The wind gradient theory has been around for a long time, is plausible and easy to explain. However on closer inspection it is full of holes. On this page I will explain why the current theories of dynamic soaring do not really explain what is going on.

Figure 20

1. What is wrong with the Rayleigh cycle?

The ‘classic’ theory of dynamic soaring is called the ‘Rayleigh Cycle’. It was first published in the journal Nature in 1883 as an attempt to explain the soaring of birds, particularly the case of pelicans circling up to great heights, which we now know to be thermal soaring. See figure 20

It works like this; The glider descends downwind and passes through a horizontal shear boundary into a layer of slower or stationary air. Actual speed is constant and airspeed increases by the same amount as the change of wind speed. The glider then turns onto an upwind heading, climbs back up through the shear boundary and the airspeed again increases by the same amount as the change of the wind speed. The glider then turns downwind and repeats the process. If the gain of airspeed in the wind gradient is equal to or greater than the losses due to drag in the turns, then the excess airspeed converts to height and height can be maintained or gained. It sounds plausible but there are many things wrong with this:

2. The theory is impractical.

If a bird passes from one layer of air to another, maintaining constant actual-velocity, and each layer is moving at constant velocity, then the speed of the bird relative to each layer, remains the same. There is no acceleration and no exchange of energy or momentum. The only change is a different value of airspeed and drag in each layer.

When descending downwind through a wind shear, airspeed may well increase suddenly but the increased drag load will cause the actual-speed to rapidly reduce at the same time. Therefore, actual speed will not be constant and airspeed will not increase by the amount that the wind-speed changes. The unbalanced drag-load will then cause the airspeed to reduce to the original point of equilibrium. Actual velocity will only be constant and airspeed increase, if the angle and rate of descent is increased, allowing gravity to overcome the extra drag.

When climbing upwind through a wind shear, actual velocity cannot be constant, because work is being done against gravity. To gain height, there has to be a gain of potential energy and therefore a loss of kinetic energy and therefore a loss of actual speed and airspeed will not necessarily increase. However, the loss of airspeed with gain of  height may not be as much as in still air.

In climbing and descending through the shear boundary, the change of height is the same but the actual speed is greater in the downwind descent compared to the upwind ascent. Therefore, the same amount of PE and KE is only worth a smaller speed increment at the greater downwind speed, because kinetic energy is a square law. Therefore, overall, actual speed (or height) is lost.

3. The theory does not explain how actual speed increases in the downwind turn

Actual speed increases when the downwind turn is made and this requires a force which is not explained. If the force is gravity then height will be lost. It turns out that with an angle of bank combined with the angle of drift, a component of lift is responsible for this acceleration.

However, continuous circling in a wind must lead to a small loss of height. This is because, in order to maintain constant average airspeed, the ground-speed in constantly changing and there is, in practice, a small variation of airspeed in counter-phase. The drag losses associated with an increase of airspeed always exceed the energy gains due to the reduction of airspeed because airspeed is a square law. Therefore drag losses when turning in a wind will always be greater than when turning in still air.

4. The theory is not a good model of the atmosphere

In Rayleighs model, to achieve height gain, the thin shear boundary would have to be repeated continuously as the bird climbs, which is not a good model of actual atmospheric wind gradients. Actual wind gradients are quite modest and marked shear boundaries normally only occur near to solid objects. Lord Rayleigh knew this and said as much in his 1883 paper;

• A priori I should not have supposed that the variation of velocity with height to be adequate for the purpose;’

The wind gradient theory as described by Rayleigh and developed by Lissaman assumes that between moving air masses there are improbably thin shear layers and therefore improbably steep wind gradients. It assumes that the bird can climb and descend through a shear boundary with no gain or loss of height, no exchange of PE and KE, with no turbulence and no excess drag. Although airspeed and drag may increase there is no increase in total kinetic energy and no extra force in the direction of flight to overcome the extra drag.

5. The theory is not a good model of bird flight

GPS tracking of albatrosses shows periods when the ground-speed and height are increasing at the same time. The wind-gradient theory does not explain this.

The theory does not explain what the pelicans are doing in thermal soaring. They are not losing height to gain height. They are gaining height continuously in rising air.

Nor is it a good model of albatross dynamic soaring in which the windward and leeward turns are of a distinctly different shape. The Rayleigh cycle assumes either 360 degree turns or alternating 180 degree turns to left and right; but albatrosses do not do that. In albatross dynamic soaring the amount of turn is approximately crosswind plus and minus 20 to 30 degrees and is nowhere near 180 degrees. Any angle off the wind reduces the headwind or tailwind component and reduces the wind-gradient by the same amount. At 60 degrees off the wind, the headwind component is proportional to the cosine of the wind-angle and is half the actual wind and the effective wind-gradient is only half the actual wind-gradient.

6. The theory does not reflect the practicalities of actual flight.

Few pilots have experience of descending downwind through wind-shears because we normally take-off and land into wind. My experience of taking-off and climbing in strong headwinds is that the normal pitch attitude yields a slightly greater airspeed than in still-air; 71 or 72 kts instead of 70 kts. This is sustained during the first 500ft of climb, during which the wind-speed might increase by 20 kts or so, until the aircraft is turned away from the wind. In other words, during this climb, the airspeed does not increase by the amount the wind changes. There is an incremental increase of airspeed (1 or 2 kts) and then a continuous reduction of ground-speed due to the unbalanced drag load caused by that increment of airspeed. I do not see constant ground-speed and continuously increasing airspeed as I climb up through a wind gradient.

When landing into wind, if the aircraft is over-rotated in the flare and starts to gain height with power at idle, the airspeed does not increase.  On the contrary, the airspeed rapidly reduces as speed converts to height. (This is known as ballooning). This happens despite climbing through the wind gradient. The loss of airspeed can result in a rapid loss of height and a hard landing, if power is not increased.

7. How much energy is in the wind gradient?

What is the minimum wind-gradient needed to overcome gravity? In other words, how much of a wind-gradient would be needed to increase airspeed by the same amount of airspeed that would be lost by climbing through the same height?

Suppose the wind gradient is a change of wind speed of w in a height band of h .The wind gradient is Wg = w / h  s-1

If the bird starts at airspeed v0 and height h dives through the wind gradient and increases to airspeed v1 = v0 + w in height of h, that excess speed converts to height according to the energy exchange given by m . g . h= ½ m . v12 - ½m . v02

therefore h = (v12 - v02) / 2 . g

For a gain of height of 1m, h=1 and say v0=10m/s the minimum necessary gain of speed will be

v12 = v02 + 2g

v12 = 102 + 2 x 9.81

v1 = 10.94ms-1

w = v1- v0

w = 0.94 ms-1

This means that if the gain of height is to be equal to the height lost gaining speed at about 10m/s, then the wind gradient would need to be 0.94ms-1 per metre of height.

That is 3.4km/hr per m of height.

8. What are real wind-gradients like?

Actual average wind gradients are reported to be more like 0.25 ms-1 per m of height but it depends on which model of wind gradient you use. In aviation terms the wind gradient occurs below about 700m, the wind approximately halves, say from 20 m/s to 10 m/s, from 700m down to 10m (the height at which the wind is measured at an aerodrome). The wind at 10m is considered to be representative for an aircraft taking off or landing, say down to a wing height of 1m. Watching smoke blowing over grass you will see a marked boundary layer below about half a metre. When landing an aircraft, the wind gradient, say 10 m/s, is most noticeable below about 100m that is 0.1 m/s per m. These effects will be less over water than over land due to less surface friction.

9. What are theoretical wind-gradients like?

Some authors advocate the wind-gradient theory using a two-layer wind-gradient model in which the lower layer is stationary. They then create a mathematical model in which the the wind gradient is the same as the wind and the bird gains airspeed equal to the wind with each passage through the shear boundary. The bird then loses this airspeed during each turn.

This is completely unrealistic. The usable wind gradient can only ever be a proportion of the whole wind. The stationary lower layer would need to be deep enough to accommodate a large bird in banked turn which could only occur occasionally in the lee of a breaking wave and not on a continuous basis. Any such calms in the troughs between swells will be moving with the swells and most probably downwind.

‘You pays your money and you takes your choice’

10. How much height could be gained from a typical wind-gradient?

Looking at it another way, how much height could be gained from a typical wind gradient? Consider a bird gliding at 10ms-1 in still air. At constant speed and rate of descent the bird is ‘propelled’ by gravity and has constant kinetic energy (KE) just like a vehicle freewheeling down a hill. All of the potential energy (PE) in its height is used up overcoming drag. At the end of the glide, there is no excess airspeed available for a zoom climb. Now consider the same bird gliding with a tailwind from a height of 21m to a height of 1m. Assume the wind at 21m above the surface is 15ms-1 and at 1m the wind is 10ms-1 .The wind gradient would then be 5ms-1 in 20m that is  0.25ms-1 per m of height. If the bird is able to increase airspeed by 5 ms-1 by transiting the wind gradient then a zoom climb would be possible. In order to gain the airspeed from the wind gradient, the ground speed in the dive must be constant and the angle of dive progressively more steep to overcome the extra drag.

The energy transfer from speed to height is given by

• KE=PE
• ½ mv12mv02 = mgh m is the mass of the bird v0 is the bird’s initial airspeed v1 is the bird’s final airspeed having gained 5 m/s in the wind gradient h is the height gain
• ½(152-102)=9.81.h
• h=6.37m

This would be a gain of height of 6.37m but minus drag losses consequent to the increased airspeed and any turning forces. If the bird continued downwind it would climb into an increasing tailwind and lose the speed gained in the dive. If it turned across the wind it would gain just the 6.37m. If it turned into a headwind it would potentially gain another 6.37m but again minus drag losses due to the turn. Anyway you look at it, the bird ends up short of the 21m height it started with. Therefore dynamic soaring cannot be only about the wind gradient. There must be a different mechanism involved.

11. A free lunch

Of course, on the other hand, if the bird encounters that 5 m/s as a random horizontal gust it could potentially gain the same 6.37m of height as a free lunch, but only if the energy transfer is achieved by direct transfer of momentum without an actual increase of airspeed and drag. If the bird reacts fast enough, it experiences the gust as a propulsive thrust.

Then again, negative gusts would have the reverse effect….

12. What is wrong with the Lissaman Loop?

The Lissaman Loop is an attempt to account for the drag losses not accounted for in the Rayleigh cycle and is a development of that hypothesis. It was published by the late Dr Peter Lissaman in April 2007 in a paper called Fundamentals of Energy Extraction From Natural Winds.

Figure 21

The proposition is that energy can be extracted from the wind in a downwind turn and in descending downwind and climbing upwind through a wind gradient. It works like this, see figure 21. The downwind turn is like flying around the inside of an alcove moving downwind at say 2 units of speed.

The bird begins with ground-speed 8 units, flying upwind with headwind 2 and therefore with airspeed 10.

The pin-ball albatross enters the alcove, loses 2 units of airspeed due to drag and emerges from the alcove flying downwind at airspeed 8, tailwind 2 and ground-speed 10

The bird now descends through a thin shear boundary into the still-air layer below, maintaining ground speed 10 and airspeed becomes 10, gaining 2.

The bird now flies around the inside of the stationary alcove below the shear boundary in still air, loses another 2 units due to drag and emerges with ground-speed and airspeed 8.

Now the bird climbs back through the shear boundary gaining 2 units from the wind so that ground-speed is 8 and airspeed is 10 again.

What is wrong with that?

Well, this is supposed to be energy neutral, that is it begins and ends with the same airspeed and height, although there is a loss of distance downwind during the downwind turn. However, it assumes that there is an improbably thin shear boundary and therefore an improbably steep wind gradient. It assumes paradoxically, that  climbing and descending through the shear boundary involves no change of height, no change of potential and kinetic energy, no turbulence and no losses due to drag in the shear boundary.

Descending, whether through a shear boundary or not, will in practice, always result in a loss of height and either a gain of speed or if no speed is gained then a dissipation of energy due to drag. When descending through the shear layer, 2 units are gained due to the wind shear but some energy must be lost due to drag. Ground-speed must then be less than 10 and, at the end of the turn, less than 8. In climbing up through the shear boundary there will again be some loss due to drag and due to the gain of PE during the climb, before gaining 2 units due to the wind gradient, so airspeed must be less than 10. This is nowhere near energy neutral.

Furthermore, as the descent is made at ground-speed 10 and the climb at ground speed 8 there is a further loss of distance downwind.

The Lissaman loop is supposed to be a model of dynamic soaring but it is clearly not that. In the Lissaman Loop the two turns are essentially the same with the same drag losses, but albatross clearly make their downwind and windward turns of different shapes and at different speeds and with much less than 180 degree turns.

To make the model work, Lissaman says that extreme values of G loading (5G+) and bank angles of 70 degrees are required but albatross are known to fly with low metabolic rates implying minimum effort. Also it can be shown that while 70 degree banked level turns, as required by the Lissaman Loop, require 3G, a steep ‘wingover’ can be flown with little more than 1G.

13. The Downwind Turn

Wind-gradient theories do not attempt to explain the increase in ground-velocity when turning downwind. If a glider is heading into wind, it has low ground-speed, momentum and kinetic energy. As it turns downwind, ground-speed, momentum and kinetic energy increase.

In practice this works as follows: The angle of bank creates a horizontal component of lift which acts as the centripetal force making the aircraft turn. The angle of drift caused by the wind creates a propulsive component of that horizontal component of lift. The propulsive component causes the aircraft to accelerate in the direction of its ground-velocity. However, that propulsive force is not necessarily enough to provide all of the acceleration, particularly at small angles of bank. In that case, the extra energy comes from an increased expenditure of potential energy, that is to say height, in other words an increased rate of descent.

14. Why is it thought that dynamic soaring is about the use of the wind gradient?

Dynamic soaring was first considered by Lord Rayleigh in 1883 and he asserted in an article in Nature that birds cannot soar in a uniform, horizontal wind. Bearing in mind that he was trying to explain the soaring of birds in general and specifically the flight of pelicans that had been observed circling and gaining height. People have been trying to explain dynamic soaring in terms of the wind gradient ever since. However, in 1883 mankind had not yet learned to fly heavier-than-air craft and had not discovered that the effect of wind gradients can be counter-intuitive. Rayleigh himself expressed doubt about his own idea in the article.

In 1889, in a further letter to Nature, he made the connection between his idea, the wind gradient over the sea and the flight of albatrosses.

Perhaps this misunderstanding about the role of the wind gradient has come about because the observers view of dynamic soaring is normally from the deck of a ship where the vertical motion of the birds is easy to see but the left and right turns are less easy to see.

The wind-gradient can improve the efficiency of the leeward turn by reducing the airspeed losses, without the wind-gradient being the actual source of energy. Ground effect can increase the efficiency of the windward turn by reducing drag. These reasons combined with the birds desire to achieve distance rather than height, is the reason the bird stays at low-level.

15. Gust Soaring                                                                                                                                                                  Figure 22

An alternative theory is called gust soaring but this is also unsatisfactory.  It is based on a misunderstanding of the glide polar diagram. See Figure 22.

A gliders rate of descent is related to its airspeed by the lift/drag ratio which varies with angle of attack and airspeed. The vertical speed (sink-rate) and airspeed can be plotted on a polar diagram at a load factor of one (1G). Each airspeed has a corresponding angle of attack so that lift always equals weight and drag increases with the square of the airspeed.

The airspeed for the best angle of glide (best glide speed) is indicated where the tangent from the origin touches the curve. The airspeed for minimum rate of descent (minimum sink speed) is at the apex of the curve.

The gust soaring theory says that when a bird encounters a vertical gust, it is the equivalent of  an increase in the birds sink rate, which results in an increase in airspeed according to the polar diagram. The excess airspeed can then be converted to height. See figure 22. According to this theory a bird gliding at say 15m/s (sink rate 0.68m/s) encountering a vertical gust of say 1m/s would then have a sink rate of 1.68m/s and would accelerate to the corresponding speed on the polar diagram of say 25.5 m/s and convert that excess speed to height gaining 21.7m. (Pennycuick 2002)

This is incorrect. The effect on the bird of the ventral gust is to increase the angle of attack, which effectively increases the load factor by increasing the coefficient of lift. The polar diagram is only valid at 1G and is therefore no longer valid. There is no reason why the airspeed will automatically increase to any particular value.

However, there is a mechanism that explains the gain of airspeed due to a vertical gust, the surge that the glider pilot experiences when he flies into a vertical gust like a thermal.

16. The Surge                                                                                                                                                                      Figure 23

.

See Figure 23. This involves vertical motion of the air and is therefore not dynamic  soaring. What happens is that the vertical gust causes an increased angle of attack which  causes an increase in the lift L and drag D components respectively normal to and in-line with the new relative wind direction, while the glider continues moving with its original velocity. The new resultant force R is now tilted forward relative to the gliders actual flight direction and can be resolved into two new forces, thrust T in the direction of flight which causes the airspeed to increase, and a force C at right angles which is like lift but is actually a centripetal force which causes the glider to pitch up in a curve. The vertical motion of the glider diminishes the effect of the gust and equilibrium is restored. The glider simply flies out of the gust having gained a little airspeed and actual speed due to the pulse of acceleration. This resolution of the forces on the wing is similar to how a helicopter rotor works during auto-rotation.

At an airspeed of 15m/s, a 1m/s vertical gust will cause a small increase of angle of attack and an increase of airspeed of about 0.08m/s but only briefly.

17. Thermal soaring

On the other hand, a sustained vertical component of wind velocity added vectorially to the gliders sink rate will give a sustained modification of its flight path.

If the vertical air current is sustained then the glider will achieve an identical rate of climb (minus its sink rate). For example, a vertical current of air of 1m/s will give a net rate of climb of

1 - 0.68 = 0.32m/s which, if sustained for 67.8 seconds would then gain the bird 21.7m of height.

Alternatively if the pilot wants to maintain height in such a 1m/s up-draught, he can dive to increase his airspeed such that his sink rate is 1m/s down at 1G, according to the polar diagram, which will then equal the vertical wind current.

18. Ocean Wave soaring

If the vertical air current is caused by a passing ocean wave then, for the albatross, this would be a valid method of soaring similar to hill soaring practiced by birds and glider pilots over land. Such an effect can indeed be seen when pelicans soar parallel to the shore, at constant height, along a swell approaching the shore with perhaps an off-shore breeze.

On the open ocean, waves and swells generally move downwind therefore the relative speed of the air and wave is less than the free air-velocity. A given slope of the wave surface will have a lesser effect than a stationary hill.

19. Conclusions

The Rayleigh/Lissaman cycle of dynamic soaring is incorrect because

1: It is impractical

2: It does not account for drag when passing through the shear boundary

3: It does not account for the depth of the shear boundary

4: It is not a good model of bird flight or of the atmosphere

5 It assumes that between moving air masses there are improbably thin shear layers and therefore improbably steep wind gradients, with no turbulence and no exchange of potential energy.

6: It is not energy neutral.

7: In turning downwind, an aircraft is not propelled by the wind, it is using a component of the lift force as a propulsive force to gain ground-speed to keep up with the wind. When this is not adequate there is an increased rate of descent.

8: Gust soaring is a mis-application of the glide polar curve.