I am a engineer working on Wankel motors, where a simil-Reuleaux triangle rotates eccentrically in a 8-shaped form:
(from Wikipedia)
Fascinated by this mechanism, I was studying the meaningful points out of sheer curiosity.
The parametric equations to define the epitroch (the external 8-shape) are for eccentricity $f$ and $0\leq t < 2 \pi$:
$$x = \cos(t)+f \cos(3t)$$ $$y = \sin(t)+f \sin(3t)$$
First and second derivatives of the parametric equations return respectively the red and fuchsia points ($f=0.2$ in my drawing):
QUESTION: Is it possible to calculate analytically the 4 blue points where the curve changes curvature?
I was wrongly assuming they were given by the second derivative, but it is not the case. I am in a cul-de-sac and perhaps I only need to find another hobby.
The signed curvature of a parametric plane curve is $$ k(t) = \frac{x'(t)y''(t) - x''(t)y'(t)}{(x'(t)^2+ y'(t)^2)^{3/2}} \,. $$ Since you are only interested in the sign of $k(t)$, it suffices to compute the numerator $$ \begin{align} x'(t)y''(t) - x''(t)y'(t) &= \bigl( \sin(t)+3f \sin(3t)\bigr)\bigl( \sin(t)+9f \sin(3t)\bigr) \\ &\qquad + \bigl( \cos(t)+3f \cos(3t)\bigr)\bigl( \cos(t)+9f \cos(3t)\bigr) \\ &= 1 + 27 f^2 + 12 f \cos(2t) \, . \end{align} $$ If the absolute value of $\frac{1+27f^2}{12f}$ is less than one then the curvature changes its sign exactly where $$ \cos(2t) = - \frac{1+27f^2}{12f} $$ and this equation has four solutions in $[0 , 2 \pi)$, one in each quadrant.
For $f= 0.2$ the first solution (in degrees) is $$ t_1 = \frac 12 \arccos\left( -\frac{52}{60}\right) \approx 75.04^{\circ} $$ and the other three solutions are $180^{\circ}-t_1$, $180^{\circ}+t_1$, $360^{\circ} - t_1$.