Calculate relative amounts of energy?

In summary, the argument is about which round, a solid copper hollow point or a normal copper jacketed lead core, has the most energy when fired from a gun. Both rounds have the same shape and physical dimensions, but differ in mass. The answer would depend on details and assumptions, but as a first approximation, the kinetic energy would remain constant regardless of the bullet weight. The propellant may alter according to bullet weight and manufacturer, but generally, Newton's second law would apply. Factors such as recoil, carry weight, and flatness of trajectory may also need to be considered.
  • #1
Stewart Mala
3
1
Hi All,

I have an argument to settle. Here's the problem. A gun has two rounds chambered. The first round is a solid copper hollow point, the second round is a normal copper jacketed lead core. Both rounds are the exact same shape and physical dimensions. Both cartridges contain the exact same quantity, quality and type of propellant. The only difference between the two rounds is the mass of each. For argument's sake, the solid copper round is 80 grams, the lead core round is 150 grams. All other parameters are equal.

On firing of each round, assuming one could measure the energy of each round as soon as it left the barrel of the gun, which round contains the most energy and why?

Is it:
a) The solid copper hollow point.
b) The full copper jacket lead core.
c) They both contain the same amount of energy.

Further, how many formulas cover this? Is there a difference between kinetic energy as measured and momentum? Or are they both the same? Should Newton's second law be used in this case, F=MxA, or should you use KE=0.5xmv2 to obtain the amount Kinetic Energy?

Many thanks,
 
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  • #2
I think the answer to this would depend on details but, if you make some assumptions, you can get an idea.
Easiest solution: If you assume that the pressure from the propellant is constant during the whole journey down the barrel then the Force times Distance is the same for both bullets. So the work done is the same, making the Kinetic Energy for both is the same.
But: it won't be constant so you would need to Integrate ∫Fdx over the distance to find the energy gained and it would be necessary to know the burning rate of the powder to know the pressure at all times.
You could think in terms of Momentum. The impulse imparted to the bullets would be Ft or more strictly ∫Fdt over the time in the barrel.
What searching have you done so far about this? I'd bet you could find a lot of stuff written about it if you spend some time on Google. I have to go out to pick up my son and grandson so I can't do much on that today, I'm afraid.
 
  • #3
Consider the limiting case when the cartridge is made very light and the powder charge remains constant. The energy of the charge is consumed almost entirely in the hot, moving exhaust gasses rushing out the barrel end and not on the projectile.

Accordingly, my expectation would be that the lighter bullet carries lower energy.

[On the other hand, consider a projectile so massive that it barely moves while the propellant cools down and leaks around. Then the expectation would be that the more massive bullet carries lower energy. Which rather makes @sophiecentaur's point. Details matter]
 
  • #4
A Momentum based argument certainly works for a very light projectile.
( in car - my wife is driving - yucky rain)
It makes me wonder why, apart from range, not have massive bullets always?
 
  • #5
sophiecentaur said:
It makes me wonder why, apart from range, not have massive bullets always?
Recoil, carry weight, flatness of trajectory.
 
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  • #6
Stewart Mala said:
On firing of each round, assuming one could measure the energy of each round as soon as it left the barrel of the gun, which round contains the most energy and why?
From a guy that used to work for Olin Corporation doing small arms ammunition development: As a first order approximation, the kinetic energy will stay constant if you change only the bullet weight.
 
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  • #7
Google "interior ballistics," and get back to us.
 
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  • #8
Bystander said:
Google "interior ballistics," and get back to us.
Nice! I learned a new term today... :smile:
 
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  • #9
berkeman said:
new term
New? You're kidding me --- aren't you?
 
  • #10
OK, thanks to all who contributed. I have sort of found an answer I was looking for. Unfortunately, in the case of my question, it seems as though the propellant does in fact alter according to bullet weight and also by manufacturer. However, I was recommended a website that works out the energy of a round for you by inputting the bullet weight and the muzzle velocity:
http://www.larrywillis.com/bullet-energy.html
So in order to use this calculator you have to research a specific bullet for its characteristics (weight and muzzle velocity).
However and generally speaking, I'm sticking with Newton's second law and thinking that the Kinetic Energy is going to be the same for either a lighter or heavier bullet, assuming the force imparted on to it is the same for either bullet. The only variation will be the bullet velocity, faster = lighter round, slower = heavier round.
 
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  • #11
jbriggs444 said:
Recoil, carry weight, flatness of trajectory.
Yes - it Engineering all over again! A compromise.
 
  • #12
Bystander said:
New? You're kidding me --- aren't you?
New to me too. It is a good term because it describes itself very well. It sounds like a blend of thermodynamics and straight mechanics.
 
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  • #13
sophiecentaur said:
New to me too.
Must be more a "gun nut" than I thought myself; glad to have added to the PF lore/vocabulary.
 
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  • #14
Bystander said:
Must be more a "gun nut" than I thought myself; glad to have added to the PF lore/vocabulary.
There is a vast 'gun divide' between those of us with NRA presence and those of us (Europe etc.) outside its influence. I was staggered when I heard a recent comment on a newsreel from an elderly woman who described a mass US shooting incident by saying that the perp "paused to reload" before resuming his barrage of shots. I cannot think of any of my elderly female (or even male) friends who would have interpreted things that way . But we all live in the UK, where guns are not the usual thing.
 
  • #15
Bystander said:
Must be more a "gun nut" than I thought myself; glad to have added to the PF lore/vocabulary.
I've been hunting and shooting since my dad taught me when I was about 6 (along with teaching me gun safety). But I've never been a reloader, so I think that's why I wasn't familiar with that term.

I'm glad you posted that link, because I was going to try to get the OP to think about how the burn rate of the powder can be tuned to the bullet weight and barrel length. But I didn't want to just give him that "answer", because we all were trying to get him to think more about it. :smile:
 
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  • #16
berkeman said:
burn rate of the powder can be tuned to the bullet weight and barrel length.
"Balanced" loads, indeed --- more "art" than science.
 
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  • #17
Additional vocabulary; "tractability."
 
  • #18
berkeman said:
I've been hunting and shooting since my dad taught me when I was about 6 (along with teaching me gun safety). But I've never been a reloader, so I think that's why I wasn't familiar with that term.

I'm glad you posted that link, because I was going to try to get the OP to think about how the burn rate of the powder can be tuned to the bullet weight and barrel length. But I didn't want to just give him that "answer", because we all were trying to get him to think more about it. :smile:

Hi Berkeman,
No, no one needs me to think more about it. The only reason I posted my question here on this forum because, when I tried the same thing on another website, "Quora", I got all sorts of gun oriented answers, that is, type of powder, bullet weights and aerodynamic characteristics, barrel lengths, etc, all of which whilst very interesting if I'd cared to go that far into it, was not really that useful to me. The reason is that I was seeking to settle a difference of opinion. I was in a conversation with a gun owner in the U.S. who stated that:

"The goal of defense (sic) when deadly force is justified is to stop an attacker as quickly as possible. So hollow points are used instead of the full metal jacket. Hollow points won't over penetrate which is a threat to bystanders. You also need less shots on target which also lessens the threat to innocent people. Hollow points expand and do more damage and copper is environmentally safer then lead. OK so who cares about that. Copper is lighter so it travels faster which increases energy at impact."

It was that last sentence which I thought seemed odd. So I thought I'd ask here. He assumes that lighter=faster=more energy at impact. But more energy than what? He clarified, meaning over a standard copper jacketed lead core round. I assumed that the propellant must be the same in either case. But my impression was that the amount of energy should be the same for each bullet, as the heavier bullet would travel slower, but its mass ensures that the energy would be the same as the lighter copper round. That's why I came here to "find out". However, what I found out from different places on the web, was that different rounds can have different amounts and types of propellant, which also varies by manufacture. So, it's become a little impossible to answer. The only way was to fix all the variables as constant or cancelling, so that the only difference that mattered was bullet weight.
 
  • #19
Stewart Mala said:
copper is environmentally safer then lead. OK so who cares about that. Copper is lighter so it travels faster which increases energy at impact.
That quote of yours, from another forum, demonstrates two rather superficial views of the situation. Not what you tend to get on PF. :smile:
But there is never a simple rule that can be applied in any complex engineering design. I guess the only generalisation is what's said about boxing: "A good big 'un will beat a good little 'un".
 

Related to Calculate relative amounts of energy?

1. How do you calculate relative amounts of energy?

To calculate relative amounts of energy, you need to know the energy value of each substance or object in question. This can be done by using a formula that takes into account the mass, velocity, and other relevant factors of the substance or object.

2. What is the unit of measurement for relative amounts of energy?

The unit of measurement for relative amounts of energy is joules (J). This is a standard unit of measurement in the International System of Units (SI).

3. Can relative amounts of energy be converted into different units?

Yes, relative amounts of energy can be converted into different units. Some common conversions include joules to calories, joules to electronvolts, and joules to British thermal units (BTUs).

4. Why is it important to calculate relative amounts of energy?

Calculating relative amounts of energy is important in many scientific fields, such as physics, chemistry, and engineering. It allows us to understand the energy changes that occur in chemical reactions, the movement of objects, and the behavior of various systems.

5. Are there any limitations to calculating relative amounts of energy?

There are limitations to calculating relative amounts of energy, as it is based on certain assumptions and ideal conditions. For example, it may not take into account factors such as friction, air resistance, or other external forces that can affect the energy of a system. Additionally, some calculations may be more complex and require advanced mathematical knowledge.

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