rjknewfire
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please help me out here. step by step, how did you derive the mv^3/4s equation? secondly, how can you asume equal penetration distance? thirdly, how is this consistant with impact equals momentum?
Power Formula: the 9mm smokes the 45ACP!!!
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Let's ignore whether the formulas themselves are correct or not. Even if they are you don't address the basic question of what measurement most accurately predicts the stopping power of a round. Is it momentum, energy, power, wound channel, penetration depth, etc.
Of all of those I think that power is the least likely to have any merit. Given two rounds with the same energy the power will go up as the depth of penetration decreases. So does that mean a super light round at hyper velocity that expends all of its energy in a 1mm deep wound channel will be devastating? Sounds like body armor to me and I think the answer is obvious.
Energy would seem like a good candidate since the body has to absorb that energy. A round that transfers all of its energy to the target would be best. The tissue has to deal with that energy with destructive consequences. But what if all of that energy is absorbed by a very small part of the body. It could absolutely destroy that tissue. But what if the destruction of that tissue doesn't cause dehabilitation? Then how effective has the round been? This would be equivalent to a shallow or narrow wound.
I tend to place more faith in hitting something vital and putting a big hole in it. The bigger the wound channel and the deeper the penetration the more likely you are to hit something vital and damage it.
So what is the real answer. How about this (and you all already knew this). It can't be boiled down to a simple mathematical equation because it is much to complex with too many variables that are difficult to control and account for.
Results from actual shootings are the best indicator. My understanding is that the 45 was requested by the U.S. Army after they had poor results from shoot .38 revolvers at hopped up natives in the Phillipines. The .38 just didn't have practical stopping power. I don't know how the loadings of that era compare to .38 Special and 9mm of today. But the Army seemed to go to the bigger is better formula based on practical experience.
Of all of those I think that power is the least likely to have any merit. Given two rounds with the same energy the power will go up as the depth of penetration decreases. So does that mean a super light round at hyper velocity that expends all of its energy in a 1mm deep wound channel will be devastating? Sounds like body armor to me and I think the answer is obvious.
Energy would seem like a good candidate since the body has to absorb that energy. A round that transfers all of its energy to the target would be best. The tissue has to deal with that energy with destructive consequences. But what if all of that energy is absorbed by a very small part of the body. It could absolutely destroy that tissue. But what if the destruction of that tissue doesn't cause dehabilitation? Then how effective has the round been? This would be equivalent to a shallow or narrow wound.
I tend to place more faith in hitting something vital and putting a big hole in it. The bigger the wound channel and the deeper the penetration the more likely you are to hit something vital and damage it.
So what is the real answer. How about this (and you all already knew this). It can't be boiled down to a simple mathematical equation because it is much to complex with too many variables that are difficult to control and account for.
Results from actual shootings are the best indicator. My understanding is that the 45 was requested by the U.S. Army after they had poor results from shoot .38 revolvers at hopped up natives in the Phillipines. The .38 just didn't have practical stopping power. I don't know how the loadings of that era compare to .38 Special and 9mm of today. But the Army seemed to go to the bigger is better formula based on practical experience.
I was given a cantalope for a brain at birth (they ran out of stock on brains, had to put something in the package to ship me). lol.
Anyhow,
KE = 1/2 M v2
To me that is the most useful equation when armchairing differences. I believe the U.S Military used something similar when it switched to the 9mm as the standard side arm (on the basis of 15 round magazine. More total energy to put on one target or mulitple targets.
Some data from remingtons site relative to this:
9MM-
124 MC: Muzzle V - 1110fps, 50 yd - 1030fps
124 MC: Muzzle KE - 339ft/lbs, 50 yd 292ft/lbs
.45-
230g MC: Muzzle V - 835fps, 50 yd - 800fps
230g MC: Muzzle KE -356ft/lbs, 50 yd - 326 ft/lbs
Take the data @ 50 yds :
9mm = 292 ft/lbs
.45 = 326f ft/lbs
The spread isn't that great and when you look at it from a total energy stand point -
.45 (7+1) has 2608ft/lbs total v.s. 9mm (15+1) 4672ft/lbs total.
This is my reasoning behind favoring 9mm guns. Aside from the fact that the ammo is cheaper to shoot on a regular basis.
-d
Anyhow,
KE = 1/2 M v2
To me that is the most useful equation when armchairing differences. I believe the U.S Military used something similar when it switched to the 9mm as the standard side arm (on the basis of 15 round magazine. More total energy to put on one target or mulitple targets.
Some data from remingtons site relative to this:
9MM-
124 MC: Muzzle V - 1110fps, 50 yd - 1030fps
124 MC: Muzzle KE - 339ft/lbs, 50 yd 292ft/lbs
.45-
230g MC: Muzzle V - 835fps, 50 yd - 800fps
230g MC: Muzzle KE -356ft/lbs, 50 yd - 326 ft/lbs
Take the data @ 50 yds :
9mm = 292 ft/lbs
.45 = 326f ft/lbs
The spread isn't that great and when you look at it from a total energy stand point -
.45 (7+1) has 2608ft/lbs total v.s. 9mm (15+1) 4672ft/lbs total.
This is my reasoning behind favoring 9mm guns. Aside from the fact that the ammo is cheaper to shoot on a regular basis.
-d
Dave,
OR, you can send to me all your 45ACP and whatever else you have there, for proper disposal.
Cheol,
Actually, I was banking on the Newton's first (or is it second) law on energy to take care of the equivalence of work and energy in the equation. You're correct that it should be work, but in the absence of such data, I deem it best to use available info instead. Much more practical, don't you agree! Also, it is the impulse component of the projectile that I'm dealing with here, not the momentum, as Hoss Cartwright mentioned. Momentum is simply the projectile's ability to push another body, altering the state of body's initial inertia, but has nothing to do with the devastation of the flesh being impacted upon.
More later...
OR, you can send to me all your 45ACP and whatever else you have there, for proper disposal.
Cheol,
Actually, I was banking on the Newton's first (or is it second) law on energy to take care of the equivalence of work and energy in the equation. You're correct that it should be work, but in the absence of such data, I deem it best to use available info instead. Much more practical, don't you agree!
Also, it is the impulse component of the projectile that I'm dealing with here, not the momentum, as Hoss Cartwright mentioned. Momentum is simply the projectile's ability to push another body, altering the state of body's initial inertia, but has nothing to do with the devastation of the flesh being impacted upon. More later...
Will Beararms
New member
Yes New Comer we all have our theories that usually are formed when we have too much time on our hands. You keep working that abacus and exercising that imagination.
Will Beararms
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GET THE STRASSBURG GOATS! GET THE STRASSBURG GOATS!
Hoss Cartwright
New member
(124 x 1450) / 3,850,000 = 179,800/3,850,000 = 0.0467 hp
I know that this # sounds low, but power is the rate at which energy is transfered and is proportional to M x V, not M x V x V, which is the case with energy.
The energy #s are high because velocity is high, and energy is proportional to the square of the velocity.
Some of these high hp #'s seem a bit off the reservation. You have to do a far amount of tweaking to get 400 hp out of a big block V-8.
For more info, see http://www.gunsandammomag.com/dynamic.asp?intSectionID=211&intArticleID=2647
Everybody seems to overlook “momentum” when comparing ballistic performance. When an object strikes another object, two things are conserved; energy and momentum. Conserved means that if the bullet comes to a complete stop, both its energy and its momentum must be transferred to the target.
Momentum may be thought of as the impulse imparted to the target - how hard the punch is. Momentum is the same as “power factor” and is simply the velocity times the mass. In relative terms, the 115gr x 1200fps 9mm has a power factor of 138 and the 230gr x 850fps 45ACP has a power factor of 195. Which one do you think imparts the greater impulse to the target?
Energy is the ability to do work. The energy part has more to do with temporary stretch cavity and tissue destruction and also work done upsetting the bullet if it expands.
Momentum may be thought of as the impulse imparted to the target - how hard the punch is. Momentum is the same as “power factor” and is simply the velocity times the mass. In relative terms, the 115gr x 1200fps 9mm has a power factor of 138 and the 230gr x 850fps 45ACP has a power factor of 195. Which one do you think imparts the greater impulse to the target?
Energy is the ability to do work. The energy part has more to do with temporary stretch cavity and tissue destruction and also work done upsetting the bullet if it expands.
its just to bad that the energy release dosn't do anything in the way of damaging tissue. Most of the horsepower (is that what were calling it now) is spent deforming the bullet, and the rest is released as heat. A small portion is spent forming the hydrostatic shock wave (temporary cavity) but it has been proven that a handgun bullet of any existing caliber dosnt produce a hydrostatic shock wave with enough force to perminently damage vital tissue. Yes, some of our body tissue can be damaged by a handgun bullets H-S shock (liver, brain, splean) but for the most part, the tissue just just stetches and returns to normal..the vital stuff in COM anyway.
Ceol Mhor,
), but definitely not for power. Also, if you prefer metric, use 1 hp=0.746 kilowatt. So a 9mm's 404 hp is roughly 300kilowatts.
Hoss / DBR,
That formula should only be used in comparing projectile momenta (on earth, that is), but has no bearing on power as it is meant to be, a rate of release of energy(or work). A constantly moving projectile cannot release energy unless it acts, or was acted upon by another body (Mr. Newton was really so brilliant). Some calcs are needed to support this, not simple substitutions...
Funny, not to flame you or anybody else, but i suddenly remembered, that is how my "row 4" classmates used to compute for power. I graduated, they didn't.
Then stop in only 12". How would you be? On the lap of a gorgeous stewardess, I hope...
Do the math. It's "liberating"...
Casey,
Now that's a lot of thump!!
Will,
"Yes New Comer we all have our theories that usually are formed when we have too much time on our hands. You keep working that abacus and exercising that imagination."
What can I say? This is a very hot weekend in Manila. Too hot to go to my favorite gunstore. And I just love to chat with you guys! You're the best!!!
More later...
Sorry, bud! Review your units. That is a formula for rate of change in power (whatever that is
), but definitely not for power. Also, if you prefer metric, use 1 hp=0.746 kilowatt. So a 9mm's 404 hp is roughly 300kilowatts.Hoss / DBR,
Sorry, Hoss. Yours is a derivative of the momentum concept, not power. Why so? Because weight is an apparent property. Go to the moon and your values simply won't hold. You'll realize that a 22LR fired here will be as powerful as a 45ACP fired on the moon. Because a 230 gr slug will weigh less than 40gr there. But mass, now there's the ticket. All the laws of physics will respect that. And we all know that whatever concept that can hold its water here must hold everywhere else.
That formula should only be used in comparing projectile momenta (on earth, that is
), but has no bearing on power as it is meant to be, a rate of release of energy(or work). A constantly moving projectile cannot release energy unless it acts, or was acted upon by another body (Mr. Newton was really so brilliant). Some calcs are needed to support this, not simple substitutions...Funny, not to flame you or anybody else, but i suddenly remembered, that is how my "row 4" classmates used to compute for power. I graduated, they didn't.
That's because the impact duration is oh so short, in the neighborhood of 0.001 to 0.002 of a second. Try this: Imagine yourself in a Concorde flying at Mach 2...
Then stop in only 12". How would you be? On the lap of a gorgeous stewardess, I hope...
Do the math. It's "liberating"...
Casey,
It comes to 768 hp!!
Now that's a lot of thump!!Will,
"Yes New Comer we all have our theories that usually are formed when we have too much time on our hands. You keep working that abacus and exercising that imagination."
What can I say? This is a very hot weekend in Manila. Too hot to go to my favorite gunstore. And I just love to chat with you guys! You're the best!!!
More later...
Hoss Cartwright
New member
New_Comer,
I actually do realize the difference between mass and weight. The 3,850,000 "finagle factor" takes the acceleration of gravity into account. I assumed that everything that we were doing was on this planet, and saw no reason to add another complication.
My 0.0467 hp answer was correct in the context of the question that was asked. Casey asked how much power that a slug of a given weight and velocity had.
As another poster has already noted, your impulse theory is flawed by the fact that deceleration is not constant. You are correct in the assumption that some calculus would be needed to get an exact answer. However, the data that would be required is missing. Using an average deceleration will not even yield a reasonable approximation in a situation such as this where deceleration is nowhere near linear and velocity is squared.
This has been fun, I was a math major in college but I minored in physics, but ,alas, that was 20 yrs. ago and some things have been forgotten .The one thing that I do remember from impulse theory is that it is very easy to accidently invent a perpetual motion machine if one is not careful.
Best Regards
Hoss
I actually do realize the difference between mass and weight. The 3,850,000 "finagle factor" takes the acceleration of gravity into account. I assumed that everything that we were doing was on this planet, and saw no reason to add another complication.
My 0.0467 hp answer was correct in the context of the question that was asked. Casey asked how much power that a slug of a given weight and velocity had.
As another poster has already noted, your impulse theory is flawed by the fact that deceleration is not constant. You are correct in the assumption that some calculus would be needed to get an exact answer. However, the data that would be required is missing. Using an average deceleration will not even yield a reasonable approximation in a situation such as this where deceleration is nowhere near linear and velocity is squared.
This has been fun, I was a math major in college but I minored in physics, but ,alas, that was 20 yrs. ago and some things have been forgotten
.The one thing that I do remember from impulse theory is that it is very easy to accidently invent a perpetual motion machine if one is not careful. Best Regards
Hoss
New Comer,
Have you attempted to invent a new mouse trap??? Your post reads as though you're convinced of the superority of the 9MM, and you have invented a formula that is designed to buttress your preconceived notion. So, are you really of the position that the 9MM is superior to the .45 ACP?
Sometimes when fleshing out used car salesmen you gotta indentify and expose their agenda lest you get stuck with a lemon. Seems like your post has a tinge of citrus.
Put me squarely in the corner of momentum.
Hasta luego,
E
Have you attempted to invent a new mouse trap??? Your post reads as though you're convinced of the superority of the 9MM, and you have invented a formula that is designed to buttress your preconceived notion. So, are you really of the position that the 9MM is superior to the .45 ACP?
Sometimes when fleshing out used car salesmen you gotta indentify and expose their agenda lest you get stuck with a lemon. Seems like your post has a tinge of citrus.
Put me squarely in the corner of momentum.
Hasta luego,
E
But when I multiplied the hypotenuese (sp?) by the coeffecient (sp?) factor, I STILL wasn't able to determine the atomic weight of Zinc...
You guys are alright...
As soon as the "magic" bullet/cartridge is found, let me know as I will chamber all my pistols in that caliber.
lol.
JMHO
You guys are alright...
As soon as the "magic" bullet/cartridge is found, let me know as I will chamber all my pistols in that caliber.
lol.
JMHO
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I'm from the experiential school of physics.
I propose a range experiment (outdoor if possible).
Attached to the from of a reasonabily sized gong a 3 lb canned ham or better yet spam.
Stand off a safe distance (see your range rules) and blast away at the canned ham / spam. Visually monitor gong displacement, sound (or noise) and the size of the mess made with the ham / spam.
Repeat for each caliber / load you wish to evaluate.
Draw your own conclusions.
I propose a range experiment (outdoor if possible).
Attached to the from of a reasonabily sized gong a 3 lb canned ham or better yet spam.
Stand off a safe distance (see your range rules) and blast away at the canned ham / spam. Visually monitor gong displacement, sound (or noise) and the size of the mess made with the ham / spam.
Repeat for each caliber / load you wish to evaluate.
Draw your own conclusions.
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