This is the one I have been fighting against.
If you figure it by wheel speed, if the plane ever moves in relation to its starting point, there will be a point where the belt is not matching the wheel speed.
This violates the Wheel speed theory. The question states that the belt matches the speed of the plane...not that it tries or catches up or that it can go faster.
Infinite speed people claim that the belt will speed up as it ATTEMPTS to match speed. This too is in violation of the wheel speed theory.
If the plane never moves from its starting point, it never has wheel speed for the belt to match.
If the plane does move from its starting point (even .5 inch) the belt can never match the speed.
Plane speed will then be ((X)+.5 inch) while belt speed will be only (X) for the distance formula of speed. Speed = distance times duration!
Wheel speed theory disproved.
Friction theory is moot as we assume that all components are capable of the function required....even the belt.
The problem is that a conveyor being sped up to "match" the wheelspeed would never be able to. It is impossible, because the thrust of the aircraft's engines WILL move the aircraft forward.
In the real world there would need to be a sensor and a computer to "drive" the conveyor. If you used the quickest sensors and the quickest computer then the conveyor might update 1,000,000 times a second. However that means that there is one millionth of a second that the airplane's wheels move without the computer adjusting the speed of the conveyor. The only effect of this is that the wheel is now spinning twice as fast as it was previously, the aircraft is still moving forward. With the next millionth of a second, the conveyor speed is again increased to match the wheelspeed of the aircraft, but this too does not move the aircraft backward, but rather just continue to increase the wheelspeed. As you should be able to see, the aircraft will continue to move forward, and the wheelspeed will increase exponentially.
Let say, for instance, that the aircraft normally reaches a "take-off" velocity of 150ft/sec (~102mph) in 15 seconds and just for argument's sake that the aircraft's acceleration is linear, gaining 10ft/sec. Then normally the wheels would see a speed of 10fps at 1 second, 20fps at 2 seconds, 30fps at 3 seconds, etc... to 150fps at 15 seconds.
However if a wheel speed sensor is used to accelerate the "conveyor belt" and this sensor updates the speed of the conveyor belt
once every second (for simplicity's sake) and assuming the "engine" driving the conveyor belt could keep up, then after the first second, the conveyor belt would be running at 10fps, doubling the 10fps of the wheel speed sensor on the aircraft. At the end of the 2nd second, the aircraft would have added another 10fps on top of the 20fps (10 for the aircraft, 10 for the conveyor belt) that the wheel speed sensors are reporting. So the conveyor will match that with 30fps for a total of 60fps.
For the 3rd second the conveyor adjusts for 70 fps (~48 mph).
For the 4th second the conveyor adjusts for 150 fps (~102 mph).
For the 5th second the conveyor adjusts for 310 fps (~211 mph).
For the 6th second the conveyor adjusts for 630 fps (~430 mph).
For the 7th second the conveyor adjusts for 1,270 fps (~866 mph).
For the 8th second the conveyor adjusts for 2,550 fps (~1,739 mph).
For the 9th second the conveyor adjusts for 5,110 fps (~3,484 mph).
For the 10th second the conveyor adjusts for 10,230 fps (~6,975 mph).
For the 11th second the conveyor adjusts for 20,470 fps (~13,957 mph).
For the 12th second the conveyor adjusts for 40,950 fps (~27,920 mph).
For the 13th second the conveyor adjusts to 81,910 fps (~55,848 mph).
For the 14th second the conveyor adjusts fo 162,820 fps (~111,014 mph).
And finally for the 15th second the conveyor adjusts for 327,670 fps or 223,411 mph.
Somewhere around the 5th second, the tires would likely give out. Of course the aircraft would only be traveling an actual 35mph, and the aircraft's landing gear would dig into the conveyor belt and then all heck would break loose...
In the hypotetical world where the conveyor COULD match the aircraft's wheels instantaniously then the instant the aircraft had any movement at all, the wheelspeed and conveyor speed would reach infinite speed. In the real world, the speed of the sensors and the speed of the computer and accelleration rate of the conveyor would determine just how quickly the wheelspeed increased. But given a conveyor capable of "keeping up" the quicker the sensor and computer cylce the quicker the accelleration rate of the wheels and conveyor. Yet, STILL the aircraft WILL move forward until the point at which the tires disintigrate. So given tires, wheels and wheelbearing capable of an infinite speed, the aircraft would still take off.
A "reality" perspective

on this theoretical problem.
1. If "speed of the plane" is taken to be the speed of the wheels on top of the conveyor, like a jogger on a treadmill, then you have just shown that as soon as wheel speed exceeds max tire speed, we no longer have rolling friction but more likely have "sliding/dragging friction", probably degrading into near static friction as parts fly. Therefore the plane would not take off.
2. If "speed of the plane" is taken to be relative to the ground, then because wheel speed will be twice the speed of the plane, as long as Vr*2 < max tire speed the plane will be able to take off. Otherwise, we again get into tire breakdown followed by increasing friction significantly slowing the plane down until it no longer can take off.
If the plane is prop driven, then a prop strike could become highly likely if the tires fly apart, thereby reducing the chance of take off. Certainly it would not be a good thing for a jet to lose ALL wheels either and you could envisions a lot of bad outcomes as a result that would likely prevent a successful takeoff.
Therefore, only if speed of the plane is taken to be relative to the ground and Vr*2< max tire speed do you have a "normal" take off.
egnorant
14th March 2006 - 06:15 PM
QUOTE (biplanepilot+Mar 14 2006, 05:47 PM)
MCHAT,
The wheel speed is the same as the conveyor speed as long as the two are in contact with each other. The whole key is airspeed. The speed of the air over the wings.
The wheel speed theory measures the speed of the belt as it relates to the ground.
Yet it measures the plane speed (via the wheels) as it relates to the belt.
See my recent post to show how this is impossible as applied to the original question.
Wiser people have the belt speed and the plane speed as it relates to the ground
(and the calm air) as a starting point.
Plane goes X in one direction..belt move X in the other direction...wheels spin at 2X. This was decided a long time ago.
No doubt....It flies!
The rest is a bunch of us teasing the silly people!
Bruce
krreagan
14th March 2006 - 06:19 PM
QUOTE (biplanepilot+Mar 14 2006, 10:47 AM)
MCHAT,
The wheel speed is the same as the conveyor speed as long as the two are in contact with each other. The whole key is airspeed. The speed of the air over the wings.
I disagree, you need to factor in the forwad motion of the plane relative to the ground that the belt itself is sitting on. You are measuring the wheel speed with a speedometer, so if the plane has any forward motion at all, then the wheel speed and belt speed can never be moving at the same speed at the same time. This is the paradox/infinity of the problem when you interpret the speed of the plane as the wheel speed. And therefore, you have to measure the speed of the plane as the speed through the air (or over the ground) and not relative to the belt (wheel speed).
Krreagan
egnorant
14th March 2006 - 06:27 PM
QUOTE (Guest+Mar 14 2006, 06:06 PM)
Therefore, only if speed of the plane is taken to be relative to the ground and Vr*2< max tire speed do you have a "normal" take off.
I love that everyone is questioning the mechanical capabilities of the plane without
questioning the mechanical capabilities of the treadmill!!!
If we assume that one is incapable then we must be able to assume that both are incapable of performing the feats we apply.
Or we must assume that both are capable!!!
Assigning lesser capability to the plane is just a trick applied unfairly by those who
have no real basis for their "no fly" theories.
The "will fly" folks would never stoop to such irrelevancies!!
Bruce
Fynlcut
14th March 2006 - 07:08 PM
If we talk wheel speed and we know the speeds must always match, then the conveyor will simply move forward at the speed of the plane.
But you say the conveyor moves in the opposite direction. True in the opposite direction with a negative speed. (two negatives make a positive)
egnorant
14th March 2006 - 07:50 PM
QUOTE (Fynlcut+Mar 14 2006, 07:08 PM)
If we talk wheel speed and we know the speeds must always match, then the conveyor will simply move forward at the speed of the plane.
But you say the conveyor moves in the opposite direction. True in the opposite direction with a negative speed. (two negatives make a positive)
So the plane would be moving in one direction at a negative speed...
the belt would be moving in the opposite direction at a negative speed...
Now if we had a really big bungee cord....and a DeLorean...
Would the belt need to sink into the ground as the plane gained altitude?
If we throw a football over the belt from side to side, what is the vector change?
Would we need to lead the plane in order to hit the plane with this football?
If an asteroid breaks in half and one half hits your car, does insurance only cover for half?
I still have a lot of boring days left to kill?
Bruce
MChat
14th March 2006 - 08:07 PM
QUOTE (egnorant+Mar 14 2006, 05:58 PM)
Even in this hypothetical world the belt could not match the wheel speed.
Question says it matches exactly!
The infinite speed scenario is NOT matching exactly..it is merely trying to catch up.
Wheel speed theory as applied to this question is only correct at the 0 speed.
Any movement by the plane to achieve wheel speed will instantly develop an environment where matching speed is impossible. ((X)+y) does not=(X)....
Use any figures you want! As long as y represents the movement of the plane to achieve wheel speed. Remember that y=0 indicates no wheel speed, so X must be 0 also.
It flies
Bruce
I think we're on the same page.
I program computers for a living so I tend to think on how I would design a system that would attempt to match the conveyor's speed to the wheelspeed of the aircraft. The reality is no system could be designed no matter the technological advances, at best we have a conveyor that is "one step behind" and never truely equal to the wheelspeed of the airplane.
Obviously the question as originally stated implied the conveyor matches the actual speed of the airplane relative to the ground. Which is a simple problem once one understands what is happening. In fact even a normal passenger car will be able to accellerate on a conveyor which matches it's ground speed; only it's topspeed (relative to the ground) will be reduced by half.
Physics Student
14th March 2006 - 08:39 PM
Has anyone said anything about the fact that any force put on the wheels is turned into torque (assuming the wheel is frictionless)? taking the cross product of the force and velocity would mean the force would be applied towards the axle (meaning perpendicular to the force of the enggines). Perpendicular forces do not act on each other.
So, the plane would move forward, gain airspeed and take off, no matter if the conveyor belt was moving a billion fps
MChat
15th March 2006 - 01:42 AM
I don't know if this has been posted here yet (I didn't read through all 312+ pages) but here is a video that proves that the conveyor DOES NOT force the aircraft to remain at zero velocity:
http://videos.streetfire.net/comment/0/35E...6BA1A43A06B.htm
Guest
15th March 2006 - 01:54 AM
QUOTE (egnorant+Mar 14 2006, 06:27 PM)
QUOTE (Guest+Mar 14 2006, 06:06 PM)
Therefore, only if speed of the plane is taken to be relative to the ground and Vr*2< max tire speed do you have a "normal" take off.
I love that everyone is questioning the mechanical capabilities of the plane without
questioning the mechanical capabilities of the treadmill!!!
If we assume that one is incapable then we must be able to assume that both are incapable of performing the feats we apply.
Or we must assume that both are capable!!!
Assigning lesser capability to the plane is just a trick applied unfairly by those who
have no real basis for their "no fly" theories.
The "will fly" folks would never stoop to such irrelevancies!!
Bruce
No one can say what the characteristics of the conveyor are other than it can always match the speed of the plane. I indicated that speed of the plane could be interpreted in two ways. However, max tire speed is as real to any wheeled aircraft as Vr and obviously more important to aircraft that land and take off at high velocities. I didn't express bias as to fly/not fly as both were possible but conditionally.
You seem rather entrenched on the issue of matching wheel speeds by the way. Don't you think that matching within a millimeter/s or so would be close enough to at least consider as a possibility? If so, then case 1 that I expressed is easily an outcome.
krreagan
15th March 2006 - 04:16 AM
QUOTE (Guest+Mar 14 2006, 06:54 PM)
QUOTE (egnorant+Mar 14 2006, 06:27 PM)
QUOTE (Guest+Mar 14 2006, 06:06 PM)
Therefore, only if speed of the plane is taken to be relative to the ground and Vr*2< max tire speed do you have a "normal" take off.
I love that everyone is questioning the mechanical capabilities of the plane without
questioning the mechanical capabilities of the treadmill!!!
If we assume that one is incapable then we must be able to assume that both are incapable of performing the feats we apply.
Or we must assume that both are capable!!!
Assigning lesser capability to the plane is just a trick applied unfairly by those who
have no real basis for their "no fly" theories.
The "will fly" folks would never stoop to such irrelevancies!!
Bruce
No one can say what the characteristics of the conveyor are other than it can always match the speed of the plane. I indicated that speed of the plane could be interpreted in two ways. However, max tire speed is as real to any wheeled aircraft as Vr and obviously more important to aircraft that land and take off at high velocities. I didn't express bias as to fly/not fly as both were possible but conditionally.
You seem rather entrenched on the issue of matching wheel speeds by the way. Don't you think that matching within a millimeter/s or so would be close enough to at least consider as a possibility? If so, then case 1 that I expressed is easily an outcome.
When considering the speed of the wheels (illogical)...
The belt can never get closer in speed to the wheels then then the velocity of the plane relative to the ground...
The wheel speed (X) is the speed of the plane over the ground (Y) - the speed of the belt (Z) in the opposite direction (and therefore a negative quantity) to the motion of the plane, which imposes additional velocity to the wheels.
so the speed of the wheels is: X = Y - Z
but the question imposes the following condition on the problem:
X = -Z
Therefore we can substitute -X into the original equation for Z and arrive at:
X = Y - (-X)
So, for any forward velocity of the plane Y the equation cannot be correct and we have our paradox.
Now in real life there is always some lag in the control system so that |Z| < |X| at any time that X is increasing.
So there is no situation that can conforms to the original question posed if you interpret the "speed of the plane" as the speed of the wheels as indicated by a speedometer and the belts speed is in relation to the ground/surroundings/air.
Krreagan
egnorant
15th March 2006 - 04:57 AM
QUOTE (Guest+Mar 15 2006, 01:54 AM)
No one can say what the characteristics of the conveyor are other than it can always match the speed of the plane. I indicated that speed of the plane could be interpreted in two ways. However, max tire speed is as real to any wheeled aircraft as Vr and obviously more important to aircraft that land and take off at high velocities. I didn't express bias as to fly/not fly as both were possible but conditionally.
You seem rather entrenched on the issue of matching wheel speeds by the way. Don't you think that matching within a millimeter/s or so would be close enough to at least consider as a possibility? If so, then case 1 that I expressed is easily an outcome.
Sorry for the hard *** attitude, but we have had a rather obnoxious fellow that
used diversions similar to yours to be a pain.
You should really go back a few dozen pages and see the tangents we traveled.
I finally asked my 0 wheel speed question of him and he has not answered.
There was even thought that he came back with a new name or as a guest.
While we (my niece is helping) have found no fault with your stepped version,
we really haven't reached the point where we have convinced each other that it has no fault.
By looking at your figures (accepted as correct) is when we started questioning
mechanical capabilities of BOTH the plane and the treadmill.
Entrenched?? Yes, had to stay focused or some yahoo would try to divert and
attempt to confuse the issue. A simple proof to disprove the "no fly" guys pet theory and a challenge to them to disprove it.
As far as considering "close enough" scenarios the answer from my niece was a resounding "NO!!! not as applied to this question!!!"
I agree with her!
Consideration as a separate question is acceptable.
But, if you can have a treadmill capable of 233,411 mph then we must consider
the possibility of wheels that are capable of 233,411 mph.
Bruce
egnorant
15th March 2006 - 05:17 AM
QUOTE (krreagan+Mar 15 2006, 04:16 AM)
When considering the speed of the wheels (illogical)...
The belt can never get closer in speed to the wheels then then the velocity of the plane relative to the ground...
The wheel speed (X) is the speed of the plane over the ground (Y) - the speed of the belt (Z) in the opposite direction (and therefore a negative quantity) to the motion of the plane, which imposes additional velocity to the wheels.
so the speed of the wheels is: X = Y - Z
but the question imposes the following condition on the problem:
X = -Z
Therefore we can substitute -X into the original equation for Z and arrive at:
X = Y - (-X)
So, for any forward velocity of the plane Y the equation cannot be correct and we have our paradox.
Now in real life there is always some lag in the control system so that |Z| < |X| at any time that X is increasing.
So there is no situation that can conforms to the original question posed if you interpret the "speed of the plane" as the speed of the wheels as indicated by a speedometer and the belts speed is in relation to the ground/surroundings/air.
Krreagan
Exactly!
But it does balance when all are zero.
That is the only time.
My niece read your post and was delighted that someone else understands.
She says you get a hug!!
She almost got in trouble because it is actually an hour past her bedtime and suddenly she screams "Hell yeah!!!" at the computer.
Bruce
krreagan
15th March 2006 - 01:26 PM
QUOTE (egnorant+Mar 14 2006, 10:17 PM)
QUOTE (krreagan+Mar 15 2006, 04:16 AM)
When considering the speed of the wheels (illogical)...
The belt can never get closer in speed to the wheels then then the velocity of the plane relative to the ground...
The wheel speed (X) is the speed of the plane over the ground (Y) - the speed of the belt (Z) in the opposite direction (and therefore a negative quantity) to the motion of the plane, which imposes additional velocity to the wheels.
so the speed of the wheels is: X = Y - Z
but the question imposes the following condition on the problem:
X = -Z
Therefore we can substitute -X into the original equation for Z and arrive at:
X = Y - (-X)
So, for any forward velocity of the plane Y the equation cannot be correct and we have our paradox.
Now in real life there is always some lag in the control system so that |Z| < |X| at any time that X is increasing.
So there is no situation that can conforms to the original question posed if you interpret the "speed of the plane" as the speed of the wheels as indicated by a speedometer and the belts speed is in relation to the ground/surroundings/air.
Krreagan
Exactly!
But it does balance when all are zero.
That is the only time.
My niece read your post and was delighted that someone else understands.
She says you get a hug!!
She almost got in trouble because it is actually an hour past her bedtime and suddenly she screams "Hell yeah!!!" at the computer.
Bruce
Ahhh shucks!
Krreagan
Fynlcut
15th March 2006 - 02:06 PM
QUOTE (egnorant+Mar 14 2006, 07:50 PM)
QUOTE (Fynlcut+Mar 14 2006, 07:08 PM)
If we talk wheel speed and we know the speeds must always match, then the conveyor will simply move forward at the speed of the plane.
But you say the conveyor moves in the opposite direction. True in the opposite direction with a negative speed. (two negatives make a positive)
So the plane would be moving in one direction at a negative speed...
the belt would be moving in the opposite direction at a negative speed...
Now if we had a really big bungee cord....and a DeLorean...
Would the belt need to sink into the ground as the plane gained altitude?
If we throw a football over the belt from side to side, what is the vector change?
Would we need to lead the plane in order to hit the plane with this football?
If an asteroid breaks in half and one half hits your car, does insurance only cover for half?
I still have a lot of boring days left to kill?
Bruce
Simply put, the wheels never turn, but the plane moves forward as the coveyor (moving at a negative speed in the opposite direction) moves in the direction of travel produced by the thrust of the engine.

or something like that.
sooks
15th March 2006 - 02:28 PM
Well thats not fair that just he gets the hug.... A lot of us have been saying that for a long long time!!!!!
MChat
15th March 2006 - 03:05 PM
QUOTE (egnorant+Mar 15 2006, 04:57 AM)
While we (my niece is helping) have found no fault with your stepped version, we really haven't reached the point where we have convinced each other that it has no fault.
By looking at your figures (accepted as correct) is when we started questioning mechanical capabilities of BOTH the plane and the treadmill.
Consideration as a separate question is acceptable.
But, if you can have a treadmill capable of 233,411 mph then we must consider the possibility of wheels that are capable of 233,411 mph.
Of course both the aircraft and the conveyor would have mechanical limitations. Since the conveyor exists only for this hypothetical situation, it's limits are whatever we define them to be. However, airplanes actually exist and I think the weakest link is the tires' speed rating. Looking online the lowest speed rating I found was 120, while the highest was in the low 200's.
So for the most logical version of this problem, where the conveyor matches the aircraft's actual speed and not some attempt at matching it's wheel speed, it is certainly possible that the aircraft is equipped with suitably rated tires and will reach departure velocity before the tires disintigrate.
For the illogical (and impossible, except the "stepped" model) version of this problem, where the conveyor matches the wheelspeed of the aircraft. The tires would disintigrate within seconds after initial movement of the aircraft. However I would think that a real conveyor would reach it's mechanical limitations around the same time and would require an extremely powerful engine just to keep up with the aircraft's wheelspeed.
egnorant
15th March 2006 - 04:19 PM
QUOTE (sooks+Mar 15 2006, 02:28 PM)
Well thats not fair that just he gets the hug.... A lot of us have been saying that for a long long time!!!!!
We may have missed it as we did not arrive until page 271.
The discussion was bogged down on speed and velocity.
ATL5p had just proclaimed that his toy plane proved something.
I had just arrived announcing that it was a word problem and
krreagen nudged us to examine the impossibility of the wheel speed theory.
I'm sure you qualify for at least a hug from a 14 year old girl!
As do many who stood fast and battled against the forces of obstinance and ignorance!!
Bruce
biplanepilot
15th March 2006 - 05:28 PM
Krreagan,
I'm one of the 'it will fly guys' but I have to disagree with you about wheel speed vs conveyor speed. The conveyor is just a big wheel, and while the two are in contact (just before flight) they have to be at equal speed. Not RPM, but speed.
If two wheels are rolling in contact with each other, their speed at the contact point has to be equal or there would be a bunch of rubber burning.
All this talk of wheels coming apart at high speeds is meaningless. As soon as the average plane reaches between 40 and 80 mph (airspeed) it's airborn and you can turn that damn conveyor off.
krreagan
15th March 2006 - 05:59 PM
QUOTE (biplanepilot+Mar 15 2006, 10:28 AM)
Krreagan,
I'm one of the 'it will fly guys' but I have to disagree with you about wheel speed vs conveyor speed. The conveyor is just a big wheel, and while the two are in contact (just before flight) they have to be at equal speed. Not RPM, but speed.
If two wheels are rolling in contact with each other, their speed at the contact point has to be equal or there would be a bunch of rubber burning.
All this talk of wheels coming apart at high speeds is meaningless. As soon as the average plane reaches between 40 and 80 mph (airspeed) it's airborn and you can turn that damn conveyor off.
Think of it this way...
Get on a treadmill with rollerblades on and stand at the far back of the belt, Now turn on the treadmill and remain where you are. The wheel speed matches that of the belt... (forward velocity equals zero) Now push/pull yourself forward to the very front of the belt while its going at a constant velocity, While you are moving from the back to the front relative to the ground, your wheels are moving at an additional speed equal to how fast you push yourself forward. So you have to add in your forward velocity relative to the ground, which is also what the treadmill is sitting on.
This is the X - (-Y) component.
When you are sitting still at the back of the treadmill, Y = 0, So the two speeds are equal. When you move forward, Y > 0 and therefore the equation becomes invalid.
By your logic, normal planes would not be able to take off because of the difference in runway/wheel speed. The runway is a belt going 0-mph and the plane is going 100-mph but there is no conflict between the surfaces (except when the wheels initially touch the runway on landing, but this is another situation).
Krreagan
MChat
15th March 2006 - 06:16 PM
QUOTE (biplanepilot+Mar 15 2006, 05:28 PM)
I'm one of the 'it will fly guys' but I have to disagree with you about wheel speed vs conveyor speed. The conveyor is just a big wheel, and while the two are in contact (just before flight) they have to be at equal speed. Not RPM, but speed.
If two wheels are rolling in contact with each other, their speed at the contact point has to be equal or there would be a bunch of rubber burning.
All this talk of wheels coming apart at high speeds is meaningless. As soon as the average plane reaches between 40 and 80 mph (airspeed) it's airborn and you can turn that damn conveyor off.
On a normal aircraft traveling at 50mph, on a normal runway, the bottom of the aircraft's tire (the part in contact with the runway) is traveling 0mph and the top of the tire is traveling 100mph; the axle in the center of the wheel is traveling (obviously the same speed as the aircraft) at 50mph. I think what most would consider "wheelspeed" is the speed at the center of the wheel, or the average between the top and bottom of the tire.
Guest
15th March 2006 - 10:18 PM
QUOTE (biplanepilot+Mar 15 2006, 05:28 PM)
All this talk of wheels coming apart at high speeds is meaningless. As soon as the average plane reaches between 40 and 80 mph (airspeed) it's airborn and you can turn that damn conveyor off.
A 737-200 has a Vr of 131 kts. Max tire speed is about 195 kts. If the speed of the plane is taken to be relative to the ground, hence wheel speed is double the ground speed, you reach max tire speed while ground speed is only 97.5 kts, well short of Vr. You'd come closer but still fall short in a Citation CJ3 for Vr(102 kts), V1(takeoff speed=105kts) and V2 (takeoff safety speed=115 kts)
Typical GA singles might have 6 ply tires with a max ground speed rating of 120pmh, or 104 kts. You'd get to the max tire speed at 52 kts. There are many singles with Vr well above that eg, Cirrus SR22 Vr=70 kts and nearly all retracts.
So again, only when you measure the speed of the plane relative to the ground and have an aircraft with Vr*2<max tire speed would you have a "normal takeoff".
NASA Engineers
16th March 2006 - 11:00 PM
This problem can be understood quite simply by considering the example of a plane with frictionless bearings in the wheels. Examine the case when no thrust is applied and the conveyer moves at any speed. The plane will remain motionless with respect to the ground while the wheels spin at the conveyer speed. Now apply some thrust and the plane will start to move forward relative to the ground because in this case the speed of the conveyer (with frictionless bearings) can have no effect on the plane motion. As you increase thrust the plane will take off just like a normal plane. In fact from an observer on the ground the take-off will look perfectly normal. Now add some friction to the bearings and the only thing you will have to do is add a little more thrust to overcome this friction. Again from an observer on the ground the take-off looks exactly the same with the exception that the wheels seem to be spinning too fast. This is different from a car because the car must have locked (not frictionless) bearings (at least in one axel) in order to move.
sooks
17th March 2006 - 01:07 AM
unfortunately atl5p cant understand that
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