Physics question + Math

[Vova]

I'm working out how much force my harness bar should be able to endure in case of a 200km/h head on collision with a tree or a wall or some other solid object.

So far I got:

I weigh 80Kg.
I travel at 200Km/h = 55m/s (meter/second).

F=ma

F=80*55

F=4400N

9.81N=1KG

4400/9.81=448KG <---- Is this right? Does this also mean that my body endures 448/80=5.6G ? Seems kinda low?

So my harness bar, in a 200km/h head on collission endures 448KG?

Another question, given the same situation, how much force again does my bar have to take when I hit another car of the same weight at the same speed of only 100km/h?

Same question, but now hitting a heavy 40T truck in the back.

[adam0bmx0]

I wouldnt say its the amount the other car weighs, more how fast your car decelerates.

[tommyb]

whack your numbers into here http://hyperphysics.phy-astr.gsu.edu/HBASE/carcr2.html 200kph head on is waaaaay more than 5g!!

[chrisdavidson152]

Does momentum not come into this also, momentum = force x distance(from pivot). I done this in techy like 14 years ago so I could be miles away. Just remembered that it's a bending moment. Either that sounds familiar to someone of I must of dreamt it like 10 years ago.

Chris..

[tommyb]

Does momentum not come into this also, momentum = force x distance(from pivot). I done this in techy like 14 years ago so I could be miles away. Just remembered that is bending moment. Either that sounds familiar to someone of I must of dreamt it like 10 years ago.

Chris..

couldn't remeber my a-level mechanics so i just hit the majic google

[s13james]

I dont think the harness will help you much at 200kph into a solid object, iv not bothered working it out in miles but i had a shunt in my 200 at 20mph into the back of a car and im having to rebuild the front end. I wouldnt like to see the aftermathe of a 200kph impact, harness or not

[tommyb]

80kg driver, 55m/s into wall/tree, seatbelt allows 1ft stopping distance. force is 43G!


shamelessly ripped from wiki

Formula One drivers usually experience 5 g while braking, 2 g while accelerating, and 4 to 6 g while cornering. Every Formula One car has an ADR (Accident Data Recovery) device installed, which records speed and g-force. According to the FIA Robert Kubica of BMW Sauber experienced 75 g during his 2007 Montreal GP crash.

Strongest g-forces survived by humans
Voluntarily:
Colonel John Stapp in 1954 sustained 46.2 g in a rocket sled, while conducting research on the effects of human deceleration.[9][10]

Involuntarily:
Formula One racing car driver David Purley survived an estimated 179.8 g in 1977 when he decelerated from 173 km/h (108 mph) to 0 in a distance of 66 cm (26 inches) after his throttle got stuck wide open and he hit a wall.[2]

Indy Car driver, Kenny Bräck crashed on lap 188 of the 2003 race at Texas Motor Speedway. Bräck and Tomas Scheckter touched wheels, sending Bräck into the air at 200+ mph, hitting a steel support beam for the catch fencing. According to Bräcks' site his car recorded 214 g.[11]

[s13james]

ouch! Id rather not test out the safety equipment at all but at least they walked away luckily.

[chrisdavidson152]

Just realized that 55 m/s is 1 Olympic swimming pool length per second, thats fast. Plus if you hit a tree or a wall or some other solid object in a head on collision it's gona hurt like a bitch.

Chris..

[DtheBUK]

use suvat

if were assuming you hit the tree and the tree doesn't move and the bonnet of you car + how much you move with the harness is 1.75m +0.5m

so

s=2.25m
V=0
u=55m/s
a=?
t=?

use v^2=u^2 + 2as

re arange to get a = (v^2 - u^2) / 2s

plug in the values gives an acceleration of -672.2 m/s^2 negative acceleration = retardation

then using f=ma



so the force exerted on you 53.776kN


think i missed something out. But if 1 g is 9.81m/s^2 and your deceleration is 672.2 m/s^s that means you would be under 68.5 g, which has been survived. Thats considerering your bonnet fully crumples ( though i think a 2.25m stopping distnace is slighhtly large, if it was 1 m you would have a deceleration of 1512.5m/s^s and you would experience 154g which has also been survived

so if you want to crash intro trees get a car with a big bonnet and lots of crumple zones

[Stu]

I would suggest swerving and hitting something less immovable if at all possible. If there are hay bails next to the tree, aim for those instead.

[s13james]

And do not try in a 200 at all due to the crumbled away zones

[Vova]

use suvat

if were assuming you hit the tree and the tree doesn't move and the bonnet of you car + how much you move with the harness is 1.75m +0.5m

so

s=2.25m
V=0
u=55m/s
a=?
t=?

use v^2=u^2 + 2as

re arange to get a = (v^2 - u^2) / 2s

plug in the values gives an acceleration of -672.2 m/s^2 negative acceleration = retardation

then using f=ma



so the force exerted on you 53.776kN


think i missed something out. But if 1 g is 9.81m/s^2 and your deceleration is 672.2 m/s^s that means you would be under 68.5 g, which has been survived. Thats considerering your bonnet fully crumples ( though i think a 2.25m stopping distnace is slighhtly large, if it was 1 m you would have a deceleration of 1512.5m/s^s and you would experience 154g which has also been survived

so if you want to crash intro trees get a car with a big bonnet and lots of crumple zones

I don't understand the formula, could you (care) to expain pleeeeezzz

Whats is

s=2.25m - s = distance of some kind?
V=0 = velocity?
u=55m/s = speed?
a=? = acceleration or speed ?
t=? = time?

Also, how did you rearrage v²=u²+2as into a = (v^2 - u^2) / 2s ?

So applying F=ma , 80*-672 = 53.000Kn

And is G force calculated using acceleration instead of force? Why?

I'm just interested in physics, that all, no arguing.

Thanks in advance

[Jon_H]

s=2.25m - s = distance of some kind?
V=0 = velocity?
u=55m/s = speed?
a=? = acceleration or speed ?
t=? = time?

s = distance travelled in the time t
v = final speed (velocity)
u = starting speed (velocity)
a = acceleration

In this case, s is the distance you cover while you're stopping - here assumed to be basically the length of your bonnet!

These formulae only work for constant acceleration... but for example, taking v²=u²+2as:

if u = 10 m/s (36 km/h, 24 mph)
a = 2 m/s² (0.2 G)
s = 100m

then after 100m your speed v will be sqrt(10² + (2 × 2 × 100))
= sqrt(100+400)
= 22.3 m/s (80.5 km/h, 50 mph)

Also, how did you rearrage v²=u²+2as into a = (v^2 - u^2) / 2s ?

Erm, no offense but this is fairly basic maths:
v²=u²+2as
subtract u² from both sides
v²-u²=2as
divide both sides by 2s
(v²-u²) / 2s = a
and there you are!

And is G force calculated using acceleration instead of force? Why?

Because it is an acceleration - "weight" is force per unit mass, which produces acceleration. 1 G is 9.81 m/s², which is the same as 9.81 N/kg (because 1 N applied to 1 kg makes it accelerate at 1 m/s²). It only exists as a force when you have a mass to apply it to (F = ma).

Also, be careful with . and , when discussing numbers with us English - we use . as a decimal, and (sometimes) , as a thousands separator. 80×672 is 53 760 N, or 53.76 kN (about 5.4 "tonnes force"!!!).

[Vova]

then after 100m your speed v will be sqrt

- SQRT? = sqrt(100+400)? Don't understand this one, is SQRT a cumulative reading of different inputs or..?

- Thanks for the maths lesson, by my understanding far from basic but that could be me.

- I understand the G thing now, thanks

- How do you know so much about diffs, bolt clamping force, physics math etc? What did you study/learn, what did you do in your free time?! (seriously).

[Jem]

I think it's a little irrelevent, hit a tree in a 200 at 120mph and you're dead

[Clouder_sx]

Why analise the maths if you dont understand it? just accept the answer

It's just some A-Level maths, which IMO everyone should be taught

[tommyb]

- SQRT? = sqrt(100+400)? Don't understand this one, is SQRT a cumulative reading of different inputs or..?

sqrt means the square root of....
theres no symbol on the keyboard

[Porker-S14a]

Bräcks' site his car recorded 214 g.[11]

****


Ing


Hell

[Vova]

Why analise the maths if you dont understand it? just accept the answer

It's just some A-Level maths, which IMO everyone should be taught

Why is that? If you don't understand something, just accept what you are told?! That is just ridiculous, sorry, if everyone was to just 'accept' what they were told there would be no innovation, no regulation (cause there is no questioning) and 90% would be all a bunch of sheep.

Or maybe I understood you wrong... no offense anyway but I'd like to know what I'm doing, why I'm doing it, why it works that way, because of what, where, when etc.

To the rest, thanks for the input, I now also know what SQRT is