Author Topic: plane on a conveyor belt?  (Read 26362 times)

Offline eskimo2

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plane on a conveyor belt?
« Reply #525 on: January 28, 2007, 02:23:19 PM »
Think about how these things work:



You pull the stick and it spins the flywheel like mad.  
You set the car down and the flywheel trades most of its rotational inertia for the forward inertia of the entire car and wheel.

Now think about the process in reverse.

Offline 2bighorn

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plane on a conveyor belt?
« Reply #526 on: January 28, 2007, 02:40:11 PM »
Quote
Originally posted by eskimo2
Now think about the process in reverse.
That's where you all nay-sayers got it wrong.

If conveyor is moving BACKWARD it helps rolling wheels FORWARD, therefore plane doesn't have to fight any excess forces but increased rolling resistance. All the wheels inertia forces are actually overcome by the conveyor.


Repeat again Eskimo, if conveyor is moving in the OPPOSING direction that of the plane, it is helping to roll wheels in the SAME direction as the plane travel.

BACKWARD --> FORWARD

Offline eskimo2

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plane on a conveyor belt?
« Reply #527 on: January 28, 2007, 03:17:49 PM »
Quote
Originally posted by 2bighorn
That's where you all nay-sayers got it wrong.

If conveyor is moving BACKWARD it helps rolling wheels FORWARD, therefore plane doesn't have to fight any excess forces but increased rolling resistance. All the wheels inertia forces are actually overcome by the conveyor.


Repeat again Eskimo, if conveyor is moving in the OPPOSING direction that of the plane, it is helping to roll wheels in the SAME direction as the plane travel.

BACKWARD --> FORWARD


2bighorn,

Are you saying that it is easier for an airplane to take off on a conveyor that travels the opposite direction (and same speed) than it is for an airplane to take off on a conveyor that travels the same direction (and same speed)?

In the first example the wheels are spinning at twice their normal rate, have absorbed and now store twice the energy they normally would.  

In the second example (plane riding with the conveyor) the wheels don’t need to spin at all and they have no rotational energy.

Offline 2bighorn

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plane on a conveyor belt?
« Reply #528 on: January 28, 2007, 03:37:44 PM »
Quote
Originally posted by eskimo2
Are you saying that it is easier for an airplane to take off on a conveyor that travels the opposite direction (and same speed) than it is for an airplane to take off on a conveyor that travels the same direction (and same speed)?
Yes, I've explained that already

Everybody hanged onto that opposite moving conveyor, which it has to be if you want wheels to rotate forward, otherwise it would be like applying the brakes.

And since the conveyor direction of movement was always opposite that of the plane, hence the wheels were always rotating forward, there is no increase in the wheels inertia for plane to fight.

If conveyor accelerate at faster rate than plane acceleration rate is, it actually helps the plane since wheels are overspeeding in direction of plane travel and conveyor alone is overcoming all the wheels rotational inertia and all the energy stored as wheels angular momentum is delivered by the conveyor.

Offline Mini D

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plane on a conveyor belt?
« Reply #529 on: January 28, 2007, 07:47:28 PM »
HT... you need to double check your math.

I = 16.1

T = 16.1*910
T = 14651

RPS = 14651/2

Feet per second at the edge of the belt:

(14651/2)*2piR
14651*pi

46,000 fps/s of acceleration

Now, you need to realize that this is a plane with less thrust than weight. Figure it out if the numbers are switched and pay particular attention to the amount of friction you can generate.

Offline hitech

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plane on a conveyor belt?
« Reply #530 on: January 29, 2007, 09:43:12 AM »
Miny D did you use pounds instead of mass to compute I? Not seeing how you got 16.1

Also you might really want to check you conversion from Radians per sec to feet per sec, because with 1 1 foot radius circle they will be equal.

Also I did make 1 typeo the Revolutions per sec should be 144 instead of 114, but the belt acceleration of 910 is correct.


HiTech
« Last Edit: January 29, 2007, 10:02:14 AM by hitech »

Offline JCLerch

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plane on a conveyor belt?
« Reply #531 on: January 29, 2007, 10:57:52 AM »
910 feet per second per second? OMG!!! :O

32 feet per second per second = 1G of acceleration., right?

So the surface of the conveyor belt is ACCELERATING at a rate slightly higher than 28g.

Or another way to look at this,

If the plane and conveyor belt are at rest with respect to one another, then the pilot opens the throttle, the conveyor belt will have to accelerate, such that the surface of the conveyor belt exceeds the speed of sound in 1.2 seconds.

Offline 2bighorn

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plane on a conveyor belt?
« Reply #532 on: January 29, 2007, 01:37:41 PM »
Quote
Originally posted by hitech
Miny D did you use pounds instead of mass to compute I? Not seeing how you got 16.1

Also you might really want to check you conversion from Radians per sec to feet per sec, because with 1 1 foot radius circle they will be equal.

Also I did make 1 typeo the Revolutions per sec should be 144 instead of 114, but the belt acceleration of 910 is correct.


HiTech

Hitech, wheel can only have one angular moment since wheel can rotate in only one direction at the time, not two. It does not matter where torque forces come, from plane or conveyor, since both help the wheel rotate in one direction only (the direction of plane forward movement) and are NOT opposing each other.

The only time worth calculating angular moment created by belt is for the case we have half the wheels sitting on the fixed runway and half the wheels sitting on the belt and rotating backwards (belt would have to move same direction as plane) and that scenario is far from original question.

Offline hitech

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plane on a conveyor belt?
« Reply #533 on: January 29, 2007, 03:04:53 PM »
2bighorn: You really need to read up on some more physics. My sketch is exactly the way Rotational inertia and forces on bodies works.

It is as simple as sum up the forces and torques involved, if force sum = 0 plane does not move.

HiTech

Offline takeda

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plane on a conveyor belt?
« Reply #534 on: January 29, 2007, 03:39:11 PM »
This is a sad day for me. Watching hitech being so wrong about this particular subject it's like being a kid discovering that Santa is really his parents.

Offline hitech

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plane on a conveyor belt?
« Reply #535 on: January 29, 2007, 04:03:56 PM »
takeda: Can you prove me wrong? If so feel free to do so.

Offline 2bighorn

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plane on a conveyor belt?
« Reply #536 on: January 29, 2007, 04:22:56 PM »
Quote
Originally posted by hitech
2bighorn: You really need to read up on some more physics. My sketch is exactly the way Rotational inertia and forces on bodies works.
Your drawing is fine so is calculation, but you missed on logic.

Quote
Originally posted by hitech
It is as simple as sum up the forces and torques involved, if force sum = 0 plane does not move.
The conveyor is not the opposing force, it just moves into opposite direction relative to the plane.

Look at the picture bellow. In order for force B to cancel force C, force B must work on the axis of rotation. But it doesn't. Due to circular body it changes it's direction and works same way as force A.
Therefore the sum of forces are not  F = C - B  but  F = A + C

Offline takeda

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plane on a conveyor belt?
« Reply #537 on: January 29, 2007, 04:27:33 PM »
No conceivable arrangement that can only spin the wheels can keep the plane from gaining forward airspeed once thrust is applied.
If the threadmill is spinning the wheels with 30g of acceleration, once those engines start pushing, the wheels have no other option but accelerate at say 32g.

I can't really follow your diagram that well, I feel that a few arrows and labels are missing, but you can't oppose the force exerted by the threadmill and plane on the wheel, because they are in fact working in the same direction, making the wheel spin forward.
The only way to stop the plane using the wheels is countering that spin, ie. brakes or blocks. A threadmill running opposite will just spin them faster and the plane will fly, a threadmill running in the same direction means the plane will fly, because it would be more aptly called a catapult.
« Last Edit: January 29, 2007, 04:32:00 PM by takeda »

Offline eskimo2

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plane on a conveyor belt?
« Reply #538 on: January 29, 2007, 04:51:21 PM »
Quote
Originally posted by 2bighorn
Your drawing is fine so is calculation, but you missed on logic.

The conveyor is not the opposing force, it just moves into opposite direction relative to the plane.

Look at the picture bellow. In order for force B to cancel force C, force B must work on the axis of rotation. But it doesn't. Due to circular body it changes it's direction and works same way as force A.
Therefore the sum of forces are not  F = C - B  but  F = A + C


2bighorn,

When you drop things on your planet, which way do they fall?

Offline eskimo2

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plane on a conveyor belt?
« Reply #539 on: January 29, 2007, 04:54:07 PM »
Quote
Originally posted by takeda
No conceivable arrangement that can only spin the wheels can keep the plane from gaining forward airspeed once thrust is applied.
If the threadmill is spinning the wheels with 30g of acceleration, once those engines start pushing, the wheels have no other option but accelerate at say 32g.

I can't really follow your diagram that well, I feel that a few arrows and labels are missing, but you can't oppose the force exerted by the threadmill and plane on the wheel, because they are in fact working in the same direction, making the wheel spin forward.
The only way to stop the plane using the wheels is countering that spin, ie. brakes or blocks. A threadmill running opposite will just spin them faster and the plane will fly, a threadmill running in the same direction means the plane will fly, because it would be more aptly called a catapult.


Think about how these things work: