Calculating Impact Force?

[druid91]

So, suppose you had a 12,000 Kilogram object moving at 2,880 Meters per second. How bad is that compared to say... a car crash? I'm sure there's some equations around somewhere, but the only ones I could find gave me rather unhelpful answers.

[Lethologica]

For ease we'll assume the car's 1200kg and moves at 28.8 m/s before the crash--a high-speed highway crash with a light mostly-empty sedan. So we're looking at 1,000x the momentum and 100,000x the kinetic energy, to give a rough idea. Impact force depends on the deformation characteristics of both the impacting object and the object being impacted, so that's a bit harder to spitball, but anything that was damaged by the car crash is basically obliterated by this.

[Aeson]

The average impact force experienced by a body during an impact event can be estimated as the change in the body's momentum due to the impact event divided by the duration of the impact event. Maximum impact forces can be considerably greater than average, however, and they're not easily computed.

[Mith]

It depends on what you mean by a car crash. And the scenario in question.

For impulse experienced on an object, the formula I would use is:

m(dv) = F(dt); m = mass of object [kg], (dv) = change in the velocity (initial velocity - final velocity) [m/s], F = Force on object [N], (dt) the time interval the impulse acts on the object [s].

This means that the shorter the change in momentum is on an object is, the greater the force is on the object. Dropping an egg onto a pillow prevents breaking the shell as the pillow compresses under the egg, spreading out the impact time, and reducing the force on the shell.

Now the numbers you gave, assuming that the object is brought to a complete stop, gives a total change in momentum of (12,000 kg)(2,880 m/s) = 34,560,000 N*s or 34,560 kN*s.

Car crashes vary in how long it takes for things to settle, but let's say it takes half a second for the initial impact and for most of the energy to be transferred through the bodies involved in the impact.

Therefore the force on the bodies involved is F = (34,560 kN*s)/(0.5 s) = 69,120 kN with the object given in your example.

Using the numbers in the previous post, let's say take a 1,200 kg car at 28.8 m/s, which is brought to a complete stop over 0.5 seconds. the force on the car is F = (1,200 kg)(28.8 m/s)/(0.5 s) = 69,120 kN or as pointed out previously, 1,000 times less severe of an impact on the car.

If you want a ratio of the forces involved, the formula is F₂/F₁ = [(m₂v₂)/(dt₂)]/[(m₁v₁)/(dt₁)]

or a bit more compactly:

F₂/F₁ =[m₂v₂dt₁]/[m₁v₁dt₂]

This gets a bit more complicated when you have two moving bodies, as this assumes essentially an object running into a wall, but the initial formula is still the same.

[jayem]

I wish I knew.
Generally in one set of simple low energy collisions it seems to be roughly proportional to initial momentum I think (I.E the duration changes slowly). Though that was a small object that bounced (so the momentum transferred also changed). I can't remember what the effect of a cushioning object was (I'm pretty sure momentum was conserved, thankfully).

[Anymage]

To highlight what Lethologica said, remember that bombs, asteroid collisions, and similarly significant events are usually measured in terms of kilotons of TNT. That's a measure of energy, not force. Similarly, the damage done by a bullet or similar object is more closely related to energy than momentum. So if you want to know how destructive something is, the energy involved is usually a better benchmark than the average or instantaneous force.

[factotum]

To highlight what Lethologica said, remember that bombs, asteroid collisions, and similarly significant events are usually measured in terms of kilotons of TNT

And in this case, a 12 tonne object travelling at the speed specified has a kinetic energy equivalent of about 12 tons of TNT...

[Aeson]

And in this case, a 12 tonne object travelling at the speed specified has a kinetic energy equivalent of about 12 tons of TNT...

While it's probably safe to assume that the speed druid91 gave is meant to be relative to the thing impacted, or at least is only marginally different from the speed relative to the thing impacted, it bears mentioning that the speed that matters is the speed relative to the thing being impacted, not the speed in some arbitrary frame of reference or a hypothetical 'absolute' speed. If we have two accidents, one a head-on collision involving two cars each moving at about 30 miles per hour relative to the Earth's surface in diametrically opposed directions and the other involving a car moving at 70 miles per hour and another car moving at 60 miles per hour in the same direction, the former accident is considerably worse than the latter insofar as the initial impact forces are concerned, even though the speeds of the cars in our chosen frame of reference (the Earth's surface) would suggest that the cars moving at 60 and 70 miles per hour should carry significantly more kinetic energy than the cars moving at 30 miles per hour, assuming all four vehicles have similar masses.

[Tyndmyr]

So, suppose you had a 12,000 Kilogram object moving at 2,880 Meters per second. How bad is that compared to say... a car crash? I'm sure there's some equations around somewhere, but the only ones I could find gave me rather unhelpful answers.

Bad for...what?

At those speeds, anything soft like a human body, it'll be essentially irrelevant. Either hit by, or doing the hitting.

For the sake of comparison, it's like driving roughly 6,500 mph. This can be safely estimated as much worse than a car crash, generally speaking. The results of most such impacts are best described as "splattering". Any exceptions will be collisions between two things of extremely different densities.

[Grey_Wolf_c]

The results of most such impacts are best described as "splattering".

Experimental results suggest the correct adjective is "pulverizing" more than "splattering".


(tongue in cheek; just wanted a reason to post the gif)
GW

[Aeson]

Any exceptions will be collisions between two things of extremely different densities.

Or collisions involving objects with sufficiently similar velocities that spitballing the objects' speeds relative to one another at the speed of one of the objects in the frame of reference is inappropriate. The ISS has a mass of about 420 metric tons and an orbital velocity of about 7660 m/s in a geocentric frame of reference, and experienced an impact event with a body massing around 80 metric tons which is also traveling at about 7660 m/s every time a space shuttle docked with it, but that doesn't mean that the ISS had to absorb the kinetic energy equivalent of ~half a kiloton of TNT or that the space shuttle has to absorb the kinetic energy equivalent of ~3 kilotons of TNT every time a space shuttle docked with the ISS.

'Absolute' speed in the frame of reference is fairly irrelevant, especially when the frame of reference is unknown; what you need for computing impact forces is the speed of the impactor relative to the impactee. As such, the answer to the question "how bad is an impact involving a 12Mg object traveling at 2880 m/s compared to a car crash?" is "how the hell should I know?" The upper bound on the relative speed between impactor and impactee at the time of impact is c; the lower bound is 0 m/s. Assuming it hits a similarly massive (say, within an order of magnitude or two) object moving in the same direction at 2879.98 m/s in a featureless void, the answer is probably along the lines of 'not bad at all;' the space shuttles and ISS survived 'worse' every time they docked with one another and were none the worse for wear. Assuming its speed is relative to the Earth's surface and it hits your car which is traveling at any speed relative to the Earth's surface which may reasonably be expected of a typical car in any direction relative to the impactor, the answer is that it looks something like what Grey_Wolf_c posted, except in real time.

[Baby Gary]

It depends on what you mean by a car crash. And the scenario in question.

For impulse experienced on an object, the formula I would use is:

m(dv) = F(dt); m = mass of object [kg], (dv) = change in the velocity (initial velocity - final velocity) [m/s], F = Force on object [N], (dt) the time interval the impulse acts on the object [s].

This means that the shorter the change in momentum is on an object is, the greater the force is on the object. Dropping an egg onto a pillow prevents breaking the shell as the pillow compresses under the egg, spreading out the impact time, and reducing the force on the shell.

Now the numbers you gave, assuming that the object is brought to a complete stop, gives a total change in momentum of (12,000 kg)(2,880 m/s) = 34,560,000 N*s or 34,560 kN*s.

Car crashes vary in how long it takes for things to settle, but let's say it takes half a second for the initial impact and for most of the energy to be transferred through the bodies involved in the impact.

Therefore the force on the bodies involved is F = (34,560 kN*s)/(0.5 s) = 69,120 kN with the object given in your example.

Using the numbers in the previous post, let's say take a 1,200 kg car at 28.8 m/s, which is brought to a complete stop over 0.5 seconds. the force on the car is F = (1,200 kg)(28.8 m/s)/(0.5 s) = 69,120 kN or as pointed out previously, 1,000 times less severe of an impact on the car.

If you want a ratio of the forces involved, the formula is F₂/F₁ = [(m₂v₂)/(dt₂)]/[(m₁v₁)/(dt₁)]

or a bit more compactly:

F₂/F₁ =[m₂v₂dt₁]/[m₁v₁dt₂]

This gets a bit more complicated when you have two moving bodies, as this assumes essentially an object running into a wall, but the initial formula is still the same.

One slight problem with your equations. The formula for force is F=MA (force=mass*acceleration), and it seems as if you did MV (mass*velocity)

so if the velocity is 2880 m/s, and lets say the thing comes to a complete stop in... 0.1 seconds. Then acceleration is 28800 m/s^2 (A=V/T). If that is the acceleration is 345600000 newtons, or 345900 kN.

[jayem]

One slight problem with your equations. The formula for force is F=MA (force=mass*acceleration), and it seems as if you did MV (mass*velocity)

so if the velocity is 2880 m/s, and lets say the thing comes to a complete stop in... 0.1 seconds. Then acceleration is 28800 m/s^2 (A=V/T). If that is the acceleration is 345600000 newtons, or 345900 kN.

He's using the momentum-impulse variation. momentum=mass*(acceleration*time)=force*time.
Which for nice numbers only means you divide by time at the end rather than the start (hence the midstep of 34,560 kN*s.) and for nasty functions means you can put off dealing with changing bits till the end (and the difference between peaks and averages), and if you don't care with how the collision occurs possibly (e.g. effectively using conservation of momentum) completely.

E.g. If the car lost 90% of it's speed in 0.09s (linearly!) and then came to a stop in 0.5seconds, then the peak force would be 345900 kN. while the average would be 69,120 kN
But in all 3 cases Aeson (0.1 second), Grey Wolf (0.5) Jayem (mix) the impulse is always 34,560 kN*s

[Grey_Wolf_c]

Grey Wolf (0.5)

I'm fairly certain my entire contribution to this thread was a quibble about adjective, and reminding everyone of an awesome Mythbuster moment, so you probably meant someone else there.

GW

[Tyndmyr]

Or collisions involving objects with sufficiently similar velocities that spitballing the objects' speeds relative to one another at the speed of one of the objects in the frame of reference is inappropriate. The ISS has a mass of about 420 metric tons and an orbital velocity of about 7660 m/s in a geocentric frame of reference, and experienced an impact event with a body massing around 80 metric tons which is also traveling at about 7660 m/s every time a space shuttle docked with it, but that doesn't mean that the ISS had to absorb the kinetic energy equivalent of ~half a kiloton of TNT or that the space shuttle has to absorb the kinetic energy equivalent of ~3 kilotons of TNT every time a space shuttle docked with the ISS.

That's irrelevant, the only speed we care about is the relative velocity. Bringing external frames of reference into this is a needless complication.

[Lord Torath]

That's irrelevant, the only speed we care about is the relative velocity. Bringing external frames of reference into this is a needless complication.

I'm pretty sure that was Aeson's point.

[Mith]

He's using the momentum-impulse variation. momentum=mass*(acceleration*time)=force*time.
Which for nice numbers only means you divide by time at the end rather than the start (hence the midstep of 34,560 kN*s.) and for nasty functions means you can put off dealing with changing bits till the end (and the difference between peaks and averages), and if you don't care with how the collision occurs possibly (e.g. effectively using conservation of momentum) completely.

E.g. If the car lost 90% of it's speed in 0.09s (linearly!) and then came to a stop in 0.5seconds, then the peak force would be 345900 kN. while the average would be 69,120 kN
But in all 3 cases Aeson (0.1 second), Grey Wolf (0.5) Jayem (mix) the impulse is always 34,560 kN*s

Thanks for the complete explanation. Due to the nature of the question, I picked the formula that would give, IMO, the simpilest explanation without too much confusion. Plus it uses the variables given, and can be modified to fit any comparison.

[Baby Gary]

He's using the momentum-impulse variation. momentum=mass*(acceleration*time)=force*time.
Which for nice numbers only means you divide by time at the end rather than the start (hence the midstep of 34,560 kN*s.) and for nasty functions means you can put off dealing with changing bits till the end (and the difference between peaks and averages), and if you don't care with how the collision occurs possibly (e.g. effectively using conservation of momentum) completely.

E.g. If the car lost 90% of it's speed in 0.09s (linearly!) and then came to a stop in 0.5seconds, then the peak force would be 345900 kN. while the average would be 69,120 kN
But in all 3 cases Aeson (0.1 second), Grey Wolf (0.5) Jayem (mix) the impulse is always 34,560 kN*s

oops, my bad. I haven't done physics in a while