Can we calclulate R0 from logistic fit Suppose we are fitting covid19 data with logistic function, 
$$\dot N(t)=rN(t)(1-\frac{N(t)}{K})$$
and get a good fit.
Can we get R0 from the optimal value of $r$ and $K$ ? If yes, how to?
 A: The SIR model is slightly different and follows the following differential equation*:
$$\dot{u}(t) = u(t) \left( C_1 - \gamma \ln u(t)+ x_0 \beta^\prime u(t) \right)  $$
where 


*

*$S(t)$, $I(t)$, $R(t)$ are transforms of $u(t)$ following: 
$$\begin{array}{}
S(t) &=& x_0 u\\
I(t) &=& \frac{\gamma}{\beta^\prime} \ln u(t) - x_0 u(t) - \frac{C_1}{\beta^\prime} \\
R(t) &=& - \frac{\gamma}{\beta^\prime} \ln u(t)
\end{array}$$

*Also of interrest is $Q(t) = I(t) + R(t)$ which is the total of people that got sick (since the logisitic fit does not differentiate between infected and recovered)
$$Q(t) = - \frac{C_1}{\beta^\prime} - x_0 u(t)$$

*The constant $C_1$ is an integration constant that can be fixed based on the boundary condition $S(t) + I(t) + R(t) = N$ where $N$ is the total people.
$$C_1 = -\beta^\prime N$$

*The constant $x_0$ is another boundary/starting condition and relates to the initial number of recovered people $R(0)$
$$x_0 = N e^{\frac{\beta\prime}{\gamma}R(0)}$$
If we start with zero recovered people then
$$x_0 = N$$

*We introduced the term $\beta^\prime$ because the differential equations for the SIR model can be written differently with a scaling $N$. The work of Harko, Lobo and Mak does not use the scaling with $N$ which makes the interpretation of $\beta$ different. We use $\beta^\prime$ for the parameter used in Harko, Lobo and Mak. We can transform as:
$$\beta^\prime = \frac{\beta}{N}$$ 
If $R(0) = 0$ and consequently $x_0 = 1$, then we can rewrite this as:
$$\frac{\dot{S}(t)}{S(t)} =  -\beta \left( 1 + \frac{\gamma}{\beta} \ln \frac{S(t)}{N} - \frac{S(t)}{N} \right)$$
or in terms of $Q(t)$
$$\frac{\dot{Q}(t)}{Q(t)} =  \beta \left(1 - \frac{Q(t)}{N} - \frac{\gamma}{\beta} \left(1- \frac{N}{ Q(t)}\right) \ln \left(1- \frac{ Q(t)}{N}\right)   \right)$$
which looks a bit like the equation for the logistic growth (the last term is different and becomes zero when $\gamma = 0$)
$$\frac{\dot{Q}(t)}{Q(t)} = r\left(1  - \frac{Q(t)}{K}\right) \hphantom{- \frac{\gamma}{\beta} \left(1- \frac{N}{ Q(t)}\right) \ln \left(1- \frac{ Q(t)}{N}\right) }$$ 

*Harko, Lobo and Mak 2014 Exact analytical solutions of the Susceptible-Infected-Recovered (SIR) epidemic model and of the SIR model with equal death and birth rates https://arxiv.org/abs/1403.2160

So you could see logistic growth as a special case of the SIR model with $\gamma = 0$ (no recovery) and $R_0 =\infty$.
Therefore based on logistic fits you can not derive the value of $R_0$ since a logistic fit is a SIR fit with fixed $R_0$.
Note that, when $Q(t) << N$ then for both the SIR model and the logistic model you get approximately exponential growth.
For the SIR model you get $\left(1- \frac{N}{ Q(t)}\right) \ln \left(1- \frac{ Q(t)}{N}\right)  \approx 1$ and
$$ \frac{\dot{Q}(t)}{Q(t)} = \beta - \gamma$$
for the logistic growth model you get
$$ \frac{\dot{Q}(t)}{Q(t)} = r $$
So fits with a logistic model, or anything else that has initial exponential growth, to data in an early phase, may be used to determine $\beta - \gamma$, but not the value of $R_0 = \frac{\beta}{\gamma}$ which requires data over a longer period. 
(If it is covid-19 that you are after, then an $R_0$ value with more data will be wrong since the data does not relate to an SIR model which is a model for homogeneous mixing and parameters that are not changing in time) 
