C-4 "Centenario" Autogiro project

Jean Claude

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As you say, the high aspect ratio of the wings was not possible at that time without the use of struts to prevent them from folding upwards. And it is still the same idea that prevails for the C4 blades: The shrouds must to avoid it folding upwards, i.e avoid any coning. This is what the picture shows.

It is only later that La Cierva has the idea to use the centrifugal force to prevent the blades from folding up too much: From the C5, he can increase the aspect ratio without the need of shrouds
 
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wolfy

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I am amazed at how stiff the rotors appear in the photo's, if it's said the blades aren't stiff enough so they used "flying wires" but look at them in droop they don't have any "landing wires" but they hardly droop at all.

wolfy
 

Jean Claude

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For the C4, La Cierva was not yet thinking about centrifugal force to keep the blades from lifting, so he designed them to be as light as possible with "flying wires."
Lightness and thickness give you the answer.
Today the blades are purposely heavy to reduce coning by centriffugal force and their stiffness is not necessary. Except at rest!

Sans titre.png
 
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Arco

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I think we are misunderstanding the picture of the C4.

The C4 blade has coning angle even the blades "looks" horizontal.

The coning angle is the angle between a plane perpendicular to the rotation axis and a line that pass through the center of gravity of the blade.

Cierva was a clever engineer but also a mathematician, he always was checking experiments versus theory.

He was aware of the centrifugal force effect from the early design, in fact, he designed the rotor flapping axis of the blade in the early models to be flat in flight even the have a big coning angle. To achieve that he located the flapping articulation far below the horizontal plane.

I re-use the picture from Jean Claude to explain the forces and the coning angle.

Later on, he changed the design of the blade root to be straight and the flapping articulation was located at the same plane than the blade.

1662216265445.png
 

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Jean Claude

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I agree with you, but it is customary to use the word "coning" for the blades angle, rather for the theoretical angle you draw.

The fact remains that with guy wires you can then design a seesaw rotor in which all mass elements of the blades are aligned with the flapping hinge. This way, the usual vibrations due to the coning disappears.

Now, if we put two rotors like this crossed at 90 degrees, then their rotating drags oppose each other and this usual source of vibration also disappears.

That's what I tried to says in posts #26
 

Aerofoam

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I think we are misunderstanding the picture of the C4.

He was aware of the centrifugal force effect from the early design, in fact, he designed the rotor flapping axis of the blade in the early models to be flat in flight even the have a big coning angle. To achieve that he located the flapping articulation far below the horizontal plane..
That explains the anhedral!
 

Arco

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I agree with you, but it is customary to use the word "coning" for the blades angle, rather for the theoretical angle you draw.
I draw this angle because I follow the definition of the coning angle I learnt at the Aeronautical engineering university, is not an invention to win the discussion.

About the idea to have wires in the blades, yes, I am agree, they will increase the rigidity in bending. The blades will behave like a rigid body and the effect and the 2 per rev vibrations due to the coning angle will disappear.
 

Arco

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There are no instruments with the appearance of the period, so we have decided to make them too. They are based on models of the period that are in the Museo del Aire in Madrid.
They are already done and are in the calibration and adjustment phase.IMG-20220330-WA0010 1.jpgIMG-20220727-WA0011.jpgIMG-20220809-WA0020.jpgIMG-20220810-WA0004.jpg
 

WaspAir

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You could probably find a market to put those into production.
 

Jean Claude

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The point of application of the centrifugal force on a blade of homogeneous mass is at 2/3 R, ie close to its lift center.
So, with the angle of 8° of the shroud and the Rrpm of 140 rpm mentioned in the book, we can suppose that the blades weighed about 15 Kg each if the total mass was 500 kg
Sans titre.png
 

XXavier

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You state that the point of application of the centrifugal force is at 2/3 R. I understand that a 'centrifugal field' is non-uniform, so the center of mass does not co-incide with the center of 'centrifugal gravity', but –excuse my ignorance– how do you arrive at the figure of 2/3...?
 

Jean Claude

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A very small blade element of width dr weighs a mass dm such that dm/M = dr/R hence dm = (M/R). dr
The centrifugal force on this element is: dFc= dm . Ω^2 . r
And the sum of these elementary forces is therefore ʃ dFc= ʃdm. Ω^2 . r = Ω^2 .(M/R) ʃ r.dr = Ω^2. (M/R) ½ r^2
Hence Fc= ½ M Ω^2. R when r varies between 0 and R.
As if all the mass of the blade (M) was concentrated at R/2 ie at its center of gravity
Sans titre.png
On the other hand the flattening moment about O due to each elementary force (assuming a small coning angle α), is :
dC = dm Ω^2. r. r Sin α or dC = dm Ω^2. r^2 . Sin α = (M/R). dr Ω^2 . r^2. Sin α
And the sum C of these elementary moments is therefore worth:
ʃdC = (M/R). Ω^2. Sin α ʃr^2. dr = (M/R) Ω^2. Sin α . ⅓ r^3
Hence C = ⅓ M . Ω^2. R^2. Sin α, when r varies between 0 and R

Now, the total centrifugal force ½ M Ω^2. R applied at a distance X from o would produce a flattening moment C = ½ M Ω^2. R. X Sin α, yet it produces the moment C = ⅓ M Ω^2. R. R Sin α
This means that X = 2R/3
 
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Jean Claude

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The original rotor of the La Cierva C4 had four guyed blades, articulated in flapping. In flight, these blades with broad chord turned slowly (140 rpm) according to a flat disc very characteristic of this model.

Double "seesaw rotors" in plane is possible
Sans titre.png

The absence of a drag hinge could seem to us today likely to produce very tiring Coriolis constraints at the blade roots. But it is not the case.

Here, it is by the cables of equi-repartition that the forces of Coriolis is transmitted from a blade to the following one. This is easy since these cables is attached far from the center (at 2/3 R) and very stretched by their own centrifugal force. Thus, the angular spacing between the blades is rigorously maintained by the cables, and the roots are no longer subjected to harmful cyclic stresses

Sans titre.png
In addition, here the centrifugal forces pass from one blade to the opposite one through the below shrouds. Thus, in flight, the roots of the blades are only subjected to the low torsional stresses due to the flapping.

Just my two cents.
 

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Interesting concept combining the drag cables into dual purpose as "Struts" to support load.
I like the double teeter and have thought about building a working model for over 20 years, but I would eliminate the cables
for 2 main reasons.
1. Klugy as hell!, yeah not very technical, but accurate. There are so many ways to achieve enough strength and, or flexibility with composites......
2. The cables are rather significant drag sources at modern rotor speeds, especially at the intersection with the rotor.
With Cierva's design, the cables are at the TE, already in dirty turbulent air flow and it is a low speed rotor, so the penalty is not significant.
With a modern 300Rpm + rotor and cables in full air flow, the penalty would be unacceptable to me. In addition, when you have an intersection
of flying surfaces, it usually magnifies the drag and turbulence immensely. Since I am not versed in high RE surfaces, I can't give examples, but I
know it is more of a problem at high RE #s. Not trying to create an argument, just an opinion that the double teeter could be highly effective, but
ditch the flying wires.....
 
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Jean Claude

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Of course! This is just my suggestion for the Arco project with an rotor looking of true C4
With this design, the blades can be designed light, as fixed cantilever wings around the point at 2/3 R

Cleaner drawing of the head:
Sans titre.png
 
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Arco

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Thanks, Jean Claude.

Yes, looks very very interesting.

The first trial will be using the conventional rotor that we know to minimize the risk, but we can do the four-blade rotor when the autogiro will be tested and the new way to fly tested and safe.

In any case, if we do the 4 blade rotor head, will be always a fix rotor, no cyclic input.

Our main concern is how to control the autogiro in the take-off and landing, as you know the autogiro will be controlled by the ailerons, and tail surfaces.

In a gyro the control is done by the cyclic so even we have cero airspeed we have control, but in our case, we need a minimum FW airspeed to control the autogiro.

From the other side, the current airstrips are runways, (longitudinal lines) with no possibility to align the take-off and landing path with the wind. In the old times were circular strips facing the wind.

This is our main concern.

How to control the autogiro in the case of cross wind?

We need a minimum speed to have “control” by the ailerons like a FW, but in a FW you can reduce the incidence angle of the wing after take-off to kill the lift. In our case the rotor will be turning, and our “wing” will have the chance to tip over the ship even when our forward speed will be cero.

We define as critical:

- The prerotation phase, with the rotor close to 30degress incidence.

- The take off run in the case of cross winds, with no lateral control.

- The rolled landing like a FW.

- The time from the airspeed is almost cero to the time the rotor will be stopped. In this time the risk of tip over is very high.

The idea we have to minimize those events, even the rotor is fix, is to have two positions: flight and ground for it. In fly mode will have the correct incidence and in ground mode will be parallel to the floor.

This is today our main concern from rotor design point of view. When will be solved we will move to a rotor closer to the C4 design.

Thanks for all your inputs.



See a video to explain the concept.
 

Jean Claude

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Arco
Air traffic controllers are there to ensure improve safety, not make it worse. They will certainly understand your unusual final path to land across the runway if the wind requires it .

In these conditions of front wind, many Pitcairn and Kellett did not seem to suffer from the control by ailerons and horizontal surfaces.
 
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Jean Claude

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Here are the results of my calculations for this four bladed rotor, taking into account the radial flow, and assuming a total mass of 500 kg, with 2° of pitch setting.
At 25 m/s steady (Mu = 0.34) Rrpm = 140.5 rpm a1= 3.4° b1 = 0.8° A.o.A shaft = 6.6° Rotor drag = 920 N
At 20 m/s " Rrpm = 139 rpm a1= 2.8° b1 = 1.1° A.o.A shaft = 9.9° Rotor drag =1130 N
At 15 m/s " Rrpm = 133 rpm a1= 2.2° b1 = 1.4° A.o.A shaft = 15.2° Rotor drag =1430 N
 

Arco

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Jean Cloude, could you do it for 4 blades, 8.5 m diameter, 22.5 cm chord for a MTWO of 350 kg.
we want to keep minimun 310 to 320 rpm, so we wich angle do we need to have in the blades?
 
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