I'm trying to find the two unknown constants of the following function:
$$f(x) = \frac{158}{1+e^{s(x-c)}}+22$$ Where s and c are constants, and 22 (or $y_{min}$) and 180 (or $y_{max}$) are the horizontal asymptotes. c is the integration constant.
This was integrated from the following ODE (Ricatti equation):
$$\frac{s}{158}\frac{dy}{dx}=a(y-22)(180-y) , f(0) = y_0 > 22 $$
I can assume that f(0) lies near the horizontal asymptote. Additionally, I have the following equations:
$$f(0) = y_0 = \frac{158}{1+e^{sc}}+22$$ Hence $$c = \frac{1}{s} log \frac{180 - y_0 }{y_0 - 22}$$
How do I find both constants mathematically? I realize that I need two equations, but am unsure of how I can find them. From plugging in this into Desmos, I have found that c is the inflection point of the graph, while s is responsible for the shape of the curve: https://www.desmos.com/calculator/hwg4vsewro
The broader context of this question is that I'm finding a logistic function to fit a set of data (x is time, y is angle).
For context of the equations, I followed this source: http://sciendo.com/article/10.2478/fman-2020-0007
This is my raw data. Left column is x value(seconds); right is y value(angle).
x y
0 22
60 23
120 24
180 27
240 32
300 32
360 36
420 36
480 39
540 41
600 42
660 45
720 46
750 51
780 54
810 57
840 63
870 79
900 103
930 124
960 155
990 177
1020 180
1080 180
1140 180
1200 180
I did not find an explanation of exactly how they find s and c, although the source has some quite complex equations involving higher order derivatives.
(One that I think might be useful involves the x-value $x_0$, where the function's second derivative takes its maximum value.
$$f(x_0) = \frac{y_{max}-y_{min}}{1+e^{s(x_0-c)}}+y_{min}$$ $$f(x_0) = y_{min} + 0.211(y_{max}-y_{min}) = 22 + 0.211(180-22) =55.338$$ No explanation why $\frac{1}{1+e^{s(x_0-c)}}=22$. From which it is somehow calculated: $$x_0 = c - \frac{1.319}{s}$$)
$$f(x) = \frac{158}{1+e^{s(x-c)}}+22$$ The given data is : $$(x_1,f_1),(x_2,f_2),..., (x_k,f_k),..., (x_n,f_n)$$ Let : $$Y(x)=\ln\left(\frac{158}{f(x)-22}-1 \right)$$ So, for each $f_k$ compute : $$Y_k=\ln\left(\frac{158}{f_k-22}-1 \right)$$ The original equation becomes : $$Y=s\:x-s\:c$$ You can compute approximate values of $(s)$ and $(-s\:c)$ thanks to linear regression (Least Mean Squares).
NOTE: Some difficulties may arrise due to eventually negative argument of the logarithm. That is why a representative example of data would be useful to check if the difficulties appear in your case and eventually suggest what to do.
NOTE: If a particular criteria of fitting is specified, you need a non-linear regression method. Then use a specialized software involving itterative calculus and guessed initial values of the parameters. The approximate values found above can be used as initial values.
UPDATED ANSWER with the given data.
Drawing $Y(x)=\ln\left(\frac{158}{y(x)-22}-1 \right)$ shows this shape of curve :
The curve is far to be linear. This proves that the chosen model $(1)$ is not convenient to obtain a good fitting whatever the method of calculus of the parameters.
$$y=a+\frac{b}{1+e^{s(x-c)}} \tag {model 1}$$
As a consequence the above proposed method is of no interest in the present case.
We have to look for another model.
For example one can try with the sum of two logistic functions :
$$y=A+\frac{b_1}{1+e^{s_1(x-c_1)}}+\frac{b_2}{1+e^{s_2(x-c_2)}} \tag {model 2}$$
The result is much better. Below, the comparison between the two models.