This note is in addition to this paper. It details the formula for determining the coefficient \(g\) using a two-point model.

The problem is this. There is a plot of current \(I\) versus time \(t\), shown in Figure 1. We can measure two points on this graph: 2 and 3. It is also known that the τ time is approximately half of the descending part of this graph, and the required coefficient can be expressed through this value: \[g = K^{\large {\tau / (t - \tau)}} \tag{1}\] Here: \(K\) is a known coefficient depending on the measurements on the graph.

If we knew exactly τ, then this problem would not arise, and we would calculate \(g\) using formula (1). But this parameter greatly affects the result even with a small error, and in the schema it is only roughly defined. In this case, we consider that τ is unknown, and the coefficient we need is determined by two points on the graph.

For a two-point model, you need to prepare the appropriate formulas. First we output τ: \[\tau = {t \over 1 + \ln(K) / \ln(g)} \tag{2}\] Since we believe that this parameter is unknown to us, but the data of two measurement points are known, we can rewrite this formula as follows: \[\tau = {t_2 \over 1 + \ln(K_2) / \ln(g)} = {t_3 \over 1 + \ln(K_3) / \ln(g)} \tag{3}\] Here: \(t_2, K_2, t_3, K_3\) - measurement data for two points. Where do we get the desired coefficient: \[ \ln g = {t_2 \ln(K_3) - t_3 \ln(K_2) \over t_3 - t_2} \tag{4}\] The same formula can be written in another way: \[ g = { K_3^{\large {t_2 \over t_3 - t_2}} \over K_2^{\large {t_3 \over t_3 - t_2}} } \tag{5}\] Thus, bypassing the τ, we got the desired result. At the same time, we can now find the τ parameter itself with maximum accuracy if we substitute expression (4) into formula (2): \[ \tau = {t_2 \ln(K_3) - t_3 \ln(K_2) \over \ln(K_3) - \ln(K_2)} \tag{6}\] Q.E.D.