1511

Item 1511

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

Objective:

To control the rotational speed of the propeller and thereby the rotational speed of the rotors; based on fixed motor speed.

Overview:

The speed controller consists of a worm & gear, with a reduction ratio of approximately 10:1. The exact reduction ratio will be that which is just sufficiently large enough to stop the gear from driving the worm.

The worm is driven by a controllable speed servomotor (or, two motors for redundancy). The speed of the servomotor is infinitely adjustable; from zero to the maximum speed of the propeller times the worm & gear ratio.

The speed of the servomotor is governed by the craft's forward velocity &/or the forward & aft position of the cyclic stick.

The friction generated by the worm & gear should be small.

The power of the servomotor is relatively small since it's function is only to brake loose from the static friction and overcome the minimal sliding friction, thereby allowing the gear to rotate. Or, add to the sliding friction and thereby resist the rotation of the gear.

For information on friction see; [Source ~ MH p.188]

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Size of Servomotor and Ratio of Worm Gear:

Let's assume that we have a worm & gear that has a very large reduction ratio, say 70:1.
Let's also assume that the axle, which passes through the gear, is applying a large torque to the gear.
Now there is no way that this gear is going to turn because it is incapable, on its own, of turning the worm.
The force of the gear tooth on the worm is just about at 90º to the wall of the worm's thread.
All that the gear is doing is apply a lot of static friction against the wall of the worm's thread.
To allow the gear to turn, it will require a VERY large servomotor to TURN the worm.

Now let's assume that we have a worm & gear that has a very small reduction ratio, say 1:1. (Actual this would be called a pair of 90º helical gears)
Let's also assume that the axle, which passes through the gear, is applying a large torque to the gear.
At this ratio, it is extremely easy for the gear to turn the so-called worm.
There is very little static friction resisting the rotation of the gear.
To NOT allow the gear to turn, it will require a VERY large servomotor to RESIST THE TURNING of the worm.

It should be apparent that as the ratio of 70:1 is reduced the size of the servomotor required to TURN the worm will be smaller.
It should be apparent that as the ratio of 1:1 is increased the size of the servomotor required to RESIST THE TURNING the worm will be smaller.

At a ratio of approximately 10:1 it become a 'flip of the coin' whether the servomotor is needed to TURN the worm or RESIST THE TURNING of the worm. In other words, at this ratio the servomotor can be quite small. This is why the ratio of 10:1 was picked for the Speed Controller.

If the forward velocity is best served by a propeller speed of 3,000 rpm, then the speed controller will turn the servomotor at (3,000 * 10) = 30,000 rpm.

Why does the motor need to be a servomotor. Could it be a high-speed motor controlled by an encoder.

Could the motor be 'embedded' in the worm? This will require a stronger shaft and bearings.

Brake Speed Controller: New Alternative to Servomotor

Consider using a worm gear with a ratio of less than 10:1. This will mean that the control device will always need to 'brake' the driving force never add to it. Now it should be possible to use a breaking mechanism instead of the servomotor. The breaking mechanism and speed sensor might be more reliable and cost less than the servomotor concept.

This mechanism does not have to be extremely accurate during cruise. The only strict requirement is that it resists any rotation during hover. To achieve this, there might be a friction brake on one end of the worm and a speed adjusted hydromatic resistance on the other end.

Concern: During cruise the worm and wheel are consuming power; whereas with the servomotor they are freewheeling and the servomotor is consuming a very small amount of power. Consider using the motor and encoder with the 9:1 ratio. The motor would only act as a brake. A strong electro-brake during hover and no electro-braking at all during cruise.

The ratio of the worm and wheel will be such that the worm can just overcome static friction when the positive braking is released. This initial overcoming of static friction might be assisted by the positive braking mechanism. This means that the variable speed controller braking mechanism must only the sliding friction. The friction and heat generated by this variable speed controller braking mechanism should be inconsequential.

Failure of the Speed Controller:

The desired failure mode is for the speed controller to freeze 'lock up'. This causes to rotors to attain the proper speed for hover and slow flight.

The actual function of the speed controller is not to control the RPM of the propeller. It is to control the RPM, from the power-train going to the propeller. By locating an overriding clutch (Sprag clutch etc.) between the speed controller and the propeller, the propeller will not come to a sudden stop should the speed controller lockup. The propeller will freewheel, at a speed based upon the velocity of the air passing through it's disk.

Perhaps a strong friction clutch will have to be located on the other side of the speed controller. Upon a failure of the speed controller, this clutch would be forced to slip for a short period of time until the rotors came up to the RRPM_HOV. The additional load required to accelerate the rotors will also slow the engine down temporarily. What will happen if the rotors come up to the RRPM_HOV while the craft is travelling at a very fast cruise??? Profile drag will probably stop the rotor from attaining RRPM_HOV until the forward speed has decreased.

Should the friction clutch ever fail, by loosing its driving force, it will put the craft into autorotation. Should the friction clutch fail by locking-up there may be problems.

Calculations:

Algorithms:

.Look into further angle of repose.

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Q =( hp * 5252) / rpm

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

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F = μ * N

Where: F is the force of friction; μ is the coefficient of friction; N is the normal force

Coefficient of static (breakaway) friction for steel on sintered bronze, lubricated (μ) is 0.13

Coefficient of sliding friction for steel on sintered bronze, lubricated (μ) is 0.025

The difference between the two coefficients of friction is about 5:1. The servomotor must be capable of overcoming both.

During hover:

The speed of the propeller is 0 rpm.
The servomotor is stopping the worm & gear from turning, therefore no friction is generated at the worm & gear.
All the power is going to the rotors.
The speed of the rotors is 550 rpm (tip speed is 550 fps, tip mach number is 0.49).
Collective changes in the rotor may effect the rotor and engine speed but, will have no effect on the propeller speed due to the locked servomotor.

During cruise:

The speed of the propeller is 2,400 rpm.
The speed of the rotors is 412 rpm.
The servomotor is turning the worm at 2400 * 10:1 =24,000 rpm.
The servomotor and worm are turning the gear at the same rpm that the drive-train wants to turn the propeller at, therefore negligible friction is generated at the worm & gear.
Collective changes in the rotor may effect the rotor and engine speed, but will have no effect on the propeller speed due to the constant speed of the servomotor.

During medium forward flight:
For Speed Controller Calculations:

Currently the below is based on mid position of all variables. Is this valid? Check

The servomotor is turning the worm at 0 + 24,000 / 2 = 12,000 rpm.

The speed of the propeller is 0 + 2,400 / 2 = 1,200 rpm.

The speed of the rotors is 550 + 412 / 2 = 481 rpm.

Collective changes in the rotor may effect the rotor and engine speed, but will have no effect on the propeller speed due to the constant speed of the servomotor.

The following is based on the engine operating at maximum power:

Total power from engine at 2,400 RPM = 160 hp.

The power at medium forward flight is probably 67% of total, which is 107 hp.

The differential sends 1/2 of the total power toward the propeller. = 53.5 hp. Is this valid???

Q =( hp * 5252) / rpm

The torque that is heading toward the propeller. 53.5 * 5252 / 1500 = 234 ft-lb = 2,810 in-lb.

If 1/3⁽¹⁾ of the 2,810 in-lb. is used to rotate the prop then 1,880 in-lb. goes to the worm & gear.

If the gear has a 3" radius then the normal force (N) at the gear tooth = 1,880 / 3 = 627 lbs.

The force of sliding friction (F) therefore is; F = μ * N; F = 0.025 * 627 lbs = 15.68 lbs.

Speed of worm is 10 * 1,200 = 12,000 rpm.

The worm has a 2" dia. = 0.52 ft. circumference.

12,000 * 0.52 = 6,240 fpm = 104 fps.

1 hp = 550 ft-lb/sec.

Friction consumes 15.68 * 104 / 550 = 2.96 hp.

2.96 / 160 = 1.9% of the TOTAL power, 2.96 / 107 = 2.8% of the CURRENT power, is the maximum consumed by the SC. This is at the medium flight speed of +/- 90 mph.

Note that this power loss is a curve that is zero at hover, climbing to 2.8% at medium forward speed, then dropping back to zero at cruise. This is probably less the any vari-speed unit.

In addition to the 2.8% will be another 2 - 5% for the many gears used in the power train, particularly the 3-way differential.

However, the above assumes that the force of 15.68 lbs is normal to the surface of the worm. In actuality, the worm has an angle of repose of 1.43º. Look further into this angle and see why it is appears to be in disagreement with Boston. Because of this 'slope' the gear is contributing to the sliding by an amount very close to the sliding friction.

(1) Guestimate

Efficiency of Worm and Gear: from Boston catalog.

Ratio = Gear teeth / Worm threads = 40 / quad thread = 10:1 Used 6 diametral pitch to get 3.33" radius. The worm has a 2" pitch dia.

Efficiency = E = (tan(γ) * (1 - f * tan(γ))) / (f + tan(γ))

where γ = Worm lead angle = 18º26' for 6 diametral pitch - quad

where f = Coefficient of friction. For a bronze worm gear and steel worm the range of 0.03 / 0.05 may be assumed. Using 0.04.

For Quad; E = (tan(γ) * (1 - f * tan(γ))) / (f + tan(γ))

E = (tan(18º26') * (1 - 0.04 * tan(18º26'))) / (0.04 + tan(18º26'))

E = (0.334 * (1 - 0.04 * 0.334)) / (0.04 + 0.334)

E = (0.334 * 0.9866) / 0.374

E = 88%~ If 2/3 of 62 hp is going to the Speed Controller and the worm & gear is 88% efficient then the lost horsepower will be 5.0 hp. This is 5/124 = 4% of the engines power. This is not good but it is better than below.

For Single; E = (tan(γ) * (1 - f * tan(γ))) / (f + tan(γ))

E = (tan(4º46') * (1 - 0.04 * tan(4º46'))) / (0.04 + tan(4º46'))

E = (0.083 * (1 - 0.04 * 0.083)) / (0.04 + 0.083)

E = (0.083 * 0.9967) / 0.123

E = 67%. ~ If 2/3 of 62 hp is going to the Speed Controller and the worm & gear is 67% efficient then the lost horsepower will be 13.7 hp.

The single thread worm are comparatively less efficient but they may be employed when self-locking is important. [Source ~ MH p.188]. Perhaps the single thread will result in a lower ratio, therefore a lower rotational speed of the worm and therefor a slower servomotor and perhaps less power is wasted because of the slower speed.

Speed Controller: Two Worms and One Wheel

One worm is located on the shaft between the motor and the propeller.
The other worm is located on a shaft that has a small speed-control motor at one end of it and/or a fan at the other end.
This fan is driven by the free-air velocity and the worm driven by the fan and/or the speed-control motor.
Between the two and intermeshed with the two is a wormwheel.
The worms and wheel have a ratio ≥ 10:1. Therefore, neither worm can be driven by its motor unless the other worm is driven by its motor.
The worms can rotate the wheel but the wheel cannot rotate the worms, due to sliding friction. Therefore, the speed of the wheel and both worms will be determined by the rpm of the slowest (retarding) motor.
The power to the main motor determines its rpm and this speed controller determines how much of this rpm goes to the propeller and how much of it goes to the rotors.
If a fixed pitch propeller is used, with its maximum tip speed is set for cruise, then the thrust of the propeller is regulated by this speed controller

Thoughts:

Would there be any advantage in connecting the speed-control servo to the worm gear via a fluid coupling and weighting the worm gear so that rotational inertia will 'smooth out' the RPM changes of the servo?

Initially displayed: July 7, 2005 ~ Posted to PPRuNe: July 9, 2005 ~ Last Revised: Friday, September 14, 2008

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.