The mathematics
of round curves on the expressway
By Zhuang Biln
Chen Huiling
From Raffles Girls’ School (Secondary)
Contents
General Introduction
The centripetal force on a round-curve road
Lateral Incline
The distribution of q and f
The changing of units
The Optimum Speed ------Freehand Speed
When the speed is higher or lower than the optimum speed
Total Length
The Graphs of the values
Calculating Cases with Specific Figures
Reference
Acknowledgements
General Introduction
Expressways are designed so that vehicles can travel quickly through them.
However, when an expressway experience obstructions, curves must be used to avoid
the obstructions. Therefore, expressways are a combination of straight lines and
curves. In order to make the expressway look pleasing and to guarantee the smooth
ride of the vehicles, we can use parabolic curves between straight lines and round
curves. So, in general, straight lines, round curves and clothoids are the basic
compositions of the expressways.
The followings are the characteristics of the compositions:
1. Straight lines: curvature = 0
2. Round curves: curvature is a non-zero constant
3. Clothoid: the curvature is changing gradually
However, in this project, only round curves will be considered. We will study the
characteristics of the round curves, as well as the relationship between the round
curve and the movement of the vehicles.
Our main objective is to link physics, especially in the field of mechanics and
engineering, to mathematics. We believe that these two subjects are closely linked,
and we hope to find out how mathematics is used in other subjects and real day life.
The centripetal force on a round-curve road
The vehicle must experience a centripetal force in order to move on a round curve.
And the centripetal force is considered as a resultant force. (Fig.1)
Centripetal force
Fig.1
In physics, we learn that the formula of centripetal force is:
F= mv2/r
Where m-----mass of the vehicle (in kg)
r------radius of the round curve (in m)
v-----speed of the vehicle (in m/s)
F-----centripetal force
Since it is more common to use km/h as unit in traffic, we can convert the above
expression as the following:
F=mv2/r=m(3.6V) 2/r=12.96mV 2/r
Where V----speed of the vehicle (in km/h)
The actual forces acting on the vehicles are weight, normal force, friction
between the wheels and the ground. Here, we are not considering the air resistance.
Their resultant force will be a centripetal force.
Lateral Incline
F
θ
Fig.2
G
Because of inertia, all the bodies in motion have the tendency to move in a
straight line. So do the vehicles on the expressways. Therefore, in round curves on the
expressway, we must give the vehicle a force to deal with inertia. We have thought of
two methods to do this. The first is that the drivers turn the steering wheel to make the
friction between the wheels of the vehicle and the road to provide a centripetal force.
The other one is making a lateral incline on the road, in order to make the weight of
the vehicle provide a centripetal force.
In expressways, since the vehicles are moving at high speed, the lateral incline is
important.
Since we are discussing the motion of the car in a round curve, we can consider
other cases of circular motions for comparison. When a stone is tied to a string and
moving in a circle, it is sure that the stone is experiencing centripetal force and the
string is experiencing centrifugal force. However, when the string is released, the
stone is moving in the direction of the tangent, instead of the direction of the
centrifugal force. Similarly, when a train is moving on the round curve, once digress,
the train will not moving in the direction of the centrifugal force but tangent. This
phenomenon is because of the inertia of the vehicles.
As shown in Figure 2, let
the lateral force causing by the lateral incline be H,
the weight of the vehicle be G,
the centrifugal force causing by the changing of velocity be F,
the mass of the vehicle be M,
the speed of the vehicle be v,
the acceleration due to gravity be g, and
the radius of the round curve be R
The balance equation of the forces is shown as the following:
Gsinθ=Fcosθ
Gsinθ= (mv2/R) cosθ
Gsinθ= (Gv2/gR) cosθ (m = G/g)
Delete G on both sides
sinθ= (v2/gR) cosθ
From the above equation, we find that the balancing equation has no relation
with the mass or the weight of the vehicle.
And,
sinθ/ cosθ= v2/gR
tanθ= v2/gR
And let the lateral gradient be q , so
tanθ= q
When we neglect the effects of the turning of the steering wheel, (lateral friction,)
we can calculate the lateral gradient, q, of the road.
Let V be the speed of the vehicle with the unit of km/h, as it is more common to
use such a unit in traffic.
q = tanθ= v2/gR = (V/3.6) 2/9.8R = V2/127R
as 3.6 m/s = 1 m/s
The above equation shows that there is a certain speed of the vehicles that can
make the parallel force causing by the weight and the parallel force causing by the
centrifugal force balanced. Now, we would like to call this certain speed the optimum
speed here.
When a vehicle is moving on a round curve with gradient q and radius R, if it is
not moving at the optimum speed, the forces will not be balanced. In this case, the
driver should turn the steering wheel to increase the parallel friction. For example,
when the speed is higher than the optimum speed, the parallel force, H, increases.
When the speed is lower than the optimum speed, H decreases. We assume that the
coefficient of parallel friction is f. When we are considering the friction causing by
the turning of the steering wheel, the balance equation should be:
q+f = V2/127R
Let Q = q+f, so
Q = V2/127R
The distribution of q and f
In the above section, we have already discuss the parallel force and find that:
Q = q + f = V2/127R
It is an indefinite equation, which has many solutions for q and f. So, we need to
discuss the value of q and f.
We can use the following two methods to distribute the value of q and f.
1. Multiplication
We can multiply Q by two different fractions, which add up equals to 1.
For example,
When q = Q, f = 0 (When the vehicle is moving at the best speed)
q = (1/2)Q, f = (1/2)Q
q = (1/3)Q, f = (2/3)Q
q = (1/4)Q, f = (3/4)Q
q = 0, f = Q (When there is no parallel incline)
2. Addition
We can define a number for f, then q = Q – f. Please note that the result can be
negative.
For example,
f=0
q=Q
f = 0.05
q = Q – 0.05
f = 0.10
q = Q – 0.10
f = 0.15
q = Q – 0.15
The changing of units
(1) Multiplication
Let the optimum parallel incline be q’
Value of q that should be set (in
m/s )
2
3
1
q = q’
q = (1/2)
q’
q = (1/3)
q’
q = (1/4)
q’
Value of q that should be set (in
km/h)
4
5
v2/gR
v2/9.8R
V2/127R
V2/127R
v2/2gR
v2/19.6R
V2/(2×127R)
V2/254R
v2/3gR
v2/29.4R
V2/(3×127R)
V2/381R
v2/4gR
v2/39.2R
V2/(4×127R)
V2/508R
Form 1
Optimum speed, V’
3
1
2
4
q=Q
127 Rq
127 Rq
V’
q = (1/2)Q
254Rq
2  127 Rq
1.41V’
q = (1/3)Q
381Rq
3  127 Rq
1.73V’
q = (1/4)Q
508Rq
4  127 Rq
2V’
Form 2
(2) Addition
Assume a certain amount of coefficient of parallel friction that is needed be f.
When we subtract it from the best gradient of parallel incline, we get the gradient
of parallel incline that we have set.
q = q’- f
q = V2/127R – f
V=
127(q  f)R = 11.269
(q  f)R
R = V2/127(q+f) = 0.00787 V2/(q+f)
The Optimum Speed ------Freehand Speed
Now, we use V’ to represent the optimum speed (in km/h), and use q’ to
represent the optimum gradient of parallel incline that is derived from it. From
mechanics we can see that the optimum speed and gradient only exist in round
curve, as the centrifugal force and centripetal force is not existent in a straight
line.
The optimum speed is the speed when the forces acting on the vehicle is
balanced and there is no parallel friction. Therefore, the forces acting on the
vehicles is the centripetal force causing by the weight of the vehicle and the
parallel incline. The parallel force causing by the weight makes the vehicles,
which tend to move in a straight line, go in a circular track.
When the vehicle is moving at the optimum speed, then radius of the round
curve, R, gradient of the parallel incline, q, and the speed of the vehicle V, have:
V=
127 Rq
= 11.269 Rq
If the vehicle is moving at this speed, the driver does not need to turn the
steering wheel.
The interesting thing is that when the vehicle is moving in a straight line, it
is impossible for the driver to leave hand from the steering wheel, as there is
also a lateral incline made for the water to flow away. However, in a round curve
it is possible for the driver to free his hand, if the vehicle is moving at the
optimum speed.
When the speed is higher or lower than the optimum speed
The equation of the centrifugal force is given as
F = v2 / R = V2 / 127 R
We can easily see that when the speed of the vehicle, v, increases, the
centrifugal force causing by inertia increases.
So, when a vehicle is moving at a speed higher than the optimum speed, a
centrifugal force that is greater than the parallel force provided by the weight is
produced. The vehicle will move outwards. The driver needs to turn the steering
wheel inward to provide a parallel friction.
On the other hand, when the vehicle is moving at a speed that is lower than the
best speed, the parallel force causing by the weight will be greater than the centrifugal
force. As a result, the vehicle will move inwards. Therefore, the driver needs to turn
the steering wheel outwards to provide an outward friction.
Total Length
The total length of the expressway is dependant on the volume of the traffic. If
the volume is large, then obviously there will be traffic congestion. Therefore, a
minimum length is required for a certain expressway.
Firstly, we need to determine the safe distance required between each vehicle.
From kinematics, we have learnt that
v 2  v02  2as
where v------final velocity of the vehicle
v0-----initial velocity of the vehicle
a------acceleration of the vehicle
s ------distance traveled by the vehicle
When a vehicle stops, its acceleration is negative. For the sake of convenience, we use
positive deceleration, d, which is equal to –a. And when a vehicle comes to a stop, its
final velocity, v equals to 0.
So we have
0  v 02  2ds
v02  2ds
s  v02 2d
Here s is the distance needed for the car to decelerate.
However, before the actual deceleration, a reaction time, a reaction time, tr is
needed for the driver. During this period of time, the vehicle is still moving at its
initial velocity. And the distance traveled just before its deceleration can be calculated
as
s' v0 t r
Where s’--------the distance traveled after reaction before deceleration
After the vehicle stops, so as to ensure it is not in contact with the vehicle in front,
a space between is needed.
So we get
S  s  s's' '1
Where S-----length required for each vehicle
L----length of the vehicle itself
Let L be the total length of the expressway (assume that there is only one
roadway). The number of vehicles that can be safely on the expressway can be
calculated by:
N=L/S
Where N-----number of vehicles
So,
S=L/N
S+s’+s’’+1=L/N
v / 2d  v0 t r  s' '1  L N
2
0
v02 / 2d  v0 t r  s' '1  L N  0
Solving the quadratic equation, we get
v0  (t  t r2  4( s ' '1  L N ) / 2d /( 2 / 2d )
v0  t r  d t r  2(s' '1  L N ) / d
v0  t r d  d 2 t r2  2d ( s ' '1  L N )
The Graphs of the values
1.The relationship between lateral gradient and lateral friction.
Lateral gradient Lateral friction
400
0.005
0.120984252
400
0.01
0.115984252
400
0.015
0.110984252
400
0.02
0.105984252
400
0.025
0.100984252
400
0.03
0.095984252
400
0.035
0.090984252
400
0.04
0.085984252
400
0.045
0.080984252
400
0.05
0.075984252
400
0.055
0.070984252
400
0.06
0.065984252
400
0.065
0.060984252
400
0.07
0.055984252
400
0.075
0.050984252
400
0.08
0.045984252
400
0.085
0.040984252
400
0.09
0.035984252
400
0.095
0.030984252
400
0.1
0.025984252
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0.1
0.08
0.07
0.05
0.04
0.02
Lateral friction
0.01
lateral friction
Speed/ (km/h) Radius/m
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
lateral gradient
2. Relationship between lateral friction and radius of the round curve.
Lateral Gradient
Speed/(km/h)
0.1
0.1
0.1
0.1
0.1
0.1
0.1
Lateral Friction Radius/m
80
0.01 458.1246
80
0.02 419.9475
80
0.03 387.6439
80
0.04 359.955
80
0.05 335.958
80
0.06 314.9606
80
0.07 296.4335
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
0.08
0.09
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.4
279.965
265.23
251.9685
239.97
229.0623
219.103
209.9738
201.5748
193.8219
186.6433
179.9775
173.7714
167.979
162.5603
157.4803
152.7082
148.2168
143.982
139.9825
136.1992
132.615
129.2146
125.9843
122.9115
119.985
117.1947
114.5311
111.986
109.5515
107.2206
104.9869
102.8443
100.7874
Radius/m
500
400
300
Radius/m
200
100
0.
01
0.
06
0.
11
0.
16
0.
21
0.
26
0.
31
0.
36
0
Coefficient of lateral firction
3. The relationship between lateral gradient and radius of the round curve.
Speed/(km/h) Lateral gradient Radius/m
0.1
80
0.005 479.9400075
0.1
80
0.01 458.1245526
0.1
80
0.015 438.2060938
0.1
80
0.02 419.9475066
0.1
80
0.025 403.1496063
0.1
80
0.03 387.6438522
0.1
80
0.035 373.2866725
0.1
80
0.04 359.9550056
0.1
80
0.045 347.5427641
0.1
80
0.05 335.9580052
0.1
80
0.055 325.1206502
0.1
80
0.06 314.9606299
0.1
80
0.065 305.4163684
0.1
80
0.07
296.433534
0.1
80
0.075 287.9640045
0.1
80
0.08 279.9650044
0.1
80
0.085 272.3983826
0.1
80
0.09 265.2300041
0.1
80
0.095 258.4292348
0.1
80
0.1 251.9685039
600
500
400
300
200
100
0
Radius/m
0.
00
5
0.
02
0.
03
5
0.
05
0.
06
5
0.
08
0.
09
5
Radius/m
Lateral friction
Lateral gradient
4.The relationship between the speed and the radius.
Lateral gradient
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
Lateral friction Speed/(km/h) Radius/m
0.2
40 50.39370079
0.2
45 63.77952756
0.2
50 78.74015748
0.2
55 95.27559055
0.2
60 113.3858268
0.2
65 133.0708661
0.2
70 154.3307087
0.2
75 177.1653543
0.2
80 201.5748031
0.2
85 227.5590551
0.2
90 255.1181102
0.2
95 284.2519685
0.2
100 314.9606299
0.2
105 347.2440945
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.05
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
110
115
120
125
130
135
140
145
150
381.1023622
416.5354331
453.5433071
492.1259843
532.2834646
574.015748
617.3228346
662.2047244
708.6614173
Radius/m
800
600
400
Radius/m
200
85
10
0
11
5
13
0
14
5
70
55
40
0
Speed/(km/h)
5. The relationship between the speed and the total length.
Length of theReserved
Reaction time Deceleration vehicle
distance
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
0.5
15
3.5
Number
cars
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
ofInitial
Velocity
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
140
145
150
Total Length
495
628.33333
778.33333
945
1128.3333
1328.3333
1545
1778.3333
2028.3333
2295
2578.3333
2878.3333
3195
3528.3333
3878.3333
4245
4628.3333
5028.3333
5445
5878.3333
6328.3333
6795
7278.3333
7778.3333
8295
8000
6000
Total Length
4000
2000
0
0
0
Initial Velocity
15
13
11
90
70
50
0
30
Total Length
10000
Calculating Cases with Specific Figures
Let:
(1) The radius of the round curve be R
(2) The coefficient of friction be f
(3) The velocity of the vehicles be v
(4) The lateral gradient of the road be q
Case 1:
Assuming that the vehicles are moving at 80km/h and the gradient is 0.02,
what is the optimum value of the radius of the arc?
q 
v2
127 R
0.02 
6400
127 R
the optimum value of R is
R
6400
127  0.02
R = 2519
Case 2:
Using the values in case 1, if the size of the area is limited, what is the
minimum value of the radius? Assume that the maximum value of f is 0.1.
q  f 
V2
127 R
In order to get the minimum value of R, we would take the maximum value of
f.
So the minimum value of R is
80 2
0.02  0.1 
127 R
R
6400
127  0.12
R= 419.95
Case 3
This area, however, gets snowy in winter. The coefficient of friction drops to
0.04. Now, what is the minimum value the radius can take?
q  f 
v2
127 R
0.02  0.04 
R
80 2
127 R
6400
127  0.06
R = 839.90
Case 4
However, because of the presence of mountains, the radius of the road is fixed
at 400m. Then what should the gradient of the road be if the drivers would like
to maintain the same speed?
q  f 
 q  0.1 
v2
127 R
80 2
127  400
q 
6400
 0.1
50800
q = 0.0260
Case 5
In recent years, because of a baby boom after a war, the numbers of people
trying for a license has increased drastically, which led to an increase in
percentage of amateur drivers on the road. If the gradient is fixed at 0.02, what
is the maximum safe velocity they can travel at?
q f 
0.02  0.1 
v2
127 R
v2
127  400
v 2  0.12  127  400
v 2  6096
v  6096
v = 78.08
Case 6
Assume that the speed of the cars is 80km/h and it decelerates at 15m/s2. After
stopping, a distance of 0.5m is needed between the car in front and it. Assuming that
the reaction time for the driver is 0.5s and that there is only one car on this portion of
the road, what is the minimum length of the road?
v02
L
 vo t r  s' '1   0
2d
N
10 2
)
30  80  10  0.5  0.5  4  L
2  15
30
80  (
L=33.39m
Case 7
If there are always 15 cars at peak hours, what should the minimum length of the road
be in order to ensure the vehicles to travel safely?
L=33.39 x 15=500.83m
Case 8
Now the length of the road is only 400m because of limited land space. If it is
impossible to reduce the number of vehicles, what will be the maximum speed of the
vehicles be now, if safety is observed?
v0  t r d  d 2 tr 2  2d ( s ' '1  L N
vo  0.5  15  0.5 2  15 2  2  15  (0.5  1  400 15
 v0  0
 v0  20.98km / h
Reference
Modern Mathematics and Dynamics, By Zhong Zhongheng, Beijing University
Publisher
Dynamics, By Zhou Kaiyuan, Xu Yanhou, Guo Changluo, Science- Technical
University of China Publisher
Dynamics, By Wu Zhen, Shanghai Traffic University Publisher
The Design of Highways, By Yang Shengfu, People Trafic Publisher
Interesting Dynamics, By Teng Zhong, Science Publisher
General Concepts of Road Designing, By Zhuang Haitao, Dongnan University
Publisher
Acknowledgement
We would like to thank
Mrs Sia Wai Leng (Teacher of Raffles Girls’ School)
Mr Kenerth Lui (Teacher of Raffles Girls’ School)
For their guidance and support.
We would also like to thank
Mr Wu Zeyu ( Physics Teacher of Shenzhen Middle School, China)
Miss Cai Ling (Student of Shenzhen Middle School,China)
For the help with the reference.