The geometric design of railway track is to calculate superelevation, gradient, etc of the railway for the smooth running of the engine on the rail.

**Gradient **

Any departure of the track from the level is known as grade or gradient. Gradients are provided to negotiate the rise or fall in the level of the railing track.

## Types of gradient

**Ruling gradient :-**The ruling gradient is the steepest gradient that exists in a section.

- In plain regions : 1 in 150 to 1 in 200
- In hilly regions: 1 in 100 to 1 in 150

**Momentum gradient:-**

The momentum gradient is steeper than the ruling gradient and can be overcome by a train because of the momentum it gathers while running on the section.

**Pusher or helper gradient:-**

The pusher gradient is a steeper gradient than the ruling gradient and needs the help of an extra engine to pull the train.

**Grade compensation on the curve:-**

If a curve is provided on a track with a ruling gradient, the resistance of the track will be increased. This curve in order to avoid resistance beyond the allowable limits, the gradients are reduced on curves. The reduction in gradient is known as grade compensation for curves.

- BG track – 0.04% per degree of curve
- MG track – 0.03% per degree of curve
- NG track – 0.02% per degree of curve

BG – Broad gauge- 1676 mm (width)

MG – Meter gauge – 1000 mm ( width)

NG – Narrow gauge- 762 mm (width)

**Rail gauge:-** It is the minimum distance between the running inner faces of two rails.

**Q) The ruling gradient on a BG track section has been fixed as 1 in 150. What should be the compensated gradient when a 3-degree horizontal curve is to be provided on this ruling gradient?**

**Solution-:**

Ruling gradient = 1/150

Grade compensation for 3-degree curve = 0.04% per degree of curve

Which is the allowable gradient to be provided.

**Centrifugal force:-**

When a body moves on a circular curve, it has a tendency to move in a straight direction tangential to the curve. This tendency of the body is subjected to a constant radial acceleration.

The radial acceleration produces a force known as a centrifugal force whose value is given as.

Where F is the centrifugal force

W is the weight of the body

V is the speed of the body

R is the radius of the curve in m

- Centrifugal force always acted away from the centre.

Superelevation on curves (cant)

- Super elevation or cant is defined as the difference I height between the inner and outer rails on the curve.
- It is provided by gradually raising the outer rail above the inner rail level.
- The inner rail is considered as the reference rail and normally maintained at its original level.
- The inner rail is known as the gradient rail.

## The function of superelevation

- It neutralizes the effect of lateral force.
- It provides the better load distribution on the two rails.
- It reduces wear and tear of rail and rolling stock.

**Relation between super-elevation, gauge, speed,** and radius of the **curve:-**

Let be the angle which the inclined plane makes with the horizontal.

Resolving the force and weight along with AC

Hence, equation (iii) is the required relation between super-elevation, gauge, speed, and radius of the curve.

Where, e is super-elevation

G is the gauge in mm + width of the railhead in mm

V is the speed of the train in kmph

R is the radius of the curve in m.

For BG track,

G = 1676mm + 74 mm = 1.75 m

For MG track,

G = 1.058 m

For NG track,

G = 0.772 m

**Maximum value of super-elevation (e):-**

The maximum value of super-elevation generally on many railways of the world has been adopted about 1/10^{th} to 1/12^{th} of the gauge.

Types

- BG track

Speed less than 120 kmph – 165 mm

Speed more than 120 kmph – 185 mm

- MG track – 90 mm ( Normal Condition)

- 100 mm ( standard condition)

- NG track – 65 mm ( Normal condition)

- 75 mm ( special condition)

**Cant deficiency (C _{d}):-** Cant deficiency occurs when a train is on a curved track at a speed higher than the equilibrium speed.

It is the difference between the theoretical cant required for much higher speeds and the actual can provided.

∴ cant deficiency (C_{d}) = C_{th} – C_{a}

Where, C_{d} = cant deficiency

C_{a} = Actual cant provided ( as per average or equilibrium speed)

C_{th} = theoretical cant ( as per maximum speed)

Allowable cant deficiency

- BG track

- 100 mm (V>100 kmph)
- 75 mm (V< 100 kmph)

- MG track – 50 mm
- NG track – 40 mm

**Cant excess (C _{e}):- **Cant excess occurs when a train travels on a curved track at a speed slower than the equilibrium speed.

It is the difference between the actual cant provided and the theoretical cant required for such lower speeds.

∴ Cant excess (C_{e}) = C_{a} – C_{th}

Where, , C_{e} = cant excess

C_{a} = Actual cant provided ( as per average or equilibrium speed)

C_{th} = theoretical cant ( as per maximum speed)

The maximum value of cant excess

BG track – 75 mm

MG track – 65 mm

**The turnout of contrary flexure:-**

A turnout of contrary flexure is one that takes off towards the direction opposite to that of the mainline curve.

## Negative superelevation:-

When the mainline lies on a curve and has a turnout of contrary flexure loading to a branch line, the super-elevation necessary for the average speed of trains running over the main curve cannot be provided.

In the figure, AB which is the outer rail of the mainline curve must be higher than the CD. For the branch line, however of should be higher than AE, or point C should be higher than point A.

These two contradictory conditions cannot be met within one layout. In such cases, the branch line curve has a negative super-elevation and therefore, speed on both tracks must be restricted, particularly on the branch line.

**Main Line**

AB>CD

A>C

**Branch Line**

CF>AE

C>A

## Calculation of negative super-elevation

The provision of negative super-elevation for the branch line and the reduction in speed over the mainline can be calculated as follows;

- The equilibrium superelevation for the branch line curve is first calculated using the formula

- The equilibrium super-elevation ‘e’ is reduced by the permissible cant deficiency ‘C
_{d}‘ and the resultant super-elevation to be provided is

X = e – c_{d}

Where x is the super-elevation

E is the equilibrium super-elevation

C_{d} is the cant deficiency ( 75 mm for BG track and 50 mm for MG track)

The value of c_{d} is generally higher than that of e and therefore x is normally a negative super-elevation of x.

- The maximum permissible speed on the main line which has a super-elevation of x is then calculated by adding the allowable cant deficiency (x+Cd). The safe speed is also calculated and smaller of the two values is taken as the maximum permissible speed on the main line.

Q) **Calculate the superelevation and maximum permissible speed for a 2 degree BG transitioned curve on a high-speed route with a maximum sanctioned speed of 110 kmph. The speed for calculating the equilibrium super-elevation is 180 kmph and the booked speed of goofs train is 50 kmph.**

Now, cant deficiency (Cd) = 190.6 – 100.8 = 89.8 mm

( which is less than 100 mm and hence permissible)

**iv) Superelevation for goods train with a booked speed of 50 kmph**

**Vi) The maximum permissible speed on the curve will be the least of the following:**

- Maximum sanctioned speed i.e. 110 kmph
- Safe speed on the curve i.e. 110.3 kmph

Therefore, the maximum permissible speed over the curve is 110 kmph and the super-elevation to be provided is 100.8 mm.

**Q) Calculate the super-elevation and the maximum permissible speed for a 3 ^{0} curve on a high-speed BG section with a maximum sanctioned speed of 110 kmph. The equilibrium speed to be 80 kmph and the booked speed of the goods train to be 50 kmph.**

**Solution:-**

**The radius of the curve for 3**^{0}of the curve,

**2. Super-elevation for equilibrium speed**

**3. Super-elevation for maximum sanction speed ( 110 kmph)**

Now, cant deficiency (Cd) = 285.83-151.2 = 134.63 mm

( which is more than 100 mm)

Take Cd as 100 mm

Now, actual cant = 185.83 – 100 = 185.83 mm

However, the actual cant is limited to 165 mm and hence this value will be adopted.

**4. Superelevation for goods train with a booked speed of 50 kmph**

Cant excess (Ce) = 165 – 59 = 106 mm

( Cant excess is more than limited value of 75 mm)

With 75 mm takes as cant excess, the actual cant to be provided now is 75 + 59 = 134 mm

**5. Safe speed for high-speed route**

**6. The maximum permissible speed on the curve will be the least of the following:**

- Maximum sanctioned speed i.e. 110 kmph
- Safe speed on the curve i.e. 99.75 kmph

Therefore, the maximum permissible speed on the curve is 99.75 kmph.

I hope this article on “**Geometric design of railway track**” remains helpful for you.

Happy Learning – Civil Concept

Read Also,

Advantages of road transport | Disadvantages of Road Transport

What is Camber in Road? Types, Advantages Disadvantages of Camber

WBM road construction procedure | Features | Application of WBM