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  • Maths/physics/sciencey question about g-forces
  • thegreatpotato
    Free Member

    I have been pondering this and already gone beyond my own knowledge so figured here is as genius a place to ask if this makes any sense.

    During yesterday’s Formula 1 race there was a crash in which drivers were subjected to an impact of 27G. Now, I know this is 27 times gravity. What I’ve been trying to do is translate this into a man-in-the-street term like double-decker buses, football pitches or Empire State Buildings.

    If the driver is around 60kg (he’s not a big chap), multiplying that by 27 gives what I reckon is a relative body weight of 1620kg. I can only remember gravity being measured in acceleration and can’t recall at all what the units would be if you multiply weight and acceleration so am completely ignoring that bit of it.

    So just going by the relative body weight thing, does that mean the crash is equatable to, say, having half an elephant dropped on him? Or is there a way of working it out in terms of “it equals falling from a [insert number] storey window”?

    stuey
    Free Member

    F=ma – so his rate of change velocity a=(v-u)/t or impulse force=mv/t – would of peaked at 60kg x 27xx 9.81N/Kg = 15000N “1.5tonne”

    I suspect the car took the 27G and the tub/seatbelts increased the deceleration time so the driver himself felt far less – I’m sure those sorts of forces would liquefy organs.
    Incidentally pre airbags etc road cars drivers would often ‘feel’ 45G in head on crashes is brought to a very abrupt standstill.

    richmtb
    Full Member

    27G is just a measure of how quickly he decelerated.

    Say he was travelling a 90m/s (which is 288km/h or 175mph) so not a bad guess.

    If he stops from that speed in 1s then he has experienced a decelaration of 90m/s2 or 9G. So 27G would be stopping from that speed in a third of a second.

    It sounds like a lot but while no doubt pretty unpleasant its not that high for an impact, the tyre walls and car crumple zones will have done a good job in lengthening the deceleration time and so reduce the G-forces involved.

    Bear in mind hitting the brakes on an F1 car gives 5G of deceleration without hitting anything

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    maxtorque
    Full Member

    If you want to learn about what the human body can withstand in terms of ultimate accelerations, then you need to read up on John Stapp:

    John Stapp on Wiki

    Lets just say, that when properly restrained, the human body is incredibly resilient to massive, but short g loadings!

    On one set of rocket sled tests, he decelerated so hard the blood vessels in his eyes burst and he went temporarily blind! Some day at work eh! 😉

    richmtb
    Full Member

    Apparently sustained g-force above 10G is fatal

    This was linked to on the g-force wikipedia page

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    Alternative to lethal injection perhaps?

    slowoldman
    Full Member

    I don’t think equating deceleration to body weight is particularly helpful in visualising anything.

    The acceleration experienced under the force of gravity is 32m/s2. A deceleration of 27G simply means (27*32)/s2 (i.e. slowing down very quickly).
    Mind you, I don’t know what the deceleration would be if you fell out of an aircraft and hit the ground without a parachute. Even with a helmet on that would have to smart a bit.

    ampthill
    Full Member

    The acceleration experienced under the force of gravity is 32m/s2.

    On this planet its 9.81 m/s/s

    You might be thinking of the American version?

    Sustained G is clearly way different to spikes lasting less than a second. Death at sustained G is due to lack of blood to the brain

    Stuey I don’t get your graph. Surely more time gives less acceleration, at any given speed?

    slowoldman
    Full Member

    Oops. Mixing my imperial and metric. 32ft/s2 or 9.81m/s2. I stand corrected. Not American though – we did used to use imperial (hence imperial). I am very old.

    maxtorque
    Full Member

    slowoldman
    Mind you, I don’t know what the deceleration would be if you fell out of an aircraft and hit the ground without a parachute. Even with a helmet on that would have to smart a bit.

    It of course rather depends on what you actually fall on!

    A “typical” human free falls at around 125mph, for a skydiver in a nice “flat” posture. If you “dive” headfirst at the ground, even at low level (where air density is high) you can reduce your drag and hit over 200mph quite easily.

    So, lets take the lower number 125mph which is approx 55 meters per second.

    So, if you decelerate to zero speed in the following distances, here are the g forces:

    dist time g
    [m] [mS] []
    200.0 3600 1.57
    100.0 1800 3.15
    50.00 900 6.29
    25.00 450 12.59
    12.00 216 26.23
    6.00 108 52.46
    3.00 54 104.91
    2.00 36 157.37
    1.00 18 314.73
    0.50 9 629.46
    0.25 5 1258.93
    0.10 2 3147.32

    So, fall into a nice tall, soft fir tree, and take over 50m to come to a halt, and there’s a good chance you’ll survive (as has be done on few occasions, most notably an airhostess who fell out of a commercial airliner and landed (alive) in a snow bank/fir tree!!)

    But, land on a concrete parking lot, and stop in half your width, say 100cm, and er, you’re gonna need a bucket……..

    legend
    Free Member

    Some scary numbers that may be of interest to show what can be survived: “he suffered a serious crash that almost cost him his life. His car locked wheels with Tomas Scheckter’s, flew into the catch fencing, and broke apart. Bräck’s crash saw the highest recorded g-forces since the introduction of crash violence recording systems, peaking at 214 g.[1][2] He suffered multiple fractures, breaking his sternum, femur, shattering a vertebra in his spine and crushing his ankles. He spent 18 months recovering from his injuries.”

    http://en.wikipedia.org/wiki/Kenny_Br%C3%A4ck

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