Free Flight, Free Speech

bomber wrote:A real life helicopter does not rotate around it's CoG. .... It swings from its rotor.

If we could remove all external forces and momentums that act on a helicopter, then the helicopter would rotate around its CoG (like all rigid bodies). Only because of the external forces and moments (especially from the rotor) we get the motion that you describe as "swinging from its rotor". Hence it's perfectly possible to simulate a helicopter with JSBSim. You "only" need to calculate the forces and momentums (especially from the rotor) and apply them to the aircraft.

IAHM-COL wrote:

mue wrote:As I wrote, the difficulty in helicopter simulation is to get what forces and moments of the rotor are applied to the aircraft.

Isn't expected that any and all momentums generated from the rotor wings to be applied to an helicopter?
How can a rotor generate a momentum that happens to not become applied to the helicopter body that is thethered to it? and if this is not possible, then why would one need to establish in JSBsim which momemtums to apply and which ones to disregard?

You are right. I should have written: ...the difficulty in helicopter simulation is to calculate the forces and momentums of the rotor (and all forces and momentums are applied to the aircraft).

So mue, what you're saying is that a helicopter does not rotate around the rotor but like all other objects rotates around it's CoG....

bomber wrote:So mue, what you're saying is that a helicopter does not rotate around the rotor but like all other objects rotates around it's CoG....

If not mue, then that would be more like what I'm saying, somehow.

Like in your pendulum example, what I see is the pendulum (sphere at the bottom of the rope), the pendulum itself has no rotation momentums. (none at least apparently from the graphics display). Instead, it carries an arc translation as a consequence of 3 translational forces applied; the gravity (down), the swinging force (perpendicularly applied to g) which acts spring like, and a 3rd force assumes a medium of oscillation (ie air) with a viscosity different to 0 that creates the friction force in equivalent opposition to the spring swinger. The consequence of that model should be: 1) it oscilates as shown, 2) it will stop (not shown) and 3) no rotation.

Now, I could spin it, of course. I could assume spin occurs independent of the tether (not transferred to it) so not to alter the spring-like coefficient described above, thus the spin acting independently of the swinging. like if I pinch the pendulum and cause it to begin to rotate. Now the way I imagine the rotation happening suggests that the 3 components of the rotation go around an x,y,z perperdiculars' axis system that cross on COG. Spinning momentums will occur in all points but on 0,0,0 the momentun is zero since distances are zero. That makes sense to me. As in all points of pendulum will spin except the COG.

The question you pose is If I can imagine any way to apply a force to that freely rotating pendulum in a manner to cause it to spin differently, as in making the non-rotational point somewhere else but in COG. I can't imagine it yet.

ok imagine this,,,,

A child picks up a toy helicopter by it's rotor blades and play flys it, swooping it through the sky. The helicopter rotates from the child hand which is holding the rotors.

A child picks up a toy helicopter by putting his hand under the rotor and holding the fuselage and play flys it, swooping it through the sky. The helicopter rotates from the child hand which is holding the fuselage.

The way I understand this problem is one in which two kinds of movements will be effected on the object; a translocation and a rotation.

In the translocation the COG shifts its location on an external xyz axis. Every other point that can be drawn in the object suffers an identical translocation as the COG, thus the objects indeed moved rigidly.

In the rotation, the COG doesn't translocate (deltax + deltay + deltaz = 0), but every other point in the object does move, and the magnitude of the displacement is correlated to the distance of given point to COG.

This is, COG just translocated, but doesnt alter by the rotate. The final location of any other point can be obtained as the summatory of the magnitude of both movements.

What's interesting here, and the bottomline of your inquiry, is what happens if you fix a reference point tau and rotate. This is what I think:
fixing a point implies no translocation/rotation occurs at the reference tau as the object is rotated; but the magnitude of the rotation correlates to the magnitude of a COG translocation. In other words, the COG now translocates a |deltax + deltay + delta z| >> 0 as the object now rotates around tau. In fact, every other point except tau will be forced to a translocation a magnitude of which varies depending to its distance from tau as well as the magnitude of the rotational move. [I think].

Furthermore, we can imagine a tau = COG. Restricting translocation on the COG means therefore an object that purely rotates but its location otherwise remains static on the external xyz axis. quite like balancing toys; or actual balances.



What's interesting on your question and hypothetical child holding the helicopter is; who's hand is fixing a RL helicopter from any tau point at all (which you seem to define as the point where rotor wing exercises forces upon the helicopter)???

Ok how about this thought....

If the helicopter is rocked, lat or lon.... It returns after some oscillations to a steady state.

For this to happen as opposed to rotating around the CoG there must be some moment of a sin pitch\roll angle equation.....

I had thought that the anti-torque on most helicopters that counters the spin given on the rotatory wings was given via the smaller tail rotor

And thus that new set of counter-acting forces provided the stability of flight.



https://www.scienceabc.com/innovation/why-helicopters-have-tail-rotor-purpose-torque-coaxial-rotors.html

Including the fact that the tail rotor torque is produced far from the COG, generating a large momentum, enough to conteract the rotor-s torque.

Is that the question?

bomber wrote:Ok how about this thought....

If the helicopter is rocked, lat or lon.... It returns after some oscillations to a steady state.

Helicopter are not as stable as you describe it. Helicopters are quite unstable and require permanent pilot input. Especially in a hover. In forward flight the fuselage can act stabilizing.

bomber wrote:For this to happen as opposed to rotating around the CoG there must be some moment of a sin pitch\roll angle equation.....

Forces and moments acting on the helicopter comes from aerodynamic forces and moments acting on the fuselage and from the forces and moments generated by the rotor.

The constant pilot input is another problem that my flight model is showing..

It's as if without the 'feeling' of being in flight, micro adjustments by the pilot are hard to judge.... I'm thinking an FCS to ease the workload of these micro adjustments might be the way to go.