Quantifying the effectiveness of direct control

[Jean Claude]

Everyone knows that the rotor of a gyrocopter is controlled by its hub: When this hub is tilted, it causes a cyclic variation of the pitch of the blades. The cyclic aerodynamic forces generated then force them to follow the desired plane, with the angular delay necessary to produce these forces.

I quickly calculated the delay θ at the orientation of the tip path plane, as a function of the pitch rate δ. The resulting formula is
θ ~ δ. (0.75 R m Ω) / 5 P (δ: rate of change of the rotation plane, R: rotor radius, m: mass of a blade, Ω: rotation rate in rad/s, P: rotor thrust).
For my light single-seat gyro project (i.e with m = 7.5 kg, R = 3.35 m, Ω= 42 rd/s, P= 2100 N the result is: θ° ~ 0.077 s . δ °/s
For example if the roll rate is 60 degrees per second, the rotation plane lags behind the hub by 0.077* 60 = 4.6 degrees. This delay shifts the thrust sideways by G and thus automatically decreases the initial roll rate (Caution: it is not stability, just damping)

Conversely, this also means that a continuous input of θ° from the hub to the structure (thanks to the stick) is required to maintain this δ°/s rate


But what about a control input starting with a zero roll or pitch rate?
In this case, the balance of moments gives I.δ' = P.H. Sin (θ - 0.077. δ) with I the roll inertia, P the rotor thrust, H the distance of the seesaw above the center of gravity, and θ the abrupt control input angle.
This equation yields the curves below, one taking into account the roll an inertia of 90 kg.m2 and the other, a pitch inertia of 190 kg.m2


This tells us about the effectiveness of direct control

 

[XXavier]

Many thanks for your very instructive posts, Jean Claude...

Could you please describe how do you arrive at the expression (0.75 R m Ω) / 5 P...?

 

[Jean Claude]

Due to the rate δ , the mass of the blade must acquire a velocity component v perpendicular to the TPP, proportional to its distance from the tilt axis: v is therefore sinusoidal during the rotation: v = r.δ Sin (Ω.t) with Vmax = r. δ
(δ = tilting speed of the hub in rd/s, r = radius of gyration of the center of gravity of the blade and Ω : rotation speed of the rotor in rd/s)

This corresponds to a sinusoidal acceleration :
acc = dv/dt = d(r.δ Sin (Ω.t))/dt = r.ω.δ.Cos(Ω.t) whose maximum has the value: AccMax = r.Ω.δ (1)
This maximum acceleration comes from the extra lift F created by the additional maximum θ.

Now A = F/m (m: mass of a blade). By replacing A by this value in (1) we obtain: F/m = r.Ω.δ (2)
Since in steady flight the force F developed by each blade is average P/2, and that average A.o.A blade is about 0.1 rad (for our gyros), I deduce that
F = ½ P.θ / 0.1 or F = 5 P.θ And, replacing F with this value, and r (radius of force) with 0.75 R in (2) we find: 5 P.θ/ m = r.Ω.δ
hence θ = δ (0.75 R m Ω / 5 P)

 

[Resasi]

I always continue to be both impressed and mystified by the dialogue often seen between JC and Xavier.

As Vance might say, it is on a higher plane than I am capable of easily comprehending, but, I remain comforted that there are those who can scientifically monitor our excursions into the air, and if needed give us advice when things could be dangerous.

 

[XXavier]

I always continue to be both impressed and mystified by the dialogue often seen between JC and Xavier.

As Vance might say, is is on a higher plane than I am capable of easily comprehending, but, I remain comforted that there are those who can scientifically monitor our excursions into the air, and if needed give us advice when things could be dangerous.

Well, it's not exactly a 'dialogue'... I sometimes place questions, or argue a bit, and he does always give solid answers...
The participation of members like JC is very valuable for us all...

 

[raghu]

Everyone knows that the rotor of a gyrocopter is controlled by its hub: When this hub is tilted, it causes a cyclic variation of the pitch of the blades. The cyclic aerodynamic forces generated then force them to follow the desired plane, with the angular delay necessary to produce these forces.

I quickly calculated the delay θ at the orientation of the tip path plane, as a function of the pitch rate δ. The resulting formula is
θ ~ δ. (0.75 R m Ω) / 5 P (δ: rate of change of the rotation plane, R: rotor radius, m: mass of a blade, Ω: rotation rate in rad/s, P: rotor thrust).
For my light single-seat gyro project (i.e with m = 7.5 kg, R = 3.35 m, Ω= 42 rd/s, P= 2100 N the result is: θ° ~ 0.077 s . δ °/s
For example if the roll rate is 60 degrees per second, the rotation plane lags behind the hub by 0.077* 60 = 4.6 degrees. This delay shifts the thrust sideways by G and thus automatically decreases the initial roll rate (Caution: it is not stability, just damping)

Conversely, this also means that a continuous input of θ° from the hub to the structure (thanks to the stick) is required to maintain this δ°/s rate

JC,
Just saw this post from a while back. There in fact is a well known formula for rotor lag that has been in use from the earliest days in the helicopter literature. The rotor lags by 16/(gamma*Omega) seconds when it is pitched/rolled
Where:
gamma is the lock number given by (bladeChord*airDensity*bladeLiftSlope*bladeRadius^4)/BladeMomentOfInertia
and Omega is the rotorSpeed in rad/sec

Assuming all the same value as you have and a blade MOI = 28.5 kg m^2 (21 slug ft^2), I get 0.068sec versus 0.077sec that you get. All in all pretty close.

 

[SSDriver]

Huh?

 

[Jean Claude]

I get 0.068sec versus 0.077sec that you get. All in all pretty close.

Yes Raghu
The discrepancy is probably due to my approximation of 0.1 rad to carry P/2, and to your approximation of the lift slope of the blades (Mach number, aspect ratio)

 

[Jean Claude]

It has often been assumed that a pusher is more agile than a tractor, due to the more concentrated distribution of the pilot's and engine's mass .
Comparing the two distributions with the same motor, the same rotor and the same pilot, I obtain the following values:
Pitch inertia: 175 kg.m2 (pusher) 190 kg.m2 (tractor)
Roll inertia: 120 kg.m2 (pusher) 70 kg.m2 (tractor)
Yaw inertia: 105 kg.m2 (pusher) 180 kg.m2 (tractor)

With regard to pitch and roll angles obtained by abrupt entry, the discrepancies are small, since they are essentially due to the rotor, as explained above #1.

So, the statement is therefore true only for yaw.