A team of engineering students is writing a MATLAB program to simulate bungee ju
ID: 3684701 • Letter: A
Question
A team of engineering students is writing a MATLAB program to simulate bungee jumping for a bungee jumping equipment company.
A new bungee cord material is made of highly specialized rubber with a linear stress vs. strain ratio throughout the entire 400% elongation.
The company wants to ensure the maximum acceleration during the jump do not exceed 2 g’s (1 g = 9.806 m / sec2). At the same time, the company wishes to sell bungee cords that can provide adventurous jumpers with an ultimate bungee diving experience - the bungee cord that can cause the peak upward acceleration to be as close to 2 g’s as possible.
Simulate the bungee jumping motion and utilize the simulation to design the length (L) of the bungee line that will cause the peak upward acceleration to be close to 2 g’s.
The bungee jumpers usually make high-altitude jumps. The purpose of the simulation is also to calculate a safe jump altitude for this new bungee cord so that the jumper will not hit the ground.
Average bungee jumpers’ mass is estimated to be 80 kg (176.37 pounds). The equation to use for the analysis is Newton’s Second Law,
F = ma
where F is the sum of the gravitational, aerodynamic drag, and bungee forces acting on the jumper, m is the mass of the jumper (which is 80 kg), and a is the acceleration. Define the distance the jumper falls as the variable x (which is a function of time, x(t)). The jumper’s velocity and acceleration are then represented as x’ and x’’, respectively. The Newton’s equation to solve for acceleration:
x’’ = F / m
Next, determines the forces making up F. The gravitational force will be the jumper’s weight, which is:
W=mg
= (80 kg) (9.806 m/s2) = 784.48 N
The aerodynamic drag, D, will be proportional to the square of the jumper’s velocity, D = c (x’)2, but the value of c, the constant of proportionality, is unknown. However, experienced skydivers know that if the jumper dives head down so that the entire body is streamlined into the wind to minimize drag, the diver can reach a speed of about 134 m/s (300 mph). At that speed, the aerodynamic drag is equal to the weight of the jumper, so c can be determined using:
c = D / (x’)2
= (784.48 N) / (134 m/s)2 = 0.04369 kg/m
Finally, after the jumper has fallen beyond the bungee cord length, the slack in the bungee will be eliminated, and it will begin to exert an arresting force, B, of 10 N for every meter that it is stretched.
The new bungee cord also has a viscous friction force, R, once it begins to stretch, which is given by: R = - 1.5 x’
Thus, there will be two regions for computing the acceleration. The first equation will be used when the distance x is less than or equal to the bungee cord relaxed length (L):
x’’ = F / m = ( W – D) / m = (784.48 – 0.04369 (x’)2 ) / 80
A second equation will be used when x is greater than bungee cord relaxed length (L):
x’’=F/m=(W–D–B-R)/m=(784.48–0.04369(x’)2 –10(x–L)–1.5x’)/80
Integrate for the interval beginning at 0 seconds to find the acceleration, velocity, and distance as a function of time from the beginning of the jump (which is assumed to occur at t = 0) to 500 seconds after the jump.
What is the bungee cord length in your design?
What are the peak values of acceleration, velocity, and distance based on your design? You may need to experiment with the simulation step size in order to choose one that will give you the peak values. A negative acceleration or negative velocity value means the jumper is going up. Remember to check the peak values for both up and down directions.
(Hint: You may want to utilize the min(x) or max(x) MATLAB functions to find the minimum or maximum values in your vector arrays.)
The company does not want the maximum acceleration to exceed 2 g’s (1 g = 9.806 m / sec2).
Is the simulation of the peak acceleration higher or lower? How close to 2 g’s does your design provides?
How far does the jumper initially fall before he starts back up?
How many seconds does he initially fall before he starts back up?
How high should the company advertise to be the minimum altitude to jump with this new bungee cord to ensure a safety factor of 2? (Hint: twice the longest distance the bungee jumper falls away from the jump off point.)
At the end of the 500 seconds simulation, what is the final length of the bungee cord? Simulation:
? Create one Simulink embedded function model to simulate the bungee jumper’s distance (x) vs. t, the velocity (x’) vs. t and acceleration (x’’) vs. t for the first 500 seconds of the jump.
? Adjust the Simulink model maximum step size to ensure the simulation calculates enough points to obtain the maximum and minimum values during the simulation.
? Use the “To File” blocks in the Simulink model to save the simulation result, distance (x) vs. t, the velocity (x’) vs. t and acceleration (x’’) vs. t, to a .mat file.
? Create a MATLAB program that will automatically activate the Simulink model and run the Simulink model. Load the .mat file generated by the Simulink to the MATLAB program.
? Use the ode45 method to simulate the bungee jumper’s distance (x) vs. t, the velocity (x’) vs. t and acceleration (x’’) vs. t for the first 500 seconds of the jump again.
? Generate one 3x2 subplots to plot and compare the Simulink solution and the ode45 solution side-by-side.
The figure should include the following:
o Plot the bungee jumper’s distance (x) vs. t chart with the Simulink Solution on the left,
and plot the x vs. t chart using the ode45 solution on the right.
o Plot the bungee jumper’s velocity (x’) vs. t chart with the Simulink Solution on the left, and
plot the x’ vs. t chart using the ode45 solution on the right.
o Plot the bungee jumper’s acceleration (x’’) vs. t chart with the Simulink Solution on the left,
and plot the x’’ vs. t chart using the ode45 solution on the right.
? Answer all the simulation analyses questions with the fprintf commands and print key
analyses parameters and answers on screen or to a text file.
? Title and label plots clearly.
? Add lots of comments. Make this simulation a demo worthy program for interviews.
Figure1 File Edit View Insert Tools Desktop Window Help Simulink Solution ODE 45 Solution a 500 500 0 0 Time Simulink Solution Time ODE 45 Solution 100 100 -100 -100 0 600 0 Time Simulink Solution Time ODE 45 Solution -20 -20 600 0 600 Time TimeExplanation / Answer
Hi below i have given a sample MATLAB code for your reference,
% bungee MQP matlab program
% plot the displacement and velocity of the body through bungee process
w(1,1)=0; % displacement of the body
w(2,1)=0; % velocity of the body
ti=0; %initial time
tf=60; % final time
n=100000; % number of steps
h=(tf-ti)/n; % change of time for each step
length=20; % string length
T=0:(tf/n):tf; % time
for i=1:n;
if w(1,i)<length; % dynamic part
t=ti+h*(i-1); % RungeKutta method of order four
k11=h*fun1(t,w(1,i),w(2,i));
k12=h*fun2(t,w(1,i),w(2,i));
k21=h*fun1(t+h/2,w(1,i)+k11/2,w(2,i)+k12/2);
k22=h*fun2(t+h/2,w(1,i)+k11/2,w(2,i)+k12/2);
k31=h*fun1(t+h/2,w(1,i)+k21/2,w(2,i)+k22/2);
k32=h*fun2(t+h/2,w(1,i)+k21/2,w(2,i)+k22/2);
k41=h*fun1(t+h,w(1,i)+k31,w(2,i)+k32);
k42=h*fun2(t+h,w(1,i)+k31,w(2,i)+k32);
w(1,i+1)=w(1,i)+1/6*(k11+2*k21+2*k31+k41);
w(2,i+1)=w(2,i)+1/6*(k12+2*k22+2*k32+k42);
else % vibration part
t=ti+h*(i-1);
k11=h*fun3(t,w(1,i),w(2,i));
k12=h*fun4(t,w(1,i),w(2,i));
k21=h*fun3(t+h/2,w(1,i)+k11/2,w(2,i)+k12/2);
k22=h*fun4(t+h/2,w(1,i)+k11/2,w(2,i)+k12/2);
k31=h*fun3(t+h/2,w(1,i)+k21/2,w(2,i)+k22/2);
k32=h*fun4(t+h/2,w(1,i)+k21/2,w(2,i)+k22/2);
k41=h*fun3(t+h,w(1,i)+k31,w(2,i)+k32);
k42=h*fun4(t+h,w(1,i)+k31,w(2,i)+k32);
w(1,i+1)=w(1,i)+1/6*(k11+2*k21+2*k31+k41);
w(2,i+1)=w(2,i)+1/6*(k12+2*k22+2*k32+k42);
end;
end;
subplot(2,1,1); plot(T,-w(1,:),'r'), grid; xlabel('Time(s)');
ylabel('Displacement(m)');hold on % plot displacement
subplot(2,1,2); plot(T,-w(2,:),'r'), grid; xlabel('Time(s)');
ylabel('Velocity(m/s)'); hold on % plot velocity
for i=1:n+1;
if (w(1,i)-length)<=4.88
k(i)=150;
else
k(i)=150;
end; end;
for i=1:n+1; i
f w(1,i)<=20;
extend(i)=0;
else
extend(i)=w(1,i)-length;
end; end;
stress=times(k,extend)/area;
plot(T,stress,'r'), grid; xlabel('Time(s)'); ylabel('Stress(N/m^2)');
hold on % plot stress
function f=fun1(t,x1,x2)
f=x2;
end
function f=fun2(t,x1,x2)
m=80;
g=9.8;
c=15;
if x2>=0
f=(m*g-c*x2^1)/m;
else
f=(m*g-c*x2^1)/m;
end
function f=fun3(t,x1,x2)
f=x2;
end
function f=fun4(t,x1,x2)
m=80;
g=9.8;
c=15;
length=20;
if (x1-length)<=4.88
k=150;
else
k=150;
end
if x2>=0
f=(m*g-c*x2^1-k*(x1-length))/m;
else
f=(m*g-c*x2^1-k*(x1-length))/m;
end
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