Car Crash Physics: Deceleration Calculations

In summary, the tool website claims that if you input the weight and speed of a person in a car crash, it will generate a deceleration in meters per second. However, the calculations given are incorrect because the data does not follow the equation. The calculator also claims that the time of the crash is 18ms, which is impossible because there are many factors that affect it.
  • #1
Jimmy87
686
17
Hi,

I was watching a documentary on car crashes and how designs are continually being optimised to keep passengers safe. They were saying how many ‘g’s’ certain crashes were based on crumple zones, seatbelt stretch etc. I was interested to do some number crunching to see how big the decelerations are and came across this tool:

https://www.omnicalculator.com/physics/car-crash-force

If you put in 80kg and 50mph it generates 127.4g of deceleration. The kinetic energy of a 80kg person at 50mph person is 19,984J which is correct if you do the calculations. If you tick the wearing a seatbelt option it says the distance it stops you over is 16cm. If you divide the kinetic energy by 0.016m to get the force you get 124,900N. If you use F=ma to find the acceleration (124,900/80) you get an acceleration of 159g.

However if you do the change in speed (50mph is 22.35m/s) divided by the collision time they give of 0.018s (18ms) you get 124g. Why are the two ways of finding the deceleration 24g out? Also how would they have calculated the collision time as surely there are many factors that affect it so where do they get 18ms from?

Thanks
 
Physics news on Phys.org
  • #2
For a constant deceleration from speed ##v##, the stopping distance ##d## and stopping time ##t## would be related by:$$d = \frac 1 2 vt$$We can see that this equation does not hold for your data.
 
  • Like
Likes russ_watters
  • #3
Jimmy87 said:
Also how would they have calculated the collision time as surely there are many factors that affect it so where do they get 18ms from?
This link should come with huge caveats but, read carefully, it's well worth looking at.
I read through parts of t hat web page. I was disturbed to read
"The driver flies forward due to the inertial force" which made me a bit doubtful about all the rest - although the calculations may be right in some respects. We all know Newton's first law of motion which tells us that a body continues with uniform velocity unless there is an impressed force. What force is involved which could make the body "fly forward"?? And what is an "Inertial Force" in this context? The driver just keeps going (with his Momentum) at 50kph until some force acts to slow him down. This is the force that can do the damage if it's too high. Dominik Czernia may have gone too far in trying to make it all approachable (as he claims) so you're only after ball park figures and should always err on the safe side.
Another problem is where is his calculated force applied? A small, brief knock on the head can really mess you up as you hit the windscreen. A steering wheel will mess you up far more than a seat belt (with or without a padded overcoat. Pressure is at least as relevant as force.

You ask about the time calculation. Nothing is constant over the impact so what time value is relevant? I've no idea. Also the calculator gives you a figure which is supposed to correspond to a mass, resting on you. That's a step too far, IMO as it's brief and goes away once you have stopped. Another misguided attempt at making things approachable.
If all this were as straightforward as the website suggests then there would be no problems associated with litigation and car design. Fact is, the only values that are worth while are those which they get with crash test dummies.
 
  • Like
Likes jack action and Lnewqban
  • #4
To estimate the distance ##x## you should model everything as a spring and get the equivalent spring constant ##k##. Then it's only a matter of equating the kinetic energy with the spring potential energy:
$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$
Or:
$$x = \sqrt{\frac{m}{k}}v$$
The maximum force is then:
$$F_{max} = kx = \sqrt{km}v$$
And the average force will happen at ##\frac{x}{2}##:
$$F_{avg} = \frac{1}{2}\sqrt{km}v$$
The corresponding accelerations (##=\frac{F}{m}##) are:
$$a_{max} = \sqrt{\frac{k}{m}}v$$
$$a_{avg} = \frac{1}{2}\sqrt{\frac{k}{m}}v$$
The forces and accelerations presented in your link correspond to the average values presented in this post.
 
  • Like
Likes Lnewqban and PeroK
  • #5
jack action said:
To estimate the distance ##x## you should model everything as a spring and get the equivalent spring constant ##k##. Then it's only a matter of equating the kinetic energy with the spring potential energy:
$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$
Or:
$$x = \sqrt{\frac{m}{k}}v$$
The maximum force is then:
$$F_{max} = kx = \sqrt{km}v$$
And the average force will happen at ##\frac{x}{2}##:
$$F_{avg} = \frac{1}{2}\sqrt{km}v$$
The corresponding accelerations (##=\frac{F}{m}##) are:
$$a_{max} = \sqrt{\frac{k}{m}}v$$
$$a_{avg} = \frac{1}{2}\sqrt{\frac{k}{m}}v$$
The forces and accelerations presented in your link correspond to the average values presented in this post.
There’s a huge snag here. If the crumple zone and the seatbelt were equivalent to springs, the car would rebound from the wall and the belt would throw the driver back into the seat. Car bodies are specifically designed not to be resilient with a Q factor as low as possible. All energy is dissipated during the crash.
 
  • #6
In the end, any object that deforms permanently is still a spring that compresses until the yield point is passed.

And for what it's worth, people (and vehicles) do bounce back in car crashes, with or without restraint systems:

 
  • Like
Likes russ_watters and PeroK
  • #7
jack action said:
In the end, any object that deforms permanently is still a spring that compresses until the yield point is passed.
Oh yes that's true over a certain range. But there is no single yield point involved and neither is there an identifiable k value; the Energy is dissipated, not stored because there is (ideally) a minimum of resilience in a crumple zone. Bounce back is just a nuisance for designers (and passengers). Could you suggest what value of spring constant could be applied here?

If a car plunges into a gravel trench at an emergency run-off, there is no identifiable spring-like action; it's just all Energy loss and that's what happens when the mild steel of the car body is deformed. Why should Hooke's Law be involved, particularly?
 
  • #8
Straightforward numerical error in the original post - 16 cm is 0.16 m, not 0.016. So the deceleration provided by the seat belt is a more survivable 15.9 g. The crash is over well before the seat belt finishes its job.
 
  • Informative
  • Like
Likes PeroK and berkeman
  • #9
sophiecentaur said:
Could you suggest what value of spring constant could be applied here?
If the energy dissipated is ##\Delta E## with an average force ##F## acting on a distance ##\Delta x##, then:
$$dE = Fdx \text{ and } dF = kdx$$
So:
$$dE = \frac{F}{k}dF$$
Integrating:
$$\Delta E = \frac{F^2}{2k}$$
Or:
$$k = \frac{F^2}{2\Delta E}$$
The fact that once the energy is stored in our spring, it is later released in some other mechanism (breaking the spring) is irrelevant: the model still holds.

You could imagine some damping (##F= cv##) and some friction (##F= \mu N##) absorbing the energy during the crash as well and then, assuming only a spring, you will get:
$$k_{eq} = k + \frac{cv}{\Delta x} + \frac{\mu N}{\Delta x}$$
I don't have any sources, but I doubt those extra forces will be prevalent in reality compared with the stiffness of the parts. (Unless some bumper is specifically designed around those concepts.)

sophiecentaur said:
If a car plunges into a gravel trench at an emergency run-off, there is no identifiable spring-like action;
I would be tempted to model this based on the friction model rather than a spring.
 
  • #10
I think we should consider momentum and not energy in the crash situation.
 
  • Like
Likes sophiecentaur
  • #11
jack action said:
You could imagine some damping (##F= cv##) and some friction (##F= \mu N##) absorbing the energy during the crash as well
If you don't absorb it all, the car hasn't stopped moving. Bouncing can happen due to the suspension but have you ever heard of a crumple zone de-crumpling? I can see how you want to use spring constant but the steel is deforming non-elastically all the time. Hooke doesn't consider that.
jack action said:
I would be tempted to model this based on the friction model rather than a spring.
What is the inherent difference between these two loss generating mechanisms? They achieve the same thing. I could also suggest that the brakes will achieve negative acceleration; there's friction there too.
 
  • #12
sophiecentaur said:
If you don't absorb it all, the car hasn't stopped moving. Bouncing can happen due to the suspension but have you ever heard of a crumple zone de-crumpling? I can see how you want to use spring constant but the steel is deforming non-elastically all the time. Hooke doesn't consider that.
Here's a stress-strain curve:

1280px-Stress_strain_ductile.svg.png
The first part is the true elastic part. After that, you get permanent plastic deformation until fracture.

But that is based on the original cross-section area, referred to as the engineering stress-strain curve. But in the necking part, the cross-section area is reducing. If you consider this area reduction, you get the true stress-strain curve. The following figure shows the differences:

main-qimg-04a1b4a9a58b967c3d40eb00efe3e396.png

As you can see, it looks a lot more like a straight line, right up to the fracture, even in the plastic deformation section.

It doesn't really matter that the spring "won't come back" because the energy has been spent instead of stored. What matters is that we can evaluate how much energy it took.
 
  • #13
sophiecentaur said:
If you don't absorb it all, the car hasn't stopped moving. Bouncing can happen due to the suspension but have you ever heard of a crumple zone de-crumpling? I can see how you want to use spring constant but the steel is deforming non-elastically all the time. Hooke doesn't consider that.
Different parts of the car are compressing during the crash; some are compressing plastically like the crush zones, and others are compressing/deforming elastically and plastically still. The frame, for example, should exhibit some elastic rebound, IMO.
 
  • #14
jack action said:
The first part is the true elastic part. After that, you get permanent plastic deformation until fracture.
The devil is in the detail here. That graph is for a simple structure. In the case of a car body (which is designed not to behave like a spring) you have a multiplicity of these curves at work - one for each part - with very different vertical scales. You don't need to do any experiments to appreciate what I'm saying - just cast your mind back when you (or a friend) dinked their car against another car or a wall. Even for a very low speed impact, did the car genuinely bounce as a result of making that dent? Even the gentlest nudge will cause (expensive) deformation, showing that the Energy was dissipated by design and not stored. The structure wants to absorb energy.

The more I think about the Hooke's Law model, the more dodgy it looks.

Incidentally, there have been attempts to replace squashy steel with dampers on the bumper / fender, with air or even water being pushed out of a small hole. If you really want me to, I will search for a reference but I heard of this system way before the Internet existed.
 
  • #15
sophiecentaur said:
You don't need to do any experiments to appreciate what I'm saying - just cast your mind back when you (or a friend) dinked their car against another car or a wall.
Thanks for dredging up old dreadful memories... :wink:

I managed to crash my small BMW 2002 into a giant Oak Tree in the Napa Valley on the first rain day of the year many years ago (very slippery roads) at about 50mph when I slid sideways in a turn and almost recovered it but lost it trying to then avoid oncoming traffic. I definitely bounced backward off of that giant Oak Tree, despite the excellent crush zones in that car (for the era -- mid 1970s).
 
  • #16
berkeman said:
The frame, for example, should exhibit some elastic rebound, IMO.
I'm sure that's how old designs would have behaved but the Crumple Zone is a design feature that's intended to cope with extreme collisions. In some modern designs, the system allows the cab to rise a bit and slide over the massive and rigid engine to protect legs. Everything works in favour of loss not storage.
 
  • #17
sophiecentaur said:
In some modern designs, the system allows the cab to rise a bit and slide over the massive and rigid engine to protect legs.
Interesting. Can you link to some discussions about this? For medics, dealing with trapped/damaged legs in front-end car collisions is common. It would be great if there were a new car design feature that would help to keep that from happening. :smile:
 
  • #18
berkeman said:
Interesting. Can you link to some discussions about this? As a medic, dealing with trapped/damaged legs in front-end car collisions is common. It would be great if there were a new car design feature that would help to keep that from happening. :smile:
A very reasonable request - I'll get digging.
 
  • Like
Likes berkeman
  • #19
sophiecentaur said:
A very reasonable request - I'll get digging.
I have had a poke around and there is a lot of stuff about crumple zones and also about the 'Safety Cage' which is a very strong cage which protects the occupants by allowing relatively weaker parts to crumple. I could find nothing about any systems for deflecting parts of the car, relative to others.
The short while that I spent on the topic was actually rather disturbing. It did make me wonder about buying a Volvo next, though. They have a very good safety record.
Apparently, Volvo patented the three point seatbelt fixing and made it available to the rest of the world. Thoroughly good chaps, IMO.
 
  • Like
Likes berkeman
  • #20
Frabjous said:
Here’s a plot of deceleration as a function of distance. They are designed for <20g for passengers.

View attachment 321497
But there are two ways of interpreting a deceleration versus distance graph:

1. Peak deceleration for a given crash against distance for the point on the car where acceleration is measured.

This will be a decreasing function of displacement.

2. Current decelleration for the passenger against distance to which the crumple has progressed so far.

This will usually be an increasing function of displacement, though pathological cases could be designed.

It is 2) that we care about for our analysis. It is 1) that the diagram shows.
 
  • #21
jack action said:
To estimate the distance ##x## you should model everything as a spring and get the equivalent spring constant ##k##. Then it's only a matter of equating the kinetic energy with the spring potential energy:
$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$
Or:
$$x = \sqrt{\frac{m}{k}}v$$
The maximum force is then:
$$F_{max} = kx = \sqrt{km}v$$
And the average force will happen at ##\frac{x}{2}##:
$$F_{avg} = \frac{1}{2}\sqrt{km}v$$
The corresponding accelerations (##=\frac{F}{m}##) are:
$$a_{max} = \sqrt{\frac{k}{m}}v$$
$$a_{avg} = \frac{1}{2}\sqrt{\frac{k}{m}}v$$
The forces and accelerations presented in your link correspond to the average values presented in this post.
This is a bit misleading. So let's start from the beginning, i.e., the equation of motion for a point mass ##m## fixed at a massless spring with the ##x##-axis chosen such that the equilibrium position is at ##x=0##:
$$m \ddot{x}=-k x=-\frac{\mathrm{d}}{\mathrm{d} x} \left (\frac{k}{2} x^2 \right).$$
Writing the force in terms of a potential shows that total energy is conserved,
$$E=T+V=\frac{m}{2} \dot{x}^2 + \frac{k}{2} x^2=\text{const}.$$
Now instead of using this first integral it's simpler to directly integrate the equation of motion, which we can write as
$$\ddot{x} = -\omega^2 x, \quad \omega^2=k/m.$$
The general solution is
$$x(t)=A \cos(\omega t-\phi).$$
Here ##A## and ##\phi## are integration constants to be determined from the intial conditions. Plugging this into the formula for the energy gives
$$E=\frac{m}{2} \dot{x}^2 + \frac{m \omega^2}{2} x^2=\frac{m A^2 \omega^2}{2}.$$
The amplitude ##A## is of course determined by ##|x|=|x|_{\text{max}}##, ##\dot{x}=0##.
 
  • #22
vanhees71 said:
This is a bit misleading. So let's start from the beginning, i.e., the equation of motion for a point mass ##m## fixed at a massless spring with the ##x##-axis chosen such that the equilibrium position is at ##x=0##:
$$m \ddot{x}=-k x=-\frac{\mathrm{d}}{\mathrm{d} x} \left (\frac{k}{2} x^2 \right).$$
Writing the force in terms of a potential shows that total energy is conserved,
$$E=T+V=\frac{m}{2} \dot{x}^2 + \frac{k}{2} x^2=\text{const}.$$
Now instead of using this first integral it's simpler to directly integrate the equation of motion, which we can write as
$$\ddot{x} = -\omega^2 x, \quad \omega^2=k/m.$$
The general solution is
$$x(t)=A \cos(\omega t-\phi).$$
Here ##A## and ##\phi## are integration constants to be determined from the intial conditions. Plugging this into the formula for the energy gives
$$E=\frac{m}{2} \dot{x}^2 + \frac{m \omega^2}{2} x^2=\frac{m A^2 \omega^2}{2}.$$
The amplitude ##A## is of course determined by ##|x|=|x|_{\text{max}}##, ##\dot{x}=0##.
It's also misleading because none oxf it appears to contain a friction term (the work done on the steel). The friction force F accounts for ∫Fxdx Energy. F will vary over the distance, of course but can the total energy lost in this way be ignored in any appropriate model?

It isn't true that "
t total energy is conserved

Unless you include the energy due to inelastic deformation. Have they done this?
 
  • #23
sophiecentaur said:
It's also misleading because none oxf it appears to contain a friction term (the work done on the steel). The friction force F accounts for ∫Fxdx Energy. F will vary over the distance, of course but can the total energy lost in this way be ignored in any appropriate model?

It isn't true that "Unless you include the energy due to inelastic deformation. Have they done this?
What is wrong with assuming the energy dissipated as heat is given by ##\frac{1}{2}k x^2##. Surely there are worse models?

Once a material yields it needs no extra force to continue the deformation, it flows. Based on that wouldn’t we expect a linear spring type force to overestimate the forces?

To me it seems like it would approximate the force to continue yielding material, and the force of bringing presently un-yielded material up to yield as the crumple progresses.
 
Last edited:
  • #24
erobz said:
What is wrong with assuming the energy dissipated as heat is given by 12kx2. Surely there are worse models?
Firstly, we are not dealing with a spring.
If so, you would have to identify which body component is 'the spring', with a particular k value.
To justify the approach, you would need to show that the decelerating force is proportional to displacement. Have you that evidence that slowly compressing a car against a wall produces a linearly increasing force and elastic deformation? The very term "crumple" implies non-springlike distortion. When the tow truck pulls the car away from the wall, does the front regain its original shape? Of course not; the very simplest model would consist of many springs with different k's and different lengths, in series and parallel. First, one spring fails, then another, then another, until the car stops.

If a simple model was sufficient then there would be much less need for expensive crash testing with slo mo photography etc. The decelerating force could be measured by a series of stress / strain measurements and that would predict the g force throughout a collision.

I'm actually surprised that this thread has survived so well when I remember the number of discussions of car collisions which have died an early death through lack of Science.
 
  • #25
sophiecentaur said:
Firstly, we are not dealing with a spring.
If so, you would have to identify which body component is 'the spring', with a particular k value.
To justify the approach, you would need to show that the decelerating force is proportional to displacement. Have you that evidence that slowly compressing a car against a wall produces a linearly increasing force and elastic deformation? The very term "crumple" implies non-springlike distortion. When the tow truck pulls the car away from the wall, does the front regain its original shape? Of course not; the very simplest model would consist of many springs with different k's and different lengths, in series and parallel. First, one spring fails, then another, then another, until the car stops.

If a simple model was sufficient then there would be much less need for expensive crash testing with slo mo photography etc. The decelerating force could be measured by a series of stress / strain measurements and that would predict the g force throughout a collision.

I'm actually surprised that this thread has survived so well when I remember the number of discussions of car collisions which have died an early death through lack of Science.
Are you calling for it to be killed? I see no unscientific discussion here? Are we not allowed to think out loud on PF?

No one is suggesting the automotive industry give up on whatever they do. This is just back of the envelope “academic” exploration.
 
Last edited:
  • #26
sophiecentaur said:
Firstly, we are not dealing with a spring.
If so, you would have to identify which body component is 'the spring', with a particular k value.
Almost all the metal material crumpling, acts like a spring until it yields, then it flows under constant force , and new un-crumpled material acts like a spring until it yields etc… this goes on until the energy of the crash has been dissipated as heat.
 
  • Like
Likes jack action
  • #27
From an engineering point of view, it would be desirable to design the crumple zone to ramp up quickly to a not-quite-deadly decelleration rate and stay there. The system we are attempting to model is one that may have been consciously engineered away from the simple idealization we want to use.

This would suggest that no simple model will be reliable without cracking open the black box.
 
  • Like
Likes cjl, sophiecentaur and erobz
  • #28
jbriggs444 said:
From an engineering point of view, it would be desirable to design the crumple zone to ramp up quickly to a not-quite-deadly decelleration rate and stay there. The system we are attempting to model is one that may have been consciously engineered away from the simple idealization we want to use.

This would suggest that no simple model will be reliable without cracking open the black box.
That may be the case that it was engineered away from what we might expect without design intervention, but trying to understand the un-engineered outcome usually comes first. I don't think anyone was looking for a reliable model (I wasn't), just a way to approach a reasonable upper bound?

Without intervention I would expect something like this, hence the linear spring.
1675472143019.png
 
  • Like
Likes jbriggs444
  • #29
sophiecentaur said:
Have you that evidence that slowly compressing a car against a wall produces a linearly increasing force and elastic deformation?
That was shown in post #12. For sure there is a yield point where if you release the input force the car will regain its original shape. Cars are actually designed that way for low-impact accidents (You'll like that, the car bounces back):



Any solid material has a stiffness with an elastic region and a yield point. A spring is only a special design that lowers substantially the stiffness of the material by playing with geometry. A solid bar is still a very stiff spring.

In the elastic region, the energy is stored and can be fully restored. Once the yield point is passed, the energy is released some other way and cannot be used to restore the original shape - but the amount of energy is still whatever was inputted by way of compressing/extending/twisting the material (stiffness times displacement).

sophiecentaur said:
the very simplest model would consist of many springs with different k's and different lengths, in series and parallel.
Yes, and you could model an equivalent spring from it. The easiest way would be to determine it experimentally by taking measurements while crushing an actual car. Even if one spring breaks before the other - therefore altering the overall stiffness - an "average" stiffness can surely be evaluated for the entire process.

A more complex model would surely involve a damping component, but a very simple model with only a spring should get you somewhere close as I'm not imagining we have an overdamped system by mostly deforming metal.

sophiecentaur said:
Even the gentlest nudge will cause (expensive) deformation, showing that the Energy was dissipated by design and not stored.
It dissipates the energy that was first stored before the yield point. There are not really other ways in the case of deforming material. (Well some have a damping effect more pronounced than others.)
 
  • Like
Likes Lnewqban and erobz
  • #30
erobz said:
Almost all the metal material crumpling, acts like a spring until it yields, then it flows under constant force , and new un-crumpled material acts like a spring until it yields etc… this goes on until the energy of the crash has been dissipated as heat.
Exactly. That force, after the yield point will be dissipating energy throughout the slowing down process. One can either insist that the lost / stored energy is kx2/2 or Fx where F is designed to be a more or less constant force. One is a spring model and the other is a friction model.
erobz said:
That may be the case that it was engineered away from what we might expect without design intervention, but trying to understand the un-engineered outcome usually comes first. I don't think anyone was looking for a reliable model (I wasn't), just a way to approach a reasonable upper bound?

Without intervention I would expect something like this, hence the linear spring.
View attachment 321704
You could design in that characteristic but , if you want a more uniform retardation, then it wouldn't be optimal as the high force would be at the end, producing way above retardation. I just don't see why this would be a good car body design.
jack action said:
Yes, and you could model an equivalent spring from it.
An equivalent Duck could be very near a real Duck so why not use a Duck? (Yes, I know - the desired loss factor.) But we don't actually want a Duck in this case.
jack action said:
It dissipates the energy that was first stored before the yield point.
And Work continues to be done on it. Actually, I think we may be using the term 'yield point' inappropriately. I think we really need it the 'elastic limit', beyond which the material maintains the stress force and, thus, dissipates the energy (similar force over the remanding distance). The fact that steel has a useful non-elastic range makes it very attractive for construction of things like car bodies; it makes small dents easy to repair. Garden tools can still be used, despite having been distorted. That hasn't taken the tool to its yield point - just past its elastic limit.

It's hard to invoke such a simple model as a spring because the dissipative force when a complicated object distorts is nothing like the basic graphs of wires and simple beams. The 'friction' comes from all over, at different times and in different directions.
erobz said:
Are you calling for it to be killed? I see no unscientific discussion here? Are we not allowed to think out loud on PF?
Sorry; I didn't mean it that way. I should have used the word "died", rather than "killed". PF does a lot of thinking aloud and that's only to be encouraged. It may be the very fact that equations appear all over the place which accounts for the thread's survival. My point is that many of those equations just don't apply here and that the method is being defended, rather than other methods considered.
 
  • #31
sophiecentaur said:
You could design in that characteristic but , if you want a more uniform retardation, then it wouldn't be optimal as the high force would be at the end, producing way above retardation. I just don't see why this would be a good car body design.
I didn't say that would be a designed characteristic force profile, I said that would be an un-designed characteristic force profile...perhaps what we would expect if we didn't intervein.

If the optimal force displacement curve was like this:

1675522454928.png


Then we can assume the car manufacturers have been able to achieve this type of response, which yields a virtually constant acceleration at some physiological threshold, and we could determine quite readily what that threshold is and/or if that assumption is correct by watching a vehicle crash and measuring the rate of acceleration.
 
  • #32
sophiecentaur said:
I think we may be using the term 'yield point' inappropriately. I think we really need it the 'elastic limit'
That is the definition of yield point (and how I was referring to it in my previous posts):
https://en.wikipedia.org/wiki/Yield_(engineering) said:
In materials science and engineering, the yield point is the point on a stress-strain curve that indicates the limit of elastic behavior and the beginning of plastic behavior.
You may be confusing ultimate strength with yield strength:
https://en.wikipedia.org/wiki/Ultimate_tensile_strength said:
[Ultimate strength] is the maximum stress that a material can withstand while being stretched or pulled before breaking. In brittle materials the ultimate tensile strength is close to the yield point, whereas in ductile materials the ultimate tensile strength can be higher.

sophiecentaur said:
And Work continues to be done on it.
And why can't we assume it still is proportional to displacement for the sake of simplicity?

One thing is sure: the force is still linked to the displacement. So let me correct my initial equation by not assuming the linearity:
$$F = Kx^n$$
Such that:
$$\frac{1}{2}mv^2 = \frac{1}{n+1}Kx^{n+1}$$
$$x = \left(\frac{n+1}{2}\frac{mv^2}{K}\right)^{\frac{1}{n+1}}$$
And:
$$F_{max} = Kx^n = \left(\frac{n+1}{2}K^{\frac{1}{n}}mv^2\right)^{\frac{n}{n+1}}$$
So, in retrospect, from good to better:
  • Using ##F = \frac{mv^2}{2x}## - as in the OP - says nothing about the vehicle; all we can do is compile typical values of ##x## for typical vehicles.
  • Using ##F_{avg} = \frac{1}{2}\sqrt{km}v## defines the force based on a - simple - design characteristic (##k##) of a vehicle.
  • Using ##F_{max} = \left(\frac{n+1}{2}K^{\frac{1}{n}}mv^2\right)^{\frac{n}{n+1}}## (too lazy to determine the average force which would compare better to the previous equations) defines the force based on two design characteristics (##K##, ##n##) of the vehicle; for the most demanding of us.
The point was that the method presented in the OP doesn't introduce any characteristics of the vehicle design other than the mass. Knowing that my neighbor's car produced a force of 100 kN with a displacement of 1 m doesn't mean my car will produce the same force with the same displacement (given equal mass and velocity). However, if my neighbor's car produces a force of 100kN with a known stiffness, I have much more chance that my car will produce the same force given my car has the same stiffness.
 
  • #33
erobz said:
Then we can assume the car manufacturers have been able to achieve this type of response, which yields a virtually constant acceleration at some physiological threshold, and we could determine quite readily what that threshold is and/or if that assumption is correct by watching a vehicle crash and measuring the rate of acceleration.
I think we can assume that they wouldn't be spending a lot of research money and produce expensive designs which would not improve crash survival. Simulating a spring characteristic would be far from optimal desirable. That 'ideal' characteristic would make the least worst of a crash situation.

In a different context, we know that stunt people use piles of cardboard boxes to land on after stunt falls from unbelievable heights. Cardboard is 'not very springy'. :wink:
 
  • #34
As I sort of said in my off-topic deleted post, 20G's is a ballpark desirable max average deceleration for the passenger compartment.
If we use the the equation ##a=v^2/2d## where d is the stopping distance
1) Using 20G for average deceleration, we get an elastic stopping distance of 1.3 cm for a 5 mph collision (assume inelastic deformation starts at 5mph)
2) Using the elastic stopping distance we get
Velocity (mph) Deceleration (G's)
10 39
15 88
20 157
25 245
30 353
35 481
3) If we limit the deceleration to 20G, we get stopping distances for the passenger compartment
Velocity (mph) Stopping Distance (m)
10 .051
15 .115
20 .204
25 .318
30 .458
35 .624
So the designers are probably designing for crushing to go something like this.

Here’s the plot I got the 20G from
158CD6C7-AA37-4649-AA51-674A3D7FD0E9.jpeg
 
Last edited:
  • Like
Likes erobz and Lnewqban
  • #35
jack action said:
That is the definition of yield point (and how I was referring to it in my previous posts):
Yes - you are right about that. Sorry.

jack action said:
And why can't we assume it still is proportional to displacement for the sake of simplicity?
That's what I have been saying all along - constant work / force with displacement. This is not the characteristic of a spring, for which the force increases in proportion to displacement.

You are suggesting that a non linear spring could be a suitable model. I guess there will be a power law which would give a characteristic near to that in post #31. The exponent would need to be fractional (a fairly small fraction). But, apart from the fact that you seem wedded to a spring model, how would you introduce actual net loss without something, somewhere, doing work on something? Perhaps a ratchet (like a catching diode in electronics) with added rebound friction (??) could absorb the energy on the way back.
 

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • General Math
Replies
1
Views
1K
  • Classical Physics
2
Replies
61
Views
2K
Replies
1
Views
2K
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top