The Math of Crashing
We need six equations to calculate the g forces from crashing. The crash is not instantaneous; but we must compute the time based on the thickness of the padding.
We use the subscript ‘c‘ when dealing with numbers based on the crash.
Let’s assume that the padding gives a smooth, constant deceleration over its full thickness. (That’s probably not true in reality). In physics, “deceleration” is just “acceleration” in the opposite direction, so we still use ‘a’. (Besides, we use ‘d’ for “distance”).
- The initial velocity at the start of the crash is the same as the maximum falling velocity: vf.
- The final velocity is zero, once the crash is complete.
- If deceleration is smooth, then the average velocity during the crash is vc = (vf – 0)/2= vf/2 .
- The distance for the crash is the thickness of the padding: dc is another “given” based on the setup of each crash test.
- We calculate the time for the crash as tc = dc / vc.
- We calculate the acceleration for the crash as ac = (vf – 0)/tc = vf / tc.
- We calculate the g forces for the crash as gc = ac / 9.8 = ac / af.
So gc refers to the impact g forces, not to the acceleration due to gravity.
Simplify the Calculations for G Forces from Falling and Crashing
Surprisingly, we can simplify the equations to something far easier to compute. Start with the ‘ac‘ formula, and substitute for previous variables such as ‘tc‘.
Using the images above, we start from equation (9); substitute from (8) and then from (6). Then we invert “x/(y/(x/2)) = x*x/(2*y)” and note that “x*x=x2“.
After that, equation (3) gives us (vf2) = 2*af*df, so that’s another substitution. But then the 2’s cancel.
Finally, gc comes from equation 10; but since we have ac = af*df / dc, we can cancel the af terms.
- ac = vf / tc = vf / dc / vc = vf / ( dc / vf/2 ) = (vf*vf) / (2*dc) = (vf2)/ (2*dc).
- ac= (vf2) / (2*dc) = 2*af*df / (2*dc) = af*df / dc.
- gc = ac / af = (af*df / dc) / af = af*df / dc*af = df / dc
The Simple Formula for the Impact G Forces Based on Distances
So we can compute the impact g forces as the ratio of the distance fallen divided by the thickness of the padding; assuming that all the padding compresses at an even rate and the object comes to a complete stop.
- gc= df / dc
Did the Mythbusters Differ from these Calculations for Falling into Bubble Wrap?
The Mythbusters did not achieve these mathematical results. The original bubble wrap was four inches thick; that’s 1/3 of a foot. The height was 35 feet. We compute gc = df / dc = 35 / (1/3) = 35*3 = 105 g‘s.
Instead, Buster’s accelerometer reading was over 300 g‘s.
At least two problems caused the discrepancy:
First of all, the bubble wrap did not compress evenly and completely through the full depth of the padding. The maximum cushioning would result in every bubble bursting during the crash. Instead, many layers of bubbles remained intact. Therefore doubling the thickness of the bubble wrap did not lessen the impact by a factor of two. How much effective padding did Buster have? Solving for dc, we have dc = df/gc = 35/300 = 0.117 feet = 1.4 inches of ideal padding. The Mythbusters addressed this problem by shaping some bubble wrap layers as crushable cones. This effectively used much more distance for the padding to cushion the crash.
Secondly, the accelerometer’s readings were lower than these equations would predict. This author speculates that Buster himself provided some of the cushioning for the accelerometer. Buster landed on his back; the accelerometer was inserted into and, apparently, attached to his chest cavity, therefore Buster nobly lessened the impact on the accelerometer.
This article discusses calculations, and is not a guide to performing science experiments. Do not try the actual bubble-wrap-drop at home, particularly on a live subject.
Discovery Network. Mythbusters Bubble Wrap Test. Referenced March 15, 2013.
Live Physics. Solve problem related to impact force from falling object. Referenced March 15, 2013.
Rowlett, Russ.How Many? A Dictionary of Units of Measurement (G). University of North Carolina at Chapel Hill. Referenced March 15, 2013.
Decoding Science. One article at a time.