Question Number 223354 by ajfour last updated on 22/Jul/25
Commented by ajfour last updated on 22/Jul/25
https://g.co/gemini/share/a41e606602a3
Commented by ajfour last updated on 22/Jul/25
Commented by ajfour last updated on 22/Jul/25
Commented by ajfour last updated on 22/Jul/25
Commented by ajfour last updated on 22/Jul/25
$${A}\:{cylinder}\:{radius}\:{r}\:{rolls}\:{a}\:{distance} \\ $$$$\:{PQ}={D}\:{on}\:{level}\:{road}.\:{Then}\:{we}\:{bend} \\ $$$${the}\:{road}\:{and}\:{place}\:{it}\:{tangentially} \\ $$$${as}\:{JK}\:{of}\:{radius}\:{R}.\:{Find}\:\theta=\frac{{D}}{\mathrm{2}{R}}\centerdot \\ $$
Commented by mr W last updated on 22/Jul/25
$${R}\theta=\left({R}+{r}\right)\:\mathrm{sin}\:\theta=\frac{{D}}{\mathrm{2}} \\ $$$${R}=\frac{{D}}{\mathrm{2}\theta}=\frac{{D}}{\mathrm{2}\:\mathrm{sin}\:\theta}−{r} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\mathrm{sin}\:\theta}−\frac{\mathrm{1}}{\theta}=\frac{\mathrm{2}{r}}{{D}} \\ $$