# The maximum speed of a train on B.G. track having a curvature of 3° and cant of 10 cm with an allowable cant deficiency of 76 mm, for conditions obtai

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The maximum speed of a train on B.G. track having a curvature of 3° and cant of 10 cm with an allowable cant deficiency of 76 mm, for conditions obtaining in India, is
1. 87.6 km/h
2. 99.6 km/h
3. 76.6 km/h
4. 65.6 km/h

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Correct Answer - Option 1 : 87.6 km/h

Concept:

For the Indian railways track

${e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}$

Where,

eth is the maximum allowable superelevation in (m)

Vmax is the maximum speed on the curve in kmph,

G is the gauge length of the track in m and

R is the radius of the curve in m where

D is the curvature of the curve in degree.

,${\rm{R}} = \frac{{1720}}{{\rm{D}}}$

${e_{th}} = {e_{act}} + CD$

Where,

eact is the actual or equilibrium superelevation

CD is Cant deficiency

Calculation:

Given,

D = 3°, CD = 76 mm = 0.076 m

eact = 10 cm = 0.1 m

As track is Broad Gauge G = 1.676 m

${e_{th}} = {e_{act}} + CD$

∴ eth = 0.1 + 0.076

= 0.176 m

Radius of Curvature  ${\rm{R}} = \frac{{1720}}{{\rm{D}}}$

$∴ {\rm{\;Radius\;of\;the\;curve}},{\rm{\;R\;}} = \frac{{1720}}{3}{\rm{m}} = 573.33{\rm{\;m\;}}$

${e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}$

$0.176 = \frac{{1.676 \times V_{max}^2}}{{127 \times 573.33}}$

Vmax = 87.6 kmph

∴ The maximum speed of the train is approximately 87.6 kmph.

Important Point:

In the formula of maximum allowable super elevation ${e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}$  all the variables should be replaced in the particular unit mentioned above, otherwise the constant term ‘127’ is not valid.