Leaving on a jet plane (or, how smart are you again?)

NEWS FLASH:

The SuperMax/zTroll has set a new land speed record ... in his garage ... on a treadmill (sitting perfectly still of course).

He argues that he isn't going to claim the record for himself, or even his car, but rather just for his tires.

JB
 
Quote from Haroki:

Ok, let's be more specific.

1-The jets provide , say, 100,000 lbs of thrust.
2- it would take, say 40,000 lbs of thrust to roll the jet along on level ground at 150mph (assumed takeoff velocity).
3- so there's 70,000 lbs of thrust that must be counteracted to negate forward movement.
4- the plane's wheels will accelerate to whatever speed necessary to achieve 150mph airspeed.

So what provides the 60,000 lbs of negative thrust against the plane?

X + (conveyor belt speed*y) = (wheel speed * y) + (100,000 lb thrust - 40,000 lb aerodynamic drag)

y = essentially equals zero since the plan'es wheels will accelerate to whatever speed necessary to achieve 150 mph.

So now the equation will be

X = -60,000lbs of thrust.

What provides the negative thrust?

In:
(conveyor belt speed*y)

What is "y" standing in for?

JB
 
force/lbs of resistance, as provided by the rolling resistance of the tires, etc.

Shouldn't this be zero, since the Q assumes that the belt can spin at whatever speed necessary, therefore the other side needs to state that the wheels will spin freely?

Or do you see it another way?
 
Haroki:
X + (conveyor belt speed*y) = (wheel speed * y) + (100,000 lb thrust - 40,000 lb aerodynamic drag)

Me:
What is "y" standing in for?

Haroki:
force/lbs of resistance, as provided by the rolling resistance of the tires, etc.

There must be one value for "y" in any single equation. I don't think the rolling resistance of the treadmill itself "(conveyor belt speed*y)" will be the same as the rolling resistance of the wheels "(wheel speed * y)".

However, if you're right and I'm wrong, you should be able to show the work solving the equation for "y". Can you?

JB
 
Quote from SuperZtroll:

their wheel has a rotational velocity to it. the very center of the AXIS of the wheel is going nowhere but any point on the wheel OTHER than that is most certainly moving.

i reiterate,
what an idiot ! :p

LOL There's a reason that circular actions are measured in *revolutions*, while linear actions are measured in *distance*.

Hint: One actually moves you somewhere -- can you figure out which one?

JB
 
Quote from Turok:

Haroki:
X + (conveyor belt speed*y) = (wheel speed * y) + (100,000 lb thrust - 40,000 lb aerodynamic drag)

Me:
What is "y" standing in for?

Haroki:
force/lbs of resistance, as provided by the rolling resistance of the tires, etc.

There must be one value for "y" in any single equation. I don't think the rolling resistance of the treadmill itself "(conveyor belt speed*y)" will be the same as the rolling resistance of the wheels "(wheel speed * y)".

However, if you're right and I'm wrong, you should be able to show the work solving the equation for "y". Can you?

JB

The only resistance that the belt can provide is a source of rolling resistance, normally provided by the asphalt. There is no aerodynamic restance provided, right?

The only interaction between the belt and the tires will be that rolling resistance. Or do you see something else?

What I'm attempting to do is explain that since the belt has no limitations, then neither should the tires, and since the interaction between the 2 will be rolling resistance, then it's logical to make that assumption. Cuz if we start discussing the situation whereby the belt has no limitations, but the plane does, then it becomes a silly one.

Thereby, by that assumption, y=0.
 
Quote from Haroki:

The only resistance that the belt can provide is a source of rolling resistance, normally provided by the asphalt. There is no aerodynamic restance provided, right?

The only interaction between the belt and the tires will be that rolling resistance. Or do you see something else?

What I'm attempting to do is explain that since the belt has no limitations, then neither should the tires, and since the interaction between the 2 will be rolling resistance, then it's logical to make that assumption. Cuz if we start discussing the situation whereby the belt has no limitations, but the plane does, then it becomes a silly one.

Thereby, by that assumption, y=0.

You and I certainly agree that compared to the engines pulling the plane, the resistance of the wheels will be minimal. It's not actually zero, but you are probably just saying "effectively zero", and not "actually zero".

JB
 
Correct.

One COULD figure out rolling resistance, but what would be the point? Cuz you'd then need to figure out the realities of a belt, etc....

The point of the Q, as I read it, is could a belt provide enough resistance , through enacting a ridiculously high rolling resistance, to prevent a plane from taking off. The tires would blow if the belt was robust enough.

I think of the experiment, in the real world, like this:

1-Put a plane on a belt with a motion sensor attached.
2-Put a stationary sensor next to it, with a feedback loop that would control the belt speed.
3- as the pilot advances the throttle, the 100,000 lbs of thrust wouldn't go to accelerating the mass of the plane initially, but accelerating the spinning mass of the tires.
4-100,000 lbs of thrust would result in the tires being spun up to 1,000mph in an instant, assuming the belt could keep up.
5-the tires blow, end of plane...

So, if you tried to include reality into the equation, where's the cutoff? An arbitrary cutoff would affect the outcome, depending on where you put it. Therefore, it's better to eliminate it.
 
This is too funny. THAT is why I continued reading! The patient teacher can't help himself/herself sometimes. lol

Quote from Turok:

Haroki, your equation makes as much sense as this one ...

Solve for X

X + blue = color of sky + engine thrust

Of course the sky is blue so X = engine thrust. OK, X = engine thrust, so what. All your equation proves is that in fact, you *don't* remember your algebra.

Both sides of an equation must be able to be simplified and in 'here and now' practical terms, "wheel speed" and "engine thrust" can't be simplfied. Mph and Lbs don't mix.

JB
 
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