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Concerning this subject, I consulted the following Wikipedia page: Universal joint

where the elementary kinematics equation of the U-joint is stated.

But, a little bit further in the article Double Cardan Shaft

where the configuration described consists of 2 U-joints, the previously mentioned equation is reversed, i.e:

$\tan \gamma _{1}=\cos \beta \tan \gamma _{2}$,

$\tan \gamma _{2}=\cos \beta \,\tan \gamma _{1}\qquad \tan \gamma _{4}=\cos \beta \,\tan \gamma _{3}$

I do not intend to doubt the validity of the info on the article, but this hit me as a pretty weird thing.

Are these relationships correct?

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2 Answers 2

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The first equation seems correct, and when it is used again it seems to be incorrectly reversed as you point out. But, it did not matter in the end because the mistake was made twice. With the correct equation we still get the same final result.

The equations $$ \tan \left(\gamma _1\right)=\cos (\beta ) \tan \left(\gamma _2\right) $$ $$ \tan \left(\gamma _3\right)=\cos (\beta ) \tan \left(\gamma _4\right) $$

together with $\gamma _3=\gamma _2+\frac{\pi }{2}$ and $\tan \left(\gamma +\frac{\pi }{2}\right)=\frac{1}{\tan (\gamma )}$, gives

$$\tan \left(\gamma _4\right) = \frac{1}{\cos (\beta )} \tan \left(\gamma _3\right) = \frac{\tan \left(\gamma _2\right)}{\tan \left(\gamma _1\right)}\frac{1}{\tan \left(\gamma _2\right)}= \frac{1}{\tan \left(\gamma _1\right)}= \tan \left(\gamma _1+\frac{\pi }{2}\right) $$

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  • $\begingroup$ Hello, thanks for your answer. Could you please enlighten me in the following? Is it possibe to achieve a sychronized input-output relationship using a double u-joint configuration? I.e with a phase of zero and same amplitude? If this is indeed true, and assuming $\gamma _{2} = \gamma _{3}$, the last equation needs to be: $\tan \gamma _{4}=\cos \beta \,\tan \gamma _{3}$ Is that contradicting the above mentioned analysis or do i miss something? $\endgroup$
    – Paraskevas Dimitris
    Sep 21, 2016 at 21:14
  • $\begingroup$ I'm sorry. I'm not understanding your follow-up question. $\endgroup$
    – Suba Thomas
    Sep 21, 2016 at 22:00
  • $\begingroup$ Let me rephrase. From my understanding, the article refers to a case where the configuration achieves an output speed that is in a 90 degrees phase with respect to the input. I was wondering if it is possible to get an output in pure harmony with the input, i.e with zero phase. And, in that case, in order for the equations to hold, the following has to be true: $\tan \gamma _{4}=\cos \beta \tan \gamma _{3}$, because in that case it has to be: $\gamma _{1}= \gamma _{4}$, $\gamma _{2}= \gamma _{3}$ From your experience, is that plausible? $\endgroup$
    – Paraskevas Dimitris
    Sep 22, 2016 at 13:28
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I think the confusion is that, for the double joint, $tan\gamma_2$ is the output angle, not the angle of the intermediate shaft relative to the source $tan\gamma_1$ . If you hack through the coordinate systems on the way from the first joint to the second joint, it should come out correctly.

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  • $\begingroup$ Hello, i don't think this is the case as the article clearly states that $ \gamma {2} $ is the speed of the intermediate shaft $\endgroup$
    – Paraskevas Dimitris
    Sep 22, 2016 at 13:29

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