By Dan Tyler

Derivation of the acceleration of an object that is flying circles within a moving air mass

 

 

Assume that the object is flying around the center of a moving circle with an angular velocity of .  Assume that the circle has a radius of  and that the center of the circle is moving with a speed of  in the  direction.  Then, the components of the position vector in the ground fixed coordinate system are given by:

Take individual derivatives of the position vector components to derive the velocity components.  You get the following:

Then, take derivatives again to get the components of the acceleration vector:

It is important to note that both components of the acceleration vector are purely sinusoidal.  This is the key to the final result.

 

 

The magnitude of the acceleration vector, , is determined by utilizing the Pythagorean theorem:

When the acceleration components are plugged into this expression we find that things simplify nicely:

The reason for this simplification is that, for any angle , the sum of the squares of the sine and cosine functions is unity: .  For those comfortable with the notion of centripetal acceleration being the square of the speed around the center of rotation divided by the radius of rotation, you may see that this is IDENTICAL, since the angular speed is .

 

 

The angle (or direction) of the acceleration vector is found by using the inverse tangent function as follow:

The minus signs have not been cancelled because they are important for determining the quadrant in which the acceleration vector resides.  However, a simple trigonometric identity for both the sine and cosine functions allows one to write an equivalent expression that allows us to remove them.  The expression is:

Thus, the angle of the acceleration vector is  radians (180^o) ahead of the angle of the position vector (the opposite direction), but rotates at the same rate.

 

 

The following image shows graphically what I derived above.  The green line segments are the radius of the moving circle in relation to the point at which the object is at time .  Note that the acceleration vector falls exactly on these line segments.  Thus, there is NO DIFFERENCE if the plane is moving within an air mass moving across the surface.