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Comment by turing | Hacker News Reader

"Of note, during validation of Model S roof crush protection at an independent commercial facility, the testing machine failed at just above 4 g's. While the exact number is uncertain due to Model S breaking the testing machine, what this means is that at least four additional fully loaded Model S vehicles could be placed on top of an owner's car without the roof caving in."

I don't have any frame of reference for how this compares to other vehicles, but damn, that is impressive.

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I wonder what sort of a drop 4g's translates too. It seems to me that the height you can drop it before the roof caves in is the more important figure, but I guess it is probably more difficult to accurately run the test that way.

Let's see if I remember my high school physics. Making all sorts of assumptions and simplifications here:

F = mv / dt (equation of impulse for zero final velocity)

F = mg (newton's laws)

mg = mv/dt

g = v/dt

9.8 = v/0.5

v = 4.9

Therefore, the car can be travelling no more than 4.9m/s at impact to survive.

mgh = 0.5mv^2 (initial gravitational potential energy = kinetic energy at impact)

gh = 0.5v^2

h = (0.5v^2) / g

h = (0.5 * 4.9 * 4.9) / 9.8

h = 1.225

So, assuming that it takes 0.5 seconds (a wild ass guess) for the car to go from freefall to rest, the car can withstand a drop of 1.2 metres without deforming the roof.

Mind you,

Ouch, units, please! Leaving the units out of your physics math is like, well, not wearing a seat belt when driving a car.

Anyway, I'm afraid that 0.5 seconds to come to rest is way too long. Toss it into the equations of motion and:

    d = 1/2 at^2
    d = 1/2 9.8m/s^2 (0.5s)^2
    d = 4.9m

In other words, decelerating at 4 gees for half a second means you decelerate over nearly five meters.

I think we're better off starting at the other end of the stick and making a wild guess that the roof can move 10cm without being permanently deformed. Then:

    d = 1/2 at^2
    0.1m = 1/2 4 * 9.8m/s^2 t^2
    t = 0.0714s

To figure out the height, we can notice that a and t will be inversely proportional for any given velocity (double the time, halve the acceleration needed), and so, for a given initial or final velocity, the subexpression at^2 is directly proportional to the change in acceleration. In other words, if we assume that the roof can withstand 10cm of deflection, you can drop the car from 40cm up.

Thanks, much appreciated. Sorry about the lack of units. If it's any consolation, I included them in the text surrounding the calculations. :)

Depends upon how squishy the roof/substrate is. The acceleration-to-height or compressive stress-to-height conversion is not well-defined in general.

If you're optimizing for dropping a car on its roof from a specified height, it's probably better to drop the car on its roof.

i'm guessing that in accidents, cars very rarely drop on their tops - they either roll and come to rest upside-down, or get hit when they're already turned over/sideways.

The one thing you learn about car accidents is to not make any assumptions at all about how they progress. Cars can get flipped up and land on their roofs easily, and it doesn't take much for this to happen either. It definitely is not the most common accident but it does happen and frequently enough that they test for it.

yeah, but from any meaningful height?

do they really hoist a car upside down to 10, 15, 20, 30 feet and then drop it on its roof? because that's what the OP was referring to.

A car could certainly flip up into the air quite a few feet while rotating and then land on the roof.

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