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.
1
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
r(x) =
0 −kx
x < 0 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 mx = mg + r(x) − ax 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. 32ft/s2 will be used for gravity.
3 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 mx = mg + r(x). When x is less than zero that reduces the equation even further to mx = mg. Adding in the parameters this becomes x = 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 x (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 x = 32 − 51 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 b e plugged into the equations for position and velocity obtained earlier. By doing this it is concluded that x(2.5) = 0 and x (2.5) = 80. The homogeneous solution is k k xh = C 1 cos + C 2 sin . 5 5
For the particular solution the guess used is x p = a. Taking the first and second derivative and substituting into the original equation, one is now able to solve for ’a’ so that 160 x p = . k This means that the general solution is 160 x(t) = + C 1 cos k
k 5
+ C 2 sin
k 5
.
After determining C 1 and C 2 by using the initial conditions the solution becomes 160 x(t) = k
−
160 cos k
k (t − 2.5) 5
+ 80
5 sin k
k (t − 2.5) . 5
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 − 51 kx = 0, which means x = 160 . If the jumper wants to just barely touch k 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 194 −
160 160 = + 80 k k
5 . 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 x = 32 − kx − x . 5 5
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
x = v 1 1 v = 32 − kx − v. 5 5 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 3
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.
4
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 5
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 Wordpress.org, March 5, 2009. Bungee Jumping 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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