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why does a bullet with less weight have more muzzle energy?
i can not figure this out. when looking at bullet weights and muzzle energies, i would guess that the heavier bullet would have more energy. for example, why does a .40 cal. 180gr. bullet have less energy and recoil? than a .165gr. bullet? can some one explain this to me?

Here's the short answer (ignoring all factors and limitations before and after the muzzle): The exponential component of higher velocity.


Lighter bullet doesn't have more muzzle energy and recoil in all cases. If you are shooting some of the action shooting sports, USPSA, IDPA for example; they require minimum power factor. Power factor is momentum, or mass times velocity. PF=Mass*Velocity. At a constant power factor, lower mass means more velocity. Energy is given by E=0.5*mass*velocity^2. Since velocity is "squared", it has a much bigger impact on energy when increased than a proportional decrease in mass. So it's only when power factor (Momentum) is held constant that a heavier bullet has less energy.
Perceived recoil is more complicated. To first order, perceived recoil is dependent on the dependent on the energy for 2 reasons. In addition to E= 0.5*m*v^2, the change in energy as the bullet goes down the barrel is given by E=Force*Distance. For the same gun, Distance, the barrel length, is fixed. If you have less energy, you have less force which means less opposing force or recoil. There is also an effect from the amount of gas and smoke that comes out of the barrel. At constant power factor, it takes more powder to get a lighter bullet going than a heavier bullet. For my competition gun, It takes almost 20% more powder to get to my desired power factor with a 124gn bullet as it does with a 147 gn bullet. This is likely a reason for faster powders generally yielding less perceived recoil than slower powders. Faster powders normally have lower charge weights for the same energy, so they have less gas going out the barrel pushing back on the gun and the shooter.

Lighter bullets move faster, but they do not hit as hard as heavier bullets so the energy created is less. Generally speaking, a heavier bullet is going to produce more energy, but if you increase the powder charge in a lighter bullet, then it can create even more energy, but doing so will definitely increase the felt recoil of the bullet. The weight of the bullet has nothing to do with the recoil. It's the powder charge that effects that. This is why +P rounds are snappier than standard or low presser rounds. But if you take a standard pressure round, a 147gr bullet will create more energy than a 115gr bullet. The 115gr bullet may be moving a little faster, b/c it is lighter, but the 147gr bullet will hit a lot harder. Sort of like a porche and a train. Going the same speed of 55mph, which is going to hit harder and cause more damage vs which one looks like it is going faster? A train moving at 55mph will still cause a lot more damage than a porche moving at say 75mph.

Jeez. The answer is : Energy equals mass times the square of the velocity. Apply it to the problem.

Originally Posted by hillman
Jeez. The answer is : Energy equals mass times the square of the velocity. Apply it to the problem.
I was waiting for someone to make it this simple. Physics 101 in application.

Originally Posted by hillman
Jeez. The answer is : Energy equals mass times the square of the velocity. Apply it to the problem.
This sounds simple, but if one does not really get "math" then it is swahili. I mean I understand the words on the page, but it really does not go anywhere beyond that. I'm not going to take the time to do the math (apply it to the problem). I just want someone to explain it to me so that I, a normal guy, can understand what "Energy equals mass times the square of the velocity" actually means in layman's terms. Train vs Porche. That's simple.

Originally Posted by GCBHM
This sounds simple, but if one does not really get "math" then it is swahili. I mean I understand the words on the page, but it really does not go anywhere beyond that. I'm not going to take the time to do the math (apply it to the problem). I just want someone to explain it to me so that I, a normal guy, can understand what "Energy equals mass times the square of the velocity" actually means in layman's terms. Train vs Porche. That's simple.
If you're not willing to learn the limited maths required here, then the question shouldn't interest you. If the question interests you, then you really should learn the maths.
E = M * V * V. Since you are multiplying by the velocity twice, then a diobling of velocity means 4 times the energy for the same mass (2 x 2 = 4)
So if your bullet is lighter, then it will leave the muzzle faster for the same powder charge. The velocitysquared thing means it will have more energy, rather than the same.
Simple, no?

Originally Posted by SailDesign
If you're not willing to learn the limited maths required here, then the question shouldn't interest you. If the question interests you, then you really should learn the maths.
E = M * V * V. Since you are multiplying by the velocity twice, then a diobling of velocity means 4 times the energy for the same mass (2 x 2 = 4)
So if your bullet is lighter, then it will leave the muzzle faster for the same powder charge. The velocitysquared thing means it will have more energy, rather than the same.
Simple, no?
So in other words, lighter bullets move faster given equal powder charge. Simpler, no?

Originally Posted by GCBHM
I just want someone to explain it to me so that I, a normal guy, can understand what "Energy equals mass times the square of the velocity" actually means in layman's terms.
As stated above, the reason is because in the kinetic energy equation K.E. = 1/2 * MASS * VELOCITY^2 the velocity has an effect that is "squared" and thus changes have more of an impact. Examining the equation, we see that a change in mass of the bullet, say 200 grains to 250 grains has an change roughly equated to "50". But a change in the velocity from that lighter bullet, say 100fps, has a factor of 100^2 or 1000 upon the equation. Changes in velocity have a much greater impact on the calculation than a change in mass. There are some conversion factors needed, but we state for example:
K.E. = 0.5* MASS (grains) * VELOCITY (fps)^2 / 450450 = xx ftlbs
Let's show an example (35 Remington):
K.E. = 0.5 * (200 gr) * (2170 fps)^2 / 450450 = 2090 ftlbs
K.E. = 0.5 * (220 gr) * (1870 fps)^2 / 450450 = 1707 ftlbs.
That change in bullet weight, 200 to 220 grains hardly effects the equation (you're multiplying by 220 instead of 200....mheh). But the change in velocity of 300 fps (squared!) has a much greater effect mathematically since it is raised to the power of 2.
Does that explain it?
Bullet Kinetic Energy Calculator

Originally Posted by SailDesign
If you're not willing to learn the limited maths required here, then the question shouldn't interest you. If the question interests you, then you really should learn the maths.
Whether they are interested in the question has no bearing if they are willing to delve into the math. It's not everybodys pieceo'pie. And usually a simple explanation is better than an admonishment.

Originally Posted by Scorpion8
As stated above, the reason is because in the kinetic energy equation K.E. = 1/2 * MASS * VELOCITY^2. Examining the equation, we see that a change in mass of the bullet, say 200 grains to 250 grains has an change roughly equated to "50". But a change in the velocity from that lighter bullet, say 100fps, has a factor of 100^2 or 1000 upon the equation. Changes in velocity have a much greater impact on the calculation than a change in mass. There are some conversion factors needed, but we state for example:
K.E. = 0.5* MASS (grains) * VELOCITY (fps)^2 / 450450 = xx ftlbs
Let's show an example (35 Remington):
K.E. = 0.5 * (200 gr) * (2170 fps)^2 / 450450 = 2090 ftlbs
K.E. = 0.5 * (220 gr) * (1870 fps)^2 / 450450 = 1707 ftlbs.
That change in bullet weight, 200 to 220 grains hardly effects the equation. But the change in velocity of 300 fps (squared!) has a much greater effect mathematically.
Does that explain it?
Bullet Kinetic Energy Calculator
Well, I think my explanation is more indicative of what I meant. Most people shut down when anyone brings complicated looking math equations into the discussion. Personally, I'm like it means absolutely nothing to me. Just tell me how it works, like in layman's terms such as planes, trains and automobiles. Granted, I understand there are a lot of folks who want to understand it mathematically, but I'm not one of them. I don't care what the math is, just tell me which bullet is better for the prescribed job.

Originally Posted by GCBHM
I don't care what the math is, just tell me which bullet is better for the prescribed job.
For 99% of things that works just fine. Until someone tries to explain why a .223 Remington has more muzzle energy than a 35 Remington, but why the 35 Remmie is a much better mediumtobig game getter than the AR bullet. You can hunt a black bear with the .35, but I wouldn't venture into the same woods with a .223 unless I had a deathwish, despite the math. There are other factors, such as why a big, slow Mike Tyson punch has more lasting value than being hit by a needle at 2000fps. Taylor Knockdown Values have another equation, but it's different than just kinetic energy.

Originally Posted by Scorpion8
For 99% of things that works just fine. Until someone tries to explain why a .223 Remington has more muzzle energy than a 35 Remington, but why the 35 Remmie is a much better mediumtobig game getter than the AR bullet. You can hunt a black bear with the .35, but I wouldn't venture into the same woods with a .223 unless I had a deathwish, despite the math. There are other factors, such as why a big, slow Mike Tyson punch has more lasting value than being hit by a needle at 2000fps. Taylor Knockdown Values have another equation, but it's different than just kinetic energy.
Agreed. I think Ivan Drago had the strongest punch, though, didn't he?

Originally Posted by Scorpion8
Whether they are interested in the question has no bearing if they are willing to delve into the math. It's not everybodys pieceo'pie. And usually a simple explanation is better than an admonishment.
I gave him the explanation as well. All I was trying to say was "If you don't want to understand what the pedals do, then PLEASE don't drive the car" or something like that.

Yup, understood. It's easy to break out my inner Engineernerd, but most folks just want the answer.

Originally Posted by GCBHM
Agreed. I think Ivan Drago had the strongest punch, though, didn't he?
I'm not sure, I just don't want my ear bitten by anyone other than my wife. I'm not sure of an equation for THAT!

Originally Posted by SailDesign
If you're not willing to learn the limited maths required here, then the question shouldn't interest you. If the question interests you, then you really should learn the maths.
E = M * V * V. Since you are multiplying by the velocity twice, then a diobling of velocity means 4 times the energy for the same mass (2 x 2 = 4)
So if your bullet is lighter, then it will leave the muzzle faster for the same powder charge. The velocitysquared thing means it will have more energy, rather than the same.
Simple, no?
Not so simple. The same powder charge? So I can cut the ends or drill holes in the bullet and make them lighter, thereby increasing muzzle energy? I don't think so.

Originally Posted by SailDesign
(2 x 2 = 4)
I like it, even I can get that one.
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