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Plane Problem

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melipone

FNG / Fresh Meat
Mar 22, 2006
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Assume the wheels can rotate at an infitite speed without breaking. Now turn on the treadmill at 500mph say. Will the plane move back at 500mph or will the wheels rotate at say 300mph (assuming some friction due to the weight) and the plane move back 200mph? Now apply an external force behind the plane at 500mph. Will it stay still or accelerate forward?

Its easier to think of the rollerskates on a treadmill analogy..Same thing
 

SnozzFartz

FNG / Fresh Meat
Jul 20, 2006
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here I go thinking again..I'm not going to say any theories, I'm still confused so I'm just going ask questions. It's honestly a hard concept for me to grasp that a treadmill that is supposed to match a planes speed actually won't due to some force that I don't know about (I didn't take physics).

melipone, you say that if the plane will speed up if you add extra force behind the plane. Well when the plane speeds up, wouldn't the treadmill speed up also to match it's newer, faster speed?
 

[CoR]MiccyNarc

FNG / Fresh Meat
Nov 21, 2005
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The treadmill will speed up yes, but can only exert force on the wheels. Since the wheels move independently of the airplane, it won't affect the airplane's movement.
 

Nestor Makhno

FNG / Fresh Meat
Feb 25, 2006
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Not the wheels cause the acceleration, the engines does. The reference point to the engines is the surrounding air. Not the treadmill.

You delink the system with the wheels.
I didn't say the wheels cause the acceleration but, being in contact with the surface they are the interface whereby the thrust is transferred. This is NOT a ballistic object, it has not taken off yet, nor will it.

Imagine that the plane is a sled that has runners made of wet ice and it is on a wet ice treadmill. (wet ice versus wet ice is the closest analogy to zero friction) you run alongside the treadmill pushing the sled and it will definitely go off the end, no matter how fast the treadmill is going in the opposite direction cos there is no friction whereby forces are transferred from mill to sled and vice versa.

But this situation isn't like that. A treadmill is a high friction surface, that's its job. A tyre is also a high friction object - that's its job.

I was trying to avoid this part as most of the people who have been participating so far have been floundering with even such basic concepts as parallelograms of force and Newtonian principles but, as most of those have fallen by the way side, I will produce a diagram showing exactly what forces are interacting with a moving wheel on a surface. Give me 10-20 minutes.
 

Ruprecht

FNG / Fresh Meat
Aug 29, 2006
211
0
0
Ulm
Code:
void PlaneTakeoff(bool TakeOff)
{
 if (TakeOff == TRUE) {cout<<"A miracle..  *swish*.";}
}


void WeelsBroken(bool Broken)
{
 if (Broken == TRUE) {cout<<"Too bad, the weels broke..";}
}


void main()
{
 int PlaneMinTakeOffSpeed = 600;
 float WeelCircumferential = 0,1;
 int AmountWeels = 4;
 int PlaneThrust = 10000;
 int PlaneIntertia = 100;
 int WindBlow = 10;
 float FrictionPerRotation = 0,0001;
 float PlaneActualSpeed = 0;
 int WeelsBreakFriction = 1000;
 float WeelSpeed;


 while(1)
 { 
   if (WindBlow >= PlaneMinTakeOffSpeed) {cout<<"Plane flys allready";}
        else {PlaneMinTakeOffSpeed = PlaneMinTakeOffSpeed - WindBlow;}

   WeelSpeed = PlaneActualSpeed + PlaneActualSpeed;

   WeelSpin = WeelSpeed / WeelCircumferential;
 
   WeelFriction = (FrictionPerRotation * WeelSpin) * AmountWeels;

   if (WeelFriction >= WeelsBreakFriction) {WeelsBroken(1); break;}
 
   PlaneAcceleration = PlaneThrust - WeelFriction - PlaneIntertia - WindBlow;

   if (PlaneAcceleration <= PlaneThrust)
           {PlaneActualSpeed = PlaneActualSpeed + PlaneAcceleration;}
   
   if (PlaneActualSpeed >= PlaneMinTakeOffSpeed) {PlaneTakeOff(TRUE); break;}
 }
}
lol, added a few more variables..
 
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Nestor Makhno

FNG / Fresh Meat
Feb 25, 2006
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and a bearings job is to reduce that friction.
No it isn't - a bearing is not there to reduce friction between the tyre/wheel and the ground - it is there to reduce friction between the wheel and its axle. You actually need friction between wheel and ground for a wheel to work.
 

worluk

FNG / Fresh Meat
Nov 21, 2005
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No it isn't - a bearing is not there to reduce friction between the tyre/wheel and the ground - it is there to reduce friction between the wheel and its axle
the difference?

there is none, it doesnt matter if the wheel slides on the ground or the wheel not having an influence on the movement of the axle.
Not for this problem.
 

melipone

FNG / Fresh Meat
Mar 22, 2006
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I'm not 100% on this, and my minds been changing about it since I first looked at the thread, but this is what I think so far..

For eg, Turn a tredmil on at 5mph. Now stand on the treadmil while wearing rollerskates (or a skateboard or something). Do you move backwards at 5mph or stay relatively still? Now you are going at 0mph. If someone pushes you at 5mph worth of force do you stay still or go forward at 5mph? If 5mph was enough for little wings on your skateboard to achieve enough upward force to counteract the gravitational force you would take off. In reality though you would always be decellerating presumably because of the friction in the wheels (and the treadmill would match the new speed), but it could take off instantly if the speed of the treadmill started high enough.
 

BuddyLee

FNG / Fresh Meat
Apr 12, 2006
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NCC 1701-D Neutral Zone
An airplanes wheels only use friction to stop(not accelerate like a car), during takeoff most of the friction is cancelled out by the bearings. There is only rolling resistance. When compared to the thrust of the turbines, the rolling resistance working against forward thrust is minimal.

If you are rollerblading on a treadmill you stay in the same spot, since the bearings are canceling out most of the friction between the wheels and the treadmill, but if I shove you from behind you will move forward.
 

Xendance

FNG / Fresh Meat
Nov 21, 2005
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Nestor, see this?



It's a ball bearing.

It is that what delinks the treadmill and wheels from the rest of the plane and it's turbines/thrusters.

As people have said, the speed of the treadmill is irrelevant. The thrust doesn't give a **** about the treadmill.
 

Nestor Makhno

FNG / Fresh Meat
Feb 25, 2006
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This is going to take some time, I can see, so please be patient. There are several things to consider when looking at the situation that this problem describes and we will need to break it down bit by bit so you can see how it all fits together.

First, let's look at the situation of a wheel attached to a rigid body which has thrust (i.e. forward force over the whole body).

The force is F in this diagram:



It is very important to understand the nature of friction - think of your own experience - you rub your hand along rough wood it opposes the motion, slowing it down.
A wheel works on the same principle - the friction is an opposing force to the main force of the moving body. As the wheel is a perfect circle (less some negligible deformation due to the plane's weight)we can say that the forward force in the wheel is centred on the axle. Two parallel forces seperated by a distance creates a couple, i.e. a rotational force.

For this couple to exist (in other words for the wheel to roll) you need to have that force of friction there. Otherwise the wheel would not spin, it would just skate over the surface.

You also need as little friction as possible at the centre of the wheel, where it joins the axle, so that the rotation of the wheel is not opposed.

That is how wheels work.

Some of you can probably see why I hesitated before introducing this part - a lot of people are going to be saying, "WTF!? You need friction for wheels to work? BS!!!" Think of this...if you are stuck on ice in your car you need to put sand or planks down for more friction/traction. Although the force in the wheel is actually rotational that doesn't matter - you need friction between the tyre and the surface for wheels to work properly (i.e. go round AND push forwards) It's just that instead of a forward linear thrust interacting witrh friction to create rotation (as in the plane) you get a rotaion interacting with friction to produce forward linear thrust (as in the stuck car).

Now I have wasted enough work time on this so far - I will put in the diagram for the moving treadmill tonight when I get home.
 
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User Name

FNG / Fresh Meat
Jan 12, 2006
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Ok, ground speed is nuetralized in the problem. Your plane can be doing 1,200 MPH ground speed. Since the tread is moving 1,200 MPH in the opposite direction. Your air speed is zero.

The plane increases thrust. This would normally make it move forward. However, the treadmill matches it so it can't move forward. Yes, it could now be doing 1,300 MPH ground speed. The treadmill has counteracted that by accelerating to the same speed.

So now the plane is doing 1,300 MPH on the tread mill. The tread mill is now going 1,300 MPH in the opposite direction. The plane still has an airspeed of zero.

It will not take off because it can't generate air speed. No matter how fast it goes on the tread mill. It's always being countered by the tread mill doing the same speed. Only the opposite direction.
 

snakedude24

FNG / Fresh Meat
Sep 16, 2006
1,042
0
0
Bernouli's Principle, anyone?

Game over. Without air pressure - i.e, without air movement below and above the wings - a plane cannot physically take off.

All other factors are irrelevant.


Only the air speed matters, and the rate at which the plane is moving through the air is 0, thusly, it cannot take off.
 

Ruprecht

FNG / Fresh Meat
Aug 29, 2006
211
0
0
Ulm
Code:
void PlaneTakeoff(bool TakeOff)
{
 if (TakeOff == TRUE) {cout<<"A miracle..  *swish*.";}
}


void WeelsBroken(bool Broken)
{
 if (Broken == TRUE) {cout<<"Too bad, the weels broke..";}
}


void main()
{
 int PlaneMinTakeOffSpeed = 600;
 float WeelCircumferential = 0,1;
 int AmountWeels = 4;
 int PlaneThrust = 10000;
 int PlaneIntertia = 100;
 int WindBlow = 10;
 float FrictionPerRotation = 0,0001;
 float PlaneActualSpeed = 0;
 int WeelsBreakFriction = 1000;
 float WeelSpeed;
 int NegWindblow = 0;

 while(1)
 { 
   if (WindBlow >= PlaneMinTakeOffSpeed) {cout<<"Plane flys allready"; break;}

   if (WindBlow <= 0) {NegWindblow = WindBlow;}
        else {PlaneMinTakeOffSpeed = PlaneMinTakeOffSpeed - WindBlow;}

   WeelSpeed = PlaneActualSpeed + PlaneActualSpeed;

   WeelSpin = WeelSpeed / WeelCircumferential;
 
   WeelFriction = (FrictionPerRotation * WeelSpin) * AmountWeels;

   if (WeelFriction >= WeelsBreakFriction) {WeelsBroken(1); break;}
   
   if (NegWindBlow == 0)
    {PlaneAcceleration = PlaneThrust - (WeelFriction + PlaneIntertia) + WindBlow);}
   else
    {PlaneAcceleration = (PlaneThrust - (WeelFriction + PlaneIntertia)) + NegWindBlow;}

   if (PlaneAcceleration <= 0)
    {cout<<"Windblow, Friction and Intertia are biger or equal as the planes thrust. You may wait for better weather or reduce the amount of weels / unload cargo."; break;}

   if (PlaneAcceleration <= PlaneThrust)
           {PlaneActualSpeed = PlaneActualSpeed + PlaneAcceleration;}
   
   if (PlaneActualSpeed >= PlaneMinTakeOffSpeed) {PlaneTakeOff(TRUE); break;}
 }
}
getting closer...
 
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