Item 0904

DESIGN: UniCopter ~ Dimension, Area & Drag - Drag - In Forward Flight

From calculations in Helicopter Performance, Stability and Control, chapter 4 & page 306

Much of the following is in: See Access database: FORM: Trim for calculations.

Get info from Hub Drag, Rotor-Wing Aerodynamics, book 2, page 66

Reducing Drag:

Forward flight:

1. Lowering the upper sheave.
1. Consider using a hypoid gear to lower the pinion vs. the crown. This may increase the drive loss excessively.
2. Locate the upper sheave closer to the lower sheave. Run its axle part way forward then via a constant velocity joint, angle the shaft up, at a steeper angle. to the intermediate reduction.
2. Lower the complete transmission slightly. This will lower the x-shaft. It will also increase the distance between the rotor and the final gears.
3. Allow air to flow through the central area of the power train.
4. The rotating blades may deflect air into the cooling openings under the rotorhubs thereby allowing this opening to be smaller and better faired.
5. Retractable Landing Gear
6. Faring the air-cooling ducts in front of the final reduction. In other words, reduce the size of the opening, or at least shift it laterally inward by wraping the skin forward an inward at the openings. The rotor hub and the blades should deflect sufficient air into the ducts to cool the engine during forward flight. The faring should also direct more air around in into the upper quadrant of the propeller.

• Sufficient Airflow to Upper Quadrant of the Propeller:
• Concerns and Potential Solutions: DESIGN: UniCopter ~ Pusher Prop - General - Pusher Prop Assist
• Parasitic Drag: To reduce the parasitic drag consider lowering the final gears and reducing their ratio by increasing the ratio of the intermediate gears. This will also probably necessitate independent blade root and tip pitch control. See drawing Drawings/Item/LowerFinalGears.dc.
• Drag of Rotors: During forward flight the rotor blades, when between 250º and 360º, are interfering with the flow of the air that is being drawn in by the upper quadrant of the propeller. Related to this is the parasitic drag of the rotor hubs and the final gear reducers, plus the disturbance of airflow to the propeller. See Task #1
• Parasitic drag of the upper sheave. Could be replaced by gear set but then a separate clutch will be required.
• Parasitic Drag of Rotorhubs and Final Gearbox, plus the Reduced Airflow to the Upper Quadrant of the Propeller:

The rotor hubs, blades, tail booms and fuselage will have detrimental effect on the flow of 'clean' air to the upper quadrant of the propeller.

Potential Solution;

• Increase the stagger between the rotorhubs by a few inches.
• Reduce the pitch of the engine and propeller slightly, by dropping the front of the engine 2 or 3 inches.
• The propeller will be drawing air from under the front half of the rotor disks and discharging it under the back half. Might this cause the craft to pitch down? If so, perhaps the prop should be aligned so as to add a nose up moment to the craft. This may not be detrimental upon an engine failure, since the loss of the nose up moment from the prop will be offset by the fact that the prop is no longer taking air from under the front half of the disk.
• When the prop is not producing thrust a folding prop should remove most of the drag resulting from the prop.
• Seriously consider lowering the secondary and final gearboxes 5 or 6 inches. The good reasons are;
• The final gears will be inward further thereby placing then more behind the firewall.
• The mid gear ratio could be increased and the final gear sets reduced in size.
• The drive train will be lower and out of the airflow ofer the top.
• The shaft to the secondary reduction from the sheave will be lower and parallel with the crankshaft of the engine with its lower pitch.
• This frame must be stronger to handle to moments, but because it is larger the mounting points will be at greater centers so the fuselage can be weaker.
• Wild idea: Consider using the large diameter mast as a radial fan (with holes and scoops) to draw air into it and down into the engine area. A bad reason might be that of having mechanical access for mechanical flight-controls.
• Electric pitch actuators.
• Relocate air scoops; in front of engine's extended cylinder heads. Particularly, if larger engine used.
• Relocate tail boom; perhaps through the prop.
• Tension torsion straps; to nigate the axial bearings.
• Aerodynamic hub faired into blade roots.
• Exhaust the 'cooling air' into this upper quadrant area.
• Note that the Interleaving configuration should be aerodynamically cleaner in the 'feed' area to the propellers upper quadrant.
• Note that the drag on the roots of the retreating rotor blades contributes to rotor rotation. Big deal.

Improved Airflow to Prop:

The jackshaft from the soft start to the rotor's mid-gearbox could have universal joints at both ends to get the sheave closer to the prop shaft. Similar to the drive shaft in a car. If the jackshaft had a sliding spline then the vibration of the engine would not effect the rotor drive train.

Fuselage:

In fast forward flight, the UniCopter will have a slight nose-down pitch. This combined with the downwash from the rotor may mean that the plan profile of the fuselage should be considered, as well.

__________________________

Total Frontal Area: (AF) = 8.31 sq-ft.; from DESIGN: UniCopter ~ Dimension, Area & Drag - View - Front, Y-Z plane

1. The cooling air intakes are 0.75 sq-ft, of the above frontal area.
2. The balance is 7.56 sq-ft.

Coefficients of Drag: (CDF); from OTHER: Aerodynamic - Drag - Parasite , [Source ~ RWP1 figure 4.17]

1. Coefficient of Drag for Fuselage - cooling air intake: (CDF1) = 0.3 My guess
2. Coefficient of Drag for Fuselage - balance: (CDF2) = 0.15 My guess
1. Equivalent flat plate area of cooling air intake fF1 = AF1 * CDF1 = 0.75 * 0.3 = 0.23 ft2
2. Equivalent flat plate area of balance fF2 = AF2 * CDF2 = 7.56 * 0.15 = 1.073 ft2
• Total Equivalent flat plate area, fF = 1.30 ft2

Rotorhubs:

For general interest overview see; http://www.cartercopters.com/gt-hub_drag-compatible_14.html

The rotor hubs are 1.25 sq-ft, of frontal area; from DESIGN: UniCopter ~ Dimension, Area & Drag - View - Front, Y-Z plane

Coefficient of Drag for Fuselage - rotor hub: (CDR) = 0.5 My guess from OTHER: Aerodynamic - Drag - Parasite , [Source ~ RWP1 figure 4.17]

Equivalent flat plate area of rotor hubs fR = AR * CDR = 1.25 * 0.5 = 0.625 ft2

Landing Gear:

Frontal Area: (ALG) = 0.2 sq-ft. Currently a guess. Will be from DESIGN: UniCopter ~ Dimension, Area & Drag - View - Front, Y-Z plane

Coefficient of Drag for Fuselage: (CDLG) = 0.7 My guess from OTHER: Aerodynamic - Drag - Parasite , [Source ~ RWP1 figure 4.26]

Equivalent flat plate area fLG = ALG * CDLG = 0.2 * 0.7 = 0.14 ft2

Horizontal Stabilizer:

Area: (AH) = 3 sq-ft

Span: (bH) = 4 ft.

Aspect Ratio: (A.R.) = b2H/AH = 5.333

Thickness Ratio: (t/c) = 0.12 NACA 0012 at present

Mean Aerodynamic Chord: (MAC) = 0.75 ft.

Reynolds number at 115 knots = RN = c x MPH x 9360 = 0.75 ft.x 132 mph x 9360. = 926,640

Estimate (CDO) from OTHER: Aerodynamic - Drag - Parasite, [Source ~ RWP1 figure 4.15] = 0.01

Estimate (CLH) from trim conditions OTHER: Aerodynamic - Drag - Parasite, [Source ~ RWP1 chapter 8] = 0.6

Estimate span efficiency factor (δ) = 1

Calculate induced drag coeff. (CDi) = (CLH2 * (1 + δ)) / (π * A.R.) = (0.62 * (1 + 1))/(3.143 * 5.333) = 0.043

Calculate root thickness (t) = 0.12 * 1 ft. = 0.12 ft.

Estimate junction drag coeff., (CDJ) from 0898.html, [Source ~ RWP1 figure 4.21] = 0.1

Compute equiv. junction drag coeff. (CDJ equiv) = 2(CDJ * (t2)/ AH) = 2 * ((0.1 * 0.12 * 0.12) / 3) = 0.0001

Total drag coeff. CDH = CDO + CDi + CDJ equiv = 0.01 + 0.043 + 0.0001 = 0.0531

Estimate qH/q = 1

Calculate (fH) = (qH/q)* (CDH * AH) = 1 * 0.0531 * 3 = 0.159 ft2

Horizontal Stabilizer: Negative lift - this has nothing to do with drag but it has to do with longitudinal static stability

Vertical Stabilizer:

Area: (AV) = 3.25 sq-ft.

Thickness Ratio: (t/c) = 0.12 NACA 0012 at present

Mean Aerodynamic Chord: (MAC) = 16"

Reynolds number at 115 knots = RN = c x MPH x 9360 = 1.33 ft.x 132 mph x 9360. = 1,600.000

Estimate (CDO) from OTHER: Aerodynamic - Drag - Parasite, [Source ~ RWP1 figure 4.15] = 0.01

Estimate qV/q = 1

Calculate (fV) = (qV/q)* (CDV * AV) = 1 * 0.01 * 3.25 = 0.0325 ft2

Main Rotors:

Get info from Hub Drag, Rotor-Wing Aerodynamics, book 2, page 66. Note that hub drag is currently included above in Fuselage.

H-force:

OTHER: Aerodynamics - Drag - H-force

Hub Concerns:

• Consider giving the two hubs flat bottoms and a bottom radius of 14.5". Then make the central surface of the fuselage, directly below the above semi-circles flat and flush with this inside portion of the hubs. Crown the tops of the hubs. Below the outer lower semi-circles of the hubs, flow the laterally extended fuselage back and down, thereby giving more clearance on the engines aft cylinder head covers.
• Consider shaping the canopy such that it aerodynamically directs the air, from fast cruise, away from the two hubs. In addition fair the hubs into the canopy so that the air that does impinge upon the hubs meets the least resistance.
• Consider having the cooling air enter through scoops that are directly in front of the cylinder heads. This might be even more advantageous if the larger engine (1" wider on each side) is used. The area directly under the hubs can now be faired better. Or, have four small entrance locations; two under the hubs and two in front of the cylinder heads. A full-time fan may be required on the shaft of the engine.

Reverse Air Flow:

If the RRPM is 480, the disc radius is 9'-6" and the forward speed is 150 kts (173 mph), the radius of zero flow at 270º azimuth is 5'-0"". The combined velocity at azimuth 90 and r = 29" / 2 (ie. the craft's longitudinal centerline) = 14.5" R will therefor be 315 fps (215 mph). This probably indicates that, when fairing the rotor hubs, the attention should be directed toward the advancing sides. See: FORM: Tangential Velocity.

Consider clading the root of the blades out to the centerline of the craft so that they have a 'D' x-section, with the flat portion on the bottom. From azimuth 20º to 130º approximately this flat portion will just clear the fuselage. From the center of the craft out to about the center of the other hub, this shape will start transitioning from the 'D' to an ellipse.

Crazy Idea :

The final drive shafts up to the hubs will probably be tubes. If the hub's top caps have a hole in their centers then it may be possible to locate non-rotating sheets between the top of each hub and the other rotor's blades, or provide a low profile axes that are inboard of the rotors axes.

Propeller:

Must consider drag during powered forward flight and during autorotation, regarding any change in pitch.

Vectored Exhaust & Cooling: No effect during autorotation.

 Area: (1) Drag Coefficient: (2) Equivalent Flat Plate Area: [f] Fuselage 8.25 ft2 0.23 (3) 1.30 ft2 Rotorhubs 1.25 ft2 0.5 0.625 ft2 Landing Gear 0.2 ft2 0.7 0.14 ft2 Vertical Stabilizer ft2 0.0325 ft2 Horizontal Stabilizer ft2 0.159 ft2 H.S. Lift (downward) ft2 Sub Total: 2.26 ft2 Roughness & Leakage 0.1 ft2 Protuberances 0.15 ft2 Cooling Losses 0.01 ft2 Total: 2.5 approx.?

Rough & wild Notes: on Pitching Momentum, (to be moved (possibly to 0902) & cleaned up)

For an airplane wing: LW = (ρ / 2) * V2 * S * CL . Where S is the area of the wing.

For an airplane wing: LW = (0.002377 [slug/ft3] / 2) * V2 * 3 [ft2] * CL . Where S is the area of the wing.

0.002377 @ ISA)

slug/ft3

-2.75ft2 - ft. / 3.0 ft2 - ft. = 0.916

Cl of -0.916 with a NACA 0012 airfoil is a Angle of attack of -8 degrees

Cl of -2.337 with a Wortmann FX 74 CL5-140 airfoil is a Angle of attack of -8 degrees

Cl of --0.916 with a Wortmann FX 74 CL5-140 airfoil is a Angle of attack of +4.5 degrees

IT, INITIALLY, LOOKS LIKE THE WORTMANN HAS MORE (LIFT) AT 0 DEGREES THAN IS EVEN NEEDED!!!!

A crude initial attempt to see if the craft, at its maximum obtainable speed will be at a nose down angle sufficient enough to cause the H.S. to stall and the craft to tumble.

For the fun of it & probably totally wrong

Assume a forward tilt on the rotors and craft of 5º

Assume a gross weight of 1150 pounds

The lift component must be 1150 pounds.

The forward component must be tan(10) * 1150 = 101 pounds.

The following algorithm is from page B319

F = 1/2 ρ CD A v2

v2 = F / (1/2 ρ CD A)

v2 = 101 / (1/2 * 0.002377 slugs/foot3 * 1.57 )

v2 = 101 / 0.0018677

v2 = 53,897

v = 232 ft/sec

v = 158 miles per hour | 137 knots

This does not include the H-force drag of the rotors.

This is based on the frontal area only whereas at 5-degrees pitch down there will be more drag

Where
F is the force [in pounds]
ρ (Greek letter "rho") is the density of air = 0.002377 slugs/foot3 (The paragraph below says to use pounds per cubic foot, which is actually = 0.0765 pounds/foot3, but it appears that the slugs/foot3 value is the correct one.)
CD is the coefficient of drag; which is 0.04 for a streamline strut (NACA 0025), 1.0 for a round wire, 2.0 for flat wire or 1.5 for a parachute. For values for helicopter component shapes see [Source ~ RWP1 p.280].
A is the area of the surface [in square feet]
v is the velocity through the air [feet per second]. (To convert from miles per hour to feet per second; multiply mph by 1.4667)

An Alternate Rough Calculation of Maximum Forward Speed: from [Source ~ RWP5 p.30]

Max Speed = 41 * cubic root( 30 min rating of engine / equiv. flat plate area)

Max speed = 41 * cubic root(124 hp / 1.66)

Max speed = 172 knots (198 mph)

Possibly Relevant Outside Web Pages:

John D. Berry, Unsteady Velocity Measurement Taken Behind a Model Helicopter Rotor Hub in Foward Flight , NASA TM-4738, March 1997 , pp. 246, (8MB). http://techreports.larc.nasa.gov/ltrs/dublincore/1997/tm/NASA-97-tm4738.html

Last Revised: March 11, 2007