 First half-success time in Grover search

For both oracle phases equal to $\pi$, the good/bad matrix is $\left(\begin{array}{cc}\cos 2\chi & \sin 2\chi \\ -\sin 2\chi & \cos 2\chi \end{array}\right)$. Acting $r$ times on the initial vector $(\sin \chi ,\cos \chi )$ yields success ${\sin }^2((2r+1)\chi )$. For $0<\chi \le \pi /4$, the first crossing of one half is the displayed integer, since the rounded angle stays between $\pi /4$ and $3\pi /4$. With ${\sin }^2\chi =m/N\ll 1$ this is asymptotic to $(\pi /8)\sqrt{N/m}$. Iterating indefinitely is incorrect: the success oscillates after its first maximum.