The correct answer is: C. Greater than $\frac{{\text{b}}}{{2{\text{h}}}}$$ and less than co-efficient of lateral friction.
The centrifugal ratio is a measure of the tendency of a vehicle to overturn when traveling around a curve. It is calculated by dividing the vehicle’s speed by the square root of the product of its wheel track and the height of its center of gravity above the road surface.
The centrifugal force acting on a vehicle traveling around a curve is proportional to the square of its speed and inversely proportional to the radius of the curve. The lateral force required to keep the vehicle from overturning is proportional to the height of its center of gravity above the road surface.
If the centrifugal force exceeds the lateral force, the vehicle will overturn. Therefore, the centrifugal ratio must be less than a certain value in order to avoid overturning. This value is given by the following equation:
$$\frac{{\text{v}}^2}{{\text{r}}} < \frac{{\text{b}}}{{2{\text{h}}}}$$
where:
- $v$ is the vehicle’s speed
- $r$ is the radius of the curve
- $b$ is the wheel track of the vehicle
- $h$ is the height of the vehicle’s center of gravity above the road surface
The coefficient of lateral friction is a measure of the maximum lateral force that can be applied to a vehicle before it skids. It is a property of the road surface and the tires.
If the centrifugal force exceeds the lateral force that can be provided by the tires, the vehicle will skid. Therefore, the centrifugal ratio must be less than a certain value in order to avoid skidding. This value is given by the following equation:
$$\frac{{\text{v}}^2}{{\text{r}}} < \mu$$
where:
- $v$ is the vehicle’s speed
- $r$ is the radius of the curve
- $\mu$ is the coefficient of lateral friction
The centrifugal ratio must be greater than $\frac{{\text{b}}}{{2{\text{h}}}}$$ in order to avoid overturning, but it must also be less than the coefficient of lateral friction in order to avoid skidding. Therefore, the correct answer is C.