Consider a 360-sided regular polygon with side length 1, rotated so that its sides are at half-degree inclinations (that is, its sides all have inclinations of $0.5^{\circ}, 1.5^{\circ}, 2.5^{\circ}$, and so on. Go to the bottom point on this polygon and then move clockwise, numbering the sides $1,2,3, \ldots, 360$ as you go. Then, take the section of 15 sides from side 31 to side 45. These sides have inclinations of $30.5^{\circ}, 31.5^{\circ}, 32.5^{\circ}$, and so on, up to $44.5^{\circ}$. Therefore, over this section, the horizontal and vertical displacements are, respectively: $H =\cos 30.5^{\circ}+\cos 31.5^{\circ}+\ldots+\cos 44.5^{\circ} V =\sin 30.5^{\circ}+\sin 31.5^{\circ}+\ldots+\sin 44.5^{\circ}$. However, we can also see that, letting $R$ be the circumradius of this polygon: $H=R\left(\sin 45^{\circ}-\sin 30^{\circ}\right) V=R\left[\left(1-\cos 45^{\circ}\right)-\left(1-\cos 30^{\circ}\right)\right]$. From these, we can easily compute that our desired answer is $\frac{H}{V}=(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})=2-\sqrt{2}-\sqrt{3}+\sqrt{6}$.