Survival under a linear force ramp ($p(F)=\exp [-(k_0{F}_0/f)(e^{F/F_0}-1)]$)

For an initially intact link with no rebinding, survival probability obeys $dp/dt=-k(F)p$. A force-tilted activation barrier with force $F=ft$, constant $f>0$ and $p(0)=1$ integrates to the displayed law. The rupture-force probability density is $k(F)p(F)/f$, not $p(F)$ itself. The no-rebinding approximation requires separated partners to leave the capture region before appreciable reassociation; it is not guaranteed by the rate equation alone.

Differentiate the logarithm of the rupture-force probability density $k(F)p(F)/f$. Its interior stationary point satisfies $e^{F/F_0}=f/(k_0{F}_0)$. The physical domain is $F\ge 0$, giving the displayed mode. This differs from the median rupture force under a linear ramp; replacing one by the other shifts a fitted intercept and can bias the inferred zero-force rate.

Solving $p(F_{1/2})=1/2$ for survival under a linear force ramp gives the displayed expression. At large $f/(k_0{F}_0)$, its slope against $\log f$ approaches $F_0$; at low loading rate the median is approximately $f\log 2/k_0$. A full fit can infer activation distance $k_{B}T/F_0$ and zero-force escape rate $k_0$. An absolute barrier additionally needs an attempt prefactor, and equilibrium binding strength additionally needs reverse-state information.