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The Downwind Turn

Examples

The problem of the downwind turn is entirely associated with the effects of wind speed. Here are four simple calculated examples based on 4 different wind speeds: zero wind, wind at 1/4 airspeed, wind at 1/2 airspeed and 'wind' experienced inside a large aircraft cruising at 570 mph.

Before we go any further it will be necessary to set some parameters. All values will be in metric units, i.e. metres, and seconds; there is no need to consider a numerical value for weight because at all times in a level turn it equals lift, which simplifies it a bit. It will be referred to as 'wt'. Also, we need to lay down a few definitions:

Kinetic energy. This is a red herring and does not concern us. The common mistake is to assume that it affects turns as if there were a law of conservation of kinetic energy. There isn't, so we can forget about it. For the record kinetic energy is 1/2 mass * velocity squared.

Momentum. This is the one that matters. There IS a law of conservation of momentum, both of linear momentum and angular momentum (rotation). The snag is that this too is not relevant to effects produced by the aerodynamic forces that cause turns. The point of the examples will be to demonstrate that the change in momentum during a turn is the result of the reversal of airspeed and that this is independent of any effects of overall speed over the ground as caused by windspeed. Momentum is defined as mass * velocity. Because velocity is a directional quantity we have to be very careful with the number values. When the direction reverses, velocity changes from a positive value to a negative one. It is probably neglect of the sign (+ or -), more than any other factor, that leads to confusion when the physics of the turn are considered.

In all the examples the initial condition is flight into wind. The airspeed in all cases will be 20 m/sec, so the groundspeed in each case will be airspeed - wind speed. The term  'speed' is not as precise as 'velocity' but it will be used for convenience because of its familiarity.

Example 1         Windspeed 0 m/sec, model airspeed 20 m/sec.

Initial state:             Groundspeed(1) = airspeed(1) - windspeed  =  20 - 0

End state:                 Groundspeed(2) = airspeed(2) - windspeed  =  -20 - 0    Note, the reversal of the direction of flight has made airspeed(2) a negative value. The change in momentum is groundspeed(1)m - groundspeed(2) and again the sign of each value is vital.

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is

                                  wt * 20 - wt * (-20)  or   wt * 40

Example 2         Windspeed 5 m/sec, model airspeed 20 m/sec.

Initial state:             Groundspeed(1) = airspeed(1) - windspeed  =  20 - 5    or   15 m/sec

End state:                 Groundspeed(2) = airspeed(2) - windspeed  =  -20 - 5   or   -25 m/sec

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is

                                  wt * 15 - (wt * -25)    which becomes   wt * 40  same as before.

Example 3         Windspeed 10 m/sec, model airspeed 20 m/sec.

Initial state:             Groundspeed(1) = airspeed(1) - windspeed  =  20 - 10   or   10 m/sec

End state:                 Groundspeed(2) = airspeed(2) - windspeed  =  -20 - 10   or   -30 m/sec

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is

                                  wt * 10 - (wt * -30)    which again becomes   wt * 40

Example 4         Windspeed' 570 mph, or 250 m/sec (well, almost).

Inside the transport aircraft within which the model is to fly, it has to be launched towards the tail, so that the mass of moving air will be coming in the opposite direction to the model's flight and thus constituting a headwind. The speed of the big aircraft therefore has to be entered as -250 m/sec.

Initial state:             Groundspeed(1) = airspeed(1) - windspeed  =  20 - 250   or   -230 m/sec

End state:                 Groundspeed(2) = airspeed(2) - windspeed  =  -20 - 250   or   -270 m/sec

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is

                                  wt * -230 - (wt * -270)    which once more becomes   wt * 40

Conclusion       No matter how you cut it the calculation, done with proper respect for vector quantities, brings the same result. The values of windspeed, however great they may be, cancel out and only the aerodynamic flight effects contribute to the change in momentum.

 

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