MythBusters, Falling, Stopping, and Integration

In a MythBusters episode some time ago, Adam and Jamie jumped off a building. There was some cool stuff in this, but I want to focus on the acceleration data they collected. Before jumping into a pit of foam, they first wanted to test the set up by dropping a dummy into it and measuring the accelerations. Lucky for me, they showed a quick screen shot of their data. Note: I previously posted the calculations for jumping and stopping off of a building.

i-263560719fcf6c75bc102d8378327403-2010-01-04_untitled_113.jpg

For me, I see this and think - numerical integration. Before that, let me look at the physics. Here is a diagram of someone jumping off a building.

i-7db1ea2cfc9daa144396ce08de91d81c-2010-01-04_integration_1.jpg

In my first analysis of this, I looked at the landing in terms of force and displacement. For this data, I have the acceleration and time. When you have force (which I kind of do if I know the mass) and time, you should think about the momentum principle:

i-d9786b21370cd16eb718b456b8d26f8d-2010-01-04_la_te_xi_t_1_6.jpg

Compare this to the work-energy principle which deals with force and displacement:

i-099b8ddc03e989f0e698e0d86e5cf6cd-2010-01-04_la_te_xi_t_1_7.jpg

So here, I am going to use both of these principles. The work-energy for falling until the faller hits the mat and then momentum principle for stopping. First for the fall. I will take the faller and the Earth as the system. This means that there is no work done on the faller while falling, but there is gravitational potential energy. I can write the work-energy principle as (using the numbers from the diagram above). Final note - I am going to let the gravitational potential be zero at the top of the mat.

i-901aa5157c0a6f90bfd34ebe9b0e7425-2010-01-04_la_te_xi_t_1_8.jpg

Now for the landing. Let me assume that the mat exerts a constant force (which it clearly does not) over the time interval. Then I can write the momentum principle (in the y-direction) as:

i-a588642e7b3b16d357cdbb130fa0ec16-2010-01-04_la_te_xi_t_1_9.jpg

The initial momentum and velocity were in the negative y-direction. This is why the change in momentum (in the y-direction) is positive. Remember, I am assuming the force is constant over this interval. So, I could re-write this as:

i-e51e57fd959d02ce99af631d209b398a-2010-01-04_la_te_xi_t_1_13.jpg

Suppose I plotted the net force as a function of time, here is a sketch.

i-edd6aabc4915765de7ba85925f3a1b53-2010-01-04_untitled.jpg

I already know what the product of F-net and &Delta t should be (this is called the impulse, by the way), but here you can see that Fnet*&Delta t would be the area under the force-time curve (clear to see since it is a box-shape). But what if it is not a box? What if it is something more complicated?

Numerical Integration

Here are the traditional options for dealing with the area under a curve:

  • If you know the force as a function of time, you could analytically determine the impulse.
  • If you have a print out of the force as a function of time, you can print out your curve on thick paper. Find the mass of the paper. Cut out the piece with the function and the part below it and find its mass. The impulse will be the max F times the max time multiplied by the ratio of the cut out mass to the total mass.
  • If you have force-time data points, you can break this integration into a whole bunch of small pieces. This is numerical integration.

Suppose a part of my force-time data looks like this:

i-61d8990f16e9b3fd59efc608c8d45c75-2010-01-11_untitled.jpg

If I take a pair of points at a time, I can find the impulse just for these two because the shape is a trapezoid. Here is another diagram.

i-956a7e881acc2bef798bf83da8667f69-2010-01-11_untitled_1.jpg

Here, the area of that piece will be:

i-1b31346dff31940dedd27b8c6d6f94f3-2010-01-11_la_te_xi_t_1.jpg

Here, I am calling the area &Delta I where I is the impulse. The &Delta means that it is just a small piece of the total impulse. Also note that for the area, the "width" is the difference in times and the "height" is the average of those two forces. It probably wouldn't be a bad approximation to call the height Fy1 and not use the average.

Getting the data

How do I go from a picture of the graph to the actual data? I used GraphClick. This is a Mac application that basically lets you load the picture of the graph and then click on the data. It will then translate the pixel data to x-y data. Very useful in this case. You could probably do something like this Tracker Video and I am sure there are other applications that do the same thing in Mac OS X as well as windows and linux.

There were two falls the MythBusters recorded. One was on an air-bag and one was in a dumpster with foam stuff.

i-fceccf2aba240f3bfaed5477b72fbae6-2010-01-11_falls.jpg

If everything is working out correctly, these two jumps should have the same impulse for landing. I can write the momentum principle as:

i-22439533015042cadbf171158654eabe-2010-01-11_la_te_xi_t_1_1.jpg

If they both have the same momentum right BEFORE landing and they both end up stopped, then both falls have the same change in momentum. This means that both falls should have the same impulse (impulse is F-&Delta t part). Ok, next question. The data is the acceleration. Is this the same as the net force? Well, it should be proportional.

Now, here is the data. I calculated the impulse two ways. The first is with the trapezoid method I showed above. The second way is just using rectangular area pieces. You can see that the two impulses are pretty close.

There are two sheets - one for each fall. The "impulse" (because it is the integration of acceleration over time and not force) for the two are about the same, 0.54 g's*s versus 0.60 g's*s. These could actually be closer. I don't have all the data, the screen shot cut it off after a little bit of time. Overall, I think it worked pretty well.

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