The proof on Calculating the time of a Brachistochrone uses the equations of a cycloid. But the equations show a cycloid that looks like enter image description here

But I thought that the brachistochrone was dependent on gravity. So shouldn't the cycloid be upsidedown? Or does it not matter?

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    $\begingroup$ Yes, the cycloid should be upside down for it to be the brachistochrone. It's still called a cycloid however, because a curve's shape is independent of its position and rotation. $\endgroup$ Jan 10, 2021 at 10:14
  • $\begingroup$ The key place where gravitational acceleration enters into the cycloid is the time-dependence. That's evident from the accepted answer to the linked question, where the time of descent is $T=\sqrt{A/g}\theta_0$. $\endgroup$ Jan 10, 2021 at 10:23
  • $\begingroup$ To frame the point in a different way: Suppose that we build a model of a brachistochrone on Earth (e.g., instructables.com/The-Brachistochrone-Curve) and transport it to the Moon. Then the claim is that it'll still work as a brachistochrone: the descent will take longer (due to the reduced gravity) but it'll still be faster than any other path. $\endgroup$ Jan 10, 2021 at 10:44
  • $\begingroup$ @Chrystomath so shouldn't the equations used be negative of the y parametric equation? $\endgroup$
    – prata
    Jan 10, 2021 at 11:38
  • $\begingroup$ @BrienLim I haven't seen the answer, but whether $y$ is negative or positive depends on which direction the $y$-axis is taken. It does not have to be vertically 'upwards'. $\endgroup$ Jan 10, 2021 at 12:29


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