1510

Item 1510

DESIGN: UniCopter ~ Pusher Prop - Variable Speed Rotors and Prop - Transmission

Criteria:

Input ~ Engine:

The Lycoming O-320 is derated for helicopter use from a base of 140 hp to 124 hp, therefor the O-360 will be derated from a base of 180 hp to 160 hp.

The speed remains the same at 2400 rpm.

The output torque is therefore Q = (hp * 5252) / rpm = 160 * 5252 / 2400 = 350 ft-lb.

Output ~ Rotors:

4-blade rotor calculations on Access

GW 1320 lbs

Rotors:

Speed during hover; 550 rpm: Tip speed; 550ft/sec : Tip Mach nbr; 0.49 .

Speed during cruise; rpm

Power to hover; 80 hp. This is quite low. Why?

Total reduction 4.35:1

Output ~ Propeller:

DESIGN: UniCopter ~ Pusher Prop - Variable Speed Rotors and Prop - Propeller

Planetary Gear Box (differential) RPMs etc:

The sun and planets have 24 teeth and the ring has 72 teeth.

This table assumes that the drag forces in the rotors and propeller are such that planets are not rotating.

RPM of planetary components during different forward velocities: Not from Access FORM since it does not seem to be working correctly.

Maximum Vertical Climb:

Maximum Forward Velocity:

Speed:

Power:

Torque:

Speed:

Power:

Torque:

Ratio:

Loss:

Ratio:

Loss:

Planet carrier [Engine]

2400

rpm

160

350

ft-lb

2400

rpm

160

350

ft-lb

Carrier to Sun

0 ⁽¹⁾

Sun Gear [Prop]

2400

rpm

(160/2)*0 = 80

175

ft-lb

Carrier to Ring

1:1.333

:-1

1.0

0 ⁽¹⁾

Ring gear

3200 ⁽³⁾

158.4

260

2400 ⁽²⁾

175

Primary: (belt)

1.16

: 1

2.5

1.16

: 1

2.5 ⁽⁴⁾

Intermediate:

2758

rpm

154.4

294

ft-lb

2062

rpm

199

ft-lb

Secondary:

1.5

: 1

1.0

1.5

: 1

1.0

X-shaft:

1839

rpm

152.9

437

ft-lb

1375

rpm

77.2

295

ft-lb

Final:

3.33

: 1

1.0

3.33

: 1

1.0

[Rotor]

550

rpm

151.4

1446

ft-lb

412

rpm

76.4 ⁽²⁾

974

ft-lb

Total:

4.35

: 1

5.5

: 1

There will not be any power loss since the planetary gears are not rotating.
50% of total power. A guestamate is that at cruise very slow turning rotors will require little or perhaps no power from the power-train. The reverse pitch on the retreating side will provide lift. Plus, the higher drag on the retreating side will contribute to rotation.
At cruise the rotor rpm is 75% of rotor rpm at hover. I.e. 550 rpm for hover and 412 rpm for cruise. This should mean that at cruse the torque, the power and the rpm of the prop and the rotors are all equal in the differential unit.
The loss to the rotors will be 1% instead of 2.5% if the sheaves are replaced by spur gears.

The above does not perfectly agree with Ratio, HP, Torque & Losses but whatever for now.

______________________________

This table is the same as above except that this one assumes that the torque is not distributed 50/50 to the sun and the ring.

I think that this is the correct one.

A/ Planet carrier and its 12" radius. (actually non-dimensional)

B/ Planet and its 12' radius. (actually non-dimensional)

C/ Ring and its 18" radius. (actually non-dimensional)

D/ Sun and its 6" radius. (actually non-dimensional)

If the torque of the Planet carrier is 350 ft-lb. then the force at both sides of the attached Planet (assumes only 1 planet) are 175 lb. each.

175 lbs force at the 6" radius of the Sun provides it with 175 / (6/12) = 87.5 ft-lb. of torque.

175 lbs force at the 18" radius of the Ring provides it with 175 / (18/12) = 262.5 ft-lb. of torque.

	Maximum Vertical Climb:						Maximum Forward Velocity:
	Speed:		Power:		Torque:		Speed:		Power:		Torque:
	Ratio:		Loss:				Ratio:		Loss:
Planet carrier [Engine]	2400	rpm	160	hp	350	ft-lb	2400	rpm	160	hp	350	ft-lb

Carrier to Sun	0	0	0				1:2		0
Sun Gear [Prop]	0		0		0		2400	rpm	40	hp	175 * (6/12) = 87.5 ⁽¹⁾	ft-lb

Carrier to Ring	1:1.333	:-1	1.0				0		0
Ring gear	3200		158.4		260		2400		120		175 * (18/12) = 262.5 ⁽¹⁾
Primary: (belt)	1.16	: 1	2.5	%			1.16	: 1	2.5	%
Intermediate:	2758	rpm	154.4	hp	294	ft-lb	2069	rpm	117.5	hp	298	ft-lb
Secondary:	1.5	: 1	1.0	%			1.5	: 1	1.0	%
X-shaft:	1839	rpm	152.9		437	ft-lb	1380	rpm	116.3		443	ft-lb
Final:	3.33	: 1	1.0	%			3.33	: 1	1.0	%
[Rotor]	550	rpm	151.4	hp	1446	ft-lb	414	rpm	114.7	hp	1455	ft-lb
Total:	4.35	: 1	5.5	%			5.79	: 1		%

These values are based on the input forces from the torque at the planets being split 50/50 between the sun and the ring. Then the torque of the sun and the ring is based on their respective diameters.

_____________________

Both tables have the same output speeds but the second table shows more torque at the rotor and less at the propeller. Now to determine which is correct? I suspect it is the latter.

Power, Torque and Rotational Speed Calculations:
The Access coding for planetary gearing does not seem to be working correctly. I cannot find an interactive one on the Internet? Therefore am using Machinery Handbook.

This is out of Machinery Handbook;

F = rotation of the Sun per revolution of the Planet Carrier.
z = number of teeth in the Sun.
S = rotation of Ring per revolution of Planet Carrier.
B = number of teeth on Sun.

F = 1 + (z x (1 - S) / B)

Examples to suit below two criteria;

Speed Controller Locked; F = 1 + (72 x (1 - 1.333) / 24) = 1 + (-1) = 0

Speed Controller fully Non-restrictive; F = 1 + (z x (1 - S) / B) = 1 + (72 x (1 - S) / 24) = ?.

Example; All three rotating at same rpm; S = 1, then F = 1

Example; for balanced torque; F = 1 + (72 x (1 - 0.5) / 24) = 1.5

Speed Controller Locked; At no forward velocity and maximum climb (max.engine power).

The engine input to the planetary carrier is 2400 rpm, 160 hp, and 350 ft-lb.
The planetary reducer's speeds are; Planet carrier (engine) = 2400 rpm, Sun (prop) = 0 rpm, Ring (to rotors) = 3200 rpm.
No power is going to the propeller therefore since the ratio between the planet carrier and the ring gear is 1.333:1 the ring gear is experiencing; 3200 rpm, 160 hp, and (160 * 5252 / 3200) = 263 ft-lb. of torque.
Since the arm length ratio between the planet carrier and the sun gear is 3:1 the sun gear is experiencing 260/3 =87 ft-lb of torque.
Therefore the Speed Controller is also experiencing 87 ft-lb of torque.

Speed Controller Fully Non-restrictive: At maximum cruise (max. engine power).

The engine input to the planet carrier is 2400 rpm, 160 hp, and 350 ft-lb of torque.
The preliminary planetary reducer's speeds are; Planet carrier (engine) = 2400 rpm, Sun (prop) = 2400 rpm, Ring (to rotors) = 2400 rpm. However, this will split the torque 25% to the sun gear and 75% to ring gear.
This provides the same torque at the rotor as when in hover, however the rotor's speed is 25% lower.

This power level may hopefully allow for a slight positive thrust through the rotor and thus a slight forward tip on the disk. In other words the disk helps the prop slightly with forward thrust.

The above gear ratios will change as the craft evolves.

The power going to the propeller is used to overcome the profile drag and the induced drag of the blades. The profile drag will increase as the square of the propeller's rotational speed. The induced drag will increase as the square of the induced thrust. However, the induced drag will not increase as fast as the profile drag since part of the induced velocity increase is due to the increasing forward velocity of the craft.

Algorithm used:
Q =( hp * 5252) / rpm

Where: Q is the torque ();hp is the horsepower; rpm is the rotational speed of the propeller.

A Look at Differential Gearing for a Better Understanding:

Information of DeGraw's Craft:

From Rotary Wing Forum ~ Dean Dolph

I don't know if it is relevant to this discussion or not but in the September '99 issue of Rotorcraft (the PRA magazine) Dick DeGraw was interviewed on the topic of the 'DeBird'; the machine he built for his wife. The 'DeBird' won Grand Champion Rotorcraft and Best New Design awards at Mentone '99 and Karol DeGraw won the "Man and Machine" award flying this machine. The award presenter commented at the time that since Karol was the first woman to win the award; they might have to change the award's name!

The interview write up is a little ambiguous in this area but Dick is driving the rotors thru a differential off of the prop with 14% of the engine power going thru a right angle gear drive to the rotor. He alludes to the fact that this is more than he provided the Gyrhino rotor. He doesn't provide any details on the differential other than "at the normal flight RRPM of 350, the differential internal parts do not turn, just like a car going down the road - both barrels turn at the same speed"

He is quoted as saying they can get 260 RRPM but usually takeoff when it reaches 220 - 230. He goes on to say that since the rotor is always powered it won't slow down 'catastrophically' (my interpretation!) during negative 'G' or in a full power push over situation and as such is an inherently safe design. He also mentions that with the driven rotor they cruise at 85 mph versus 70 mph when it is not driven. Top speed is around 100 mph but the fuel burn at 90 mph is 6 gph vs. 5 at 85 mph. The DeBird's powered rotor flies at a 4 degree angle of attack vs. 8 - 10 degrees unpowered like a standard gyro. The power to the rotor increases the RRPM by only 10 rpm over unpowered.

See Chuck Beaty's article about Dick DeGraw's partially powered rotors in Aug 2005 issue of Rotorcraft. Have hard copy in Prop and rotor folder.

Chuck Beaty's talks about a balance of torque between the rotor and the prop. This may be true at the differential, however as the rpm to the rotor is reduced beyond the differential I suspect that the rotor will be experiencing more torque than the propeller and the propeller will be experiencing more speed than the rotor. Look further into the unbalanced differential above and the inequality in a planetary reduction, which is used for splitting power.

The following supports what Chuck said and what I believed. Obviously I did not read his article slowly and deeply enough. The bottom line is that by adding a further reduction on the rotor output after the 'differential' the power to the rotor can be increased visa vie the power to the propeller. Alternatively, it appears that some of this power balancing can be done inside the differential if the differential is a planetary reduction.

	Rotor:	Propeller:
Rotational speed: [RPM]	300 rpm	3000 rpm
Torque: [Q]	100 ft-lb	100 ft-lb
HP = RPM x Q / 5252	300 x 100 /5252 = 5.7 hp	3000 x 100 /5252 = 57.1 hp
Radius to center of Force [R]	1 ft.	1 ft.
Velocity [V] = RPM x 2 * (22/7) x R	300 x 2 x 22/7 = 1885 fpm = 31.4 fps	3000 x 2 x 22/7 = 18857 fpm = 314 fps
Force (drag) [F]= Torque / Radius	100 / 1 = 100 lbs	100 / 1 = 100 lbs
Area [A] = F / V²	100 ft-lb. / 31.4² fps = 0.10 sq-ft	100 ft-lb. / 314² fps = 0.001 sq-ft

The following algorithm calculates the areas more accurately. However the proportionality does not change from the calculated values in the above table.

Working in (English) units

F = 1/2 ρ C_D A v²

A = F / (1/2) x ρ x C_D x v²

Where
F is the force [in pounds]
ρ (Greek letter "rho") is the density of air = 0.002377 slugs/foot³

C_D is the coefficient of drag; which is Flat plate =1.28

A is the area of the surface [in square feet]
v is the velocity through the air [feet per second].

A = 100 / (1/2) x 0.002377 x 32.16 x 32.16 lbs/foot³ x 1.28 x x 31.4 x 31.4 = 0.06 sq-ft.

A = 100 / (1/2) x 0.002377 x 32.16 x 32.16 lbs/foot³ x 1.28 x 314 x 314 = 0.00064 sq-ft.

The following is just a bunch of related algorithms.

The drag force [lb] depends on the square of the Velocity [fps * fps] * Area [ft * ft]

Power [lb-ft/min] = Force [lb] * Arm [ft] * Velocity [fpm]

Torque [ft-lb] = Power [lb-ft/min] / Rotational speed [rpm]

Torque [ft-lb] = HP * 5252 / Rotational speed [rpm]

Power = Torque * Velocity

Velocity [fpm] = Arm [ft] * 2 * pi * Rotational speed [rpm]

Power = Work / Time

Work = Force * Distance

Speed = Distance / Time

Power = Force * Distance / Time

_______________________

Power = Work / Time

Power = Force * Distance / Time

Power = 1/2 ρ C_D A v² * Distance / Time

v is the = rpm * Arm [ft] * pi * 2

Power [lb-ft/min] = Area [ft * ft]* Velocity [fpm] * Velocity [fpm]* Arm [ft] * Velocity [fpm]

Torque [ft-lb] = Power [lb-ft/min] / Rotational speed [rpm]

Torque [ft-lb] = Area [ft * ft]* Velocity [fpm] * Velocity [fpm]* Arm [ft] * Velocity [fpm] / (Velocity [fpm] / Arm [ft] * pi * 2)

Torque [ft-lb] = Area [ft * ft]* Velocity [fpm] * Velocity [fpm]* Arm [ft] * Arm [ft] * pi * 2

Torque [ft-lb] = Area [ft²]* Velocity² [fpm] * Arm² [ft] * pi * 2

Torque [ft-lb] = Area [ft²] * Velocity² [fpm] * Arm [ft] * Rotational speed [rpm]

Torsen Differential: Mentioned by Quadrirotor as being used on Dick DeGraw's Rhinogyro. Is it really a Torsen differential?

The invention on this web page and the Torsen differential may two features in common. They both involve a differential action and they both make use of the fact that the ability for reverse power transmission through a worm & wheel is dependent upon the ratio of the gear. Beyond these two features, they are totally different.

Some Outside Information on the Torsen Differential, as use on Road Vehicles:

"Torsen (Quaife) The Torson differential works via a special arrangement of gears inside the housing, which in a nutshell exploits the fact that worm gears like to be driven in one direction and not another - don't ask to explain in any more detail than that, as I don't rightly know myself.

Because this is a 100% mechanical solution with no wear surfaces (aside from normal gear->gear wear) this is a very progressive and consistent solution that provides easily predicable handling response."

A reference from US patent 6,491,126, Straddle-type all terrain vehicle with progressive differential, Bombardier Inc.

"The Torsen.TM. differential, manufactured by Zexel-Gleason, is a second example of a torque-sensing diff. The Torsen.TM. uses three pairs of worm gears meshed with a pair of perpendicularly mounted worm gears that are each splined to a drive axle. The principle of operation relies on the fact that worms transmit torque in essentially only one direction."

".... objectives are accomplished by associating the function of differentiation with a proportioning torque between drive axles.

Technical Information on Torsen Differential

May be of some value in determining the rotor and prop relationship.

Most of the drag seen by the current recreational gyros is induced drag.
Max L/D on an autorotating rotor occurs at ~35% of peripheral tip speed. My gyro rotor spins about 340 RPM, for a tip speed of 410 fps. Max L/D is about 143 fps, or 97 mph.

Above that, profile and parasitic rotor drag starts picking up.

Unfortunately due to my Bensen's airframe drag, I can't even touch Max L/D . I top out at about 85 mph.

More streamilined machines could likely cruise at their most efficient rotor speed though. Andy Keech routinely cruises above 90 mph in his Little Wing gyro.

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Initially displayed: July 7, 2005 ~ Posted to PPRuNe: July 9, 2005 ~ Last Revised: July 8, 2007

The above utility invention is openly and publicly disclosed on the Internet to negate an entity from patenting it, to the exclusion of all others whom may wish to use it. ~ Reference patent law 35 U.S.C. 102 A person shall be entitled to a patent unless - (a) the invention was known ... by others in this country, ..., before the invention thereof by the applicant for patent.