= Solution
Write $A_i=\phi_{i1}$, $b_i=\phi_{i2}>0$ and $k_i=-\phi_{i3}=\theta_{i3}$. Then
$$
m_i(t)=\frac{A_i}{1+b_ie^{-k_it}}.
$$
For $k_i>0$, $A_i$ is the limiting circumference and $1/(1+b_i)$ is the fraction of that asymptote attained at the centered day $x=\overline x$. Under the code's transformations,
$$
\boxed{A_i=e^{\theta_{i1}},\qquad\frac{m_i(0)}{A_i}=\frac1{1+e^{-\theta_{i2}}}.}
$$
The latter expression is the <logistic function>. The curve also satisfies the <logistic growth equation> $dm_i/dx=(k_i/s_x)m_i(1-m_i/A_i)$. Its inflection point is at half the asymptote. The original prior permits $k_i\leq0$, in which case it does not describe increasing growth.
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