How High Can You Jump?

Model a high jumper at the top of the arch as a semicircle of uniform mass, draped symmetrically over the bar. Let $a$ be the vertical distance from the body’s centre of mass down to its lowest points, and $b$ the vertical distance from the centre of mass up to its highest point. What is the ratio $a/b$?

The Fiddler, Zach Wissner-Gross, August 16, 2024(original post)

Solution

Take the body to be the upper half of a circle of radius $R$, a thin uniform arc from $(-R,0)$ over $(0,R)$ to $(R,0)$. Its lowest points are the two ends at height $0$, its highest point is the top at height $R$, and the centre of mass of a uniform semicircular arc sits at height $2R/\pi$ above the centre. So $$a=\frac{2R}{\pi}-0=\frac{2R}{\pi},\qquad b=R-\frac{2R}{\pi},$$ and $$\frac{a}{b}=\frac{2/\pi}{1-2/\pi}= \boxed{\ \frac{2}{\pi-2}\ }\approx1.752 .$$ This is what lets a jumper clear a bar higher than their centre of mass ever rises: the mass hangs well below the peak of the arch, so $a>b$.

The computation

Encode the body, not the centroid formula: spread points uniformly along the semicircular arc, average their heights to get the centre of mass, and read off $a$ and $b$.

import numpy as np
th = np.linspace(0, np.pi, 200_000)              # uniform points on a unit semicircular arc
com = np.sin(th).mean()                          # centre-of-mass height (= 2/pi)
a, b = com - 0.0, 1.0 - com                       # ends at height 0, top at height 1
print(a / b, 2 / (np.pi - 2))                     # 1.7519  1.7519

Extra Credit

Now the body is an arc of total angle $2\varphi$ centred over the bar. As $\varphi\to0$, what does $a/b$ approach?

Solution

For a uniform arc of half-angle $\varphi$, averaging height over the arc gives a centre of mass at $R\sin\varphi/\varphi$, the ends sit at $R\cos\varphi$, and the top at $R$, so $$\frac{a}{b}=\frac{\sin\varphi/\varphi-\cos\varphi}{1-\sin\varphi/\varphi}.$$ Expanding for small $\varphi$, the numerator is $\bigl(1-\tfrac{\varphi^2}{6}\bigr)-\bigl(1-\tfrac{\varphi^2}{2}\bigr)=\tfrac{\varphi^2}{3}$ and the denominator is $1-\bigl(1-\tfrac{\varphi^2}{6}\bigr)=\tfrac{\varphi^2}{6}$, so $$\frac{a}{b}\longrightarrow\frac{\varphi^2/3}{\varphi^2/6}=\boxed{2}.$$ A nearly flat jumper, almost a straight bar, has its centre of mass twice as far below the ends as the ends are below the peak.

The computation

The same averaging, over an arc of shrinking half-angle, drives the ratio to $2$.

for phi in (0.5, 0.1, 0.01):
    al = np.linspace(-phi, phi, 200_000)         # uniform points on the arc
    com = np.cos(al).mean()                       # centre-of-mass height (= sin(phi)/phi)
    a, b = com - np.cos(phi), 1.0 - com
    print(phi, round(a / b, 4))                   # -> 2