Research website of Vyacheslav Gorchilin
2023-05-23
Two-point model for determining the coefficient g
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.
 Fig.1. Graph of current I versus time t
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.