Today is the birthday of Joseph Petzval (1807-1891).
Today’s Problem: Thumb at arm’s length
If our thumb at arm’s length (\(61\) cm from eye) subtends an angle of \(2\) degrees, and it exactly covers the height of a \(183\) cm tall person a distance away, what is that distance? and how wide is the thumb?
The distance \(s\) along an arc subtending an angle \(\theta\) in radians along a circle of radius \(r\) is \(s = r\theta\). So the distance to the \(183\) cm tall person is
\[r=s/\theta = \frac{183\;\mathrm{cm}}{2\;\mathrm{degrees}}\left(\frac{180\;\mathrm{degrees}}{\pi}\right) = 5240\;\mathrm{cm} = 52.4\;\mathrm{m}\]
For the width of the thumb \[\theta = \frac{s_1}{r_1} = \frac{s_2}{r_2}\]
where \(s_1\) and \(r_1\) are the width and distance of the thumb, and \(s_2\) and \(r_2\) are respectively those for the person. Solving for \(s_1\): \[s_1 = \frac{r_1}{r_2}s_2 = \frac{61\;\mathrm{cm}}{5240\;\mathrm{cm}}183\;\mathrm{cm} = 2.13\;\mathrm{cm}\] which is the width of the thumb.
© 2026 Stefan Hollos and Richard Hollos