This theoretical analysis began as an attempt to write a program to simulate the wind gradient model of dynamic soaring 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. The problem here is that any attempt to analyse turning flight would normally involve the whole 360 degree circle whilst it appears the albatrosses use only part of the circle - the crosswind parts. They only gain advantage from two segments of the circle, that is they fly crosswind plus and minus about 20 to 30 degrees. The other 240 odd degrees of the circle are useless to them.
For a more generalised illustration of the effect of the wind on an circling powered aircraft take a look at the Downwind Turn page. The maths is the same.
So the Windward Turn Theory evolved. There are a large number of terms and variables and so I have not attempted to reduce the maths to a single equation. It was a bit like juggling with jellyfish.
Successful dynamic soaring depends on a minimum wind velocity as well as a particular flight technique so, proving a theory of dynamic soaring is probably impossible The best we can achieve is an illustration of what an animal does given a certain wind. The albatross model is known to work, so its data were used as a starting point. The data used is that of an albatross in typical wind conditions as published in Avian Flight by J J Videler 2010. Many of the nominated parameters are estimated by observation of film of albatross in flight.
The object of this exercise is to see whether albatross flight can be modeled using the forces acting on the bird, the triangle of velocities and the equations of motion. An Excel spreadsheet is used with a single set of equations to model the windward and leeward turns. The end result depends on not just the equations but also certain parameters such as the range of wind-angles and the wind speed. By accepting a limited range of wind-angles, typically crosswind plus and minus 20 to 30 degrees, we can calculate for each wind-angle, the effect of angle of bank and angle of pitch. The angle of bank will give a centripetal force which will act with the drag force to give a rate of turn and therefore a rate of change of headwind or tailwind component. The aerodynamic forces will also control the rate of tangential acceleration, which acts opposite to the rate of change of the head/tailwind component to control the airspeed.
The motion of the bird is defined by airspeed, lift/drag ratio, load factor and angles of pitch and roll combined with the wind speed and wind-angle. The roll and pitch profile is intended to give a flat windward turn and a leeward wing-over. Using a spread sheet, different angles of bank can be tested to see which produce an airspeed ground-speed and height profile similar to those seen in nature. The important point being to test whether the result gives at least as much airspeed and height at the end of the dynamic soaring manoeuvre as at the beginning.
The following diagram shows an example of the results of the calculations. See figure 8. The horizontal axis is crosswind distance in tens of metres, an overall distance of 450m and lasting about 20 seconds. The left side is the windward turn and the right side is the leeward turn, with the bird flying from left to right, reversing the direction of turn at about 370m. The model bird has a mass of 10kg, and a lift/drag ratio of 20. The program works with a starting point of airspeed 20m/sec and height 2m. The wind-velocity is a uniform 10m/s with no wind gradient or any vertical motion.
A profile of angle of bank (10 degrees left in the windward turn and 70 degrees right in the leeward turn) and angle of pitch is then applied at 10m intervals. The program uses the triangle of velocities, the laws of motion and the forces acting on the bird, to calculate the load factor, the centripetal and tangential accelerations, the rate of turn, airspeed and wind-angle (heading relative to the wind), ground speed and track, the rate of climb or descent and height.
The first diagram shows variation of airspeed and ground-speed against distance flown. Starting at 20 m/s the airspeed increases and the ground-speed decreases during the windward turn from 0 to 370m. During the leeward turn from 370m to 450m, the airspeed decreases as the bird gains height in the wing-over and then increases slightly to about 27m/s as the bird descends, leaving a slight gain of airspeed overall . The ground-speed decreases slightly as the bird climbs and then increases as it descends with a slight gain of ground-speed at the end.
In the middle diagram of height against distance, the height scale is expanded for clarity. The height at the beginning is 2 metres. The pitch profile gives a slight loss and then a slightly greater gain of height in the windward turn. The leeward turn is then a wing-over rising to about 5 m before descending to 3m, leaving a slight gain of height overall. Notice that from about 300m to 370m the model albatross gains both height and airspeed just as seen in the data from actual albatross GPS tracking. (See the Data section) The small kink in the line at 370m is the point at which the model albatross reverses direction.
Bear in mind that, in straight flight with a lift/drag ratio of 20, at constant speed, the model glider will descend from 2m to the surface in a distance of only 40m whereas in the windward turn the bird is able to fly 300+ m without losing airspeed or height.
The lower diagram is a plan view showing the air-path, the motion of the bird within the air mass, in blue. The heading is the same as the wind-angle with the wind coming from the top. It shows the range of wind-angles, turning left during the windward turn from 120 through 90 to 60 degrees and then turning right during the leeward turn from 60 through 90 to 120 degrees. The ground track, the birds actual path over the ground is shown in orange. The gap between the ends of the lines on the right is the distance lost downwind, about 200m, the effect of the wind. The same bird is is shown at two places on both lines. The bird is not drawn to scale but does illustrate the angle of drift.
These results are an illustration of how it is possible for an albatross to dynamic soar in a uniform horizontal wind maintaining average airspeed and height. It requires the typical profile of roll and pitch angles which albatrosses are seen to perform. Any other profile results in a loss of airspeed or height.
The viability of the theory depends on there being a single basic mechanism, controlled primarily by the angle of bank, which will enable the bird to maintain airspeed and height throughout the windward and leeward turns. That mechanism is the tendency of the airspeed to change due to two effects. Firstly, the changing headwind component caused by the changing wind-angle, that is the rate of turn. Secondly the opposite acceleration of ground-speed caused by the aerodynamic forces acting on the bird.
The rest of this page is an explanation of the maths used to produce these results.
Angle of bank and Rate of turn in level flight
The angle of bank is measured from the vertical. The centripetal acceleration it provides is horizontal. See figure 9 in which a level turn is depicted where the vertical component of lift is equal to the weight.
Angle of bank x Figure 9
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
Tan x = v2 / g . r
r = v2 / g .Tan x (1)
Rate of turn Ry
The angle of bank x gives the rate of turn Ry
Ry = 360 . v / C Ry is rate of turn deg/sec
C is circumference of circle radius r
Ry = 360 . v / 2 . pi . r (2)
Ry = 360 . v . g . Tan x / 2 . pi .v2 (1&2)
Ry = 180 . g . Tan x / pi . v (3)
So the rate of turn Ry is a function of tangential speed v and an angle of bank x assuming a level turn in which the vertical component of lift is equal to the weight.
Load factor Lf
Load factor is the ratio of lift to aircraft weight. In straight and level flight, lift equals weight and Lf is one (1G). In a straight pitch-up the load-factor is increased by the increased angle of attack, so that in a loop for example the load factor might be three (3G). In a level turn the vertical component of lift must remain equal to weight and to achieve this the actual lift is increased by increasing the angle of attack, to correspond with the increasing angle of bank, so that the load factor is equal to 1/ cosine of the angle of bank. See figure 9. In a 30 degree banked level turn the load factor is 1.15G
In a wing-over at any given angle of bank, the load-factor can be whatever we want but the vertical component of lift will not necessarily be equal to the weight. In this case we can introduce the load-factor to equation (3) by saying that tan x = sin x / cos x and Lf = 1/cos x therefore tan x = Lf . sin x Substituting in equation (3)
Rate of turn
Ry = 180 . g . Lf . sin x / pi . v (4)
Rw Rate of change of the headwind component H versus the wind angle y
Rw is a function of trigonometry and simply depends on the aircraft heading relative to the wind, the wind-angle. Given wind-speed W there is a headwind component H corresponding to each wind angle y.
H = W . cos y
Taking a given interval between wind-angles, (say 1 degree), then in a turn 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) (5)
(m/ s per deg) where H1 and H2 are headwind components at points y1 and y2 separated by a small angle (y1 - y2)
Note that Rw , in units m/s per deg, does not depend on the rate of change of y with time. It simply depends on the heading during the turn. The time element is introduced by the aircraft rate of turn Ry in units deg per sec.
RH rate of change of headwind component with time
The rate of change of the headwind component RH in units m / sec2 is the product of the rate of change Rw of the headwind component with respect to the wind angle and the rate of change Ry of the wind-angle, which is the same as the rate of turn.
RH =Rw . Ry (6)
Looking at the units here we get:
m / sec2 = (m / sec . deg) x (deg / sec)
The Drag Force F1
The drag force depends approximately on the lift divided by the lift/drag ratio Ld. The lift is equal to the weight (m.g) at 1G. In a turn it is then multiplied by the load factor Lf which, in level flight, is inversely proportional to the cosine of the angle of bank x. Note that, for aircraft, the lift/drag ratio varies with the angle of attack which also changes the load factor but in birds the relationship is unknown, so I will assume constant L/D ratio over the small range of airspeeds and angles of attack used in dynamic soaring.
F1 = m . g . Lf / Ld (7)
The Centripetal Force F2
The centripetal force F2 in level flight (corresponding to acceleration a in figure 9) depends on the lift (weight) multiplied by the tan of the angle of bank or, in a wing-over, by the sine of the angle of bank times the load-factor
F2 = m . g . Lf . sin x (8)
The minimum angle of bank
Now we can calculate the minimum angle of bank x, which will give constant airspeed while turning through any particular wind-angle.
For simplicity we will use F1 for the tangential load and Lf .sin x for the centripetal acceleration.
In figure 12 it can be seen that airspeed is the sum of the headwind component H and the ground-speed component K. The acceleration of airspeed is the sum of the acceleration RH of component H and the acceleration RK of component K. When airspeed is constant these are equal and opposite. Therefore, while RH is a function of the rate of turn, it is the same magnitude as RK, the acceleration caused by aerodynamic force, when the acceleration of airspeed is zero.
RH can therefore be substituted as the acceleration term in F = m . a
RH = F1 / m (9)
m . g . Lf / (Ld . m) = Rw . Ry (6,7)
RH = g . Lf / Ld = Rw . Ry
g . Lf / Ld = Rw . 180 . g . Lf . sin x / (pi . v) (4)
1 / Ld = Rw . 180 . sin x / (pi . v)
sin x = pi . v / (Ld . Rw . 180 ) (10)
Rw = (H1 - H2) / (y1 - y2) (5)
So, for each wind-angle y, at say one degree increments, there is a value of Rw (m/s per deg) combined with a rate of turn represented by the angle of bank x, which will give a rate of change of tail/headwind component which will make the airspeed increase. The load factor and mass have cancelled-out and there remains a drag load, represented by the lift/drag ratio Ld , which causes the airspeed to reduce. Together they give an angle of bank x which gives the a rate of turn at which airspeed is constant.
The result gives small angles of bank which correlates quite well with the small angles of bank observed in film of albatross. The nominated values of airspeed and wind-speed can then be modified to see, for example, what the minimum wind speed needs to be and what effect the airspeed /windspeed ratio has on the useable range of wind-angles.
Given wind-angle 80 to 79 deg, bird airspeed 20m/s, bird L/D 20, wind 8m/s
Rw = H1 - H2 = (8 .cos 80) - (8 .cos 79)
Rw = 0.13 m/s per deg
Angle of bank x = 7.7 deg
This is the minimum angle of bank needed to maintain height and airspeed at one point in the windward turn, at a wind angle of 80 degrees. In practice, the GPS data suggest that the airspeed increases during the windward turn with a gain of height before rolling into the leeward turn. To achieve this, the angle of bank is made slightly greater than the minimum.
This is somewhat counter-intuitive because it means that airspeed can be constant, during a windward turn, even though there is an unbalanced drag load. Fortunately it demonstrates the viability of the Windward Turn Theory. In the windward turn it is easy to understand how the drag load causes the ground-speed to reduce even as the airspeed increases. However, in the leeward turn the ground-speed increases and it is less easy to see how this is possible given an unbalanced drag load.
To construct a model of the complete windward/leeward turn cycle, it is necessary to use the same maths applied to both turns and therefore a slightly different approach is required.
Force and acceleration components
Now things get complicated. On the face of it there are two horizontal force components: the drag force F1 and the centripetal force F2. In the windward turn it is easy to see how the drag force causes the airspeed and ground-speed to reduce, while the centripetal force provides a rate of turn which makes the head-wind component and therefore the airspeed increase. However, in the leeward turn it is necessary for the ground-speed to increase and it is less easy to see how the drag and centripetal forces make this possible. The best way to explain this is to say that the drag force and the centripetal force are really components of a single horizontal force, the horizontal resultant F3 and this is the starting point for our calculations.
Horizontal resultant F3 and components
Figure 10 is a plan view of a windward turn with the wind coming from the top of the diagram and the bird is flying left to right in a banked left turn. The horizontal component of lift F2 is normal to the direction of the air velocity and combines with the drag F1 to create the horizontal resultant F3.
We can calculate F3 using F1 and F2 and pythagoras which means that F3 is always positive.
F3 =sqrt (F12 + F22)
Forces and angles
Force F3 is not exactly in line with the ground velocity and can be resolved into components Fgt the ground tangential force and Fgc the ground centripetal force. It can be seen that when the wind is a large proportion of the airspeed, the drift angle d is also large. In albatross dynamic soaring, the drift angle is on the same side in both windward and leeward turns and is taken to have a positive sign. The sign of the angle of bank is negative in the windward turn, because it is on the opposite side to the angle of drift. It changes to positive in the leeward turn, because then it is on the same side as the angle of drift. This is important because it ultimately determines whether the tangential forces are positive or negative, that is, propulsive or retarding.
Forces are positive in the direction of flight, that is the direction of the air-velocity, therefore force F1, the drag force, has a negative sign. F2 the centripetal force also has a negative sign because of the negative angle of bank in the windward turn (opposite side to the angle of drift). The angle of drift d is the angle between the air-velocity and the ground velocity. Angle d is also the angle between the accelerations of air-velocity and ground-velocity. The angle b is the angle between F3 and the direction of the air-velocity. Angle e is the angle between F3 and the direction of the ground-velocity.
Force components relative to the ground-velocity
In the windward turn (figure 10) a tangential force component Fgt acts opposite to the direction of the ground-velocity.
Fgt = F3 cos e (11)
Note that angle e is not the internal angle between F3 and Fgt but is the external angle measured from the forward direction. Angle e has a large negative value which ensures that Fgt is negative and therefore a retarding force.
The centripetal component Fgc provides the centripetal acceleration which creates the curved path relative to the ground.
Fgc = F3 sin e
Once again Fgc has a negative sign because e is negative.
Acceleration or equilibrium
The logical inference is that under acceleration, the aerodynamic forces cause acceleration of the ground-velocity and then that acceleration causes acceleration of the air-velocity. This is somewhat different to the situation when the aircraft is under equilibrium, when the ground-velocity is the vector sum of air-velocity and wind-velocity. In fact, the only difference is that under equilibrium the accelerations are all zero.
(It should be noted that this process is the same as that which causes ground speed to change during turns in a wind by a normal powered aircraft where thrust is equal to drag. F1 is then effectively zero and F3 equals F2. However, because of the angle of drift, F3 still produces values of Fgc and Fgt which cause the ground speed and direction to change.)
Angles b, d & e
Angle b is the angle between F3 and the glider heading sin b = F2 / F3
Angle e is the angle between F3 and the ground track (the direction of the tangential acceleration)
The tangential acceleration of the ground-velocity is : Agt = F3 cos e / m (12)
The centripetal acceleration of the ground-velocity is Agc = F3 sin e / m
Sorting out the positives and negatives
Angles b, d and e are illustrated in figure 10 for the windward turn.
By inspection in the windward turn, angle e = b + d Angle d is the angle of drift and has a positive sign. Angles b and e are on the opposite side and are therefore negative.
(-e) = (-b) + (+d)
e = b - d
In the leeward turn in figure 11, by inspection, e = b - d but all the angles are on the same side and have the same sign (positive) therefore
e = (+b) - (+d)
e = b - d
So that in the calculations, we use the same formula e = b - d in both the windward and leeward turns. Angle e then has a large negative value in the windward turn (fig 10) and a small positive value in the leeward turn (fig 11). This results in ground tangential force Fgt and acceleration Agt being retarding in the windward turn and propulsive in the leeward turn
The sign of the drag force F1 is negative, opposite to the direction of flight. The value of centripetal force F2 can be negative or positive depending on the angle of bank which is measured left or right from the vertical and is negative for a windward turn and positive for a leeward turn.
Note that when the angle of bank is zero, then F2 is zero, F3 is the same as F1 and angle b is 180 degrees.
This seems rather complicated but it is necessary treat all the forces equally and account for the acceleration of ground speed. If the albatrosses did not exist you might think that this is all a bit far-fetched but this treatment of the force components is the only way to produces a realistic model of albatross flight as illustrated in the diagrams of the dynamic soaring manoeuvre at the top of the page.
Components of airspeed and ground-speed
There is a component of the ground speed K = G . cos d and a component of the wind-speed H = W . cos y Both are parallel with the air-velocity which is the sum of the two components. See figure 12
Airspeed V= K + H.
The rate of change of airspeed RV is the sum of the rate of change RH of the headwind component H and the rate of change RK of the ground-speed component K.
RV = RH + RK
In order for the airspeed to be constant, the ground-speed component K must reduce and the wind component H must increase at the same rate RH
RH = - RK
But force F1 does not directly affect component H, it only affects component K by causing a tangential acceleration of the aircraft.
(For comparison, in straight and level flight in a uniform wind, headwind component H is constant. Therefore, an unbalanced drag force F1 would cause the airspeed V to reduce but in reality it is only making component K reduce. Therefore we use component K in the acceleration calculations and not airspeed V)
Acceleration of component K
The tangential acceleration RK of the component K will be the acceleration Agt of groundspeed G reduced by the cosine of the angle of drift. The ground tangential acceleration is caused by the ground tangential force Fgt which in turn, is a component of force F3
RK = Agt . cos d
= Fgt . cos d / m
= F3 . cos e . cos d/ m.
As mentioned before, the calculation gives e a large negative value in the windward turn which makes RK a retarding tangential acceleration.
Rate of Turn
The rate of turn depends on the centripetal force F2. This gives a good result in the spreadsheet. Any other treatment of the centripetal forces will reduce the rate of turn and reduce the rate of change of the headwind component and the rate of increase of airspeed in the windward turn.
Referring to equation (4) the rate of turn is Ry = 180 . g . Lf . sin x / pi . v The term g . Lf sin x is the centripetal acceleration caused by the angle of bank in units m/s2 . In the spreadsheet this is substituted with centripetal acceleration F2 / m which depends on angle of bank x. The velocity term v is substituted with component K
Ry = F2 . 180 / m. pi . K
The story so far
The aerodynamic forces F1 and F2 are combined into a single horizontal resultant force F3 which causes tangential accelerations of the ground-velocity. This then combines with the rate of change of the head-wind component to cause acceleration of the air-velocity. The centripetal component of F3 is F2, normal to the air-velocity giving the rate of turn. Figure 13
The effect of angle of bank
Figure 13 shows the effect of different angles of bank on acceleration of airspeed at one point in the windward turn, using a modified version of the spreadsheet that produced the diagrams at the top of the page. It consists of RK , the rate of change of component K and RH , the rate of change of the headwind component H which are nearly equal and opposite versus angles of bank from 0 to -12. (Negative angles of bank in the windward turn). The grey line in the middle is Rv , the rate of change of airspeed which is the sum of the other two lines Rv = RK + RH
Rv is negative at small angles of bank but becomes positive at about 6 degrees angle of bank. This shows that 6 degrees is the minimum angle of bank needed to maintain airspeed (and height) at this particular point in the windward turn (with a certain combination of airspeed and wind-speed). Also, it shows that by increasing the angle of bank, a slightly greater rate of increase of airspeed is achieved, which will enable the albatross to control its airspeed by varying its angle of bank. This will enable the albatross to climb and descend to follow the rising and falling surface of the sea without much change of airspeed and to gain some excess airspeed prior to commencing the leeward wing-over.
The Leeward Turn
Having lost ground-speed in the windward turn, but having maintained airspeed and height, the albatross now has to recover that ground speed in the same range of wind angles and end up with at least the original airspeed and height. Figure 14 shows a plan view of the leeward turn with the wind coming from the top. Drag force F1 and centripetal force F2 combine vectorially to give the Horizontal Resultant F3. Force F3 then provides a propulsive force in the direction of the ground-tangential-velocity. Fgt = F3 . cos e This force produces an acceleration of the ground velocity Agt = F3 . cos e / m This tangential acceleration of the ground-speed then appears as a tangential acceleration RK of component K but reduced again by the cosine of the drift angle
RK = F3 cos e . cos d/ m.
The centripetal component of acceleration giving the rate of turn is
In the calculation e has a small positive value, RK is positive and therefore propulsive. Centripetal acceleration F2 /m is also positive and gives a turn to leeward
However, whereas in the windward turn the increasing headwind component tends to increase airspeed, in the leeward turn the reducing headwind component tends to reduce airspeed. See figure 14. Therefore the propulsive effect of F3 needs to be substantial and requires a large angle of bank but without an increase in drag.
Airspeed components in the Leeward turn
In figure 14 can be seen a plan view of the leeward turn with the wind coming from the top. The airspeed V consists of a ground-speed component K and a wind-speed component H (which could be a headwind or a tailwind). As the wind angle y increases, the headwind component reduces and then becomes an increasing tailwind with the same effect. Acceleration of airspeed is the sum of the acceleration of the ground speed component K and the acceleration of the wind-speed component H. When these two accelerations are equal and opposite the airspeed is constant. (The albatross tracking GPS data suggest that airspeed reduces slightly in the leeward turn balanced by an increase of airspeed in the windward turn).
The maths are the same as the windward turn except that in the leeward turn, the bank angle and the drift angle are on the same side and have the same sign, 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 as a part of the horizontal resultant F3. When the angle of bank and the angle of drift are big enough and on the same side, the propulsive component makes the ground speed increase which in turn makes component K increase. Airspeed remains approximately constant because of the balance of accelerations; a combination of increasing ground-velocity and reducing headwind component, the ground 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. In the case of albatross dynamic soaring there is a very small drag force and a large horizontal resultant which is only possible in a low G, low drag wing-over.:
Component K is part of the airspeed. Therefore tangential acceleration component RK causes tangential acceleration of the air-speed, while the other component of the airspeed, the wind component H is reducing due to the bird turning downwind. As in the windward turn, under acceleration, the aerodynamic forces cause acceleration of the ground velocity and then the ground acceleration causes acceleration of the air-velocity.
This boils down to the fact that we can use the same set of equations to describe the bird motion in both the windward and leeward turns. The only difference is that the angle of bank has a negative sign in the windward turn and a positive sign in the leeward turn
Flat turns and wing-overs
In the windward turn the vertical component of lift is equal to the weight of the bird and therefore the actual lift is increased proportionate to the load factor. The angle of bank is quite small and therefore the load-factor is small and the drag increase is minimal. However, in the leeward turn, because of the steep angle of bank, the vertical component of lift is less than the weight of the bird and we can assume (guess!) that the load factor is one and therefore the actual lift is equal to the bird weightThe rate of change of the headwind component Rw versus the wind angle y is the same at any particular wind-angle in both turns but the rate of change of the headwind component with time RH depends on the rate of turn Ry and is therefore greater in the leeward turn.
Rate of turn is proportional to load-factor but increasing the load-factor will increase the drag load. To avoid increasing the drag load, we keep the load-factor at one (1G) and the rate of turn is then only proportional to the sine of the angle of bank. Figure 15
However at 48 degrees angle of bank and load factor of 1G the vertical component of lift will be cos 48 = 0.7G which is not enough to sustain level flight. So, in the leeward turn the rate of turn can be increased at a load-factor of one without increasing the drag loading but only by doing a wing-over and not a level turn.(A level turn at 48deg would require a load factor of 1.5G, which is relatively high-stress. 70 degrees angle bank would require 3G which is jet-fighter territory!
How does a wing-over differ from a level turn?
In level flight, when an aircraft banks, the lift force is tilted off the vertical and the horizontal component provides the centripetal acceleration causing the aircraft to turn. (See figure 15) 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. This increase in load-factor will increase the drag loading.
In dynamic soaring the leeward turn is a wing-over, at a load-factor of about one, in which total lift is not increased and remains approximately equal to weight. The vertical component of lift partly supports the weight whilst the horizontal component provides the centripetal acceleration which makes the bird turn. Drag is not shown. The flight path is partly ballistic as the bird climbs and descends and with a large centripetal force and a quick rate of turn the bird has enough time to complete the turn before running out of height.
Effect of Load Factor
Figure 15 shows, in the middle, the forces during a level 45 degree banked turn at a load-factor of 1.4 (1.4G) with the vertical component equal to the weight. On the right is an 80 degree wing-over at a load-factor of one and therefore without an increase of drag but with the vertical component of lift less than the weight.
Figure 16 shows the acceleration effects at different angles of bank from 0 to 80 in the leeward turn with, a load-factor of 1/cos angle of bank as for a level turn.
Up to about 60 degrees angle of bank, both RK and RH increase gradually in opposite senses and then increase rapidly with corresponding increases of drag. The sum of RK and RH gives Rv the acceleration of airspeed, the grey line, which is negative at all angles of bank and the loss of speed increases greatly at angles of bank greater than about 60 degrees.
In figure 17 the load-factor is equal to one at all angles of bank, as for a wing-over. Note the different vertical scale. The negative acceleration in the 1G wing-over is about -2.8 m/s2 at 70 degrees angle of bank compared to -8.2 m/s2 at 70 degrees angle of bank and 2.9G in a level turn. Thus in a wing-over, large angles of bank and therefore a quick rate of turn, can be achieved without a big increase in deceleration.
Exchanging speed and height
During the leeward turn the bird climbs and descends in a wing-over. An angle of climb will give an additional retarding force equal to the weight times the sine of the climb angle.
In a climb the bird gains potential energy as it loses airspeed. However, the propulsive acceleration will reduce the airspeed losses and the height energy then converts to speed in the descent.
Vertical Acceleration Figure 17
Whenever the vertical component of lift is less than the weight, the bird will accelerate downwards. Its rate of descent will then be the sum of the effect of gravity and the effect of its angle of climb or descent. This limits the time and length of the leeward wing-over.
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. 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
Doing a wing-over is simply a way of gaining time to gain ground-speed before running out of height. 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 birds cross-country course has to be fairly straight to maintain a suitable range of wind-angles relative to the wind direction. The effects described so far diminish when the heading gets within about 50 degrees of an upwind or downwind heading. Therefore 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. A suitable range of wind-angles is found to be not less than about 50deg and not greater than about 130 deg (probably less than this in nature). In effect, the bird is flying a crosswind heading +/- 20 to 30 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
As the bird turns, it climbs and descends and airspeed converts to height and back to airspeed again. 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, may improve the efficiency of the wing-over by decreasing the loss of airspeed. This, combined with the reduction of drag due to the ground-effect, may be the reasons why the albatross stays close to the surface
However, it should be noted that the overall effect of the wind gradient is to reduce the average wind near the surface which will reduce the birds ability to dynamic soar. Also bear in mind that according to GPS data logging, dynamic soaring does not appear to involve 180 degree turns and the effect of any wind gradient will be diminished by the angle off the wind achieved at the end of each turn. Therefore, wind-gradient effects will be relatively small.
(For simplicity, and to make a point, I have left the wind gradient effect out of the calculations as the results seem to work without it.)
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
It is not possible to soar upwind using the Windward Turn Theory. Although the mechanism allows the bird to use wind energy to maintain average airspeed and height, the geometry of the windward and leeward turns means that there will always be a fairly large downwind drift angle and it only works when the wind-speed is a large proportion of the birds airspeed. On the other hand, the manoeuvre is inherently low-G and therefore requires minimum effort from the bird.
If the bird needs a wind speed equal to half its airspeed to maintain height while dynamic soaring, then its maximum angle of drift will be 30 degrees at the cross wind position and it never gets within about 60 degrees of an upwind heading. If the bird flies an air-distance of 290 m at 20m/s in 14 sec with a 10m/s wind, the distance lost downwind will be 140m. If the bird tries to recover that distance by flying upwind at a ground speed of 10m/s that distance would take 14 secs. Thats a lot of flapping!
It is not possible to do the dynamic soaring manoeuvre with 180 deg turns. The bird cannot extend the windward turn because the reducing wind-angle means that it cannot maintain airspeed and height. It cannot extend the leeward turn because again the reducing drift-angle means it loses the propulsive effect and it cannot gain any more ground momentum. The turns have to join up in terms of speed, height and wind-angle. Although it appears the bird could gain height slowly by trading airspeed for height, it would lose the advantage of ground effect and also the ability to judge whether it is climbing or descending. So, again the bird remains close to the surface.
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.
Upwind dynamic soaring is possible using a different technique and this is discussed on the Upwind Dynamic Soaring page
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 purposes 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 also moving in a curve. This means that velocity is really angular velocity and momentum is really angular momentum.
Any change of velocity of the bird is caused by forces acting on the bird, modified by the wind. It is the bird which is accelerating and not the ground, therefore the ground can be regarded as being stationary. Energy in the wind is derived from pressure and temperature gradients in the atmosphere caused by solar heating. Energy in the wind is gained because it is the wind and not the ground, which gains the momentum