Deriving the equations of motion for and airfoil using the method of Lagrange's Equations. Two degree of freedom system.
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@kaursingh6375 жыл бұрын
YOU ARE THE ONLY ONE ON THIS TOPIC ---EXCELLENT LECTURE
@paralelepip92 жыл бұрын
Sadly! But yeah, great lecture!
@mertgunes98546 жыл бұрын
I don't understand how these videos don't have more views and likes, they're brilliant
@Freeball996 жыл бұрын
Thank you for your feedback. It's very encouraging to hear.
@hakankarakurt11003 жыл бұрын
Yeah couldn’t agree more :)
@Manuel484376 жыл бұрын
You are one of the best web content creators of the world, thank you very much.
@nishantduttmishra29032 жыл бұрын
Thank you much for this. I was having trouble understanding how to incorporate the torque in the forcing function in the same way as we used generalized force Q.
@shanmukhansmmunagala23416 жыл бұрын
Excellent explanation .Got clarity on the equations ,really helpful thank you
@Omar-sj7wl5 жыл бұрын
Great channel!
@user-gz6kk3cw7zАй бұрын
How can we write the equation if it is written at CG for the same case
@user-gz6kk3cw7zАй бұрын
If we want to include both the translational and torsional damping along with the springs, then what would be the modifications?
@Freeball99Ай бұрын
I'm assuming you want this relative to the CG again... First construct the Rayleigh's Dissipation function, R, where R = (1/2)c(ḣ_cg + eα̇)² + (1/2)c_t(α̇)² This looks very similar to the potential energy expression, except that the k's become c's and the displacements become velocities. Substituting this into the extended form of Lagrange's equation (take a look at: kzfaq.info/get/bejne/hLxnlsdenLjeXX0.html ) will yield a damping matrix that looks like [c] = [ [c, ce], [ce, (ce²+c_t)] ]
@user-gz6kk3cw7zАй бұрын
@@Freeball99 Yes, I wanted at CG. Thanks a lot for the reply.
@dariorafael_oficial31466 жыл бұрын
Hi, I really liked the video. I am wondering about the drag effect, there is not an horizontal force component in this type of movement?
@ironpine58154 жыл бұрын
For the kinetic energy term, why is the rotational KE just 0.5*J*alpha^2? Since the CG is in general plane motion (translation and rotation), don't we have to consider the components of its velocity (ie. like you did for the pendulum's mass in your pendulum on a cart video)?
@Freeball994 жыл бұрын
The choice of coordinates is somewhat arbitrary. I could have modeled this problem that way. My rule-of-thumb is that when dealing with gravity, generally use x-y coordinates (which makes one less prone to careless errors). But, to be clear, I could have used x-y coordinates for this problem too. My suggestion is to try it both ways and compare the results.
@ironpine58154 жыл бұрын
Got it - thanks. When you wrote the velocity of the CG, you just included the vertical component: h dot - e*alpha dot. Is this because the CG velocity in the horizontal direction as it pitches is negligible? If my understanding is correct, KE in general plane motion is 0.5*m*V^2 + 0.5*J*omega^2, where V is a vector (with two components in 2-D).
@Freeball994 жыл бұрын
@@ironpine5815 Yes, the displacement in the horizontal direction is negligible due to the assumption of small displacements (and small rotations). In this problem, I assumed small displacements from the start (probably should have made that clearer). For the pendulum problem I never made the small displacement assumption.
@ironpine58154 жыл бұрын
@@Freeball99 Awesome - thanks for the clarification and prompt replies!
@brianwashingtonmusungwa75672 жыл бұрын
Hello sir thank you for the whole work but The lift acts on CP not CG ... So did we make an assumption that the CP and CG is at the same point ?
@Freeball992 жыл бұрын
There is no mention of any external force in the problem. This is a free vibration problem.
@johnm33982 жыл бұрын
For the derivation of kinetic energy (around the 2 minute mark), why is it heave rate minus pitch rate?
@Andy-hy8px2 жыл бұрын
This is a function of the coordinate system I have used. I chose the h-direction to be positive downwards and the α-direction to be positive when the airfoil pitches upward. As a result, the C.G. moves in the NEGATIVE h-direction when α is positive - as drawn. As a result, we subtract the pitch effect. IT SHOULD BE NOTED HOWEVER, that for a typical, real airfoil, the C.G. is usually AFT of the elastic axis (so e would be negative based on my coordinates). In this case, the sign would be flipped and the pitch effect would be added.
@ThomasCynthiaMomo Жыл бұрын
You have excellent handwriting and a fascinating accent. You should do professional voiceover work.
@Freeball99 Жыл бұрын
Maybe one day!
@user-gz6kk3cw7zАй бұрын
I want to write mass decouple equations,.. please suggest the method
@Freeball99Ай бұрын
EDIT: I realize that I gave you bad information here. I wasn't thinking clearly...if you write the EOM's about the CG (we would then assume that h describes the plunge of the CG rather than the elastic axis), then the velocities are decoupled and so the mass matrix would become diagonal, however the stiffness matrix would then become coupled. Your energies would become T = (1/2)m(dh/dt)² + (1/2)J_CG(dα/dt)² V = (1/2)K(h + eα)² + (1/2)K_t(α)² Plugging these into Lagrange's equations will give you the EOMs that you are after. [m] = [ [m , 0], [0, J_CG] ] and [k] = [ [K, Ke ], [Ke, (Ke² + K_t)] ]
@user-gz6kk3cw7zАй бұрын
@@Freeball99 Thank you very much for the reply.
@McTreestump5 жыл бұрын
Should there not be a potential energy term associated with the mass moving up or down? Is it because the weight is supposed by the spring at the top?
@Freeball995 жыл бұрын
SHORT ANSWER: You are not told that gravity is present in the problem, so the potential energy associated with the mass moving up and down is due to the potential energy in the spring only. LONGER ANSWER: If gravity were present in the problem, then you could still ignore the change in potential energy due to the motion of the mass up and down (translation) IF YOU ASSUME that the mass is oscillating about it's equilibrium point then you can ignore gravitational effects. I have explained this effect in this video (kzfaq.info/get/bejne/itaKgLegkrLNl2g.html). If gravity were present in the problem, you would, however, need to include the rotational potential due to gravity (i.e. gravitational force on the mass will produce a moment which cannot be ignored). The rotational component of this basically acts like a pendulum. In the case of a pendulum, you CANNOT ignore gravity.
@McTreestump5 жыл бұрын
@@Freeball99 That makes perfect sense, thank you. I may try the problem with a gravity component just to compare the behaviour.
@Freeball995 жыл бұрын
@@McTreestump That is definitely a worthwhile exercise. What you should find is that by including gravity, the x location is increased by a constant offset x0 = F/k. NOTE: Gravity WILL have a effect on the torsional/rotational part of the system (since this is a pendulum motion). The effects on the rotational motion on the airfoil due to gravity CANNOT be ignored by considering oscillations about an equilibrium position (since it changes the effective stiffness properties of the torsional system - much like a pendulum). Going through this exercise should make this all obvious.
@ShivangShekharRai3 жыл бұрын
@@Freeball99 Hi, great video and explanation! I know it has been a while since these comments but I just wanted to point out that in this case since the equilibrium position is close to horizontal (because of the torsional spring kt), the gravity term CAN be ignored even for rotational case. You are right though that it shows up in the pendulum but I think that is because the pendulum is vertical at its equilibrium position and hence for small oscillations the effect of gravity is like stiffness. Let me know if I misunderstood something.
@nathanharshman80332 жыл бұрын
Big Brain
@mechgyan42003 жыл бұрын
Why you considered J about c.g. ,not about the o??please sir
@Freeball993 жыл бұрын
The motion of the cg with respect to O was already captured in the velocity expression (the - eα_dot term). This is what give rise to the me^2 expression which is the moment of inertia of the mass about point O.
@mohemahmoude33565 жыл бұрын
sir plz i need help on how to get Lagrange's Equations for 2DOF helicopter
@90s2kgamer86 жыл бұрын
em means?
@Freeball996 жыл бұрын
m is the mass of the airfoil (which is assumed to be concentrated at the c.g.)
@ThomasCynthiaMomo Жыл бұрын
Ah, I'm trying to understand flutter and all of the examples stop short of a rigorous explanation of the right hand side.
@Freeball99 Жыл бұрын
It is certainly my intention to get to flutter eventually. I have a video showing a very simple example of aeroelastic instability using a 1DOF model (kzfaq.info/get/bejne/l7FhZdOrz6yah6s.html). Flutter analysis, however, requires at minimum a 2D model. To perform a full 2D work up will require a lot of math. I will make a video for it eventually when I can figure out the best way to keep the video (reasonably) short.
@DEChacker10 ай бұрын
@@Freeball99 a video about fluttering would be great Sir