What is the ideal banking angle for a gentle turn of \(1.20{\rm{ km}}\) radius on a highway with a \(105{\rm{ km}}/{\rm{h}}\) speed limit (about \(65{\rm{ mi}}/{\rm{h}}\)), assuming everyone travels at the limit?

To supply the centre gravity force required for a vehicle to perform a safe turn, the outside edge of curved roadways is raised above the inner edge. This is referred to as road banking.

For a gentle turn, the optimal banking angle is

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

where \(v\) is the vehicle's velocity, \(R\) is the radius of the curved turn, and \(g\) is the acceleration due to gravity.

Substitute \(105{\rm{ km}}/{\rm{h}}\) for \(v\), \(1.20{\rm{ km}}\) for \(R\), and \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\),

\(\begin{aligned}{c}\theta = {\tan ^{ - 1}}\left[ {\frac{{{{\left( {105{\rm{ km}}/{\rm{h}}} \right)}^2}}}{{\left( {1.20{\rm{ km}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}} \right]\\ = {\tan ^{ - 1}}\left[ {\frac{{{{\left[ {\left( {105{\rm{ km}}/{\rm{h}}} \right) \times \left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ hr}}}}} \right)} \right]}^2}}}{{\left( {1.20{\rm{ km}}} \right) \times \left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}} \right]\\ = 4.14^\circ \end{aligned}\)

As a result, the ideal working angle for gentle rotation is \(4.14^\circ \).