Question

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 ______

Answer 1

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,

V_avg 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

 V_max = 100 kmph, V_avg = 80 kmph

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

Taking V_max 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.