Calculating the lift created by the wings of Mi-24

[AeriaGloria]

   Hello, I am trying to figure out the data for amount of lift created by the wing of the Mi-24. Using lift force calculators that take into account air density, lift coefficient, wing area, and airspeed. 

Wing area is taken from Yefim Gordon’s book on Russian Gunship helicopters, a figure of 6.25 meters squared

Now looking at the LUA, there is a Cy value of .0668. And a Cy Flap value of .35. I did some searching to see in what way is Cy related to CL lift coefficient, and it was hard to tell if it was ED’s chosen term for CL or some relation of CL to AOA slope angle. 
 

Anyways just using Cy in the lift calculation at 260 kmh (normal cruise) ended up with 139 kg, extremely small, and both using Cy Flap and Cy Flap + Cy equaled values less then 10% of max takeoff weight of 11,500 kg. Which is believable, but I expected much higher given the much larger then normal wing and 19 degree incidence angle. 
 

This paper seems to show that in flight testing, at a speed of 260 kmh and weight of 11,500 kg, the rotor is giving about 9,500 kg vertical force in level flight. This is about 17% of the lift generated for level flight, which is much closer to the 20-25% figures often apocryphally/anecdotally thrown around for Mi-24 wing lift compared to total lift. 

   This Embry Riddle paper on compound helicopters analyzed several configuration types, one of them is the S-67 which bears the most similarities to our Mi-24, in a section called “pure lift compound helicopters.” They find that the wing, which is at 8 degrees of incidence, begins to be at negative AOA and produce negative lift once speed increases enough and enough negative pitch angle is needed, thus decreasing performance. At such high speeds. With the Mi-24’s 19 degrees of incidence, and by observing that it never seems to come close to that in level flight all the way up to its 330 kmh top speed, I do not think it would suffer from the same effect. 

Anyways, I don’t know if @Yo-Yo can help as this is your area of expertise, but I would very much appreciate any guidance in being able to figure out the age old adage of if the Mi-24 wing really does produce 20-25% of the lift at normal cruise speeds or other speeds, and wether the Cy numbers in the LUA or any other numbers can be used for it

I was using sites such as these to calculate lift, which gave identical numbers, using the common 1.225 kg per meter cubed figure for dry air density, I see that another site seems to default to 1.2754 kg per meters cubed, but they are close enough to give similar results https://www.ajdesigner.com/phpwinglift/wing_lift_equation_force.php#ajscroll

https://www.vcalc.com/wiki/force+of+lift

Thank you for any help given

[AeriaGloria]

It appears my link for the Mi-24 rotorhead Force calculations study are gone, here they are 

preprints202201.0128.v1 2.pdf

[AeriaGloria]

Someone on Reddit recently brought to my attention that multiple first hand sources say wing lift is 25% of weight. I realized this myself when translating the aerodynamic manual for the Mi-24 about a year ago but did not update this post 

However, I was also interested in how high the lift can get, and how it might contribute in other flight regions, or what regions it may be stalled. One thing I still would love to learn more about is the influence of the winglets, the contribution of the elevator to lift, and force of the vertical stabilizer. 

In addition, I realize that when I made this post so long ago, that I did not include all my figures and calculations. 
 

So using the Polish rotor head force study, this was what I came up with for the speeds tested 

140 kmh: 

Wing: 500 kg Force, 6-4.3% of lift

Tail rotor: 3.6% of thrust

 

260 kmh: 

Wing: 2000 kg force, 23.5-17.4% of lift

Tail rotor: 5.2% of thrust 

 

300 kmh: 

Wing: 2,170 kg Force,  25.5-18.8% of lift

Tail rotor: 7% of thrust 

The difference in % of lift is to compensate for how the change from empty to max takeoff weight might effect things. If CG stays the same, and thus angle of attack, wing lift in this case should be little changed, but the rotor would need to match its lift to the remaining weight. Tail rotor figures are from the study using the force on the tail rotor shaft  in the left direction compared to vertical force on the main rotor shaft 

Edited August 12, 2024 by AeriaGloria

[Mr_sukebe]

Interesting numbers.
Whilst the increase in lift at higher speed makes sense, I'm a little surprised at an increase in the power required by the tail rotor.

I would have through that a decrease in lift required would slightly reduce the torque effect of the main blades and that the aero implications of the body of the helicopter would tend to stabilise where it's pointing.  

Thoughts?

[AeriaGloria]

On 8/12/2024 at 3:46 AM, Mr_sukebe said:

Interesting numbers.
Whilst the increase in lift at higher speed makes sense, I'm a little surprised at an increase in the power required by the tail rotor.

I would have through that a decrease in lift required would slightly reduce the torque effect of the main blades and that the aero implications of the body of the helicopter would tend to stabilise where it's pointing.  

Thoughts?

I do not know. The numbers might be less correct for tail rotor. I am only comparing the vertical force on the rotor shaft to the tai rotor lateral force. The study says the transducer was fixed to the shaft, so I would not think that the vertical stabilizer lift would be included 

According to the aerodynamic manual, you reach lowest pedal, about the 1/4th left at around 140 kmh. Pedal moves to around middle at 260 kmh. At 280 kmh, you might need 1/4 right pedal. 

For hover, you would similarly be around 1/4-1/3 right pedal. 
 

Above 140 kmh, the engine torque needed increases. 140 is your bucket speed, Lowest power for flight needed. So even though you are more fuel efficient over distance at 150-280 kmh, your powertrain still needs to output more power, more anti torque to fight the torque. 
 

IIRC, the vertical stabilizer is supposed to produce half the anti torque force needed at cruise (260 kmh), and power increase (looking at torque/EPR) is about 50% there compared to 140 kmh. 

I bet if the vertical stabilizer did not produce the lift it does, you would see the tail rotor doing 10-15% of the main rotor power at cruise speed 

That being said, the point that the tail rotor has 0 pitch is about 1/2 left pedal. 

Edited August 13, 2024 by AeriaGloria

[Mr_sukebe]

Cool, thanks for that.

One question, which I think you partly alluded to, is about the optimum efficiency, which I assume needs to take into account the jets always running, drag etc.

For the Mi24 and Mi8, would you know what those optimum speeds are?  I tend to fly at around 200kph, such that I've got some leeway either way.  I've no idea if that's the most efficient.

[AeriaGloria]

13 hours ago, Mr_sukebe said:

Cool, thanks for that.

One question, which I think you partly alluded to, is about the optimum efficiency, which I assume needs to take into account the jets always running, drag etc.

For the Mi24 and Mi8, would you know what those optimum speeds are?  I tend to fly at around 200kph, such that I've got some leeway either way.  I've no idea if that's the most efficient.

Yes, for Mi-8, your bucket speee, best loiter/least fuel burn is around 120 kmh. This will stay largely static as you go higher. Its cruise (best range) is 205-220 kmh bellow 1000m depending on exact altitude and weight. At ceiling, this is 100 kmh and likely limited by top speed

For Mi-24 its bucket speed/least power regime is 140 kmh; and this only decreases slightly with altitude. 

It is 260 kmh for cruise speed, but manual stresses that 10-20 kmh difference in speed only yields 2-3% change in efficiency, and since the range tables are calculated for a constant 4x rocket and 4x ATGM load (which also hurts efficiency by 2-3%), it is within the margin of error. Like Mi-8, this speed becomes smaller as you go up and will eventually be limited by max speed. 
 

In surprised these two speeds aren’t talked more about by DCS players. They are very important for aircraft also, many people cruise too fast or climb to slow because it’s not often shared, and the value of saving fuel is often not seen. Less fuel you need, the lighter you can be and get more performance 

For both Mi-8 and Mi-24, you also have a “turbine adjust rpm” switch that changes target rpm for governor. This is 95-96% default; but by lowering it to 92% you get 2-3% efficiency gain. Similarly, you can increase it to 98-99% for more rotor lift for 2-3% fuel efficiency loss. The manual mentions increasing the rotor rpm this way to increase tail rotor authority at high altitude, but also works anywhere. I often use it as a “combat” mode 

[Mr_sukebe]

Wow, that is fabulous info, thanks.

I think I’ll create a kneeboard to capture that.