Measure time in units of $L_K/R$ and currents in units of $I_c$,
where $L_K = \hbar / 2e I_c = \Phi_0 / 2\pi I_c$ is the kinetic inductance at zero bias.
The dimensionless voltage becomes $v_x = V_x (L_K/R) (2\pi / {\Phi_0}) = V_x / R I_c = \dot \theta$, where the dot now indicates the time derivative wrt the rescaled time $t/\tau_{L/R}$.
The dimensionless current $i_x = I_x^\text{tot}/I_c$ becomes
The dimensionless impedance is $Z_0 / R = \sqrt{L_K/R^2 C_0} = \lambda/Q$.
The RC and L/R times are $\tau_{RC} = Q^2$ and $\tau_{L/R} = 1$, respectively.
The velocity of the Mooij-Schön mode is $L_K / R/\sqrt{L_K C_0} = \sqrt{L_K/C_0 R^2} = \lambda / Q = Z_0 / R$ junctions / dimensionless time,
where $\lambda = \sqrt{C/C_0}$ is the charge screening length (in units of $\xi$).
In dimensionless units we can therefore set
: Ic = 1
: R = 1
: Rqp = ?
: C = Q^2
: C0 = Q^2 / lambda^2
: Rterm = Rterm / R
: Rshunt = Rshunt / R
: Cterm = Q^2 Cterm / C
: T = k_B T / E_J = k_B T / L_K Ic^2
: Vgap = v_g = 2Delta/e R Ic = 4/pi or slightly less, maybe 1.
Note that $I_c$, $NR$, and $Z_0$ should be easy to determine experimentally.
Then $Q/\lambda = (R/Z_0)$, so we can set the dimensionless $C_0 = (R/Z_0)^2$
The velocity is $c_0 = Z_0 / R$ junctions / time unit.
In dimensionfull units it is $1/\sqrt{L_K'C_0'} = (Z_0/L_K) \xi= \xi/Z_0 C_0$.
Possibly one can read off the sum gap voltage $N V_g = N 2 \Delta/e$ from an IV curve?
From the litterature we know $\Delta$, and then we may estimate $N$?
Microscopically, for Josephson juncions
Using the Ambegaokar-Baratoff relation $I_c = \frac{\pi \Delta}{2 e R} \tanh \frac \Delta {2k_BT}$.
The gap voltage $V_g = 2\Delta / e ; \Rightarrow , v_g = 2\Delta / e I_c R = 4 / \pi \tanh (\Delta/2k_B T)$.
Assuming $T \lesssim 0.4 T_c$, we take $\Delta(T) \approx \Delta_0$ and $\tanh(\Delta/2k_B T) \approx 1$.
Superconducting nanowires
Discretize using a lattice constant $\approx \xi =$ the superconducting coherence length.
Note that in the discrete model the kinetic inductance is $L_K(i) = \hbar / 2e \sqrt{I_c^2 - I_b^2} = L_K(0) / \sqrt{1-i_b^2}$ is larger than $L_K \equiv L_K(0)$, especially close to the critical current. For a continuous superconducting nanowire we similarly expect from GL that the supercurrent density obeys
$$
J_s = 2e n_s (1 - (v/v_0)^2) v
$$
and therefore that $L_K$ increase near $I_c$.
Depairing current
London limit depairing current density: $J_d = H_c/\lambda_L = \Phi_0/2\pi \mu_0 \lambda_L^2 \xi$, where $2 e n_s = 1/\mu_0 \lambda_L^2$.
The corresponding $I_c = J_d s = \Phi_0 / 2\pi L_K = \hbar / 2e L_K$ (as above!).
GL depairing current density differs by a numerical factor: $J_d = \Phi_0/ 3 \sqrt 3 \pi \mu_0 \lambda_L^2 \xi$.
Up to a numerical factor of order one these relations agree with $J_d = \sigma \frac {2\Delta} {e \xi} ; \Rightarrow ; I_c = V_g/R$.
Quasiparticles
The shunt resistors in the model describe the dissipation caused by quasiparticles in a two fluid model of superconductivity.
At low temperatures the density of quasiparticles is strongly suppressed,
$n_{qp}(T) = n e^{-\Delta / k_B T} \ll n$, which motivates using a nonlinear shunt resistor in the model.
For small currents or eqivallently $|V_{x}-V_{x+1}| < V_g = 2 \Delta/ e$ we thus ignore their contribution, while for $|V_{x}-V_{x+1}| > V_g$ we assume that the quasiparticle current is ohmic with resistance $R$.
This suggests that the model should be applicable also for continous wires. A possible refinement would be to include a subgap resistance.
Microscopic model of nanowire
(See Golubev, D. and Zaikin, A. (2001)Quantum tunneling of the order parameter in superconducting nanowires, Physical Review B 64, 014504. doi: 10.1103/PhysRevB.64.014504.)
