This theoretical analysis began as an attempt to write a spreadsheet illustrating the wind gradient model of dynamic soaring but including drag losses. It soon became clear that the wind gradient model did not work because we have to take into account the acceleration of the bird during the windward and leeward turns. So the windward turn theory evolved. There are a large number of terms and variables and so I have not attempted to write a single equation. It was a bit like juggling with jellyfish. The albatross model is known to work so it’s data were used as a starting point. Thus, this analysis is not a proof but rather an illustration of dynamic soaring using representative data.
The data used is that of typical wind conditions and that of an albatross published in Avian Flight by J J Videler 2010
The flat windward turn and a leeward wing-over are defined in terms of airspeed, wind speed, load factor and angles of pitch roll and yaw. Many of the nominated parameters are estimated by observation of film of albatross in flight. The manoeuvre is then optimised by trial and error so that the two turns fit together with smooth rates of change of the parameters and no net height loss. Some of the nominated parameters and results are then produced in graph form.
Also on this page is an explanation of why upwind soaring is not possible by the windward turn theory but may be possible using the lee-soaring technique. At the end there are conclusions and references.
The viability of the theory depends on there being a mechanism which will enable the bird to maintain airspeed and height throughout the windward turn. That mechanism is the tendency of the airspeed to increase due to the increasing headwind component caused by the changing wind-angle, that is the rate of turn, and balancing the loss of ground-speed due to drag. So, what minimum angle of bank will provide a rate of turn which will enable the bird to just maintain airspeed and height and maximise the distance flown?
The Windward Turn
Angle of bank
The angle of bank is measured from the vertical and the centripetal acceleration it provides is horizontal.
In still air the ground-velocity is the same as the air-velocity and the tangential and centripetal accelerations are the same in both frames. The effect of the wind is to change the ground-velocity and therefore the tangential and centripetal components are different in the two frames but the total acceleration is the same.
The angle of bank x (figure 7) depends upon what centripetal acceleration is needed to achieve the rate of turn Ry. The rate of turn will be the same relative to the ground and relative to the air.
Tan x=a/g x is angle of bank
a=V2/r a is centripetal acceleration m/s2
V is tangential air-speed m/s
r is radius of turn m
Ry=360V/C Ry is rate of turn deg/sec
C is circumference of circle radius r
Tan x=V.Ry.pi /180g (3)
So for each point of the turn there is an angle of bank x which is a function of tangential speed V and rate of turn Ry.
(See Figure 8a) There is a component of the ground speed VT = G*cos d and a component of the wind-speed H = W*cos y
Both are parallel with the air-velocity
In order for the airspeed to be constant, the ground-speed component VT must reduce and the wind component H must increase at the same rate RH
In straight and level flight the force F1 would cause the airspeed V to reduce but, in the windward turn, the airspeed is constant because RH is equal to dH/dt and equal to dVT/dt. Therefore the effect of the drag force F1 according to F=ma is:
Force components under acceleration
(See Figure 8b) When there is a wind, there is a drift angle. Relative to the air-velocity, the drag force and the horizontal component of lift combine to create a horizontal resultant F3. Force F3 then resolves into different tangential and centripetal relative to the ground velocity.
In a glide, the ratio of lift to drag is the same as the glide ratio. For an albatross the ratio is 1:20 So the drag force F is equal to weight m.g divided by the lift/drag ratio L
Rate of change of the headwind component RH
Force is mass times acceleration and in straight and level flight, in still air, the drag force would cause an acceleration of airspeed (and groundspeed). Because airspeed is constant in the windward turn, the acceleration due to the drag force is equal to the rate of change of headwind component RH
RH=m.g/L.m (4 & 5)
RH=g/L (6) RH is the acceleration
L is the lift/drag ratio
This means that, whilst maintaining height in the windward turn, the rate of increase of headwind component is inversely proportional to the lift/drag ratio and independent of the mass of the glider.
Rate of Turn Ry
The rate of change of headwind component RH depends on the rate of change of the wind-angle Ry which is the rate of turn of the glider.
Ry (deg /sec) is calculated by the rate of change of the headwind component per sec RH (m/s2 ) divided by the rate of change of headwind component per degree of wind-angle Rw (m/ s per deg)
Therefore at each wind angle y
Ry=RH / Rw (7)
Ry= g/L. Rw (8)
Rate of change of the headwind component Rw versus the wind angle y
Rw is a function of trigonometry. Given air-velocity and wind-velocity there is a headwind component H corresponding to each wind angle y.
Taking a given interval between wind-angles, (say 1 degree), then between each pair of wind angles there is a corresponding change of headwind component. Therefore we get a rate of change of headwind component with respect to the wind-angle.
The rate of change of head-wind component per degree of wind-angle
Rw= (H1 -H2) / (y1-y2) (m/ s per deg) where H1 and H2 are headwind components at points y1 and y2 separated by a small angle (y1-y2)
Angle of bank x
The angle of bank at each wind-angle point in the turn
Tan x =V . Ry . pi./180g (3)
Ry= g/L . Rw (8)
Tan x = V. pi / L. Rw . 180 (9) (3 & 8) L is the lift/drag ratio.
V is tangential velocity
Rw is rate of change of head-wind component per degree of wind-angle
So for each point in the windward turn there is a wind-angle and a value of Rw, and there is an angle of bank which will provide a rate of turn and a rate of change of tail/headwind component to negate the tendency of the airspeed to reduce due to drag. This calculation is repeated for each point in the windward turn to generate a profile of angle of bank versus wind-angle. The nominated values of airspeed, wind-speed and range of wind-angles are modified by a fairly crude iterative process and the result correlates quite well with angles of bank observed in film of albatross.
Given Bird airspeed 20m/s Bird L/D 20 Wind 20m/s Wind angle 69 to 70 deg
Rw = H1 - H2 = (20*cos 70) - (20*cos 69)
Rw = 0.33m/s per deg
Tan x = V* pi / L* Rw*. 180 (9)
Tan x = 20*3.142 / 20*0.33*180
Angle of bank x = 3.06 deg
The trouble is, of course, that the angle of bank affects the load factor which in turn affects the lift/drag ratio, so that some form of optimisation process is needed to find the best angle of bank and load factor at each point of the turn. One way to do this by trial and error, is to pick an arbitrary range of wind-angles where the angle of bank is less than say 10deg (load factor =1.02). This seems to fit with what the albatross does in terms of minimum effort
Top of page
In a theoretical windward turn manoeuvre how much height or distance can be gained?
In a straight glide a bird maintains airspeed and ground-speed and therefore maintains kinetic energy but loses height and potential energy according to its glide ratio. In the windward turn the albatross maintains height and airspeed as shown previously but loses ground-speed and therefore loses kinetic energy. The acceleration of ground-speed due to the drag force is the same as the change of ground-speed due to the bird turning relative to the wind. That change of KE is equivalent to an amount of potential energy and when converted to an equivalent height and multiplied by the glide ratio, gives the distance flown. Note that the bird does NOT gain potential energy because it does not gain height. The KE is lost to drag equivalent to an amount of PE. So we can sample potential and kinetic energy states at each end of the windward turn to see what the equivalent gain of PE and therefore gain of height or distance would be.
In Figure 9, the lower diagram (a plan view) W is the average wind at the height of this manoeuvre. There are tail and head-wind components, a tailwind at G and a headwind at H. The bird’s kinetic energy depends on it’s airspeed plus or minus the component of the wind in line with it’s flight path. Therefore, in maintaining airspeed, kinetic energy in the ground-velocity, equivalent to potential energy, is expended due to the drag force.
The wind components along the bird’s flight path are derived from the wind W factored for the dive and climb angles a and b (not shown and only ever very small) and for the wind angles c and d . The wind angles are the horizontal angles between the wind and the bird’s flight path.
Wind component at G wg = W cos a. cos c
Wind component at H wh = W cos b. cos d
Airspeed is V1 at the beginning and end of the turn. The kinetic energy at the beginning of the turn depends on airspeed plus wind component and at the end of the turn on airspeed minus wind component. Therefore, in maintaining airspeed, kinetic energy in the wind is converted to potential energy and expended overcoming drag. The angles of descent and climb make no difference to the energy balance except in the way they affect the wind components
Total speed (inertial speed) at G is V1 + wg Total speed at H is V1 - wh
PE + KE @ G = PE +KE @ H
0 + ½ m (V1 + wg )2 = m.g.h1 + ½ m (V1 - wh )2
m.g.h1 = ½m (V1 + wg )2 - ½m (V1 - wh )2
h1 = (V12 + 2 V1 wg + wg2 ) - (V12 - 2 V1 wh + wh2 ) / 2g
h1 = (V12 + 2 V1 wg + wg2 - V12 + 2V1wh - wh2) / 2g
h1 = ( 2 V1 wg + wg2 + 2V1wh - wh2) / 2g (1)
If wg is equal to wh this becomes h1 = 2V1w/g (2) h1 is the equivalent gain of height V1 is the airspeed w is the wind component g is gravity
The equivalent height gain is proportional to the product of the airspeed and the wind component.
For example: say if the bird dives and climbs at 5o at 12ms-1 at 60o to the wind and the wind is 10ms-1 with no wind gradient; g is 9.81ms-2
Wind component at G and H w = W cos a. cos c
w = 4.98ms-1
From (2) h1 = 2V1w/g
h1 = 12.18m
The height gain with a uniform wind will be 12.18m . This is much greater than the potential height gain from just the energy in a typical wind gradient.
If we allow a margin of 3m at the end of the turn and a glide ratio of 20 to 1 the distance flown is
This compares quite well with observed times (from film) of 10 to 15 secs.
Note that this energy exchange will work in a uniform horizontal airstream. Also, the bird does not need to make any significant changes to it’s airspeed. It is not trading airspeed for height . The manoeuvre extracts energy directly from the bird’s ground-velocity and therefore from the wind-velocity.
Note also that this is a simplified theoretical explanation of how the bird can maintain height and airspeed in the windward turn. In practice there can be a trade off between height and airspeed or a gain of height or airspeed at the expense of an increased rate of loss of ground-speed and of ground-distance flown. This is shown in the Data section in which GPS data from tracked albatross suggests a slight gain of airspeed in the windward turn.
It makes sense that the height gain is proportional to the wind speed, but it seems paradoxical that the height gain depends on the bird’s airspeed when the bird is not sacrificing any airspeed during the manoeuvre. The answer is simply that KE is a square law and increasing either airspeed or wind-speed will increase the total KE value of the product.
What is the effect of the wind gradient?
You are probably thinking, hang on a minute, if the bird is gaining height energy in the windward turn by increasing it’s headwind component by turning towards the wind, then surely, as it gains height, it will gain additional energy from the wind gradient as the wind component increases with height?
No, the effect of the wind gradient is not to increase the wind speed from the surface upwards but rather to reduce the wind speed from the free airstream velocity at height downward to the reduced wind at the surface. This is an important distinction.
The kinetic energy available is proportional to the difference between the wind component (squared) at the beginning of the turn and that at the end of the turn. In a uniform wind, the wind components at the beginning and end of the windward turn are the same. If the tailwind component is +w and the headwind component is -w the difference between +w and -w is 2w. However, the effect of the wind gradient is to reduce the wind at the lower level to less than +w. If the tailwind is less than the headwind the difference is less than 2w. Therefore if the bird gains height, there is a reduced wind at the beginning of the windward turn compared to the end of the turn, this will work against the efficiency of the dynamic soaring manoeuvre.
What happens if there is a wind gradient?
For example: say if there is a wind gradient and the wind at the beginning of the turn is 8ms-1 and at the end of the turn is 10ms-1
Wind component at G: wg = Wcos a. cos c wg =8.cos 5.cos 60 wg=3.98
Wind component at H wh = Wcos b. cos d wh =10.cos5.cos60 wh=4.98
From (1) h1 = ( 2V1wg + wg2 + 2V1wh - wh2) / 2g
The height gain with a wind gradient is only 10.50m compared with 12.18m in a uniform wind, showing that a wind gradient reduces the height gain in the windward turn. This is because the wind at the end of the manoeuvre is close to the free air-stream velocity and the wind at the lower height at the beginning of the manoeuvre is reduced by the wind gradient.
The effect of the wind gradient, compared to a uniform wind, is to reduce the height gain in the windward turn because the beginning of the turn is lower than the end of the turn but only if actual height is gained. However, it appears that the albatross does not gain much height in the windward turn, so the only effect of the wind gradient is that the wind is less strong than at height and there is less total energy available.
This does not matter as long as a sufficient wind blows continuously. (Within the leeward wing-over turn the effect of the wind gradient is to offset the loss of airspeed and therefore improve the efficiency of the turn.)
Therefore for maximum efficiency overall, a modest wind-gradient during the leeward turn and no height change during the windward turn, would seem to be an advantage. Dynamic soaring works so well over the oceans because the wind is stronger and steadier and the wind gradient is presumably more consistent and less like the random turbulence found over land. The birds can therefore manoeuvre in a more systematic way.
1. In a Windward Turn, the albatross can maintain airspeed and height because the tendency to lose airspeed due to drag is balanced by the tendency to increase airspeed caused by the changing tail/head-wind component. These changing tail/headwind components are due to the bird turning relative to the wind.
2. The bird can maintain height because, rather than losing potential energy due to losing height, instead it loses an equivalent amount of kinetic energy due to deceleration of ground-speed.
3. The loss of ground-momentum is caused by the unbalanced drag force due to maintaining height.
4. The efficiency of the windward turn is improved by flying in ground-effect.
Top of page
The Leeward Turn
How does a wing-over differ from a level turn?
When an aircraft in level flight banks, the lift force is tilted off the vertical and the horizontal component provides the centripetal acceleration causing the aircraft to turn. (See figure 10) The total lift force must be increased (normally by increasing the angle of attack) in order that the vertical component can balance the weight of the aircraft and prevent a loss of height.
Figure 11 is a simplified diagram of the forces in a wing-over. In 1G flight, lift equals weight. Normally in a level turn lift is increased so that the vertical component equals the weight and the horizontal component provides the turning effect.
In dynamic soaring the leeward turn is a wing-over in which the vertical component of lift partly supports the weight and the horizontal component provides the centripetal acceleration which makes the bird turn. This means that lift is not increased and remains approximately equal to weight. Drag is not shown.
Figure 12 is a plan view of the leeward turn with the wind coming from the top. It can be seen that when the wind is a large proportion of the airspeed, the drift angle ( the angle between the aircraft heading and its actual track across the ground) is also large. The horizontal component of lift is normal to the direction of the air velocity and combines with the drag to create the horizontal resultant.
A component of the horizontal resultant is propulsive in the direction of the ground velocity which provides the acceleration, allowing the bird to ‘keep up with the wind’ as it turns downwind. The centripetal component of the horizontal resultant provides the centripetal acceleration which creates the curved path relative to the ground.
It should be noted that without the propulsive component of lift, only gravity is available to provide the ground acceleration, with consequent loss of height.
Acceleration of airspeed is zero
In figure 13 it can be seen that airspeed consists of a ground-speed component and a wind-speed component (which could be a headwind or a tailwind). The headwind component reduces as the wind angle increases. Acceleration of airspeed is the sum of the acceleration of the ground speed component and the acceleration of the wind-speed component (which could be a reducing headwind or an increasing tailwind). When these two accelerations are equal and opposite the airspeed is constant.
The math is the same as the windward turn except that in the leeward turn, the bank angle and the drift angle are on the same side, so that the horizontal resultant has a propulsive component in the direction of ground-velocity.
Although it seems as if drag should reduce airspeed, in fact the drag force only exists on its own when the wings are level. At all other times when there is an angle of bank, the horizontal force acting on the mass of the bird is the vector sum of drag and the horizontal component of lift. Airspeed remains approximately constant because of the balance of accelerations; a combination of increasing ground-velocity and reducing headwind component.
What is the propulsive force?
Figure 14 shows triangles of velocity at sample points in the leeward turn. The turn is a wing-over and therefore there are vertical and horizontal components. The wind-angle y is the horizontal angle between the air-velocity and the wind velocity. The middle triangle shows the one point where all the velocity components are horizontal and shows the direction of the lift vector and the horizontal components of the lift vector one of which, Pf, provides acceleration of the ground-speed Gh.
The leeward turn is assumed to have the same range of wind-angles as the windward turn.
For each point, given airspeed V , and pitch angle p , the horizontal components of airspeed Vh ,
Vh =V*cos p
and given the wind speed W and wind angle y, the horizontal component of actual speed Gh
Gh =sqrt(Vh2+W2-(2*Vh*W*cos y)
The drift angle d is then
W2 =Vh2+Gh2-(2*Vh*Gh*cos d)
cos d = (Vh2+Gh2-W2)/(2*Vh*Gh*)
Assuming a load factor of 1G The propulsive force Pf is
Pf = m*g*sin r*sin d m is mass r, d are angles of roll, drift
During the leeward turn the bird climbs and descends in a wing-over. The propulsive force then has vertical and horizontal components.
The horizontal component Pf cos p provides acceleration of ground speed.
The vertical component Pf sin p increases the potential energy (height) gain for a given loss of kinetic energy. That extra potential energy is used anytime the bird is losing height to overcome drag or reduce the angle of descent.
This is not a proof, it is merely an illustration of what a particular albatross could do with a particular wind and drift angle.
What is the centrifugal force in the leeward turn?
Using a nominated airspeed 11m/s , duration of turn 2.0 secs, angles of climb and descent (same as the range of wind-angles 40 deg) and mass of the bird 10 kg we can estimate the centrifugal force in the middle of the turn. This comes out slightly less than the weight of the bird and the balance is the vertical component of lift. Using this value of lift and an assumed load-factor of 1.1 the angle of bank comes out at about 78 deg
This means that an albatross can fly a low-stress wing-over at 78deg angle of bank, Load-factor 1.1G , Airspeed 11m/s taking 2.0 seconds (observed performance from film). This in turn, means that in a leeward turn wing-over the albatross can turn downwind and gain the necessary downwind velocity without losing height, other than a small drag penalty which is recovered in the windward turn. (A level turn at 78deg would require a load factor of 4.8G, which is high-stress jet-fighter territory!)
Return to Leeward turn
Between each turn there is a roll reversal with associated drag losses. The gain of height in the leeward turn will have to compensate for this. Also a little height can be gained by briefly increasing the angle of bank and rate of turn at the end of the windward turn. The penalty is a small loss of the distance flown in the windward turn.
Limits on the energy gained
The amount of energy gained in this way will be limited by the amount of time the bird can spend in the wing-over in a steeply banked attitude (about 2 to 5 seconds) and this will depend on how much excess height or air-speed can be gained first.
The amount of turn in the leeward turn has to be the same as that in the windward turn in order to maintain an approximately straight average course relative to the wind and a suitable range of wind-angles is found to be not less than about 50deg and not greater than about 130deg (probably less than this in nature). In effect, the bird is flying a crosswind heading +/- 40 deg. Even with this reduced range of wind-angles, the bird can still gain sufficient momentum because the drift angle and the propulsive force, is greatest in the middle part of the leeward turn.
The effect of the wind gradient
In the wing-over, airspeed converts to height and back to airspeed, as the bird turns. For simplicity the airspeed is shown as constant. In practice the airspeed will reduce and then increase by a small amount as height is gained and lost.
If there is a wind gradient, the wind will change with height and the effect of this will be to reduce the change of airspeed. During the climb the reducing airspeed will be offset by the increasing headwind. During the descent the increasing airspeed will be assisted by the decreasing tailwind. Thus the wind-gradient, by adding a little airspeed energy, may improve the efficiency of the wing-over by increasing the duration of the turn. This, combined with the reduction of drag due to the ground-effect, may be the reasons why the albatross stays close to the surface.
(For simplicity I have left the wind gradient out of the calculations as the results seem to work without them)
Also bear in mind that according to GPS data logging, dynamic soaring does not appear to involve 180 degree turns and so the effect of any wind gradient will be diminished by the minimum wind-angle achieved between windward and leeward turns.
The bird can maintain height using dynamic soaring but can only do so at the expense of drifting downwind due to the consistently large drift angles. This is probably why albatross circumnavigate the Antarctic continent with the prevailing wind and fly downwind around the high pressure patterns of the North Pacific
1. In a leeward turn, aerodynamic forces combined with a large angle of bank and a large angle of drift provide a downwind acceleration enabling acquisition of KE without gaining airspeed or losing PE. Provided that the bird gains more KE in the leeward turn than it loses in the windward turn, then it can maintain or gain height.
2. Airspeed is approximately constant due to a balance of accelerations. The ground-speed component increases at the same rate as the wind component decreases.
3. The efficiency of the leeward turn is improved by the wind gradient but does not depend upon it.
4. The Windward Turn Theory helps to explain some aspects of bird physiology such as tube nostrils and high aspect ratio wings and of behaviour such as the direction of large scale flight patterns the shape of the dynamic soaring manoeuvre and flight close to the surface.
5. The bird is both pilot and airframe combined and as such has a relationship with aerodynamic and physical forces which are mostly beyond human experience.
Top of page
Can an albatross use dynamic soaring to travel against the wind?
Not by using the Windward Turn Theory. If the albatross is only just maintaining average height, then it always has a large drift angle in both turns and will therefore always be drifting downwind but what if the bird gains some height ? In the time it takes the bird to complete the dynamic soaring manoeuvre, the wind will have pushed the whole thing downwind but can the bird convert the height gained into distance by gliding into the wind at reduced ground speed and gain ground overall?
Say if the airspeed, V=12 Wind, W=10 wind angles y = 45 Wind component Wc = Wcos y L/D ratio,L=20
Height gain in the windward turn
Drag loss in the windward turn say 10 secs at 12m/s
Height loss in Leeward Turn about 2m taking about 2.0 secs
Total height gain
Total time 12secs
Distance lost due to the wind
Dw=10*12 = 120m
Rate of descent
Rod=12/20 = 0.6m/s
Time to lose 9.3m at 0.6m/s
T = 8.4/0.63 = 15.5sec
Distance flown upwind
Du=15.5*2 = 31m
The bird can fly upwind 31m but has lost 120m downwind so even if the bird gains height in the windward turn it cannot use the height to gain distance upwind. If the bird wants to maximise distance flown rather than gain height then it will not have any spare height at all and it can only maintain height by flying crosswind headings plus or minus about 40deg with large downwind drift angles.
But even if the bird cannot fly directly into the wind, surely it can tack upwind like a boat can?
Unfortunately not using the windward turn theory. It only works on crosswind headings plus and minus about 40 dgrees. To maintain height, you need the minimum wind. Because of the wind, you now lose ground downwind relative to your starting point but you make crosswind distance. The resolution of the crosswind and downwind velocities gives the track made good (TMG) and a minimum downwind drift angle (DDA).
To maximise the crosswind distance the bird makes the windward and leeward turns in opposite directions. It is not possible to do the dynamic soaring manoeuvre with 180 deg turns.
The bird cannot extend the windward turn because it cannot maintain height and it cannot extend the leeward turn because it cannot gain any more ground momentum. It cannot shorten the beginning of the turns because they would no longer join up in terms of speed and height. Can it reverse direction after each dynamic soaring manoeuvre? No, it cannot make 180 degree turns and any turns linking opposite handed dynamic soaring manoeuvres will lose height. Of course the bird cannot fly directly downwind either, but it can tack downwind if it has a particular objective.
Alternatively if the bird can find some vertical component in the wind for example on the upwind side of a swell or a ship, then it might be possible to make distance against the wind but this would not be dynamic soaring and of course a swell will normally be moving with the wind.
Have albatross been seen dynamic soaring upwind?
There is anecdotal and data logged evidence of upwind soaring by albatross. This is explained later and should be characterised by roll reversals during the leeward and windward turns.
It is not possible to soar upwind using the windward turn theory because of the downwind drift angle but there is an optical effect which can lead the observer into thinking that they have seen albatross dynamic soaring upwind. There is an explanation for this optical illusion and it is down to those pesky triangles of velocity again!
See Figure 18. Here is a diagram showing how the bird appears to fly upwind. The ship velocity is CB, the wind velocity is AB and the wind velocity relative to the ship is AC . A glance at the flag at position C suggests the ship is steaming approximately into the wind. The bird is seen dynamic soaring along path DH relative to the ship (average direction DG) and is actually keeping up with the ship. The observer only sees the bird following path DH relative to the ship. He cannot see the motion described by triangle DEF. The observer on the ship concludes that the bird is making distance upwind.
In reality the bird is flying the dynamic soaring pattern along a mean air-velocity DE (that’s the path relative to the air). Applying the wind velocity EF (same as AB) to that air-velocity, the bird’s track-made-good (path relative to the ‘ground’ ) is DF. The track-made-good resolves into components DG, lateral to, and GF parallel to the motion of the ship. Velocity component GF means the bird keeps up with the ship. Angle d is the bird’s actual drift angle while angle tmg is the track-made-good relative to the wind. This angle is greater than 90 degrees and the bird is actually losing distance downwind.
Top of page
How can the albatross soar upwind?
It is not possible to soar upwind using the windward turn theory. It is a mechanism for the bird to gain and use ground-speed at nearly constant airspeed and the geometry of the windward and leeward turns means that there will always be a fairly large downwind drift angle. It only works when the wind-speed is a large proportion of the bird’s airspeed. On the other hand the manoeuvre is inherently low-G (small load-factor) and therefore requires minimum effort from the bird.
However, there is anecdotal and data-logged evidence of upwind soaring but with a suggestion of higher metabolic rates on upwind legs compared with downwind or cross wind-soaring. This suggests that a different technique is being used to achieve upwind soaring and if the bird is not actually flapping then one reason for this greater effort may be higher-G manoeuvring.
To glide upwind without losing height, the bird must gain airspeed greater than the wind and excess height which it can sacrifice. Before the upwind glide commences there must be a force, other than gravity, in the direction of flight to provide the acceleration of airspeed.
The lee-soaring model provides the force by a process similar to auto-rotation. In normal 1G flight at the best lift/drag ratio angle of attack, the total aerodynamic force resolves into lift and drag respectively normal to and opposite to the direction of flight. If a vertical gust causes the angle of attack to increase, the aerodynamic force is increased and tilted forward of the aircraft vertical and then resolves into lift and thrust. The aircraft accelerates forward and upward (the surge experienced by glider pilots flying into rising air). This motion reduces the angle of attack and the aircraft returns to a state of equilibrium so that the effect can only be brief.
See Figure 18. A similar effect can occur if the albatross catches a horizontal gust or penetrates a shear boundary whilst in a steep angle of bank, during the leeward wing-over turn. The sudden increase in angle of attack will allow a component of aerodynamic force to act as a pulse of thrust instead of drag and will be inherently high-G. Having gained an increment of airspeed, the bird then drops below the shear boundary and reverts to normal angles of attack.
The bird can use the excess airspeed in any of three ways:
1 - It can gain height.
2 - It can continue the turn and drop below the shear boundary in the downwind trough and gain distance downwind.
3 - It can roll-reverse the direction of turn and drop below the shear boundary into the upwind trough and gain distance upwind, gliding as far as the next wave crest. If the swell is deep enough to create an air-flow separation in the troughs between wave crests, it may be possible to find still air, although the swells may be moving downwind.
The increased load-factor due to the gust and the reversal of turn for upwind progress will take more effort than rolling out of the turn into the downwind trough, hence the greater effort and metabolic rate.
Figure 19 shows how the aerodynamic forces change as the bird penetrates a shear boundary in a steeply banked, and initially, low G, turn. The sudden increase in angle of attack and load factor (G) changes the lift and drag components briefly into thrust and lift (which becomes a kind of centripetal force). The bird thus gains a pulse of acceleration of airspeed (and ground-speed). This model depends upon there being a gust or a wind shear which will most likely occur in the lee of a breaking wave but cannot be guaranteed especially in light winds or in the absence of steep waves.
Top of page
1: There is not enough kinetic energy in the wind gradient to account for dynamic soaring (See the wind gradient page)
2: In a Windward Turn, at constant airspeed due to a decreasing tailwind/ increasing headwind component but with reducing ground-speed, kinetic energy is expended on drag.
3: In a leeward turn, aerodynamic forces provide a downwind acceleration enabling acquisition of KE without gaining airspeed or losing PE.
4: Provided that the bird gains more KE in the leeward turn than it loses in the windward turn then it can maintain or gain height.
5: The bird can maintain height in a uniform horizontal wind rather than the wind gradient in a series of diving, climbing, S turns without excessive speed or high G
6: The wind gradient does not add energy to the manoeuvre but may improve the efficiency of the leeward turn. Ground-effect will reduce drag in the windward turn. These two effects plus the bird’s desire to make distance rather than height explain why it stays at low level.
7: Dynamic soaring probably works better over the oceans than over the land because the wind gradient is smaller and the wind is stronger and steadier
8: Birds use the Lee-soaring technique to travel against the wind with greater effort due to higher load-factors.
9: The Windward Turn helps to explain some aspects of albatross and petrel physiology such as tube nostrils and high aspect ratio wings and of behaviour such as the direction of large scale flight patterns and the shape of small scale repetitive turns. The relatively low G forces help to explain the albatross’ low in-flight metabolism
10: If the model is correct then albatross will be seen to do low, wide turns with small angles of bank from tailwind to headwind and high, steep leeward turns. Airspeed will be approximately constant and will only reduce slightly in the leeward turn. Ground speed will be gained during the leeward turn and lost in the windward turn. They will fly approximately crosswind +/- 40 degrees always losing distance downwind. and will need a wind speed of about 10m/s to maintain height. (see the graphs).
Flight of Birds... Rayleigh 1883
Mechanics of Flight... AC Kermode 1972
Avian Flight... J J Videler 2010
Natures Flyers... D E Alexander 2002
Birdflight…Otto Lilienthal trans Markowski 1889
Birds in Flight... C L Henderson 2008
Gust Soaring…. CJ Pennycuick 2002
Minimum shear wind strength….G Sachs 2005
In-Flight Measurement of Dynamic Soaring in Albatross…G Sachs et al 2010
Experimental verification of dynamic soaring in albatrosses G. Sachs et al 2013
Albatrosses...WLN Tickell 2000
Manual of Ornithology... NS Proctor PJ Lynch 1993
Ornithology... FB Gill 2007
Eye of the Albatross... C Safina 2002
Birdflight... G Ruppell 1975
Modelling the Flying Bird... CJ Penycuick 2008
Click here to return to the top of this page
Click here for the graphs illustrating dynamic soaring