On this page I will explain why the current theories of dynamic soaring do not really explain what is going on. The main reason why dynamic soaring has been misunderstood for so long is that there is simply more to dynamic soaring than just the wind-gradient.
What is wrong with the Rayleigh cycle
The ‘classic’ theory of dynamic soaring is called the ‘Rayleigh Cycle’. It was first published 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 the losses due to drag in the turns, then airspeed converts to height and height can be maintained or gained. There are many things wrong with this:
1: The theory is impractical, it does not account for the depth of the shear boundary
If a bird, maintaining constant actual-velocity, passes from one layer of air to another 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 gain of energy. 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 airspeed to rapidly reduce to the original point of equilibrium. Actual velocity will only be constant if the angle and rate of descent is increased, allowing gravity to overcome the extra drag. In fact the actual velocity would need to increase to balance the loss of actual velocity in the upwind climb and that will lead to a greater loss of height than needed simply to pass through a thin shear boundary.
When climbing upwind through a wind shear, actual velocity cannot be constant because work is being done against gravity. There has to be a gain of potential energy and a loss of kinetic energy and therefore a loss of actual speed and airspeed even as the airspeed is increasing due to the wind shear. The gain of airspeed must be less than the difference of velocity between the layers. Again, the airspeed may increase suddenly but the increase in drag must cause the actual speed to reduce more quickly
If the change of height is the same in climbing and descending through the shear boundary, but the actual speed is greater in the downwind descent, then the same amount of PE and KE is only worth a smaller speed increment at the greater speed, because kinetic energy is a square law. Therefore, overall, actual speed or height is lost.
2. 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 a component of lift combined with the angle of drift, 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.
3: The theory is not a good model of the atmosphere or of bird flight
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;
The theory does not explain what the pelicans are doing in thermal soaring and is not a good model of albatross dynamic soaring in which the windward and leeward turns are of a distinctly different shape and the amount of turn is nowhere near 180 degrees.
4: 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 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 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.
How much energy is in the wind gradient?
What is the minimum wind-gradient needed to overcome gravity?
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, then the wind gradient would need to be 0.94ms-1 per metre of height.
That is 3.4km/hr per m of height.
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 useable 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.
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.
‘You pays your money and you takes your choice’.
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 is constant and the angle of dive progressively more steep to overcome the extra drag.
The energy transfer from speed to height is given by
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 less 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.
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….
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.
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 ‘windward’ alcove 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?
1: 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 an improbably thin shear boundary and therefore in improbably steep wind gradient with no change of height, no change of potential 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 is less than 10.
2: Furthermore as the descent is made at groundspeed 10 and the climb at ground speed 8 there is a further loss of distance downwind. This is nowhere near energy neutral.
3: Looking at the downwind alcove: the airspeed reduces due to drag by 2 units. That means that the ground-speed also reduces by 2 units (as the windward alcove) and then the ground-speed increases by 2 units due to the tailwind. So 8 - 2 +2 = 8 We are missing 2 units of ground-speed and the Loop does not join up.
4: 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, but albatross clearly make their downwind and windward turns of different shapes and at different speeds and with much less than 180 degree turns.
5: 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
The Downwind Turn
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. 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.
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. People have been trying to explain dynamic soaring in terms of the wind gradient ever since. In 1883 mankind had not yet learned to fly 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, 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. In 1889 he made the connection between his idea, the wind gradient 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 without the wind-gradient being the actual source of energy. The ground-effect can improve the efficiency of the windward turn and combined with the wind gradient and the birds desire to achieve distance rather than height, is the reason the bird stays at low-level.
An alternative theory is called gust soaring . 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. See Figure 22. The polar diagram relates sink rate and airspeed at a load factor of one (1G). Each airspeed has a corresponding angle of attack so that lift equals weight and drag increases with the square of the airspeed. The polar diagram tells you the sink rate corresponding to each airspeed at 1G. The airspeed for the best angle of glide is indicated where the tangent from the origin touches the curve. The airspeed for minimum rate of descent is at the apex of the curve.
The gust soaring theory says that when a bird encounters a vertical gust, it is the same as an increase in the birds sink rate, which results in an increase in airspeed according to the polar diagram, which excess airspeed can then be converted to height. 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 because of the increase of load-factor and there is no reason why the airspeed will automatically increase to any particular value. An increase of airspeed as in the example above would require a sustained force in the direction of flight. The only force available is gravity and that will require a loss of height.
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. See Figure 23 What happens is that the increased angle of attack causes an increase in the lift and drag 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 is now tilted forward relative to the gliders actual flight direction and can be resolved into two new forces, thrust in the direction of flight which causes the airspeed to increase, and a force 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 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.
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). A vertical current of 1m/s will give a net rate of climb of 1-0.68=0.32m/s which, if sustained for 67.8 s would then gain the bird 21.7m of height.
Alternatively if the bird wants to maintain height in such a 1m/s up-draught, it can dive to increase its airspeed such that its sink rate is 1m/s down at 1G according to the polar diagram, which will then equal the vertical wind current.
Ocean Wave soaring
If the vertical current is caused by a passing ocean wave then this would be a valid method of soaring similar to hill soaring practiced by birds and glider pilots over land. Such an effect can be seen when pelicans soar at constant height, along a swell approaching a shore with perhaps an off-shore breeze.
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.