The Felix Baumgartner Equation

21 10 2012

Introduction

We want to describe, even in not with a fully realistic accuracy, the motion of Felix Baumgartner while…falling from the sky as it happend on October 14, 2012. To do this, we well make use of Wolfram’s Mathematica.

The Austrian daredevil jumped from a helium ballon at an altitude of 39,014 m and opened the parachute at 1,524 m after 4 minutes and 16 seconds. During the first minute he broke the record previously held by Kittinger reaching a maximum speed of 1,342 km/h .

The Equations of motion

First of all, let’s see what happens if we model the free fall ignoring the friction due to air: in this case, the only force is gravity. To a first approximation, we consider the acceleration due to Earth’s gravity as a constant equal to 9.8 m/s^2:

m y'' = -m g

As you see, without air Felix would land after just 80 seconds at an incredible speed of 800 m/s or 2880 km/h (and btw probably choking while attempting to remove his helmet, since he’s in the vacuum). However, to make this estimation a bit more realistic we can introduce an additional force due to the presence of air, called drag:

drag(v) = \frac{1}{2}A K \rho v^2

which depends on the square of the velocity and on a set of parameters: the cross-section of the body [m^2], the density of the medium Ρ[kg/m^3] and the “drag”coefficient K which is a pure number representing the shape of the moving object.

From direct measurements we know that the density of the atmosphere does not drop smoothly as predicted by simple thermodynamical calculations (a falling exponential). Instead, it can be seen as a collection of exponentials, each describing one different layer of the atmosphere.

 

At this point we can put the average value of the density of the atmosphere in the drag formula. We also model the body of Felix as a long cylinder (k = 0.82) with a cross-section of 1 m^2 (considering all the equiments).

m y'' = - m g + \frac{1}{2}A K \rho y'^{2}

In this case, the drag force makes the velocity to reach a maximum, called limit velocity. After that, our falling guy can land (almost) safely after 6 minutes and a half at a speed of 100 m/s or 360 km/h. A bit too slow.

In most of the cases, the descriptions of this kind of motion end here. But we know that Felix set the new record in the first minutes. How was this possible? Are we ignoring some facts in our model? I think we are, definitely.

For instance, we are ignoring the not-so-small effect that the air density is not constant! If we want to add it to our model, unfortunately we cannot expect to find an analytical solution to the differential equation, but we can still make it using a numerical integration:

m y'' = - m g + \frac{1}{2}A K \rho (y) y'^{2}

In this case, a bump in the velocity can be clearly seen: this explains why the record can only be broken in the first minute and not at the end of the free fall! The maximum velocity in this case is 330 m/s or 1190 km/h, quite close to Mach 1 and in my opinion a rather good estimate of Felix’stunning feat.

One can still ask: what about gravity? Is it really constant throughout the falling? Of course not, as it has a “famous”1/r^2 dependence. However, given the relatively short path, the rate of variation in the accelaration due to gravity is very small. Anyway, we still want to check if this effect is really so small as expected:

m y'' = - m g(y)+ \frac{1}{2}A K \rho (y) y'^{2}

The final answer is: yes, the effect is very small and the two solutions are basically identical.

Conclusions

The sky dive peformed by Felix Baumgartner was a good excuse to learn something more about classical dynam- ics. We proved that ignoring the drag to to the presence of air leads to an unrealistic solution (a too short falling time). Including the drag force, we see that the diver reaches a limit velocity but we can’t still answer to the question why the record was broken in the fist minute. Then, taking into account the dependence of the air density on the altitude, we saw that the velocity reaches a maximum during the first minute and then drops even below the limit calculated with a constant density. The expected velocity of 1190 km/h can be considered a rather good estimated of what happened in reality. Unfortunately, no analytical solution can be found with standard methods in such a case.

Last but not least, we also proved that the variation of the accelaration due to gravity as a function of the altitude can be safely neglected.

Acknoledgements

Thanks to Ettore Ferranti (ABB) for the useful discussion on the interpretation of the results.

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7 responses

14 11 2012
Sergio Gonzalez

Hi!
I’m doing a proyect for my differential equations subject and im having trouble with the Felix Baumgarner model… in part it is because I’m bad with Wolfram Mathematica…
My model does not include that the gravity and air density are not constant…
Could you please send me the .nb files?? You will be my hero if you do…
My mail: sergio.xga@gmail.com

27 12 2012
Electronics Club Fans

in part it is because I’m bad with Wolfram Mathematica…http://www.hqew.net

11 05 2013
Remi

Hi,
I’m doing a project on Felix Baumgartner.
I have a problem with the equation: my” = -mg + 1/2A * K * rho (y) * (y ‘) ^ 2
I don’t know how to plot the graph of this equation with Matlab.
Can you help me please?
(it’s been several days that I try to find solution..)
My mail : remi.desmet@hotmail.com

11 05 2013
Riccardo Di Sipio

Uhm, unfortunately my knowledge of Matlab is very limited. Sorry, I think I can’t help you.

11 05 2013
Remi

No luck! Thanks for your answer.

30 11 2013
omar

what does my”=-mg represent

1 04 2014
Riccardo Di Sipio

Oh, it’s gravity! It points downward (minus sign) with an acceleration equal to g = 9.8 m/s^2

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