If the width of the highway is 10 m and its outer edge is 40 cm higher, the super elevation is 1 in
25
Super elevation, also known as cant, is the transverse slope provided to the outer edge of a highway curve relative to the inner edge. This is done to counteract the centrifugal force acting on vehicles negotiating the curve, thereby ensuring stability and preventing skidding.
The super elevation (e) is typically expressed as a ratio, which is the rise (height difference between outer and inner edge) divided by the width of the roadway. It can also be expressed in the format "1 in X", where X is the ratio of the width to the rise.
To perform the calculation, we need to ensure all units are consistent. The width is given in meters, and the height difference is in centimeters. We will convert the height difference to meters.
Height difference (e) = 40 cm = $\frac{40}{100}$ m = 0.4 m
The super elevation (e) can be calculated as the ratio of the height difference to the width of the highway:
e = $\frac{\text{Height difference}}{\text{Width}}$
e = $\frac{0.4 \text{ m}}{10 \text{ m}}$
e = 0.04
The question asks for the super elevation in the format "1 in X". This means we need to find X such that:
e = $\frac{1}{\text{X}}$
Substituting the calculated value of e:
0.04 = $\frac{1}{\text{X}}$
To find X, we can rearrange the equation:
X = $\frac{1}{0.04}$
X = $\frac{1}{\frac{4}{100}}$
X = $\frac{100}{4}$
X = 25
Therefore, the super elevation is 1 in 25.
| Parameter | Value |
|---|---|
| Highway Width (B) | 10 m |
| Height Difference (e) | 40 cm = 0.4 m |
| Super Elevation Ratio (e/B) | 0.4 m / 10 m = 0.04 |
| Super Elevation (1 in X) | 1 in (B/e) = 1 in (10/0.4) = 1 in 25 |
Based on the calculation, the super elevation is 1 in 25.
| Concept | Definition | Formula |
|---|---|---|
| Super Elevation (e) | Transverse slope on highway curves. | e = $\frac{\text{Rise}}{\text{Width}}$ (ratio) or 1 in $\frac{\text{Width}}{\text{Rise}}$ (1 in X format) |
| Purpose | Counteracts centrifugal force, prevents skidding, ensures comfort and safety. | |
| Maximum Limit | Usually restricted based on location (plain, hilly, snowy) and road type (urban, rural). Indian Roads Congress (IRC) guidelines specify max 'e'. |
The amount of super elevation provided on a highway curve is influenced by several factors:
The theoretical formula for super elevation considering design speed (V) and radius (R) is often given by:
e + f = $\frac{\text{V}^2}{127\text{R}}$ (for V in kmph, R in meters)
However, the current problem directly provides the rise and width, allowing for a direct calculation of the super elevation ratio.
The rate of super-elevation for a horizontal curve of radius $500 \text{ m}$ in a national highway for a design speed of $65 \text{ kmph}$ is:
Consider the following statements about grade compensation:
(i) Grade compensation is given up to the maximum value of '75/R', where R is the radius of circular curve in metres.
(ii) According to Indian Roads Congress, grade compensation is not necessary for gradients flatter than 4 percent.
Which of the above statement/s is/are correct?
The type of transition curve that is generally provided on hill road is
The minimum design speed adopted where hair-pin bends are provided at hill roads is _________.
The rear wheels do not follow the same path as that of the front wheels. This phenomenon is called: