Engine power for a ultralight

XXavier

Member
Pertaining to the quest for building a very light yet strong gyro...

Helicopters have a minimum flight weight (minimum solo pilot weight) that's needed, not just for weight and balance, but also for the minimum potential energy needed to keep rotor RPM up during a forced, engine-out landing.

Doesn't a gyro have to obey that law of physics too? I'd hate to know I got a golden trophy for a 150# gyro with a 50 HP engine and then
when that 2-stroke dies one day at 2000 ft. I can't autorotate down because I don't weigh enough to keep the rotor going.
Very little mass is required. Remember the 'Rotachute'...

Jean Claude

Junior Member
The minimum mass for the helicopter concerns the mass of the blades, not that of the load. It is simply a matter of giving the pilot enough time to reduce the collective pitch before the Rrpm decreases below that the stall of the blades.

The rotor of a gyroplane does not have this concern, since it is already at the right collective step before the engine failure.
Some say that minimal inertia is necessary to pass through severe low g turbulence. But in my opinion, this can be avoided by a suitable empennage instead of a increase in blade mass

bryancobb

Junior Member
The minimum mass for the helicopter concerns the mass of the blades, not that of the load. It is simply a matter of giving the pilot enough time to reduce the collective pitch before the Rrpm decreases below that the stall of the blades.

The rotor of a gyroplane does not have this concern, since it is already at the right collective step before the engine failure.
Some say that minimal inertia is necessary to pass through severe low g turbulence. But in my opinion, this can be avoided by a suitable empennage instead of a increase in blade mass
I have always been taught that a 100kg mass that is 1000 meters above the ground has more potential energy to consume on the way down, than a 20kg mass that is 1000 meters above the ground. That seems true to me.

Let's assume STEADY STATE descent and not the helicopter entry phase. Both the hypothetical gyro and the hypothetical helicopter are in a stable descent where all numbers are steady. For a meaningful comparison, lets say the helicopter pilot does not change collective during the stable descent or to cushion the touchdown. Let's say he only flares like the gyro pilot does.

It takes a certain amount of energy to slow the descent rate of the two machines to a survivable number. The descent rate can only be slowed by spinning the rotor faster or slower. The only energy source to overcome the rotor's drag that is trying to slow down the rotor RPM is from the potential energy that was available early-on because of the MASS AND HEIGHT OF THE ENTIRE FLYING MACHINE.

If you decrease all-up mass, that hurts your ability to maintain rotor RPM. At some light weight, the energy available to use is not enough to overcome rotor drag and keep RPM up.

This is pretty simple physics. If I drop a brick on my toe from 12," I might say "ouch." If I drop that same brick on my toe from 10 meters, I may have to ride in an ambulance. Which had the most energy? If the amount of "squish" of my toe represents my ability to keep the rotor spinning, would a heavy brick or light brick smush my toe more if dropped from the same height? Certainly you would agree, the heavy brick would do more "work?"

Jean Claude

Junior Member
During a gliding descent, aircrafts are not like a brick which regularly increases its speed of fall during the descent. To say that the descent is stabilized means that the potential energy is dissipated as it goes.
To cancel the impact speed of a helicopter when descending vertically, it is necessary to use part of the kinetic energy of its rotor by slowing down his RPM ie increasing the collective pitch at the last moment ( the rotor mass is important for available energy )
To cancel the impact speed of an autogyro downhill flying forward, it is necessary to absorb its kinetic energy in the increased drag of the nose-up rotor at the last moment (the changing of rrpm is neglectible and the rotor mass about without effect )

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WaspAir

Supreme Allied Gyro CFI
I'm finding it hard to imagine a gyro that can climb to altitude in autorotation but then can't descend in autorotation, no matter what the weight might be. Rotor diameter, blade pitch, loading, etc. are the same both ways and so are the flight dynamics.

By the way, gyros have an H-V avoid region, too, although there is no low altitude/high speed segment as in helicopter diagrams. You need a minimum combination of kinetic and potential energy for a successful cyclic flare to arrest descent.

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Doug Riley

Platinum Member
The role of fore-aft cyclic tilt of an autogyro rotor is, I think, widely misunderstood. For example, many gyro pilots believe that tilting the rotor aft (with back stick) only increases drag in the short term, and later increases lift, via increased RRPM, in the longer term. But there's another, immediate effect.

This means that a cyclic-only flare can slow the aircraft's descent in the very short term, even before the rotor has a chance to speed up. The effect is strongest with high forward airspeed, and is zero during a vertical descent.

Helos can accomplish a micro-hover, in addition to the cyclic flare, at the bottom of an autorotation by pulling collective while RRPM is still high. Gyros without collective can do a little of this if their blades are massive enough or the pilot aggressively overspeeds the rotor, but it's generally less impressive than the momentary collective hover at the bottom of an auto in a helo.

Aviator168

Member
Tilting the rotor back increases the mechanical AoA of the advancing blade at the location of 3 o'clock and reduces the AoA of the retreating blade at the location of 9 o'clock. At the same time, it increases aero force componment of for rotation. This last for less than 1 revolution and the blades will settle on a bit less AoA increment and a bit less AoA reduction due to flapping.

In short, higher the AoA, higher the rotation force for the rotater. For vertical descent, if the gyro descent too fast relative to rotor rpm, the blades will see an increase in AoA which will cause it to turn faster. Eventunally, it will settle on a descent rate at which the lift force and the downward drag of the rotor equals to the total weight of the gyro. In the gyro I fly, 1500ft/min.

bryancobb

Junior Member
I'm finding it hard to imagine a gyro that can climb to altitude in autorotation but then can't descend in autorotation, no matter what the weight might be. Rotor diameter, blade pitch, loading, etc. are the same both ways and so are the flight dynamics.

By the way, gyros have an H-V avoid region, too, although there is no low altitude/high speed segment as in helicopter diagrams. You need a minimum combination of kinetic and potential energy for a successful cyclic flare to arrest descent.
Keep in mind...during climb, energy from the burned gasoline becomes prop thrust which pushes the gyro upward through the air as it climbs, with the rotor tilted aft for the climb. The energy in the additional airflow, put there by gasoline spins the rotor even though you are going uphill.

In an unpowered, steady-state, autorotational descent, the only energy you can use to spin the rotor is the energy of falling. If you stopped the rotor, you'd fall like a brick. When airflow from falling spins the rotor, it acts like a parachute. There is definitely a low weight that cannot autorotate, in a helicopter or a gyro.

And Jean Claude, The masses in the tips of the blades do not have anything to do with the amount of energy available to spin the rotor during a steady-state, autorotational descent. That POTENTIAL energy is only a function of entire mass of the machine and its passengers. What blade mass DOES do for you...is it makes the rotor more resistant to RPM changes when some outside force tries to slow or speed up RPM. Another thing a high inertia (heavy) rotor does for you in helis and gyros, is during the last flare before touchdown, the RPM stays up a little longer and ads a margin of safety.

bryancobb

Junior Member
Robinson uses "HorsepowerSeconds" as the units during all of their training they offer on autorotations. Old man Frank said the reason he used that weird measure unit is people's lives would be saved because it's sort of easy to see how critical seconds are at getting the collective down when the horsepower the rotor needs is being consumed quickly.

Vance

Gyroplane CFI
Tilting the rotor back increases the mechanical AoA of the advancing blade at the location of 3 o'clock and reduces the AoA of the retreating blade at the location of 9 o'clock. At the same time, it increases aero force componment of for rotation. This last for less than 1 revolution and the blades will settle on a bit less AoA increment and a bit less AoA reduction due to flapping.

In short, higher the AoA, higher the rotation force for the rotater. For vertical descent, if the gyro descent too fast relative to rotor rpm, the blades will see an increase in AoA which will cause it to turn faster. Eventunally, it will settle on a descent rate at which the lift force and the downward drag of the rotor equals to the total weight of the gyro. In the gyro I fly, 1500ft/min.
The Predator’s rotor tachometer that not will show me what happens in one revolution.

I find during a vertical descent in The Predator the rotor turns slower than in forward flight at 50kts indicated air speed.

The disk is relatively flat during a vertical descent in The Predator and she descends between 1,400 and 1,600 feet per minute.

XXavier

Member
The Predator’s rotor tachometer that not will show me what happens in one revolution.

I find during a vertical descent in The Predator the rotor turns slower than in forward flight at 50kts indicated air speed.

The disk is relatively flat during a vertical descent in The Predator and she descends between 1,400 and 1,600 feet per minute.
In a vertical autorotation, the rotor disk acts as a parachute, but in forward flight, it generates a downward-directed net flow of air in order to achieve the required lift. The extra power involved needs more revs...

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Vance

Gyroplane CFI
In a vertical autorotation, the rotor disk acts as a parachute, but in forward flight, it generates a downward-directed flow of air in order to achieve the required lift. The extra power involved needs more revs...
In my opinion in a vertical descent a gyroplane rotor still has a driving region and a driven region and does not act as a parachute.

It is my observation in The Predator that the rotor rpm increases marginally as the indicated airspeed increases.

XXavier

Member
In my opinion in a vertical descent a gyroplane rotor still has a driving region and a driven region and does not act as a parachute.

It is my observation in The Predator that the rotor rpm increases marginally as the indicated airspeed increases.
It has a driven and driving regions, but the disk doesn't direct any net flow downwards, and hence it works like a parachute.

A change in revs, however small, is significant concerning the power involved, since angular speed has an exponent of three or four...

Vance

Gyroplane CFI
It has a driven and driving regions, but the disk doesn't direct any net flow downwards, and hence it works like a parachute.

A change in revs, however small, is significant concerning the power involved, since angular speed has an exponent of three or four...
In my opinion the airflow directed downward by the driven region in the vertical descent is what slows the descent.

Significant how?

In my opinion the horsepower consumed by the rotor goes down as the indicated airspeed increases.

XXavier

Member
In my opinion the airflow directed downward by the driven region in the vertical descent is what slows the descent.

Significant how?

In my opinion the horsepower consumed by the rotor goes down as the indicated airspeed increases.

In vertical autorotation, and from the reference plane of the descending rotor disk, there's no downward airflow

As JC writes in a recent post:

A possible formula for profile power when forward speed is nul is: P = ¼ Cd.ρ.c.ω^3. R^4 with Cd = 0.011

If you add forward speed, the power involved will be even bigger... But remember that we're talking about vertical autorotation....

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Vance

Gyroplane CFI
In vertical autorotation, and from the reference plane of the descending rotor disk, there's no downward airflow

As JC writes in a recent post:

A possible formula for profile power when forward speed is nul is: P = ¼ Cd.ρ.c.ω^3. R^4 with Cd = 0.011

If you add forward speed, the power involved will be even bigger... But remember that we're talking about vertical autorotation....
You are welcome to your opinion and your semantics.

In my opinion if the rotor is going around near flight rpm the driven region is directing air downward.

Parachute | Definition of Parachute by Merriam-Webster

1 : a device for slowing the descent of a person or object through the air that consists of a fabric canopy beneath which the person or object is suspended.

2 : patagium.

3 : a device or structure suggestive of a parachute in form, use, or operation.

XXavier

Member
Wayne's Helicopter Theory, pages 109-110:

XXavier

Member
(...)

In my opinion if the rotor is going around near flight rpm the driven region is directing air downward.

If the gyro is on the ground, yes.
If the gyro is in vertical autorotation, NO.

XXavier

Member
From page 11 of 'Flugphysik der Tragschrauber', with a partial translation:

Jean Claude

Junior Member
And Jean Claude, The masses in the tips of the blades do not have anything to do with the amount of energy available to spin the rotor during a steady-state, autorotational descent. That POTENTIAL energy is only a function of entire mass of the machine and its passengers.
Yes, but my words were "To cancel the impact speed ..."
During this phase of slowing down, the state is no longer "steady" and the helicopter can use part of the kinetic energy of a heavy rotor, while the gyro cannot: A heavy rotor don't help him

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