Airplane physics question

Well, I will agree we disagree. My flying and float plane experience is first hand, not the internet. If the opposite speed/flow of water/ground/ice is equal to or exceeds the thrust of the plane, it will negate any forward movement and I can not take off (been there, done that). Sure, if I am in a plane that can overcome that (like a fighter with a locked wheel), I can achieve forward movement to get airflow on my lift surfaces.

My ground speed needs to exceed the speed of the ground because there is the period that my interaction with the ground is relevent. The treadmill puzzle appears to counter achieving ground speed on my way to air speed, regardless of how much thrust I can generate.

 

Blue, think about what would happen: the plane tries to go forward. The belt goes back just as fast. The wheels spin faster and faster. No matter how fast the plane makes the wheels go forward, the belt makes them go backward. So what gives first? the irresistable force or the immovable object? That is the essence of the question.

It's only when you actually analyse the forces that you see that no matter how fast the belt goes backward, at some point the belt cannot generate enough force to keep the plane back because the backward force depends on the friction generated between the tire and the belt. That friction force is finite, and defined by F=(weight of plane x coefficient of friction between tire and belt) The belt isn't just rolling, it's actively TRYING to keep the plane from moving, by matchng the wheel speed. THAT is the definition of the problem. How hard can the belt work to stop the plane moving? Only as hard as the friction between the belt and tire allows it to work. Just like stopping your car. Brake as hard as you can, you can only stop at a certain rate. The belt can only retard the plane at a certain rate.

Meanwhile I can build more and more powerful engines that will overcome the friction force of the belt. SO if I built the motor, the plane takes off!

 

Good morning campers....are we STILL crunching our noodles on this silly query????

CMY/Chris clearly pointed us to the write-up article that solved this. The plane takes off, it even happens to take off in the same distance it would IRREGARDLESS of whether it was on a treadmill or not.

The plane takes off because it is propelled by the thrust of the engine, the engine pushes the whole plane forward...when enough lift is generated by means of having enough air-flow move over the wings, the plane takes to the air.

Hey SMG2...I think I figured a way so planes can take-off on shorter runways: use smaller wheels on their landing gear, the higher rpm could save hundreds of yards of asphalt

THE PLANES FLIES, DAMMIT BEAVIS!!!!!!!

 

That's because this friction IS horrendous. This force is the weight of the plane (875,000 lbs) mulitplied by the coefficient of friction for rubber on concrete (let's say its .7) (and that's being generous. Rubber on rubber would be higher). That means that the engines would have to produce 612,500 lbs of thrust to move the aircraft forward.The PW4062 makes 63,300 lbs each, for a total of 253,200 lbs of thrust. That sucker ain't moving.

EDIT: Sorry, I deleted my opening paragraph about a loaded 747 sitting on the runway with its brakes locked.

 

You might want to re-look at the last 2 pages, and check out the neato force diagrams posted. The plane only moves with warp engines.

 

Again, Mule you are relating real-word experience, which is valuable. For a given engine power, there is a backward force that will stop the plane. And I agree. without enough power, the plane will not take off. (see above-it depends) but imagine those times when you couldn't take off. What of you had double or triple the power? Hey! No problem! I can overcome the friction/drag of the water/ice!

So in case I didn't make it clear above:

THE ANSWER IS:

IT DEPENDS!

It depends on the power of the plane's engine, the weight of the plane, and the coefficient of friction between the plane's tires and the belt. The solution is left as an excercise for the student.

Good night all!

 

Stubborn as a.....


Your spinning prop pulls the plane forward, just think about that. Unless your wheels are locked, what happens below you doesn't matter.

 

A Mig 29M (typical 33,000 lbs take off weight) produces 41,450 lb thrust with the RD-33K engines. Even if the coefficient was 1, it would still take off with locked wheels! Sweet, I want to see a video of this!

 

Quite true. I've done that with MS FlightSim a couple of times It could also climb at a 90 degree angle, as long as the thrust was greater than the weight.

 

No....that's just tangenting off into extremes. Go back to page 2 or 3, CMY gave us the link to the article & answer of this question. The plane is thrust-driven by the engines. In this case the wheels would just be spinning faster, twice as fast to be accurate. It's not that complicated.

I'm really shocked at how confused people got over this. I'm gonna go visit my old Hummer forum, bet those knuckleheads are arguing about whether the landing gear is 4-wheel drive or not

 

Hi folks,

mxblue23 is correct.

I wouldn't have been able to keep as much patience as him in my explanations.

If you can't see it by now, there's a chance you probably never will.

From static (both plane and treadmill), the ONLY way the treadmill would start to move and get up to any reasonable speed is IF THE PLANE MOVES FORWARDS, because (as stated by the problem the belt moves backwards at the same speed the wheels move forwards). To clarify what I understand this to mean, if the centre point of the wheel is moving forwards at 50mph, the belt is moving backwards at 50mph, and so the wheel is spinning as if the plane was doing 100mph (which, in relation to the treadmill, it is, but in relation to the ground, its doing 50).

The wheels are just a [very low friction] way of keeping the belly of the plane off the deck and have nothing to do with the means it achieves its forward speed. Throttle up, roll forwards - who cares if the wheels below you are doing the equivalent of 200mph if your take off speed is 100mph?

When are people gonna realise that the engine(s) push AIR (or exhaust gases in the case of jets, for the picky). Engine pushes gas backwards and plane goes forwards as a reciprocal of the displaced mass of gas.

This is a good problem for seeing whether people jump to immediate conclusions when reading a problem - I think I'll give it as an interview question in future!!

Cheers!

Rich.

 

Scot,
Good analysis. I am thinking in real terms and applications, but I see that once we go past current technology, science provides a solution.

So I alter my opinion to - the plane will not take off in terms of a modern plane and a current technology "treadmill", but could take off once we are way past those limits to the realm of math and science formulas. Somewhere where the treadmill counteraction has a limit and the thrust does not.

 

LOL, how about we use some real world calculations there. Your theory has no provision for wheel speed? Lets break it down here. To take off say a 777 has to achieve 140mph of air speed to create lift, this is just a random number. Now in this equation the only difference would be that the wheel speed would be 280mph instead of 140. Now how much is the difference in friction between a wheel rolling at 140mph and 280mph. If this slight change in friction would cause the plane to not be able to take off, well heck that is no plane I want to be in.

Your equation doesn't take into account any speed, just rolling friction. What you said states that the thrust of the plane wouldn't allow it to move what so ever which is completely wrong.

And your coefficient of friction is EXTREMELY high.

Here's something I looked up:

When a cylinder rolls on a surface the force resisting motion is termed rolling friction. Rolling friction is generally considerably less than sliding friction. If W is the weight of the cylinder converted to force, or the force between the cylinder and the flat surface, and R is radius of the cylinder and F is the force required to overcome the rolling friction then.
F = f x W/R

f is the coefficient of rolling friction and has the same unit of length as the radius R -in the example below m (metres)

Typical values for f are listed below
Note: Values for rolling friction from various sources are not consistent and the following values should only be used for approximate calculations.

* Steel on Steel f = 0,0005m
* Wood on Steel f = 0,0012m
* Wood on Wood f = 0,0015m
* Iron on iron f = 0,00051m
* Iron on granite f = 0,0021m
* Iron on Wood f = 0,0056m
* Polymer of steel f = 0,002m
* Hardrubber on Steel f = 0,0077m
* Hardrubber on Concrete f = 0,01 -0,02m
* Rubber on Concrete f = 0,015 -0,035m

To say the CF was .7 is obsurd.. that that to push a car that weighs 2500lbs would require 1750lb of force... but then why can I alone easily push it? your calculations are WAY WAY off here.

 

How did you know that's how I got that name...

 

Thankyou.

Now in this equation the only difference would be that the wheel speed would be 280mph instead of 140. Now how much is the difference in friction between a wheel rolling at 140mph and 280mph.

Thats all there is to it.

 

+1 The speed of the plane with respect to the Earth is only equal to the speed the wheels turning when they're actually ON the Earth. Once you put it on something other than the actual ground, you can control how fast the wheels turn.

Put a treadmill in the sky that is turned off and right below a 747 with landing gears down. The 747 is moving at 550mph, but the wheels are stationary on the treadmill. If you turn on the treadmill and set it to 50mph, the plane is still moving at 550mph, but the wheels are moving at 50mph. The wheels are not connected to the drivetain (engines)!


Again using the qutoe from above:

 

But Seve if the wheels spin twice as fast, the belt instantly spins twice as fast, that's what it designed to do. no it's not a realistic question, but given the parameters, the belt goes back as fast as the plane goes forward until equilibrium is reached. That equilibrium is when the friction force of teh belt equals the forward thrust of the engines. Any thrust over that moves the plane forward. The article neglects to account for the belt matching the speed of the wheels at any given time.

Yes it's a silly question, but it has an answer, and doubling the wheel speed was my first thought as well, but that doesn't work if you note that doubling the wheel speed doubles the belt speed, which doubles the wheel speed again, etc....

The answer is the belt can only generate so much friction (see above) and that is the limiting factor. Have an engine with more power than that friction force and you take off, otherwise no.

 

mxblue:

er, considerably less friction than is building up in this thread?

I once had to explain to a fellow engineering student how the outside end of a clock/watch hand moved faster than the centre.

Even though I could show him this on his own alarm clock, he wouldn't believe me.

Strangely, he failed all his exams and left University.

Cheers!

R.

 

mxblue23, I think the 0.7 was the coefficient for kinetic friction (dragging locked wheels). The 0.035 and other such numbers you posted were rotational (unlocked wheels). We were talking about taking off with locked wheels, which I showed a Mig 29M can do.

 

LOL some people have the math smarts but just can't comprehend the real world application.

I'm gonna sit back and see what other examples people can conjure up here, it's rather entertaining really. I forwarded this thread to a few of my engineer friends, they are getting quite a kick out of it to say the least.

 

Hope you aren't in charge of hiring engineers!

Best answer from an interviewee would be "I have no idea." Because this may seem simple to some, but there is a real twist in its tail. Took me three pages of this thread to appreciate the "it won't take off" argument, and get out the old thinking cap and sort it out properly.

And again, the correct answer is : it depends on how much power the airplane has. As some have pointed out: the average modern plane coudn't do it, with the exception of some fighters. Harriers in VTOL mode are disqualified, but they could clearly do it on a straight take off, given their power to weight ratio.

 

How is it going 50 in relation to the ground? The plane has no relation to the ground, only the treadmill which is moving in reverse at the same speed as the wheels. The only time I've moved forward when jogging on a treadmill is when I've fallen off.

 

And that's the end of my discussion for a while. If I want to hang out with my engineering friends tonight, I'll have to spend some time in my engineering office this afternoon.

 

For **** sake man do you not get it? If we are talking about rolling wheels, ANY, let me repeat this again, ANY airplane that is capable of achieveing flight would EASILY overcome this. You need to go SERIOUSLY look at the CF of friction forces on a rolling wheel. The only additional force is the speed difference in the wheel from normal takeoff speed. If X plane takes off at 140mph, then we know it has plenty of power to move at that speed. The only difference here is that the wheels will be turning at 280mph... now how much more force does it take to turn the wheels an additional 140mph? I'll tell ya, it isn't that much. Your not talking overcoming air friction on the plane, NOTHING other than that little bit of ROTATIONAL friction. Even spinning 16 of the HUGE tires on a 747 my guess is it would be LESS than 100hp in friction loss to spin them. Remember your NOT changing the air speed, not changing the rotation of the earth, JUST spinning the wheels.

Sheesh, I need a beer, and you guys need to go sit in on an elementary physics class.

 

Dubai Vol, me old spud.

Thanks for the drawings earlier. Yes, I did understand the bit about the couple in the second drawing.

I sort of see how you're thinking, but could you clarify your thought process for me (stop me where you don't agree anymore):

You are confident that the plane stays in one place (relative to the Earth and the still air, seeing as there is no wind in the problem statement). Because of this no air flows over the wings, and so, unsurprisingly it doesn't take off.

You are confident that the backward speed of the treadmill is what makes the plane stay in one place.

From the initial static position, lets say the plane's intended acceleration is known and the treadmill is programmed the do the same in reverse (this removes from the problem anyone attributing an effect to the treadmill having to 'react' (always at a slight time lag) to the plane).

The plane's engine throttle up, shoot a load of air out the back, and by plain ordinary Newton's laws, the plane starts to accelerate forwards, and at the same rate the treadmill starts to go backwards.

Brief aside:
You are confident that if this was a car, because the wheels are the only thing pushing the car forwards, and the wheels are planted on the treadmill, it would in fact stay stationary in relation to the Earth.

So far, I think we're together, but...

In the case of a plane, you are confident that if the plane does indeed move forwards (with respect to the Earth/Sky) then the treadmill will have to suddenly zoom up to a really high speed, because the wheels of the plane will have 'gained' on the treadmill and it will have to redress this balance by going even faster, but the plane is moving forwards still because its pullings it way against the air (which is still). Your conclusion is that if there is any forward motion of the plane, the treadmill will just blast ever-faster until the limit of the [extremely low but still existing friction] in the planes wheel bearings will 'drag' the plane back and some equilibrium will be reached.

This is the sense in which I understand your argument - am I right or way off base?

Best,

Rich.