December 1, 2010
The 12a’s wings each have seven pinion feathers and progressive washout. The angle of attack of each progressively distal pinion feather lessens in one degree increments toward the wing tip. The change from a membrane covered wing to feathers was made for several reasons; the main one being to gain a better glide ratio than could be expected from the membrane style wing. To enable the craft to perform at the low airspeeds needed for a foot-launched takeoff and landing, thin, highly chambered airfoils, set to a high angle of attack, are proposed.
To estimate the amount of lift that these wings could produced, each wing element (each pinion feather, etc.), is approximated by a small, trapezoidal, swept-back wing with the proper airfoil and angle of attack (see the Falcon 12a Wing Data .png). The lift production of each wing element is then determined using NASA’s FoilSim II airfoil simulator.
Given the many power and energy relationships involved and the complex geometry of the wing motions, an Excel spreadsheet---the Falcon 12a Performance Evaluation Spreadsheet---was built to keep track of the processes and perform the calculations. The spreadsheet automatically updates all related values when an input variable such as the pilot’s weight or the hydraulic system pressure is changed. If the wing or airfoil geometry is changed, however, the new values must be run through the airfoil simulator before further processing by the spreadsheet. Placing the mouse cursor over a title box in the spreadsheet will show the particular equation used and give a brief explanation of the calculation.
To simplify matters, the simulator is used to establish a “baseline lift” at 10 MPH for each wing element. All other lift values are then calculated from the baseline values. In this way, it is necessary to run the numbers through FoilSim only once for each proposed wing and airfoil configuration.
Seven-pinion design with progressive washout
Span: 47.75 feet (14.48 meters)
Wing area: 150 square feet (13.93 square meters)
Weight, total: 288 lb (131 Kg)
Pilot: 175 lb (79.5 Kg)
Aircraft: 113 lb (51.4 Kg)
Glide ratio: 20:1
7 HP Halbach Array electric motor, 1.4 lb (0.64 Kg) driven by
two 5.4 lb (2.45 Kg) Li-poly battery packs (0.9 KWH total)
Pressure: 3400 PSI
Power-stroke actuators (one for each wing): 2” (5.04 cm) diameter piston with a 1.57” (4.0 cm) stroke
Pitch-control actuators: dimensions to be determined
Mechanical coupling to the wings:
Power-stroke actuators: 3” (7.62 cm) lever arms at the shoulder joints
Pitch-control actuators: 3” (7.62 cm) lever arms at the wing roots
Power-stroke duration: 1.5 seconds
Backstroke duration: 1 second
The minimum energy required for any aircraft to keep flying at a constant altitude is that which will just compensate for the change in gravitational potential energy as the craft loses altitude during its glide. For an ornithopter, this is the minimum amount of lift energy that the wings must produce during each flapping cycle.
Given a total weight, a glide ratio and an airspeed, the potential energy change (PE) during the glide is (in the English system):
PE = weight x change in altitude
Where the change in altitude is just the airspeed / glide ratio.
We begin with an assumed glide ratio of 20:1 and an estimate of 288 pounds for the craft and pilot. The glide ratio of the typical vulture is something like 15.5:1 and that of a modern sailplane approaches 70:1 so it seems plausible that, with careful engineering, our craft could achieve a ratio of 20:1. Most of the weight estimates are pure speculation and will be refined as the design process proceeds.
For a glide speed of 20.56 ft/s (14 MPH) and the glide ratio of 20:1 we have a continuous altitude loss of:
20.56 ft/s / 20 = 1.028 ft/s.
For a total weight of 288 lbs, we get an ongoing power loss of
288 lbs x 1.028 ft/s = 296.1 ft-lb/s or 401.4 Watts.
For the above power loss, the energy loss per second is 401.4 Watt-Seconds (WS). If the stroke cycle---the power-stroke plus the backstroke---duration is 2.5 seconds, we get an energy loss per cycle of
401.4 WS/second x 2.5 seconds = 1003.5 WS or 1.0035 Kilowatt-Seconds (KWS).
This energy loss must be made up for during the 1.5 second power-stroke.
VG = VB x sqrt ((LG/2) / LB)
VB = baseline-speed
LG = glide-lift
LB = baseline-lift
For the calculation of the glide-lift (LG) see the Glide-speed Lift section that follows below.
As mentioned previously, the Falcon’s wings are approximated by substituting for each wing element, a small trapezoidal, swept-back wing.
The Mean Aerodynamic Chord (MAC) and center of lift (CL) of each trapezoid is calculated based on formulae adapted from the NASASCALE.ORG online calculator.
The MAC and the geometry of each wing element is then entered into the FoilSim II airfoil simulator to determine its lift contribution (LB) at a baseline-airspeed (VB) of 10 MPH. These lift values are then summed to get the baseline-lift generated by the entire wing. Using the baseline-lift values, lift at any other speed is obtained by simple calculation (see below).
The lift is determined for the glide-speed alone (with no flapping) and for a power-stroke initiated while moving at the glide-speed.
Since, during a fixed-wing glide, all the wing elements are moving at the same speed, the total lift produced, the glide-lift (LG), is just the sum of that produced by each wing element at the baseline-speed, increased by the following formula:
LG = (VG/VB)^2 x LB
VG = glide-speed
VB = baseline-speed
LB = baseline-lift
The whole wing swings as a unit through a predetermined arc (the sweep angle) at an angular velocity measured in radians per second. Each wing element, however, moves at a different sweep-speed depending on its distance from the shoulder joint.
To find the lift produced by the power-stroke we first determine the speed, at the center of lift, of each wing element and use that to calculate its lift. The lift values for the elements are then summed to get the lift generated by the entire wing.
The airflow over a wing element is the vector sum of the stroke-velocity and the overall velocity of the craft. The resultant vector gives the average airflow speed over the center of lift of the element. That average, in turn, is used to obtain the average lift produced by the particular wing element.
The actual sweep-speed during the stroke is not the constant, simple average, however. The stroke begins at zero sweep-speed, increases to some maximum and then slows down to zero again. Because lift is proportional to the square of the speed, the lift produced while the sweep-speed is above the simple average is disproportionately greater and consequently skews the total lift to be larger than that found by using the simple average. Never the less, for reasons explained below, the simple average for the stroke-speed was used to find the lift produced during the power-stroke:
Simple average stroke-speed (in feet per second) = the sweep-angle (in radians) x the distance (in feet) from the shoulder joint to the center of lift of the element / the power-stroke duration (in seconds).
To see how much the simple average lift differs from that produced by more realistic sweep motions, a detailed analysis of the power-stroke is made (see Falcon 12a Power-Stroke Analysis .png). As shown in the illustration, the analysis uses three different approximations for the motion. In the first case, the wing swings at the simple average speed mentioned above. In the second case, the wing’s speed changes linearly with time up to its maximum then decreases linearly back down to zero. The third approximation, the most realistic one, uses a sinusoidal sweep-speed.
The lift produced by the latter two motion-profiles are obtained, as explained in detail in the illustration, by integration of their speed curves over one-half of the sweep interval and multiplying by two. Comparing the results to the lift obtained using the simple average speed (Figure 4 of the illustration) shows that the difference is not significant (a factor of 1.026 or less) so it was decided to use the simple average speed to calculate the power-stroke lift values.
The minimum energy needed to maintain level flight is found as described previously. Next, we choose a power-stroke duration that seems reasonable and proceed to determine whether or not the lift produced by such a stroke would be sufficient to equal or exceed the minimum energy requirement. Finally, in order to find out how much hydraulic power is needed, we determine the torque required at the shoulder joints to swing the wings through that stroke. The shoulder-joint torque is found for two different cases: the glide mode by itself and for the power-stroke initiated while moving at the glide speed.
Finding the torque values requires finding the center of lift (CL) of the whole wing and its center of gravity (CG). To find the wing’s CL, the lift produced by each wing element is multiplied by the distance from its center of lift to the shoulder joint (its moment arm) and those torque values are used to obtain the CL for the entire wing. The center of lift for our wing design is found to be 9.89 ft (3.01 meters) from the shoulder joint. The center of gravity of the wing is not known at this point so a rough estimate of five feet was used.
The torque produced by the entire wing while gliding is just the distance from the shoulder joint to the wing’s CL times the total lift of the wing. This is the torque needed to hold the wings out flat while gliding. The glide-torque is found to be 1425 ft-lbs.
The power-stroke torque has two components: that which is necessary to produce lift (the “lift-torque”) and that which is necessary to overcome the inertia of the wing structures themselves (the “inertia-torque”). The total of these two torques must be applied to each shoulder joint by the hydraulic system.
The lift-torque is found in a manner similar to that used to get the glide-torque. The difference being that the vector-sum speed of each wing element is used instead of the glide-speed. The vector-sum lift is that produced by the simple average sweep-velocity added, vector wise, to the glide velocity. The lift of each element is multiplied by its lever arm distance to the shoulder joint and all are summed to get the total. As shown in the spreadsheet, the lift-torque is found to be 1876 ft-lbs.
Above and beyond the issue of lift production, it takes a certain amount of torque to simply swing a bare, uncovered wing structure through its arc. In our case, the wing sweeps through an angle of 30 degrees (pi/6 radians). If the stroke duration is 1.5 seconds, the wing rotates at an average of 20 degrees per second (pi/9 radians per second). The CG of the 75 square foot wing is located about 5 feet from the shoulder joint so the CG swings at an average speed of
pi/9 radians/second x 5 feet (radius) = 1.745 feet/second or 1.19 mph.
For a wing to obtain this average speed, requires the wing to accelerate at twice that rate, per second, or 3.49 ft/s^2. At the end of 0.75 seconds---halfway through the stroke---the CG of the wing is moving at its maximum speed of 3.49 ft/s.
The force required to produce this acceleration in a 25 lb wing is
F = ma = 25 lb mass x 3.491 ft/s^2 = 87.28 lb force
The torque needed at the shoulder joint to rotate the wing structure is, therefore,
87.28 lb x 5 ft = 436.37 ft-lb
To get the total torque required for a power-stroke, this torque is added to the lift-torque found above for a total of 2312 ft-lbs.
The total energy (in ft-lbs) produced during the power-stroke is the sum of the lift (in lbs) produced by each wing element multiplied by the distance the center of lift of each element moves during the stroke. All of this is duly accounted for in the spreadsheet under the FoilSim Lift Simulator and Energy Output sections. In the spreadsheet, the energy in foot-pounds is converted to Kilowatt-Seconds.
We see from the spreadsheet that the total lift-energy produced by the wings during a 1.5 second power-stroke is 1.56 KWS. This is more than one and a half times the 1.004 KWS of gravitational energy lost during the flapping cycle.
We also see that, with a pressure of 3400 PSI, the hydraulic system will produce 1.87 KW of power which is more than sufficient to satisfy the 1.83 KW of power needed to drive the wings through the stroke.
Furthermore, we see that the 2670 ft-lbs of torque produced at the shoulder joint by the hydraulic system is more than the total of 2312 ft-lbs required to drive the power-stroke.
Finally, we see that the 7 HP, Halbach Array motor is more than twice as powerful as needed to drive the hydraulic system pump-pressure of 3400 PSI.
If the excess lift-energy found above is turned into gravitational potential energy, it would mean a net height gain of 1.44 feet per cycle as shown in the Performance Results section of the spreadsheet.
Given a battery pack capacity of 0.90 KWH, we have sufficient energy to flap the wings about 823 times during 34 minutes of constant flapping. This will result in a total altitude gain of about 1187 feet. When the unaccounted-for energy requirements mentioned previously are determined, the battery packs can be enlarged accordingly. Doubling the battery capacity, for instance, adds only 11 lbs to the total weight of the craft.
In summary, if the machine can be built to the weight specifications assumed, it will perform sufficiently well to justify its construction. In addition, since it is expected that a large portion of the time aloft will be spent soaring, actual flight times will greatly exceed the 34 minutes of powered flight. While gliding, the wings are fixed in a horizontal position and, like a car supported by a hydraulic lift, the craft will require no energy input at all except for the small pitch-control actuators and the various electronic accessories.
It should be kept in mind, also, that the craft will gain airspeed as well as altitude with each power-stroke and so the lift numbers will improve accordingly. In short, the results of this study represent the worst-case scenario of conditions at the beginning of the flight.
During the ongoing development of the design, many of the input variables will change as needed but the bottom line is that, if the assumed weight constraints can be adhered to, it appears that a Halbach Array electric motor and Li-poly battery packs could, indeed, be used to power the Falcon long enough to achieve reasonable flight times.
To download your own copy, click here: http://www.grc.nasa.gov/WWW/K-12/FoilSim/index.htm
 Nature’s Flyers, David E. Alexander, John Hopkins University Press, 2004