Calculating the difference in force between crumple zone cars and bull bar car

[Noobert]

I want to calculate the difference in force between a car with a bull bar an a car with crumple zones hitting a pedestrian.
Assume that when the car with bull bar collides with pedestrian, the transfer of velocity is completely transferred

Known variables:
Cars speed: 60kmph ; (16.67 m/s)
Weight of pedestrian: 60kg
Width of pedestrian: 20cm ; (0.2m)
Crumple Zone: crumples 10cm ; (0.1m)

Attempt at Solution

(Bull Bar)
t=d/v
t=0.2/16.67
t=0.012 sec

a=vf-vi /t
a=16.67/0.012
a=1389 m/s

f=ma
f=60*1389
f=83340(N)
This shows that it takes 0.012 seconds for the impact on the pedestrian to be completed, in which time the pedestrian accelerates at 1389 meters per second, which would exert a force of 83340 Newtons.

(Crumple Zone)
t=d/v
t=0.2+0.1/16.67
t=0.018 sec

a=vf-vi /t
a=16.67/0.018
a=926 m/s

f=ma
f=60*926
f=55560(N)
This shows that it takes 0.018 seconds for the impact on the pedestrian + the time for the crumple zone to cave, in which time the pedestrian accelerates at 926 meters per second, which would exert a force of 55560 Newtons.

Would this be correct for calculating the force exerted on a pedestrian when being hit by a bull bar or crumple zone car?

 

[PeterO]

I want to calculate the difference in force between a car with a bull bar an a car with crumple zones hitting a pedestrian.
Assume that when the car with bull bar collides with pedestrian, the transfer of velocity is completely transferred

Known variables:
Cars speed: 60kmph ; (16.67 m/s)
Weight of pedestrian: 60kg
Width of pedestrian: 20cm ; (0.2m)
Crumple Zone: crumples 10cm ; (0.1m)

Attempt at Solution

(Bull Bar)
t=d/v
t=0.2/16.67
t=0.012 sec

a=vf-vi /t
a=16.67/0.012
a=1389 m/s

f=ma
f=60*1389
f=83340(N)
This shows that it takes 0.012 seconds for the impact on the pedestrian to be completed, in which time the pedestrian accelerates at 1389 meters per second, which would exert a force of 83340 Newtons.

(Crumple Zone)
t=d/v
t=0.2+0.1/16.67
t=0.018 sec

a=vf-vi /t
a=16.67/0.018
a=926 m/s

f=ma
f=60*926
f=55560(N)
This shows that it takes 0.018 seconds for the impact on the pedestrian + the time for the crumple zone to cave, in which time the pedestrian accelerates at 926 meters per second, which would exert a force of 55560 Newtons.

Would this be correct for calculating the force exerted on a pedestrian when being hit by a bull bar or crumple zone car?

I don't understand what you mean by the width of the pedestrian, and your use of it.

 

[Noobert]

Width of pedestrian as in measurement from their back to their stomach. Used in the t=d/v equation, to find the time it takes for the car to travel the distance of the pedestrian's body; (width). This is because the person is being accelerated to the same speed of the car in the time it takes for the car to travel the distance of the pedestrians body.

 

[PeterO]

Width of pedestrian as in measurement from their back to their stomach. Used in the t=d/v equation, to find the time it takes for the car to travel the distance of the pedestrian's body; (width). This is because the person is being accelerated to the same speed of the car in the time it takes for the car to travel the distance of the pedestrians body.

Who told you "This is because the person is being accelerated to the same speed of the car in the time it takes for the car to travel the distance of the pedestrians body" ?

The first part of their body to be struck will almost instantly take on the speed of the car/bull-bar - like an insect hitting a windscreen.

 

[kcrayback]

please help me calculate hydrostatic load strength of mild steel plain sheet 4.5mm 4'x8', mild steel flat bar 2"x 3/16", mild steel pipe 1" diameter sched 80. submerged into 4 feet depth seawater with density of 1.025 mt/m3.

 

[Noobert]

Here is where i found the stuff on the width or thickness of a persons body affecting the acceleration and force on a pedestrian...

Impact on a pedestrian

"Because the pedestrian, Sam, is so much lighter than the car, he has little effect upon its speed. The car, however, very rapidly increases Sam's speed from zero to the impact speed of the vehicle. The time taken for this is about the time it takes for the car to travel a distance equal to Sam's thickness – about 20 centimetres"

Source:http://www.science.org.au/nova/058/058key.htm