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)