Wednesday, December 22, 2004

Oh no, a physics post! It even has... formulas!

Our day-to-day experience with weights and speeds is limited to a small range. We don't expect the behaviour of objects to change very much, even when the object is far outside this limited range of weight or speed.

We know very well that a small piece of lead (between 2 and 30 grams), travelling very fast (between 300 and 1,000 meters per second), can do a great deal of damage - i.e. a bullet.

However, when we think of a large piece of polystyrene foam (about 30 cm across), then all bets are off. After all, it's hard to think of being able to hurl a large piece of foam fast enough to do damage to anything, right?

I went to a lecture the other day by Doug Osheroff that demonstrated this very clearly, using some high-school level physics.

Doug has a Nobel Prize. He's a low-temperature physicist who works at Stanford, and who recently served on NASA's Columbia Accident Investigation Board. So I paid attention to what he was saying.

He outlined the basic calculations behind the conclusion that a 0.75 kg chunk of foam from the Shuttle's external tank (ET) had irreparably damaged the leading edge of Columbia's left wing.

Video of the launch taken from various angles shows the foam breaking off and striking the wing - this allows an estimation of the source of the foam (the -Y bipod ramp, and therefore its size & weight, +/- 30cm radius, 0.75 kg), and also the amount of time it took to travel the distance from the ramp to the leading edge (19 meters in about 0.16 seconds).

The key here is to realize that since the foam fell off the external tank, it was decelerating relative to the air rushing by at Mach 2.5, and accelerating relative to the shuttle (and its onrushing wing).

So:
d0 = ½at02


Where d0 is the distance travelled, a is the acceleration of the foam, and t0 is the time between the foam breaking off and the wing strike.

Substituting the values:

19 m = ½ a (0.16 sec)2


gives

a = 1,480 m/s2


or about 150 g's.

How fast was that foam travelling when it got to the wing?

v0 = a t0


substituting values:

v0 = 1,480 m/s2 · 0.16 sec


gives

v0 = 238 m/s


Converting to something we understand better: 855 km/h or 530 mph. Pretty fast.

But it's still hard to comprehend a chunk of foam doing such tremendous damage.

In terms of kinetic energy,

Ek = ½ m v02


plugging in our values:

Ek = ½ 0.75 kg· (238 m/s)2


gives

Ek = 21,240 N·m (or J)


Say we had a bullet with equivalent kinetic energy - how fast would it have to be going? Let's choose the a fairly standard .22 bullet of about 40 grains weight, or about .0026 kg

Then:

21,240 J = ½ 0.0026 kg · v12


gives

v1 = 4,040 m/s


or about 13,260 fps, which is about five times faster than a bullet coming out of a rifle - or about twenty-five times the energy a regular 0.22 bullet would pack.

OK, obviously were talking something serious here.

And what was it crashing against? Aren't those wings tough? After all, they can take flying through plasma, right?

Umm, well, no. In fact, the leading edge material is really fragile. The RCC panels on the shuttle have to be covered with protective padding when the shuttle is being serviced to avoid damage from dropped tools or even dropped bolts.

The plasma-resistant wing never stood a chance against a chunk of foam.

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3 Comments:

At 8:13 AM, Anonymous Rajesh said...

good One. Never worked out with physics On columbia.

 
At 8:14 AM, Anonymous christopher riccard said...

interesting

 
At 8:29 PM, Anonymous hechizos said...

I need help understanding the formulas your supposed to use for the "work kinetic energy theorem" and the "conservation of mechanical energy theorem" the physics book doesn't explain jack diddly squat an i can't find any web pages that explain it simply enough for me to understand. HELP!!!!!!! |

 

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