6
$\begingroup$

I wonder how I can solve the Brachistochrone problem for 3 points? The matter starts from point A that is the highest point and it must pass from B and must finish with point C. (No any friction in the system) I wonder what the shortest time path for that problem?

Is it enough to use Johann Bernoulli's solution from A to B and then Use the solution from B to C?

Or do I need to follow a different way to solve the shortest time path problem for 3 points?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

Yes, but one first has to generalize the classical 2-point Brachistochrone problem $A \to B$ where the initial speed $v_A$ traditionally is zero, to the case where the initial speed $v_A$ may be non-zero but fixed. The solution to this initial speed Brachistochrone problem (assuming no friction) is still a cycloid.

Now consider the 3-points Brachistochrone problem $A \to B\to C$ with initial speed $v_A$. The speed $v_B$ is given by energy conservation alone. Thus the two segments $A \to B$ and $B \to C$ are completely decoupled, and they can be optimized as two independent 2-point Brachistochrone problems with initial speeds $v_A$ and $v_B$, respectively, leading to two corresponding cycloids $A \to B$ and $B \to C$.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ what about y'? You would still need continuity there, this is not guaranteed with two patched together solutions. $\endgroup$
    – tmac
    Commented May 5, 2012 at 22:13
  • $\begingroup$ Here I'm considering the idealized Brachistochrone problem where solution curves are allowed to be only piecewise smooth. Instantaneous change of velocity is provided by an idealized (infinitely big) normal force, which does no work. $\endgroup$
    – Qmechanic ♦
    Commented May 6, 2012 at 6:46
  • $\begingroup$ Notes for later: $1+ (\frac{dx}{dy})^2=\frac{k^2}{y}$. Rescale $x$ and $y$ so that $k^2=2$. Then $\frac{dx}{dy}= +(\frac{2}{y}-1)^{-1/2}=\sqrt{\frac{y}{2-y}}$ with antiderivative $x=\arctan \frac{\sqrt{y(2-y)}}{1-y}-\sqrt{y(2-y)}=\theta-\sin\theta$, where $1-y=\cos\theta$. $\endgroup$
    – Qmechanic ♦
    Commented Aug 2, 2019 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.