What is the g force experienced?

May 5, 2008 4:43 PM

How would I calculate the G-Force of this impact?

These are the things I know.

Start speed: 150 mph
End speed: 0 mph
Start position: 0 ft
End position: 8 ft

In other words, a car went from 150mph to 0mph in 8 feet.

What is the g force experienced?

Assume uniform deceleration.

posted by gummo to science & nature (19 comments total)

Is this a homework question?

Deceleration is change in speed over change in time (dv/dt for the calculus-loving types). Velocity is change in distance over change in time (dx/dt). Given those two things, you should be able to figure out the deceleration. If you then divide by 9.8m/(s^2) or 32 feet/(s^2) (the value of g, or the acceleration caused by gravity at sea level), that's the g-force.

If you have a plausible explanation other than homework for needing this, I'll do it for you/show you how. Otherwise, take a stab yourself and let me (us?) know when you run into trouble.
posted by JMOZ at 4:52 PM on May 5

It's from watching Formula 1 racing. We were trying to calculate the g forces experienced during a crash. I took a stab a few times. Not sure if I'm in the ballpark. If I come up with 9 gs, I believe it, if I come up with 40, it seems high, but I'd have to believe it. And it's been a while since I was in shool. So a) I don't have homework, and b) I can't remember what I'm doing.

Here's my best guess.

Is this the right ballpark?

32f/s/s for gravity.
=115200ft/hr/sec
=21.818miles/hr/sec

Accel of gravity in mph/s = 21.818 (sounds reasonable)

I calculated .07 seconds for the deceleration time using

t = delta x / average velocity (.07 seconds also sounds reasonable, if not a little quick)

So.

1 second would only get me down to 0 from 21.8, So I figure multiply that by 6.88 gives me 150 down to 0 in 1 second. I want .07 seconds. So multiply that by 14. to make it happen in 1/14th the time. Gives me 1 g X 6.88 X 14.

96 G. Sounds WAY too high.
posted by gummo at 5:00 PM on May 5

On preview... what JMOZ said. Your posting profile makes me think this isn't your homework.

I would use one of Newton's second law equations that I remember from school

v^2 = u^2 + 2as

Where
v = final velocity = 67.05m/s
u = starting velocity = 0
a = acceleration = 9.8m/s^2
s = distance = 2.4384m

Which gives me... 92g. Which doesn't sound correct but tallies with gummo.
posted by NailsTheCat at 5:02 PM on May 5

I'm guessing this is the crash gummo is talking about.
posted by foodgeek at 5:11 PM on May 5

50--100G sounds about right to me for a hard F1 crash. People have survived crashes with higher g-loads when suitably protected.
posted by ROU_Xenophobe at 5:15 PM on May 5

You are quite the sleuth foodgeek. Kovalainen's crash was indeed what sparked the conversation. 150mph and 8 feet were guesses, but I think pretty close.

Can anyone verify the answer of 95+ gs? Sounds too high.
posted by gummo at 5:43 PM on May 5

Update: News outlets are reporting 26 gs for that crash. So my math is bad. Anyone?
posted by gummo at 5:58 PM on May 5

Just an aside... Captain John Paul Stapp pulled 42.6 negative G's on a rocket sled. His research helped lead to mandatory seatbelts in cars.
posted by Marky at 6:10 PM on May 5

It's apparently survivable. The wiki says that David Purley, a largely forgotten Formula One driver, survived a force of 180g, decelerating from 108 MPH to zero in two feet and change, and actually raced again. . .eventually. And this was in the seventies, when carbon fiber monococks, crumple zones, and H.A.N.S. devices were entirely unheard of.
posted by Capostrophe at 6:10 PM on May 5

Also, my math is apparently way off. I'm getting:
150 mph = 41.67 meters per second
8 ft = 2.44 meters

Assuming linear deceleration, average speed during deceleration = 20.835 m/s, meaning that total deceleration took .12 seconds (2.44 metres / 20.835 metres per second). To go from 41.67 meters per second to zero meters per second in .12 seconds requires deceleration of 347 meters per second per second (41.67 meters per second / .12) . 1g = 9.8 meters per second per second, so this would be a 35g impact.
posted by Capostrophe at 6:28 PM on May 5

My math is this: assuming constant deceleration,
the distance x = (at²)/2
the speed v = at
so a = v²/2x
which, being v=67m/s and x=2,44m, gives you a=920m/s², or about 93.8 g.

The guy came out of the crash unharmed because, as you see in the youtube video, the tyre bursts at the moment he enters the curve, add .1s for reaction (possibly less) and the car is braking from that moment until it crashes, so the distance in which the deceleration takes place is way longer than 8ft.

On top of that, while the car is apparently skidding on gravel, it is being "braked" very efficiently by the gravel itself, which is about 1,5 ft. deep.
posted by _dario at 6:45 PM on May 5

monococks

It's "monocoque," an impossible word to spell correctly, including for me.
posted by maxwelton at 7:52 PM on May 5

Best way to find out the G force of that impact would be to have access to the in-car telemetry records. Somebody on the team is sure to have a precise figure for you, and might tell you if you ask nicely.
posted by flabdablet at 8:32 PM on May 5

monococks

There's an oxymoron for you.
posted by davejay at 8:53 PM on May 5

92 G is roughly normal for a 100mph car crash, but G tells you nothing. Links and math here.

You ask about G-force, but you are in fact only attempting to calculate deceleration. G (32 ft/s/s) is a measure of acceleration, not force (F=ma).

Surviving is a function of force, not acceleration. If you fall into a hard chair from a standing position, you experience about 10G's on impact. But the force is spread over a half square meter of the tissue of your ass, which absorbs much of the force before passing it on to bones, etc. But 10g's spread over 1 square cm will pass right through you.

Car impact analysis is massively complicated. Watch that car crash video and notice the front left wheel collapse and bits of metal flying off the car. Those breakage events represent a force applied to that part of the car that exceeds that part's mechanical limits. At breakage, the force isn't transferred through the part to the rest of the car, it's absorbed entirely by the part which snaps. Commensurate with that breakage is a reduction in acceleration.

People die in car crashes not because the driver hits a wall or a tree or another car. They die (usually) because the cabin of their own car hits them (usu. wheel or dash, and the impact area is about 2cm^2). The purpose of crumple zones and material choices in the body are designed to model a shock absorber between the point of impact and the position of the driver. In other words, to ensure that the driver's acceleration curve is different than the car's and to ensure that the cabin is able to traverse a slightly longer distance than the front of a car. Even an additional meter of travel makes a huge difference. More distance = lower acceleration = lower force. An inverse relationship between the car surviving and the driver surviving.

Cars in the fifties weighed about as much as they do now, believe it or not. The reason they look like tanks is because they were built like that. Cars were built to survive impacts. So the steel was think and heavy, and everything was bolted together as strongly as possible. The result of this is that cars survived crashes, but drivers didn't.

Car mfgs had to design in breakage points. Lift the hood of your car and look at the underside. You'll see the steel rolled into a box around the perimeter of the hood that gives the hood it's rigidity. But look closely and you'll see notches taken out of it. That notch is precisely where the hood is going to bend on impact. They know exactly how much force that notch can take before the hood bends around it.

As the body crumples, acceleration is lost, and force is absorbed by the car body and not the driver. Thus, the proper way to model a collision is at minimum as a three-body problem: the wall, the cabin, the driver. You can see that as you add collapsible steering columns and breakaway dashes, axles, engine parts, etc, the number of bodies in the collision becomes huge.

Also, note that he doesn't actually hit a wall, he hits a tire wall and appears to extend into it. The tire wall also absorbs the force of impact both through friction (car slides between tires and tires slide on road) and as less-than-ideal springs (the tires squish). This combines to help reduce the a in F=ma. This is also why abutments on US highway exit ramps are often filled with sand or water.

Furthermore, as dario noted, the deceleration is very non-linear and non-uniform, especially with the car falling apart as it heads towards the wall. Secondly, think of the length of an F1 car. If the front of the car crumpled by one meter on impact, it means the driver moved an additional meter after the front of the car stopped.

So assuming the front of the car crumples by 1m, that actually gives you a distance of 3.44 meters that the driver travels, not 2.44m. This drops the G's to about 66.6 from 92, almost 30%. That's 30% less force he feels. Factor in the skidding, the smashed tire, the dirt, etc, and it's very easy to see how his speed was far below 150mph when he hit the wall. That means that the force he felt was well within the range of minimal bruising. You can see he walked away from this crash.

With all that said, you can now appreciate why the single most important safety feature in a car, and especially an F1 car, are the seat belts. Without seat belts, the deceleration of the car and the human are independent. The car decelerates first on impact, which means the driver starts moving in the cabin at a speed equal to the difference between the speed of the car at the moment of impact and a moment later. Even if the car was going 60 and a moment later was going 40, the driver is moving 20mph.

The seat belt links the human's acceleration curve to the midpoint of the car body, allowing the entire front half of the car to be destroyed in the process of absorbing the impact force and slowing the midpoint down before any part of the car touches you.
posted by Pastabagel at 9:17 AM on May 6 [4 favorites]

I realize the question would have been much more complicated had I asked about the driver's survivability in a crash. So I didn't. I was in fact just looking for the deceleration (in g's) of the 150mph over 8 feet event I created.

My hypothetical was only based on the F1 crash. But I was specifically interested in the correct way to arrive at an answer to my simple math problem. Sounds like my guess of 92 might be right, but I'm still not sure. Looks like there have been 6 answers. Ranging from 35g to 100g.

As for the crash itself, the 150mph was my estimate of speed when he hit the wall, not when he was on the track (before scraping on the road, scraping through gravel, etc...). He didn't look like he scrubbed off that much speed before he hit. And the 8 foot estimate was the distance I guessed his body traveled while it decelerated that 150mph. These estimates could admittedly be wildly wrong, but they should work fine for a math problem.

And the reason for assuming linear deceleration, despite knowing it would be somewhat inaccurate, was just to achieve an 'at least' value for Gs experienced. If the linear deceleration were 92g, nonlinear deceleration is only going to average that, but have higher spikes. So assuming linear should give us the 'at least' figure for Gs experienced.

Anyway, if anyone still comes across this, and is confident they can provide a correct solution for the above simple problem, please go for it.
posted by gummo at 4:46 PM on May 6

The correct way to do your simple problem (start at 150mph, end at 0mph after travelling 8 ft) is how _dario described it.

The problem is plug-n-chug equations of motion.

final velocity = (initial velocity)^2 + 2 * acceleration * distance.

Final velocity = 0
initial velocity = 150 miles per hour = 67.0560 meters per second
distance = 8 ft = 2.4384 m
solve for acceleration

accel = - (67.0560^2) / (2 * 2.4384) = -922.02 meters per second

A 'g' is 9.80665 meters per secong, so expressed in g's, this is -922.02/9.80665 = 94.1, approximately. The difference between this answer and others is the rounding of the decimals, but you get the picture.
posted by Pastabagel at 5:30 PM on May 6

Let me add one thing. When you say "93G sounds way too high", that's because it is. Princess Diana was killed in an approximately 100G crash from the back seat of a much safer car. But as you acknowledge, the physics of the actual crash and your simple model are not even remotely similar, so you can't do a gut check on a model that isn't modelling what your gut knows from experience.

More precisely, in reality the car and driver and experiencing acceleration that changes over time. Just like rate of change of distance over time is velocity, and rate of change of velocity over time is acceleration, the rate of change of acceleration over time is called 'jerk', measured in m/s^3. Jerk and its accompanying force analog, dF/dt. These concepts are useful in collision mechanics in order to determine the conditions under which necks start snapping and ribs start breaking under seatbelts.

If the news reports are reporting 26G's based on telemetry data, that's the number to use, because that is the actual number that accounts for all the decelerations, crumpling, force absorption, etc.

My example was to show how a crumple zone that buys the driver another meter shaves 60% off the deceleration to get down to 65G. If we factored in everything else in the crash, you can see how we'd get down to 26G.

Capostrophe's 35G number is wrong simply because the starting velocity is a bit off. 150 mph is 67 meters per second, not 41.67m/s. Google is your friend.
posted by Pastabagel at 5:50 PM on May 6

Surviving is a function of force, not acceleration. If you fall into a hard chair from a standing position, you experience about 10G's on impact. But the force is spread over a half square meter of the tissue of your ass, which absorbs much of the force before passing it on to bones, etc. But 10g's spread over 1 square cm will pass right through you.

Most of your comment is right on, but this section is... strange.

If you're interested in expressing damage to tissue, pressure (force/area) is more interesting than force. Your example of sitting down on a chair not hurting because "the force is spread out" is exactly to say that force doesn't hurt on its own; force over a small area hurts.

By "10g's spread out over 1 square cm", do you mean the equivalent force required to accelerate a person's mass by 10g, spread out over 1 square cm?
posted by Jpfed at 5:38 PM on May 7