Felix Baumgartner Jump – October 2012
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Felix Baumgartner Jump – October 2012 In a giant leap from more than 24 miles up over Roswell, NM, a daredevil skydiver shattered the sound barrier on October 14, 2012 while making the highest jump ever – a tumbling, death- defying plunge from a balloon to a safe landing in the New Mexico desert. Felix Baumgartner hit Mach 1.25, or 844 mph, and became the first person to reach supersonic speed without traveling in a jet or a spacecraft. After hopping out of a capsule that had reached an altitude of 128,100 feet above the Earth, Baumgartner was timed at 4 minutes and 20 seconds in free fall. Landing on his feet in the desert, the man known as "Fearless Felix" lifted his arms in victory to the cheers of jubilant friends and spectators. A worldwide audience watched live on the Internet via cameras mounted on his capsule as Baumgartner, wearing a pressurized suit, stood in the doorway of his pod, gave a thumbs-up and leapt into the stratosphere. Baumgartner's descent lasted just over nine minutes, about half of it in a free fall of 119,846 feet. During the first part of Baumgartner's free fall, anxious onlookers at the command center held their breath as he appeared to spin uncontrollably. "When I was spinning first 10, 20 seconds, I never thought I was going to lose my life but I was disappointed because I'm going to lose my record. In that situation, when you spin around, it's like hell and you don't know if you can get out of that spin or not. Of course it was terrifying. I was fighting all the way down because I knew that there must be a moment where I can handle it." Baumgartner said traveling faster than sound is "hard to describe because you don't feel it." The pressurized suit prevented him from feeling the rushing air or even the loud noise he made when breaking the sound barrier. With no reference points, "you don't know how fast you travel," he said. He activated his parachute at 5,000 feet as he neared Earth, gently gliding into the desert about 40 miles east of Roswell and landing smoothly. He then was taken by helicopter to meet fellow members of his team, whom he hugged in celebration. The dive was, in fact, more than just a stunt. NASA is eager to improve its blueprints for future spacesuits. He broke the sound barrier; the highest manned-balloon flight record; and the highest altitude jump. As Baumgartner ascended, so did the number of viewers watching on YouTube; company officials said the event broke a site record with more than 8 million simultaneous live streams at its peak. After Baumgartner landed, his sponsor, Red Bull, posted a picture of him on his knees on the ground to Facebook, generating nearly 216,000 likes, 10,000 comments and more than 29,000 shares in less than 40 minutes. On Twitter, half the worldwide trending topics had something to do with the jump, pushing past seven NFL football games. Among them was this tweet from NASA: "Congratulations to Felix Baumgartner and RedBull Stratos on record- breaking leap from the edge of space!" See video: http://www.youtube.com/watch?feature=player_detailpage&v=FHtvDA0W34I Felix Baumgartner Jump Model Some models of the Baumgartner Jump have used very extensive computer fluid simulation tools in order to determine details of the flow and drag behavior. The figure on the right illustrates an example output from SolidWorks Flow Simulation. FD For our class we will pursue a simpler model employing the basic drag equation from fluid dynamics that is used to calculate the drag force experienced by an object moving through a fluid. The relation is given by 1 2 FD v CD A (1) 2 where FD is the drag force, is the fluid density, v is the velocity of the object (relative to the fluid), A is the reference area (often taken as the projected area) and CD is the drag coefficient related to the object’s geometry. Normally the drag coefficient will vary with velocity, shape, and fluid density and viscosity through a dimensionless group called the Reynolds number. However, at high speeds CD can often be assumed to be constant, and we will use this fact in our modeling. Applying Newton’s second law to Felix’s body dv F ma mg F m (2) D dt and this yields the following non-linear ordinary differential equation in terms of his body’s downward velocity v dv 1 g v 2C A (3) dt 2m D dv dv ds dv Rather than using the time dependent form, we will use the fact that v , and re- dt ds dt ds write equation (3) as dv g 1 vC A (4) ds v 2m D We choose the following values for model parameters: g = 32.2 ft/s2, m = W/g = 180/32.2, 2 A = 5 ft , CD = 0.8, Free Fall Height = 120,000 ft. Since the fluid (atmosphere) density will vary significantly over the jump, we will use a cubic function approximation for its behavior with distance s measured from the starting point. The relationship and graph are shown below. Note the earth’s surface air density is about 2.5 x 10-3 3 slugs/ft . -3 x 10 Variation of Atmospheric Density 2.5 2 3 (slugs/ft 1.5 =(9*s.2-0.00005*s.3)*0.55e-013 1 Atmospheric Density, 0.5 0 0 2 4 6 8 10 12 Distance (ft) 4 x 10 Our non-linear model equation (4) can be easily integrated numerically using the MATLAB ode45 routine, and the results are shown in the figure below. Numerical Simulation of Baumgartner Sky Diving Problem 1100 Model Prediction 1000 No Drag Prediciton 900 Mach 1 = 1100 ft/s 800 700 600 500 Velocity, v (ft/s) 400 300 200 100 0 0 2 4 6 8 10 12 Distance (ft) 4 x 10 The graph compares model predictions with the case of neglecting air resistance, v 2gs . Notice that the model does not quite predict the Mach 1.25 speed that Baumgartner reached. Perhaps our simple model over-predicts the drag force. The MATLAB code that does the calculations and makes the plots is shown below. % MCE 372 Fall 2013 % Fearless Felix Sky Diving Problem function main % Calculate and plot air density behavior clc;clear all;clf s=0:100:120000; d=(9*s.^2-0.00005*s.^3)*0.55e-013; figure(1) plot(s,d,'Linewidth',1.5) xlabel('Distance (ft)');ylabel('Atmospheric Density, \rho (slugs/ft^3'); title('Variation of Atmospheric Density');grid on text(6000,.00125,'\rho=(9*s.^2-0.00005*s.^3)*0.55e-013') % Integrate non-linear governing ODE % Need to start with a small non-zero initial condition [s,v]=ode45(@ODE,[0:100:120000],0.01); figure(2) plot(s,v,'k-','Linewidth',2) axis([0,120000,0,1100]) xlabel('Distance (ft)'),ylabel('Velocity, v (ft/s)'),grid on text(84000,850,'Mach 1 = 1100 ft/s') title('Numerical Simulation of Baumgartner Sky Diving Problem'); hold on V=sqrt(2*32.2*s); plot(s,V,'--r','Linewidth',1.2) legend('Model Prediction','No Drag Prediciton') display('explanation . ') function dvds=ODE(s,v) g=32.2;m=180/g;A=5;Cd=0.8; d=(9*s.^2-0.00005*s.^3)*0.55e-013; dvds=g/v-(d*v*Cd*A)/(2*m); For a humorous re-enactment of the jump using Lego building blocks see – http://www.youtube.com/watch?v=yFU774q6eVM .