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Calculating impact force

by klubo
Tags: force, impact
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klubo
#1
Dec28-04, 01:47 PM
P: 3
Hi, sorry if this is posted someplace else - I couldn't find a search function.

I want to know the force of impact of a Porsche 911 impacting a standing pedestrian at 95 mph.

I understand that F=m*a

But what I don't get is this: acceleration is measured here in mph. But the impact occurs over an instant of time. Doesn't that affect the calculation?

Here's my equation, in which I'm trying to move toward foot-pounds of force - (is that the direction I should be headed?):

Porsche 2940 lbs
Fuel 8 lbs more or less
Passenger 183 lbs
Luggage 18 lbs more or less
Total Weight 3149 lbs
Force = Mass x Acceleration

In this case the acceleration is equal to the speed of the vehicle since the pedestrian was traveling at zero miles per hour.

So the Porsche was traveling at 95 mph, which equals 501,600 feet per hour, 8,360 feet per minute, 139.33 feet per second
I realize this is basic stuff but I've managed to confuse myself anyway. Thanks for any help you can give me here -
Bill
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Integral
#2
Dec28-04, 03:16 PM
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You question is not easy to answer. We would need to know the maximum speed attained by the pedestrian, the mass and even speed of the car are of no consequence in calculating the force felt by the pedestrian. It is simply his initial speed, his max speed and the time required to attain it.

Next problem. let me tell you a story. Something over 30 yrs ago I was driving a nearly new Toyota Corolla, along a poorly lit road , on a dark rainy January evening at about 40 mph and slowing, as I was entering town. In a pull out, on the left side of the road, was a car with lights on high beam, on the right side of the road was a second car, with only dim parking lights showing. To avoid the bright lights of the car on the left and out of concern for activity around the car on the narrow shoulder on my side of the road, I was watching more to the right. Suddenly a pedestrian ran from the car on the left, across the center line in front of me, at much less then my stopping distance. In an instant he was on my hood, through my windshield, then as I was braking hard, back on to my hood, and off the hood onto the road in front of me. Fortunately this happened just as I rolled to a stop so I did not run over him a second time. His main injury was a broken leg.

Now, what happened here? My small car had a relatively low bumper, which first struck and broke his leg, but since I hit him well below his center of gravity I essentially swept him off his feet, spinning him around his CM so his shoulder struck and broke my windshield (we found half of his glasses on the floorboard of my car the next morning!) He was then laying on my hood as I quickly decelerated, at some point while he was crashing through my windshield he must have finally been accelerated to the velocity of my car, because as I slowed he continued forward with his new velocity. Which is why as I came to a stop he slid off of my hood.

Now, let's try to use this real life example to get some ideas about your collision. First like my old '73 Corolla the bumper on a Porsche is pretty low, so you can bet that initial energy will go into knocking the feet out from beneath the ped. It seems like a fair estimate could be made of the acceleration experienced by the peds leg. One could assume that by the time the Porsche has moved forward 1ft from the point of impact the foot will be off the ground and the lower leg will be moving at very near the velocity of the car. The peds CM may or may not have even moved. If we know the height of the ped, and the height of the cars bumper we can calculate his angular acceleration.

The question of what happens to the CM is more difficult? How high is his CM? How high is the top of the Porsche? Will the collision with his leg bring his CM down? If so will there be a second impact (as I experienced) or will the high speed of the car mean that it has passed completely underneath the ped without a second impact.

I do not know how to answer those last critical questions.
klubo
#3
Dec28-04, 06:27 PM
P: 3
Integral thank you for a much more detailed question than I ever deserved. Sorry to hear about your accident.

Can we simplify this? Clearly there's more to the motion than I gave it credit for as you pointed out so well.

Let's say I just wanted to compute in the simplest terms the force of an object traveling at 95 mph hitting a stationary object. How should I compute that impact force?

And how do I handle that the acceleration portion of the equation is expressed in "miles per hour" or "kilometers per hour" but the impact doesn't take an hour, it takes an instant -

also what units should I be using, since the result will come out differently if I use miles per hour than if I use kilometers per hour? given the equation F=m*a.

thanks much.
B

dextercioby
#4
Dec28-04, 06:42 PM
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Calculating impact force

Quote Quote by klubo
Integral thank you for a much more detailed question than I ever deserved. Sorry to hear about your accident.
I believe it was a "story",an invention,at least,i hope so.

Quote Quote by klubo
Let's say I just wanted to compute in the simplest terms the force of an object traveling at 95 mph hitting a stationary object. How should I compute that impact force?
As i recall:
[tex] F=\frac{\Delta p}{\Delta t}=m\frac{v_{f}-v_{i}}{t_{f}-t_{i}} [/tex]
Chose the initial moment of time to be zero.If u're given final time (the moment of time when the body loses contact with the car) and no energy is lost in the collision (perfect elastic collision,not really possible,though),then u can compute the force of impact quite easily.

Quote Quote by klubo
And how do I handle that the acceleration portion of the equation is expressed in "miles per hour" or "kilometers per hour" but the impact doesn't take an hour, it takes an instant -
also what units should I be using, since the result will come out differently if I use miles per hour than if I use kilometers per hour? given the equation F=m*a.
thanks much.
B
As u can see,there is no acceleration involved.U're given the car's velocity in mph which should be transformed into SI units,to recover the force in Newtons.
[tex]1Mph \sim 1609 m\cdot hour^{-1}=\frac{1609}{3600} ms^{-1}[/tex]

Daniel.
klubo
#5
Dec28-04, 07:17 PM
P: 3
thanks Daniel!
Yes it's a story. Your help is much appreciated. Bill
krab
#6
Dec28-04, 07:19 PM
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Quote Quote by klubo
And how do I handle that the acceleration portion of the equation is expressed in "miles per hour" or "kilometers per hour" but the impact doesn't take an hour, it takes an instant -
Hi klubo: "miles per hour " is a speed, not an acceleration. An acceleration is how quickly your speed changes, for example, in miles per hour per second. For example, the acceleration due to gravity is 32 fet per second per second.

Now, back to the collision. The information you have given is insufficient to calculate the force. If the car collides with something soft and light, the force will be small, maybe 100 pounds. If it is heavy and solid like a concrete divider, the forces can be tens of thousands of pounds.

In a sense, the Porsche is continuously colliding with something soft and light, namely air. The force of collision with the air at 95 mph is something like 320 pounds. (How did I calculate this? I know it takes roughly 80 hp to go 95 mph. 80 hp = 44000 ft-lbs/sec. 95 mph = 139 ft/sec. Divide the two and you get 316 lbs. F=P/v.)

So I guess the main message is that even though you know the weight and speed of the car and the object, and you know a formula (F=ma), you don't know enough to calculate the force. You need at least one other thing, like the time interval of collision, or the distance traveled during the collision.
Integral
#7
Dec28-04, 07:22 PM
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I believe it was a "story",an invention,at least,i hope so.
Absolute fact, every word of it. What I left out was that the car on the left side of the road with the bright lights, was the State Police, so I had State police witness that it was unavoidable on my part. The man I hit, had parked on the right shoulder of the road, ran across the road to to talk to the policeman, then turned and dashed back across the road to his car, unfortunately I was there. The only damage to my car was a broken windshield, to him a broken leg and bruises. Hitting and breaking the windshield may have prevented more serious injuries. It gave, the grill of a Mack truck,for instance, would not.
lsmJones
#8
Jan10-11, 11:07 PM
P: 2
In your equation, isn't the expression (vf - vi) / (tf - ti) simply equal to the average acceleration during the give time interval (ti to tf) ? Just an observation, which may be a solid argument against the "acceleration isn't a factor" comments.


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