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A cycloidal pendulum is isochrone, meaning it's period is independent of it's amplitude. But a cycloidal pendulum - as usually depicted - doesn't do a full 360 degrees swing. Why is that? Is there a limit on how high a pendulum can swing, so that it is still isochrone? Or is it possible to build a isochrone pendulum that does a full 360 degrees swing? And if so, what kind of curve would it need to follow? Is it just an extension of a cycloid as shown in the picture below.

enter image description here

Or would a 360 degree isochrone pendulum maybe follow a different curve, e.g. maybe something like a Cardioid?

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    $\begingroup$ Isn't the upper limits unreachable due to boundary conditions? $\endgroup$
    – Kyle Kanos
    Jul 23, 2017 at 21:36
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    $\begingroup$ I don't understand the close votes on this one. It's a perfectly reasonable, and perfectly clear, question: given a time period $T$, is there an upper bound to the maximal velocity that an isochronal gravity-powered pendulum can achieve? The standard construction suggests that there is, but it is not a trivial question at all. $\endgroup$ Jul 24, 2017 at 15:47

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This sketch became too long to be a comment so I will post it as an answer anyway.

Let us say we have a wire forming a cycloid in the vertical plane as in the figure bellow

enter image description here

A friction-less bead into this wire will oscillate with known period $T$ independently of its releasing point. Assume we can symmetrically continue the wire above the dashed line. If we release the bead from $A'$, which is infinitesimally close to $A$ but above the dashed line, then it will take a time $\Delta t_{AB}<T$ to complete the segment ${AB}+BA$ since it travels faster. If we want the same period $T$ for the bead to oscillate between $A'$ and $B'$, then the time taken along $A'A$ shall be $d t_{A'A}=(T-\Delta t_{AB})/2$. The idea is to use energy conservation and the parametric equation of the cycloid to compute $\Delta t_{AB}$ and then obtain $d t_{A'A}$.The last step is to suitably chose the slope of the straight segment $A'A$ so that the interval $dt$ for the bead to slide from $A'$ to $A$ matches $dt_{A'A}$. I think you will find that $dt_{A'A}<dt$ even when $A'$ is vertically above $A$ (which gives the least $dt$). This means that the tautochrone curve cannot be continued above the dashed line. You can also recall that this construction can be actually used to obtain the tautochrone. The slope of the infinitesimal segments are always increasing, from zero at $C$ to $\pi/2$ at $B$. One increment further and we can no longer increase the slope.

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