Translational lift

Aviator168-aircraft

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Just curious. Anyone here studied the translational lift generated by a gyroplane rotor system. Like percent of total lift attributed to the forward motion of the gyroplane.
 

XXavier

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100% of the lift generated by a gyro rotor in level flight is due to the deflection of the 'relative wind' downwards. Here's a drawing due to Jean Claude:

Captura de Pantalla 2022-09-10 a las 22.18.40.png
 

XXavier

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What about during a vertical descent in which a gyroplane has no forward speed?
That's an special case. The rotor is powered by the relative wind blowing from below the disk, and the rotor does indeed produce a certain amount of lift, that has the same value and opposite direction as the weight of the gyro. Hence, the machine falls at a constant velocity...
 

Jean Claude

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It is enough to consider the speed in relation to the upstream infinite, and the angle of the disc in relation to this speed.
For example in vertical descent at Vz =-10 m/s, we have Vupstream= +10 m/s and A.o.A = 90 degrees.
This case can be treated like any "forward" flight.

Of course since the drag is always the force opposite to this speed and the lift is always the force perpendicular, then in vertical descent, it is the drag that will oppose to the weight, while the lift will be zero
 

XXavier

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It is enough to consider the speed in relation to the upstream infinite, and the angle of the disc in relation to this speed.
For example in vertical descent at Vz =-10 m/s, we have Vupstream= +10 m/s and A.o.A = 90 degrees.
This case can be treated like any "forward" flight.

Of course since the drag is always the force opposite to this speed and the lift is always the force perpendicular, then in vertical descent, it is the drag that will oppose to the weight, while the lift will be zero

True, if we consider the disk as a whole, but –in vertical autorotation– the blades do indeed produce lift (and drag, of course...). At any blade section, the aerodynamic force has a lift component...
 

Aviator168-aircraft

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It is enough to consider the speed in relation to the upstream infinite, and the angle of the disc in relation to this speed.
For example in vertical descent at Vz =-10 m/s, we have Vupstream= +10 m/s and A.o.A = 90 degrees.
This case can be treated like any "forward" flight.

Of course since the drag is always the force opposite to this speed and the lift is always the force perpendicular, then in vertical descent, it is the drag that will oppose to the weight, while the lift will be zero
The blades of a gyroplane rotor work exactly the same way as the wings of a glider. Both lift and force to spin the rotor, as well as drag are produced.

XXavier,
I do not agree verticle descent is a special case. The rotor works exactly the same way regardless of whichever direction the gyroplane is flying in.
 

XXavier

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The blades of a gyroplane rotor work exactly the same way as the wings of a glider. Both lift and force to spin the rotor, as well as drag are produced.

XXavier,
I do not agree verticle descent is a special case. The rotor works exactly the same way regardless of whichever direction the gyroplane is flying in.

With a forward component of the rotor's movement, there's an additional downwash component that is lacking in pure vertical autorotation. Hence, the sink speed in a glide is always less than in vertical autorotation, and –given enough horizontal velocity– the rotor's downwash becomes strong enough to generate a force that allows for level flight.

Besides –in vertical autorotation– the aerodynamic force generated by the blades has a symmetrical, circular distribution that is different to the unsymmetrical distribution observed in forward flight.
 
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Jean Claude

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Until A.o.A about 50-60 degrees, it's just a matter of angle:

The induced speed is always axial to the rotor, so when the angle of attack of the disk is low, the forward speed combines with the induced speed to give a thrust mainly by deflection, secondarily by slowing down of the flow through the rotor (V1~ V0)

And when the angle of attack of the disc is large, then the combination with the induced speed, gives a thrust mainly a slowing down, secondarily by deflection of the flow passing through the rotor (V1<< V0)

Until then, there is continuity in the formulas
Sans titre.png

Only when Vi ~ V0 (i.e. A.o.A >70°), then the flow through the disc is impeded and produces a reduction in disk CL (i.e a sink rate increased)
 
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If you do a blade element analysis, everything is the same and what is changing are the environmental parameters. Better yet, put the rotor into a CFD simulation, and you can see how autorotation works. Need to stop looking at the rotor as a disk.
 

XXavier

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If you do a blade element analysis, everything is the same and what is changing are the environmental parameters. Better yet, put the rotor into a CFD simulation, and you can see how autorotation works. Need to stop looking at the rotor as a disk.

I see the problem in a slightly different way as Jean Claude, but do also contemplate the rotor as a disk. It may be a fiction, but it's a useful fiction.

Instead of the 'parachute' approach held by JC, I prefer to believe that –in vertical autorotation– the disk does indeed produce a a downwash that creates a certain amount of vertical thrust, a force having the same value and opposite direction as the weight of the gyro. So, the gyro sinks at a constant velocity...

ssf01k.jpeg

When the motion is not strictly vertical, but there's also an horizontal velocity component, we have an additional downwash due to the disk working as a wing. A drawing by Jean Claude follows, taken from one of his postings in a French forum. I have dared to add a translation...

Captura de Pantalla 2022-10-03 a las 13.54.04.png
 

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Xavier,
Due to the high forward speed the environmental parameters of the rotor are very similar to those of a solid disk i.e lift mainly due to deflection, as showed in this sketches.
But the difference with the solid disk becomes progressively more pronounced as the angle of attack increases, due to the thrust due to slow down of the airflow through the rotor.
Only beyond 70 degree, the slowdown flares the air flow so much that a certain discontinuity appears.
Sans titre.png
 

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Vance

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Just curious. Anyone here studied the translational lift generated by a gyroplane rotor system. Like percent of total lift attributed to the forward motion of the gyroplane.
In my opinion a gyroplane does not experience the effect of translational lift because a gyroplane does not takeoff from a hover.

In my opinion a vertical descent is not the same as a hover.

The thread seems to have developed into a different question that I will not attempt to answer.
 

Jean Claude

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If you do a blade element analysis, everything is the same and what is changing are the environmental parameters. Better yet, put the rotor into a CFD simulation, and you can see how autorotation works. Need to stop looking at the rotor as a disk.
If you send me your e-mail address, I can send you my calculation file of a rotor in forward flight with 11 elements per blade and 24 azimuth positions
 

XXavier

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Xavier,
Due to the high forward speed the environmental parameters of the rotor are very similar to those of a solid disk i.e lift mainly due to deflection, as showed in this sketches.
But the difference with the solid disk becomes progressively more pronounced as the angle of attack increases, due to the thrust due to slow down of the airflow through the rotor.
Only beyond 70 degree, the slowdown flares the air flow so much that a certain discontinuity appears.
View attachment 1155996


I appreciate and respect, of course, your knowledge of the matter, but I'm not fully convinced...

The sketch reminds me of an actuator disk, but one working in reverse... An 'actuator' that slows down the air stream (as if it were a porous disk, a wire mesh one, for example...). But the rotor 'disk' of a gyro is not a merely passive element, but an active one, a 'disk' that –in vertical autorotation– generates a certain thrust due to the downwards-directed airstream caused by the action of the rotating blades. Of course, the downwards velocity of that stream is 'absorbed' by a much faster column of ascending air (as seen by an observer on the sinking disk) so that the net flow is upwards...

I believe that the disk behaves like thrust-generating element, and not as a drag-generating parachute. Yes, I know that the drag coefficients are similar, but disk and parachute are essentially different. In my opinion...

Concerning the discontinuity that you mention, it would be interesting to see, in a numerical simulation, if it really exists. Intuition leads me to believe that it doesn't. An indication of that discontinuity could be a sharp variation of the RRPMS at a certain value of the disk's AoA...
 

Aviator168-aircraft

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I believe that the disk behaves like thrust-generating element, and not as a drag-generating parachute. Yes, I know that the drag coefficients are similar, but disk and parachute are essentially different. In my opinion...
Yes XXavier. One cannot simply look at the rotor as a disk. You can have a rotor with more than 2 blades but still, have the same disk area. I took the data of Cavalon at gross weight. If the rotor is considered a flat plate, descending from 3000 ft, the vertical descent rate is more than 2000ft/m.

Cavalon gross weight 560kg = 5488N
Cavalon rotor diameter 8.8m -> 60.79 square meters
Flat plate drag coefficient 1.98
Air density at 3000 ft 0.84

Vertical velocity = ((5488 * 2) / 0.84 / 1.98 / 60.79)^0.5 = 10.42 m/s = 2062ft/minute
*Note descent rate without the passenger (100kg, person + luggage + fuel) is about 100ft/minute less.

Can someone confirm this?
 

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My recollection is a typical Cavalon near maximum takeoff weight will descend somewhere around 1,400 feet per minute in power off vertical descent.

Generally speaking I do not do zero airspeed vertical descents in a Cavalon during training because all of the Cavalons I have flown (7) will spin in a zero airspeed vertical descent.
 
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XXavier

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That density of 0,84 looks too low. It corresponds, probably, to 3000m, not 3000ft.

Once, years ago, I made a couple of tests of vertical autorotation with an ELA, but got scared and gave up when passing the 1000 ft/min mark, more or less. The gyro yawed perceptibly, too. The 'terminal velocity' for a pilot-only ELA is probably 1500 ft/min...
 
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Aviator168-aircraft

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My mistake. It was meters. That will put the plane at 10k ft. It would be 1.08ish at 3000ft. Re-work the number to about 1800ft/minute.

Generally speaking I do not do zero airspeed vertical descents in a Cavalon during training because all of the Cavalons I have flown (7) will spin in a zero airspeed vertical descent.
Hope you don't mean SPIN as the way a fixed wing does.

Once, years ago, I made a couple of tests of vertical autorotation with an ELA, but got scared and gave up when passing the 1000 ft/min mark, more or less. The gyro yawed perceptibly, too. The 'terminal velocity' for a pilot-only ELA is probably 1500 ft/min...
A T-10 parachute descents at about 1400ft/minute
 

Jean Claude

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I believe that the disk behaves like thrust-generating element, and not as a drag-generating parachute. Yes, I know that the drag coefficients are similar, but disk and parachute are essentially different. In my opinion...
I fully agree with this, and this is the reason why the maximum lift coefficient of the rotor relatively to the swept disk area is much higher than that of a solid disk in forward flight.

Nevertheless, during a axial descent, assuming an ideal rotor with no losses due to the friction of the blades in the air, the correct adjustment of the pitch of its blades could, at best, totally prohibit the crossing of the air flow, like a flat solid disc, and thus its drag not would higher (Cd ~ 1.2).
It is in these last degrees (From 60° to 90°) that the discontinuity of functioning of which I spoke occurs.
Sans titre.png
 
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