As MarkFL correctly deduced, we have 2 phases:
- Move to a point from which we can escape within a quarter of the radius, where our angular velocity is higher than the velociraptor's.
- Run straight to the edge.
Without loss of generality, let's assume that the radius of the circle is $R=1$.
And assume that our speed is 1 as well.
That means that the velociraptor moves at speed $v=4$.
Denote our position in polar coordinates as $(r, \theta)$.
Let the velociraptor start at $\theta=\pi$, and let us start in the direction of the x-axis.
After an initial feint to the right, we'll move to the left so that our angle increases from $\theta=0$, while the velociraptor runs in an increasing angle as well.
In the first phase every step we take must keep the center between us and the velociraptor.
So if the velociraptor moves by an angle $d\theta$, we need to move by that angle as well.
The distance the velociraptor covers is $Rd\theta=d\theta$, while we cover $\sqrt{(dr)^2 + (rd\theta)^2}$.
So:
$$
d\theta = v\sqrt{(dr)^2 + (rd\theta)^2} \quad\Rightarrow\quad
\frac{vr'}{\sqrt{1-(vr)^2}}=1 \quad\Rightarrow\quad
\arcsin(vr) = \theta + C \quad\Rightarrow\quad
r = \frac 1v\sin \theta
$$
Therefore our cartesian position will be:
$$
\frac 1v\sin \theta(\cos\theta, \sin\theta)
= \frac 1{2v}(\sin(2\theta),1-\cos(2\theta))
$$
This is a circle with center $(0,\frac 1{2v})$ and radius $\frac 1{2v}$.
And we move through this circle at double the angle that the velociraptor covers.
If at angle $\theta$ we switch to the second phase and run straight to the edge, we have to cover the distance:
$$1-r(\theta)= 1 - \frac 1v\sin\theta$$
So our total distance $d$ is:
$$d = \frac 1{2v}\cdot 2\theta + (1 - \frac 1v\sin\theta) = \frac 1v(\theta + v - \sin\theta)$$
And the velociraptor covers a distance of $vd$, although he started at $\pi$.
So when we exit, the distance $\Delta$ between us and the velociraptor is:
$$\Delta = R\theta - (vd - R\pi) = \theta - ((\theta + v - \sin\theta) - \pi) = \pi - v + \sin\theta$$
To maximize that distance, we need $\theta=\frac\pi 2$, so that $\Delta = \pi - v + 1 = \pi - 3 \approx 0.14$.
The corresponding escape path is:
\begin{tikzpicture}[>=stealth,mnode/.style={circle,draw=black,fill=black,inner sep=0pt,minimum size=2pt}]
\def\angle{90}
\def\nsteps{6}
\def\msteps{11}
\draw[ultra thick,green!70!black,dashed] circle (4);
\draw[help lines] circle (1);
\draw[help lines] (0,4/8) circle (4/8);
\draw[help lines] (180:4) -- (0:5);
\draw[help lines] (270:5) -- (90:5);
\draw[help lines] ({\angle+180}:4) -- ({\angle}:5);
\draw[gray,->] (0:1.2) arc (0:{\angle}:1.2);
\draw[gray] ({\angle/2}:1.2) node[above right] {$90^\circ$};
\draw[thick,blue] (0,0) foreach \i in {1,...,\nsteps} {
-- ({\i*\angle/\nsteps}:{sin(\i*\angle/\nsteps)})
node[mnode] {} } -- (0,4) node[mnode] {};
\draw[thick,blue] ({\angle}:{sin(\angle)}) foreach \i in {1,...,\msteps} {
-- ({\angle}:{sin(\angle)+\i*\angle*pi/180/\nsteps})
node[mnode] {} };
\draw[ultra thick,red] (180:4) {} foreach \i in {1,...,17} {
arc ({180+(\i-1)*\angle/\nsteps}:{180+(\i)*\angle/\nsteps}:4)
node[mnode] {} } -- ({3*180/pi-90}:4) node[mnode] {};
\draw[fill,red] (180:4) circle (.06) node
{Velociraptor};
\draw[fill,blue] circle (.06) node[below right] {You};
\draw[fill,gray] (0,4/8) circle (.04) node
{\small$\frac 18$};
\draw[fill,gray] (0,1) circle (.04) node
{\small$\frac 14$};
\draw[fill,gray] (0,4) node[above left] {\small$1$};
\end{tikzpicture}
To make our earliest escape, we have $\Delta=0$, or:
$$\Delta = \pi-v+\sin\theta = 0 \quad\Rightarrow\quad \theta=\arcsin(v-\pi)$$
The corresponding escape path is:
\begin{tikzpicture}[>=stealth,mnode/.style={circle,draw=black,fill=black,inner sep=0pt,minimum size=2pt}]
\def\angle{asin(4-pi)}
\def\nsteps{4}
\def\msteps{12}
\draw[ultra thick,green!70!black,dashed] circle (4);
\draw[help lines] circle (1);
\draw[help lines] (0,4/8) circle (4/8);
\draw[help lines] (180:4) -- (0:5);
\draw[help lines] (270:5) -- (90:5);
\draw[help lines] ({\angle+180}:4) -- ({\angle}:5);
\draw[gray,->] (0:1.2) arc (0:{\angle}:1.2);
\draw[gray] ({\angle/2}:1.2) node[above right] {$\arcsin(4-\pi)$};
\draw[thick,blue] (0,0) foreach \i in {1,...,\nsteps} {
-- ({\i*\angle/\nsteps}:{sin(\i*\angle/\nsteps)})
node[mnode] {} };
\draw[thick,blue] ({\angle}:{sin(\angle)}) foreach \i in {1,...,\msteps} {
-- ({\angle}:{sin(\angle)+\i*\angle*pi/180/\nsteps})
node[mnode] {} } -- ({\angle}:4) node[mnode] {};
\draw[ultra thick,red] (180:4) {} foreach \i in {1,...,16} {
arc ({180+(\i-1)*\angle/\nsteps}:{180+(\i)*\angle/\nsteps}:4)
node[mnode] {} } -- ({\angle}:4) node[mnode] {};
\draw[fill,red] (180:4) circle (.06) node
{Velociraptor};
\draw[fill,blue] circle (.06) node[below right] {You};
\draw[fill,gray] (0,4/8) circle (.04) node
{\small$\frac 18$};
\draw[fill,gray] (0,1) circle (.04) node
{\small$\frac 14$};
\draw[fill,gray] (0,4) circle (.04) node[above left] {\small$1$};
\end{tikzpicture}
Finally, if the velociraptor would move at speed $v\ge\pi+1\approx 4.14$, we would not be able to escape.
