OTHER: Helicopter - Inside - Single Rotor - Rotor X Wing - Improving Lift During Transformation:

• Criteria for Calculations:
• JAR-VLR allowable maximum GW = 1320 lb. Divided by 4 blades = 330 lb per blade.
• Ref: UniCopter RRPM in Hover is 550 rpm, Rotor radius is 9'-6", Tip speed is 547 fps and NACA 0012 airfoil requires 94 hp.
• Rotor X Wing:
• Rotor Radius = 12'.
• Root chord = 24".
• Taper =1" per foot
• Twist = -12º
• Airfoil NACA 0012 symmetrical.
• Transformation forward velocity is 80 kts = 92 mph = 135 fps. This is also the reverse airflow at the root of the blade.
• At the tip of the retreating blade (@ 270º ψ) the air speed is set to the above 135 fps. The RRPM is therefore 215 rpm.
• The above will hover with a blade pitch of 24º and 61 HP. (86 hp by momentum theory)
• This places the point of zero velocity at 0.5 R.
• This retreating blade will now be considered as a 2-blade rotor with a 6' radius and a tip speed of 135 fps at both tips.
• The thrust of this '2-blade rotor' is only 73-lb. Increasing the pitch does not help much.
• This means that because of lateral symmetry of lift the advancing blade (@ 90º ψ) can only deliver 73-lb of thrust. However, the tip speed ratio of mu = 0.5 and the larger taper may, as a guestimate reduce the mean moment arm from 0.75 R to 0.6 R. Since the moment arm on the retreating blade is at 0.5 R the lift on the advancing blade may have to be less than 73-lb of thrust. Perhaps a very large twist on the advancing blade may bring its moment arm down to 0.5 R, or even a little less.
• A moment arm of 0.5 on the advancing and the retreating blade means that the fore and aft blades must each produce ((330 * 2) - 73) = 587 lb of thrust.
• For each of these 2 blades to produce 587-lb of thrust, with 12º twist, their root pitches must be a theoretical 34º.
• NOT GOOD ENOUGH.

• The reaction to the torque of the propeller might contribute some, assuming that it can rotate in the desired direction.
• Moment from propeller:
• Say 50% of crafts power is 165/2 = 82 hp. and the PRPM is 2400 then the torque will be 5252 * HP / RPM = 179 [ft-lb] at full power. At .50 R this is 179 / 6 = 30 lb of additional lift on the rotors retreating side.
• However, this torque will exist when going from hover to cruise but it will not exist when going from cruise to hover.
• It might exist when going from cruise to hover if the craft was temporarily placed in a nose high position.
• A temporary high propeller pitch would help. Actually the propeller will have a fixed high pitch for cruise. Therefore at the much slower transformation speed the torque should be higher.
• The above assumes that the transition is not taking place during entry into autorotation.
• This would be a greater problem if the propeller is unable to deliver torque. Perhaps the propeller could be given a relatively high reverse pitch and it's rotation braked.
• Consider having two propellers with on as a tractor and the other as a pusher. They could be counterrotating or not, depending on optimization throughout the flight envelope.

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