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Post by Br. Marius on May 7, 2015 15:17:15 GMT
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Derive ground resonance equations for a two-bladed rotor with three-degree of freedom system: Cyclic lag \((\zeta_{1c},\zeta_{1s})\) and roll motion \((\alpha_x)\).
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Post by matthorr on May 11, 2015 0:36:53 GMT
So, what DOFs do you think he wants, 2 with /(/zeta_1/) or 3 with /(/zeta^{(1)}/) and /(/zeta^{(2)}/)?
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Post by Br. Marius on May 11, 2015 12:46:53 GMT
In my notes for 2-bladed rotors, Chopra talks about a cyclic lag in terms \(\zeta_{1s},\zeta_{1c}\)...I just assumed that was what he was asking for here since he made the same typo last week. Really, it seemed to me as if this week's derivation is roughly the same as last week's, just with a specific number of blades and body motion in a different direction.
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Post by matthorr on May 11, 2015 13:08:43 GMT
Yes, I think what he is looking for is to put the lag motion into the fixed frame - \(\zeta_1\) - and then separate the cos and sin terms into \(\zeta_{1c}\) and \(\zeta_{1s}\) since \(\zeta_1 = \zeta_{1c}\cos\psi + \zeta_{1s}\sin\psi\). So, in the end, you should not have periodic terms in the two \(\zeta\) equations.
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Post by Br. Marius on May 11, 2015 13:57:22 GMT
I think you do end up with a periodic term in the lag equations, though, because of the moment given to the blades by the rolling motion of the aircraft (the effect of which varies with blade azimuth). Rather, I don't see how the periodic term in the rotating frame vanishes when the equation is transformed into cyclic fixed-frame equations for a 2-bladed rotor. Physically speaking, the rolling motion of the aircraft should create a moment in both, just varying in magnitude by the appropriate trigonometric term. On the other hand, I'm not entirely sure what you mean. Doesn't FCT take care of separating out the fixed-frame cyclic lag equations? Attachments:HW12 P2.pdf (219.26 KB)
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Post by matthorr on May 11, 2015 15:15:10 GMT
I'm still working it out so it's a mess at them moment (I will post when I have it done), but I am converting \(\zeta_1\) in the fixed frame to \(\zeta_1c\) and \(\zeta_1s\) in the fixed frame. It should remove periodic terms from the coefficients in the two \(\zeta\) equations but will introduce more coupling. I start with \(\zeta_{1c} = \zeta_1\cos\psi\) and \(\zeta_{1s} = \zeta_1\sin\psi\), find the derivatives, back substitute until everything on the left hand side is \(\zeta_1\) & derivatives times \(\cos\) or \(\sin\) and the right hand side is all \(\zeta_{1c}\) and \(\zeta_{1s}\) and their derivatives. Similar to what we did in Aero to get the periodic solution to the blade flapping equation.
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Post by Deleted on May 11, 2015 15:30:10 GMT
For body equation for roll I have Mx-hY as the body forcing. I am using the side force instead the drag force you are using. This was looking at pdf post by Br. Marius
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Post by matthorr on May 11, 2015 15:32:59 GMT
Yes, the signs and sin / cos are different but it should be side force and roll moment (/(M_x/) or /(L/).
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Post by Deleted on May 11, 2015 15:36:45 GMT
Matt, I am sort of taking your approach by treating ζ1 as two separate cases. One with ζ1=ζ1c cosψ and ζ1=ζ1s sinψ. That lets me have 3 DOF like he wants. Still working on it.
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Post by matthorr on May 11, 2015 15:44:38 GMT
That's what I'm m working on. When you take the derivatives you will get some more terms when you substitute back into the /(/zeta_1/) equation and then you can group terms to split it into two equations with /(/zeta_{1c}/) and /(/zeta_{1s}/). Matt, I am sort of taking your approach by treating ζ1 as two separate cases. One with ζ1=ζ1c cosψ and ζ1=ζ1s sinψ. That lets me have 3 DOF like he wants. Still working on it. |
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Post by Br. Marius on May 11, 2015 15:49:24 GMT
For body equation for roll I have Mx-hY as the body forcing. I am using the side force instead the drag force you are using. This was looking at pdf post by Br. Marius Yeah, that's my mistake for forgetting to update my notation as I was writing. I believe that the azimuth references are still correct as I have them written if you use Mx-hY instead of Mx+hH, though. At any rate, I think I'm starting to get what you're saying, Matt. Still not 100%, but I'm going to look back through my notes and see what comes out. |
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Post by matthorr on May 11, 2015 18:36:27 GMT
I feel like I'm running a fool's errand running around in circles - I can get the lag equation to come out nicely with \(\zeta_{1s}\) and \(\zeta_{1c}\) terms, but then my body equation becomes really nasty...
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Post by matthorr on May 12, 2015 1:07:09 GMT
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