### Centrifugal Force – The “Imaginary” Force (or, The Physics of Coordinated Turns)

### by John Koser

**from On Final February 2005**

The diagram shown on page 45 of the August 1997 Issue of “Flight Training,” also in part of several other flight training aids (Jeppeson Pilot Manual, FAA Flight Training Handbook, & ASA Private Pilot Test Prep), is basically incorrect from the outside observer’s (as they illustrate it) point of view. Two features about the diagram need to be addressed. The diagram looks much like the one shown below, where the two horizontal vectors are the same length, and the upward pointing vertical vector is longer than the downward one.

A. The vectors shown all seem to be real from the point of view of the pilot aboard the aircraft (shown heading toward the reader), but from the point of view of an observer in the position of the reader, suspended in space, there is one vector shown that shouldn’t be there – the one labeled “Centrifugal Force.”

B. The relative lengths of the vectors shown also need to be addressed, as vector addition is a scale process, and the vectors shown aren’t drawn to correct scale.

**Analogy – Ball on a String With Vectors**

To examine the two ideas, look at a simple analogy – a ball suspended on a string, which is whirled around in a horizontal circle. The airplane shown in the diagram is supposedly flying in a horizontal circle, so from our position in front of it as readers, we would be in relatively the same position with respect to the ball, and in the same horizontal plane.

If one considers a ball being swung on a string so its path is a horizontal circle, and asks, “What forces act on the ball?” one could see that the two forces (ignoring air friction) acting in the plane of the paper are: **weight W**, and **string tension T**. See Figure 2, noticing that *no other forces act on the ball.*

Since the ball is not accelerating in the vertical direction (no unbalanced forces, therefore no acceleration – Newton’s Second Law), the vertical component of the vector **T** must just be balanced by the downward pointing vector **W**. See Figure 3.

Notice that the vector **T** has *no component pointing to the right*. Its only components in the **x** and **y** directions (horizontal & vertical) point up and to the left (toward the center of the horizontal circle in which it is moving). The component of **T** which points to the left **Tx**, which is also toward the center of the horizontal circle of the ball’s movement is a *center-pointing* force, or ** centripetal** force. It is an unbalanced force, therefore it produces acceleration (Newton’s Second Law again). The direction of this resulting acceleration must be in the same direction as the unbalanced force, which is to the left in the diagram, and this direction is along a radius, which points toward the center of the circle. In effect, the ball, going in a horizontal circle is always being accelerated toward the center of the horizontal circle.

**Is there an Outward (Centrifugal) Force?**

Where does the idea of a centrifugal (outward-pointing) force come from? If you are the pilot of the aircraft coming toward the reader (Figure 1), you feel an outward pointing force, just as does the driver of a car going around a horizontal curve. If you were riding on the ball in Figure 2, you would perceive that same outward pointing force. You think you feel this force, because the car, or the ball, or the airplane, to which you are secured by the seat/shoulder belt system, is pushing you inward. It does this because it’s an accelerated system, not an inertial system. In our everyday experience, we tend to think of all forces as balanced by equal and opposite forces (Newton’s Third Law). We tend to think there must be a force opposing this inward – pointing force, but that’s not true. It is this inward pointing force that makes you go in a circle. If you suddenly could remove it, you would continue in a straight line tangent to the curve at that point, except that gravity would make you begin to accelerate downward (an unbalanced force).

**Comparison of Ball on String with the Airplane**

The tension vector T on the ball is analogous to the lift vector provided by the airplane’s wings. The lift is produced because the wings are moving toward us through the air, whether the wings are oriented horizontally or at some angle (angle of bank). This lift vector describes a cone as the airplane banks at a constant angle about a vertical axis, just as does the string suspending the ball. The airplane then, generates lift perpendicular to its path, and that lift vector has two components, vertical and horizontal. The vertical component must be equal and opposite to the airplane’s weight and the horizontal component is completely unbalanced, therefore generating acceleration in the direction it points – toward the center of the horizontal circle.

**Vector Addition and Vector Lengths**Notice in Figure 3, the vertical component of lift and the weight vector are the same length. That must be true since there is no vertical acceleration. When a vector is resolved into its components along coordinate axes, the components are defined to the same scale. If the angle of bank to the horizontal is Ã˜, and if we define the Lift Vector as L, we use trigonometry to define the components as:

(Vertical Component) Ly = L(sin Ã˜), and

(Horizontal Component) Lx = L(cos Ã˜).

That’s why the “Resultant Lift” vector shown in Figure 1 is too long for the vector diagram’s scale.

**Conclusion**

There is no centrifugal force – only a perceived such sensation as felt by the pilot because the pilot is accelerating toward the axis of the horizontal circle, but it’s really the airplane exerting an inward force (which he/she perceives as an outward force) on the pilot.

The only force that’s unbalanced is the center-pointing force, which causes the acceleration toward the center of the circular path.

Anything going in a horizontal circular path is constantly being accelerated toward its center by an unbalanced force, the centripetal force.

**Corrected Diagram**

The diagram in Figure 1 can easily be corrected by making the vectors of correct length to represent components of lift L and by removing the fictitious centrifugal force vector. See Figure 4 below.

In this diagram, L has been divided into its components, Ly and Lx. These two components replace L. Since Ly is countered by the airplane’s weight W as it moves in a horizontal circle (not accelerating up or down), Ly & W disappear, and the only remaining force is Lx, the net force, which is a centripetal force.