Wooden Blades and the Stability of Ken Wallis' gyros

[kolibri282]

Thank you for your patience and your thorough explanation, Jean-Claude! People now have two different views on the subject in this thread so hopefully everyone will find one that is easier to understand for her or for him.

 

[XXavier]

Juergen, it is not necessary to decompose. The physical reality is simply a regular rotation in the Tip Path Plane, without any reason to take into account its inclination to the shaft except for its cyclic pitch control action in our case.
Moreover, in the case of the helicopters pitch control by the usual swashplate, you can even totally ignore the shaft position and conclude very simply that, in the TPP, the pitch cyclic variations speed/air cancels the cyclic pich variations
View attachment 1153404
To incorporate the drag hinges, no mental contortion is necessary: It is enough to observe that this constant rotation in the TPP is no longer constant in the drive plane, for a simple reason of geometric projection on this oblique plane in which the perfect circle becomes elliptical. So, even the invocation of the law of conservation of momentum becomes a complication useless in my model.

It is unfortunate that due to an inappropriate reference plane, many engineers still do not understand why Doman or Doblhoff rotors can do without any joints and generate no Coriolis stress.

https://vtol.org/files/dmfile/AHS-History-Brief-by-Glid-Doman_Aug20131.pdf

That mention of the tilted circular tip-path plane that becomes an ellipse when projected on the plane perpendicular to the shaft is a very clear way to understand the need for the lead-lag hinges in a fully articulated rotor...
One learns a lot of new things in this forum...

 

[Jean Claude]

If αav is the average aerodynamic angle of attack of a blade element, a1 the longitudinal flapping angle, Vc its circumferential velocity, and Vf the forward velocity of the rotor, then on the advancing side its air speed is about (Vc +Vf) and its angle of attack is (αav - a1). While on the retreating side, its air speed is about (Vc -Vf) and its angle of attack is (αav + a1).

Since in the Tip Path Plane the cyclic flapping aerodynamic forces cancel out, we can write for azimuth 3 o'clock - 9 o'clock:
(αav +a1).(Vc -Vf)^2 = (αav - a1).(Vc +Vf)^2
Solving this equation gives a1 = 2 αmed . Vc .Vf / (Vc^2 +Vf^2)

For example, for an ELA07 weighing 3650 N in flight and whose rotor (8.28 x 0.22 m) rotates at 350 rpm at 120 km/h , we have : Vc = 110 m/s (at ¾R), Vc = 33 m/s, αav = 5 degrees, and the result is a1 = 2.7 degrees. This value is well confirmed by Mike's recordings

 

[XXavier]

If αav is the average aerodynamic angle of attack of a blade element, a1 the longitudinal flapping angle, Vc its circumferential velocity, and Vf the forward velocity of the rotor, then on the advancing side its air speed is about (Vc +Vf) and its angle of attack is (αav - a1). While on the retrating side, its air speed is about (Vc -Vf) and its angle of attack is (αav + a1).

Since in the Tip Path Plane the cyclic flapping aerodynamic forces cancel out, we can write for azimuth 3 o'clock - 9 o'clock:
(αav +a1).(Vc -Vf)^2 = (αav - a1).(Vc +Vf)^2
Solving this equation gives a1 = 2 αmed . Vc .Vf / (Vc^2 +Vf^2)

For example, for an ELA07 weighing 3650 N in flight and whose rotor (8.28 x 0.22 m) rotates at 350 rpm at 120 km/h , we have : Vc = 110 m/s (at ¾R), Vf = 33 m/s, αav = 5 degrees, and the result is a1 = 2.7 degrees. This value is well confirmed by Mike's recordings

Very neat and instructive, thanks...

 

[XXavier]

If αav is the average aerodynamic angle of attack of a blade element, a1 the longitudinal flapping angle, Vc its circumferential velocity, and Vf the forward velocity of the rotor, then on the advancing side its air speed is about (Vc +Vf) and its angle of attack is (αav - a1). While on the retreating side, its air speed is about (Vc -Vf) and its angle of attack is (αav + a1).

Since in the Tip Path Plane the cyclic flapping aerodynamic forces cancel out, we can write for azimuth 3 o'clock - 9 o'clock:
(αav +a1).(Vc -Vf)^2 = (αav - a1).(Vc +Vf)^2
Solving this equation gives a1 = 2 αmed . Vc .Vf / (Vc^2 +Vf^2)

For example, for an ELA07 weighing 3650 N in flight and whose rotor (8.28 x 0.22 m) rotates at 350 rpm at 120 km/h , we have : Vc = 110 m/s (at ¾R), Vc = 33 m/s, αav = 5 degrees, and the result is a1 = 2.7 degrees. This value is well confirmed by Mike's recordings

My own, clumsy and involved calculation from first principles, threw a figure of 2,63º for my ELA (3724N, Rotor 8,5 x 0,2m RPM= 335 @ 120 km/h). With Chuck Beaty's 'Rotocalc', it's 2,81º...

May I ask how did you calculate the value of αav...? It's my understanding that it's a little above the pitch value, since there's an upward flow that increases the AoA of the blade, but how do you compute that...?

Many thanks for your instructive pieces of information...

 

[Jean Claude]

The air flow through the rotor need for its rotation directly depends on the power drawn by to the profile drag of the blades.
With our usual drag coefficient this requires about S disk x 3.5 m/s. Therefore, the average angle of attack αav is ATAN 3.5 / Vc to added to the pitch setting.
This is, of course, an approximation but it gives us a better idea of how it works

 

[Jean Claude]

May I ask how did you calculate the value of αav...? It's my understanding that it's a little above the pitch value, since there's an upward flow that increases the AoA of the blade, but how do you compute that...?

The upward flow through the rotor need for its rotation directly depends on the power drawn by to the profile drag of the blades.
With our usual drag coefficient this requires about S disk x 3.5 m/s. Therefore, the average angle of attack αav is ATAN 3.5 / Vc to added to the pitch setting.
This is, of course, an approximation but it gives us a better idea of how it works
If Cd0 was zero, then no upward flow would pass through and the angle of attack of the rotor disk (TPP) would be just ATAN (Vi/Vf)
But with usual Cd0 (say 0.011), then the angle of attack of the disk must be increased by about ATAN (3.5/Vf)

 

[XXavier]

The upward flow through the rotor need for its rotation directly depends on the power drawn by to the profile drag of the blades.
With our usual drag coefficient this requires about S disk x 3.5 m/s. Therefore, the average angle of attack αav is ATAN 3.5 / Vc to added to the pitch setting.
This is, of course, an approximation but it gives us a better idea of how it works
If Cd0 was zero, then no upward flow would pass through and the angle of attack of the rotor disk (TPP) would be just ATAN (Vi/Vf)
But with usual Cd0 (say 0.011), then the angle of attack of the disk must be increased by about ATAN (3.5/Vf)

From your post, I understand that if the blade drag for zero lift Cd,0 were zero, no lift would be produced by the blades, and with the induced velocity due to the rotor at zero, the AoA of the TPP = ATAN (Vi/Vf) = 0 , so the disk would be be parallel to the airflow... But, with the real Cd,0 = 0,011 the TPP would be inclined to the airflow by ATAN (3,5/Vf).

All that, in case I got you right, and I'm not sure...

 

[Jean Claude]

What I meant was that if the air friction on the blades was zero (i.e. Cd0 =0) then there would be no need for any flow through the disk to keep its rotation, and the rotor lift would be obtained with just disk A.o.A = ATAN (Vi/Vf)
But to obtain the extra power when Cd0 = 0.011 it is necessary to increase this angle of about ATAN (3.5/Vf), in order to have a sufficient flow through

 

[JetRangerJohn]

Everyone who fly's a non-certificated gyroplane is a test pilot. Those who realize this fly very conservatively. Those who choose to tickle the unknown envelope could become statistics, and often do.

 

[Tyger]

Or flies one...