Basic physics: Why do lighter bullets have exponentially more energy?

[Mustey]

I've done alright in physics at school and remember all my formulas, including kinetic energy: E = (mv^2)/2

My question is not how to calculate kinetic energy but what is it in the barrel that yields more energy for a lighter bullet, ceteris paribus?

EG:
A) I load a 45acp with a 230-grain bullet and it does 850fps = 370 ft-lb of energy
B) I load the same shell with the SAME powder charge with a 185-grain and it does 1000fps = 411 ft-lb of energy

I understand that smaller bullet means that the same pressure will result in higher acceleration and therefore higher speed... And higher speed will result in exponentially higher energy - I get it!

But from an energy conservation point of view: The kinetic energy of the bullet comes from the chemical energy of the powder charge - surely there is no other source of energy...

So, how come the same powder charge is converted to different kinetic energies?

I know there's no magic, Energy=Energy.
Just wondering if anyone studied internal ballistics and can spare a few words on the mechanics of the barrel, to explain how lighter bullets convert energy more efficiently than heavier bullets?

(I'd actually think it should be the other way around: Heavier bullets spend more time inside the barrel and therefore have more time to pick up kinetic energy from the gas build up)

[Aguila Blanca]

Because the formula for energy is E = MV^2, where

E = Energy
M = Mass
V = Velocity

V^2 means squared -- multiplied by itself. This means if you double the bullet weight (mass) without changing the velocity, the resulting energy is double. If you double the velocity without changing the bullet weight, the resulting energy is four times as great.

[44 AMP]

I think your confusion is the result of a fundamental misunderstanding in the way firearms work.

You are correct that using the formula for KE higher speed does result in higher energy numbers. This is the result of using velocity squared in the formula. The KE formula puts more "weight" on velocity than it does on mass.

But from an energy conservation point of view: The kinetic energy of the bullet comes from the chemical energy of the powder charge - surely there is no other source of energy...

So, how come the same powder charge is converted to different kinetic energies?

You are correct that all the energy comes from the powder charge, there is no other source.

The reason the same powder charge gets converted to a larger KE number is the velocity difference between bullets due to the weight (mass)

In your example of .45ACP, the lighter bullet is moving faster, and it is the increased velocity of the lighter bullet, plugged into the formula that results in a larger energy number.

The powder charge's "push" on both bullets is the same. There is no change there. The heavier bullet moves slower than the lighter one, with the same amount of "push".

A 55gr .22-250 bullet and a 405gr .45-70 bullet can be loaded to speeds that produce identical Kinetic Energy numbers. Here you have a case of a small light bullet, weighing close to 1/8 the amount of the heavy bullet producing exactly the same energy (per the formula) because it is moving nearly 3 times faster.

Do consider that the energy numbers are something useful only for relative comparison, and not the only, or the most significant factor on the effectiveness of a given bullet. Again, with the example of a high speed 22 vs a much slower (and heavier) .45 bullet, while the calculated energy can be identical, which one do you think would be a better choice to stop an angry buffalo from stomping you to paste? SAME energy on paper, vastly different effect possible in the field.

Does this help?

[LeverGunFan]

Not knowing what data you used in your example, I ran two test cases through Quickload. One was the 44 Special, the other the 45 ACP. In both cases the only change was the bullet weight, I selected the same type bullet from the same manufacturer and used the exact same powder charge. In both cases, the heavier bullet had higher kinetic energy than the lighter bullet.

For your example, what was the powder and charge weight?

I can think of a couple of reasons why in your example the 185 grain bullet had higher energy. The first is that there is difference in the resistance of the bullet in the bore, the 230 grain bullet likely has a longer bearing surface and that retards the muzzle velocity. However I would expect the pressure to be higher in this case which should mitigate some of the effects of the longer bullet. The second is that not all of the energy of the powder is expended in accelerating the bullet, some energy is lost in a larger muzzle blast for one projectile than the other. In other words, the powder is more optimum for one bullet as opposed to the other.

[Mustey]

Does this help?

It looked like you understood my question and I was especially excited to have one of my many misunderstandings about how firearms work rectified

And you proceeded to show me a calculation of K.E...
And then some thought experiment about what KE is good for anyway.

I sincerely thank you for every second of your life that you put down to answer.

But that was not my question.

My question is really simple:
For the same powder charge (same chemical energy), lighter bullets come out with more KE than heavier bullets.
Clearly, heavier bullets are less efficient in converting the explosion into kinetic energy.
Where is the loss happening?

In particular, what are the losses that are directly related to bullet weight?

[2damnold4this]

I think that LeverGunFan did a good job answering your questions with the first point made being that lighter bullets don't always come out with more KE for the same powder charge.

I think LGF's point that machines that convert chemical energy to kinetic energy aren't 100% efficient is also important. Identical powder charges do mean identical chemical energy but I'd be surprised if half of that is converted to KE in the bullet.

[Aguila Blanca]

Originally Posted by mustey

In particular, what are the losses that are directly related to bullet weight?

Except in the case of some rifles with long barrels, it's extremely unlikely that ALL the powder will have burned and been converted into pressure on the base of the bullet by the time the bullet exits the barrel. That's why many handgun rounds display impressive fireballs when fired.

As you have noted, a lighter bullet accelerates faster, so it spends less time in the barrel. As soon as the bullet has left the barrel, any further combustion of the powder is mostly wasted energy, because it's just making fire and noise, not acting on the bullet to increase its velocity or energy.

So there's a significant loss right there.

[HHollow]

lighter = more efficient

[Nathan]

There is “magic”. Energy is meaningless. Energy is used to do work. What you care about is how much energy you can transfer to work.

Think of energy as food. If you have a bunch of food and you feed it to a person who races out and builds you a barn, you think wow….I need more food(energy). If you feed it to my kid, they go to bed without putting their plate away….no work.

Now bullets. You want to maximize the work done on the target. Light bullets tend to over expand and blow up, fall apart or stop working effectively. A heavy Bullet has more bullet to expand and keeps expanding and driving through until it exits.

If a light bullet can exit and create massive wound channels, it would be ok. Typically, they expand, blow up and fail to drive through to exit.

Even switching to monos which are better users of energy, still tend to create slightly smaller wound channels, but almost always drive through at nearly full weight.

What you want to think about is how much velocity do I need to expand fully without explosion? Will it drive through? If you have both and they are consistent, you are good.

BTW, what caliber and target medium are we talking?

[44 AMP]

My question is really simple:
For the same powder charge (same chemical energy), lighter bullets come out with more KE than heavier bullets.
Clearly, heavier bullets are less efficient in converting the explosion into kinetic energy.
Where is the loss happening?

Ok, I'll try to come at this a different way, and try to keep to simple explanations and general things.

My question is really simple:
For the same powder charge (same chemical energy), lighter bullets come out with more KE than heavier bullets.

Yes, because lighter bullets are moving faster. And the KE formula uses mass as a single factor, but squares the velocity factor.

Consider this,
Run the KE formula with a heavy and a light bullet at the SAME speed. Which one has more KE? The heavier one will have the higher KE number.

Now run the formula again with one weight of bullet at two different speeds. The faster bullet will have the higher KE number.

That's the way the formula works. Mass (weight) is A factor, but speed (velocity) is a factor squared, so a change in velocity has a larger effect on the result than a change in mass.

I know this makes sense to me, I'm writing it but is it making sense to you??

[tangolima]

In every energy transfer process, the efficiency is often function of the load. Zero load or infinite load lead to little or no energy got transferred. There almost always exists optimum load at which energy transfer is maximum. This principle applies in different disciplines of engineering, electrical, mechanical, or even chemical.

Same here in powder burning efficiency. Very often, but not always, heavy projectile has higher energy, even with reduced powder charge to avoid over pressure. But there are exceptions, although it is quite rare in my limited experience. Your case is one of them.

-TL

Sent from my SM-N960U using Tapatalk

[JohnKSa]

But from an energy conservation point of view: The kinetic energy of the bullet comes from the chemical energy of the powder charge - surely there is no other source of energy...

So, how come the same powder charge is converted to different kinetic energies?

Because lighter masses are easier to accelerate.

The burning powder produces pressure and that pressure acting on the back of the bullet generates a force. With everything other than the bullet weight being identical, the pressures are very much the same and therefore the force on the back of the two bullets is also about the same.

Push on a 10lb wagon with a given amount of force for a second and then push on a 3000lb car with the same amount of force for a second. Which will go faster?

Force = Mass x Acceleration

Let's take some made up numbers just to illustrate the point.

100N = 20kg x 5m/s/s

Keep Force the same. Make Mass less--let's say it's now 10. How do you maintain the equality? You have to increase the Acceleration.

100N = 10kg x 10m/s/s

Higher acceleration means a higher velocity is achieved. In our simple example, if we push on both our masses for 1 second at 100N of force, the 10kg mass will be going 10m/s at the end of a second while the 20kg mass would only be going 5m/s.

The same force for the same time resulted in double the velocity.

So the same force applied to a lighter mass means the mass will achieve a higher velocity and higher velocity means more energy.

In our simplified example, we applied a fixed amount of force for a fixed amount of time. It's more complicated than that in a firearm because the force (due to pressure) decreases as the bullet moves down the barrel and the amount of time the bullet is in the barrel will decrease the faster it is accelerated. But the general principles apply.

Rather than trying to work out a budget for the energy, it's easier to view it from a pressure standpoint. The burning powder creates a certain amount of pressure that's going to act on the bullet to create a force. If one bullet is lighter than the other, all else being equal, it will accelerate faster, achieve higher velocity and therefore more energy.

You could work it out in terms of energy, but it's a lot more complicated. Much more complicated than I would want to try.

[Shadow9mm]

Crunching the numbers myself with GRT I'm getting very different results from what you got.

45 auto, 4in barrel, both using Hornady's max listed charge of 6.9g of CFE Pistol for 230g XTP for both rounds, and Hornady's listed COL for the respective loads I got the following.

240g XTP, 861fps
185g XTP 862fps

If you were to use a DIFFERNT charge weight with the 185g bullet, say Hornady's listed max of 8.0g of CFE Pistol, which would contain more energy, that would give you 966fps which is fairly close to the listed velocity of 1000fps.



[Note], while I am using GRT to get simulated velocities, the load data is from publish data, Hornady's 11th. This is why I did not add the general warning pertaining to GRT and load data.

[LeverGunFan]

Originally Posted by Shadow9mm

Crunching the numbers myself with GRT I'm getting very different results from what you got....

I agree... I ran some numbers earlier in Quickload and I could not find a case where the lighter bullet had higher energy than the heavier bullet with identical powder charges. None of my several reloading manuals show a 45 ACP 185 grain bullet at 1000 fps with the same powder charge as the 230 grain bullet at 850 fps. Every manual shows the 185 grain bullet with a greater powder charge to get to 1000 fps.

The OP hasn't provided the powder type and charge weight, or the bullet type/manufacturer. I'd like to have that data to investigate this further.

[JohnKSa]

My explanation assumed the same pressure in both cartridges.

Generally, in the real world, a lighter bullet is shorter and if loaded to the same COAL as a heavier bullet it leaves more space inside the case which lowers the discharge pressure if the same amount of powder is used.

Less pressure, of course, means less force on the bullet.

[tangolima]

This most certainly is not a simple physics problem. It requires solving differential equations either analytically, as in books on internal ballistics, or numerically, as in such software as quickload.

Lighter bullet accelerates quicker, the volume in the bore increases faster, making the pressure drops faster, etc. There are different opposing factors to include.

Several posters have pointed out, same as my own experience, heavy bullet usually gives higher energy, even with reduced charge to avoid over pressure.

230gr versus 185gr bullets, there is 24% difference in mass. The lighter bullet needs to have 12% higher MV to achieve the same energy. 12% in MV is quite a bit. That is about the same as from min. to max. powder charge, if not more than. I have been assuming the figures given by op are correct. That would be one of the rare cases that lighter bullet has higher energy.

-TL

Sent from my SM-N960U using Tapatalk

[MarkCO]

To answer the OP...Energy goes as velocity squared. Absolutely simple HS physics.

[natman]

Why do lighter bullets have exponentially more energy?

Because lighter bullets go faster. While the increase in velocity is linear, energy increases exponentially with velocity.

It really is that simple. It's all due to the way energy is calculated, not some mystery of internal ballistics.

[tangolima]

Originally Posted by natman

Why do lighter bullets have exponentially more energy?



Because lighter bullets go faster. While the increase in velocity is linear, energy increases exponentially with velocity.



It really is that simple. It's all due to the way energy is calculated, not some mystery of internal ballistics.

Why 230gr at 850fps and 185gr at 1000fps? That's the key. The pressure can't possibly be the same. Not the chamber pressure alone, but the whole pressure curve.

Calculating energy with given speed is trivial. Determining the speed... is hard.

-TL

Sent from my SM-N960U using Tapatalk

[Aguila Blanca]

I think those who have responded to this question -- myself included -- are missing the point.

Originally Posted by Mustey

I understand that smaller bullet means that the same pressure will result in higher acceleration and therefore higher speed... And higher speed will result in exponentially higher energy - I get it!

The OP apparently understands that the velocity component is squared, so increasing velocity has more effect than increasing bullet mass. What we're not addressing is the INPUT side of the equation: both projectiles are fired from cartridges loaded with the same charge of the same powder. Whatever bullet weight is used and whatever muzzle velocity is produced, it is the result of the SAME AMOUNT of energy created by the ignition of the powder charge. In a closed system, the same amount of energy on the input side should result in the same amount of energy on the output side.

But if one projectile produces more muzzle energy than the other ... something must be getting lost somewhere in the firing cycle. I think the OP is attempting to grasp where the loss(es) occur(s).

My first thought is that the major issue is that a firearm is NOT a closed system. As I suggested in one of my posts above, in most firearms at least some of the powder is still burning and still pushing on the base of the bullet when the bullet leaves the barrel. At that point, any residual energy/pressure still behind the bullet gets spread out around the bullet and dissipates. A lighter bullet, since it accelerates more rapidly, clears the barrel sooner, thereby (I think) allowing more of the combustion product to be wasted.

However, I'm not sure this is an adequate explanation. I see a potential fallacy in my argument, but I'll wait a bit to see who picks up on it.

[LeverGunFan]

Originally Posted by Aguila Blanca

....What we're not addressing is the INPUT side of the equation: both projectiles are fired from cartridges loaded with the same charge of the same powder. Whatever bullet weight is used and whatever muzzle velocity is produced, it is the result of the SAME AMOUNT of energy created by the ignition of the powder charge. In a closed system, the same amount of energy on the input side should result in the same amount of energy on the output side.

But if one projectile produces more muzzle energy than the other ... something must be getting lost somewhere in the firing cycle. I think the OP is attempting to grasp where the loss(es) occur(s)....

The problem is that we can not find data that supports the premise that the OP presented in his question, that the same powder charge produces more energy with the lighter bullet. Quickload, GRT and every reloading manual that I have shows that the heavier bullet will produce more energy with the same powder charge, or in other words, a larger powder charge is required for the lighter bullet to produce more energy.

The OP has not provided the details of the powder, charge weight, or bullet type to support his velocity numbers for the two bullet weights. I conclude that there is no mystery here, the two bullet weights perform just like we expect them to behave with the heavier bullet having higher energy than the lighter bullet with the same powder charge.

[44 AMP]

The OP hasn't been back in a couple days, maybe he's just busy, or maybe he figured out what he wanted to know, or perhaps, he's been scared off by the well meant but effusive "technobabble" going into more detail than he was looking for, and I think a bit off into the weeds.

This is his question, as of his most recent postl

My question is really simple:
For the same powder charge (same chemical energy), lighter bullets come out with more KE than heavier bullets.
Clearly, heavier bullets are less efficient in converting the explosion into kinetic energy.
Where is the loss happening?

In particular, what are the losses that are directly related to bullet weight?

Let's discuss this a bit, and see if we can all get on the same page for what we think he means, and is asking.

First, I think that the example he is using is not about powder charge, and certainly not about what specific powder do in real world cartridges. I think what he is asking about is bullets launched with equal force, and how their energy numbers are different, and why.

IF different weight bullets were launched with a spring (mechanical energy) and that energy were constant, his question would still be the same. So I think focus on the powder charge isn't answering the question.

Next point, one that confuses me a bit, is that while he has said he understands that the difference in KE if the result of the difference in velocity, he feels heavier bullets are less efficient converting the energy driving them into KE and that there is a "loss" that is the cause of this, and he wants to know where that loss is.

Having read the replies, and seeing where this has been explained to him by several different people a few different ways, I'm not sure what he isn't getting.

TO me, in order for there to be a "loss" you have to have had the thing in the first place, in order to lose it. As I see it, heavier bullets with less velocity than light ones haven't lost anything, they never had the speed of the light bullet to begin with.

Thoughts??

[LeverGunFan]

Originally Posted by 44 AMP

First, I think that the example he is using is not about powder charge, and certainly not about what specific powder do in real world cartridges. I think what he is asking about is bullets launched with equal force, and how their energy numbers are different, and why.

Thoughts??

In this case. the same powder charge does not produce the same force. If the 185 grain bullet is loaded to the same OAL as the 230 grain bullet, then there is more volume in the cartridge behind the bullet. Less volume results in lower pressure, and less pressure results in lower force.

So while you changed the discussion to equal force, the question the OP raised in his original post was "So, how come the same powder charge is converted to different kinetic energies?"

To get to equal pressure/force, we need to increase the powder charge for the 185 grain bullet compared to the 230 grain bullet. That is supported by data in every reloading manual that I have. There is no mystery here, the premise of a 185 grain bullet having more KE than the 230 grain bullet with the same powder charge with all other factors held constant is inconsistent with actual data. Perhaps the OP assumed that the same powder charge would result in the same pressure, but that is an incorrect assumption. We really need the OP to explain how the velocity numbers in his post were generated.

[Bob Willman]

The calculation of KE is the mass times velocity squared. Period. No other values matter. How the bullet got to that velocity does not matter. Whether it was chemical reaction as in gun powder, magnetic rail gun or something threw it at that velocity does not matter.
The amount of energy required by whatever means to get it to that velocity is a separate matter. The relative amounts of energy between the two will depend on the efficiency of the system.

NRA Benefactor

[44 AMP]

In this case. the same powder charge does not produce the same force. If the 185 grain bullet is loaded to the same OAL as the 230 grain bullet, then there is more volume in the cartridge behind the bullet.

IF. If loaded to the same COL, yes. But not all are. If the bullets are loaded to the same depth in the case, then the powder space will be the same, and the COL will be different.

I think the OP was saying "same powder charge" for both bullets, meaning the same amount of "chemical energy" being applied to both bullets. And, I think the way he worded it is causing us a bit of confusion about what he's actually trying to find out. Both of his posts use terms in ways that lead me to think he doesn't understand some things the way we do.

I suggested considering spring driven bullets as a way to illustrate a constant and equal force, and avoid all the many different factors that can change the pressure of powder charges, in order to focus on what I think the OP was asking about, why "equal force" gave different KE results.

I think all the detailed explanations of powder behavior in the thread have been accurate and correct, but, so far, it seems the OP has not been able to grasp that and we won't know until he posts again.

I think his question is essentially, "if you push them equally, why do you get such different results", and I think he just assumed equal powder charges push them equally and is not aware of the factors that can result in the same powder charge yielding different pressures.

To get to equal pressure/force, we need to increase the powder charge for the 185 grain bullet compared to the 230 grain bullet. That is supported by data in every reloading manual that I have. There is no mystery here, the premise of a 185 grain bullet having more KE than the 230 grain bullet with the same powder charge with all other factors held constant is inconsistent with actual data.

I disagree. I just don't see things that way, absent more information relating to the specific loads. Here's my viewpoint, and, its going to contain that dreaded word "if".

IF the powder space in the case is the same (and it may not be, but IF it is) how can the same amount of powder, in the same amount of space produce different pressure??

COL is not an accurate representation of the amount of powder space. I have a Hornady book showing 2 different 185gr and 2 different 230gr bullets and the COL for each one is different. Hornady might be seating these bullets so the depth of the bullet base in the case is the same (which would make the available powder space the same) but there is no information given saying that is what they do. I guess it is implied, but not stated??

Loading different length bullets to the same COL, obviously means the bases are at different positions inside the case, and so the powder space will be different. SO, as I see it, COL can give you an indication, but not an actual amount without additional information.

Also, with all other factors held constant, won't the lighter bullet have higher velocity, and therefore the formula will result in a higher KE number due to the difference in mass not being as much of a factor in the formula as the difference in velocity, squared?

Perhaps the OP assumed that the same powder charge would result in the same pressure, but that is an incorrect assumption. We really need the OP to explain how the velocity numbers in his post were generated.

this, I can absolutely agree with.