
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 pitchflap 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. 219. 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.



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.



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 \(\nu_\beta^2 = 1+\frac{3}{2}\frac{e}{Re}\). If you use the approximation \(\nu_\beta^2 = 1+\frac{3}{2}\frac{e}{R}\) I get \(e = 1.648\). This give a flaphinge distance of \(21.522 = 0.478\) and \(\delta_3 = \tan^{1}\left(\frac{0.478}{1}\right) = 25.5^\circ\).



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.



Post by matthorr on Apr 6, 2015 11:42:04 GMT
That makes sense. Here would \(\omega_{\beta_0}\) be the same as the nonrotating frequency? He gave \(\gamma\) but I don't think we used it in our solutions.



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.



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 \(\nu_\beta^2\).

