, can not be seen as opposing force.

In our case angular moment of the wheel and friction created by conveyor do not cancel each other.

[2bighorn]

Originally posted by hitech
Bighorn: Please take my diagram and analyze all forces involved .
I assume you still belive that the friction force on the belt can not be transfered to the frame?

It can, but is limited to the max amount of friction between wheel and the frame.

Originally posted by hitech
Based on my sum of forces being 0 argument? Which you belive to be incorect?

I did look at it again. I just don't see it as opposing forces.

EDIT:
Your 910ft/lb of torque should be added to angular moment of the wheel not opposed to plane's thrust
As you did it, it would be handy when calculating necessary brake force (friction between conveyor and wheel + friction between frame and the wheel opposing the thrust)

[WhiteHawk]

The plane will take-off, so long as it has the power needed to overcome the additional drag generated by the conveyors.  

Proof:  Lets weld the wheels.  now conveyor speed=0; wheel speed =0; it is just a matter of how much drag/friction is generated by the wheels along the conveyor and how much power the plane has.  Drag/friction being a combination of incalculabel intangibles that is not given in this question.  This rule holds through the entire domain.  DO not assume that the wheels have to start with the plane and it all becomes clear.  This proves that it is possible for a plane to take off with WS=CS, therefore it is not impossible.  The proof that a plane connot take off under some conditions where WS=CS is obvious and left as an excersise for the student.

 

[eskimo2]

2bighorn,

Did you watch the movies I posted?  Did you see the way the balls, hollow cylinder and wheel traveled under rotational acceleration from the treadmill?  Have you ever seen what happens when a golf ball (or any ball) hits the ground with backspin or topspin?  Some of the things that you have stated are exactly opposite from reality.  You need to observe the real world a bit more closely.

Get on Ebay and buy one of those SSP cars, be sure to get one with a pull-strip/Tee-strip cord thing.  They are pretty cool and you will learn more from playing with it than you will on this board.  They’re collector’s items so you’ll be able to double your money on one in a few years.

[2bighorn]

Originally posted by eskimo2
You need to observe the real world a bit more closely.

LOL Eskimo, since when did you become an expert on my life and its relation to the real world? But if you wanna go that way, ok. In real world, long ago, part of my career was work on gyro stabilizers, therefore I'm pretty confident that you should drop golf balls more often in order to understand why and when direction of rotation matters. Fair enough?

Discussion between me and Hitech was about applying certain torque to the wheel and in which case the torque would cancel thrust out.

If conveyor belt is running opposite to the direction of plane, torque as a result  of belt/wheel friction is added to angular momentum (which in simplified form can be expressed as torque) and does not oppose the plane's thrust.

Hitech's exercise becomes true if the belt would run in the same direction as plane travels. Torque as result of belt/wheel friction would oppose plane's thrust.

It's just a matter of simple direction reversal and I'm sure that Hitech, sooner or later, will realize where and why very simple error occurred.

[eskimo2]

2bighorn,

Did you watch the movies I posted? Did you see the way the balls, hollow cylinder and wheel traveled under rotational acceleration from the treadmill?

[2bighorn]

Originally posted by eskimo2
2bighorn,
Did you watch the movies I posted? Did you see the way the balls, hollow cylinder and wheel traveled under rotational acceleration from the treadmill?

Eskimo, your movies show how single external torque works when applied to objects with different mass and different mass distribution, we however are talking about how external torque relates to other forces in play like thrust, specifically if and when they are opposing each other and when not.

Unfortunately, the way the wheel is attached to the airframe (no coupling other than bearings which allows wheel to rotate in either direction) it would not matter how you turn it around, the plane would take off, so our question is hypothetical one.

Many understood correctly the weak link (airframe/wheels) and how that coupling is limiting factor like Mietla, MiniD and few others.

So lets summarize (please correct every statement if you think it is wrong)
 
a) Thrust is responsible for plane forward movement.
b) External torque applied to the wheel (belt) would translate into wheel rotation (translational rotation).
c) If you increase the external torque (belt speed), you would increase the angular velocity of the wheel
d) The maximum amount of torque acting on airframe (due to angular momentum of the wheel)) can not be greater than bearing friction allows it to.
c) If you reverse the belt direction the wheel would rotate in opposite direction but it would not prevent the take off (angular velocity and direction of rotation would change however).
d) plane takes off


Hitech:
Your belt/wheel friction opposes wheel/airframe (bearings) friction not the thrust.
Excess torque applied to the wheel is added to its angular momentum.

[lukster]

Originally posted by 2bighorn
d) The maximum amount of torque acting on airframe (due to angular momentum of the wheel)) can not be greater than bearing friction allows it to.

You do realize that in almost every (probably all of them) plane built, bearing friction is quite capable of locking up the wheel, preventing it from rotating about the axle, right?

Let me anticipate your argument. The bearing is destroyed and the wheel then continues to rotate, rather roughly, in it's absence.

My answer to that is that the wheel itself will be ground away in short order as was the bearing. As I said many posts ago you are then left with the plane dragging itself on the wheeless gear.