Tilting torques/forces

Controlling the gyro when it's in the air is pretty vital obviously...how do we take into account the gyroscopic effects of the rotor?

how much torque is required to tilt the rotor? how is this calculated? do the hinges(lead-lag, flap) make it easier?

The pilot doesn't actually apply forces to tilt the rotor directly, but applies control forces that cause cyclic blade pitch changes, and those pitch changes cause lift-distribution imbalances in the spinning rotor disk which causes to rotor to tilt. So, the force to tilt a rotor is aerodynamic and comes from the airstream, not form the pilot.

you're talking about helicopter type (fully articulated) rotors where blade pitch can be changed....right? that's what it sounds like.

i'm just talking about a gyroplane (3 bladed) rotor.

thanks
mo

Several years ago I made a 20 inch model of a two bade teeter rotor attached to a stick. I would let it spin up in the wind. Until it reached some critical rpm I could feel the gyroscopic effect. The forces were significant. Suddenly it felt as though I had power steering. The slightest movement with the smallest force would make the disk tilt.

I feel the same thing with the rotor on the Predator. I start to spin up as I cross the hold short line. As I turn to the centerline and the wind changes relative direction I need to change the tilt of the disk. the stick force is substantial. Suddenly at somewhere under 100 rpm it becomes very easy to tilt the disk. In flight the stick forces are fairly light.

I had trouble grasping this concept; I am no longer troubled by it.

I would expect it to work the same way with a three blade system.

Thank you, Vance

Maybe this will help:

wisemo said:

you're talking about helicopter type (fully articulated) rotors where blade pitch can be changed....right? that's what it sounds like.

i'm just talking about a gyroplane (3 bladed) rotor.

thanks
mo

Click to expand...

No, I'm talking about the standard two blade teetering rotor like what is on most gyrocopters. Although each blade doesn't have the ability to individually pitch change like in a helicopter because they are bolted to the hub bar fixing them in place, they do cycle for cyclic control. Study how a fixed pitch R/C helicopter works, gyros work similar.

Wisemo, Do you have the August/October 2008 PRA magazine, look at page 18 & 19 it may help explain what Alan and Chuck were alluding too.

Explained in another way..

Explained in another way..

Alan_Cheatham said:

The pilot doesn't actually apply forces to tilt the rotor directly, but applies control forces that cause cyclic blade pitch changes, and those pitch changes cause lift-distribution imbalances in the spinning rotor disk which causes to rotor to tilt. So, the force to tilt a rotor is aerodynamic and comes from the airstream, not form the pilot.

Click to expand...

I have just grasped this a few months ago - I think so I would like to explain it a bit more or in another way:

Rotor blades are heavy, long and spinning fast so one would expect a big force to tilt this ‘disk´- and it does take a big force!
BUT the pilot only add the force needed to tilt the rotor blades around their long axis – the rotor head make sure, that the spinning blades are tilted back and forth a few dg. around their long axis 2 times per rev. = cyclic change of blade pitch = cyclic pitch.

In illustration 1 the rotor head is tilted for a left turn by a little control force, and the rotor blades are perpendicular to the main frame.
At this exact position there is equal lift from each rotor blade.

In the next 90 degree there is an increase of tilt of the rotor blades around their long axis – see illustration 2.

This tilt gives an increase of pitch on one blade, and a decrease of pitch on the other!

The result is a lower lift on the front half of the ‘disk’, and a higher lift on the back half.

NOTE: For a spinning ‘disk´’, gyroscope, the force is in 90 degree - so the ‘disk’ tilts to the left.

I'm in a bit of a pickle - I'm automating the control of the tilt using actuator motors. But I'm clueless as to how much torque/force I'd need... here's just the rotor setup...

wisemo,

To your question.

Here is the rotorhead of the Breguet G-11E coaxial helicopter.

It might be applicable to a gyrocopter by eliminating the pitch bearings and perhaps locating the CVJ (uniball) above the flapping hinge elevation. The zero flapping hinge offset should significantly reduce the required control forces.

The rotor has delta3 hinges that might, or might not, be an advantage.

This rotorhead might also allow experimentation, in that stiffening the flapping hinges will give greater control authority; at the expense of increased control force.

People on this forum with more experience in the design and history of gyrocopters may elaborate on the above.


I would ask the question; why do you wish to use a 3-blade rotor instead of a 2-blade rotor?

Dave

The control force is simple to calculate. It is proportional to the height from teeter bolt to universal pivot and the following rate of the rotor, its velocity of precession.

The angular velocity of precession is proportional the precessing torque divided by the product of rotational velocity and moment of inertia of the gyro wheel: Ω = T/ωI if you like Greek letters*. Units are radians/sec, ft-lbs and slug-ft².

The precessing torque is generated aerodynamically by cyclic pitch input.

Say you want to tilt the rotor at the rate of 0.1 radian/second (~5.7º/second), the rotor rotational velocity is 40 radians/second and its moment of inertia is 50 slug-ft² (382 rpm and 2 ea 10 foot blades of uniform mass distribution weighing 24 lb each).

Then 200 ft-lb of precessing torque is required (0.1 x 40 x 50)

If you did all the math, you would come up with a partial differential equation that could not be solved on paper except by plotting and counting squares due to the non-linear variables. I don’t have the foggiest notion of how computers do it but I presume they count squares too.

But the bottom line is 70% of rotor radius is the mean value for the entire rotor.

So if each blade has to generate a 100 ft-lb moment of force, we can say everything is concentrated at 70% radius and the lift increment required is 100/7 = 14.3 lb.

The mean velocity of the rotor blade is ωr*0.7 = 280 fps.

The lift of an airfoil is: ½*p*V²*A*Cl where p is air density, slugs/ft³, ~0.0023 at sea level, V = air velocity, fps, A = rotor blade area, ft² and Cl = lift coefficient, close to 1/10 of angle of attack for normal airfoils.

Say the chord is 6” for a blade area of 5 ft².

The differential pitch angle required is: (14.3/0.5*0.0023*280²*5)*10 = 0.3º

This means the rotor spindle must be tilted 0.3º relative to the tip plane axis to keep the rotor tilting at a 0.1 radian/sec rate.

If the suspended weight is 400 lb and the distance from teeter bolt to gimbal pivot is 6”, the control moment is: sin0.3º*6”*400 lb = 12.6 in-lb.

These approximations are simple but a little longwinded.

This assumes the blades have zero pitching moment coefficient and are balanced about their aerodynamic centers. If not, things go to hell in a handbasket.

*This forum doesn't like Greek letters either. I had to change some back to their English equivalents.

Its always a challenge for a SCG to compriehend wot your tech posts say CB, and this one is no exception.

I was follown it pretty well, and thinkn " but wot about..... and wot about..." then you said; This assumes the blades have zero pitching moment coefficient and are balanced about their aerodynamic centers. If not, things go to hell in a handbasket.
You answered my question spoton before i even asked it.

Thats why i never take much notice of any of these flash harry computer modle thingys. They are only as smart as the smartest one feedn it.

Wot’s a “flash harry computer modle thingy”, mate? You confuse me when you talk Austraaylian.

Oops!, Wisemo, I had overlooked your very nice CAD graphics. The stuff I ran through in post #11 only works for central flap hinges; teetering or zero offset as in a Sikorsky S-51.

The centrifugal pitching couple, sometimes known as the “T” bar effect, pretty much rules out tilt head control if the flap hinges are offset from the center of rotation by any appreciable amount.

“flash harry computer modle thingy”
Anythn thats looked at as gospel by the plastic punchers.
IOW, people who have no idea of the real world, and recon anythn and everythn that comes outa a computer program is absolute truth.

C. Beaty said:

Oops!, Wisemo, I had overlooked your very nice CAD graphics. The stuff I ran through in post #11 only works for central flap hinges; teetering or zero offset as in a Sikorsky S-51.

The centrifugal pitching couple, sometimes known as the “T” bar effect, pretty much rules out tilt head control if the flap hinges are offset from the center of rotation by any appreciable amount.

Click to expand...

Ah I think that answered another of my questions.

I wondered, why we do not see more tilting head rotor control in helicopters. Juan de la Cierva experienced massive vibration when he made his first tests in that direction. He ended up with a direct per blade pitch change as we know today.

The norm for tip jet helicopters is a tilting head.

With torque driven rotor, a component of drive torque shows up in the cyclic controls; no different from the sideways kick of the stick on a gyro when getting aggressive on the prerotator.

Imagine a drive shaft with a universal joint in its center mounted on a flat surface via pillow blocks and running at an angle. A torque applied to the input not only causes the output to rotate; it tries to lift the output shaft off the surface or push it down, depending on direction of rotation.

The control force moment of a torque driven tilt hub will be: tan of ½ tilt angle x drive torque.

birdy said:

“flash harry computer modle thingy”
Anythn thats looked at as gospel by the plastic punchers.
IOW, people who have no idea of the real world, and recon anythn and everythn that comes outa a computer program is absolute truth.

Click to expand...

Thank goodness. For a moment, I thought you were talking about Harry S. and his antique Oldsmobile Toronado with genuine fake wire wheel hubcaps.

Kai, the Cierva machines that did not include collective pitch had offset gimbal tilt-heads, same as we use today. They did experience a bit of T-bar effect, since the flap hinges weren't quite on the spindle axis. This added some control "heaviness".

I haven't seen a close-up of Cierva's jump-takeoff heads. These necessarily incorporated collective pitch, which can be awkward (but not impossible) to accomplish with a tilting head.

Chuck, Doug,

I just read through Bruce Charnov's book and the problem of massive stick vibration in the first "direct control" models was mentioned, I'm trying to make ends meet (which apparently weren't meant to meet).

But in anything BUT a teetering assembly you always have some leverage because the teetering hinges have to be off center. So it is merely a matter of how much.

If you introduce this element of constant changing load on the control mechanism, this seems like negligence, opening the door to fatigue on metal as well as pilot.

How did they manage in tip-jet helis, i wonder?

Kai