For many two closed curves $c, d: \mathbb R \to \mathbb R^n$ that are regularly homotopic to each other, the map $$ \tag{1} h: \mathbb R \times [0, 1] \to \mathbb R^n, \ (t, u) \mapsto (1 - u) c(t) + u \cdot d(t) $$ is a regular homotopy between them. An example is $c(t) := (\cos(t), \sin(3t))$ and $d(t) := (\cos(t), \sin(t))$.
Question. How can we characterise the pairs of two closed curves $c, d: \mathbb R \to \mathbb R^n$ that are regularly homotopic to each other, but for which the map $(1)$ is not regular?
An example for two closed parametrised curves (unfortunately, they are the same closed curve) is $(c, 2 \pi)$ and $(d, 2 \pi)$ with $c(t) := (\cos(t), \sin(t))$ and $d(t) := \left(\cos\left(t \cdot e^{t - 2 \pi}\right), \sin\left(t \cdot e^{t - 2 \pi}\right)\right)$, which are both the positively traversed unit circle but traversed with different "speed", so they are regularly homotopic, but $(1)$ is not regular for $u \approx 0.27$ and $u \approx 0.83$ (haven't calculated the exact values yet).
Another example are the following curves with period $2 \pi$ $$ c(t) := (\cos(t), \sin(t)), \qquad d(t) := (\sin(t), \cos(t)), $$ which represent the unit circle being traversed negatively resp. positively. They are regularly homotopic, but the homotopy $(1)$ is a straight line for $u = 1/2$, which is not closed.
Definition 1. A closed parametrised curve is a pair $(c, p)$ where $c: \mathbb R \to \mathbb R^n$ is parametrised curve with period $p$, i.e. $c(t+p)=c(t)$ holds for all $t \in \mathbb R$.
Definition 2. A closed curve is an equivalence class of closed parametrised curves, where $(c,p) \sim (d,q)$ if and only if there exists a bijective smooth map $\phi: \mathbb R \to \mathbb R$ such that $d = c \circ \phi$ and $\phi'(t) > 0$ and $\phi(t + p) = \phi(t) + q$ hold for all for all $t \in \mathbb R$
Definition 3. Two closed curves $(c, p)$ and $(d, q)$ are regularly homotopic if there exists a regular homotopy between them, i.e. a map $$ h: \mathbb R \times [0, 1] \to \mathbb R^n $$ such that $h(t,0) = c(t)$ and $h(t,1) = d(t)$ hold for all $t \in \mathbb R$, all curves are regular, i.e. $\frac{\partial}{\partial t} h(t,u) \ne 0$ holds for all $t \in \mathbb R$ and all $u \in [0, 1]$ and all curves are closed, i.e. there exists a continuous map $u \mapsto \tau_u$ such that $h(t + \tau_u, u) = h(t,u)$ holds for all $t \in \mathbb R$ and all $u \in [0, 1]$ and we have $\tau_0 = p$ and $\tau_1 = q$.