BANKING ON HIGHWAY CURVES
Ideal Angle of Banking
Consider a car weight W that makes a horizontal turn on a curve of radius r while traveling
at velocity v. The curve is banked at an angle θ with the horizontal so that there is no
tendency to slide up or down the road. This angle is known as the ideal angle of banking.
W v2
Assuming the centrifugal inertia force ― ― applied through the center of gravity to
g r
create dynamic equilibrium, the free-body diagram is as shown in Figure. The resultant normal
pressure against the wheels is represented by Ν .
(a) Ideal angle of banking, neglecting friction
v2
tanθ ＝ ―
gr
It should be consider that the dimensions of the car are negligibly small in comparison to
the size of the path, so that the car may be considered as a particle.
The ideal angle of banking, neglecting friction defines in terms of the velocity of the car
and the radius of the turn and is independent of the weight of the car. The velocity in this
case is often termed the rated speed of the curve.
(b) Ideal angle of banking, considering friction
If the car is on the point of slipping up the plane of banking:
vmax 2
tan ((θ + ϕ)) ＝ ――
gr
If the car is on the point of slipping down the plane of banking:
vmin 2
tan ((θ - ϕ)) ＝ ――
gr
Sample Problem-1: A boy running a foot race rounds a flat curve of 12 m radius. If he runs
at the rate of 6 m/s, at what angle with the vertical will he incline his body.
SOLUTION:
m
v＝6 ― r ＝ 12 m
s
v2
tanθ ＝ ―
gr
⎛ 62 ⎞
θ ＝ tan -1 ⎜―――― ⎟
⎝ 9.81 ((12)) ⎠
θ ＝ 17.0° ((Answer))
Sample Problem-2: The rated speed of highway curve 300 ft radius is 40 mph. If the
coefficient of friction between the tires and the road is 0.6, what is the maximum speed at
which a car can round the curve without skidding
Sample Problem-2: The rated speed of highway curve 300 ft radius is 40 mph. If the
coefficient of friction between the tires and the road is 0.6, what is the maximum speed at
which a car can round the curve without skidding
SOLUTION:
v ＝ 40 mph r ＝ 300 ft μ ＝ 0.6
v2
tanθ ＝ ―
gr
40 mi ⎛ 5280 ft ⎞ ⎛ 1 hr ⎞ ⎛ 1 min ⎞
v ＝ ――― ⎜――― ⎟ ⎜――― ⎟ ⎜――― ⎟
hr ⎝ 1 mi ⎠ ⎝ 60 min ⎠ ⎝ 60 s ⎠
ft
v ＝ 58.67 ―
s
-1
⎛ 58.67 2 ⎞
θ ＝ tan ⎜―――― ⎟
⎝ 32.2 ((300)) ⎠
θ ＝ 19.61°
tanϕ ＝ μ (formula for angle of friction)
tanϕ ＝ 0.6
ϕ ＝ tan -1 ((0.6))
ϕ ＝ 30.96°
vmax 2
tan ((θ + ϕ)) ＝ ――
gr
vmax ＝ ‾‾‾‾‾‾‾‾‾‾‾‾
tan ((θ + ϕ)) ((gr))
vmax ＝ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
tan ((19.61 + 30.96)) ((32.2 ⋅ 300))
ft
vmax ＝ 108.39 ―
s
vmax ＝ 73.9 mph ((Answer))
Sample Problem-3: A daredevil drives a motorcycle around a circular vertical wall 100 ft in
diameter. The coefficient of friction between tires and wall is 0.6.
a) What is the minimum speed that will prevent his sliding down the wall?
b) At what angle will the motorcycle be inclined to the horizontal?
Sample Problem-3: A daredevil drives a motorcycle around a circular vertical wall 100 ft in
diameter. The coefficient of friction between tires and wall is 0.6.
a) What is the minimum speed that will prevent his sliding down the wall?
b) At what angle will the motorcycle be inclined to the horizontal?
c) What is the effect of travelling at a greater speed?
SOLUTION:
Part a:
μ ＝ 0.6 D ＝ 100 ft r ＝ 50 ft
ΣFh ＝ 0
Wv 2
N ＝ ――
gr
f＝F
f ＝ μN
⎛ Wv 2 ⎞
f ＝ 0.6 ⎜――⎟
⎝ gr ⎠
ΣFv ＝ 0
f＝W
⎛ Wv 2 ⎞
0.6 ⎜――⎟ ＝ W
⎝ gr ⎠
gr
v 2 ＝ ――
0.6
‾‾‾‾‾‾‾‾
32.2 ((50))
v ＝ ――――
0.6
ft
v ＝ 51.8 ― or v ＝ 35.3 mph ((Answer))
s
Part b:
ft
v ＝ 51.8 ―
s
Part b:
f
tanθ ＝ ―
N
0.6 N
tanθ ＝ ――
N
tanθ ＝ 0.6
θ ＝ tan -1 ((0.6))
θ ＝ 30.96° ((Answer))
Part c:
Considering the effect of traveling at a greater speed. As the speed increases,
the centripetal force required to keep the motorcycle moving in a circular path
also increases. If the speed exceeds the maximum velocity calculated earlier,
the friction force will not be able to supply the necessary centripetal force, and
the motorcycle will slide down the wall. Hence, it is not safe to travel at a
greater speed than the maximum velocity calculated in this scenario.
