Escape from the velociraptor

[I like Serena]

You are stuck in a circular area with a fence around it, while there is a velociraptor out there. At least he can't get in. You can pass through the fence anywhere you want though. And if you can reach the perimeter before the velociraptor reaches you, you can escape. Unfortunately the velociraptor can run 4 times as fast as you can, and whenever you move, he will always run at 4 times your speed to your point of escape.
\begin{tikzpicture}[>=stealth]
\draw[ultra thick,green!70!black,dashed] circle (4);
\draw[thick,red,->] (0,4.2) arc (90:270:4.2);
\draw[thick,blue,->] (0,0) -- (0,-{4*pi/4});
\draw[fill,blue] circle (.04) node

{You};
\draw[fill,red] (0,4.2) circle (.04) node[above] {Velociraptor};
\end{tikzpicture}View attachment 7997

How can you escape?​

 

[Dethrone]

Wait until the velociraptor sleeps, then run out.

Maybe I'm thinking about this in the wrong way, but is the motion of the velociraptor deterministic? That is, if I'm at the center of the circle and move along the line connecting the velociraptor and the human, does the velociraptor move left or right or is it dumb and not move at all?

 

[I like Serena]

Wait until the velociraptor sleeps, then run out.

I tried that. And just when I was sure it was asleep and started out carefully, it jumped up and intercepted my escape anyway. It must have some weird subconscious reflex. Then again, do velociraptors even sleep? No one seems to know. They are just always there, and always alert.

Maybe I'm thinking about this in the wrong way, but is the motion of the velociraptor deterministic? That is, if I'm at the center of the circle and move along the line connecting the velociraptor and the human, does the velociraptor move left or right?

When you start moving away from the middle, the velociraptor has some sense that tells it that you're either going slightly to the right, or slightly to the left. As a result, it makes a decision on the spot, and immediately starts running.

 

[MarkFL]

Spoiler

I am assuming the velociraptor heads for the point on the circle pointed to by the person's velocity vector at any given instant. Then, if the human moves to the fence, but doesn't cross it, the velociraptor will be just on the other side of the fence. The person then walks a small circle, whose radius is less the 1/4 of the circular area in which he/she is trapped. This will cause the velociraptor to run the entire perimeter, but as this distance is more than 4 times the circular path taken by the person, the person gets back to the fence first, crosses and escapes?

 

[I like Serena]

Spoiler

I am assuming the velociraptor heads for the point on the circle pointed to by the person's velocity vector at any given instant. Then, if the human moves to the fence, but doesn't cross it, the velociraptor will be just on the other side of the fence. The person then walks a small circle, whose radius is less the 1/4 of the circular area in which he/she is trapped. This will cause the velociraptor to run the entire perimeter, but as this distance is more than 4 times the circular path taken by the person, the person gets back to the fence first, crosses and escapes?

Indeed, that's how someone can escape.
Good job! ;)

Spoiler

Nitpick: suppose we walk to the fence and turn around. And then head (slowly) to the other side. Then for some reason the velociraptor just keeps sitting there, and won't allow us to trick him.
The velociraptor actually heads for the point on the circle that is closest to us, which is not necessarily the point we're headed at.

Now let me add a bonus question.
What is the shortest path we can take to (just) escape, and where will we exit?

 

[I like Serena]

Now let me add a bonus question.
What is the shortest path we can take to (just) escape, and where will we exit?

Spoiler

As MarkFL correctly deduced, we have 2 phases:

1. 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.
2. 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);
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\draw[gray] ({\angle/2}:1.2) node[above right] {$90^\circ$};
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-- ({\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] {} };
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node[mnode] {} } -- ({3*180/pi-90}:4) node[mnode] {};
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{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);
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\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.
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