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Post by Br. Marius on Mar 31, 2015 13:36:29 GMT
For a hingeless rotor, the pitch bearing is located at 2 ft. from the rotation axis and the pitch link is positioned 1 ft towards the leading edge. The nonrotating flap frequency was determined experimentally to be 1 Hz. Calculate approximately the effective pitch-flap coupling (\(\delta_3\) angle) for a 40 ft. diameter rotor with Lock number \(\gamma=8\) and rotation speed \(\Omega=300RPM\). Assume uniform blade and you may use Fig. 2-19. Attachments:
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Post by Deleted on Apr 2, 2015 18:03:30 GMT
I got:
\(\nu_\beta = 1.06\) from Fig. 2.19
\(e = 1.11 ft\)
and finally:
\(\delta_3 = 41.7^o\)
I not sure what kind of tolerance there is on the answer though, because I think it can vary quite a bit depending on the \(\nu_\beta\) chosen.
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Post by Br. Marius on Apr 3, 2015 15:50:43 GMT
I got 1.06 from the chart and got the same answers. I'm agreed about tolerances, though...I chose 1.08 at first and got a crazy answer (~17 deg). Going back with a digitizer confirmed 1.06 as a better selection.
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Post by matthorr on Apr 5, 2015 2:12:44 GMT
I got \(\omega_{NR} = 0.2\) and I also chose \(\nu_\beta = 1.06\) from the chart. But I got \(e = 1.522\) using the equation \(\nu_\beta^2 = 1+\frac{3}{2}\frac{e}{R-e}\). If you use the approximation \(\nu_\beta^2 = 1+\frac{3}{2}\frac{e}{R}\) I get \(e = 1.648\). This give a flap-hinge distance of \(2-1.522 = 0.478\) and \(\delta_3 = \tan^{-1}\left(\frac{0.478}{1}\right) = 25.5^\circ\).
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Post by Br. Marius on Apr 5, 2015 19:52:53 GMT
I used \(\nu_\beta^2=1+\frac{3e}{2R}+\frac{\omega_{\beta0}^2}{\Omega^2}\) to get my solution...that might explain things.
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Post by matthorr on Apr 6, 2015 11:42:04 GMT
That makes sense. Here would \(\omega_{\beta_0}\) be the same as the non-rotating frequency? He gave \(\gamma\) but I don't think we used it in our solutions.
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Post by Br. Marius on Apr 6, 2015 13:35:33 GMT
Yes, I believe that is so concerning \(\omega_{\beta0}\). I thought the same thing about \(\gamma\)...it doesn't seem like giving extraneous information is a standard mode of operation around UMD's grad school, but it would appear to be the case here. The information we do have does not seem to necessitate its use.
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Post by matthorr on Apr 6, 2015 18:29:45 GMT
My numbers agree with yours now. I get \(e=1.056\) ft and \(\delta_3=43.4^\circ\) using the more exact equation for \(\nu_\beta^2\).
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