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Posted: 3/20/2017 1:55:44 PM EDT
Or Impulse divided by area? I was thinking about internal ballistics, wondering if there is a way to express the force acting on a bullet in the barrel. Obviously there is gas pressure, and it is exerted over some small time. Is there a name for this?
Or say someone is hanging something from a rope. How would you express the tension force in the rope, divided by the area, multiplied by the length of time it is hanging there? 


I think the poiseuille (Pl) is what you're looking for.


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I'm guessing the unit you seek is a Watt  which is defined as newtonmeters per second.
Pressure exerted over surface area reduces to newtonmeters, and adding a time element.... 

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To answer the title of your thread, Josh is right. Units of dynamic viscosity are in those terms.
To answer the question as described in the OP, Circuits is right. What you are trying to characterize is an energy (force over time). 

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Sounds like work (force over a distance creating motion) that imparts kinetic energy (mv^2/2to the bullet.



I was thinking more like the total accumulated stress on a part over time. No work is being done in my rope example, no energy is exerted.



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Originally Posted By JoshAston:
That's where you're wrong. View Quote View All Quotes View All Quotes 


Originally Posted By JoshAston:
That's where you're wrong. View Quote View All Quotes View All Quotes 


Originally Posted By TacticalGarand44:
Okay, there's no change in potential energy from gravity, no change in elastic energy in the rope... what am I missing? View Quote 

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Originally Posted By JoshAston:
There is a change in energy from when the rope was at rest. Once you hang something from it has a constant force being applied. View Quote 


That doesn't mean


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Originally Posted By JoshAston:
That doesn't mean work isn't being done or that energy isn't being exerted. Energy is just a constant in this case. View Quote 


No work is being done, but energy is being expended maintaining a nonrigid static system.


"The only real difference between the men and the boys, is the number and size, and cost of their toys."
NRA Life, GOA Life, CSSA Life, NRA Certified Instructor 
Originally Posted By TacticalGarand44:
Work is force over a distance. The rope is not exerting any energy on the weight, which is obvious by Conservation of Energy. Where is the energy coming from that the rope is exerting in your proposition? View Quote 

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So how much work is work is done to maintain a static system? Seriously asking, I don't remember this being covered in any physics courses.



Originally Posted By TacticalGarand44:
So how much work is work is done to maintain a static system? Seriously asking, I don't remember this being covered in any physics courses. View Quote You hold a 100lb bag on your arm, extended. Which is expending energy? No work is done either way, if no movement occurs. 

"The only real difference between the men and the boys, is the number and size, and cost of their toys."
NRA Life, GOA Life, CSSA Life, NRA Certified Instructor 
Originally Posted By TacticalGarand44:
So how much work is work is done to maintain a static system? Seriously asking, I don't remember this being covered in any physics courses. View Quote Been a few years since physics. 

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How is the energy quantified? How do I measure the amount of energy I expend holding up a 100 pound bag with my arm extended? How about over my head?



PSI? I'm really not certain. The rope is being maintained in an unnatural state though, that requires energy.
as for an individual holding it up, I'm not sure the unit of measurement, but again you're muscles are not at rest. 

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There is a dynamic system being described  a rigid arm versus a flesh and bone arm, resisting gravity.
The amount of energy expended to maintain a flexible arm as a rigid system is related to the description of the entire system. There is no net expenditure of work to maintain the rigid system, but the amount of energy expended to maintain the nonrigid system is nonzero. I'm not even sure how this gets described, but it is intuitively obvious that there is a difference between a rigid arm, and a nonrigid system being maintained in semirigid equilibrium. 

"The only real difference between the men and the boys, is the number and size, and cost of their toys."
NRA Life, GOA Life, CSSA Life, NRA Certified Instructor 
Okay, so rigidity is a factor? We never discussed anything like this in physics.



A bookshelf does not dissipate ANY energy (work) holding objects up.
It is a static system without a time term. Your body is NOT a rigid object. Holding an object requires some energy to be used. Statics and Dynamics are NOT the same thing. If their is any motion (even vibration) you no longer have a statics problem and work can be nonzero. The stressstrain behavior of materials varies enough to require more than simple considerations though. Steel does NOT typically exhibit classic 'fatigue' failure in low level loading well below the yield point. It is possible to construct steel alloys that CAN exhibit 'fatigue' type failure modes. Aluminum DOES exhibit 'fatigue' at loading well below yield point. Just the repeated application of a load to aluminum 'works' the material on the stressstrain curve and leads to crack formation. The first pressurized planes failed after repeated cycles. Airframes have a life span limited by the vibration and stressstrain environment they operate in. Because movement under stressstrain is not required it is NOT a work related phenomenon. The motion in stressstrain is so small it is not useful t indicate anything. You have to define your free body model correctly to evaluate work in a system. 


Thank you. That makes sense. I do recall the fatigue limits from my materials science course. That was one of the most fascinating classes I've ever taken, perhaps second only to thermodynamics.



Originally Posted By TacticalGarand44:
Okay, there's no change in potential energy from gravity, no change in elastic energy in the rope... what am I missing? View Quote Pressure = force / area Pressure * time = force * time / area force * time is known as impulse Impulse = change in momentum momentum = mass * velocity Play with these relationships and see where it takes you. 



Originally Posted By Trollslayer:
This is not correct. View Quote View All Quotes View All Quotes Originally Posted By Trollslayer:
Originally Posted By TacticalGarand44:
Work is force over a distance. The rope is not exerting any energy on the weight, which is obvious by Conservation of Energy. Where is the energy coming from that the rope is exerting in your proposition? 


Originally Posted By TacticalGarand44: View Quote The rope is not exerting any energy on the weight, You do not "exert energy". The rope certainly is exerting a force to stop the bag from falling. which is obvious by Conservation of Energy. . This is jibberish Where is the energy coming from that the rope is exerting in your proposition? You need to learn about strain energy. Go read up on energy stored in springs. Then, remember, that everything is a spring. Things respond to forces by stretching or compressing. That movement has associated with it a force and energy. 


As I said in other responses, I am considering the system in equilibrium, once the rope has stretched a little. Exert was the wrong word. 'Dissipate' was what I was looking for.



And the answer to the initial question of force over area is energy.
The gas pushes the bullet and the bullet moves. Now you have a force over a distance and are doing work. With a whole lot of unit conversions and integration over the length of the barrel you will get a KE expressed in ftlbs(force). KE = 0.5* m*v^2 You can also do changes like SCFM at a stated pressure and convert that directly into horsepower. It just takes a lot of careful unit conversions and paying attention to mass and weight. Keep in mind that in PSI it is pounds of force NOT pounds of mass. In the one common conventional system weight is in pounds and mass ins in slugs. There are other systems. There are also lbs (mass) with acceleration carried separately. In metric kg (mass) are not Newtons (force). The gas did work on the bullet as it moved down the barrel. The things that complicates the math is that the pressure is NOT constant but after a brief peak continuously decreasing as the volume behind the moving bullets increases. Once the powder has turned to a gas (a very hot gas) the moving bullet increases the volume trapping the gas. 


Originally Posted By brickeyee:
And the anser to the initial question of force over area is energy. The gas pushes the bullet and the bullet moves. Now you have a force over a distance and are doing work. With a whole lot of unit conversions and integration over the length of the barrel you will get a KE expressed in ftlbs(force). KE = 0.5* m*v^2 You can also do changes like SCFM at a stated pressure and convert that directly into horsepower. It just takes a lot of cared full unit conversions and paying attention to mass and weight. Keep in mind that in PSI it is pounds of force NOT pounds of mass. In the one common conventional system weight is in pounds and mass ins in slugs. There are other systems. There are also lbs (mass) with acceleration carried separately. In metric kg (mass) are not Newtons (force). The gas did work on the bullet as it moved down the barrel. The things that complicates the math is that the pressure is NOT constant but after a brief peak continuously decreasing as the volume behind the moving bullets increases. Once the powder has turned to a gas (a very hot gas) the moving bullet increases the volume trapping the gas. View Quote I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? 


Originally Posted By TacticalGarand44:
As I said in other responses, I am considering the system in equilibrium, once the rope has stretched a little. Exert was the wrong word. 'Dissipate' was what I was looking for. View Quote Even in equilibrium, the strain energy remains manifest in the system. It is a potential energy source. 


How about this one?
From earlier: pressure * time = impulse / area A bullet's section density (SD) is defined as SD = mass / area (crosssectional area) Expanding then simplifying: pressure * time = impulse * SD / mass = Pressure * time = velocity * SD Is that of interest to you? 


Originally Posted By Trollslayer:
How about this one? From earlier: pressure * time = impulse / area A bullet's section density (SD) is defined as SD = mass / area (crosssectional area) Expanding then simplifying: pressure * time = impulse * SD / mass = Pressure * time = velocity * SD Is that of interest to you? View Quote 


Pressure * time / SD = velocity



Originally Posted By TacticalGarand44:
Force over area is pressure. I am asking about pressure over time. I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? View Quote View All Quotes View All Quotes Originally Posted By TacticalGarand44:
Originally Posted By brickeyee:
And the anser to the initial question of force over area is energy. The gas pushes the bullet and the bullet moves. Now you have a force over a distance and are doing work. With a whole lot of unit conversions and integration over the length of the barrel you will get a KE expressed in ftlbs(force). KE = 0.5* m*v^2 You can also do changes like SCFM at a stated pressure and convert that directly into horsepower. It just takes a lot of cared full unit conversions and paying attention to mass and weight. Keep in mind that in PSI it is pounds of force NOT pounds of mass. In the one common conventional system weight is in pounds and mass ins in slugs. There are other systems. There are also lbs (mass) with acceleration carried separately. In metric kg (mass) are not Newtons (force). The gas did work on the bullet as it moved down the barrel. The things that complicates the math is that the pressure is NOT constant but after a brief peak continuously decreasing as the volume behind the moving bullets increases. Once the powder has turned to a gas (a very hot gas) the moving bullet increases the volume trapping the gas. I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? Once you have a velocity number you can calculate a Kinetic Energy. You are overthinking this. 


Originally Posted By brickeyee:
You do understand that QuickLoad integrates that pressuretime curve to find velocity? Once you have a velocity number you can calculate a Kinetic Energy. You are overthinking this. View Quote View All Quotes View All Quotes Originally Posted By brickeyee:
Originally Posted By TacticalGarand44:
Originally Posted By brickeyee:
And the anser to the initial question of force over area is energy. The gas pushes the bullet and the bullet moves. Now you have a force over a distance and are doing work. With a whole lot of unit conversions and integration over the length of the barrel you will get a KE expressed in ftlbs(force). KE = 0.5* m*v^2 You can also do changes like SCFM at a stated pressure and convert that directly into horsepower. It just takes a lot of cared full unit conversions and paying attention to mass and weight. Keep in mind that in PSI it is pounds of force NOT pounds of mass. In the one common conventional system weight is in pounds and mass ins in slugs. There are other systems. There are also lbs (mass) with acceleration carried separately. In metric kg (mass) are not Newtons (force). The gas did work on the bullet as it moved down the barrel. The things that complicates the math is that the pressure is NOT constant but after a brief peak continuously decreasing as the volume behind the moving bullets increases. Once the powder has turned to a gas (a very hot gas) the moving bullet increases the volume trapping the gas. I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? Once you have a velocity number you can calculate a Kinetic Energy. You are overthinking this. 


Originally Posted By TacticalGarand44:
I did not understand that, because I have never heard of QuickLoad. View Quote View All Quotes View All Quotes Originally Posted By TacticalGarand44:
Originally Posted By brickeyee:
Originally Posted By TacticalGarand44:
Originally Posted By brickeyee:
And the anser to the initial question of force over area is energy. The gas pushes the bullet and the bullet moves. Now you have a force over a distance and are doing work. With a whole lot of unit conversions and integration over the length of the barrel you will get a KE expressed in ftlbs(force). KE = 0.5* m*v^2 You can also do changes like SCFM at a stated pressure and convert that directly into horsepower. It just takes a lot of cared full unit conversions and paying attention to mass and weight. Keep in mind that in PSI it is pounds of force NOT pounds of mass. In the one common conventional system weight is in pounds and mass ins in slugs. There are other systems. There are also lbs (mass) with acceleration carried separately. In metric kg (mass) are not Newtons (force). The gas did work on the bullet as it moved down the barrel. The things that complicates the math is that the pressure is NOT constant but after a brief peak continuously decreasing as the volume behind the moving bullets increases. Once the powder has turned to a gas (a very hot gas) the moving bullet increases the volume trapping the gas. I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? Once you have a velocity number you can calculate a Kinetic Energy. You are overthinking this. It models the pressure dependent behavior of gunpowder. Like any model is has limits. I have used it to develop loads for obsolete cartridges that often do not have measured (CUP, LUP. or PSI) data. With the use of strain gauges you can tweak parameters in the model to more closely match actual measured data. I managed to get an old 'pin fire' gun up and running this way. It was a real PITA though. It models the burning of powder, the engraving of a bullet into rifling, and then the expansion of the powder gasses as the bullet is driven down the barrel. 


Thanks.



The weight is exerting a force on the rope, but no work is done. Same think if you held a weight at arms lengths. It may FEEL like work is being done, but that requires movement.



Originally Posted By euclid:
The weight is exerting a force on the rope, but no work is done. Same think if you held a weight at arms lengths. It may FEEL like work is being done, but that requires movement. View Quote A spring may have stored energy (from compression or extension) but if nothing is moving it is not doing any work. The stored energy is just that. Stored. It has the potential to do work if movement occurs. There are many ways to store energy. Compress it, stretch it, pressurize it. Until something moves or a load (from gravity say) changes there is no work or power present. Springs get warm from repeated cycling. Ropes do also. A rope going over a pulley gets warm. If only from its internal friction. A steel cable on a pulley can get far hotter than a rope ever would. There are real reasons stranded cables are loaded with grease. 


Originally Posted By TacticalGarand44: Force over area is pressure. I am asking about pressure over time. I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? View Quote View All Quotes View All Quotes Originally Posted By TacticalGarand44: Originally Posted By brickeyee: And the anser to the initial question of force over area is energy. The gas pushes the bullet and the bullet moves. Now you have a force over a distance and are doing work. With a whole lot of unit conversions and integration over the length of the barrel you will get a KE expressed in ftlbs(force). KE = 0.5* m*v^2 You can also do changes like SCFM at a stated pressure and convert that directly into horsepower. It just takes a lot of cared full unit conversions and paying attention to mass and weight. Keep in mind that in PSI it is pounds of force NOT pounds of mass. In the one common conventional system weight is in pounds and mass ins in slugs. There are other systems. There are also lbs (mass) with acceleration carried separately. In metric kg (mass) are not Newtons (force). The gas did work on the bullet as it moved down the barrel. The things that complicates the math is that the pressure is NOT constant but after a brief peak continuously decreasing as the volume behind the moving bullets increases. Once the powder has turned to a gas (a very hot gas) the moving bullet increases the volume trapping the gas. I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? @TacticalGarand44 I realize this is an old thread. Bore pressure is used simply because it is a measurable characteristic. What you are really interested in, I think, is net force acting on the bullet. You would simply multiply the pressure values at every point along the curve by the base area of the bullet, less the force of friction (whatever that is). This is now your new "forcetime curve." The area under this curve is the impulse, which will be numerically equal (or nearly so) to the momentum that the bullet will have upon exiting the barrel. Be advised: the propellant gases have mass, velocity, and momentum of their own. I believe the Imperial unit for momentum is slugft/sec. 


Originally Posted By brownbomber: @TacticalGarand44 I realize this is an old thread. Bore pressure is used simply because it is a measurable characteristic. What you are really interested in, I think, is net force acting on the bullet. You would simply multiply the pressure values at every point along the curve by the base area of the bullet, less the force of friction (whatever that is). This is now your new "forcetime curve." The area under this curve is the impulse, which will be numerically equal (or nearly so) to the momentum that the bullet will have upon exiting the barrel. Be advised: the propellant gases have mass, velocity, and momentum of their own. View Quote View All Quotes View All Quotes Originally Posted By brownbomber: Originally Posted By TacticalGarand44: Originally Posted By brickeyee: And the anser to the initial question of force over area is energy. The gas pushes the bullet and the bullet moves. Now you have a force over a distance and are doing work. With a whole lot of unit conversions and integration over the length of the barrel you will get a KE expressed in ftlbs(force). KE = 0.5* m*v^2 You can also do changes like SCFM at a stated pressure and convert that directly into horsepower. It just takes a lot of cared full unit conversions and paying attention to mass and weight. Keep in mind that in PSI it is pounds of force NOT pounds of mass. In the one common conventional system weight is in pounds and mass ins in slugs. There are other systems. There are also lbs (mass) with acceleration carried separately. In metric kg (mass) are not Newtons (force). The gas did work on the bullet as it moved down the barrel. The things that complicates the math is that the pressure is NOT constant but after a brief peak continuously decreasing as the volume behind the moving bullets increases. Once the powder has turned to a gas (a very hot gas) the moving bullet increases the volume trapping the gas. I've looked at pressure curves in barrels, they are pretty simple. Quick spike then a decay. Seems like it would be pretty simple to do a numerical integration of a pressure vs time curve using Matlab, what I am asking here is what would the physical significance of that quantity be? @TacticalGarand44 I realize this is an old thread. Bore pressure is used simply because it is a measurable characteristic. What you are really interested in, I think, is net force acting on the bullet. You would simply multiply the pressure values at every point along the curve by the base area of the bullet, less the force of friction (whatever that is). This is now your new "forcetime curve." The area under this curve is the impulse, which will be numerically equal (or nearly so) to the momentum that the bullet will have upon exiting the barrel. Be advised: the propellant gases have mass, velocity, and momentum of their own. I was thinking more in terms of the cumulative stress than a part has experienced over its lifetime. Basically putting a number to fatigue stress. 

China LIED. Americans DIED.

Ok, I misunderstood you. I believe your question defies an easy answer, and is well beyond my ability.



Originally Posted By Captain_Morgan: To answer the title of your thread, Josh is right. Units of dynamic viscosity are in those terms. To answer the question as described in the OP, Circuits is right. What you are trying to characterize is an energy (force over time). View Quote Force over time (as in force x time) is actually impulse (change in momentum). Force over a distance, newtonmeters, lbft, etc., is energy. Mike 


You're right on the heels of the whole field of material fatigue. Keep in mind, there's a bit of black magic and guesswork in the whole field.
You can plot what's called an "SN" curve for any given material, which means you stress (S) a given material so many times (N) until you get it to fail by fatigue. The higher the stress, the lower the N value tends to be. There is a limit, I forget what the "good practice" value is, but under something like 50% of a material's yield strength, N becomes so large that the material will not fail by fatigue. However, as you start getting over that 50% limit, you can failure by fatigue. A rifle barrel will experience a stress cycle every time a round is fired. If this stress is under 50% of the yield strength of the barrel, the barrel will probably not fail by fatigue. If higher than that, you may only get so many rounds out of it before fatigue cracking sets in. Here's a PDF with a presentation very similar to how I learned about fatigue in college, you may find some of the first bits interesting: https://www.efatigue.com/training/Chapter_4.pdf Note that this is an oversimplification of fatigue, but could get you started. Even without plastic deformation, a material can "remember" how many times it's been stressed, and a life cycle can be assigned to a given part in a mechanical system. 

"Oh look, Edward. That guy with the rage face comic avatar on the assault rifle forum is speaking ill of the salade I so enjoy. That settles it. I'm not voting for Ted Cruz."  AlamTX

Originally Posted By Chapman: You're right on the heels of the whole field of material fatigue. Keep in mind, there's a bit of black magic and guesswork in the whole field. You can plot what's called an "SN" curve for any given material, which means you stress (S) a given material so many times (N) until you get it to fail by fatigue. The higher the stress, the lower the N value tends to be. There is a limit, I forget what the "good practice" value is, but under something like 50% of a material's yield strength, N becomes so large that the material will not fail by fatigue. However, as you start getting over that 50% limit, you can failure by fatigue. A rifle barrel will experience a stress cycle every time a round is fired. If this stress is under 50% of the yield strength of the barrel, the barrel will probably not fail by fatigue. If higher than that, you may only get so many rounds out of it before fatigue cracking sets in. Here's a PDF with a presentation very similar to how I learned about fatigue in college, you may find some of the first bits interesting: https://www.efatigue.com/training/Chapter_4.pdf Note that this is an oversimplification of fatigue, but could get you started. Even without plastic deformation, a material can "remember" how many times it's been stressed, and a life cycle can be assigned to a given part in a mechanical system. View Quote Great info, thanks. 

China LIED. Americans DIED.

I've personally never heard of a small arms barrel failing by fatigue, or what I thought was fatigue.
Springs, bolts, extractors, yes. Even hard chrome plating, which has a fairly large residual tensile stress, apparently doesn't accelerate fatigue for some reason. Normally, plating stressed components in chrome doesn't do you any favors in that department, and it effectively reduces the allowable stress level or the number of cycles. Nitriding, shot peening, burnishing, and carburizing usually have the opposite effect. I think larger weapons, like howitzers and tank guns, probably have firm "retire this component by this round count" restrictions. Having something break on something like that can kill a lot of people. 


Originally Posted By brownbomber: I've personally never heard of a small arms barrel failing by fatigue, or what I thought was fatigue. Springs, bolts, extractors, yes. Even hard chrome plating, which has a fairly large residual tensile stress, apparently doesn't accelerate fatigue for some reason. Normally, plating stressed components in chrome doesn't do you any favors in that department, and it effectively reduces the allowable stress level or the number of cycles. Nitriding, shot peening, burnishing, and carburizing usually have the opposite effect. I think larger weapons, like howitzers and tank guns, probably have firm "retire this component by this round count" restrictions. Having something break on something like that can kill a lot of people. View Quote Different material have different responses to repeated stressstrain cycles. Most steels are 'well behaved' and as long as the strain does not result in permanent deformation they remain at the same strength. Aluminum (and a few other metals) is completely different. Application of stress below yield stress results in cracks forming from the repeated cycles. When enough of these cracks form and join together you have a stress induced failure. Some alloys can be heated to close the cracks Others will behave this way. The internal force from heat expansion close the cracks and the surface actually rejoin. Impulse like force like firing a fun are relatively difficult to model and analyze, and many times must be done numerically in small time steps to develop a picture of what is going on. Gun barrels are below the yield level, but are large enough to deform the material. That is how strain gauges can be used on barrels to measure pressure. The strain gauge is actually stretched as the material is is attached too deforms under pressure. Large gun barrels are very different critters. The typical procedure is to heat the exterior shell, cool the interior piece, and then quickly force them together. That is what all those 'steps' on a large barrel are from. This allows the exterior piece to create a large compressive force on the interior piece. The pressure can now exceed the yield strength of the interior piece since the internal pressure is 'cancelled' by the massive external compressive force present. 


And the integral of pressure time bullet base area is velocity of the bullet.



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