Question

A $5^o$ curve diverges from a $3^o$ main curve in an opposite direction in the leg out of a B.G. yard. The speed on the branch line is restricted to $30kmph$. The speed on the main line will be ___$kmph$.
[Assume permissible cant deficiency as $76mm$]

• A. 72.22

Answer 1

The correct option is A 72.22
Super elevation on a diverging track (branch line) is given by the following relations:
$c=\frac{G{V}^2}{127R}$
Here,
$G=1676,mm$
$V=30kmph$
$R=\frac{1750}{5}$
$\therefore e=\frac{1676\times {30}^2}{127\times \left(\frac{1750}{5}\right)}=34mm$
$\therefore$ Negative cant $=$ Equilibrium cant $-$ Cant deficiency
$=|34-76|=42mm$
But Negative cant $=$ Maximum permissible super elevation on the main line
$=42mm$
$\therefore$ Theoretical super elevation which can be provided on the main line $=42+76=118mm$
$\therefore$ Speed on the main line is given by,
$e=\frac{G{V}^2}{127R}=\frac{1676\times V^2}{127\times \left(\frac{1750}{3}\right)}$
$V^2=\frac{127\times (\frac{1750}{3}\times 118)}{1676}$
$V=72.22kmph$