Sunday, October 14, 2012

le Saut de Baumgartner (Baumgartner's Jump)

          "And He will raise you up on eagle's wings,
          bear you on the breath of dawn,
          make you to shine like the sun,
          and hold you in the palm of His hand."
                                                 -- Josh Grobman "On Eagle's Wings"

Felix Baumgartner completed his much pre-hyped jump from 128,100 feet, achieving a top speed 1.24 times the speed of sound.  It must have been an amazing ride, with the bill footed by Red Bull and everything.  I am neon green with envy.  Sure, it could be death to try, but it would be worth it just to see the blue-glowing world one time.  Besides, the life insurance is paid.

Gravity creates constant downward acceleration.  Friction with air produces drag that runs the opposite direction in which one is moving, and is proportional to velocity squared.  So the sum of all the forces on Felix, as with any skydiver, during his fall were equal to his mass times his actual acceleration.  Since drag is proportional to the square of velocity, the drag force and the gravitational pull become equal, and acceleration reaches zero.  This is known as terminal velocity.  Not nearly as exciting as the name sounds.  I know; I was disappointed, too.  In equation form it looks like this:

The drag coefficient includes half the surface area of the guy in the suit and a proportionality constant.  Based on his reported time in freefall and an area of 4.3354 square meters I found this to be 1.15, which is actually quite reasonable.

The next issue to deal with in the problem is the air density.  Anybody who's ever been up a tall mountain or a plane ride knows air gets thinner with elevation.  As a result, drag forces are higher near the ground than at the elevation Baumgartner jumped from.  Without that included, we're only left with the beautiful sky, without any interesting math.  So we can expect that Felix's speed went way up, and then it actually decreased until he pulled his parachute at 270 seconds.  Then it really dropped.

There's a ton more I just didn't have time to do.  For example, the drag coefficient is dependent on the velocity and density of the air, the gravitational force decreases as elevation increases, etc, etc.  But I just didn't have time.  Sigh!

Enough of all that.  I drew it out on paper and coded it out in Matlab.  Let's see what it looks like.  First, his position with time:
The scale of this is in tens of thousands of feet, so you can see where he pulls the cord at 270 seconds and about 8400 feet.  At that point, his descent slows way down.  Now, the velocity of the fall:
Two important spots here.  On the right, you can see where his chute deploys.  On the left, you can see where he reached terminal velocity in the upper atmosphere about 50 seconds in.  This was his terminal velocity.  After that, the rising density of the atmosphere continually slowed him down.  Kinda like being married.

As you can see, the model is a little off; the maximum velocity should be closer to 850 feet per second, while I've got him maxing out a little over a thousand.  That is probably due to my crude modeling of gravitational and temperature gradients at high elevation.  What can I say?  There's only so many hours in a Sunday afternoon.  Oh--one last thing.  The modeling on this is highly non-linear, so I saw no way to punch out an analytical solution.  Really, I don't think there is one, but maybe someone out there knows better than I do.
The equations and code are posted here:  http://thesexyuniverse.com/2012/12/08/baumgartners-jump/


Cheers!

4 comments:

  1. farukunker@gumushane.edu.tr

    Hello;
    Could you send the detailed equations or the code by @mail,please?

    ReplyDelete
  2. Hello. Could you send me too the detailed equations( Matlab Code)? Thank you so much. It is important. Thank you.

    sn-rocks@hotmail.com

    ReplyDelete
  3. Hi, great piece of work, what solver did you use in matlab to get those graphs?
    would you mind sending your code to me too?
    645048@swansea.ac.uk

    Thanks
    Chris

    ReplyDelete
  4. Hi, really impressive work :) please could you send me your code? up713081@myport.ac.uk
    Thanks,
    Jo

    ReplyDelete