confused: right pedal causes additional lift

[Rongor]

I tested while leaving all the AP channels off.

While hovering a few meters above the field, you can clearly notice how giving left pedal (easing power demand) induces descend, giving right pedal (demanding power) causes a slight climb.

 

Why?

 

The anti-torque pedal (right pedal in the Mi-24) is demanding more power. Regularly this would draw away power from the main rotor for some moment, until the engines match the new power setting and deliver the increased power. Giving right pedal would cause a small descend, not a climb.

Giving left pedal actually is reducing the power demand of the tail rotor. Hence the main rotor will benefit from the surplus power and generate some more additional lift and a slight climb, until again the engine control notices the decreased power demand and decreases engine power to match the new situation.

 

It seems mixed up here in the Mi-24. Doesn't make sense...

Edited June 29, 2021 by Rongor

[VampireNZ]

Fair question - I remember the same question coming up for the Mi-8 Hip, nice comprehensive reply here from Frank Lombardi:

 

Question:

Why does a helicopter with rotors turning clockwise (e.g., Lama or Mil Mi-8) gain height in a steady spot turn to right and lose height in a left spot turn?

 

Answer:

In helicopters with a conventional main/tail rotor, any increase in pedal application that opposes the torque effect of the main rotor will increase the total amount of power demanded from the drivetrain. With main rotors that turn clockwise when viewed from above, this is a right pedal input (left pedal in those turning counter-clockwise). Now consider a turbine-powered helicopter such as the SA315 Lama with governed engine/rotor RPM (N2/NR). Upon applying right pedal during a “spot turn” in the clockwise-rotating Lama, the increase in power required is sensed by the governor as a slight droop in the N2/NR. The governor then tries to maintain the N2/NR by adding fuel to raise the RPM back to normal. Keep in mind that the governor is sensing rotor RPM through a system that is fixed to the airframe, so as the aircraft yaws right, it follows the direction of the spinning rotor shaft, and the governor senses a relative rotor RPM that still appears lower than desired. The response is a continued increase in fuel flow by the fuel control, and a resulting higher-than-necessary RPM relative to the air mass you’re flying in. It does not take much yaw rate to cause this. Yawing at 30 degrees/second (that’s 12 seconds for a full 360-degree pedal turn) will change the relative rotor speed by 5 RPM. Since lift is proportional to the square of NR, even this small change in RPM will bring an increase in main rotor thrust enough cause the helicopter to climb. Of course if you add left pedal the opposite is true, and the helicopter will descend.

 

https://www.rotorandwing.com/2011/11/29/height-variations-in-rightleft-spot-turns-with-rotors-turning-clockwise/

[S. Low]

Learned stuff today 

[tusler]

Plus there is a whole bunch of aerodynamic stuff going on cause the main rotor disk in flight is never flat and the tail is also putting a twisting motion on the airframe. look at the tail rotor on a ch-53e and it does all kinds of things besides just push the tail around.

[ARM505]

As a fixed wing pilot, this is why helicopters, while fun, are actually highly reliant on witchcraft to keep them in the air. There's a lot of stuff happening that's mathematically far above my simple mind, so I just know 'if this, then that', and assume that the gods of earth-repulsion technology (ie. primary gods governing rotary winged flight) are appeased for the moment. Until they're suddenly not, and there's a pile of spare parts - this also explains why helicopters utterly disintegrate with even the smallest crash - the gods weren't pleased/pilot didn't pray to the machine spirit enough, the magic fails, and the craft is struck down into it's component pieces.

 

In Chapter 2 of my book, I explain electronics, and the 'magic smoke'. Thank you for subscribing.

[Wychmaster]

Since helicopters are extremely complex and I havn't been part in the construction of one, I can not tell if my answer is THE answer, just a part of the whole story or has no effect at all. This out of the way:

 

The hind has a govenor that tries to maintain a constant RPM of the rotor. If you measure the RPM directly at the rotor shaft, you get it in relation to the airframe, but the relevant part for lift is the RPM in relation to the air. In a hover, with no wind at all, you could take the ground as reference.

 

Now lets say, the engine govenor is extremely good at is job in maintaining the RPM and you can do whatever you want with your pedals, the RPM remains constant. The hind has a clockwise turning rotor, forcing the airframe to rotate counter clockwise (nose turns left) which is cancelled by the tail rotor. Lets say the maintained RPM is 240. Now you apply right ruder so that the airframe turns with 1 RPM in clockwise direction (like the main rotor). Now your total RPM in relation to the ground (and hence the air) is 241, causing additional lift. If you would do it in the opposite direction it would be 239 and produce less lift.

 

Lets overexaggerate a bit and say you have a extremely powerful, magical tail rotor that can turn your airframe with an RPM of 240 to the left. Neglecting a lot of aerodynamic effects and other physics that prevent this, it would leave you in a state where the rotor has no RPM in relation to the ground/air. Hence it would produce 0 lift and you would just drop out of the air.

 

As I said, I can't tell if this is truly the answer, but at least it makes sense to me