1. No category

Modeling a Bungee Jump
Nicole Kelly
May 17, 2013
Abstract
The purpose is to model a bungee jump using differential equations and Matlab.
Given a set of predetermined parameters, an appropriate k-value, or a value
representing the relative stiffness of a spring, will be determined for a bungee
rope. The model will be examined with and without a damping force and with
different k-values.
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Introduction
When a bungee rope hangs straight it is at it’s equilibrium length. In this
example the bridge is 300 feet tall and the rope is 100 feet long. The equilibrium
length to be at x=0, therefore the bridge is at x=-100 feet and the river below
the bridge is located at x=200 feet.
For the first part of the jump, the jumper is in free fall. This means that
the only forces acting upon the jumper are that of gravity and air resistance.
However, once the jumper reaches the equilibrium point the rope itself is also
acting on him/her to pull the jumper in the opposite direction of gravity. This
is modeled by the equation
0
x<0
r(x) =
−kx
x≥0
where k is the spring constant of the rope used in the bungee jump.
The sum of the forces acting on the jumper can be written as the differential
equation mx00 = mg + r(x) − ax0 where m is the mass of the jumper, g is the
force of gravity, r(x) is 0 or -kx depending on the conditions listed above, and
-ax’ represents the damping force acting on the jumper.
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Parameters
The assumption is that the person attempting the bungee jump weighs 160 lbs.
and is 6 feet tall. 32f t/s2 will be used for gravity.
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3.1
No Damping
x<0
When there is no damping present ’a’ in the differential equation becomes 0
and the equation is left as mx00 = mg + r(x). When x is less than zero that
reduces the equation even further to mx00 = mg. Adding in the parameters this
becomes
x00 = 32.
Assume that the jumper falls straight off the bridge without jumping, in which
case the initial velocity is 0. Given the initial conditions x(0) = −100 and
x0 (0) = 0 one can integrate twice to easily find the solution
x = 16t2 − 100.
Next is to find at what time, ’t’, the equation is equal to 0, or in other words
at what time the rope reaches equilibrium and begins to act on the jumper.
Setting x equal to 0 gives
0 = 16t2 − 100
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, which can be solved algebraically for t to find that t = 2.5. Therefore the
conclusion is reached that the rope reaches its equilibrium point 2.5 seconds
into the jump.
3.2
x≥0
When x ≥ 0 the differential equation becomes x00 = 32 − 15 kx. One method of
solving this inhomogeneous, linear, autonomous differential equation is to find
a homogeneous and particular solution and combine them to obtain a general
solution. Initial conditions will be needed later on, and for this t=2.5 will be
plugged into the equations for position and velocity obtained earlier. By doing
this it is concluded that x(2.5) = 0 and x0 (2.5) = 80. The homogeneous solution
is
r !
r !
k
k
+ C2 sin
.
xh = C1 cos
5
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For the particular solution the guess used is xp = a. Taking the first and second
derivative and substituting into the original equation, one is now able to solve
for ’a’ so that
160
.
xp =
k
This means that the general solution is
r !
r !
160
k
k
x(t) =
+ C1 cos
+ C2 sin
.
k
5
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After determining C1 and C2 by using the initial conditions the solution becomes
!
!
r
r
r
160 160
k
5
k
x(t) =
−
cos
(t − 2.5) + 80
sin
(t − 2.5) .
k
k
5
k
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The jumper wants to just barely touch the top of his head to the river below
at x=200. When the jumper comes to rest the forces are in balance so that
32 − 15 kx = 0, which means x = 160
k . If the jumper wants to just barely touch
the water under the bridge, the difference between this value and how far the
rope needs to stretch must be determined. The bridge is 300 feet tall, and with
a 100 foot rope and a 6 foot jumper the rope will need to stretch 194 feet. The
k
amplitude of the oscillation then becomes 194 − 160
. This can now be used to
solve for k. Substituting all of the values into the equation we get
r
160
160
5
194 −
=
+ 80
.
k
k
k
Using various algebraic techniques we find that k=2.5. This means that to make
a successful jump the rope will need to have a k-value of at least 2.5 pounds/foot.
2
4
With Damping
Air resistance and the bungee cord itself will dampen the oscillation of the
jumper, causing him to eventually come to a resting position instead of bouncing
on the bungee cord indefinitely. The dampening force will be impacted by many
different factors and in reality is hard to pinpoint an actual value for it. To avoid
making the problem unnecessarily complicated it is assumed that the dampening
constant is 1. This means that the equation with a=1 is
1
1
x00 = 32 − kx − x0 .
5
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Using the k-value obtained for the situation without damping, the jumper
will still be able to make a safe jump. This is because in the case without
damping the jumper just barely touches their head to the stream on the very
first osciallation and each succeeding bounce. With damping present the jumper
will not come as close to the water on the first jump and each succeeding bounce
will be closer to the equilibrium point. In order to illustrate the position and
velocity of the jump Matlab’s pplane8 function will be used. To use the pplane
the function must be converted from a second order differential equation to a
system of first order differential equations. The conversion results in
x0 = v
1
1
v 0 = 32 − kx − v.
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Using the parameter k = 2.5 ∗ (x >= 0) + 0 ∗ (x < 0) and the initial condition
x(0) = −100 and v(0) = 0 the figure illustrated below is obtained.
Now it will also be useful to see the graph of position and velocity separately
in relaltion to time. Pplane8 gives the option to plot both the position and
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velocity with repsect to time on the same graph, which is how figure 2 was
obtained.
From the graph it can be seen that this k-value allows the jumper to stay below
x=200 (or above the stream below the bridge). As time goes on the oscillations
get smaller until the jumper essentially comes to rest at approximately 160 feet
below the bridge.
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More Matlab Graphs
Now that the model of the bungee jump with damping has been obtained, it will
be useful to compare what the jump would look like if there were no damping,
as was examined in section 3. This case is shown in the graph below, where
one can see that the jumper just barely misses hitting the water at x=200.
In addition this case is different because the oscillations all have a constant
amplitude. Without damping the jumper would bounce continually.
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It would also be useful to see what effect different k-values would have on
the model. Using the case with damping we will use pplane8 to graph the path
of the jumper when the k=1.5, k=3.5, and k=6.
The above graph illustrates the position of the jumper with a bungee cord having
a k-value of 1.5 pounds/foot. All other parameters remain the same as those in
section 4. From the graph it is evident that the lower spring constant allows the
jumper to fall below the x = 200 mark. This means that the jumper would hit
the water below instead of just barely touching it. More than likely this would
not be ideal in a jump, especially when the jumper is plunging 300 feet off a
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bridge. A miscalculation that produces a k-value that is too low for a given
jump could result in serious injury or fatality.
The above figure illustrates the same jump with a k-value of 3.5. This is only
slightly higher than the predetermined value of k=2.5 from section 4. In this
case, the jumper would still be able to make a safe jump, but having the higher
k-value makes the cord stiff enough that the jumper does not fall as far as the
other two cases that have been examined. The jumper will only make it to
approximately x = 110, 90 feet short of the stream below.
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The graph on the bottom of the previous page illustrates the jump when k = 6
pounds per foot. In this case the jumper does not even reach 100 feet below the
equilibrium point. While this is still a relatively safe bungee jump, the jumper
may be looking for more of an adrenaline rush than this bungee cord would
provide. If the k-value was much higher the bungee cord might be so stiff that
not only would the jumper not fall very far, but it would pull the jumper back
too quickly, making for an unpleasant jump.
Looking at each case one can deduce that the higher the k-value, the stiffer
the rope. With a very low k-value, such as 1.5 pounds per foot, the jumper
would hit the water below, which could be dangerous and unpleasant. With
a very k-value, such as 6 pounds per foot, the jumper will hardly go past the
equilibrium point of the rope before being yanked upwards again, which might
not sound appealing to the jumper. For an optimal jump a rope with a kvalue around the predetermined 2.5 pounds per foot or slightly higher would be
appropriate.
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Conclusion
In the case with no damping, given the parameters, a bungee with a k-value
of at least 2.5 pounds per foot would be necessary for a safe jump. However,
this case is not realistic because without damping the jumper would oscillate at
the same amplitude continuously. Realistically there is air resistance and other
forces that will essentially bring the jumper to rest at some time. Using the
k-value determined in the no damping case one can see that the jumper can still
make a safe jump and eventually be brought to equilibrium. Looking at other
values of k can determine what values are too low or too high for a jump that
is both safe and exhilarating.
References
[1] The
Mathematically
Inclined
Blog
A
Guide
to
Bungee
Jumping
Wordpress.org,
March
5,
2009.
http://mathematicallyinclined.wordpress.com/2009/03/05/a-guide-tobungee-jumping/
[2] Internet Differential Equations Activities Bungee Jumping National Science
Foundation http://www.idea.wsu.edu/Bungee/
[3] Adkins, Samantha A Mathematical Representation of the Motion of
a Bungee Jumper Oklahoma State University, October 14, 2002
http://www.math.okstate.edu/ wolfe/Classes/BUNGEE.pdf
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