• Negative Superelevation Figure,

• When main line is on a curve and a branch line takes off from a main line with a curve which is in contrary flexure to the main line, the outer rail of the branch line instead of being raised, has to remain lower than the inner rail upto a certain distance from take off point.

• The amount by which outer rail of branch line is lower than the inner rail is called negative superelevation.

  Problem:-

• $V_{\max ,B T}=35 \mathrm{kmph}$

• Theoretical Cant, $e_{\text {th }}=\frac{G V_{max}^{2}}{127 R}$

• ethBT $=\frac{1.676 \times 35^{2}}{127 \times \frac{1720}{5}}=4.7 \mathrm{~cm}$

Since,

$e_{t h}, B T=e_{a c t, B T}+CD$

• $\Rightarrow \quad 4.7 \mathrm{~cm}=$ eact, BT $+7.6 \mathrm{~cm}$ $\Rightarrow$ eact, BT $=-2.9 \mathrm{~cm}$

• As, For negative Superelevation,

• eact, $MT=$ - eact, BT$=2.9 \mathrm{~cm}$

$\therefore e$ th,MT = eact, MT$+C D$ $e_{th,MT}=2.9+7.6=10.5 \mathrm{~cm}$

• eth, $M T=\frac{G V_{\operatorname{max}}^{2}}{127 R}$

$\Rightarrow \quad 0.105=\frac{1.676 \times V_{\text {max,MT }}^{2}}{127 \times \frac{1720}{3}}$ $V_{\max ,MT}=67.54 \mathrm{kmph}$

Check, As per martin's formula,

$$\begin{aligned} V_{\text {max, } M T} &=4.35 \sqrt{R-67} \\ &=4.35 \sqrt{\frac{1720}{3}-67} \\ &=97.88 \mathrm{kmph}\gt67.54 \mathrm{kmph} \\ (OK\mathrm) \end{aligned}$$