# why does a bullet with less weight have more muzzle energy?



## donk123 (Jun 6, 2013)

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?


----------



## TurboHonda (Aug 4, 2012)

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


----------



## ponzer04 (Oct 23, 2011)

Veloities


----------



## Smitty79 (Oct 19, 2012)

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.


----------



## GCBHM (Mar 24, 2014)

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.


----------



## hillman (Jul 27, 2014)

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


----------



## slayer61 (Aug 4, 2014)

hillman said:


> 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.


----------



## GCBHM (Mar 24, 2014)

hillman said:


> 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.


----------



## SailDesign (Jul 17, 2014)

GCBHM said:


> 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 velocity-squared thing means it will have more energy, rather than the same.

Simple, no?


----------



## GCBHM (Mar 24, 2014)

SailDesign said:


> 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 in other words, lighter bullets move faster given equal powder charge. Simpler, no?


----------



## Scorpion8 (Jan 29, 2011)

GCBHM said:


> 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 ft-lbs

Let's show an example (35 Remington):

K.E. = 0.5 * (200 gr) * (2170 fps)^2 / 450450 = 2090 ft-lbs
K.E. = 0.5 * (220 gr) * (1870 fps)^2 / 450450 = 1707 ft-lbs.

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


----------



## Scorpion8 (Jan 29, 2011)

SailDesign said:


> 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 piece-o'-pie. And usually a simple explanation is better than an admonishment.


----------



## GCBHM (Mar 24, 2014)

Scorpion8 said:


> 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 ft-lbs
> 
> ...


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.


----------



## Scorpion8 (Jan 29, 2011)

GCBHM said:


> 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 medium-to-big 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 death-wish, 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.


----------



## GCBHM (Mar 24, 2014)

Scorpion8 said:


> 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 medium-to-big 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 death-wish, 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?


----------



## SailDesign (Jul 17, 2014)

Scorpion8 said:


> Whether they are interested in the question has no bearing if they are willing to delve into the math. It's not everybodys piece-o'-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.


----------



## Scorpion8 (Jan 29, 2011)

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


----------



## Scorpion8 (Jan 29, 2011)

GCBHM said:


> 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!_


----------



## TurboHonda (Aug 4, 2012)

SailDesign said:


> 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)
> 
> ...


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.


----------



## denner (Jun 3, 2011)

SailDesign said:


> (2 x 2 = 4)


I like it, even I can get that one.


----------



## hillman (Jul 27, 2014)

Well shucks. E=M times V times V used to be grade school math back in 1949. That was before "New Method" though, eh?


----------



## SailDesign (Jul 17, 2014)

TurboHonda said:


> 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.


Well, let's see (disclaimer: I am NOT a ballistics expert - just an engineer/designer) if you lighten a car with the same powerplant, it accelerates faster, so over a given barrel length a lighter bullet using the same charge SHOULD be going faster. if a half-weight bullet accelerates to twice the speed (not likely, given drag on the rifling, etc.) then it will have twice the energy since E = 1/2 *M*V-squared. M is a half of what it was, V-squared will be four times, and if my math still works, 4 divided by 2 = 2. Reality will be less than that, sure, but the principle is there. Momentum won't have changed in this case, but that's not energy....

Any ballistics guys here want to pitch in? I know I'm not totally right, but I don't think i'm totally wrong, either.


----------



## Scorpion8 (Jan 29, 2011)

TurboHonda said:


> 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.


Powder charge is not germane to the calculation because it's just "whatever" gets you the V (in velocity). Could be electro-magnetic rail, could be a chemical reaction. How you get yourself a bucket full of V for your bullet isn't considered, just how much V you get yourself. All things being equal, if you drill a hollow-point and make the bullet lighter it will go faster. The question for you is then if that charge is now over-charged for the lighter bullet, resulting in bad pressures? But if you have a test barrel that can take the pressure, then yes, a lighter faster bullet follows the same math as above. The math care not how you get there, just what the digits are.


----------



## denner (Jun 3, 2011)

donk123 said:


> 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?


It doesn't in general, but there is an exception for the 40 S&W cartridge(case) using a 180 grain bullet, compared to using a 165 grain bullet which is explained below. Basically the 180 grain bullet is too long and seats too deeply in the case to allow for maximum efficiency of the powder charge and if you overcharge the case it can result in dangerous overpressures using the deeper seated 180 grain projectile. "For this reason, most factory .40S&W 180gr ammunition is loaded a little on the weak side." In other words the 165 grain bullet can be used to the most efficiency in the 40 S&W case allowing for higher energy levels and velocity over the 180 grain bullet.

CALIBERS -- Why the 180gr Bullet is a Bad Choice for .40 S&W


----------



## Scorpion8 (Jan 29, 2011)

SailDesign said:


> M is a half of what it was, V-squared will be four times, and if my math still works, 4 divided by 2 = 2.


Yes, but ..... chances are you won't double the speed at half the bullet weight. Remington's latest catalog shows for the .223 Remington:
45 grain .......3550 fps muzzle.......1259 ft-lbs muzzle
77 grain........2788 fps muzzle.......1329 ft-lbs muzzle
So for a 41% drop in bullet weight they got a 27% increase in muzzle energy at a 5% increase in speed.


----------



## SailDesign (Jul 17, 2014)

Scorpion8 said:


> Yes, but ..... chances are you won't double the speed at half the bullet weight. Remington's latest catalog shows for the .223 Remington:
> 45 grain .......3550 fps muzzle.......1259 ft-lbs muzzle
> 77 grain........2788 fps muzzle.......1329 ft-lbs muzzle
> So for a 41% drop in bullet weight they got a 27% increase in muzzle energy at a 5% increase in speed.


Hmmm... 27% increase in* speed*, 5% drop in energy, by my calculator. 

Edit: Which goes too show that a lighter bullet will travel faster, and the loss in energy is VERY slight. Somewhere in between *may* have increased energy.


----------



## Scorpion8 (Jan 29, 2011)

Yea, just noticed that. Too wrapped up in the math.


----------



## SailDesign (Jul 17, 2014)

Scorpion8 said:


> Yea, just noticed that. Too wrapped up in the math.




If you go down to the 45-70 n that same table (couple of pages further) you'll find the two samples there are lighter/faster/more energy as I would have expected (there are other examples along the way, but that one stands out  ). i suspect the tables re NOT showing any difference in charge, due to chamber pressure limits, etc., that may mess up the math.


----------



## Scorpion8 (Jan 29, 2011)

Well, an even more "telling" story of data is the .30-06 Springfield. Again, using Remington's data we see that at the muzzle we have:

125-grain....3140fps......2736 ft-lbs
220-grain....2410 fps.....2837 ft-lbs.

Taking statistical variation into account, for a bullet "almost" half the bigger one, if we drive it fast enough we get _basically_ the same kinetic energy. And yet you'd hunt moose with the 220-grainer, but not with the 125-grain load. The difference comes in that 600+fps of V that we get a bonus from being "squared". And all that time back in high school wearing corduroy pants you thought being square wasn't cool.  To paraphrase Thulsa Doom (Conan the Barbarian): This is the Riddle of Speed.


----------



## TurboHonda (Aug 4, 2012)

Scorpion8 said:


> Powder charge is not germane to the calculation because it's just "whatever" gets you the V (in velocity). Could be electro-magnetic rail, could be a chemical reaction. How you get yourself a bucket full of V for your bullet isn't considered, just how much V you get yourself. All things being equal, if you drill a hollow-point and make the bullet lighter it will go faster. The question for you is then if that charge is now over-charged for the lighter bullet, resulting in bad pressures? But if you have a test barrel that can take the pressure, then yes, a lighter faster bullet follows the same math as above. The math care not how you get there, just what the digits are.


Powder charge (propellant) is certainly germane, because it is not a fixed value. A given amount of propellant will burn and build pressure depending on its burn rate and the volume of the chamber it is burning in. Typically a heavy bullet is slower to start moving and slower to continue moving. Therefore a high pressure buildup can be expected. If the bullet weight is miraculous reduced, it will start moving quicker and continue moving faster. The burn chamber is now expanding more rapidly and the pressure will probably be at a lower value when the bullet departs the muzzle. Ergo, the increased velocity of the lighter bullet won't necessarily overshadow the reduction in mass.

Check any loading manual. You'll see that for any given brand of propellant a higher quantity will be listed for lighter bullets.


----------



## paratrooper (Feb 1, 2012)

So.....no one has brought up the factor of limp-wristing yet? :watching:


----------



## Steve M1911A1 (Feb 6, 2008)

...And also nobody has even begun to discuss shape factor, otherwise known as ballistic coefficient. 

Further, lighter bullets _may_ deliver more energy...at the muzzle. But what do they deliver at practical distances?
And blunter, heavy bullets lose velocity more quickly. They, too, may deliver less energy at longer ranges.


----------



## SailDesign (Jul 17, 2014)

Steve M1911A1 said:


> ...And also nobody has even begun to discuss shape factor, otherwise known as ballistic coefficient.
> 
> Further, lighter bullets _may_ deliver more energy...at the muzzle. But what do they deliver at practical distances?
> And blunter, heavy bullets lose velocity more quickly. They, too, may deliver less energy at longer ranges.


That's coz the discussion was about *muzzle*-energy, Steve 

(See thread title for details)


----------



## Scorpion8 (Jan 29, 2011)

TurboHonda said:


> Powder charge (propellant) is certainly germane, because it is not a fixed value.


No, you're getting too wrapped up in HOW you get there. The equation careth-not what propellant that you select, how much, or it's burning rate. It only gives you a "V" value. Like I said, you could use an electromagnetic rail gun to achieve the same V and that's all that matters, no powder! The equation has two values, M and V. It works for steel-core bullets, flechettes, arrows, solids, whatever. How you get to that V value is a whole different problem tackled over there in that corner, where folks argue over this powder or that.

Steve - all true, but now we're talking about a subset of flight dynamics, wind drags, coefficients of friction, surface drag numbers and so forth. Those ballistics are a subset of the physics of understanding true flight of the bullet. The physics equation is made in an absolute state, and does not consider those things.


----------



## SailDesign (Jul 17, 2014)

Scorpion8 said:


> No, you're getting too wrapped up in HOW you get there. The equation careth-not what propellant that you select, how much, or it's burning rate. It only gives you a "V" value. <snip-de-dip!>


But, but, but.. If the propellant has no relation to the "V", then what DOES make the boolit accelerate? It absolutely HAS to have a bearing on Velocity. It doesn't matter which powder, No. But it DOES matter how much of it.


----------



## Scorpion8 (Jan 29, 2011)

SailDesign said:


> It absolutely HAS to have a bearing on Velocity.


Of course it does, but it has no spot of consideration in the kinetic energy equation. As long as you have a bucket or a bag full of V, it doesn't matter where it came from. Now if you can use a different powder and get a better V, then all the well and good. That bucket is better than your old bucket. The bullets resulting KE is only a factor of that V value, not where it came from. Try not to confuse "how do I get a bullet to go _this_ fast" with "okay, a bullet of this size going this fast has how much energy?" If two powders give you the same V from whatever loads, then all-else-being-equal, they will give you the same KE.


----------



## TurboHonda (Aug 4, 2012)

Scorpion8 said:


> No, you're getting too wrapped up in HOW you get there. The equation careth-not what propellant that you select, how much, or it's burning rate. It only gives you a "V" value. Like I said, you could use an electromagnetic rail gun to achieve the same V and that's all that matters, no powder! The equation has two values, M and V. It works for steel-core bullets, flechettes, arrows, solids, whatever. How you get to that V value is a whole different problem tackled over there in that corner, where folks argue over this powder or that.
> 
> Steve - all true, but now we're talking about a subset of flight dynamics, wind drags, coefficients of friction, surface drag numbers and so forth. Those ballistics are a subset of the physics of understanding true flight of the bullet. The physics equation is made in an absolute state, and does not consider those things.


The question I posed was this: If I cut or drill a bullet (in a loaded round) will the muzzle energy increase? I contend that the pressure within the gun will change. Furthermore, I believe the pressure will be decreased and the resultant muzzle energy will be lower. If it was an absolute problem I would expect an absolute answer. But, it's not.


----------



## Scorpion8 (Jan 29, 2011)

Assuming that all else is held equal....powder charge, etc. then a lighter bullet should go faster for the same charge. If it goes faster, and depending on how much mass you removed, the KE will change. For example, I looked up two similar 7x57 Mauser listings, both using 45.0 grains of IMR4350. A 160-gr bullet will leave the barrel at approximately 2526 fps. A 175-grain bullet will depart at 2502 fps. There's not enough spread in there to truly get good values (i.e. they are statistically too close) since the difference in weight and speed is not enough for the calculation to account: it's only 0.6%. But the lighter bullet does go faster. In this case it just has less KE because the drop in mass is not sufficiently accounted for by the slightly faster speed: it doesn't go significantly faster enough to gain from the V^2 factor.

You would need to find a load that you could shoot a 100-grain bullet and a 200 grain bullet using the same powder charge, without blowing something up.


----------



## Steve M1911A1 (Feb 6, 2008)

SailDesign said:


> That's coz the discussion was about *muzzle*-energy, Steve
> 
> (See thread title for details)


I was making a joke by going to extremes, piggybacking on paratrooper's "limp wristing" comment.
I guess it was just a little too arcane.


----------



## TAPnRACK (Jan 30, 2013)

Fooled me... stop being so esoteric Steve, lol.


----------



## SailDesign (Jul 17, 2014)

Steve M1911A1 said:


> I was making a joke by going to extremes, piggybacking on paratrooper's "limp wristing" comment.
> I guess it was just a little too arcane.


I wasn't going to mention anyone being limp-wristed, but if you're going there....


----------



## SailDesign (Jul 17, 2014)

Scorpion8 said:


> Of course it does, but it has no spot of consideration in the kinetic energy equation. As long as you have a bucket or a bag full of V, it doesn't matter where it came from. Now if you can use a different powder and get a better V, then all the well and good. That bucket is better than your old bucket. The bullets resulting KE is only a factor of that V value, not where it came from. Try not to confuse "how do I get a bullet to go _this_ fast" with "okay, a bullet of this size going this fast has how much energy?" If two powders give you the same V from whatever loads, then all-else-being-equal, they will give you the same KE.


I'm just trying to be practical, is all. You can have as much V as you want, but you have to take charge into account when you're figuring out how MUCH faster that lighter bullet is going to go. Otherwise you're just pissing in the wind. (with all due respect)


----------



## Scorpion8 (Jan 29, 2011)

SailDesign said:


> ... but you have to take charge into account when you're figuring out how MUCH faster that lighter bullet is going to go. Otherwise you're just pissing in the wind. (with all due respect)


True, which is why many 44 Rem Mag pistol bullets are totally wrong for 444 Marlin velocities, likewise the same in cast bullets. Just because you "can" drive a bullet to a certain speed doesn't mean you "should". Again, _not_ a consideration of the simple KE equation. Bullet quality doesn't enter into the equation.


----------



## Steve M1911A1 (Feb 6, 2008)

SailDesign said:


> I wasn't going to mention anyone being limp-wristed, but if you're going there....


[comment deleted by poster: self-censored]


----------



## KeithC. (Dec 24, 2013)

Why does the heavier bullet penetrate deeper when it has less energy?


----------



## hillman (Jul 27, 2014)

KeithC. said:


> Why does the heavier bullet penetrate deeper when it has less energy?


Inertia, other factors being equal (_which they aren't_). The thing about the energy is, it gets dissipated; 'shock', secondary (temporary) would channel, etc. The heavier bullet has more mass, hence more inertia at a given velocity. This is the same inertia situation as the straight overhand right delivered by a heavyweight compared to a featherweight. One of them doesn't stop at the chin.


----------



## Scorpion8 (Jan 29, 2011)

KeithC. said:


> Why does the heavier bullet penetrate deeper when it has less energy?


The other name for inertia is momentum, which has a physical equation of P = M * V. So it's simply related to mass and velocity. The deeper penetrating bullet will not only depend on which factor rules that equation, mass or velocity but also other factors in the "real world" such as bullet shape, substance that it strikes (ballistic gelatin?) and so forth.


----------



## hillman (Jul 27, 2014)

Scorpion8 said:


> The other name for inertia is momentum, which has a physical equation of P = M * V. So it's simply related to mass and velocity. The deeper penetrating bullet will not only depend on which factor rules that equation, mass or velocity but also other factors in the "real world" such as bullet shape, substance that it strikes (ballistic gelatin?) and so forth.


Inertia and momentum are synonyms technically (so I hear), but they associate differently in my mind. Inertia has a 'meaning-lean' toward weight, momentum 'leans' toward speed. Probably tied to how the terms show up in everyday life (hah, which ain't often).


----------



## SailDesign (Jul 17, 2014)

hillman said:


> Inertia and momentum are synonyms technically (so I hear), but they associate differently in my mind. Inertia has a 'meaning-lean' toward weight, momentum 'leans' toward speed. Probably tied to how the terms show up in everyday life (hah, which ain't often).


For me, inertia is always associated with a resistance to movement (whether a body being pushed or a beam being bent) and momentum is resisting being stopped.


----------



## pic (Nov 14, 2009)

very informative, good information here


----------



## SailDesign (Jul 17, 2014)

pic said:


> How about if I'm shooting out of an automobile going 65 mph? And there is an additional wind velocity of 30 mph not created by the movement of the automobile . Would that be a mute point?
> 
> should I be self censored?


"Yes" (it is "moot", by the way...) and "Definitely". Pretty much in that order. 

The muzzle energy is unaffected by external conditions, except maybe temperature minimally. Your 65 mph may make a small difference (if you're not shooting straight sideways) though.


----------



## TurboHonda (Aug 4, 2012)

SailDesign said:


> "Yes" (it is "moot", by the way...) and "Definitely". Pretty much in that order.
> 
> The muzzle energy is unaffected by external conditions, except maybe temperature minimally. Your 65 mph may make a small difference (if you're not shooting straight sideways) though.


If it were possible to exceed the speed of light, would you be able to see where you're going?


----------



## SailDesign (Jul 17, 2014)

TurboHonda said:


> If it were possible to exceed the speed of light, would you be able to see where you're going?


Probably Yes (since the "light" would be coming at you twice as fast) but you couldn't see where you'd been


----------



## pic (Nov 14, 2009)

I


SailDesign said:


> "Yes" (it is "moot", by the way...) and "Definitely". Pretty much in that order.
> 
> The muzzle energy is unaffected by external conditions, except maybe temperature minimally. Your 65 mph may make a small difference (if you're not shooting straight sideways) though.


Since I am "MOOT". I will remain "MUTE"


----------



## SailDesign (Jul 17, 2014)

pic said:


> I
> 
> Since I am "MOOT". I will remain "MUTE"


That would take care of the self-censorship issue, anyway.


----------



## Couch Potato (Jun 3, 2010)

SailDesign said:


> 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 doubling of velocity means 4 times the energy for the same mass (2 x 2 = 4)


It is not doubling of the velocity, that would be 2V; it is squaring the velocity. You are correct that when the velocity is 2 it doubles to make M x 4, but when the velocity is 3 the equation becomes M x 9 rather than a simple doubling to M x 6. The long and short of it is velocity is far more important than mass in determining the amount of kinetic energy in the bullet. What happen to that energy in flight and at the target is the most important thing for defensive use.


----------



## Goldwing (Nov 5, 2014)

Okay math wizards, lets talk about the other side of the coin. What happens 100 yards down range with regard to bullet weight increasing and the charge remaining the same. (I suspect but do not know the answer and want to learn!)
GW


----------



## SailDesign (Jul 17, 2014)

Couch Potato said:


> It is not doubling of the velocity, that would be 2V; it is squaring the velocity. <snippage>.


Which is why I wrote "E = M * V * V" My tablet doesn't "do" superscripts, but V * V is generally the same as V-squared.


----------



## SailDesign (Jul 17, 2014)

goldwing said:


> Okay math wizards, lets talk about the other side of the coin. What happens 100 yards down range with regard to bullet weight increasing and the charge remaining the same. (I suspect but do not know the answer and want to learn!)
> GW


In general, with a moving body, mass is king. A lighter bullet at a greater speed will lose velocity faster, resulting in less down-range energy. I'm not a ballistics guy, however, just a yacht designer, so anyone who IS a ballistics guy would be interssting to hear from.

Edit: drag is also proportional to speed-squared, hence the lighter faster bullet losing energy/speed faster


----------



## Steve M1911A1 (Feb 6, 2008)

Couch Potato said:


> ...The long and short of it is velocity is far more important than mass in determining the amount of kinetic energy in the bullet. What happen to that energy in flight and at the target is the most important thing for defensive use.


Well... There really is something else that needs to be considered.

If the velocity is too high, and thus producing relatively high energy, a smaller-diameter bullet may be induced to punch all the way through its "meatbag" target.
When a bullet goes all the way through, it leaves relatively little of its energy behind. But it's that energy which does the damage and stops the fight.
So a compromise must be reached, energy versus bullet diameter and weight, such that the bullet stays inside the target and delivers all of its energy to it.

One of the oldest, and most valid, arguments about the 9mm versus the .45, is that the 9mm _round-nose_ bullet is much more likely to go all the way through, while the .45 _round-nose_ bullet is much more likely to remain inside. Thus, the .45 RN bullet will be a better fight stopper than a comparable 9mm bullet.
However, when speaking about modern hollow-point bullets all of the parameters can change. In that case, the 9mm HP bullet can be engineered to stop and deliver, just as well as a .45 HP bullet can.


----------



## Groundhog34 (Mar 20, 2015)

You got to be Einstein to figure this out.


----------



## Steve M1911A1 (Feb 6, 2008)

Groundhog34 said:


> You got to be Einstein to figure this out.


Nope.
All you need is some high-school math and physics.

Oh...Wait a minute...Einstein flunked high-school math and physics!


----------



## SouthernBoy (Jun 27, 2007)

There is another "problem" with high velocity, light weight hollow point bullets. Assuming their orifice doesn't get plugged by clothing or some other material, when they enter a body, they expand much more quickly. And since they are of a lighter weight, this increased resistance to their travel slows them down faster than a heavier bullet, which can result in less penetration and a shallower wound.

However, there are a lot of variables to all of this. The bullet design and the manufacturer who loads it may have intended for this to take place, the idea being a frangible bullet. On the other side of the coin would be bonded bullets with harder cores to offer more expansion control. Just a huge amount of variables involved in all of this, but it is rather easy to comprehend.


----------



## SailDesign (Jul 17, 2014)

SouthernBoy said:


> There is another "problem" with high velocity, light weight hollow point bullets. Assuming their orifice doesn't get plugged by clothing or some other material, when they enter a body, they expand much more quickly. And since they are of a lighter weight, this increased resistance to their travel slows them down faster than a heavier bullet, which can result in less penetration and a shallower wound.
> 
> However, there are a lot of variables to all of this. The bullet design and the manufacturer who loads it may have intended for this to take place, the idea being a frangible bullet. On the other side of the coin would be bonded bullets with harder cores to offer more expansion control. Just a huge amount of variables involved in all of this, but it is rather easy to comprehend.


Yup! Every upside has its downside, and every bullet is bad at something. You just have to pick based on your requirements.
+


----------



## PT111Pro (Nov 15, 2014)

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


That's the answer. No doubt.


> Train vs Porche. That's simple


That is not simple, that is bull and reflects the problem with the today's public school system.

Soso this is how it is!!?
A Train traveling 55 MPH and a Porsche traveling 75 MPH create the same or a different damage to a glass wall, a Box of Matches, to a shoe box, a passenger car, a brig wall, a plastic drinking cup or a stone wall....... That creates assumptions that create all kind of problems based on misunderstandings and false opinions.

That's exactly the problem today. They don't learn anything valuable in school anymore, don't comprehend the minimum facts on science and believe everything that someone put in front of them. Facts get replaced by opinions depending who said it like school teachers, the Media, political or religious leaders. They became the perfect citizen. "You keep them poor and I keep them stupid. This was not invented yesterday. And why not, they don't know the first thing about it, can not falsify anything what is told to them but like to look smart. 
W Bush said "No Child left behind" and the liberals lowered the requirements even more down, so a cocker-spaniel could make it successful through College.

If someone can not even comprehend the basics of the basics like: Energy equals mass times the square of the velocity, and is unable to apply that to anything that moves, like a Train, 18 wheeler, Porsche, Soccer Ball, Rock or a three leaf, has a different problem than only the comprehension of basic math and why a bullet behaves differently based on weight and energy. This person should comprehend immediately to do something against it before s/he becomes a old sucker that cant change anymore.

Taking my 5 cents back in my pocket.


----------



## berettatoter (Sep 1, 2011)

Velocity. Yup, I have to say velocity. :watching:


----------



## Couch Potato (Jun 3, 2010)

Steve M1911A1 said:


> Nope.
> All you need is some high-school math and physics.
> 
> Oh...Wait a minute...Einstein flunked high-school math and physics!


Actually, the topic is typically covered in middle school these days.


----------



## Glock37 (Nov 14, 2014)

SailDesign said:


> Hmmm... 27% increase in* speed*, 5% drop in energy, by my calculator.
> 
> Edit: Which goes too show that a lighter bullet will travel faster, and the loss in energy is VERY slight. Somewhere in between *may* have increased energy.





KeithC. said:


> Why does the heavier bullet penetrate deeper when it has less energy?


The actual answer is sectional density. The 77 grain 223 and 45 grain will be acting very different as range gets further from the muzzle. I know this is about muzzle energy, but no one shoot point blank so muzzle figures are inane. As distance gets longer the 77 grain will be going faster and drop less. One projectile, the 45 grain is designed for rapid explosive expansion for 15# critters and the other 77 grain is designed for long range precise hits.

Why does the heavier projectile penetrate more? Sectional density. Raw weight don't mean a 185 grain 45 will penetrate deeper than a 147 grain 9mm though. Take a similar constructed projectile in both like say a Goldensaber. The 9mm 147 grain has a higher sectional density (SD) and will out penetrate the faster and heavier 185 grain from the 45. To match penetration values you need a 45 with the same SD. this is also where projectile construction comes in. As the bullet penetrates and expands it loses SD and its penetrating ability is shed. Projectiles that expand at a slower rate like XTP retain SD over a longer period while expanding and penetrate more than faster expanding projectiles in the same caliber and weight and speed. It's why all projectiles of the same speed weight and caliber do not all penetrate the same exact amount.

If I had two choices and they were a 185 grain 45 and a 9mm 147 grain I Would pick the 147 9mm because handguns do not generate enough "energy" to make any kind of difference in performance. You need penetration to do the job in a pistol. The slower lighter 9mm will penetrate more, so it will be my choice.

There is more to it than raw muzzle energy and it should be the last thing you look for. A deeper penetrating designed projectile proven to expand with a higher SD will be my choice every time no matter the velocity.


----------



## SailDesign (Jul 17, 2014)

Glock37, you toss around the words "Sectional Density" but you haven't explained what they mean... Can you give us a definition? I'm an engineer, but boats, not bullets.


----------



## Glock37 (Nov 14, 2014)

SailDesign said:


> Glock37, you toss around the words "Sectional Density" but you haven't explained what they mean... Can you give us a definition? I'm an engineer, but boats, not bullets.


It is a weight to length ratio. It is far more important than energy or caliber. It's main fault is it is not as glamorous as snazzy words like energy that inspire people to buy ammunition.

Here is a link to a longer more detailed explanation.

Sectional Density for Beginners


----------



## TurboHonda (Aug 4, 2012)

@Glock37

Sectional density is a static term. A Hornady #35580 HP-BT-XTP 147 grain has a sectional density of 0.167, whether it's still in the box or penetrating something downrange. 

The OP was about muzzle energy. Dynamic and variable. 

Maybe sectional density deserves a thread of it's own.


----------



## Glock37 (Nov 14, 2014)

TurboHonda said:


> @Glock37
> 
> Sectional density is a static term. A Hornady #35580 HP-BT-XTP 147 grain has a sectional density of 0.167, whether it's still in the box or penetrating something downrange.
> 
> ...


It does. However someone asked within the thread, so I answered. I also noted muzzle energy or one of its similar names is inane when it comes to pistols. It does nothing. It's more like bow hunting. You want deeper penetration with more flesh coming in contact with that projectile making the deepest hole you can.

When that .167 SD projectile penetrates and starts to get shorter and fatter it sheds SD, that's also why I mentioned projectile construction. Slower expanding projectiles stay longer as they penetrate, so they get more of it. It don't end with a .167 after full expansion. So no, it is not .167 as it penetrates.


----------



## SailDesign (Jul 17, 2014)

Glock37 said:


> It is a weight to length ratio. It is far more important than energy or caliber. It's main fault is it is not as glamorous as snazzy words like energy that inspire people to buy ammunition.
> 
> Here is a link to a longer more detailed explanation.
> 
> Sectional Density for Beginners


Cool! Thanks. It is a bit like the Displacement/Length ratio that we use for sailing boats (200 and up is "Heavy" while 100 and under is "ultralight") except that ours is non-dimensional while SD is a weight/square inch.

Definitely can see why it is important after impact, and less so but still there before that. It is a factor in the drag calcs, as Reynold's number is just as important in incompressible flow (supersonic) with bullets as it is in water for boats (water is considered incompressible at ANY speed  ).


----------



## Glock37 (Nov 14, 2014)

SailDesign said:


> Cool! Thanks. It is a bit like the Displacement/Length ratio that we use for sailing boats (200 and up is "Heavy" while 100 and under is "ultralight") except that ours is non-dimensional while SD is a weight/square inch.
> 
> Definitely can see why it is important after impact, and less so but still there before that. It is a factor in the drag calcs, as Reynold's number is just as important in incompressible flow (supersonic) with bullets as it is in water for boats (water is considered incompressible at ANY speed  ).


Yes sir! It is also very important during flight. Higher SD projectiles retain more velocity as range gets longer. It will reach a point where the heavier higher SD slow MV projectile is going faster than the lower SD faster lgheter one was at the muzzle.


----------



## SailDesign (Jul 17, 2014)

Glock37 said:


> Yes sir! It is also very important during flight. Higher SD projectiles retain more velocity as range gets longer. It will reach a point where the heavier higher SD slow MV projectile is going faster than the lower SD faster lgheter one was at the muzzle.


Sounds good - I'm not used to dealing with decaying speed - usually boats have some sort of propulsion, so we ignore those effects unless you're a Ship Guy dealing with docking situations.


----------



## TurboHonda (Aug 4, 2012)

Glock37 said:


> When that .167 SD projectile penetrates and starts to get shorter and fatter it sheds SD, that's also why I mentioned projectile construction. Slower expanding projectiles stay longer as they penetrate, so they get more of it. It don't end with a .167 after full expansion. So no, it is not .167 as it penetrates.


Of course you are correct. I should have said "just prior to penetration".


----------

