Table 1.

Algorithmic parameters

Fig/System	Algorithm	Parameters
Fig. 2B/hENM	HB	$k_{\varphi }={10}^3$, ${10}^4\frac{\text{kJ}}{\text{mol}{\text{rad}}^2}$, $\overline{\varphi }=-6.32°$
Fig. 2C/hENM	SGD	$A=10k_{B}T$, $\overline{\varphi }=-6.32$τ̅avg = from 1–100 ps
Fig. 2D/hENM	SGD	$A=10k_{B}T$, ${\tau }^{\text{avg}}=10\text{ps}$. $\overline{\varphi }:-9.16°\text{to}-25.2°(\langle \varphi \rangle \approx -17°,{\sigma }_{\varphi }\approx 4°$
Fig. 3A (top)/hENM	SGD	τavg = 10 ps, A = 10kBT, $\overline{\varphi }=-6.32°$, $\overline{d}=20.62\text{Å}$
Fig. 3A (top)/hENM	Covar	See above, but A = 1kBT
Fig. 3A (bottom)/hENM	SGD	τavg = 100 ps, A = 10kBT, $\overline{\varphi }=-6.32°$, $\overline{{\varphi }^2}={40.86}^{{(\circ )}^2}$, $\overline{d}=20.62\text{Å}$, $\overline{d^2}=425.03{\text{Å}}^2$
Fig. 3A (top)/hENM	Covar	See above, but A = 1kBT
Fig. 3B/hENM	Covar	τavg = 100 ps, A = 1kBT, $\overline{\varphi }=-6.32°$, $\overline{{\varphi }^2}={40.86}^{{(\circ )}^2}$, $\overline{d}=20.62\text{Å}$, $\overline{d^2}=425.03{\text{Å}}^2$
	LM	See above, with γ = 0.1,0.01
	Adaptive LM	See above, with γ0 = 0.1, m = 3, l = 1.2, LM_stride = 10
Fig. 4A (top), 4B/G-monomer	HB	$k_d=k_{\varphi }=5000\frac{\text{kcal}}{\text{mol}{\text{red}}^2}$, $\overline{\varphi }=-6.32°$, $\overline{d}=20.62\text{Å}$
	Adaptive LM	τavg = 100 ps, A = 10kBT, γ0 = 0.1 $\overline{\varphi }=-6.32°$, $\overline{{\varphi }^2}={40.86}^{{(\circ )}^2}$, $\overline{d}=20.62\text{Å}$, $\overline{d^2}=425.03{\text{Å}}^2$, m = 3, l = 1.2, LM_stride = 100
Fig. 4A (bottom)/F-monomer	Adaptive LM	See above, except γ0 = 0.1
	Frozen bias (after 80 ns of above)	${\lambda }_{\varphi }=-10.2\frac{\text{kcal}}{\text{mol}\text{rad}}$, ${\lambda }_{{\varphi }^2}=484.0\frac{\text{kcal}}{\text{mol}{\text{rad}}^2}$, ${\lambda }_d=2.2\frac{\text{kcal}}{\text{mol}\text{Å}}$, ${\lambda }_{d^2}=6.63\frac{\text{kcal}}{\text{mol}{\text{Å}}^2}$
Fig. 5B,C	LM	τavg = 100ps, A = 1kBT, γ0 = 0.01 $\overline{{\varphi }_{A1}}=\overline{{\varphi }_{A2}}=\overline{{\varphi }_{A3}}=-6.32°$, $\overline{d_{A1}}=\overline{d_{A2}}=\overline{d_{A3}}=20.6\text{Å}$