Probability of getting out of a circular area

Question

An airplane is moving (straight) within a circle of radius $R$ with constant speed $V$ for $t$ seconds. It can start at any place within the circle and move in each direction (uniform distributions). What is the probability that it gets out of the circle (as a function of $V$, $t$ and $R$)?

Answer 1

Denote the center of our circle $A$, the disk bounded by it $D(A, R)$ and the velocity vector by $\vec{V}$. We know $|\vec{V}| = V$. Finally, denote the event that our plane leaves the circle $L$.

The key observation is that the locus of starting points such that the plane doesn't leave $D$ is congruent for any possible direction: our problem is symmetric with respect to the direction. Therefore, $\mathbb{P}[L|\vec{V}] = \mathbb{P}[L]$, and we can consider the direction fixed.

Let's fix the direction of $\vec{V}$ and find this locus. Define $f(x) = x + t\vec{V}$ be the function that, given a takeoff point, returns the plane's landing.

Let $D_1 = \{x:f(x) \in D\}$ be the set of points, both inside and outside $O$ such, that the plane starting from there would end up inside $D$. Then $D_1$ is also a disk with radius $R$ and center $f^{-1}(A) = A - t\vec{V}$.

Of course, the starting point of the plane must be in $D$, so it must be in $D \cap D_1$.

Thus, the probability that our plane leaves $D$ equals the probability that our starting point is in $D \cap D_1$.

The probability that a uniformly random point is inside a region is proportional to the area of that region, so we need to find $\frac{\mathbb{A}(D \cap D_1)}{\mathbb{A}(D)}$ where $\mathbb{A}$ denotes area.

$\mathbb{A}(D \cap D_1)$ is twice the area of a circular segment.

If $tV > 2R$ the disks don't intersect, and the probability to leave $D$ is 1. Let's assume $tV \leq 2R$.

Our central angle is $\theta = 2\arccos(\frac{tV}{2R})$, so $\sin(\theta) = \frac{tV}{R}\sqrt{1 - \big(\frac{tv}{2R}\big)^2}$

Since we have two segments, the total intersection area is

$$\mathbb{A}(D \cap D_1) = R^2(\theta - \sin\theta)$$

Therefore, the probability that the plane leaves the circle (denote this event $L$) is

$$\mathbb{P}[L] = 1 - \frac{\mathbb{A}(D \cap D_1)}{\pi R^2} = 1 - \frac{\theta - \sin(\theta)}{\pi} = 1 - \frac{2\arccos(\frac{tv}{2R}) - \frac{tV}{R}\sqrt{1 - \big(\frac{tV}{2R}\big)^2}}{\pi}$$

Answer 2

HINT:

assuming $r=Vt < 2R$, and once starts not changing direction.

pick a point $c$ inside the circle for the plane starting point, draw a circle or radius $r$, find the chord angle $\theta$ of the intersections. The probability of the plane getting out of the big circle is $\frac{\theta}{2\pi}$.

In case $r>2R$, there is no intersection, so probability is 1.

Answer 3

Let us simplify the problem and reduce it to its core properties. Let the radius of the original "bounding" circle be $1$, and the (linear) flight distance be $vt =r$. Our problem can be characterized entirely by the value of $r$.

If $r>2$, then every airplane escapes the bounding circle, no matter its takeoff point; the probability of escape is thus $P_{r>2} = 1$.

Here is a figure illustrating a more general case for a given $r<1$. If the airplane takeoff point is inside the annulus of inner radius $1-r$, then the airplane will never escape, regardless of its flight direction. That is shown by the white core of the disk. The probability of the plane not escaping is the ratio of the inner (white) disk region divided by the area of the bounding disk, i.e., $P = {\pi r^2 \over \pi 1^2} = r^2$. But of course that is not the final solution. We must calculate the probability that if the airplane's takeoff point is in the pink annulus, that it escapes.

Now look at the more interesting case:

Here $x$ is the distance of the takeoff point from the bounding disk center (in any direction). The airplane will land somewhere along the perimeter of the green disk. The (one-dimensional) length of the purple arc divided by the total circumference of the green disk is the probability the airplane will land outside the bounding disk. This, in turn, depends upon the angle $\theta$, as shown.

We use the law of cosines for the dashed triangle, which has sides of known length:

$$1^2 = x^2 + r^2 - 2 x r \cos (\theta),$$

or

$$\theta = \arccos \left( {x^2 + r^2 - 1 \over 2 x r} \right).$$

The probability the airplane lands inside the bounding disk is $P = \theta/\pi$, and thus the probability the airplane lands outside is $1 - \theta/\pi$.

Now we must integrate all possible values of $x$, as:

$$\int\limits_{x=1 - 2 r}^1 2 \pi x\ {1 \over \pi} \arccos \left( {x^2 + r^2 - 1 \over 2 x r} \right) \ dx$$

One then performs this integral, collects terms and simplifies.

Answer 4

Consider a circle of radius $Vt$ with center r away from the center of the first circle, itself having radius R. The chances Escape is 1 minus the ratio of the area of overlap between the two circles to the area of the second circle.

Let the radius of the circle be R. Let initial distance from center of circle be r. Let u=V=Velocity and t=time. $\theta$ is angle of velocity vector with respect to displacement vector from center of circle.

By symmetry, the probability of escape is the same for positions the same distance from the center, so probability is a function of r, $P(r)$.

The final distance from the center is $d^2=r^2+u^2t^2-2rut\cos{\theta}$ and we have our crossing points at $d=R$.

So: $$\cos{\theta}=\frac{r^2+u^2t^2-R^2}{2rut}$$

Solve for the range of $\theta$ and divide by $2\pi$. This will give you P(r). We can approximate with a Taylor Series:

$$1-\frac{\theta^2}{2}=\frac{r^2+u^2t^2-R^2}{2rut}$$

So $\Delta \theta /2 \pi \approx \frac{1}{\pi}\sqrt{\frac{R^2-(r-ut)^2}{rut}}\approx P(r)$

$r$ can take on values uniformly from $r$ to $R$ giving a probability of $\frac{2r}{R^2}$.

$$P\approx\int_0^R\frac{2r}{\pi R^2}\sqrt{\frac{R^2-(r-ut)^2}{rut}}dr$$