Plane Puzzle – My Solution

This post is in response to a puzzle posted by a friend of mine.

The puzzle is as follows:

“If a plane is placed on a giant treadmill, and the treadmill is programmed to automatically match the plane’s forward speed (in the opposite direction), can the plane take off?”

Before I begin to solve the solution, here are a few assumptions:

  1. The wheels of the plane do not slip on the surface of the giant treadmill (in other words, friction between the wheel and the ground is constant and static)
  2. The wheels of the plane will not explode, burn up, slide off, or otherwise be adversely affected by any forces, frictional or otherwise.
  3. The wheels can rotate an any speed.
  4. All thrust force from the plane acts horizontally (no vertical components).
  5. Lift is generated by the plane’s motion, not the air’s. In other words, a strong gust is not going to spring up and carry off the plane, air flow due to shear from the treadmill will not occur, there will be no Dyson bladeless fan effects from plane turbofans/turbojets, etc.
  6. There is no acceleration in the system since the velocities of the plane and the treadmill are constant. Acceleration is the derivative of velocity, and the derivative of a constant is zero.
  7. The surface of the treadmill is level.

The first thing to recognize is that if the plane does not move relative to the air around it, it cannot become air-borne. The upward force (lift) on a plane’s wings can only occur when the plane is moving through air. In other words, if the speed of the plane relative to the ground (and the air around it) is zero, then the plane will not fly. If you have ever seen an airplane take off, this should be very intuitive.

Consequently, for the plane to remain at the same location relative to the ground, the velocity of the plane relative to the treadmill (v_p) must be equal to the velocity of the treadmill relative to the ground (v_t), assuming that each is in an opposite direction than the other.

Really, the question answers itself. If the velocities of both the velocity and treadmill are equal and opposite, then motion cannot occur since the plane will stay in the same place relative to the ground. However, I believe that the intent of the question was as follows:

Is there ever a case where the force due to the treadmill’s velocity can counter the force due to the velocity of a plane, thus grounding the plane by keeping it from moving relative to the air around it?

In order to answer this question, I will start by making an equivalent model of the plane to simplify things, then do a little physics.

Here is my simplified model of the plane:


In this model, everything which impacts the problem is included. A single, equivalent wheel represents all wheels on the plane. F_thrust is simply the thrust force from the plane’s turbofans/turbojets.

Here’s a free body diagram of the equivalent wheel:


 In this diagram representing forces acting on the wheel, c is the coefficient of damping due to bearing friction of the wheel/landing gear interface, r is the radius of the wheel,  and θ represents the direction of angular movement (and angular velocity, ω). F_t is the lumped tangential force at the interface of the wheel and treadmill. This value includes both the constant force due to static friction acting against the wheel and the resistive force due to the plane’s thrust.

All vertical forces are equal and opposite and thus cancel.

To begin, we are going to sum the wheel forces in the horizontal direction. This value will be equal to the mass times acceleration, which is zero (since velocity is constant):

We get

F_t = F_thrust.

Simple enough. Now, we sum the moments about the center of the wheel. This value will be equal to the angular acceleration of the wheel times its inertia, which is also also zero (angular acceleration is proportional to linear acceleration by way of the wheel’s radius).

This gives us

r*F_t = c*ω

where ω is the angular velocity of the wheel. Solving for F_t and plugging it into the previous expression, we get


Note that


where v is the relative velocity between the wheel and the treadmill. This equation can be written as

ω=(v_p + v_t)/r.

Note: I should note that I have been using speed rather than velocity in this post, while calling it otherwise. This is shameful. Velocity is a vector, a quantity having both direction and magnitude. Speed is a quantity having magnitude, but not direction. Since I have been using magnitudes of velocity, I take into account direction by adding treadmill speed to plane speed. This is the same as subtracting treadmill velocity from plane velocity, since the treadmill velocity is negative relative to plane velocity.

Plugging this into the expression for F_thrust, we get

F_thrust=c*(v_p + v_t)/r^2


v_p + v_t= (1/c)*(r^2)*F_thrust.

The expression (1/c)*(r^2) is a constant value for the plane at at a given velocity (since the radius of the wheel won’t change and neither will the damping coefficient from bearing friction). Replacing with a representational constant, K, we see that

v_p = F_thrust*K - v_t.

Note that for efficient bearings c is very small. Consequently, K is a very large value (it is inversely proportional to c).

Subtracting v_t from both sides gives

v_p - v_t = F_thrust*K - 2*v_t.

For the plane to go nowhere, v_p -v_t  must equal zero, or

F_thrust*K - 2*v_t = 0.

Solving for v_t, we see that for this to happen,

v_t = F_thrust*K/2

Assuming that my calculations are correct (I am human), this proves that there is always a value for the velocity of the treadmill that will result in the plane remaining in place relative to the air around it.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s