SPECIFICATIONS NOTE: As of August 2004, as a result of survey responses, the empty weight is increased from 1700 lbs. to provide
For performance calculations, such as weight and drag effects for flight, the design uses the following specifications:
An example of available powerplants is the Eggenfellner conversion of a Subaru auto engine
The Wankel type engines may be modified from two-rotor to three-rotor versions to provide more power.
ECONOMIC ANALYSIS WITH AERODYNAMIC DRAG (Rev. 2010) The reader can appreciate the effect of these calculations on the design when the it's realized that the power
As an example, for the operation of the Dragon, one can use the typical data for a best cruise fuel consumption
Drag is based on wind tunnel results and extrapolating them with consideration for the streamlining, flip-tips,
The drag equation for a flat plate is the simple D = Cd*S*q where Cd is the flat plate drag of unity,
Using the horsepower equation where 1 hp=550 ft-lbs/sec, the result is that a square foot of surface at 100 knots
Working backwards from cruise thrust horsepower, the Magic Dragon can pull 117/10= 12 sq.ft. flat plate area,
Another calculation determines the drag due to the wing's lifting action, called the induced drag. The formula is based on the
(90^2)/25.6 pi , or 101 lbs. at 100 mph; and
Another simple calculation using a maximum lift/drag ratio of the usual 10:1 for conventional airplanes and the Magic Dragon
The cost per pound of drag per hour would then be $ 20/300 lbs. = $ 0.067 per pound per hour. For an annual usage of
As a practical example of using this trade formulae, one may examine the landing gear. If one were to pull the wheels up
A simple calculation based on parasite drag comparison would show a gain of about 12 knots, if one ignored the
One could also envisage a simpler 'KISS' 1G method whereby locks would be inserted into the suspension as the
AEROJET SYSTEM EFFICIENCY
The overall aircraft system efficiency depends on the interaction of the aerojet flow with the fuselage and wing root.
The efficiency of the duct may be analyzed as a matter of drag. The drag is mainly a matter of skin friction.
Consideration of engine internal drag for cooling and refinement of the inner wall area for the cowling and the
Consideration should be given to the reduced skin friction drag of the duct walls, cowling, and throttles when
The scenarios of idling power for descent and for total loss of power require further study.
Calculation of performance is based on estimated:
LAUNCH & TAKE-OFF
Since the Dragon's four-wheel landing gear is unusual, calculations are made to ensure that
Acceleration: Thrust is based on the commonly-accepted horsepower-to-thrust ratio, that is, 3:1.
With the four-wheel braking, stopping distance for the balanced field length may be reduced from that
Rotation: The pilot may raise the nose to the horizon about 8 degrees to give the wing
Four-seat version Rotation: Moments are summed around the rear wheel contact point at station 175, wing lift
Moments are summed as positive for nose-down and negative for nose-up for weight, wing lift, elevator
+[3,500lbs. * (175-99)] {weight}
= [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. After rotation, the elevator down-force is partially relieved, except for trimming load of about 200 lbs. (depending
L = [(0.6 * 120 * 12) + (1.2 * 128 *12) L = 864 + 1,843 = 2,707 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.
The four-seat version adds 800 lbs. to the figures above.
CRUISE
For cruising, estimated lift coefficients are 0.3 for the wing and 0.1 for the body-fuselage and 0.15 for the tailplane
With 180 hp, using the same drag figures, with parasite drag increase offset by induced drag decrease,
APPROACH AND LANDING
AUTOMOBILE MODE WEIGHT DISTRIBUTION
ROAD TURNING RADIUS
The Excel Spreadsheet has weight and balance data at * Master Plan
for larger wheels and tires (to 14" diameter) that are retractable, air conditioning system, fan clutch, and
chassis fittings, to 2103 lbs. Payload weight is increased to provide for 390 lbs. for pilot and passenger, an increase
of 50 lbs., plus an increase in baggage weight from 80 lbs. to 132 lbs., and an increase of 24 lbs. for fuel.
Gross weight is increased to 2900 lbs.
40 gallons of fuel with 45 minute reserve of 7 gallons ;
SPECIFICATIONS:
Height: 80” [203 cm]
Width: 92” [234 cm]
Length: 19’ 4” [589 cm]
Cabin width: 60” [152 cm]
Cabin height: 54” [137 cm]
Front wheel tread: 66’ [168 cm]
Rear wheel tread: 56” [142 cm]
Wheelbase: 119” [302 cm]
Weight ratio front/rear: 45/55
Turn radius: 24’ [7.32 m]
Baggage space: 12” h x 60”d x30”w
[30 cm h x 152 cm d x 76 cm w]
Weight empty: 2130 lbs [968 kg]
Weight loaded: 2900 lbs. [1,318 kg]
Curb weight: 2390 lbs [1,086 kg]
Wingspan: 29’ [8.84 m]
Wing chord (avg.): 130 cm]
Wing area: 95 sq ft [8.7 sq m]
Body area: 135 sq ft [12.4 sq m]
Wing + body area: 230 sq ft [21 sq m]
Wing + body loading: 12.6 psf [603 Pascals
Aspect Ratio: 6.7
Span loading: 100 lb/ft [150 kg/m]
Tailplane area: 33 sq ft [3 sq m]
Tail volume: 316 cu ft [8.8 cu m]
Dihedral: 12 deg [0.2 r]
Fuel capacity: 40 gals [151]l
Pilot plus passenger weight: 390 lbs [177 kg]
Baggage weight allowance: 132 lbs [60 kg]
Performance with Eggenfellner Subaru::
Cruise speed: About 140 kts [260 kph]
at 8,000 ft, burning about 7 gals/hr
at 65% power.
*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 220 h.p. supercharged version; and
(3) it has simplified controls without mixture control, carburetor heat, magneto switch,
or primer controls and is very economical on fuel burn.
Performance is shown below.
Alternatives include the Canadian *Crossflow conversion of Subaru and
conversions of Mazda rotary engines by *Atkins and *Mistral.
Likewise, the Subaru conversions may be acquired in six-cylinder versions, rather than four-cylinder versions.
For the four-seat version, with the more powerful 260 hp turbocharged engine, the cruise speed for 75% power,
with 10% losses to give a net of 175 hp, is estimated to be around 140 KTAS. However, the main advantage
of the higher power is a capability of cruising at higher altitudes to be above the weather.
Some two-seat version users may choose the higher power engine for performance.
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.
From an environmental energy viewpoint, the StrongMobile uses a small amount of electricity for
reduction of conversion time versus burning fuel for higher cruising speeds.
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.
based on the usual specific fuel consumption rate of 0.5 pounds of fuel per horsepower per hour.
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 2700 lbs., one would then need 338 lbfs. 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
automobile engines 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.
S is equivalent flat plate area and q is aerodynamic pressure. q uses sub-factors of half air density, equal to
.0012 slugs/ft3, 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.
For cruise speed of 140 KTAS, this is doubled to 58 lbf/sq. ft..; at 8,000 ft., 46 lbf/sq. ft..
requires 29*169/550 = 8.9 HP/ sq. ft.. For 140 KTAS, this is 12.5 HP/sq. ft.
At 8,000 feet., at 80% air density and drag, this is roughly equivalent to 10 HP/sq. ft..
similar to a 4 foot diameter disk. For checking, the fuselage/body has a cross-section of about 30 sq.ft. and
an assumed drag coefficient (compared to similar shapes) of about 0.2, for an equivalent flat plate area of 6 sq.ft..
Note that this method assumes a uniform ambient flow field, whereas the duct has a higher q
for the 15 sq. ft. wetted cross-section and the other cross-section does not see the fanwash.
The flight surfaces and duct are estimated to contribute about 1 sq. ft., [ total of 7 sq.ft.].
Skin friction, based on wetted area, surfaces exposed to airstream, is estimated to contribute about 1.5 sq.ft. [8.5 sf]
*Skin Friction Data - *Reynolds Number Calculator
The wetted area is about 24*20 = 480 sq. ft. for the body-fuselage plus about 300 for the flight surfaces,
for a total of 800 sq.ft.. With a drag coefficient of 0.002, the drag is about 800 * 0.001 * q = 0.08 * 58 = 92 lbf.
This results in an equivalent flat plate area of about 1.5 sq. ft. [ 10 sf ]
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:
101/0.36= 280 lbs at 60 mph, or, 76 lbs at 100 knots, or 38 lbf at 140 KTAS, about 1.25 sq.ft. .
One may use the Oswald efficiency factor that would increase the drag by about 20% to 1.5 lbf. [ 11.5 sf ]
One may estimate that the horizontal tail would add another 1- sq. ft. of flat plate equivalent for a total of 12.5 sq.ft..
The 9 sq.ft. times the 46 lbf/sq.ft. gives 414 lbf of drag at 140 knots, the desired cruise speed at the desired altitude.
Since this is around the 12 sq. ft. from power available, the design is confirmed.
as a 2900 lb. airplane at maximum weight gives 290 lbs. of drag.
1,000 flight hours per year, this gives $ 67 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 * $ 67 = $ 2,000 per pound of drag.
Designers may use the StrongWare to calculate present value * Structured Problem-Solving .
into their wells to the extent of the spring deflection of 6" and cover the wheel well with sliding doors, then
one would reduce the almost-flat plate area by one square foot, or 29 lbs. of drag, or about 8% of cruise drag.
One might implement this with Commercial Off-The-Shelf, COTS, electric actuators.
This simple estimate ignores the drag of the holes and is conservative, so the drag reduction would be more.
weight of the retracting and extending mechanism and the reduction of induced drag.
The life-cycle cost would then be 29 lbs*$2,000/lb. = $58,000. One could then either:
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 1 G 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.
This method would eliminate the gear-up landing hazard risk.
The aerojet flow is only directed along the sides of the fuselage and the top and bottom of the fuselage only see
ambient free stream flow. The duct exhaust doors will block about 90% of the flow; the perimeter gaps of about an
inch will reduce the expected whistle noise from the gap leaks. The landing gear strut drag is less than conventional
tractor configurations, particularly for nosewheel drag.
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 270 mph or 400 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 240 hp, the rule of thumb thrust would be 240*3=720 lbs., so the net thrust would be
720 - 150 = 570 lbs. 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 8 lbs. and an efficiency of 0.78.
throttle vanes would further reduce the duct efficiency to an estimated 0.67 for a net maximum thrust of 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..
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%.
thrust, T;
air density, R;
duct cross-sectional area, A;
Initial velocity, Vo; and
added velocity, Va.
Where Vo + Va = Total exit velocity.
Using the relationship T = 2RA * (Vo + Va) * Va , then
T = 675 lbs = 0.05 * (220 + 50) * 50 as an approximation, showing duct exit velocity as 270 ft./sec.
or 160 knots.
Advice given to the inventor by the CalTech wind-tunnel test experts (circa 1970s) 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.
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
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 2900 lbs.. Design speed for take-off is 57 knots, or 96 fps.
With 2900/540 = 0.19 g acceleration, about 3.5 kt/sec, for 16 seconds to accelerate to 57 kt, take-off distance at an
average speed of 48 fps, distance would be 770 feet. Aerodynamic drag and wheel drag would extend
this to an estimated distance of about 1,000 feet for sea level, standard day, no-wind conditions.
An alternative technique with the road drive engaged may be used to increase thrust and acceleration;
however, since this requires precise coordination, it is to be discouraged.
For "chop'n'stop", maximum braking from 60 knots at 20 kts/sec would be about 350 feet.
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.
of conventional lightplanes.
an effective angle of attack with flaps down 20 degrees of about 18 degrees.
Fanjet at take-off velocity, Vo = 110 ft/sec, Va = 70 fps, T = 630 lbf.
Calculation of the vectored lift effect of the fan exhaust at 180 ft./sec. for a q factor of 38 on the 2.5-feet of span equivalent and
4-foot chord of the blown wing roots would add a moment of
[-(1.3*10*38)] - [-(1.2*10*10)] = -506 lbs.*(144” – 110”) = - 17,200 lb.-in.
Aerodynamic pressure, q, at 75 mph is calculated as 14 psf. Wing area is 4.3*24 = 104 sq. ft.
Moments are summed as positive for nose-down for weight and negative for nose-up for wing lift, elevator download and drag,
in in.-lbs. , in sequence as follows:
+[2,900lbs. * (144” – 92”)] for aircar weight
- {[(1.3 * 104 * 14 lbs.) * (144”- 111”)]} - 18,633 for wing lift
= [2,900 lbs. * 52"] – {[1,893 lbs. * 33"]} - 18,633
= 150,800 –80,585 - 18,633 = 51,582 lb-in. nose-down moment
= or 4,300 lb.- ft. nose-down moment to rotate before elevator is deflected.
When elevator is deflected for rotation, elevator area is (12'*3') = 33 sq. ft., less tip horn reliefs. Elevator drag of about 40 lbs. at a
height of 6.5 ' will contribute -260 lb.-ft. to the nose-up moment.:
- [(1.8 *33 * 14 lbf.) * (144” - 208.5”)] – (40 lbf * 78”) for elevator downlift and drag
- [832 lbs. * (64.5")/12] - [260] = - 4,440 – 260 = - 4,700 lb.-ft.
For rotation, 4,300 – 4,700 = - 400 lbf-ft. nose-up moment
The wing and tailplane moments with section moment coefficients of about 0.1 would add a nose-down moment of about 100 lb.-ft.
for a net of about – 300 lb.-ft., nose-up, which is sufficient to raise the nose. Fuselage lift is based on the wing and tailplane moments
acting on the wheel springs to raise the nose slightly. The moment on the 113" wheelbase gives forces at the springs and an angle of
more than 2 degrees angle of attack. Fuselage lift area is based on an average width, calculated as 6 feet and a length of 20 feet and
a lift coefficient of 0.3 and a center of pressure at station 60 at 0.25 fuselage chord.
- [(0.3 * 120 * 14 lbs.) * (144” – 60”)] = 42,336 lb-in = 3,528 lb-ft
The summation of moments then gives -3,500 – 300 = - 3,800 lb.-ft. of nose-up moment, momentarily,
to relieve most of the elevator download and drag. As velocity increases, wheels are retracted and slippered,
flaps are raised and wing lift decreases, alpha is increased, body lift increases, then tail drag decreases.
The upward sweep of the bottom aft fuselage-body may have a reduced pressure due to negative dynamic pressure,
since it is partially sealed by the runway surface and the ventral fins. This may result in a significant nose-up moment,
during acceleration, up to -1,000 lb.-ft. at rotation speed, at the expense of about 100 lbs. of drag.
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 of 6 feet and a length of 24 feet and a lift coefficient of 0.2 and a c.p. at station 72.
NOTE: Fuselage lift is based on the wing and tail moments acting on the wheel 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.
down-load, and fuselage lift, in sequence as follows:
- [(1.8 * 152 * 11) * (175-149)] {wing lift}
- [(1.8 *36 * 11) * (175-246.5)] {elevator}
- [(0.2 * 144 * 11) * (175-72)] {fuselage}
Calculation of the vectored lift effect of the duct exhaust at 270 ft./sec. for a q factor of 85 on the 2.5-foot of
combined span equivalent and 4-foot chord of the blown wing roots would add a net effect of
[ -(1.8*10*85 ) ] - [ - ( 1.8*10*14 ) ]
= -1,278 lbs.*( 137-110 )
= -34,506 lb.-in.. = -2,876 lb.-ft.
With the vectored slipstream effect, the total moment is
8,678 – 2,876 = 5,893 lb.-ft.
The following calculation is based on the use of retractable canards extending outwards from the nose.
The canards will contribute 200 lbs. on an arm of 175-24=151 in.
151/12=12.5 ft.)= 2,500 ft.-lbs., leaving a deficit of
5,803 - 2,500 = 3,303 ft.-lbs.
The 2009 design replaces the canards with a belly-mounted Gurney flap as shown for the same simpler effect.
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.;
3,303 – 1,200 = 2,103 lb.-ft.
The upward sweep of the bottom aft fuselage-body may have a reduced pressure due to negative dynamic pressure,
since it is partially sealed by the runway surface and the ventral fins. This may result in a significant nose-up moment,
about -1,000 lb.-ft., at the expense of about 100 lbs. of drag at rotation speed.
2,103 lb.-ft. - 1,000 lb.-ft. = 1,103 lb.-ft.
Therefore, there appears to be insufficient moment for rotation. This may be easily accomplished by increasing
rotation speed to 80 mph, or 70 KIAS to increase q from 14 PSF to 16 PSF for the calculations.
on fuel and baggage), and the fuselage lift is increased due to the increased angle of attack. With flaps up,
wing lift coefficient is decreased to an estimated 1.2; fuselage lift coefficient is estimated as increased to 0.6.
q for 60 knots is calculated as 12 psf. Without considering the boosted wing root lift, lift is calculated as follows.
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 initial climb with wheels down and Cl/Cd of only 6, drag is increased to 450 lbs,
so power required is 450*135/550 = 110 hp and rate is reduced to 34*550*60/2700 = 415 ft/min.
Obstacle clearance distance of 50 feet requires about 415 ft/min*min/60 sec = 7 secs.,
7 secs. at 135 ft/sec = 945 ft..
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 induced drag is also increased, so drag for an L/D ratio 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.,
ratio of 8 for 3500 lbs. is 437 lbs.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.
download. The wing lift is composed of the root and the main wing, where the root is estimated to contribute
about 300 lbs. and the main wing about 1800 lbs. The body lift is estimated as about 900 lbs. and the tailplane
download about 300 lbs. Net lift is estimated as 2700 lbs.
65% power cruise is estimated as 140 KTAS, or 160 mph.
Using the Subaru conversion figures of 0.3 lbs of fuel per hp per hour for the 220 hp version,
0.65*220*0.3 = 43 lbs => about 7 gals per hour, giving 160/7 = 23 mpg.
The more efficient automobile engines would be expected to better this figure significantly.
As examples, the * Eggenfellner website gives a burn rate for cruise at 65% power of 6.5
gallons per hour. Allowing for a half hour of driving before and after flying and an hour to get to alternate,
and a 40 gallon fuel supply, this gives about about 4.5 hours of cruise, which, at 120 KTAS, gives
a no-wind range of about 540 miles.
The design is based on the assumption that the pilots will aim for keeping the
nose on the horizon during approach and while landing.
This technique will allow the ground effect to naturally flare and cushion the touchdown.
The airplane mode c.g. at gross weight is at station 92. With the 190-lb. wings stowed to give a c.g. at
station 94, the front wheels at station 26, rear wheels at station 137, the weight is mostly on the rear wheels.
The distribution is calculated at 44 % on the front wheels and 56% on the rear wheels.
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.
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.
