Banking of roads is a crucial aspect of modern roadway design, particularly on curves, where the road surface is tilted at an angle to the horizontal. This technique is employed to enhance safety and improve vehicle handling by counteracting the lateral forces that act on vehicles when they navigate a bend.

When a vehicle rounds a curve, centrifugal force pushes it outward, which can lead to skidding or tipping over if the road is flat. By banking the road, engineers can use the inclined surface to create an inward component of the normal force that helps counterbalance the outward pull of centrifugal force. This not only allows vehicles to maintain better traction but also enables them to traverse curves at higher speeds more safely. Learn more about Banking of road, its concept, formula and examples

Banking of roads, also known as road superelevation, is the practice of raising the outer edge of a road curve higher than the inner edge. This creates an inclined surface, which helps vehicles navigate curves more safely and efficiently. The primary goal of banking is to counteract the lateral forces that vehicles experience when making a turn, thereby enhancing stability and reducing the risk of skidding or overturning.

Centrifugal Force:

When a vehicle moves along a curved path, it experiences a centrifugal force that pushes it outward, away from the center of the curve. On a flat road, this force can cause the vehicle to skid or even overturn.

Inclined Surface:

By banking the road, engineers create an inclined surface where the outer edge is higher than the inner edge. This incline helps redirect the forces acting on the vehicle towards the center of the curve.

Normal and Frictional Forces:

On a banked road, the normal force (the perpendicular contact force between the vehicle and the road) has a horizontal component that provides the necessary centripetal force to keep the vehicle on its curved path.

Friction between the vehicle’s tires and the road surface also contributes to the centripetal force, but banking reduces the reliance on friction alone, making curves safer under various weather conditions.

Banked Turn: A turn or turn in which the vehicle banks inwards.

Bank Angle: The bank angle of the vehicle. At this angle, the vehicle is therefore tilted about its longitudinal axis with respect to the plane of its curved path.

The formulas related to banking of road are mentioned below:

Certainly! Here’s a table that outlines the key formulas related to the banking of roads, including explanations for each parameter involved:

Formula Explanation

$\tan(\theta) = \frac{v^2}{rg}$

Ideal Banking Angle: Calculates the ideal banking angle (

$\theta$

) for a road curve, where

$v$

is the vehicle speed,

$r$

is the radius of the curve, and

$g$

is the acceleration due to gravity.

$\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)$

Banking Angle (Degrees): Provides the banking angle in degrees by taking the arctangent of

$\frac{v^2}{rg}$

.

$F_c = \frac{mv^2}{r}$

Centrifugal Force: The outward force (

$F_c$

) acting on a vehicle of mass

$m$

moving at speed

$v$

$r$

.

$N = mg \cos(\theta)$

Normal Force: The normal force (

$N$

) acting perpendicular to the inclined surface, where

$m$

is the vehicle mass,

$g$

is the acceleration due to gravity, and

$\theta$

is the banking angle.

$f_s = mg \sin(\theta)$

Frictional Force (Sliding): The frictional force (

$f_s$

) parallel to the inclined surface, helping to prevent sliding.

$F_{centripetal} = N \sin(\theta) + f \cos(\theta)$

Centripetal Force (Banked Curve): The inward force required to keep the vehicle moving in a curved path, which is a combination of the horizontal component of the normal force and the frictional force (

$f$

).

$v_{max} = \sqrt{r g \left( \mu + \tan(\theta) \right) / (1 – \mu \tan(\theta))}$

Maximum Speed on Banked Road: The maximum speed (

$v_{max}$

) a vehicle can maintain on a banked curve without skidding, where

$\mu$

is the coefficient of friction between the tires and the road.

Safety: It reduces the likelihood of skidding and overturning by balancing the centrifugal force.

Comfort: It provides a smoother and more stable ride through curves.

Efficiency: It allows vehicles to maintain higher speeds on curves, improving overall traffic flow and travel times.

Highways and Freeways: Banking is widely used in designing curves on high-speed roads to ensure safety and efficiency.

Race Tracks: In motorsports, tracks often feature significant banking to enable high-speed turns.

Railways: Similar principles are applied in railway design, where the tracks are tilted to help trains navigate curves safely.

Solved Examples For Banking Road Formula

Example 1: A road curve has a radius of 50 meters. If vehicles are expected to travel at 20 meters per second on this curve, what is the ideal banking angle?

Solution:

Given:
Radius of the curve ($$r$$): 50 meters
Speed of the vehicle ($$v$$): 20 meters per second
Acceleration due to gravity ($$g$$): 9.81 meters per second squared

$\tan(\theta) = \frac{v^2}{rg}$

$\tan(\theta) = \frac{20^2}{50 \times 9.81} = \frac{400}{490.5} \approx 0.815$

To find $$\theta$$, take the arctangent:
$\theta = \tan^{-1}(0.815) \approx 39.4^\circ$

Answer: The ideal banking angle is approximately $$39.4^\circ$$.

Example 2: A car is traveling on a banked curve with a radius of 100 meters and a banking angle of 30 degrees. The coefficient of friction between the tires and the road is 0.3. What is the maximum speed the car can maintain without skidding?

Solution:

Given:
Radius of the curve ($$r$$): 100 meters
Banking angle ($$\theta$$): 30 degrees
Coefficient of friction ($$\mu$$): 0.3
Acceleration due to gravity ($$g$$): 9.81 meters per second squared

$v_{max} = \sqrt{r g \left( \mu + \tan(\theta) \right) / (1 – \mu \tan(\theta))}$

First, calculate $$\tan(\theta)$$:
$\tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577$

Then, substitute the values into the formula:
$v_{max} = \sqrt{100 \times 9.81 \left( 0.3 + 0.577 \right) / (1 – 0.3 \times 0.577)}$

Calculate the numerator:
$100 \times 9.81 \times (0.3 + 0.577) = 100 \times 9.81 \times 0.877 \approx 860.337$

Calculate the denominator:
$1 – 0.3 \times 0.577 = 1 – 0.1731 = 0.8269$

Now, divide the numerator by the denominator:
$\frac{860.337}{0.8269} \approx 1040.32$

Finally, take the square root:
$v_{max} = \sqrt{1040.32} \approx 32.24 \, \text{m/s}$

Answer: The maximum speed the car can maintain without skidding is approximately 32.24 meters per second.

Example 3: A motorcycle is traveling on a curved road with a radius of 75 meters. The road is banked at an angle of $$20^\circ$$. Given that the coefficient of friction between the tires and the road is 0.25, determine the maximum speed the motorcycle can maintain without skidding.

Solution:

Given:
Radius of the curve ($$r$$): 75 meters
Banking angle ($$\theta$$): $$20^\circ$$
Coefficient of friction ($$\mu$$): 0.25
Acceleration due to gravity ($$g$$): 9.81 meters per second squared

$v_{max} = \sqrt{r g \left( \mu + \tan(\theta) \right) / (1 – \mu \tan(\theta))}$

First, calculate $$\tan(\theta)$$:
$\tan(20^\circ) \approx 0.364$

Then, substitute the given values into the formula:
$v_{max} = \sqrt{75 \times 9.81 \left( 0.25 + 0.364 \right) / (1 – 0.25 \times 0.364)}$

Calculate the numerator:
$75 \times 9.81 \times (0.25 + 0.364) = 75 \times 9.81 \times 0.614 = 455.86$

Calculate the denominator:
$1 – 0.25 \times 0.364 = 1 – 0.091 = 0.909$

Now, divide the numerator by the denominator:
$\frac{455.86}{0.909} \approx 501.65$

Finally, take the square root:
$v_{max} = \sqrt{501.65} \approx 22.41 \, \text{m/s}$

Answer: The maximum speed the motorcycle can maintain without skidding is approximately 22.41 meters per second.

1. What is banking of roads?

Banking of roads, also known as road superelevation, refers to the angling of the road surface on curves. This design technique raises the outer edge of the road higher than the inner edge to help vehicles navigate curves more safely and efficiently.

Roads are banked to counteract the centrifugal force experienced by vehicles when traveling around curves. By tilting the road surface, engineers can create an inward force component that helps keep vehicles on the curved path, reducing the risk of skidding or overturning.

3. What happens if a road is over-banked or under-banked?

If a road is over-banked (too steep), vehicles may experience excessive lateral forces, leading to discomfort or loss of control. If it’s under-banked (too shallow), vehicles may still experience significant lateral forces, reducing the effectiveness of the banking in improving safety and stability.