Blake Cicenas
Physics 710
Back to the Traffic Circle
I am going to restart work on the traffic circle by first coming up with a zeroth order
model of a one lane roundabout. I first have to start out with some assumptions. I will assume
the radius of the roundabout is 50.00 m and that the roundabout is banked at 5.00 degrees. The
road is an asphalt surface and I will assume that the coefficient of static friction is around 0.800
and that the rolling is smooth. I will also assume that the normal force acts on all four tires
equally. The force body diagram is located below where FN is the normal force fs is the static
frictional force, Fc is the centripetal force, m is mass and g is gravity.
I then solve the Force Body Diagram in the y-direction to get FN cos   mg   s FN sin  .
mv 2
Solving in the x-direction yields
 FN sin    s FN cos  . Solving the y-direction equation
r
for FN and plugging it into the x-direction equation and then solving the whole system for
velocity yields the equation v 
gr (  s cos   sin  )
. Plugging in the given values yields 21.6
cos    s sin 
m/s or 48 mph for the velocity. This is of course, under dry conditions. And since most highway
engineers are forced by the big bad federal government to tell the citizens how to be live their
lives, the value should be calculated under wet conditions. Assuming a coefficient of static
friction of 0.300 would then yield 13.8 m/s or 30.8 mph. This is a much more reasonable value
for cars in the roundabout. However, this is the maximum velocity that can be achieved without
the car sliding off so we would probably want to lower the maximum speed to the nearest value
divisible by 5 mph so that skidding will not occur if the traffic signs are followed.
So I will simply round down the speed to 30 mph and set that as the temporary speed
limit for my roundabout. Now, I am really only concerned with the time it takes the car to go
through the roundabout. To aid myself, I will turn my linear equations to rotational equations so
the max velocity becomes  
g (  s cos   sin  )
where ω is the angular velocity of the
r (cos    s sin  )
vehicle. Now that I am working with angular variables, then the maximum velocity would then
become 0.276 rad/s. I also know that the maximum angular distance traveled will be ¾ π
(assuming that someone does not make a complete circle and go back the opposite direction).
But what I do not know is the acceleration of the vehicle. So I will assume a max linear
acceleration of 1 m/s/s which yields an angular acceleration of 0.020 rad/s/s. So that begs a
question; will a lead-footed driver be able to reach the maximum acceleration before exiting the
roundabout? Well, if I plug in the values into the rotational kinematic equations I get a
maximum time of 15.3 seconds which then yields a maximum potential angular velocity of 0.307
rad/s which exceeds my speed limit. So I will need to take this into account when finding the
times the cars will stay in the roundabout.
For those cars that are traveling for ¾ π the equation will have to combine   0  t1
(where ω is maximum angular velocity, ω0 is the initial angular velocity, α is the angular
acceleration and t1 is the time traveled with a constant acceleration) and  2  c t 2 (where ωc is
the constant angular (maximum) velocity, θ2 is the angular displacement traveled after the speed
limit is reached and t2 is the time traveled at ωc). This means that the first leg of the travel to the
3rd exit would be have a time of
c
2
and a displacement of 1  c . The second leg of the travel
2

to the 3rd exit would have a displacement of  2 
t2 
3  c2
and would then have a time of

4 2


3
 c . The first time would then be t1  c so the total time would be
4 c 2

tT 3 
3 c c
where tT3 is the total time in the circle for a car taking the third exit. If I


4 c 2 
plug in my known variables I get a total time of 15.4 seconds, only slightly longer than if I let the
car break the speed limit. However for consistency, I will use this time anyhow.
The time for the second exit will be much simpler since an imposed terminal velocity will
not be reached. If I plug in ½ π for displacement then the time will simply becomes tT 2 


where tT2 is the total time to the second exit. This total time is then 12.5 seconds.
Finally, the time for the 1st exit will also have a simple equation in the form of
tT 1 

where tT1 is the time spent in the circle while traveling to the first exit and ¼ π was
2
plugged in for the displacement. This yields a maximum travel time of 8.86 seconds.
Now I that I know how long the maximum time the cars will be in the roundabout, I can
start to move on to other, more complex endeavors. Everyone knows that a typical traffic circle
will not have equal probabilities of cars exiting and entering from each side. The probabilities,
however, are higher that a car entering from the south will want to proceed north and a car
entering from the east will want to proceed west. So I created a picture (where the base of the
picture was taken from KDOT) of the situation below.
This picture has tiny graphics next to each outlet to represent what I will assume to be
located there. On the north side of the roundabout is a casino, and obviously, many cars will be
driving to and from this destination. On the west side of the roundabout is farmland so very few
people will be driving to or coming from this direction. On the south side is the interstate where
a decent number of people will to be traveling to and fro. Finally, the east side of the roundabout
is a small town where a fair number of people will be driving to or coming from this direction.
In addition, I will assume that all cars not in the roundabout will yield to the cars already in the
circle, a method traditionally seen in Kansas.
While this is all clearly arbitrary, I can make some probability assumptions about what
the chance might be of a car traveling from one inlet to a specific exit. Below is a probability
chart of those assumed occurrences.
Direction
S to E
S to N
S to W
E to N
E to W
E to S
N to W
N to S
N to E
W to S
W to E
W to N
Probability
0.25
0.60
0.15
0.40
0.15
0.45
0.05
0.60
0.35
0.50
0.45
0.05
Notice, however, I did make the probabilities all in terms of 5 to make the calculations a
bit easier. Also notice some of the probabilities are opposites such as North to South and South
to North, while others are not such as South to East and East to South. This was done because
the percentage of people going to town from the interstate will not be the same as the percentage
of people going from the town to the interstate because the presence of the casino-goers and
farmers will definitely skew those results. However, I chose to make the North-South / SouthNorth identical values because many casino-goers will have no interest in the small town that lies
beyond the interstate. They simply want to exit, hit the casino, get something to eat at the casino
restaurant, maybe stay a night at their hotel, and then leave the way they came.
Now that I have some basic probabilities, I can then calculate the average time that a car
will spend coming from a particular direction by multiplying the probabilities by the previously
calculated times as described by P(T 1)tT 1  P(T 2)tT 2  P(T 3)tT 3  t avg . This will then give me
an estimate of who is taking the most time to go through the traffic circle given my basic
assumptions. A chart outlines the results below.
From the direction of…
South
East
North
West
Average Time (s)
12.00
12.30
13.30
10.80
This chart shows the casino-goers from the north clearly spend the most time in the
roundabout while the farmers from the west spend the least time in the traffic circle. Now I
would like to try to estimate how often cars will come into contact with each other in the
roundabout. My assumptions have been that the cars will move through freely without any
hindrances to hold back their times. I therefore will then make have to make some more basic
assumptions.
I will assume that the roundabout sees an average of 100 cars per hour. Out of the 100
cars I will estimate that 35 of those cars are coming from the north, 35 of those cars are coming
from the south, 25 of those cars are coming from the east, and 5 of those cars are coming from
the west. I will now make a giant leap and create a value Ct that will stand for Car times which
is simply the number of cars from that direction times the average time that the car spends in the
loop. A chart outlines my results below.
From the direction of…
South
East
North
West
Car Times (s)
420.00
308.00
466.00
54.00
Weight
0.34
0.25
0.37
0.04
I also added a column I called “Weight” to show how much weight each direction had
based on the average time the cars would spend in the loop and the average number of cars that
would come from that direction. I did this because although the townsfolk from the east have a
higher average time, they may not have the sheer numbers that the casino-goers from the south
have. So I made up arbitrary figures and factored that in to give the cars from the north and
south greater weight, which is exactly what occurred in the chart above. But this is all leading
into my next problem; that there will be times when a car from a particular direction will have to
yield to a car from another direction. This will then affect their average time and I want to figure
out how it will do so.
So I devised a simple random-number generated system based on the weight of the
direction and when the car will be in the loop. (See attached Excel Document- sheet 2). I
essentially created a random number generator to output a digit from 0 to 100. Then, based on
that value it will give a direction that a car is coming from. For example, since the weight of
south was .34, I then assigned all random numbers that were 34 of less to output “south”. I then
created another random number generator to create a value between 0 and 3600. Then based on
whether that value occurred below or above 1248I could make a guess of when the car would be
in the loop (1248 seconds is the sum of all of the car times). I made the assumption that if the
value was below 1248, then the chances were that the car was currently located within the circle.
I then created an if statement that outputted “In” If that value occurred below 1248. I then
created another If statement to output the random time if “In” was outputted. But in order to
scale the value correctly, I redivided by the chance of any car being in the circle in order to scale
the time back to an hour. Finally, I hid the “In” column to make it easier to glance at each to see
the direction and the time the car was in the loop. I then decided to check each number manually
to see if any of the numbers were within 11 of each other. I chose 11 because that the was
approximately the lowest average time a car would spend in the roundabout. A chart (a larger
bar chart can be found on sheet 3 of the excel sheet) below shows the cars that were in the loop
together and from what direction they were coming as well as the time that it occurred.
With
this data, I can then find out if there was a possibility of an oncoming car having to yield to a car
already in the traffic circle. I can then use this information to recalculate the average time for
each car spent in the roundabout depending on their direction.
In the Direction of…
North
North
North
East
North
South
South
East
South
South
East
North
North
North
South
North
North
South
Time (s)
3051
3046
2576
2570
2570
2518
2510
2498
2495
2489
2446
2435
2247
2247
1498
1489
858
847