# For a 2° curve on a high speed Broad Gauge (BG) rail section, the maximum sanctioned speed is 100 km/h and the equilibrium speed is 80 km/h. Consider

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For a 2° curve on a high speed Broad Gauge (BG) rail section, the maximum sanctioned speed is 100 km/h and the equilibrium speed is 80 km/h. Consider dynamic gauge of BG rail as 1750 mm. The degree of curve is defined as the angle subtended at its center by a 30.5 m arc. The cant deficiency for the curve (in mm, round off to integer) is ______

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Concept

For the Indian railway's track

Theoretical Cant is given by:

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

Actual cant or equilibrium cant provided is given by

${e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R\;}}$

Where,

eth is the maximum allowable super-elevation in (m)

Vmax is the maximum speed on the curve in kmph,

Vavg is the average speed on the curve

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 super-elevation

CD is Cant deficiency

Calculation:

Given,

Degree of curve = 2°

Type of Gauge is Broad Gauge (BG) G = 1750 mm

Vmax = 100 kmph, Vavg = 80 kmph

Radius of curve  $R = \frac{{1720}}{D}$ = 1720/2 = 875 m

Taking Vmax theoretical cant is

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

${e_{th}} = \frac{{1.750 \times {{100}^2}}}{{127 \times 875\;}}$ = 157.480 mm

Taking Veq equilibrium cant is

${e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R\;}}$

${e_{eq}} = \frac{{1.750 \times {{80}^2}}}{{127 \times 875\;}}$ = 100.787 mm

∴ Cant Deficiency = 157.480 - 100.787 = 56.693 mm

∴ Cant Deficiency is 57 mm as in Question it is asked to roundoff to integer.