Calculating kinetic energy turned into Pressure (Pascals) during an impact

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Hello,
A silly question I'm sure, if a projectile moving at 13.59 m/s with a KE = 527.47 J hits an area of 3 cm^2, what is the force transfer to that area and the resultant pressure applied expressed in Pascals?
Do I divide the KE by a volume instead of area to Pa?
 

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  • #2
berkeman
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Welcome to the PF. :smile:

Is this related to your work? (Physiotherapist, Kinesiologist)

The force upon impact will be applied over time over some distance as the object is brought to rest. Pressure is Force/Area, and in this case will be varying during the deceleration of the object. The "pressure" or force as a function of time will vary depending on the duration of the impact and how big the deceleration distance is.

Do you have an example in mind? Maybe we could try to help you ballpark the forces and "pressures" if we had more information about what that projectile hits.

Also, have you read about the concept of "Impulse"?
 
  • #3
russ_watters
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Hello,
A silly question I'm sure, if a projectile moving at 13.59 m/s with a KE = 527.47 J hits an area of 3 cm^2, what is the force transfer to that area and the resultant pressure applied expressed in Pascals?
Do I divide the KE by a volume instead of area to Pa?
It's not a silly question, it is actually a very difficult question. There is no direct conversion from kinetic energy to pressure. Pressure at impact isnt even a single number over time or across an impact area.

The best you might be able to do would be to way simplify by choosing a penetration distance, solving for average deceleration rate. From that you calculate average force and over the cross sectional area of the projectile, pressure.
 
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  • #4
anorlunda
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Wouldn't the peak pressure depend on the shapes? I'm thinking of a pointed bullet.
 
  • #5
Nugatory
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Wouldn't the peak pressure depend on the shapes? I'm thinking of a pointed bullet.
It will depend on the shape because that affects the area and hence the pressure for a given force at any given moment. The force in turn will depend on the rate of deceleration, which will depend in a very complicated way on the way in which the two objects deform, the forces required to produce that deformation, the energy absorbed by the deformation, and how quickly the impacting object comes to a stop (relative to the impacted object, which may itself be accelerated by the collision). Like Russ says, this is a very difficult problem - there is general or simple relationship between the initial kinetic energy and the peak pressure.
 
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  • #6
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This is actually a problem I have been chipping away at for a bit in my spare time.
We can assume the projectile was a sphere with a volume of 0.0021 m^3.
I have thought about how much displacement of the target, maybe it moved a couple of centimetres after impact.
I am trying to establish if the energy transfer from the projectile to the target placed sufficient energy into the target to damage structures inside that have shear strength of 0.112 MPa and 0.130 MPa.
I sincerely appreciate all the interest and help!
 
  • #7
russ_watters
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I am trying to establish if the energy transfer from the projectile to the target placed sufficient energy into the target to damage structures inside that have shear strength of 0.112 MPa and 0.130 MPa.
Now it's even harder. If it were a mass impacting a spring, we could calculate an exact answer. A rubber ball impacting a floor that is assumed not to "give", maybe there would be a shot. But for damage to a structure, outside of a sophisticated computer model I doubt it can be done.

There is a reason the NTSB still does actual crash testing.
 
  • #8
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Now it's even harder. If it were a mass impacting a spring, we could calculate an exact answer. A rubber ball impacting a floor that is assumed not to "give", maybe there would be a shot. But for damage to a structure, outside of a sophisticated computer model I doubt it can be done.

There is a reason the NTSB still does actual crash testing.
I'd love to do an experiment but for now just have to see if there was sufficient energy into the system to create a sufficient shear to the internal structures.

I thought about just taking the KE and transferring into force as KE/d, approximating about a 2 cm distance of movement but thought I should try a couple of things to be more precise.
 
  • #9
berkeman
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We can assume the projectile was a sphere with a volume of 0.0021 m^3.
I have thought about how much displacement of the target, maybe it moved a couple of centimetres after impact.
I am trying to establish if the energy transfer from the projectile to the target placed sufficient energy into the target to damage structures inside that have shear strength of 0.112 MPa and 0.130 MPa.
This sounds like something our local ballistics expert @Dr. Courtney can help with. He may be able to point you to some of his publications on the subject...
 
  • #10
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This sounds like something our local ballistics expert @Dr. Courtney can help with. He may be able to point you to some of his publications on the subject...
Awesome. Thank you.
 
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  • #11
Dr. Courtney
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There are no exact answers in impact ballistics. There are a few principles that will give order of magnitude estimates that are often better - factor of 2 or 3 instead of 10. But usually, the experiment needs to be done if you want accuracy on the order of 20% or better, because material properties are much different at the high strain rates of ballistic impacts.

You can reckon the pressure as the impact energy divided by the volume. But what volume do you use? For a hard sphere impacting modeling clay and for the related phenomena of behind armor blunt trauma, it makes sense to use the volume of the displaced clay. But what if both the projectile and the target are deformed? What if the target is set into motion by the impact?

In these cases, it makes more sense to relate the impact energy to the impact force and displacement (stopping distance) using the work energy theorem. That will give you the average impact force in an accurate and straightforward manner every time if you know the stopping distance. This works even if the target is set into motion and if both target and projectile are deforming. But often, it is the _peak_ force rather than the average force that is of interest. Having looked at hundreds of these real events carefully, the peak to average ratio is usually close to 3 - so simply multiply the average by 3 to estimate the peak impact force of a ballistic event.

Pressure is then simply the force divided by the area. But which area to choose if there are multiple possibilities - deforming projectile, deforming target, cross sectional or full frontal, etc? Most of the inaccuracy comes from the error in the peak to average ratio and in an inability to choose which area you really need - which is tough because you don't know the time sequence of the deformations. Still, you should be able to put bounds on your error bars by considering the most likely candidates and realize your SWAG is likely within a factor of 2 or 3. With enough experience comparing SWAGs with experiments, your SWAGs will get better. My SWAGs are usually better than the best numerical models, because the finite element models still don't have very good material properties for the high strain rate events.
 
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