How to start this.... 'i need a lift formula for the main rotor to start with anything'

After reading few docs all i can say there are too muh integrals involved which i am unable to read thus i can't understand what its all about.

My bigges issue is on how i could calculate a lift for each rotor blade (without involving airflow from other blades so far) so imagine a simple one blade (instead of 4 as for standard helis) and rotating over and over. The segment near the engine would spin slower than a segment on the end.... Is this why those integrals are used? Should i divide each blade into smaller pieces and calculate lift force for each of them or is there an ideal eqauation that deals with it?

Why do you need to do this?

For a helicopter sim, I'd think all you need is a curve translating RPM of the rotor for a particular helicopter rotor into a lift. You can then translate that into a force applied to the centre of the rotor and it'll be functionally identical for almost any purpose. You can apparently calculate that lift force using this method.

Quote

v2 signifies true airspeed squared, which is the square of the speed of the helicopter relative to the oncoming air, expressed in meters per second. 

Which says if heli travels with 0 m/s and rotor airspeed is equal to Light speed heli has no motion x)

Looks like they assume rotating blades form a solid circulical disc that could be treated as a wing...

For now i dont see how this could work ( when calculating the drag of a rotor blades that form such kind of plate)

I'm just starting with heli physics so maybe ill try to make a game, anyway i would like to avoid any movement stiffeness, and apply proper forces (like in reallife) so any movement would deaccelerate smoothly without involving voodo functions

Well, perhaps think of it like this: if you were making a jet plane sim would you bother with the drag on each rotor blade or would you just deal with the thrust of each engine?

I think you want to think of the rotors the same way: they're just things that produce thrust and rotation. The details of each blade are not worth modelling.

14 hours ago, Irusan, son of Arusan said:

Why do you need to do this?

For a helicopter sim, I'd think all you need is a curve translating RPM of the rotor for a particular helicopter rotor into a lift. You can then translate that into a force applied to the centre of the rotor and it'll be functionally identical for almost any purpose. You can apparently calculate that lift force using this method.

 

I think at some point / within some set of operating conditions, you can basically pretend that the blades aren't rotating and simulate it the same way as a fixed wing, too

But, there's a lot of subtle details with helicopter sims to actually make them feel like a helicopter.

e.g. you can switch the engine off (or have it be disabled by your enemy) and still safely land a helicopter by using the upwards-motion of the air during your fall to keep the rotors spinning, and use the spinning rotors to create drag similar to a parachute, and then when close to the ground, use the spinning rotors to generate enough lift to save your arse and land softly.
Simulating aerodynamic drag over the body would be important to make the tail-rotor controls feel like a helicopter too.
Or even just maneuvering - you tilt forward by having an unbalanced amount of lift on the left and right sides of the helicopter (and gyroscopic effects rotate that by 90 degrees ).
I think if you skip over a lot of these details, it's going to feel like an arcade flight game, but not a helicopter.

Yeah, once again about that main rotor so assume it rotates at 30 m/s and heli travels.at 20 m/s would this mean the true airspeed is 50 m/s (without involving direction of flight, rotor rotation and airflow direction)). ?

One side is (where the blade move from back to front if you're flying forward), the other side is 30-20=10. The rotor in the back of the tail compensates this difference, I think (never studied helicopters).

Note that m/s has little meaning for a blade, it rotates around an axis, so speed is essentially an angular speed (radians / second), which translates in different amounts of m/s and thus different amounts of lift, depending on distance from the axis.

On ‎3‎/‎3‎/‎2019 at 8:27 AM, Alberth said:

One side is (where the blade move from back to front if you're flying forward), the other side is 30-20=10. The rotor in the back of the tail compensates this difference, I think (never studied helicopters).

That's a good point! AFAIK the main factors in aerodynamics are wing surface area (and profile), speed and angle of attack. While, with airplanes during takeoff the angle of attack is different on each side of the engine, the blade at the left has different angle against movement direction than the right side, thus generating different amount of thrust. With helicopters this could be an issue of forward and backward moving blades, generating different amount of lift, I've never thought of it before 

Also the angle of attack plays here too, so as you move (tilting the rotor) the angle of blades change with regard to movement direction. And as you can change the rotor blade angle manually (with the "handbrake") the RPM has no relation to the actual lift.

The bigger concern should be the engine reaction torque, when you get airborne the rotor turns the whole helicopter in the opposite direction. That's why you need the back rotor, which also has adjustable blades (use your pedals)

I'm just stretching the surface of aviation dynamics, (not sure about helicopter controls though), but I hope I gave some ideas

As an addition, as some meaningful info, that you might already know

I used this formula for my race car wing simulation, probably something similar could do it for rotor blades:

F_down = -0.5 * cord * span * down_coeff * angle * air_density * velocity^2

If you invert it, (using negative angle, or +0.5 at the beginning) you get F_Lift

And here comes the "how much is down_coeff" question. No idea. My best practice was, knowing that a specific car has that much down force at that much speed, so I could calculate it back, or at least have an educated guess. For a helicopter I'd tweak until it feels good.

And yes I'd also go for dividing up the blade to some reasonable sizes, and handle them as a "couple of wings", but for first maybe one sample at the center could also do it...

The same formula can be used for the drag, just change the coeff to something else
I don't know how much impact that would have on the engine, like some maneuver could make more air resistance and drop the rpm, that could end up in less lift...etc. And as @Hodgman mentioned in autorotation the blades would provide some drag that works against gravity and also would spin the rotor.

It's really interesting topic

13 hours ago, bmarci said:

With helicopters this could be an issue of forward and backward moving blades, generating different amount of lift, I've never thought of it before

Gyroscopic effects cause the forces to be applied with a 90deg offset, so instead of that unbalance causing the helicopter to roll, it causes it to pitch. This actually causes it to be self-limiting - if you go too fast, the helicopter will level itself out and slow down!

13 hours ago, bmarci said:

The bigger concern should be the engine reaction torque, when you get airborne the rotor turns the whole helicopter in the opposite direction. That's why you need the back rotor, which also has adjustable blades (use your pedals) 

At high speeds, the air rushing past the sides of the body, causing drag, is enough to overcome that torque -- if the torque causes the body to turn, it will cause an increase in drag on the side that's now moving forwards, which causes it to spin back the other way and balance out!

If your tail rotor malfunctions, you will spin out of control at low speeds, but, if you're at altitude, you can rock the helicopter back and forth while it's spinning until you get moving in a direction, and once that happens the drag from flying forward will be enough to stop it spinning and regain control. You can then land with minimal engine usage to avoid losing control again ?