For a roller coaster loop, if it were perfectly circular, we would have a minimum speed of $v_{min} = \sqrt{gR}$ at the top of the loop where $g=9.8 m/s^2$ and $R$ is the radius of the 'circle'. However, most roller coaster loops are actually not circular but more elliptical. I've been looking for ways to calculate the min. speed at the top for an elliptical loop, but so far I haven't been able to. How could I go about that?

up vote 2 down vote accepted

As it turns out, you actually can use that same formula $ v_{min} = \sqrt{gR} $. However $R$ is the radius of curvature at the top of the loop, which is simply equal to the radius in the case of a circle. See here for more information on finding the curvature of an ellipse.

In general, the curvature of a plane curve given by $ (x(t),y(t)) $ can be found using the formula $ \kappa = \frac{|\dot x \ddot y - \dot y \ddot x|}{(\dot x^2 + \dot y^2)^\frac 3 2} $ (or one of many other formulae) where $ R = \frac 1 \kappa $

  • I'm a little confused bc your answer and the other answer seem to differ, and I was wondering if there's any reason why... – herminny Dec 6 at 0:13
  • @herminny I'm just explaining the concepts without getting into all of the math behind it. If you want to ask about the other answer, you should comment there. – Sandejo Dec 6 at 0:33
  • okay, thank you. Be R (the radius of the curvature at the top of the loop), do you just mean the longer radius of the ellipse? – herminny Dec 6 at 0:45
  • @herminny No, the radius of curvature is the reciprocal of curvature, which is a quantity that describes how curved or flat a curve is at a given point. It is typically covered in multivariable calculus courses, which you may not have taken yet. Since it is not simply one value for the whole ellipse, you need to find the radius of curvature at the top, regardless of whether that is on the major or minor axis. – Sandejo Dec 6 at 1:03
  • in the link you gave me in the answer, there's a formula I could use to calculate the curvature. However, my equations are actually $x=20\cos(2\pi t)+150$ and $z=-40sin(2\pi t)$ (and $y=-15t+5$)... but I'm not sure how to modify the given formula to fit my equations. could you please help me out? – herminny Dec 6 at 1:49

The ellipse equation in polar coordinate is:

$\begin{bmatrix} x \\ y \\ \end{bmatrix}=\left[ \begin {array}{c} r\cos \left( \varphi \right) \\ r\sin \left( \varphi \right) \end {array} \right] $

$r={\frac {ba}{\sqrt {{a}^{2} \left( \sin \left( \varphi \right) \right) ^{2}+{b}^{2} \left( \cos \left( \varphi \right) \right) ^{2 }}}} $

where $2a$ is the major axis and $2b$ the minor axis

Kinetic energy

$T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$

with $\begin{bmatrix} \dot{x} \\ \dot{y} \\ \end{bmatrix} =\left[ \begin {array}{c} -{\frac {b{a}^{3}\sin \left( \varphi \right) \varphi p}{ \left( {b}^{2} \left( \cos \left( \varphi \right) \right) ^{2}+{a}^{2}-{a}^{2} \left( \cos \left( \varphi \right) \right) ^{2} \right) ^{3/2}}}\\ {\frac {{b }^{3}a\cos \left( \varphi \right) \varphi p}{ \left( {b}^{2} \left( \cos \left( \varphi \right) \right) ^{2}+{a}^{2}-{a}^{2} \left( \cos \left( \varphi \right) \right) ^{2} \right) ^{3/2}}}\end {array} \right] $

and $\varphi p=\dot{\varphi}$

Potential Enegry

$V=m\,g\,y$

we solve the equation $T=V$ for $\dot{\varphi}$ and get for $v_{max}=r\,\dot{\varphi}|_{(\varphi=\frac{\pi}{2})}$

$v_{max}=\sqrt{2\,b\,g} $

For a circle ($b=a=R$) with radius $R$ we get:

$v_{max}=\sqrt{2\,R\,g} $

Simulation result

enter image description here

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