"muzzle energy" versus ft./lbs. of energy

[moooose102]

well, whatever the case, the current measurement system has been used for longer than i have been alive (52 years), and i for one do not want to start learning another system now! whether the current system is right or wrong, it is the standard that we all have learned to use for comparison of calibers and power. if i read that xxx cartridge has 3200 ft pounds of muzzle energy, i KNOW that it will kill a deer fine, and then some. if you tell me that the same cartridge has 232 pounds of momentum, i have no clue what that means. and in your first example, it seems pretty prposterous that a 45 acp would be a better deer gun than a 308!

 

[jim in Anchorage]

The old hands look at MV/ grains bullet. No math necessary.

 

[Ned Kelly]

Just another question.

Since Kinetic energy in a .223 is significantly higher than a .45 acp yet they both are similiar in momentum (both in the 20s) where does the kinetic energy of the .223 go so that energy is conserved?

Does it go into a greater tendency to explode, transfer heat or penetrate more than deliver momentum to an object?

I admit I am a layman on the subject. I see no reason why I should be dressed down over it. 90 percent of gun shooters probably never even think about what the numbers mean.

If I knew all the answers I wouldn't have posted the thread.

 

[Navy_Guns]

http://www.thehighroad.org/showthread.php?t=301126

If an object has a kinetic energy of 1,000 foot-pounds, what it means is that if you were magically able to capture all of that energy and magically convert it to a force that lifted an object against the force of gravity, the energy COULD lift a 1,000 pound mass through a distance of 1 foot working against a standard gravity. I say MAGICALLY because no process is 100% efficient.

There is an absolutely horrible article in the 17th edition of Handloader's Digest written by Lee Saunders called "Thoughts on Energy" where the man boldly tries to wrap his brain around the observation that when a 1,000 pound object is shot by said projectile it never does fly a foot in the air. It's one of those laws like "heat only flows from hot to cold" or "you can't push a rope" - there is no perfect process by which you can capture all the kinetic energy of a moving object and use it to do work (force x distance, lift a weight for example).

If you take a small piece of metal and smash it with a big hammer, the thing will be hot - sometimes REAL hot. Plenty of energy goes into deforming the bullet, target, and produces heat.

 

[Officers'Wife]

Hi Ned,

Kinetic energy is either used to overcome the target's inertia or is converted to heat. Friction contributes far more to global warming than the petroleum industry.

 

[RyanM]

Since Kinetic energy in a .223 is significantly higher than a .45 acp yet they both are similiar in momentum (both in the 20s) where does the kinetic energy of the .223 go so that energy is conserved?

Conservation of energy and conservation of momentum are not actually closely related at all, from a layman's perspective. When you get into relativity and string theory and unified field theory, I suppose they are though.

But conservation of momentum means that, if a bullet hits something and doesn't exit, 100% of the bullet's momentum is "transferred" to the target, thus accelerating it in the direction the bullet was going.

Conservation of energy only means that energy cannot be created or destroyed (except by nuclear reactions and the annihilation or creation of antimatter ), it can only change form. And that's what happens with any process. Eventually, 100% of the bullet's energy (really, the gunpowder's energy) turns into heat.

Like a .223 with 1,000 ft-lbs, the gunpowder will probably have about 3,000 ft-lbs of energy in it. When it's burned, 2,000 of that goes to the bullet, and the other 2,000 ft-lbs are used to heat various parts of the gun, and make a loud sound and bright flash, and propel bits of powder and smoke at high speed. But eventually, all of that energy turns inot heat. Like the flash heats things up a bit from the light; everyone knows strong lasers can burn things. And the sound heats things, too. You could boil a cup of water by yelling at it for a few decades, if it were well insulated enough.

Then the bullet hits the target. It's force, or more accurately strain, which does damage, not "energy." Though force will correlate somewhat with energy, it doesn't tell you the whole picture, as less force is necessary to destroy tissue which the bullet directly contacts, which is why a .45 makes a bigger hole than a 9mm, even though they both have the same energy. For a given amount of energy, a larger and heavier bullet makes more efficient use of it (however, don't make the mistake of confusing efficiency with effectiveness).

But most rifles 1,000 ft-lbs and up can, with a softpoint bullet, exert enough force on an organic target to make a big temporary cavity (the so-called hydrostatic shock effect), which stretches and tears and shreds tissue, making a hole many times larger than the bullet.

In any case, some of the energy is used to make a hole, some of it remains as kinetic energy for awhile, moving tissue around, some of it goes straight to heat, some to sound, etc. But eventually, 100% of the bullet's energy turns into heat, as well.

 

[d2wing]

No matter how you explain it, some guys insist that big bullets are always better than small ones regardless of velocity. How old you are has nothing to do with it. How much you are willing to ignore reseach and science is.
Since animals are largely fluid, displacement is a factor in bullet performance, but velocity creates more energy than mass does. The shock wave and it's damaging effect increases in relation to velocity. A bigger bullet will tear more tissue at total penetration but may do less total damage.
I don't think that ammo companies are hiding any secrets from the public. If they recommend a cartridge for a particular use, I tend to trust that they know what they are talking about. That's my unscientific pronouncement.

 

[brickeyee]

If all you have is big and slow, mass matters.

If you can get small and fast, speed is to your advantage.

Handgun calibers are sort of right in the middle.
A .45 ACP is big and slow, a .357 magnum can be light and fast.

But conventional handgun calibers cannot approach a typical rifle caliber.

(Yes, for the engineers out there, the U.S. pound is a unit of mass, defined as exactly 0.45359237 of the mass of the SI kilogram/IPK.)

To keep a consistent set of units, the pound is treated as a weight, and the slug used as the unit of mass.

While there are systems and conditions that use pounds (mass) the pound weight and slug mass are what is needed here.

 

[benEzra]

Quote:
(Yes, for the engineers out there, the U.S. pound is a unit of mass, defined as exactly 0.45359237 of the mass of the SI kilogram/IPK.)
To keep a consistent set of units, the pound is treated as a weight, and the slug used as the unit of mass.

While there are systems and conditions that use pounds (mass) the pound weight and slug mass are what is needed here.

There is more than one way to keep a consistent set of units. Defining the pound as a weight measure, even though it officially is and has always been defined as a unit of mass (and has been defined in terms of the kilogram for roughly a century), is one way; keeping the pound as a mass measure and using the poundal as the force unit also works.

http://en.wikipedia.org/wiki/Poundal

Another way (and perhaps the most common) is to define both pound-mass AND pound-force as separate units, use them as appropriate, and insert a conversion constant into the F=ma equation when necessary to ensure the units come out correctly.

But it is moot, since for calculations that matter, SI units would generally be used, and the pound mass vs. pound weight ambiguity is one reason for it.

 

[Sir_William1]

All Those Numbers, Now I have a headache, and I'm Confused.

 

[Ned Kelly]

Why is it when I use 1/2 mass in pounds x velocity squared I end up with a higher number than it should be?

Say a projectile has 350 ft.lbs kinetic when you calculate using the above equation you end up around 11,000.

Now if I use the grains x velocity squared devided by 450, xxx I get the kinetic energy as shown in ballistic tables.

 

[JohnKSa]

Why is it when I use 1/2 mass in pounds x velocity squared I end up with a higher number than it should be?

Because (the above debate notwithstanding) the pound is not a measure of mass in the U.S. system and the formula calls for mass. You have to convert pounds to slugs, the measure of mass in the U.S. system. Divide by about 32.174 to get the correct result.

When you use the formula I gave (with grains), the 450,437 already contains that conversion.

450,437 = 2 * 7000 * 32.17407...

2 is from the 1/2 in the formula, 7000 converts grains to lbs and 32.17407 converts lbs to slugs.

 

[benEzra]

Because (the above debate notwithstanding) the pound is not a measure of mass in the U.S. system and the formula calls for mass. You have to convert pounds to slugs, the measure of mass in the U.S. system. Divide by about 32.174 to get the correct result.

Actually, in the U.S. system, the pound is indeed a mass measure, and has been defined as a fraction of a kilogram since 1893. You are thinking of the British Gravitational System, an old engineering convention in which the pound is redefined solely as a force measure and the slug is the mass unit.

The everyday approach in the United States is more along the lines of the English Engineering System, another old engineering convention in which the pound mass is retained, an additional unit known as the pound force is derived, and you use pound mass or pound force depending on what you are talking about. When using Newton's Second Law in this formulation, though, you have to insert a correction constant into the formula, i.e.

E = (1/2 * m * v^2) / 32.174

http://www.engineeringtoolbox.com/mass-weight-d_589.html

The kludges necessary in the dual use of the pound are a major reason why most anyone who does energy calculations outside of the firearms realm uses SI units. No kludges are needed if one uses kilograms, meters per second, and express energy in joules.

For example, the kinetic energy of a 3.6-gram (0.0036 kg) 5.56x45mm bullet at 1000 meters per second is

1/2 * 0.0036 * (1000 * 1000) = 1800 joules or 1.8 kJ

 

[ghoster]

i use a very compclated method.

LITTLE bullet---22lr, small charge, tiny hunk of lead = goes bang, little birdie falls down. kick ( muzzel energy ) about as hard as a good kiss.

HUGE bullet---500 nitro, lots of powder, big hunk of lead = goes keeerbooom, 13,000 lb elephant flops down. kick ( muzzel energy ) is like walking up and smacking a rino.

they have been doing balistics for 100years, kinda doubt they been doing it wrong all along.

 

[JohnKSa]

When using Newton's Second Law in this formulation, though, you have to insert a correction constant into the formula...

A "correction constant" equal to the acceleration of gravity. In other words, it's not really a correction constant, it's a conversion factor to convert from pounds to mass. Since the formula calls for mass...

You are thinking of the British Gravitational System, an old engineering convention in ...
The everyday approach in the United States is more along the lines of the English Engineering System...

Yeah, when I said the "U.S. System", I meant the "everyday approach in the United States", more specifically the one which requires that you convert pounds to mass by dividing by the acceleration of gravity.

 

[benEzra]

A "correction constant" equal to the acceleration of gravity. In other words, it's not really a correction constant, it's a conversion factor to convert from pounds to mass. Since the formula calls for mass...

Numerically the same, yes, as it would have to be (they are two different ways of arriving at the same numerical result). But the English Engineering formulation uses the pound mass as a fundamental unit and the pound force as a derived unit; slugs are BGS convention only and were not part of the EE convention.

The fundamental problem, of course, is that if you define the pound force as the weight of a pound mass at one standard gravity but still want to measure velocity in feet per second, you have to either define a new energy unit compatible with both, invent a new mass unit (slug) and ignore the pound mass a la British Gravitational, invent a new force unit (poundal) and express energy in foot-poundals (Absolute English, yuk), or build conversion factors into the equations themselves and use pound mass and pound force freely (English Engineering). None of those approaches are very good.

In the 1950's and 1960's, a lot of aerospace engineering (including NASA, IIRC) went with the British Gravitational approach (slugs-mass and pounds-force), but that is by no means the only way to do it, and it is not consistent with the U.S. definition of the pound.

To me, the best solution is the one that NASA and everyone else has adopted in the last two decades, and that is to ditch the vestiges of the ancient Roman/medieval/British/U.S. customary/Imperial hodgepodge entirely and exclusively use SI in technical fields. I personally prefer joules/kJ to ft-lb even in firearm comparisons, but unfortunately most ballistics tables are done in the former.

 

[JohnKSa]

To me, the best solution is the one that NASA and everyone else has adopted in the last two decades, and that is to ditch the vestiges of the ancient Roman/medieval/British/U.S. customary/Imperial hodgepodge entirely and exclusively use SI in technical fields.

I concur.