plane on a conveyor belt?

[2bighorn]

Originally posted by hitech
This one had me for a sec, now I see what you are missing, you are not seeing the change in force between the surface of the wheel and the runway when the moment of inertia of the wheel changes.
 

I did not forget. lets say for a moment plane goes down the runway at constant speed and it does not accelerate so all forces would remain constant including angular momentum.
Angular momentum stays constant unless there's an external torque applied to it. Torque is the rate at which angular momentum is transferred from or to plane and/or runway. According to Newton's third law runway is exerting a force of the same magnitude on the plane as the plane is exerting on runway. Since plane is moving and runway not, torque forces acting on wheel are opposing and cancel each other out.

If the planes accelerate so does velocity and with it angular wheel momentum and so do opposing torque forces. But net amount of torque is 0.

Kinda like cavemen moving big block on logs.

[Mini D]

Eskimo,

Get rid of the lame analogies. They don't work.

Instead, look at something you do know. Does increasing the acceleration used to pull the table cloth out from under place settings cause the place settings to move more or less?

Then explain your answer.

[2bighorn]

Originally posted by lukster
Any rotational inertia stored in the wheels is not countering the planes thrust. Spin the wheels to 100,000 RPM, instantly stop the belt and the plane will launch like a UFO (you probably don't want to do any banking for a while though). However, an accelerating belt is applying a force to the plane through the wheel. If the plane's thrust is great enough to overcome the frictional coefficient of the tire/belt contact the plane can move forward. The thrust of the plane, the frictional coefficent of the tire/belt, and the acceleration of the belt are variables, any two of which is capable of determining the outcome.

You got most if it right except that (without going into materials mechanical and thermal properties) coefficient of rolling friction is constant.

[eskimo2]

 - C Johnson, Physicist, Physics Degree from Univ of Chicago
…For a 747 airliner, which weighs around 400,000 pounds at landing, and which lands at about 130 mph, the numbers are all bigger but the effect is very similar:
400,000 pounds weight is equal to 12,500 slugs of mass. 130 mph is equal to about 191 ft/second. Therefore, the kinetic energy of the airplane just before touchdown is 1/2 * m * v2 or 0.5 * 12500 * 1912 or about 227 million ft-lb of energy.
That size tire is around 8 feet in diameter, and the tire and rim probably weighs around 1000 pounds. The Rotational Inertia is equal to 1000/32 * 2.52 or 195 slug-ft2.
Once the tire stops skidding, it will be spinning at 191/25 revolutions per second and so omega is 48 rad/sec. (The giant wheels actually spin more slowly than the small aircraft tires do!)
The kinetic energy of rotation that the wheel/tire will eventually have is then 0.5 * 195 * 482 or 225,000 ft-lb. The aircraft has sixteen main tires/wheels, so this total is 3.6 million ft-lb if kinetic energy needed to up-spin all the main landing gear tires/wheels..
So, if the aircraft has 227 million ft-lb of kinetic energy the moment before touchdown, it will have around 223 million ft-lb left after fully spinning the wheels/tires up! Again, an almost irrelevant effect as regarding stopping the aircraft.
Say the 747 normally takes 5,000 feet of runway to completely stop. We can easily calculate the deceleration that occurs. Another Physics formula is 2*a*d = v2. We know everything but a, 2 * a * 5000 = 189.062. Solve for a and get 3.5744 ft/second, a gentle deceleration of around 1/10 G.
Let's see how far that exact same aircraft would have taken to stop if it had pre-spun the wheels/tires, and applying the same deceleration! Same equation:
2 * 3.5744 * d = 1912. This gives 5103 feet as the needed landing distance, roughly one hundred feet longer, half the length of the aircraft. That also is certainly not any "20% longer landing distance"!
 
That necessary tangential force for upspinning the wheels/tires on impact must entirely be provided by friction with the ground. This gives a value that indicates how much heating and wear is likely to occur to a tire under those circumstances.
In a very small fraction of a second, the heavy wheel and tire assemblies must be spun up to the 130 mph (191 feet per second) tread speed. From the lengths of runway skid marks (seemingly under 20 feet), this seems to occur in well under 1/10 second. …

http://mb-soft.com/public/planetir.html

So, a 747 transfers 1.8% of it’s kinetic energy into the rotational inertia of its wheels in less than 1/10th of a second!  If it landed on a treadmill that accelerated the tires at the same rate in 1 second it would transfer 18% of its it’s kinetic energy into the rotational inertia of its wheels and in 5.6 seconds (or less) it would transfer 100% of its it’s kinetic energy into the rotational inertia of its wheels and would be stopped.  
Here’s some RPM math: please check my work, I may be rusty!
Plane is traveling at 191 fps.
Wheel diameter = 8 feet, when the wheel makes 1 revolution it covers 25 feet.
191 fps divided by 25 feet is 7.64 revolutions per second.
7.64 revolutions per second X 60 seconds = 458.4 RPM
Converting 1.8% of the aircraft’s kinetic energy into rotational inertia brought the wheel RPM up to 458.4
1.8 is 1/56th of 100.
458.4 RPM X 56 = 25,670.4 RPM
Final RPM of the tires…  25,670.4 RPM

Treadmill’s Final Speed:
25,670.4 RPM X 25 feet (circumference of the tire) = 641,760 feet per minute
641,760 feet per minute X 60 Seconds = 38,505,600 Feet per hour.

38,505,600 Feet per hour divided by 5,280 (1 mile) = 7,293 Miles Per Hour

**********************************************************

Final RPM of the tires:  25,670.4 RPM
Treadmill’s Final Speed: 7,293 Miles Per Hour
747 brought to a stop from 130 MPH in 5.6 seconds on the super treadmill, no breaks, no reverse thrust (also no forward thrust).

So, the question is: can a 747 accelerate up to 130 MPH in under 5.6 seconds?  

If so, it may be able to take off on the super treadmill.

eskimo

[lukster]

Originally posted by 2bighorn
You got most if it right except that (without going into materials mechanical and thermal properties) coefficient of rolling friction is constant.

Huh? The coefficient of friction is dependent on many factors not the least of which is material characteristics which in most materials on this planet are very much dynamic as conditions vary.

For example, the wheel bearing. Under normal operating conditions it will offer one coefficent. This will change dramtically as the bearing is overheated beyond the intended operating range.

[eskimo2]

Originally posted by Mini D
Eskimo,

Get rid of the lame analogies. They don't work.

Originally posted by hitech
eskimo2:
Not a bad analogy eskimo. And obviously they would accelerate at different speeds do to the rotational inertia stored in the wheels. Since 1 has 0 mass, 1 has 1/2 mass, the 0 would be in front , followed by the 1/2 mass.  Followed by the original plane.

Originally posted by Mini D

Instead, look at something you do know. Does increasing the acceleration used to pull the table cloth out from under place settings cause the place settings to move more or less?

Then explain your answer.

Of course not, they are sliding, not rotating.  I think that you are confusing the two and need to answer the question in the story:  

Where did the rotational inertia and energy in Bob’s and Al’s wheels come from?

[lukster]

Perhaps I misunderstood you bighorn. The thrust of the plane and the belt acceleration capabilites we are dealing with are unknowns. You are the one who brought a super powered engine capable of infinite thrust into this and so I am therefore anticipating your obstacle. Are we dealing with the tires of a dragster or solid rubber thin bicycle tires? How much does your plane weigh?

[2bighorn]

Originally posted by lukster
Perhaps I misunderstood you bighorn. The thrust of the plane and the belt acceleration capabilites we are dealing with are unknowns. You are the one who brought a super powered engine capable of infinite thrust into this and so I am therefore anticipating your obstacle. Are we dealing with the tires of a dragster or solid rubber thin bicycle tires? How much does your plane weigh?

I have suggested to ignore materials used because on one side we had super unlimited acceleration capable conveyor and darn stinky Cessna on another.

I don't mind going into details but the limits have to apply to both, plane and conveyor, or none at all.
Otherwise, in case of a plane taking off at 100 mph which landing gear is rated maybe up to 200 mph, I wanna have down to earth real conveyor to which same laws of physics and material properties would apply and has finite amount of energy to burn.

If playing field is not level, discussion about applied physics is useless and the whole exercise becomes matter of fiction and impossibility.

[2bighorn]

Originally posted by eskimo2
So, the question is: can a 747 accelerate up to 130 MPH in under 5.6 seconds?
If so, it may be able to take off on the super treadmill.

You forgot few bits and pieces tho.

a) 747 tyre measures 49 inches and tyre wheel assembly weighs about 480 lb (190 + 290). That alone reduces angular momentum several times and considerably lowers the percentage of total energy needed for spin up.
(If you wanna do some math here's the relevant take-off data:
MTOW of 474-400 is 875,000lb, take-off speed with 30deg flaps at MTOW about 188mph, take-off runway length roughly 10,000ft, V1 time to abort incl., total thrust 250,000lbf with PWs)

b) Any wheel RPM above that of the normal take-off is consequence of conveyor belt moving in opposite direction. All the extra work going into wheel rotation above the plane's acceleration rate is done by the CONVEYOR and NOT the plane.


Once you understand point B little detail you may start seeing apples for apples.

[WhiteHawk]

We are getting way to complicated.  Lets start the conveyor and run it at take-off speed +1 mph.  Now the wheels are 0 friction so the plane stays still.  Therefore the conveyor and wheels are moving at take-off+1, and the plane is moving at 0mph.  The fact that the plane is capable of staying still due to 0 friction transferred from the belt to the plane means that the plane speed is only related to the conveyor/wheel speed by the friction variable, which now is 0%.  Now lets fire the engines at take-off thrust.  newton says that the thrust must push the plane at take-off speed, and the plane takes off.  Now, lets say the friction variable of the wheels is such that the plane moves back at 10mph when the conveyros are at take-off speed +1.  Again fire the engines at take-off thrust +the extra needed to compensate for, what amounts to a stiff headwind.  And we take-off, no problem.  In fact, we can say that the plane in question will take off so long as it has the power is available to overcome the friction generated by the belt racing below the wheels.

[Mini D]

Originally posted by eskimo2
Of course not, they are sliding, not rotating.  I think that you are confusing the two and need to answer the question in the story:  

Actually, you are getting confused.

Friction is friction. Does the presence of a bearing (in this case.. the tire) reduce or increase friction.

Because... you seem to think if there were wheels on the bottom of the dinner plates this would be impossible.

You are wrong.

[Max]

Originally posted by rabbidrabbit
A plane is standing on a runway that can move like a giant conveyor belt.  

Why?

[hitech]

Did some simple estimations of my RV on the conveyor.

Some how I knew I was going to have to do a drawing.

Also here is the reference page.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html


HiTech

[Mustaine]

Originally posted by Max
Why?

been watching this thread...




I got the same question

[lukster]

Originally posted by Max
Why?

Why not? Same question, different pespective. Which exerts the greater force?