Why do heavier projectiles tend to have higher momentum?

[skh149]

I am wondering why heavier bullets have a higher momentum than lighter bullets when using similar powder charges?

At the muzzle, a typical 150 grain bullet fired from a 30-'06 will travel at around 3000 ft/s.
A 200 grain bullet from the same rifle will travel around 2600 ft/s.

(The velocities might vary slightly, but they are fairly typical)

The mass of the heavier bullet is 33.3% greater than the lighter bullet, but the lighter bullet travels only travels 15% faster.

Since momentum is equal to mass multiplied by velocity, the heavier bullet will carry more momentum. That heavier bullets have greater momentum is generally accepted to be true in the shooting community.

Why is it that the lighter bullet doesn't travel 33.3% faster, equalizing the momentum between the two?

Is it a quirk of the rifle system? Is it unreasonable to expect them to have similar momentums? Am I misunderstanding or miscalculations something?

Greater velocities exponential increase kinetic energy, so if lighter projectiles had proportional momentum it would seem crazy not to use the lightest projectile one could...

I can't find any explanation online, and it seems like it might be a bit too physics oriented for most shooting forums, although I will try there next if no one knows the answer. (Also, I'm neither a physicist nor an expert marksman)

Thanks in advance for any replies

 

[jrmichler]

Given a particular cartridge, and the same powder charge, two bullets of different weights will have the same kinetic energy. The muzzle energy of both of your examples is 3000 ft-lbs.

The science behind this is the field of interior ballistics (search the term).

 

[phinds]

... if lighter projectiles had proportional momentum it would seem crazy not to use the lightest projectile one could...

No, it would not, unless your objective was to put a hole all the way through something instead of having stopping power

 

[skh149]

Thanks for the reply Jr, having a term to search is useful enough.

Would it be right to say that the velocity doesn't increase "proportionately" because it's contribution to kinetic energy is squared, and so if it did increase proportionately it would cause the two to no longer have equal muzzle energies?

 

[Lnewqban]

Interesting problem!
Same propellant, diameter and length of muzzle should produce about similar work on both masses.
Could the increased aerodynamic drag (around 33% higher) of the lighter bullet be a reason for the lack of proportionality in final average velocities?

 

[Delta2]

Same propellant, diameter and length of muzzle should produce about similar work on both masses.

I agree so far.

Could the increased aerodynamic drag (around 33% higher) of the lighter bullet be a reason for the lack of proportionality in final average velocities?

I disagree here. I think if both bullets have the same shape (but one is from denser material than the other) then the force of drag (but not the acceleration) will be the same, hence the work of drag will also be the same. So same total work in both bullets, from work energy theorem we conclude same final kinetic energy.

I think it is a not so hard algebraic excercise to prove that if two masses have the same kinetic energies, then the bigger mass will also have bigger momentum.

 

[sysprog]

Given a particular cartridge, and the same powder charge, two bullets of different weights will have the same kinetic energy. The muzzle energy of both of your examples is 3000 ft-lbs.

With the same propellant, although the momentum is the same for a lighter and a heavier bullet ($p=mv$), a lighter bullet at a higher speed has more kinetic energy than a heavier bullet at a lower speed ($KE = \frac 1 2 m v^2$).

 

[sysprog]

Here's an example:

10g bullet at 100 m/s: $\mathtt{KE= 50J}$
5g bullet at 200 m/s: $\mathtt{KE= 100J}$

From https://www.calculatorsoup.com/calculators/physics/kinetic.php:

 

[Delta2]

@sysprog Yes ok what you saying is correct but you assume that the two bullets will have the same momentum.

This assumption is not correct according to my opinion.

The correct assumption is that the two bullets will have the same kinetic energy. This is because the total work (the work from expansion gases inside the barrel of the gun minus the work from the drag) is the same if the shape of the bullets are the same and we fire them with the same gun.

 

[sysprog]

@sysprog Yes ok what you saying is correct but you assume that the two bullets will have the same momentum.

Both momentum and kinetic energy are conserved in internal ballistcs. Given the same cross-section, and the same gas pressure (produced by a burn that is fast enough to complete before the less massive bullet leaves the barrel) the momentum will be about the same (if the burn rate is too slow, the less massive bullet will leave the barrel before the burn completes, and less of the gas expansion energy will be imparted to the bullet).

This assumption is not correct according to my opinion.

Why would the momentum not be about the same?

The correct assumption is that the two bullets will have the same kinetic energy. This is because the total work (the work from expansion gases inside the barrel of the gun minus the work from the drag) is the same if the shape of the bullets are the same and we fire them with the same gun.

A less massive bullet with the same propellant will leave the muzzle at a higher velocity, and velocity is an exponential component of kinetic energy, whereas mass is a linear component of kinetic energy − doubling the muzzle velocity of a bullet quadruples its kinetic energy, whereas doubling the bullet mass only doubles its kinetic energy.

$$\mathtt{KE=\dfrac 1 2 mv^2}$$
To get the muzzle kinetic energy of the bullet in Joules, we multiply half the bullet mass in grams (instead of kilograms, because bullet mass is more conveniently measured in grams rather than in kilograms) by the square of the muzzle velocity in meters per second, then divide by 1,000 (because we used grams instead of kilograms).

A 10g bullet fired at 300 m/s will thus have a kinetic energy of

$$\mathtt{\frac {5g×300^2(m/s)}{1,000}=450J}$$
The momentum imparted to the bullet and gases leaving the barrel will be the same as that imparted in the opposite direction to the shooting apparatus and the person operating it, but the kinetic energy of the bullet will be much higher than will that of the person − the velocity of the bullet is much higher and the mass of the person is much greater, so for the same momentum imparted to the bullet and the person, the energy of the bullet will be much greater.

 

[A.T.]

Given the same cross-section, and the same gas pressure (produced by a burn that is fast enough to complete before the less massive bullet leaves the barrel) the momentum will be about the same

The same pressure force applied along the same barrel length gives you the same work, and thus the same kinetic energy. Not the same momentum.

 

[nasu]

@sysprog If kinetic energy were conserved during the "internal ballistics" the bullet will newer leave the gun. It's initial KE is zero and conservation means that it does not change. Same for momentum.
The fact that two bullets acquire the same KE is not conservation of energy but results from the work-energy theorem.
Edit. Also, the fact that the momentum is not the same results from the impulse -momentum theorem. The two bullets travel the same distance along the barrel but not the same time. It is just a matter of basic physics even if we call it "internal ballistics". ,

 

[Lnewqban]

... I disagree here. I think if both bullets have the same shape (but one is from denser material than the other) then the force of drag (but not the acceleration) will be the same, hence the work of drag will also be the same. So same total work in both bullets, from work energy theorem we conclude same final kinetic energy.

I think it is a not so hard algebraic excercise to prove that if two masses have the same kinetic energies, then the bigger mass will also have bigger momentum.

Let me go back to the beginning of the original question.
OP believes that both bullets receive the same impulse from the propellant (I am not sure about that since propellant force acts on heavier bullet a little longer).

Because the difference of masses, the leaving velocities should be different, if each momentum is aproximately the same.
According to his data, that difference of (leaving or flying) velocities is not proportional to the difference between the two masses.

My reasoning, perhaps incorrect, is that the faster flying (lighter) bullet should encounter a higher resistive force from air friction, killing that theoretical proportionality of velocities.

The slowing down effect of aerodynamic drag, and subsequent degradation of the kinetic energy of the lighter bullet, should be more significant as the traveled distance increases.
Much less important for short range impacts, I believe.
Therefore, the actual flying distance may be relevant in this discussion about destructive power of each bullet.

 

[DrStupid]

I think if both bullets have the same shape (but one is from denser material than the other) then the force of drag (but not the acceleration) will be the same, hence the work of drag will also be the same.

Drag increases with speed and with the same shape and same kinetic energy the lighter bullet is faster.

 

[A.T.]

OP believes that both bullets receive the same impulse from the propellant

I see no reason to assume this.

 

[skh149]

Thank you for all the replies.

Just to clarify, I am not too concerned with terminal ballistics or the practical effects on a target; although interesting, I understand there are trade-offs. What I am trying to figure out is why those trade-offs occur, and I think this discussion will help that (and already has quite a bit)

So far it seems like there are three main explanations:

1) The heavier bullet stays in the barrel for a longer period of time, and thus experiences more force and is accelerated disproportionately, compared to the lighter one which exits before it gets to its "potential" velocity. With this I would expect that if they were in the barrel for the same period of time the numbers would be closer to expected

2) The faster bullet experiences more drag since drag increases with speed, thus it is slowed down (would it travel the appropriate speed in a vacuum or frictionless barrel?)

3) The muzzle energies are actually (about) the same, meaning the velocity must be disproportionately low in the faster bullet, since velocity ends up being counted exponentially. I'm having difficulty articulating this point, but as someone already mentioned, solving the equations algebraically shows the heavier bullet with greater momentum if you assume equal KE. And calculating my example numbers, they have basically equal KE.

So far I am most convinced by #3, but it could also be some combination of these three or something not discussed

Interestingly, in the guns/shooting community, light weight, higher velocity cartridges are often touted as superior since most people know velocity squares in the KE equation. What people might not realize is that (if #3 is correct) the velocity increase doesn't translate to increased energy when the only thing changed is projectile mass. Energy seems to mainly comes down to powder charge. One caveat here is I do not know for sure that my example numbers are using the same charge. My biggest assumption is that my example numbers are an "all else equal" comparison.

Thanks again S.

 

[A.T.]

Interestingly, in the guns/shooting community, light weight, higher velocity cartridges are often touted as superior since most people know velocity squares in the KE equation. What people might not realize is that (if #3 is correct) the velocity increase doesn't translate to increased energy when the only thing changed is projectile mass.

Higher velocity means less vertical drop for the same target distance, and less required lead for moving targets. But a faster projectile will lose it's energy quicker due to drag, so even if muzzle energies are the same, the energy on target will be less for the lighter projectile.

 

[jbriggs444]

1) The heavier bullet stays in the barrel for a longer period of time, and thus experiences more force and is accelerated disproportionately

Same force, longer time. Greater "impulse". Impulse being a physics term meaning roughly "change in momentum during a brief event".

, compared to the lighter one which exits before it gets to its "potential" velocity. With this I would expect that if they were in the barrel for the same period of time the numbers would be closer to expected

It is not clear what potential velocity is supposed to mean. But yes, projectiles exposed to the same force for the same time will accumulate the same impulse.

2) The faster bullet experiences more drag since drag increases with speed, thus it is slowed down (would it travel the appropriate speed in a vacuum or frictionless barrel?)

Yes. Though it is not clear what "appropriate speed" is supposed to mean.

3) The muzzle energies are actually (about) the same, meaning the velocity must be disproportionately low in the faster bullet, since velocity ends up being counted exponentially.

Not exponentially. Quadratically.

 

[skh149]

Hi Jb444,

Thanks, I hadn't heard of impulse in terms of a physics concept, I will look that up as it seems like it might be helpful for understanding the subject.

When I said '"potential" velocity', I didn't mean anything technical by it, just that the bullet could have reached a higher velocity were it allowed to acquire more of the energy by staying in the barrel for a longer amount of time. Similarly, by 'appropriate speed' I just meant speed that it would reach the 33.3% higher rather than the 16%, as I mentioned in my original post. I used these as shorthand to try to keep my last post a bit shorter, and hoped people would understand through context, but in the future I will write it out fully to avoid confusion.

Finally, although I have the quadratic formula memorized and can solve the equations, I have no idea why it is quadratic as opposed to exponential. I'm not sure if that's just me, or the education system!

 

[jbriggs444]

Hi Jb444,

Thanks, I hadn't heard of impulse in terms of a physics concept, I will look that up as it seems like it might be helpful for understanding the subject.

When I said '"potential" velocity', I didn't mean anything technical by it, just that the bullet could have reached a higher velocity were it allowed to acquire more of the energy by staying in the barrel for a longer amount of time. Similarly, by 'appropriate speed' I just meant speed that it would reach the 33.3% higher rather than the 16%, as I mentioned in my original post. I used these as shorthand to try to keep my last post a bit shorter, and hoped people would understand through context, but in the future I will write it out fully to avoid confusion.

Finally, although I have the quadratic formula memorized and can solve the equations, I have no idea why it is quadratic as opposed to exponential. I'm not sure if that's just me, or the education system!

An exponential function takes the form f(x) = k^x.
A quadratic takes the form f(x) = kx².

Energy is given by $E=\frac{1}{2}mv^2$ That means that it is quadratic in v, not exponential.

 

[jrmichler]

What I am trying to figure out is why those trade-offs occur

When the powder burns, it turns into a hot gas at high pressure. The peak pressure is on the order of 50,000 PSI in a high power rifle. The pressure decreases as the bullet travels down the barrel. The pressure vs bullet travel distance is, to a first order approximation, the same for bullets of different weights. Friction between the bullet and the barrel is small compared to the force of the hot gas, so is ignored for simplified calculations.

The force on the bullet is the pressure times the cross sectional area of the bullet. A 30 caliber bullet has cross sectional area of 0.0707 square inches, so the peak force is 0.0707 X 50,000 = 3500 lbs. If the force vs distance is the same for heavy and light bullets, and the gun barrel length is the same, then the work is the same. Work is force times distance, so is not a function of bullet weight or time in the gun barrel.

Two different weight bullets, and the same work, results in the same kinetic energy for each. Search work energy theorem to learn more. Two different masses, each with the same kinetic energy, will have different momentums.

All of the above is a simplified first order approximation that explains the approximate bullet weights and velocities in the OP. Reality is more complex.

 

[skh149]

When the powder burns, it turns into a hot gas at high pressure. The peak pressure is on the order of 50,000 PSI in a high power rifle. The pressure decreases as the bullet travels down the barrel. The pressure vs bullet travel distance is, to a first order approximation, the same for bullets of different weights. Friction between the bullet and the barrel is small compared to the force of the hot gas, so is ignored for simplified calculations.

The force on the bullet is the pressure times the cross sectional area of the bullet. A 30 caliber bullet has cross sectional area of 0.0707 square inches, so the peak force is 0.0707 X 50,000 = 3500 lbs. If the force vs distance is the same for heavy and light bullets, and the gun barrel length is the same, then the work is the same. Work is force times distance, so is not a function of bullet weight or time in the gun barrel.

Two different weight bullets, and the same work, results in the same kinetic energy for each. Search work energy theorem to learn more. Two different masses, each with the same kinetic energy, will have different momentums.

All of the above is a simplified first order approximation that explains the approximate bullet weights and velocities in the OP. Reality is more complex.

Hi there, of course I understand that it's an approximation, and that there will be more complexity to to consider in reality, but your explanation makes the most sense to me.

I appreciate the dementional analysis especially.

I was always lead to believe (not by physicists mind you) that muzzle energies were quite different for a given load when changing bullet mass. And that lighter bullets traveling faster had more energy despite their lower mass.

It seems either I was misinterpreting the people who said this, or they were misinformed, or perhaps trying to favour their preferred smaller cartridge/caliber. In defense of this view: because heavier bullets tend to take up more space, one might be able to theoretically load a cartridge with more powder when using a smaller bullet, leading to greater energies; and the lighter bullet with less momentum might "dump" its energy faster in a target. But that's all for another post

Knowing that the muzzle energies should basically be equal, as per your explanation, makes things very clear, thank you.

Thanks for all the replies -- I think I have a satisfactory answer

 

[Delta2]

Drag increases with speed and with the same shape and same kinetic energy the lighter bullet is faster.

I am afraid you are correct here :D. Well my original statement that the work of the drag will be the same apparently doesn't hold. Thanks for the correction.

But I think that the work of the drag will be small in comparison to the work of explosion gases, hence the total work will be approximately the same in the two cases.

 

[nasu]

Drag increases with speed and with the same shape and same kinetic energy the lighter bullet is faster.

This is true no matter what the drag is. Or without drag. If the KE is the same, less mass means more speed. By definition of KE.

 

[vanhees71]

What's the formula to calculate the kinetic energy in terms of momentum? Note that it's always much more elucidating to write down symbolic formulas first and only at the end plug numbers in!

 

[erobz]

The kinetic energy of the projectile at the end of a barrel of length $\ell$ will have some dependency on its mass via a drag force $\propto A \beta v^2$:

examining an analytically solvable case where the gas expands basically at constant pressure $P$ over distance $x = \ell$, Newton's law implies the following ODE:

$$ P A - \beta A v^2 = m v \frac{dv}{dx} \tag{1}$$

Making the substitution that $u = P A - \beta A v^2$ transforms $(1)$ into:

$$ - \frac{m}{2 \beta A} \frac{du}{dx} = u \tag{2}$$

The solution to $(2)$ for $v^2$ at the end of the barrel $x = \ell$ is given by:

$$ v^2 = \frac{P}{\beta} \left( 1 - e^{-\frac{2 \beta A \ell }{m}} \right) $$

Thus, the kinetic energy of a projectile of mass $m$, drag proportionality $\beta$ , and projected area $A$ would be:

$$KE = \frac{m P}{2 \beta} \left( 1 - e^{-\frac{2 \beta A \ell }{m}} \right) $$

I don't know the magnitude of the dependency on $m$, but it seems clear the exit kinetic energy will have some non-linear dependency on the mass of the projectile.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

After deleting a spam necropost and replies, the thread is now back open.

 

[Mark44]

From post #1:

Since momentum is equal to mass multiplied by velocity, the heavier bullet will carry more momentum. That heavier bullets have greater momentum is generally accepted to be true in the shooting community.

This is also generally accepted to be true in the physics community, assuming that both bullets leave the muzzle at the same velocity.

 

[sophiecentaur]

Why would the momentum not be about the same?

I quoted this from an early post. Have we sorted this out now? None of this stuff is a matter of opinion, of course and, when the variables have all be specified (not trivial) the results will be predictable. Non Scientists /Mathematicians do not qualify to have an opinion in these matters - any more than they can do rocket Science. "Doesn't mean you're a bad person."

This is also generally accepted to be true in the physics community, assuming that both bullets leave the muzzle at the same velocity.

If those are your specified conditions then fair enough but the two cartridges would have to contain different amounts of potential energy or just be designed badly so is that a worthwhile comparison? A good big one will beat a good little one but where's the Engineering?