
Post by Br. Marius on Feb 28, 2015 17:01:57 GMT
Calculate the higher harmonic flap response of a blade caused by a multicyclic pitch input \(\theta(\psi)=q_{3s}sin(3\psi)\). For the given flap equation and definitions, assume \(\beta(\psi)=\beta_{3c}cos(3\psi)+\beta_{3s}sin(3\psi)\).



Post by Br. Marius on Mar 3, 2015 15:20:02 GMT
That was some nasty algebra. Insofar as simplification goes, I figured that it would not be appropriate to make assumptions about terms being "small" (such as higher order advance ratios) as it seems to alias the higher harmonics...so here goes. I've sorted by harmonic in hopes of greater clarity: \(\beta_{3s}=q_{3s}\left\{\frac{72\nu_{\beta}^2+72\mu^2\nu_{\beta}^2648\mu^2648+cos(\psi)[12\mu+18\mu^3]+sin(\psi)[1728\mu+192\mu\nu_{\beta}^2]+cos(2\psi)[648\mu^272\mu^2\nu_{\beta}^2]+sin(2\psi)[9\mu^4+25\mu^2]+cos(3\psi)[18\mu^3]+sin(4\psi)[4.5\mu^4]}{80\mu^24.5\mu^4+576\nu_{\beta}^410368\nu_{\beta}^2+46737+cos(\psi)[192\mu\nu_{\beta}^21728\mu]+sin(\psi)[216\mu]+cos(2\psi)[64\mu^2]+cos(4\psi)[4.5\mu^4]}\right\}\)
\(\beta_{3c}=q_{3s}\left\{\frac{27+75\mu^2+sin(\psi)[108\mu+54\mu^3]+cos(2\psi)[75\mu^2]+cos(3\psi)[18\mu^3]}{4.5\mu^464\mu^2+576\nu_{\beta}^410368\nu_{\beta}^2+46737+cos(\psi)[1728\mu192\mu\nu_{\beta}^2]+sin(\psi)[216\mu]+cos(2\psi)[64\mu^2]+cos(4\psi)[4.5\mu^4]}\right\}\)Attachments:HW4 P2.pdf (453.54 KB)


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Post by Deleted on Mar 3, 2015 17:47:21 GMT
Hmmm, so our answers are wildly different. Though I think that's mainly due to the fact that I eliminated periodic terms since I don't think \(\beta_{3c}\) and \(\beta_{3s}\) are supposed to be functions of \(\Psi\). Right?



Post by Br. Marius on Mar 3, 2015 19:05:45 GMT
That would make sense to me, but, because all of the harmonics are mathematically intertwined as you go through the problem, they are. So how do we justify dropping terms?

