Buffon's needle problem: Difference between revisions

Now consider the case where the plane contains two sets of parallel lines orthogonal to one another, creating a standard perpendicular grid. We aim to find the probability that the needle intersects at least one line on the grid. Let <math>a, b</math> be the sides of the rectangle that containcontains the midpoint of the needle whose length is <math>\ell</math>. Since this is the short needle case, <math> \ell < a, \ell < b </math>. Let <math>(x,y)</math> mark the coordinates of the needle's midpoint and let <math>\varphi</math> mark the angle formed by the needle and the ''x''-axis. Similar to the examples described above, we consider <math>x, y, \varphi</math> to be independent uniform random variables over the ranges <math> 0\leq x \leq a, \quad 0\leq y \leq b, \quad \frac{-\pi}{2}\leq \varphi \leq \frac{\pi}{2} </math>.

where <math>F(\varphi)</math> is the region where the needle does not intersect any line given an angle <math>\varphi</math>. To determine <math>F(\varphi)</math>, let's first look at the case for the horizontal edges of the bounding rectangle. The total side length is <math>a</math> and the midpoint must not be within <math>\frac{l}{2}\cos(\varphi)</math> of either endpoint of the edge. Thus, the total allowable length for no intersection is <math>a - 2(\frac{l}{2}\cos(\varphi))</math> or simply just <math>a - l\cos(\varphi)</math>. Equivalently, for the vertical edges with length <math>b</math>, we have <math>b \pm l\sin(\varphi)</math>. The <math>\pm</math> accounts for the cases where <math>\varphi</math> is positive or negative. Taking the positive case and then adding the absolute value signs in the final answer for generality, we get

F(\varphi)d\varphi = \pi ab-2bl-2al+l^2 </math>.