Pushovers/low-g semi-rigid systems

[heliflyknow]

The question I have been wondering for a while is how does a pushover cause a low g condition?

I can understand if you are in a climb and you jam the cyclic forward that the disk will be pulling you forward, and the fuselage will still be moving upwards leading to a low g condition.

What I can't understand is how in level flight abrupt forward cyclic can cause a low g condition. Is it because in level flight the rotor is providing 1G(weight of AC) and rotor tilt decreases this lift? Even if it is it seems like the rotor would still be pulling the fuselage, but instead of up it would be pulling up and forward.

Another follow up question is, why is the recovery aft cyclic and not up collective? The only reasons I can think of is that collective can cause a nose up(up collective) or nose down(down collective) because of precession, and that cyclic input results in more rapid disc loading. Is this the reasoning,nor is there something I am missing?

[HawaiiCFII]

It's like a car on level ground that goes down over a hill and you get that feeling in your stomach. The aircraft is flying down. In this case the rotor is going down and the fuselage has inertia moving it into the rotor system. Normally the fuselage is hanging from the rotor head but in an extreme low G condition it is unsupported.

[iChris]

The question I have been wondering for a while is how does a pushover cause a low g condition?

 

What I can't understand is how in level flight abrupt forward cyclic can cause a low g condition.

 

Another follow up question is, why is the recovery aft cyclic and not up collective?

 

It's like that car going over and down the hill HawaiiCFII wrote about in the post above. The rate at which that happens is important.

 

Like any physical system, the helicopter must obey the basic laws of physics. The low-g was caused by the fact that the pushover caused a change in velocity of the helicopter from its state of equilibrium (1g flight state). Equilibrium is a state of zero acceleration, meaning there’s no change in velocity (speed or direction). Therefore, an aircraft moving in straight and level flight, at a constant speed and direction is in a state of equilibrium (1g flight state).

 

In this state of equilibrium the helicopter is producing lift equal the weight of the helicopter, which we also call a 1g-flight state. An extension of Newton’s law states, weight = mass x g, were g equals the acceleration force of gravity. The acceleration force of gravity near sea level is approximately g=32 ft./s². G-force = (a + g)/g, were “a” equals the rate of change in velocity/time, acceleration.

An aircraft in equilibrium, straight and level flight, experiences a g-force equal to (a + g)/g, or

(0 ft./s² + 32 ft./s²)/ 32 ft./s² = 1g

 

An aircraft pushover approaching a downward acceleration of 16 ft./s² during the first few seconds experiences a g-force equal to (a + g)/g, or

(-16 ft./s² + 32 ft./s²)/ 32 ft./s² = .5g

 

An aggressive pushover during the first few seconds, may approach a downward acceleration of 32 ft./s² in which case, the aircraft experiences a g-force equal to (a + g)/g, or

(-32 ft./s² + 32 ft./s²)/ 32 ft./s² = 0g

 

In a far more aggressive pushover of 48 ft./s² during the first few seconds, you would have a downward acceleration in excess of the 32 ft./s², the force of gravity, in which case, you would experiences a negative (-) g-force equal to (a + g)/g, or

(-48 ft./s² + 32 ft./s²)/ 32 ft./s² = -.5g

 

The recovery is aft cyclic and not up collective. In order to do work we need to get the energy from some source. Collective increases the pitch of all the blades collectively, but is less effective in regaining rotor thrust or changing the downward acceleration of the helicopter. What we do have is the kinetic energy in our forward airspeed which we can trade-off for lift just as we do in the flare in autorotation with aft cyclic. The aft cyclic immediately reloads the rotor and increases rotor angle-of-attack; restores rotor thrust, and change the line of flight back toward equilibrium.

 

The low-g condition is important to the semi-rigid rotor because to develop a control moment, the rotor must first develop thrust. If a control force or moment is needed to change fuselage attitude, the rotor system must first develop thrust. Therefore, any control moment becomes a by-product of thrust. Cyclic control power (the ability to change fuselage attitude) decreases under flight conditions below 1g and is non-existent at 0g. The rotor itself responds to cyclic input regardless. Some two-bladed rotor systems like the Bell 222, 230, and AH-1 incorporate hub springs, allowing better control power in low-g conditions and help guard against mast bumping.

Edited December 30, 2014 by iChris

[heliflyknow]

Thanks for the thorough explanation iChris, I can see the relationship between downwards acceleration and "G's", and now I have the equation to use to find it.

What I am having a hard time visualizing is at what point the fuselage would be at 0 G's. Using force vectors to try to draw what is happening is what is hanging me up.

If you are in an out of ground effect hover with no wind (rotor parallel to the horizon) and you then tilt the rotor forward with the cyclic at a 45* angle the lift is cut in half and you accelerate forward while descending(.5G), by that same logic it seems that you would have to have the rotor perpendicular to the horizon in order to be falling at the speed of gravity(0G).

Is this the attitude you have to be in to reach 0G? It seems you would reach 0G's much sooner. I've seen videos of airplanes giving the sensation of weightlessness, but it seems they are flying a parabolic flight path(upside down U).

[iChris]

Thanks for the thorough explanation iChris, I can see the relationship between downwards acceleration and "G's", and now I have the equation to use to find it.

 

What I am having a hard time visualizing is at what point the fuselage would be at 0 G's. Using force vectors to try to draw what is happening is what is hanging me up.

 

If you are in an out of ground effect hover with no wind (rotor parallel to the horizon) and you then tilt the rotor forward with the cyclic at a 45* angle the lift is cut in half and you accelerate forward while descending(.5G), by that same logic it seems that you would have to have the rotor perpendicular to the horizon in order to be

falling at the speed of gravity(0G).

 

Is this the attitude you have to be in to reach 0G? It seems you would reach 0G's much sooner. I've seen videos of airplanes giving the sensation of weightlessness, but it seems they are flying a parabolic flight path(upside down U).

 

At what point would the fuselage be at 0g? It would happen within a few second after the aggressive pushover. During that first few seconds when the downward acceleration rate reaches 32 ft./s², the aircraft would experiences 0g during that short time frame. Shortly after drag will begin increasing with the square of the velocity and lift will start to return and increase until the net forces will again return to zero. Since the net forces on the helicopter are zero, the helicopter has zero acceleration and is back in equilibrium.

 

It may not be obvious, but free fall rates (32 ft./s²) after the pushover is obtainable for only a few seconds. What 32 ft./s² means is during each second of the helicopters descent, the rate would have to increase by 32 ft. per second, until terminal velocity. In other words v = g x t:

 

1 sec velocity acquired 32 ft./s

2 sec velocity acquired 64 ft./s

3 sec velocity acquired 96 ft./s

4 sec velocity acquired 128 ft./s

5 sec velocity acquired 160 ft./s

 

The airplane parabolic flight you’re talking about generates longer free fall periods by following a trajectory wherein the acceleration of the aircraft cancels the acceleration due to gravity along the aircraft vertical axis. As the plane goes over the top of the arc, the centripetal force exerted on the plane and everything in it cancels out the gravitational force. At this point, passengers experience "0 g"

 

Those flights typically consist of a number of parabolic trajectories, each only about 25 seconds of free fall. However, we’re getting into another type of motion, curved path. Curved path or ballistic trajectory represents another form of accelerated motion.

 

The parabolic flight begins when the aircraft starts accelerating to gain velocity before pulling up to convert horizontal velocity into vertical velocity. During the pull-up the g level increases. When a sufficient upward velocity is achieved, the pilot pushes-over and reduce thrust so that the aircraft and occupants fall together. At the end of the parabola the pilot pulls up and the cycle is then repeated.

 

Contrary to popular misconception, the 0g free fall phase of flight begins as the pilot pushes-over and reduce thrust, and does not occur solely as the aircraft descends. Although the aircraft has upward velocity during the initial 0g phase, its net acceleration is downward: the upward velocity is decreasing.

 

During those first few second the helicopter is just another object with a vertical component of descent that’s experiencing free fall for a few seconds.

 

Edited December 31, 2014 by iChris

[iChris]

What I am having a hard time visualizing is at what point the fuselage would be at 0 G's. Using force vectors to try to draw what is happening is what is hanging me up.

 

Any motion in a curved path represents accelerated motion, and requires a force directed toward the center of curvature of the path. This force is called the centripetal force, which means "center seeking" force, the force that pulls the aircraft up, over, and down the curved path.

 

Weightlessness (0g) is met when the downward acceleration of your seat is equal to the acceleration of gravity. Considering the path of the roller-coaster or an aircraft going over the top to be a segment of a circle so that it can be related to the centripetal acceleration, the condition for weightlessness is;

 

v²/r = g or v²/r = 32 ft./s² or v_weightless = √(32 ft./s²)r

 

Again, we’re getting into other types of motion, curved path. Curved path or ballistic trajectory represents different forms of accelerated. From straight and level flight a far more aggressive pushover is need to reach 0g since the full centripetal acceleration advantage of the parabolic trajectory is not available.

 

You’ll see all this explained from one of two concepts, Centrifugal Force, which is an apparent force that draws a rotating body away from the center, or Centripetal Force, which is the force that makes a body follow a curved path.

 

So, when Centrifugal Force equals weight, we’re at 0g, or as we have here, when Centripetal Force, the force making a body follow a curved, equals the force of gravity, we’re also at 0g. Like opposite sides of the same coin, when the body is held at the same radius, both forces are equal and opposite.

 

 

Edited December 31, 2014 by iChris