Relation between settling time and natural frequency for a critically damped system
You need to be careful when using settle time approximations that relate \$\omega_n\$ and \$\zeta\$ to a  \$T_s\$. 

First, \$T_s\$ is defined as the time where the signal remains within \$\pm x\$% of the final value. Typically the settle limits are \$\pm 2\$% or \$\pm 5\$%, and these limits impact all second order system settle time approximations. Yes, every 2nd order settle time equation is an approximation, and they all differ as a function of \$\zeta\$.

Next, $$\omega_n \approx \frac{5.8335}{T_s},$$ is only a numerical approximation to the normalized unit step response for a critically damped system (i.e., \$\zeta=1\$), which is $$\theta(t)= \left(1-(1+\omega_n t)e^{-\omega_n t}\right)u(t),$$ where \$u(t)\$ is a unit step (aka Heaviside step).  The settle time approximation for this is eloquently derived in your "this answer" link using a \$\pm 2\$% window to numerically arrive at \$\omega_n \approx \frac{5.8335}{T_s}\$.

When I look at the graphs of the data you shared, it looks like the 2% settle time for the critically damped trace is closer to 35.  This results in \$\omega_n\approx\frac{5.8335}{35}= 0.167\$ rad/sec.  A result is closer to the actual \$\omega_n\$ in the article. Better graphs, and clarification on which trace is actually the critically damped one, will improve this estimate.

Sorry, but I do not have free access to your article, so cannot answer your question directly, but most likely the authors knew \$\omega_n\$ directly.