Yes, we are re-cycling old chestnuts!

**Introduction**

Well, we may as well grasp this particular nettle because, if the albatrosses can gain airspeed by turning to windward (turning upwind) then there must be a loss of airspeed when turning to leeward (turning downwind). This has been a bone of contention for years and is widely known as the Downwind Turn Myth.

There are two sides to the argument: the proponents think there is obviously an effect whereby airspeed tends to reduce when an aircraft turns downwind because in the early days of aviation (and even recently) aircraft did stall and spin when turning back to the airfield after a power failure. The opponents think there is no effect because ground-speed has no effect on the aircraft and accidents are caused entirely by pilot error. Specifically when turning downwind, the pilot sees ground-speed and drift increasing and applies up-elevator and opposite-rudder leading to a stall and spin.

This analysis will show that it can be said that both sides are correct: there is an effect but it is very small and easily over-looked in normal flight operations. It will only become significant when the aircraft is on the verge of stalling and the wind is a large proportion of the aircraft airspeed, which was likely the case in the early days of aviation.

The explanation is not simple but here we go:

**1 Forces acting on the aircraft **

**1.1** Consider a powered aircraft where thrust equals drag and airspeed is constant. When the aircraft is circling in __still-air__ there is a small increase in drag due to the increased load-factor; therefore, airspeed reduces slightly and is then constant. Ground-velocity is the same as air-velocity. The path through the air is the same as the path over the ground and is a circle.

**1.2** Now, consider the same aircraft circling in a wind. The airspeed indicator appears to show constant air-speed and the direction indicator (compass) shows a steady rate of turn. We look over the side of the cockpit and see a track which is no longer a circle but instead is a stretched-out spiral. The on-board inertial navigation system plots the same spiral. We infer the direction of the stretch of the spiral is the direction of the wind. The INS tells us the tangential-speed along the spiral track is increasing and decreasing. We infer that this is tangential acceleration of our ground-speed.

**1.3** If you think, at this point, that the wind is irrelevant to an aircraft in flight, remember that the spiral ground track is the vector sum of aircraft velocity relative to the air and wind-velocity relative to the ground. The acceleration of the aircraft is caused by forces acting on the aircraft and is seen as acceleration components of air-velocity and ground-velocity. In this context, the __acceleration__ of the aircraft is affected by the wind.

**1.4 **Our aircraft has mass and is experiencing a tangential acceleration (as well as a centripetal acceleration) What force is causing that tangential acceleration? It is a component of the centripetal force. As follows:

*Figure 1*

* 1a 1b 1c*

**1.5** In figure 1a an aircraft is in a banked turn to the right. The lift force is increased by increasing the angle of attack to ensure the vertical component Fv equals the weight for level flight. This is the load-factor which is the reciprocal of the cosine of the angle of bank. The horizontal component of lift Fc is the centripetal force which makes the aircraft turn.

**1.6** In figure 1b, seen in plan-view, the aircraft is in a banked turn to the right. Va is the air-velocity, Vw is the wind-velocity and Vg is the ground velocity. The wind blows from left to right, so that the angle of drift d is in the same direction as the angle of bank and therefore this is a down-wind turn. The difference between the direction of the air-velocity and the direction of the ground-velocity is the angle of drift. The angle of drift is also the difference between the directions of the __acceleration__ of air-speed and the __acceleration__ of ground-speed.

**1.7** Note that the total acceleration caused by force Fc is the same in both the air and the ground frames of reference (FoR). The total acceleration in each FoR is the sum of the tangential and centripetal accelerations in each FoR.

**1.8** The centripetal force Fc, produces two components relative to the direction of the ground-velocity. Force component Ft is tangential to the ground-velocity and results in tangential acceleration of the ground-speed, in this case an increase of ground-speed. The other component, the third side of the triangle, is a centripetal force Fgc which provides the curvature of the ground-track.

**1.9** In figure 1c the wind is the same as figure 1b and the aircraft has turned through 180 degrees. The aircraft is still in a starboard bank (to its right) but the angle of drift is on the opposite side to the angle of bank and this is an upwind turn. Force Ft is now opposite to the ground-velocity and causes ground-speed to reduce. (As the aircraft turned from crosswind to downwind to crosswind again, the force Ft reduced to zero at the downwind heading and then increased in the opposite direction).

**1.10** The tangential force component Ft causes the ground-speed to increase during the downwind turn (from upwind to downwind) and decrease during the upwind turn (from downwind to upwind)

**1.11** So far so good. We have explained why ground-speed changes when circling in a wind. It is not simply a case of turning in a moving air-mass. The acceleration actually involves force components acting on the mass of the aircraft, as required by Newtons laws of motion. But when the ground-speed is changing why doesn’t the airspeed change as well in the same sense? When the ground-speed increases why doesn’t the airspeed increase?

**2 The effect of wind-components**

**2.1** On the face of it, the answer is that the centripetal force Fc is normal to the air-velocity and is therefore normal to the tangential acceleration of the air-velocity and therefore has no direct effect on the acceleration of airspeed. *However, there is a tangential acceleration of ground-velocity and a component of that acceleration acts in the direction of the air-velocity, purely because of the angle of drift.* Does that cause an acceleration of air-speed? Or is there an effect which negates it? Yes, opposite to that effect, it is the rate of change of the headwind or tailwind component.

*Figure 2*

*Downwind Turn*

* 2a 2b 2c*

**2.2** Figure 2 shows a plan view of the downwind turn. The airspeed Va can be represented by the sum of a ground-speed component K and a head/tailwind-speed component H. Airspeed is then the sum of K and H. By itself this is meaningless. However, during the turn, the headwind/tailwind components are varying due to the changing wind-angle y. Compare figures 2b and 2c; During the downwind turn the headwind component H is decreasing (or the tailwind component is increasing, which has the same effect) because the wind angle y is increasing. So component H reduces while at the same time component K is increasing. *Therefore, the acceleration of airspeed is the sum of the acceleration of component H and the acceleration of component K. *

**2.3** Conversely, during the upwind turn, (see figure 3) the headwind component H is increasing (or the tailwind component is decreasing) because wind-angle y is decreasing and this tends to make the airspeed increase. Ground speed Vg reduces due to the retarding action of force Ft and therefore component K reduces which makes the airspeed tend to reduce.

*Figure 3*

*3a 3b 3c*

**3 The balance of the effects**

**3.1** Now, (read this carefully) during circling flight in a wind, ** IF** the rate of change of ground-speed component K is

**3.2** The rate of change of the headwind component H depends on the wind, the wind-angle and the aircraft rate of turn. The rate of change of ground-speed component K depends on tangential force components which depend on the drift-angle. Both H and K vary during the circle. H being at a minimum on the crosswind headings but the __rate of change of H__ is at a maximum whilst turning through the crosswind headings. K also varies cyclically during the circle, having maximum effect close to cross-wind headings. However, the calculation of the acceleration of these two variables is obviously different; so are they exactly equal and opposite? That is as far as logical reasoning can take us. However, we can apply some numbers to the problem, plot the rate of change of components H, K and airspeed V and find out. Let’s say wind-speed 10, angle of bank 30 and airspeed is initially 50.

*Figure 4*

**4 Results**

**4.1** Figure 4 plots rates of change of headwind component H, ground-speed component K and airspeed V against wind-angle during a 360 degree circle.

**4.2** The graphs confirm our reasoning so far, the acceleration of the headwind component H (blue line) and the acceleration of the groundspeed component K (orange line) are indeed greatest on the crosswind headings (90 and 270) and appear to be equal and opposite. However, the * acceleration* of airspeed (the grey line which is the sum of the two other lines) is NOT exactly zero, being slightly positive around the downwind headings (180), slightly negative around the upwind headings (0 or 360) and about zero close to the crosswind headings (090 and 270).

**4.3** So the effect of turning in a wind is that the acceleration of airspeed is not exactly zero and therefore the * airspeed* is not exactly constant. Airspeed is slightly less than average during the downwind turn and slightly more than average during the upwind turn. This is seen more clearly in figure 5 which shows airspeed and ground-speed versus wind-angle and in figure 6, which shows just airspeed, with an expanded vertical scale.

*Figure 5*

**4.4** While ground-speed (orange line) varies by plus and minus the wind-speed (+/- 10), the variation of airspeed (blue line) is much less (+ 0.5/-1.5, which is barely 2%. This is less than the speed that will be lost due to the increased load factor compared to straight flight and so is easily overlooked in normal flight operations.

**4.5** Note that airspeed is less than average during the downwind turn from wind-angle 0 to 180. This reduced airspeed will reduce the drag loading and the change of drag will reduce the speed change which caused it. The calculation includes a drag compensation factor but the overall effect is a net loss of energy seen as a loss of airspeed at constant height (in this case)

**4.6** We can infer that the effect of the changing headwind component H is slightly greater than the effect of the changing ground-speed component K.

**4.7** Does this mean that average airspeed is the same as circling in still air? No. When airspeed varies cyclically, the drag also varies. Drag is a square law, therefore the drag increase associated with a given increase in airspeed is greater than the drag reduction from the same reduction of airspeed. This can be seen in figure 6 where the starting airspeed is 50 but the average airspeed is about 49.

*Figure 6*

* Same data as fig 5 but with expanded scale*

(The effect is the same as when flying in turbulence. For a given thrust setting, the average drag load in turbulence is slightly greater than in still air and the average airspeed is slightly less).

**4.7** So, if airspeed varies cyclically, even by a small amount, when circling in a wind, then average drag will be greater and average airspeed will be less compared with circling in still air. Airspeed could be made to be exactly constant by varying thrust or accepting a variation of height. For a genuinely constant airspeed, there is a loss of height or, for a glider, a slightly increased rate of descent.

**5. Derivative method**

**5.1** Using the components K and H is a good way of visualising what is going on but it does not lend itself to a definitive mathematical solution. So far the maths has been based on a trigonometrical solution which can only be approximate because the various effects do not happen simultaneously in the calculation. There is another way of looking at this situation which is to take the triangle equation and differentiate with respect to time. This will give a value for the rate of change of airspeed Va using the rates of change of the internal angles and the rate of change of groundspeed Vg but with uniform wind Vw

Va^{2} = Vg^{2} + Vw^{2} – [2.Vg.Vw.cos z]

2.Va.(dVa/dt) = (2Vg.Vg/dt) + ~~(2.Vw.dVw/dt)~~ – [d(2.Vg.Vw.cos z)/dt] dVw/dt = 0

2.Va.(dVa/dt) = (2Vg.dVg/dt) – [(2Vg.Vw.(d cos z/dt)) + (~~2.Vg.cos z.(dVw/dt~~)) + (2.Vw.cos z.(dVg/dt))]

~~2~~.Va.(dVa/dt) = (~~2~~.Vg.dVg/dt) – [(~~2~~.Vg.Vw.(-sin z.dz/dt)) + ((~~2~~.Vw.cos z.(dVg/dt))]

Va.(dVa/dt) = Vg.(dVg/dt) + Vg.Vw.sin z.(dz/dt)) - Vw.cos z.(dVg/dt))

__dVa/dt = [(dVg/dt.(Vg - Vw.cos z)) + (dz/dt.(Vg.Vw.sin z.))/ V__ Figure 7

**5.2** All of the variables are now acting simultaneously. We can now say that the acceleration of ground-speed, dG/dt is a function of the tangential force component Fgt. The rate of turn of the ground track, dz/dt is a function of the centripetal force component, Fgc. The aircraft rate of turn, dy/dt is a function of the centripetal force Fc. If dy/dt is positive, then dz/dt is negative and since the sum y+z+d is always 180, the sum of dy/dt, dz/dt and dd/dt is zero. The rate of change of the drift angle d, dd/dt flips between positive and negative twice in 360 degrees of turn as the drift flips from right to left and back.

**5.3** This formula is used in a spreadsheet to give a graphical output of ground-speed and airspeed versus wind-angle. See figs 7 and 8. The spreadsheet works on intervals of 10 degrees of wind angle; the wind-angle divided by the rate of turn gives the time interval which is then applied to dV/dt to give the next airspeed V at the next wind-angle y. In the output the ground-speed varies by two times the wind-speed from V-W going upwind to V+W going downwind. The airspeed shows a slight variation with the speed reducing during the downwind turn and increasing during the upwind turn. The derivative result is slightly different and should be more accurate than the trig version.

**5.4 **The model assumes a powered aircraft in which thrust equals drag in straight and level flight. In a turn in still-air the drag load increases proportionate to the load-factor which is equal to the reciprocal of the cosine of the angle of bank. This results in a slightly reduced airspeed in a steady turn in still-air and it is relative to this speed that the airspeed varies during a turn in a wind. This variation of airspeed will cause a change of drag which will reduce the change of airspeed which caused it. This effect is included in the model and has the effect of moderating the rate of change of airspeed so that for example, an increase of airspeed will cause an increase of drag which will reduce the airspeed.

Figure 8

**6 Conclusions **

**6.1** When turning downwind there is a * very slight* tendency to lose airspeed or height. This is a very small effect and will only be significant when airspeed is close to the stall and the wind-speed is a large proportion of the airspeed; which may well have been the case in the early days of aviation especially during the First World War when the pressures were great to go flying regardless of the weather. This is what lead to the warnings about the dangers of the downwind turn.

**6.2** In modern times airspeeds are greater, training is better and under normal circumstances the change of airspeed in a downwind turn is going to be less than the loss of airspeed due to the increased load-factor and therefore easily over-looked. The effect will only be critical when airspeed is within 2% of stalling speed which should not be the case with correct flying technique. Therefore the biggest risk in turning downwind is pilot error caused by confusing visual cues.

**6.3** The gain of airspeed in the windward turn is again a very small effect but, in certain circumstances, there is a gain of airspeed even when thrust is zero and there is a small unbalanced drag force. This is the case for the albatrosses who are able to gain a small amount of airspeed during the windward turn and are able to minimise the negative effect of the downwind turn by flying a wing-over. This is dynamic soaring in which the smallest gain of airspeed or height in each and every dynamic soaring manoeuvre, means that they can keep flying for thousands of kilometres with minimum effort.

As Dickins’ Mr Micawber said, the difference between joy and despair is a mere half-penny!

**7** **Frames of Reference**

Some people have objected to this hypothesis because velocity, momentum and kinetic energy are frame-specific. In other words these parameters can be measured relative to any Inertial Frame of Reference (IFR) and one IFR is as good as another. An IFR is a frame of reference which is moving with uniform velocity that is, not subject to acceleration or rotation.

However, while the ground can be treated as an *approximate* IFR for the puposes of these calculations, the fact is that the ground (the surface of planet Earth) is a rotating curved surface and any object moving across the surface is subject to gravity and is moving in a curve. This means that velocity is really angular velocity and momentum is really angular momentum.

In that sense, ground-speed is real as opposed to any other speed measured relative to an imaginary IFR

**8 Calculations**

** The graphs** figures 4, 5 and 6 are produced in an Excel spreadsheet by plotting acceleration or speed against wind angle simulating a turn from 0 to 360 degrees:

**Figure 4:** acceleration versus wind-angle;

R_{H} the rate of change of headwind component H

R_{K} the rate of change of component K and

R_{V} the rate of change of airspeed V the sum of R_{H} and R_{K}

_{versus wind angle }

**Figure 5:** speed versus wind-angle;

Airspeed V

Groundspeed G

versus wind angle

**Figure 6:** speed versus wind-angle Note the expanded y axis compared with figure 5;

Airspeed V

versus wind angle

** Initial conditions are:**

Airspeed 50

Uniform Wind-speed 10

Uniform Angle of bank 30

Wind-angle 0 increasing to 360 (starting with a full headwind and turning right)

**Equations**

Ground speed G =√ (V^{2}+W^{2}-(2.V.W.Cos y))

Drift angle d = cos^{-1} ((V^{2}+G^{2}-W^{2})/(2.V.G))

Component K = G.cos d

Component H = W.cos y

Rate of change of headwind component H versus wind angle y

R_{w}= W.(cos (y+1) – cos y) m/sec/deg for wind W at wind-angle angle y (1)

Angle of bank x = 30

Rate of turn Ry = Tan x.180.g/K.pi (2)

I am not using airspeed V but rather component K. This is the result of trial and error whilst developing a dynamic soaring simulation. The justification is that K is the actual speed of the aircraft whereas V is the speed of the relative airflow.

Rate of change of headwind component H versus time at each wind-angle increment

R_{H} = R_{W}.R_{Y} (1,2)

Load factor L_{F} = 1/cos.x

Centripetal force F_{C }= m.g. L_{F}.sin x

F_{C }= m.g.tan x (3)

Ground Tangential force F_{T} = F_{C}.sin d (4)

Rate of change of component K

R_{K} = F_{T }.cos d/m (5)

R_{K} = m.g.tan x .sin d. cos d/m (3,4,5)

Rate of change of airspeed R_{V} = R_{K} + R_{H}

Time increment for each 1 degree wind-angle increment d_{T} =1/Ry (6,2)

New airspeed at the next wind-angle increment V_{2} = V_{1}+(R_{V} . d_{T})