2008 Behind the Scenes: Athletic Events
Competition Scenarios
This scenario provides a preliminary overview of the question content that will be posed
during the 2008 TEAMS competition. Scenarios may be edited and expanded upon in the
questions of the actual TEAMS competition; however the overall content will remain.
Students are encouraged to research the general topics and terms in order to prepare for
the competition. Check the JETS website frequently for additional information about the
TEAMS competition that may become available.
Scenario 4:
NASA engineers apply aerodynamic principles to assist members of American
teams to help increase their gold-medal yield in international athletic
competitions.
2008 is an Olympic year and nations will gather in Beijing, China to celebrate the Olympic
Games. To help America “bring home the gold,” NASA has been asked to provide
mathematical and technical assistance to various Olympic sports committees. Applying
aerodynamic principles to the discus, the track, the shot-put and the corner kick in soccer
competitions should assist USA teams in their understanding of the properties of the events.
This will enable the team members to improve their performance and bring home more
metals.
2008 TEAMS Competition Scenarios
Explore…Assess…Experience Engineering
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Behind the Scenes:
Athletic Events
Part I
TEAMS 2008
Sponsors:
S.D. Bechtel, Jr.
Foundation
Shell
Tyco Electronics
JETS Sponsors:
Bechtel Group
Foundation
CH2M Hill
Motorola Foundation
NISH
Shell is the exclusive sponsor of the
TEAMS 2008 T-Shirt Design Challenge
Winning T-Shirt Design submitted by Teresa Lieh,
Beavercreek High School, Beavercreek, OH
Rockwell Collins
SolidWorks
United Engineering
Foundation
18
{
}
© JETS, Inc. TEAMS Competition 2008
problem # 4
Aerodynamic Principles Applied to
Olympic Games
Submitted by Tom Benson, NASA Glenn Research Center, Cleveland, OH
The Discus
Problem Statement
2008 is an Olympic year and nations will gather in Beijing,
China, to celebrate the Olympic Games. To help America
“bring home the gold,” NASA has been asked to provide
mathematical and technical assistance to various Olympic
sports committees. Your engineering team has been asked
to help NASA and the U.S. Olympic team by applying
aerodynamic principles to explain certain aspects of the
games. In particular your group will study the discus, track,
the shotput, and soccer competitions.
The current world record for the men’s discus throw is 74 m
and is held by a German athlete. To help break this record,
the United States is improving its training facilities. The old
discus field was a 40-degree, pie-shaped field that was 75 m
on one side. See Figure 4-1.
Background
You will need some mathematical equations to solve the
following problems.
Figure 4-1.
Assumptions and Givens
•
Rather than building a new discus field, the committee
has decided to modify the existing field by increasing its
size. The length will now be increased to 90 m on one
side to increase the landing area for the discus.
31. The additional turf needed to be installed to increase the
size of the discus field is most nearly:
a. 25,446 sq m
b. 17,671 sq m
c. 7,775 sq m
d. 2,827 sq m
e. 864 sq m
Additional Assumptions and Givens
•
An Olympic discus is a circular disc with a diameter equal
to 220 mm and a mass of 2 kg. Its weight is therefore
19.6 Newtons. For safety during training, a bright red
donut is painted on the top and bottom of the disc to
make it more easily visible in flight.
© JETS, Inc.
TEAMS Competition 2008
19
32. If the internal radius of the donut is 10 mm, and the
external radius is 100 mm, the amount of paint required
to cover both the top and the bottom is most nearly:
a. 38,013 sq mm
b. 76,026 sq mm
c. 62,202 sq mm
d. 62,831 sq mm
e. 31,415 sq mm
•
d. 22.9 mph
e. 34.5 mph
Additional Background
The discus, like any other object, generates an
aerodynamic force as it moves through the air. The
aerodynamic force can be divided into two components,
lift and drag. The amount of lift is described by a lift
equation:
L = .5 D V2A cl
34. The current record for the men’s 100-m dash is
9.77 seconds. The average speed of the runner expressed
in miles per hour (mph) is most nearly:
a. .218 mph
b. .381 mph
c. 10.2 mph
Additional Assumptions and Givens
In the women’s 4 x 100 relay, there are four runners. Each
woman runs 100 meters carrying a baton that she passes
to the next runner. If the baton is dropped, the team is
disqualified.
Here is a table of the best times for each runner to go 100 m:
(4-1)
Runner A :
11.23 sec
where,
Runner B :
10.71 sec
L
D
V
A
Runner C:
10.59 sec
Runner D:
10.41 sec
=
=
=
=
the lift
the density of air (1.22kg/m3)
the velocity of the object in m/sec
a reference area (equal to the surface area of the top of
the disc) in m2
cl = the lift coefficient (assumed equal to 1 for the discus in
flight)
33. The speed an athlete would have to throw the discus for
the lift to equal the weight of the discus (19.6 N) is most
nearly:
a. 846 m/sec
b. 29.1 m/sec
c. 14.53 m/sec
d. .029 m/sec
e. 206 m/sec
Additional Background
Track
Americans have always done well in the track events.
Your team will be doing some calculations related to the
100-m dash and the 4 x 100 relay race. You will need the
following information:
1m
12 in
5,280 ft
=
=
=
39.37 in
1 ft
1 mile
}
35. If each runner can match her best time, the average
speed of the baton expressed in miles per hour (mph) is
most nearly:
a. .240 mph
b. 9.31 mph
c. 20.84 mph
d. 34.51 mph
e. 83.38 mph
Additional Background
The Shotput
The shotput event requires strength and coordination.
The athlete must throw a 16-lb steel ball from a pit out
into a field. The longest distance from the edge of the
pit to the first landing point of the shot is the winner. For
these calculations, we are going to neglect the effects of
aerodynamics on the flight of the shot. The only external
force acting on the shot in flight is the pull of gravity. In
Imperial units, the acceleration of gravity, g, is 32.2 ft/sec2. If
we neglect aerodynamics, the flight path of the thrown shot
is a parabola as shown in Figure 4-2. For an optimum height
and distance, the shot must be thrown at a
45-degree angle to the vertical or the horizontal.
20
}
© JETS, Inc. TEAMS Competition 2008
plane parallel to the ground, much like a big league curveball,
or a golf hook or slice. Soccer players call this flight path a
“bend” and it is produced by aerodynamic forces on the ball.
To produce the bend, the ball must be kicked off-center so
that the ball is made to spin. The equation for the force on
the ball is given by:
Figure 4-2.
(4-3)
The total time of the flight is given by the following equation:
t = 2 × Vv / g
(4-2)
where,
Figure 4-3.
Vv = the vertical velocity component
g = the gravitational acceleration
D = the distance traveled and is equal to the time of
the flight multiplied by Vh the horizontal velocity
component
36. The equation that relates the launch velocity V to the
distance D is most nearly: (sqrt indicates the square root
function.)
a. V = D × g / 4
b. V = sqrt ( 2 × D × g)
c. V = sqrt (D × g / 2 )
d. V = sqrt (D× g)
e. V = D× g
37. To set a new men’s Olympic and world record distance of
80 ft, the shot must leave the athlete’s hand at a speed
(expressed in feet/sec) of most nearly:
a. 50.75 ft/sec
b. 644.0 ft/sec
c. 25.37 ft/sec
where,
F
r
s
d
V
The diameter of a soccer ball is 9 in., and
the density of the air in Imperial units is .00237 slug/ft3
1 slug = lb-sec2/ft.
38. The amount of force in pounds that is generated on a
soccer ball if the ball is flying 50 mph, and spinning at
60 revolutions per minute is most nearly: (Be very
careful about your units!)
a. .33 lbs
b. .48 lbs
c. 1.28 lbs
d. 32.2 ft/sec
e. 80.0 ft/sec
Additional Background
Soccer
One of the most exciting plays in Olympic soccer is the corner
kick. This play occurs when a member of the defending team
kicks the ball beyond their own end line. The ball is then
given to the attacking team and placed on the ground in the
corner formed by the sideline and the end line. One of the
attackers then kicks the ball to the front of the goal where
her teammates attempt to head or kick the ball into the goal.
During the corner kick, the ball can be made to curve in a
= the side force
= the radius of the ball
= the spin rate
= the air density
= the velocity of the ball
d. 3.86 lbs
e. 29.9 lbs
Additional Assumptions and Givens
•
The force on the ball causes the ball to accelerate in the
direction of the force according to Newton’s second law
of motion:
Force = mass × acceleration
F = m × a
(4-4)
21
© JETS, Inc. 39. If the weight of a soccer ball is 1 pound, and the side
force is 2 lbs, the amount of acceleration the ball
experiences is most nearly:
a. 16.1 ft/sec2
b. 32.2 ft/sec2
c. 64.4 ft/sec2
R=
V2
a (4-5)
where,
d. 96.6 ft/sec2
e. 9.8 ft/sec2
V = the velocity of the ball
a = the acceleration as computed in question #39
R = the radius of curvature
40. If the side force on the ball is ½ lb, and the velocity is
50 mph, the radius of curvature of the flight path
expressed in feet is most nearly:
Additional Background
The ideal flight path of the ball is a circular arc as shown in
Figure 4-4:
a. 4.55 ft
b. 155 ft
c. 77.6 ft
Figure 4-4.
}
TEAMS Competition 2008
d. 334 ft
e. 166 ft
2008 Competition
Behind the Scenes: Athletic Events
Part 1 Solution
Aerodynamic Principles Applied
to Olympic Games
Problem #4
Aerodynamic Principles Applied to Olympic Games
31.
Answer: E
Original Radius = 75m
Area of Circle = (3.14)(75m) 2 = 17,671m 2
⎛ 40 ⎞
2
2
Area of Segment = ⎜
⎟(17,671m ) = 1,963m
360
⎝
⎠
New Radius = 90m
Area of new circle = (3.14)(90m) 2 = 25,446m 2
⎛ 40 ⎞
2
2
Area of new segment = ⎜
⎟(25,446m ) = 2,827m
360
⎝
⎠
Additional Area = 2,827m 2 −1,963m 2 = 864m 2
32.
Answer: C
Outer Radius = 100mm
Area-1 = (3.14)(100mm) 2 = 31,415mm 2
Inner Radius = 10mm
Area 2 = (3.14)(10mm) 2 = 314mm 2
Area of One Donut
= (Area-1) – (Area-2)
= 31,415mm 2 −314mm 2
= 31,101mm 2
Area of Two Donuts = 62,202mm 2
33.
Answer: B
Weight = Lift (L) = 19.6 N
L = (0.5)(D)(V) 2 (A)(cl)
where,
D = 1.22 kg/m 3 = Density of air
A = surface area of top of disk:
=
(π )(d) 2 (3.14)(0.22m) 2
=
4
4
= 0.038m 2
cl= coefficient of lift = 1
V=
19.6N
(0.5)(1.22 kg/m 2 )(0.038m 2 )
V = 29.07m/s
34.
Answer: D
⎛ 100m ⎞⎛ 39.37in ⎞⎛ 1ft ⎞⎛ 1mile ⎞⎛ 60s ⎞⎛ 60min ⎞
⎟⎜
⎜
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 9.77s ⎠⎝ m ⎠⎝ 12in ⎠⎝ 5,280ft ⎠⎝ min ⎠⎝ hr ⎠
Average Speed = 22.89 mph
35.
Answer: C
Speed =
Distance
Time
Average Speed =
400m
(11.23s) + (10.71s)(10.59s)(10.41s)
= 9.315m/s
⎛ 60s ⎞⎛ 60min ⎞⎛ 39.37in ⎞⎛ 1ft ⎞⎛ 1mile ⎞
(9.315m/s) ⎜
⎟
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎝ min ⎠⎝ hr ⎠⎝ m ⎠⎝ 12in ⎠⎝ 5,280ft ⎠
Average Speed = 20.837 mph
36.
Answer: D
Key factor here is the 45 degrees angle.
Therefore,
V h = Vv
V = Vh⋅ 2
V = V v⋅ 2
2
2
V 2 = 2V = 2 V
h
v
2
2 V2
V =V =
h
v
2
t=
(2)(Vv )
g
D = (V h )(t) =
D=
2(Vh )(Vv )
g
(2)(Vh ) 2
g
(V h ) 2 =
(D)(g)
2
V 2 = (D)(g)
V=
37.
(D)(g)
Answer: A
Based on the derivation of question #36:
V=
(D)(g)
where,
D = 80ft
g = 32.3ft/s 2
Therefore,
V=
(80ft)(32.3ft/s 2 )
V = 50.75 ft/s
38.
Answer: B
F = (4/3)(4)( π 2 )(r 3 )(d)(V)
r=
(9in)(1ft/12in)
= 0.375ft = Ball radius
2
s = spin rate = 60 revs/min = 1 rev/sec
d = air density = 0.00237 slug/ft 3
V = ball velocity
= (50mph)(1hr/3,600s)(5,280ft/mile)
= 73.33ft/sec
F = (1rev/sec)(4/3)(4)(3.14 2 )(0.375ft) 3 (0.00237 slug/ft 3 )(73.33ft/sec)
= 0.48 slug – ft/sec 2
= 0.48 (lb-sec 2 /ft)(ft/sec 2 )
F = 0.48 lb
39.
Answer: C
Force = (Mass)(Acceleration)
= (m)(a)
= (W/g)(a)
a=
(F)(g)
W
F = 2 lb
W = 1 lb
g = 32.3 ft/s 2
Therefore,
a = 2g = 64.4 ft/s 2
40.
Answer: D
V = 50 mph = 73.3 ft/s
R = radius of curvature
a=
=
(F)(g)
W
(0.5lb)(32.3ft/s 2 )
1lb
= 16.1 ft/s 2
Therefore,
R=
V2
a
R=
(73.3 ft/s) 2
16.1ft/s 2
R = 333.7 ft
Behind the Scenes:
Athletic Events
Part II
TEAMS 2008
Sponsors:
S.D. Bechtel, Jr.
Foundation
Shell
Tyco Electronics
JETS Sponsors:
Bechtel Group
Foundation
CH2M Hill
Motorola Foundation
NISH
Rockwell Collins
Shell is the exclusive sponsor of the
TEAMS 2008 T-Shirt Design Challenge
SolidWorks
United Engineering
Foundation
Winning T-Shirt Design submitted by Teresa Lieh,
Beavercreek High School, Beavercreek, OH
1
{
extension to problem # 4
Aerodynamic Principles Applied to
Olympic Games
Submitted by Tom Benson, NASA Glenn Research Center,
Cleveland, OH
Problem Statement
Background
2008 is an Olympic year and nations will gather in
Beijing, China to celebrate the Olympic Games. To help
America “bring home the gold,” NASA has been asked
to provide mathematical and technical assistance to
various Olympic sports committees. Your engineering
team has been asked to help NASA and the U.S.
Olympic team. The Olympic Soccer Team has requested
some additional help in making better corner kicks.
Let’s develop some theories about the physical
mechanics of the corner kick using algebra and some
geometry. We can then make intelligent suggestions
about improving the kick.
An ideal corner kick produces a curved flight path due
to aerodynamic forces on the ball. Figure 4-1A below
shows a sketch of the corner kick with the flight path
given by the thick dashed line. The solid vertical and
horizontal lines denote the side line and end line of the
field. The corner and the near goal post are also noted
on the figure. The kicker strikes the ball off-center to
impart a spin to the ball. As the spinning ball moves
through the air, an aerodynamic force is generated that
bends the flight path.
Figure 4-1A.
5
There are mathematical equations that describe various
aspects of the problem. The radius of curvature of the flight
path is related to the velocity of the ball and acceleration
produced on the ball by the aerodynamic force, as shown on
the figure. The acceleration is the result of Newton’s second
law of motion when an external force is applied to a mass.
The aerodynamic force is defined by a complex equation
relating the spin, size of the ball, velocity, and air density.
Some of the parameters in these equations are constants
and some are variables. The size and mass of the ball does
not change. The spin rate, density of the air, and velocity can
change during a soccer game.
Task 1
Derive a single equation that relates the radius of curvature
to the problem variables. What does the single equation tell
you? How does the radius of curvature change with velocity?
With spin of the ball? With density of the air?
6
Task 2
The density of the air decreases with altitude. Beijing is
located at a mean height of 145 ft. above sea level. If the
American soccer team trained in Denver (5,280 ft. altitude)
how would the practice corner kicks differ from game
conditions?
Task 3
Because the ball curves in flight, the kicker aims at a spot
in front of the goal during the kick. The angle between the
initial flight path and the end line is designated angle b as
shown in the figure. Using concepts from geometry, derive
the equation that relates the angle b to the length L of the
kick and the radius of curvature R. (The radius of curvature is
perpendicular to the flight path.)
Task 4
What are the limits imposed on the relative size of the
variables from the equation derived in Task 3?
2008 Competition
Behind the Scenes: Athletic Events
Part 2 Solution
Aerodynamic Principles Applied
to Olympic Games
Part II
Extension to Problem #4
Aerodynamic Principles Applied to Olympic Games
TASK 1:
Since parameters r and m along with constant π do not change, we group them into
single constants k 1 , k 2 and k3.
F=
4
(4π 2 r 3 s d V)
3
= k1 s d V
Also,
a=
F
= k2 s d V
m
R=
k V
V2
V2
=
= 3
a
k2 sd V
sd
Therefore, the radius of curvature (R) varies linearly with velocity.
The higher the velocity, the larger the radius of curvature.
The higher the velocity the straighter the path of the ball.
The radius of curvature (R) varies inversely with the spin of the ball (s).
Therefore, the greater the spin, the smaller the radius of curvature, and the greater the
curve of the path.
The radius of curvature (R) varies inversely with the density of the air.
The greater the density, the smaller the radius of curvature, and the greater the curve of
the path.
TASK 2:
The lower air density at Denver produces a smaller aerodynamic force and a larger radius
of curvature.
A larger radius of curvature (R) means that the ball would go straighter in Denver for the
same amount of spin and the same velocity.
In Beijing, the ball will curve more.
For a very precise game, like soccer, this could cause some problems for the team.
The ball won’t go where one has been practicing for it to go.
The team would do better to train at a location near sea level…like Beijing.
TASK 3:
The angle (b) that a player strikes the ball is perpendicular to the other angle (b) of the
radius of curvature (R). See figure above.
Construct a line perpendicular to the end line that also forms an equal angle (b) with the
radius.
The line through the center of the circular are, perpendicular to the end line, cuts the end
line in half and produces two (b) angles.
Since the constructed line through the center is perpendicular to the end line, we have a
right triangle formed and therefore:
sin(b) =
L
2
R
=
L
2R
TASK 4:
Since sin(x) can never be greater than 1,
sin(b) ≤ 1
L
≤1
2R
R ≥ L/2
This means, that if the combination of spin, air density and velocity (as shown in Task 1)
produces a value for the radius of curvature that is smaller than L/2, then the ball can
never go into the goal area.