Max Turning Velocity on a Banked Curve

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Equation / Last modified by Administrator on 2015/08/11 07:06
MichaelBartmess.Max Turning Velocity on a Banked Curve

This equation computes maximum velocity a car could achieve on a banked surface.

The inputs to this equation are:

  • r - the radius of the curve
  • `phi` - the angle of the banked surface
  • `mu` - the coefficient of friction characterizing the two surfaces interaction.


In the illustration, we see the back end of a car turning to the left.  The weight of the car due to the force of gravity pushes downward.  That force is equal and opposite to the vertical component of the force of the road pushing up on the car's tires.  The frictional force, `F_"Friction"`, is toward the direction of the turn and parallel to the banked surface and keeps the car from sliding outward during the turn. 

The force of the road pushing up (perpendicular to the road surface) on the car is is the force `F_"Road"`.

We can determine the maximum velocity, v,  by assuming the horizontal forces sum exactly to zero.  We first compute the force exerted on the car's tires by the road, `F_"Road"`:

Eq. 1: `sum F_"Vertical" = F_"Road" * cos(phi) - F_"Road" * mu * sin(phi) - m*g = 0`

Thus we can compute `F_"Road" * mu = F_"Friction"` from Eq. 1 as:

Eq. 2: `F_"Road" * (cos(phi) - mu  * sin(phi)) =  mg`

Eq. 3: `F_"Road" = mg`/`(cos(phi) - mu  * sin(phi))`

We then substitute `F_"Road"` into the following equation:

Eq. 4: `sum F_"Horizontal" = F_"Road" * sin(phi) + mu*F_"Road" * cos(phi) = m*v^2/r`

which provides us:

Eq. 5: `((m*g)/ (cos(phi) - mu  * sin(phi)))  * sin(phi) + ((m*g)/(cos(phi) - mu  * sin(phi))) * mu * cos(phi) =m*v^2/r`

Eq. 6: `((r*g) / (cos(phi) - mu  * sin(phi)))  *  (sin(phi) + mu * cos(phi)) = v^2`

Eq. 7: `v = sqrt(    (r*g * (sin(phi) + mu * cos(phi))) / (cos(phi) - mu  * sin(phi))    )`