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PRELIMINARY PERFORMANCE CALCULATIONS
TABLE OF CONTENTS
The Excel Spreadsheet has weight and balance calculations at * Master Plan
SPECIFICATIONS
NOTE: As of August 2004, as a result of survey responses, the empty weight is increased from 1700 lbs. to provide for larger wheels and tires (to 14" diameter) that are partially retractable, air conditioning system, fan clutch, and chassis fittings, to 1933 lbs. Payload weight is increased to provide for 395 lbs. for pilot and passenger, an increase of 35 lbs., plus an increase in baggage weight from 80 lbs. to 132 lbs., and an increase of 24 lbs. for fuel. The previously-designed STOL version with the added wingtips is now considered as the standard to accommodate the added weight. Gross weight is increased to 2700 lbs., an increase of 12.5%.
For performance calculations, such as weight and drag effects for flight profiles, the Dragon design uses the following specifications as goals:
An example of available powerplants is the Eggenfellner conversion of a Subaru auto engine *Eggenfellner. Its features are that it: (1) may be purchased as a "firewall forward" package; (2) the price is reasonable at about $30,000 for the supercharged version; and (3) it has simplified controls without mixture or carburetor heat or magneto switch controls. Performance is shown below.
ECONOMIC ANALYSIS WITH AERODYNAMIC DRAG
The reader can appreciate the effect of these calculations on the design when the it's realized that the power required varies as the 5th power of the speed. The intent of the Dragon and other roadable aircraft is to trade off the speed for time saved by using an autoplane as contrasted to a combination of automobile and airplane and the necessary conversion between the two modes. The Dragon will be slower than conventional fast lightplanes when cruising; however, it will be quicker for the trip on a door-to-door basis.
Cost of operation can be simply calculated from engine performance test curves and propulsor estimates. Propulsion performance can be translated into costs with figures for fuel consumption and fuel costs.
As an example, for the operation of the Dragon, one can use the Franklin O-300 data for a best cruise fuel consumption based on a specific fuel consumption rate of 0.5 pounds of fuel per horsepower. Similar data may be obtained for a variety of aircraft or automotive engines, such as a Lycoming IO-360, perhaps with liquid cooling. The preferred powerplants for now are automotive, due to maintenance requirements, such as the Atkins Mazda Wankel rotary engine or the Eggenfellner Subaru conversions.
Drag is based on wind tunnel results and extrapolating them with consideration for the streamlining, flip-tips, and so on, one may assume a maximum economy cruise lift-to-drag ratio, CL/CD|max, of 8. For cruising at a weight of 2700lbs., one would then need 338 lbs. of thrust. For the four-seat version at 3500 lbs, the drag would be 3500/8 = 437 lbs.
The cruise horsepower of a 180 hp engine, at 65% of rated power, is 117 h.p., so the consumption would be 58.5 lb./hr., or 9.75 gallons per hour, gph. (Modern electronic fuel injected engines would likely provide more efficient operation.) For fuel costs of around $2.00 per gallon, the cost of fuel would be about $ 20 per hour. Cruising speed at maximum weight would be about 120 knots, improving as fuel is burned. Fuel capacity will be 40 gallons with full baggage of 132 lbs,, providing a no-wind range of about 360 nautical miles or three hours, plus :45 for reserves. Some users may prefer to install auxiliary fuel tanks under the seat for trips without a passenger.
The drag equation for a flat plate is the simple D = Cd*S*q where Cd is the flat plate drag of unity, S is equivalent flat plate area and q is aerodynamic pressure. q uses sub-factors of half air density, equal to .0012 at sea level, and of V^2, where V^2 is velocity in feet per second, squared. For estimate calculations, the drag of one square foot of flat plate at 100 knots, or 169 ft/sec, is 29 pounds.
Using the horsepower equation where 1 hp=550 ft-lbs/sec, the result is that a square foot of surface at 100 knots requires 29*169/550 = 8.9 HP/ sq. ft..
Working backwards from cruise horsepower, the Magic Dragon can pull 117/8.9 = 13 sq.ft. flat plate area, similar to a 4 foot diameter disk. For checking, the fuselage/body has a cross-section of about 34 sq.ft. and an assumed drag coefficient (compared to similar shapes) of about 0.15, for an equivalent flat plate area of 5 sq.ft.. The flight surfaces and duct are estimated to contribute a similar amount for a total of 10 sq.ft.. Skin friction based on the wetted area, the surfaces exposed to the airstream, is estimated to contribute a negligible amount.
Another calculation determines the drag due to the wing's lifting action, called the induced drag. The formula is based on the weight per foot of wingspan, squared, and is influenced by the inverse of q multiplied by pi. For preliminary calculations for the Dragon, it is equal to:
(90^2)/25.6 pi , or 101 lbs. at 100 mph; and
101/0.36= 280 lbs at 60 mph and, at 100 knots, 76 lbs., or the equivalent of about 3 sq.ft. of flat plate area. One may refine the calculation by using the Oswald efficiency factor that would increase the drag by about 20%.
One may estimate that the horizontal tail would add another 1- sq. ft. of flat plate equivalent for a total of 13 sq.ft.. The 13 sq.ft. times the 29 pounds/sq.ft. gives 377 pounds of drag at 100 knots, the economy cruise speed.
Another simple calculation using a maximum lift/drag ratio of the usual 8:1 for conventional airplanes and the Magic Dragon as a 2700 lb. airplane at maximum weight gives 338 lbs. of drag.
The cost per pound of drag per hour would then be $ 20/377 lbs. = $ 0.053 per pound per hour. For an annual usage of 1,000 flight hours per year, this gives $ 53 per year. Although this may seem to be a small cost on this basis, one can figure the life cycle costs for flying about 1,000 hours per year, 20 hours per week, for thirty years and get a life-cycle cost of 30 * $ 53 = $ 1,590 per pound of drag.
Designers may use the StrongWare to calculate present value * Structured Problem-Solving .
As a practical example of using this trade formulae, one may examine the landing gear. If one were to pull the wheels up into their wells to the extent of the spring deflection of 6", one would reduce the almost-flat plate area by one square foot, or 29 lbs. of drag, or about 8%. One might implement this with Commercial Off-The-Shelf, COTS, electric actuators, similar to those envisioned for the wing swinging.
A simple calculation based on parasite drag comparison would show a gain of about 12 knots, if one ignored the weight of the retracting and extending mechanism and the reduction of induced drag. The life-cycle cost would then be 29 lbs*$1,590/lb. = $46,000. One could then either:
One could also envisage a simpler 'KISS' 1G method whereby locks would be inserted into the suspension as the vehicle sits ready for take-off and so limit the downward movement when the ground loads are relieved in flight. This method would not require any powered actuators to raise the wheels and would provide about half of the drag reduction. The solenoid locks would carry the spring compression and flight loads. The pilot could simply lock the wheels in the 1G position prior to take-off roll and unlock them prior to landing. One could then reduce drag by about half of that of the above method and save on the life-cycle cost and weight.
AEROJET SYSTEM EFFICIENCY
The overall aircraft system efficiency depends on the interaction of the aerojet duct flow with the fuselage and wing root. The aerojet flow is only directed along the sides of the fuselage and the rest of the fuselage only sees ambient free stream flow. The landing gear strut drag is less than conventional configurations, particularly for nosewheel drag.
The efficiency of the duct may be analyzed as a matter of aerodynamic drag. The drag is mainly a matter of skin friction. An estimate may be based on a factor of 1% of flat plate drag for the skin area wetted. The inner wall area is 36" times the length of the duct. The length is estimated by comparing the ogive curve to a line from the splitter nose to the exit plane. The length is 60-18 = 42" and the width is 33", therefore, the line is the root of the sum of the squares and is 63". Allowing for the curves with a factor of 20% gives 76" as an approximation. The area would then be 36*76/144=19 sq. ft. At a maximum operating speed of 250 mph or 367 fps, the drag of the inner walls would be 30 lbs. each for a total of 60 lbs.. The outer walls would have a height of pi*36 and a length of 76 for an area of 60 sq. ft. and a drag of 93 lbs.. This results in a total skin drag of 93+60=153 lbs. With an engine of 180 hp, the rule of thumb thrust would be 180*4=720 lbs., so the net thrust would be 720-150=570lbs. for an efficiency of 570/720=0.80. If the intake maw area is considered, then the added area would be estimated for a length of about 18" and a circumference of about 42" for an area of about 5 sq. ft.. This gives a delta drag of 8lbs. and an efficiency of 0.78.
Consideration of engine internal drag for cooling and refinement of the inner wall area for the cowling and the throttle vanes would further reduce the duct efficiency to an estimated 0.67 for a net maximum thrust of about 480 lbs.. For the aircraft system with a maximum lift/drag ratio of 8 and a gross weight of 2700 lbs., this gives a margin of 480-338=142 lbs..
Consideration should be given to the reduced skin friction drag of the duct walls, cowling, and throttles when operating at cruising speed and reduced power, with a lift/drag ratio of 8 requiring 338 lbs. of net thrust. This requires further study. Assuming that the airflow speed is roughly the same for full power static and 75% power cruising, 360 lbs. of thrust would balance 338 lbs. of drag for a margin of 22 lbs. or 7%.
The scenarios of idling power for descent and for total loss of power require further study.
LAUNCH & TAKE-OFF
Advice given to the inventor by the CalTech wind-tunnel test experts was to "...keep the c.g. about a foot in front of the wing leading edge..." due to the lift from the body with its quarter-chord center of pressure at station 55. This would translate in the current design to about station 84. This is also the approximate location of the pilot's and passenger's laps, which is most comfortable when the airplane is pitching in turbulence.
Since the Dragon's four-wheel landing gear is unusual, calculations are made to ensure that launch/take-off is feasible. The process consists of three phases: acceleration as thrust-to-weight ratio and distance; rotation as moments about the rear wheel contact point; and lift-off as lift-to-weight ratio.
Acceleration: Thrust is based on the commonly-accepted horsepower-to-thrust ratio, that is, 3:1. With 180 hp, where a 200 hp engine is assumed to lose 10% for muffling, thrust is then 3*180 or 540 lbs.. Maximum gross weight is specified as 2700 lbs.. Design speed for take-off is 57 knots, or 96 fps. With 2700/540 = 0.2g acceleration, about 4 kt/sec, For 15 seconds to accelerate to 57 kt, take-off distance at an average speed of 48 fps, distance would be 715 feet.However, aerodynamic drag and wheel drag would extend this significantly to an estimated distance of about 1,000 feet for sea level, standard day, no-wind conditions. For the four-seat version at 3,500 lbs., the engine would be more powerful, that is, about 260 hp, with corresponding thrust of about 780 lbs., less 10% for muffling.
Note: If the hybrid system is installed, then it may be used to augment thrust for a shorter take-off distance.
Note: With the four-wheel braking, stopping distance for the balanced field length may be reduced from that of conventional lightplanes.
Rotation: Moments are summed around the rear wheel contact point at station 137, wing lift at the quarter-chord point at station 105, and elevator down-lift at station 208.5. The c.g. is estimated at station 77 at gross weight with two 197-lb. people at station 84, full fuel load of 40 gallons or 240 lbs. at station 118, and baggage load of 132 lbs. at station 120. Without fuel nor baggage, the c.g. is around station 69. The wing airfoil section may be similar to a Clark Y as modified by Higgins and further modified with 0.25% slotted flaps and with slats. Lift coefficient for the wing with flaps deflected 12 degrees is estimated as 1.8 and for the elevator when deflected 30 degrees is estimated as 1.8. Aerodynamic pressure, q, at 65 mph is calculated as 10 psf. Wing area is 4*24=96 sq. ft.. Elevator area is (12'*3') = 36 sq. ft.. Fuselage lift area is based on an average width, calculated as 6 feet and a length of 20 feet and a lift coefficient of 0.2 and a center of pressure at station 60. (However, while in ground effect, the center of pressure may be farther aft.)
NOTE: Fuselage lift is based on the wing and tail moments acting on the wheel suspension springs to raise the nose slightly. The actual calculations are that the 92,040"lbs. moment on the 113" wheelbase gives forces at the springs of 814 lbs., for a deflection of 4" and an angle of more than 2 degrees angle of attack.
Moments are summed as positive for nose-down and negative for nose-up for weight, wing lift, elevator down-load, and fuselage lift, in sequence as follows:
+[2,700lbs. * (137-77)] {weight}
- [(1.8 * 128 * 10) * (137-111)] {wing lift}
- [(1.8 *36 * 10) * (137-208.5)] {elevator}
- [(0.2 * 120 * 10) * (137-60)] {fuselage}
= [2,700lbs. * 60"] - [2,304 lbs. * 26"] - [648 lbs. * (71.5)"] - [240lbs. * (63")]
= 162,000 - 59,904 - 46,332 - 15,120
= 40,664 "lbs. , or 3,387 lb. ft. moment to overcome to rotate.
Calculation of the vectored lift effect of the duct exhaust at 250 mph for a q factor of 156 on the 2-foot of combined span equivalent and 4-foot chord of the blown wing roots would add a net effect of
[-(1.8*8*156)]-[-(1.8*8*10)] = -2,102lbs.*(137-105)= -67,264 lb.-in.. = -5,605 lb-ft
With the vectored slipstream effect, the total moment is -2,218 lb.-ft.
Consideration of the wing and tailplane moments with section moment coefficients of about 0.1 would add another nose-down moment of about 1,000 lb.ft. for a net of about -1,000 lb.ft., which implies that rotation speed and take-off distance may be decreased slightly for a shorter take-off roll.
Elevator drag at a height of 78" will also contribute to the nose-up moment.
Four-seat version Rotation: Moments are summed around the rear wheel contact point at station 175, wing lift at the quarter-chord point at station 143, and elevator down-lift at station 246.5. The c.g. is estimated at station 99 at gross weight Lift coefficient for the wing with flaps deflected 12 degrees is estimated as 1.8 and for the elevator when deflected 30 degrees is estimated as 1.8. Aerodynamic pressure, q, at 59 knots or 70 mph is calculated as 11 psf. Wing area is 4*38= 152 sq. ft.. Elevator area is (18'*3') = 54 sq. ft.. Fuselage lift area is based on an average width, calculated as 6 feet and a length of 24 feet and a lift coefficient of 0.2 and a center of pressure at station 72.
NOTE: Fuselage lift is based on the wing and tail moments acting on the wheel suspension springs to raise the nose slightly. The actual calculations are that the 92,040"lbs. moment on the 113" wheelbase gives forces at the springs of 814 lbs., for a deflection of 4" and an angle of more than 2 degrees angle of attack.
Moments are summed as positive for nose-down and negative for nose-up for weight, wing lift, elevator down-load, and fuselage lift, in sequence as follows:
+[3,500lbs. * (175-99)] {weight}
- [(1.8 * 152 * 11) * (175-149)] {wing lift}
- [(1.8 *36 * 11) * (175-246.5)] {elevator}
- [(0.2 * 144 * 11) * (175-72)] {fuselage}
= [3,500lbs. * 76"] - [3,009 lbs. * 26"] - [713 lbs. * (71.5)"] - [317lbs. * (103")]
= 266,000 - 78,234 - 50,980 - 32,651
= 104,135 in.-lbs. , or 8,678 lb. ft. moment to overcome to rotate.
Calculation of the vectored lift effect of the duct exhaust at 250 mph for a q factor of 156 on the 2-foot of combined span equivalent and 4-foot chord of the blown wing roots would add a net effect of
[-(1.8*8*156)]-[-(1.8*8*10)] = -2,102lbs.*(137-105)= -67,264 lb.-in.. = -5,605 lb-ft
With the vectored slipstream effect, the total moment is 8,678 - 5,605 = 3,073 lb.-ft. The canards will contribute 200 lbs.on an arm of (175-24=151, 151/12=12.5 ft.)= 2,500 ft.lbs., leaving a deficit of 3,073 - 2,500 = 573 ft.lbs. Elevator drag of about 10 lbs. at a height of 6 ft. will also contribute to the nose-up moment with about 200 lbs. of drag at 6 ft. = -1,200 lb.ft.; therfore, there appears to be sufficient force for rotation.
After rotation, the elevator down-force is partially relieved, except for trimming load of about 200 lbs., and the fuselage lift is increased due to the increased angle of attack. With flaps up, wing lift coefficient is decreased to an estimated 1.0; fuselage lift coefficient is estimated as increased to 0.8. q for 60 knots is calculated as 12 psf. Without considering the boosted wing root lift, lift is calculated as follows.
L = [(0.8 * 120 * 12) + (1.0 * 128 *12)
L = 1,152 + 1,536 = 2,688 lbs., adequate for sustained flight and climb at 60 knots.
Initial climb and climb-out depend on the excess power available over the power required.
Initial climb rate after acceleration to best climb speed is estimated to be about 80 kts., or 135 fps. From 100 kt. cruising speed drag calculations, this reduces parasite drag by a factor of 0.64 and increases induced drag similarly by its inverse, for a total of 324 lbs. of drag. This assumes that Cl/Cd|max is about the same. The required power would be 324*135/550=80 hp. With engine 180 hp reduced by 80% efficiency to 144 hp, then this gives excess power available as 144-80 = 64 hp, which then gives a climb rate of:
64*550ft-lb*60/sec*2700lb = 782 ft/min.
For sustained cruise climb at 75% power, 135 hp @ 80% for 108 hp and 80 knots, for 108-80= 28 hp, the rate would be reduced to:
28*550*60/2700 = 342 fpm.
The four-seat version adds 800 lbs. to the figures above. The induced drag is also increased, so drag for an L/D of 8 for 3500 lbs. is 437 lbs. The 260 hp is reduced by 10% for muffling to 234 hp. At 80% fan efficiency, a further reduction leaves 187 hp. The required power is then 437*135/550 = 107 hp., leaving an excess hp of 187-107 = 80 hp. Lifting the 800 lbs. with 80 h.p. gives a rate of 80*550*60/3500=~ 754 fpm. For the cruise climb rate at 75% power, or 140 hp, the rate would be 33*33,000/3500 = 311 fpm.
For cruising, estimated lift coefficients are 0.3 for the wing and 0.1 for the body-fuselage and 0.15 for the tailplane download. The wing lift is composed of the root and the main wing, where the root is estimated to contribute about 700 lbs. and the main wing about 1700 lbs. The body lift is estimated as about 600 lbs. and the tailplane download about 300 lbs. Net lift is estimated as 2700 lbs.
DUCT AND FAN
An advantage of the duct is that the fuselage sees only free-stream velocities, except along the sides, reducing overall drag.
The front end of the engine is restrained by a four-point stabilizing harness to maintain the fan position and provide for a close fit in the fan ring. Whenever the engine must be removed, the grill, spinner, fan, and harness are removed. The plan of the duct walls is quasi-ogive. The flow is presumed to exit and follow the contour of the fuselage-body to the rear by Coanda effect. The material is composite Fiberglas for the outer walls and aluminum alloy inner walls. The inner walls are made so that they may be removed for engine servicing; the gills would be removed before extracting the inner walls to the exits. Soundproofing may be applied to the walls if desired.
The fan has nine blades with a four- to six-inch chord and pitch of about 30 degrees at the tips to 60 degrees at the roots, relative to the plane of rotation. The hub will be solid aluminum alloy and the flades are of 0.125 inch stock. The mockup is shown at * Fan. The engineering model vehicle may have adjustable, changeable flade mounting flanges. The duct has a nominal 36-inch to 38-inch diameter with constant cross-section except for reliefs at the forward bottom to accomodate the suspension springs and power steering mechanism and the engine cowling. Production Dragons may use a design criteria based on bird strikes.
The cross-sectional area at the fan hub plane is 904 to 1136 sq. in. for the 36-in. and 38-inch diameters, respectively, then expanding to the exit planes to 1,017 to 1,134 sq.in., with a velocity of about 250 mph for the static case. Assuming a horizontally-opposed engine similar to the O-300, the area decreases at the engine cowling to 466 to 668 sq.in., increasing the local velocity to about 424 to 546 mph.
AUTOMOBILE MODE WEIGHT DISTRIBUTION
The airplane mode c.g. at gross weight is at station 77. With the 200-lb. wings stowed to give a c.g. at station 79, the front wheels at station 26, rear wheels at station 137, the weight is mostly on the rear wheels. The distribution is calculated at 52% front, 48% rear. Without baggage and full fuel, the weight would be more evenly distributed. When the rear bumper weight and c.g. shift are included, there would be a slight forward shift.
ROAD TURNING RADIUS
The requirement is defined as capability to perform a 180 degree turn on a four-lane highway, where each lane is 12 feet wide.
Allowing for starting in the center of an outside lane, the radius at the centerline of the vehicle must be 1.5*12=18' = 216".
The wheel tread is designed as 54"; therefore,
the radius for the outside wheel is 216 + 27=243".
The radius of the inside wheel is 216-27=189".
The wheelbase is designed to be 106".
Solving for arctangent 106/189 gives 29.3 degrees. The vehicle will therefore be designed for a front wheel turning angle of 30 degrees.
WING BENDING AND LOCKING
The wing bending moment can be easily calculated as a function of g and/or lift.
It is assumed that the flaps are down for one case and that flaps are up for the other case. The spanwise lift distribution consists of the three factors: the body-fuselage; the wing roots that are blown by the duct exhaust; and the wings outboard of the roots. It is assumed that the body and roots carry one-third of the weight.
For the g-loading case of 6.0 gs for Aerobatic Category, to be conservative even though the design is for Utility Category that would only require 4.4 g capability, the wing load is then 2700*0.5*0.67*3=2700lbs..
For a spanwise rectangular distribution (worse than elliptical), the load may be determined most simply as the semi-span of 6'. The result is then 14,400 ft-lbs., or 172,800 inch-lbs. To estimate the loads at the hinge, where the hinge has shear planes 6" apart at top and bottom, the force is then 172,800/12= 14,400 lbs. The hinge may be fabricated of aluminum or steel material. For long-term fatigue resistance, aluminum with a strength factor of 48 kips is selected. For a cylindrical hinge pin with adequate safety margins of a factor of 2, the diameter required is about 1.25 inches.
ELECTRIC ROAD DRIVE
The aircraft engine's electrical generation will be kept intact to provide utility DC power. Very preliminary figures for 120 VAC electrical power generation initially uses data for off-the-shelf, belt-driven units:
120 lbs., 8kW, 66.7 Amps, 16+ h.p. @ 3600 r.p.m., $849 from 2kstore.com.
Electric Vehicles of America website is http://www.ev-america.com/
Assume that the batteries are charged overnight. Assuming that the two 20 kW motors are used to accelerate for 15 seconds per cycle, the 0.18 kWh drain from the batteries can be replaced by charging from the generator in 0.18/8 = 0.0225h = 81 seconds, plus allowances for efficiency losses. Longer times at lower rates will be studied to determine an optimum for the overall system in typical scenarios.
For a typical scenario of city driving, one can assume a one-mile stretch of road with traffic lights every two blocks, where there are 12 blocks per mile. Also assume that cruising speed is 35 m.p.h., or 50 feet per second, acceleration is 5 m.p.h. per second, and braking is 10 m.p.h. per second. This gives 10 seconds to accelerate that would drain the batteries, then 10 seconds for cruising and 5 seconds for braking that would charge the batteries.
Assuming a 50-50 chance of getting a red light and that they're red for the typical 30 seconds, the waiting time is 15 seconds, during which time the batteries would be charged. The scenario gives a deficit of 6 Ampere-hours per mile. With 3 batteries of 50 Ah capacity each for an 80% Depth Of Discharge, DOD, for a weight of 132 lbs., they will provide for a range of this kind of driving of 150/6 = 25 miles with a 66 lb., 4 kWh generator drawing 6+ h.p.. For the 120 lb. 8 kWh generator, the deficit would only be 3 Ah per mile, so, the range would be doubled to 50 miles. Additional batteries could be added for chronic, severe driving conditions. The scenario could be refined for operating electrical load for headlights and so on and for regenerative braking. For country driving, the range is limitted only by human capacity; for a 30-gallon supply and drawing 15 h.p. at 65 m.p.h. cruise with a specific fuel consumption of about 0.5 lb./h.p./hr. for a burn rate of 7.5 lb./hr. , gives about 1.25 g.p.h., or 52 m.p.g. for a range of 30 * 52 = 1,560 miles in 24 hours.
Ideally, one would desire enough capacity to be able to drive home from a standard mission trip, a distance of about 400 road miles. Given a mixture of city driving and country driving, between 50 and 1,560 miles, it appears that capacity is sufficient to perform this task.
The generator may be equipped with a cut-out switch during take-off and climb and approach to landing and in-flight emergencies requiring maximum power.
For extended hill-climbing on mountain roads at highway speeds, the power required for climbimg will be about 20-25 HP, leaving 15-20 HP available for maintaining speed. In severe cases on long grades, the duct exhaust vanes may be opened to provide thrust.
Users may tailor the generators and batteries to suit their particular scenarios, possibly trading baggage and/or fuel weights for greater driving capabilities.
