How big was the aerodynamic force on Felix Baumgartner at 1342 km/h? (Or on Alan Eustace at 1321 km/h?)

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Did you watch Felix Baumgartner jump out of a balloon gondola at an altitude of 39 km? Here's a question for your physics students, or even your physics teacher! How big was the drag force on Felix Baumgartner at his maximum speed of 1342 km/h? Pretty enormous, right? Wrong.

Big Ideas: 
  • Free fall
  • Maximum velocity means zero acceleration
  • Drag

Felix Baumgartner recently jumped out of a balloon gondola at an altitude of 39 km. He attained a maximum speed of descent of 1342 km/h.

A object falling in a vacuum accelerates in the gravitational field. The acceleration is simply g. Falling in air, the object experiences a drag force that rises as the velocity rises, until the drag force equals the weight of the object and the object's acceleration falls to zero.

So the simple answer to the question posed in the title is "his weight". In an atmosphere of uniform density, when the object reaches a steady terminal velocity which it will maintain until it hits the ground. However, when free falling from an altitude of 39 km, as Felix Baumgartner did recently, maximum velocity is reached quite early in the thin upper atmosphere. Then the velocity slows as denser, lower air is reached. It is during this deceleration phase, that the maximum aerodynamic drag is experienced. To determine how big the maximum force is, we need a mathematical model of the fall. Making no pretense at a detailed understanding of the aerodynamics of a body of ill-defined shape (i.e. a human) crossing the sound barrier, let us try what we call a "toy model" with the intention that it be just sophisticated enough to understand how the different physical effects involved play against each other and to give results "in the right ball-park". We will need known data or an educated guess at the following: The initial conditions of the fall - Altitude = 39.045 km (precisely known) - Mass of human and suit = 118 kg - Approximate projected area for the attitudes planned during the fall, say 1 m2 for the initial dive to maximum velocity, 1.5 m2 thereafter.

The forces at play: #1 Gravity

The acceleration due to gravity g' at altitude h is somewhat less than that at the surface, g. Newton tells us how to model this, if we know the Earth's radius RE:

 $ g' = g(\frac{R_E}{R_E+h})^2 $

 The forces at play: #2 Aerodynamic Drag 

At subsonic speeds the form of the aerodynamic drag Fdrag on a body of projected area S, drag coefficient CD, at speed v in a fluid of density ρ is straightforward:

 $ F_{drag} = \frac{1}{2} C_D \rho S v^2 $

 However, CD and S are hard to get a handle on in this case as the human shape and attitude is not well defined. For the initial high speed fall, we choose CD = 0.2 (the value for a sphere the size of the human at around 10 m/s and up), and S = 0.85 m2. As the speed approaches that of sound (300 m/s at high altitude) CD rises rapidly. Consulting White's textbook, we let CD rise smoothly from the subsonic 0.2 under Mach 0.7 to CD=0.9 at Mach 0.9 and above. For the subsequent subsonic fall, as Baumgartner adopted a more traditional sky-diving attitude, we use the value of CD for a flat plate, with a surface at 90o to the flow, of 1.0. These numbers yield values of CDS that closely resemble those given in a Colino and Barbero's more detailed analysis. To find ρ we use a simple double exponential model of the Earth's atmosphere:

$ \rho/\rho_0 = exp(-h/h_0) $

with a "scale height" of h0 = 10 km below 8.5 km altitude and 6.45 km above. The values of ρ0 are 1.2 kg/m3 below 8.5 km and 2 kg/m3 above.

Modelling

To calculate altitude, velocity and acceleration as a function of time t dropping in a varying atmosphere with velocity-dependent drag we need to invoke a finite-difference method, specifically the conceptually simple Verlet integration. Using this approach, the position at any time can be found if the positions at two previous times (separated by a small time interval Δt) and the acceleration are known. The initial position is just the balloon altitude, and the first step can be found with straightforward Physics 11 kinematics as the velocity and therefore the drag force are negligibly small. With the first two positions and therefore the velocity found, the acceleration can be determined from the velocity-dependent drag force. This gives position 3, and so position 4 and all subsequent positions can be found by iteration. Through the magic of modern computers, we can rapidly step second by second through the fall, calculating everything we want to know along the way. How do we know the calculation and our assumptions are at least reasonable? To check the results, we know from press reports a few important details about Baumgartner's fall

- reached Mach 1 after 34 seconds

- maximum velocity: 1342 km/h

- time to parachute opening: 259 seconds; altitude of parachute opening: 2.5 km.

Results

   

     

As can be seen in the above figures, the model does a reasonable job of reproducing the three pieces of numerical data available. Of course, we could fine-tune the drag coefficients to reproduce the data to arbitrary precision, but this would be a pointless exercise. The model gives a maximum aerodynamic force of about 1900 N, or 1.6g, non-trivial but hardly any worse than rounding the bottom of a roller-coaster ride.

But if the size of the force doesn't surprise, look at the power dissipation; it peaks at over a half a megawatt. Baumgartner had almost 50 MJ of potential energy to get rid of in a only a few minutes. Its a good thing he had a stiff breeze to keep him cool.

Update October 2014

Alan Eustace's recent fall beat Baumgartner's altitude record, but not his speed (Eustace topped out at 1321 km/h). Eustace's total mass was larger (181 kg), so his frontal area must have been larger (according to this model, c. 1.5 m2).

References

http://www.huffingtonpost.com/2012/10/14/felix-baumgartner-jump-redbull-...

http://en.wikipedia.org/wiki/Verlet_integration

Frank White, "Fluid Mechanics", McGraw-Hill.

Donald G. Miller and Allan G. Bailey, "Sphere drag at Mach numbers from 0-3 to 2-0 at Reynolds numbers approaching 10^7", J. Fluid Mech. (1979), vol. 93, part 3, pp. 449-464

J. M. Colino and A. J. Barbero, "Quantitative Model of stratospheric freefall", Eur. J. Phys. 34 (2013) 841-848.

http://www.bbc.com/news/world-us-canada-29766189 (Eustace fall).

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