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MrL_JaKiri · 20 Dec 2006, 17:03

Lets go over this in detail.

Let us first assume that there is a first order relationship between friction and velocity (that's y=mx + c, where y is friction, x is velocity and m and c are the coefficients of kinetic and static friction respectively, for all you GCSE maths types out there)..

This is a pretty safe assumption, because it's true.

Next: if you have a static force on an object, then it will take the same amount of time and distance to accelerate to a velocity as the object would to come to a stop from that velocity when the force opposes the movement.

For the next bit, I'll explain it as if the forces were constant (which is fine, the net impulses aren't changed by assuming this, and we're only effectively dealing with the plane at two points, when it's stopped and when it's taking off, so we're not losing any particularly useful information).

Lets call the force from the engines E.

Let us call the (constant in this example) friction force F.

The net force when taking off is E - F.

The net force when landing, from the wheels alone (no breaks, or flaps or whatnot) is F.

Now, let us assume that the plane doesn't need breaks or flaps when landing. It just stops of its own accord.

In this case E - F = F

E = 2F.

Would it take off on a rolling runway made of the same materials? Lets investigate!

The friction between the wheels and the track works as if the velocity was twice what it usually is, so y(2) = 2mx + c, not just mx + c. However, if c is greater than 0, which it is, y(2) is less than y, because 2mx + c is less than 2mx + c + c, obviously.

So, the friction in that case would be less than 2F, so there would still be some resultant force on the plane, so it would still accelerate. It would still take off, albeit over a very long time if c is small.

Remember that this is assuming that a plane will just roll to a halt once you've got it down on the ground.

Do I really have to go into the real world, where you have to use breaks and whatnot, in which case E - F = F + K, where K is the retardation force provided by the breaks (etc), and is in all likelihood larger than F?

Please say I don't.

20 Dec 2006, 17:03	#34
MrL_JaKiri The Twilight of the Gods Join Date: Jan 2001 Posts: 23,481	Re: A Plane Problem Lets go over this in detail. Let us first assume that there is a first order relationship between friction and velocity (that's y=mx + c, where y is friction, x is velocity and m and c are the coefficients of kinetic and static friction respectively, for all you GCSE maths types out there).. This is a pretty safe assumption, because it's true. Next: if you have a static force on an object, then it will take the same amount of time and distance to accelerate to a velocity as the object would to come to a stop from that velocity when the force opposes the movement. For the next bit, I'll explain it as if the forces were constant (which is fine, the net impulses aren't changed by assuming this, and we're only effectively dealing with the plane at two points, when it's stopped and when it's taking off, so we're not losing any particularly useful information). Lets call the force from the engines E. Let us call the (constant in this example) friction force F. The net force when taking off is E - F. The net force when landing, from the wheels alone (no breaks, or flaps or whatnot) is F. Now, let us assume that the plane doesn't need breaks or flaps when landing. It just stops of its own accord. In this case E - F = F E = 2F. Would it take off on a rolling runway made of the same materials? Lets investigate! The friction between the wheels and the track works as if the velocity was twice what it usually is, so y(2) = 2mx + c, not just mx + c. However, if c is greater than 0, which it is, y(2) is less than y, because 2mx + c is less than 2mx + c + c, obviously. So, the friction in that case would be less than 2F, so there would still be some resultant force on the plane, so it would still accelerate. It would still take off, albeit over a very long time if c is small. Remember that this is assuming that a plane will just roll to a halt once you've got it down on the ground. Do I really have to go into the real world, where you have to use breaks and whatnot, in which case E - F = F + K, where K is the retardation force provided by the breaks (etc), and is in all likelihood larger than F? Please say I don't. Last edited by MrL_JaKiri; 20 Dec 2006 at 17:16.