energy

04 March 2011, 13:49

303Guy

energy

someoldguy is right!

A object - in this case a bullet - cannot have momentum without energy (or vise versa). But - too much energy with too little momentum results in an explosive impact on the surface which on a big animal is not going to put it down. At very high bullet velocities a huge amount of energy gets converted into heat and elastic shockwaves which carries the energy away (which might do significant damage elsewhere but might not - I have seen an animal disembowelled by a chest shot from a 22-250. The abdominal skin split open and all the entestines where blown out as a in the intestinal sack). A lot of energy gets used up in creating temporary wound cavities which close up and don't always kill. The energy or momentum is not what counts. It's the RATE OF ENERGY TRANSFER which equates to power. It's the power the bullet transmits to the animal and area it is delivered to and that will often include a deep wound channel i.e penetration. The faster the bullet the higher the possibility of reduced penetration. On a small animal that means more energy transfer and a quicker death. On a larger animal ... ? Well, it depends how far the bullet has penetrated before it 'explodes'. I'm refering to different wound channel profiles for different bullets at different velocities.

04 March 2011, 14:23

someoldguy

quote:
A 55 grain .22 caliber bullet at 3,600 FPS has a calculated 1,583 FPE but only 28.29 pound feet of momentum. A 360 grain .45 caliber bullet from a .454 at 1,400 FPS has a calculated energy of 1,567 FPE, yet produces 72 pounds feet of momentum. We know from expereince that the 360 grain .45 caliber bullet will out penetrate the 55 grain .22 caliber bullet at 3,600 FPS despite having less kinetic energy.

Sorry I didn't come up with a more appropriate response earlier.
OK, then...
That .22 caliber bullet I would guess is a pointy-type expanding bullet which is likely going to lose a lot of mass. I doubt seriously that you'd get a net 17 inches of penetration out of it in ballistic gelatin. My program says 17.7", but that is straightline, with expansion, and no mass lost. (This is, after all, a short program. No, it won't do your taxes for you.

)
For the .454 bullet, I would say this is one of those flatnosed deals which all the kids have to have now. I would guess that it would do around 62" in ballistic gel, maybe some better.

There are two factors which go into what I call the resisting force. The resistance on the bullet caused by the target's density and the resistance on the bullet caused by the target's strength. The resistance caused by the target's density is the one which is velocity dependent. (The square of the velocity in fact.) Calculating it looks similar to the familiar equation for drag. What I get is a value for deceleration (negative acceleration.) The other factor is a shortcut, but it also gives me another deceleration component. It involves the strength (in pounds-force per square foot) times the penetrating surface area divided by retained mass. It's also a deceleration in feet per second squared. After adding these two deceleration components together, I multiply by the retained mass and, voila, there is the resisting force.
<whew>
After I find the kinetic energy, the rest is just simple arithmetic.

quote:
someoldguy is right!

Thanks for the support, 303guy. We're undoubtedly related, as our last name is "guy."

04 March 2011, 18:24

WhatThe

quote:
Originally posted by someoldguy:
quote:
A 55 grain .22 caliber bullet at 3,600 FPS has a calculated 1,583 FPE but only 28.29 pound feet of momentum. A 360 grain .45 caliber bullet from a .454 at 1,400 FPS has a calculated energy of 1,567 FPE, yet produces 72 pounds feet of momentum. We know from expereince that the 360 grain .45 caliber bullet will out penetrate the 55 grain .22 caliber bullet at 3,600 FPS despite having less kinetic energy.

Sorry I didn't come up with a more appropriate response earlier.
OK, then...
That .22 caliber bullet I would guess is a pointy-type expanding bullet which is likely going to lose a lot of mass. I doubt seriously that you'd get a net 17 inches of penetration out of it in ballistic gelatin. My program says 17.7", but that is straightline, with expansion, and no mass lost. (This is, after all, a short program. No, it won't do your taxes for you. )
For the .454 bullet, I would say this is one of those flatnosed deals which all the kids have to have now. I would guess that it would do around 62" in ballistic gel, maybe some better.

There are two factors which go into what I call the resisting force. The resistance on the bullet caused by the target's density and the resistance on the bullet caused by the target's strength. The resistance caused by the target's density is the one which is velocity dependent. (The square of the velocity in fact.) Calculating it looks similar to the familiar equation for drag. What I get is a value for deceleration (negative acceleration.) The other factor is a shortcut, but it also gives me another deceleration component. It involves the strength (in pounds-force per square foot) times the penetrating surface area divided by retained mass. It's also a deceleration in feet per second squared. After adding these two deceleration components together, I multiply by the retained mass and, voila, there is the resisting force.
<whew>
After I find the kinetic energy, the rest is just simple arithmetic.

quote:
someoldguy is right!

Thanks for the support, 303guy. We're undoubtedly related, as our last name is "guy."

Mr. Oldguy hello and good day,

You have some interesting formulations herein and about. It's clear you are discussing terminal results but I'm a bit confused when you bring in "momentum". Momentum as we know it provides positive acceleration across any given spectrum (rate of acceleration is known) and has no negative rate and is = to C=M (a+)/E=MC Sq. However we are provided momentum values with a crony and can determine deceleration values based on those results using angle of flight/velocity/projectile weight/and of course atmospheric conditions. But now we are discussing ext. ballistics. Hence the confusion. At the terminal point or "exchange" we are only interested in E=Mc Sq.(v). to provide Ke. values. Penetration is another subject dealing with impact (target) structure, projectile design and of course using the solution of our above values.

Interesting indeed, your comments.....

04 March 2011, 19:26

jwp475

quote:
Originally posted by someoldguy:
quote:
A 55 grain .22 caliber bullet at 3,600 FPS has a calculated 1,583 FPE but only 28.29 pound feet of momentum. A 360 grain .45 caliber bullet from a .454 at 1,400 FPS has a calculated energy of 1,567 FPE, yet produces 72 pounds feet of momentum. We know from expereince that the 360 grain .45 caliber bullet will out penetrate the 55 grain .22 caliber bullet at 3,600 FPS despite having less kinetic energy.

Sorry I didn't come up with a more appropriate response earlier.
OK, then...
That .22 caliber bullet I would guess is a pointy-type expanding bullet which is likely going to lose a lot of mass. I doubt seriously that you'd get a net 17 inches of penetration out of it in ballistic gelatin. My program says 17.7", but that is straightline, with expansion, and no mass lost. (This is, after all, a short program. No, it won't do your taxes for you. )
For the .454 bullet, I would say this is one of those flatnosed deals which all the kids have to have now. I would guess that it would do around 62" in ballistic gel, maybe some better.

There are two factors which go into what I call the resisting force. The resistance on the bullet caused by the target's density and the resistance on the bullet caused by the target's strength. The resistance caused by the target's density is the one which is velocity dependent. (The square of the velocity in fact.) Calculating it looks similar to the familiar equation for drag. What I get is a value for deceleration (negative acceleration.) The other factor is a shortcut, but it also gives me another deceleration component. It involves the strength (in pounds-force per square foot) times the penetrating surface area divided by retained mass. It's also a deceleration in feet per second squared. After adding these two deceleration components together, I multiply by the retained mass and, voila, there is the resisting force.
<whew>
After I find the kinetic energy, the rest is just simple arithmetic.

quote:
someoldguy is right!

Thanks for the support, 303guy. We're undoubtedly related, as our last name is "guy."

You can make the .22 caliber bullet any shape that you want and of course make it non-expanding and it will not penetrate anywhere close to as far. I have no idea what type of program you have, but if it bases the outcome on FPE it will not be 100% accurate.

Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this:

04 March 2011, 23:39

someoldguy

All right, jwp475. Thanks.

05 March 2011, 00:23

303Guy

Have a look at this link. This is what we are talking about. Check out the pictures first (unless you are a prolific reader).

http://www.chuckhawks.com/energy_transfer.htm

quote:
Duncan MacPherson's book Bullet Penetration makes it quite clear that damage is done by stress, not energy . Stresses cause damage only if they strain body tissues above their elastic limits.

Itallics are mine.

05 March 2011, 00:56

Captain Finlander

Energy is just a number derived from velocity and bullet weight. Obviously there is a minimum energy level but it is still just a factor of the other two.

05 March 2011, 01:22

someoldguy

quote:
Duncan MacPherson's book Bullet Penetration makes it quite clear that damage is done by stress, not energy . Stresses cause damage only if they strain body tissues above their elastic limits.

What is stress? A unit of force per a unit of area. Right? Wrong?
What is energy? A unit of force times a unit of distance. Right? Wrong?
Now, am I alone in noticing similarities and possible relationships between the two?

Thanks for posting that article, 303guy. The original illustrations are at firearmstactical.com. Great site!

05 March 2011, 02:22

Hot Core

quote:
Originally posted by jwp475:
... Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this: ...

I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

05 March 2011, 02:28

rcamuglia

quote:
Originally posted by Hot Core:
quote:
Originally posted by jwp475:
... Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this: ...
I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

05 March 2011, 02:39

someoldguy

quote:
Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this:

What works 100 percent of the time?
Anybody?
Speak up, dammit! I can't hear you!
OK, never mind then.

quote:
I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.

05 March 2011, 03:01

crbutler

Come on guys...

Energy, momentum, et al are measurements.

They have meaning.

Whether or not that measurement tells you what you want to know is a different matter. You need to define what it means in the context of the question before you can say how valuable the answer is.

And no model is 100%. Again, context.

05 March 2011, 03:08

465H&H

quote:
Originally posted by someoldguy:
quote:
Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this:

What works 100 percent of the time?
Anybody?
Speak up, dammit! I can't hear you!
OK, never mind then.

quote:
I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.

It seems like it would be easier to just go shoot a few animals and measure the results than try to predict those results from some formula.

465H&H

05 March 2011, 03:18

Ghubert

Easier how?

05 March 2011, 03:18

someoldguy

quote:
Interesting indeed, your comments.....

Thanks, WhatThe.
Of course I can't guarantee 100 percent accuracy out of my results, just that they get reasonably close (for my taste anyway.) After all, this is a very simple formula I'm talking about. And terminal ballistics has a lot of exceptions to practically every rule. I think if you've studied terminal ballistics for any length of time, you know that nothing is ever 100 percent because not always do you know what a bullet will do inside a target. Stability is always the number one gremlin and that can't always be predicted.

05 March 2011, 03:22

someoldguy

quote:
It seems like it would be easier to just go shoot a few animals and measure the results than try to predict those results from some formula.

465H&H, thanks just the same, but I don't want to go out and shoot an elephant with a .223 just to see how it works.
I really don't intend to be turned into tembo toejam if I can help it.

05 March 2011, 03:59

jwp475

quote:
Originally posted by Hot Core:
quote:
Originally posted by jwp475:
... Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this: ...
I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

Explain how the science changes depending on the target
.. This I want to hear

05 March 2011, 04:03

jwp475

quote:
Originally posted by someoldguy:
quote:
Duncan MacPherson's book Bullet Penetration makes it quite clear that damage is done by stress, not energy . Stresses cause damage only if they strain body tissues above their elastic limits.

What is stress? A unit of force per a unit of area. Right? Wrong?
What is energy? A unit of force times a unit of distance. Right? Wrong?
Now, am I alone in noticing similarities and possible relationships between the two?

Thanks for posting that article, 303guy. The original illustrations are at firearmstactical.com. Great site!

The wound channel is created by the frontal area and the amount of direct tissue crushed, and the amount stretched past the elasticity. A handgun does not create anywhere near as much hydraulic pressure as a rifle does and the hydraulic pressure can disrupt and rip tissue apart in excess of the direct crush of tissue by the frontal area of the bullet.

In an INELASTIC COLLISION ENERGY IS MOSTLY TRANSFORMED INTO OTHER FORMS OF ENERGY, MOMENTUM IS ALWAYS TRANSFERED IN ALL COLLISIONS.

05 March 2011, 04:57

465H&H

quote:
Originally posted by someoldguy:
quote:
It seems like it would be easier to just go shoot a few animals and measure the results than try to predict those results from some formula.

465H&H, thanks just the same, but I don't want to go out and shoot an elephant with a .223 just to see how it works.
I really don't intend to be turned into tembo toejam if I can help it.

Yep, that would be an extremely stupid thing to do. But why not test it on a coyote?

465H&H

05 March 2011, 04:59

303Guy

quote:
MOMENTUM IS ALWAYS TRANSFERED IN ALL COLLISIONS

Been thinking about that. There have to be losses. The transfered momentum is the momentum remaining after some of it has been transformed into heat and other forms. Most of the momentum is transfered in lower velocity impacts but in higher velocity impacts much of it is transformed. (This does not contradict the quoted statement which does not say that all of the momentum is transfered). Momentum = Velocity X Mass while Energy = half the Mass X the square of the Velocity. This means that if there is a transfer of energy there will be a velocity loss which means there will be a loss of momentum. However, from this one can see that the more energy is carried by mass rather than velocity the greater will be the reserve of momentum to supply the energy as the bullet penetrates. Also, a slower bullet with equal energy will tend to fragment less so keep its energy supplying momentum more intact. Remember what I said about power in an earlier post?

All this by the way, is before bullet design and construction has entered the fray!

JBM offers a 'killing value' as the product of the Momentum and the Energy of the bullet which actually makes a lot of sense.

05 March 2011, 05:01

jwp475

"Energy" is tranformed into other forms of energy mostly "thermal" (heat), not the momentum

05 March 2011, 05:03

303Guy

By the way, this thread is thought provoking.

05 March 2011, 06:15

jwp475

quote:
Elastic and Inelastic Collisions
A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy

05 March 2011, 06:29

jwp475

http://www.physicsclassroom.co...a/momentum/cthoi.cfm

quote:
Inelastic Collision

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum change of the individual objects are equal in magnitude and opposite in direction.

Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

The animation below portrays the inelastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.

In the collision between the truck and the car, total system momentum is conserved. Before the collision, the momentum of the car is +20000 kg*m/s and the momentum of the truck is -60000 kg*m/s; the total system momentum is -40000 kg*m/s. After the collision, the momentum of the car is -10000 kg*m/s and the momentum of the truck is -30 000 kg*m/s; the total system momentum is -40000 kg*m/s. The total system momentum is conserved. The momentum change of the car (-30000 kg*m/s) is equal in magnitude and opposite in direction to the momentum change of the truck (+30000 kg*m/s) .

An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 800000 Joules (200000 J for the car plus 600000 J for the truck). After the collision, the total system kinetic energy is 200000 Joules (50000 J for the car and 150000 J for the truck). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.

05 March 2011, 07:03

Cross L

So the question seems to be:
Do I use hollow point ergs or solid ergs?

If an erg is a measure of energy whigh erg is better at becoming a non-erg?

SSR

05 March 2011, 08:33

303Guy

Thanks jwp475

It's a tricky subject. Like fr'nstance the momentum of a rifle firing a bullet is equal to the momentum of the bullet but the energy of recoil is NOT equal to the energy of the bullet.

05 March 2011, 08:33

someoldguy

quote:
If an erg is a measure of energy whigh erg is better at becoming a non-erg?

I don't know, Cross, but I prefer my ergs scrambled.

quote:
Yep, that would be an extremely stupid thing to do. But why not test it on a coyote?

Ah, I knew what you meant, 465. I was just kidding. I've been just a real comedian lately, not just here but around the house. When my family gets tired of me and my sh!t, I have to come here and spread it around.

The reason I look to formulas is because I don't always have the means to do testing of my own, either through shooting wet paper or through hunting. I have a place to shoot, but it's not in a convenient place where I can cart a load of newspapers to. So I have to settle for milk jugs or a row of empty plastic 2 liter Pepsi bottles. But I do love to shoot! I wouldn't be here if I didn't!

But I know my 8mm will take deer. It might even do larger animals with the right bullet, but I don't get the opportunity to do hunting of anything larger than whitetail.
I guess I'm just curious and want to learn. Just in case I might be able to go after bigger game one fine day!

05 March 2011, 08:37

303Guy

quote:
I guess I'm just curious and want to learn.

05 March 2011, 16:50

Wstrnhuntr

The less energy your weapon has, the more relevant energy becomes..

05 March 2011, 17:13

Hot Core

quote:
Originally posted by someoldguy:
Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.

I thought so. My hero "alf" used to bring it up in nearly every thread he posted to.

For people who have first-hand experience of seeing how both people and Game react to shots taken, there is simply ZERO comparison. The Freedom Freeloaders of the world will never understand it though because they have no comparison base and are unable/unwilling to learn from those that do have it.
-----

By the way SomeOldGuy, though I've not mentioned it previously, I really enjoy your humor.

05 March 2011, 17:20

Hot Core

Hey 303Guy, Glad to see none of the buildings fell on you.

05 March 2011, 19:02

Ackley Improved User

quote:
Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

This is the most insightfull post on this thread! Brilliant.

05 March 2011, 19:41

vapodog

quote:
Originally posted by Ackley Improved User:
quote:
Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

This is the most insightfull post on this thread! Brilliant.

AMEN to that.....

The choir has sung!

05 March 2011, 21:00

jwp475

quote:
Originally posted by Hot Core:
quote:
Originally posted by someoldguy:
Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.
I thought so. My hero "alf" used to bring it up in nearly every thread he posted to.

For people who have first-hand experience of seeing how both people and Game react to shots taken, there is simply ZERO comparison. The Freedom Freeloaders of the world will never understand it though because they have no comparison base and are unable/unwilling to learn from those that do have it.
-----

By the way SomeOldGuy, though I've not mentioned it previously, I really enjoy your humor.

No, MacPhearson work has to do with modeling "HANDGUN ROUNDS" in ballistic gelitin.

THe science does not change as reguard to media and what is treansfered, transformed, etc. MacPhearson spends an entire chapter explaining why any model based on FPE is doomed to faliure

05 March 2011, 21:01

jwp475

quote:
Originally posted by Hot Core:
quote:
Originally posted by someoldguy:
Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.
I thought so. My hero "alf" used to bring it up in nearly every thread he posted to.

For people who have first-hand experience of seeing how both people and Game react to shots taken, there is simply ZERO comparison. The Freedom Freeloaders of the world will never understand it though because they have no comparison base and are unable/unwilling to learn from those that do have it.
-----

By the way SomeOldGuy, though I've not mentioned it previously, I really enjoy your humor.

More BS from the one that constatntly posted about the calibers required for taking ELK, without ever shooting at one, much less ever taking one

05 March 2011, 21:03

jwp475

quote:
Originally posted by vapodog:
quote:
Originally posted by Ackley Improved User:
quote:
Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

This is the most insightfull post on this thread! Brilliant.
AMEN to that.....

The choir has sung!

You only take into account "energy" with no reguard to "momentum" which much more important and you also seem to ignore the type of projectile which can make or break how effective a cartirdges is or is not

05 March 2011, 21:05

jwp475

quote:
Originally posted by Ackley Improved User:
quote:
Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

This is the most insightfull post on this thread! Brilliant.

I'd call BS on this statement. In the example that of the 22 caliber 55 grain bullet at 3600 FPS it has more FPE than a 45 caliber bullet at 1400 FPS, but will be much less effective with a shoulder shot on a mature Bison yet the 45 call bullet will break the shoulder and continue through the vitails.

06 March 2011, 08:35

someoldguy

OK. This isn't the physics problem I was looking for, but it's quite similar, so I'll use it. With it I will prove that kinetic energy, resisting force, and penetration depth definitely have a relationship. (But momentum also, yes indeed.)

Here is the problem itself.
It's located on this page:
http://answers.yahoo.com/quest...0100302215610AAmDFEy

quote:
A 9 g bullet hits a tree with impact speed of 250 m/s. The bullet makes a 8 cm long hole in the wood, then stops. Find the average force resisting bullet motion through the wood.

For my purposes, I need to convert these values to English units because this is what I'm most comfortable using. I have also calculated the kinetic energy, in ft-lbs.

9 g bullet = 139 grains
250 m/s = 820 feet per second
8 cm penetration = 3.15 ins (0.2625 ft) [Must be some pretty strong wood!]
KE = 820^2 x 139 / 7000 / 32.2 / 2 = 207 ft-lbs
(What kind of bullet is this? Maybe a .38 Special handgun bullet?)

In the answer, the poster used momentum to arrive at the force, expressed in Newtons. Of course, the answer is in metric units.

quote:

impulse = change in momentum

FΔt = p1 - p2
F = (p1-p2)/Δt

p1 = mv = (.009 kg)(250 m/s) = 2.25 kg*m/s
p2 = 0

v^2 = v0^2 +2a(x-x0)
a = (v^2 - v0^2) / 2d
a = (0-(250 m/s)^2) / 2(.08 m)
a = -390625 m/s^2

v = v0 + at
t = (v-v0)/a = (0-250 m/s) / -390625 m/s^2
t = .00064 sec

F = (p1-p2)/Δt = (2.25 kg*m/s - 0) / .00064 sec
F = 3515.63 N

So they've arrived at the answer:
Force = 3515.63 Newtons

Converting to English units, this quantity is
3515.63 x 0.2248 = 790 pounds-force

But guess what? If you're using English units, there's a quick and easy shortcut to arrive at this same answer.
It's simply the kinetic energy divided by the penetration depth in feet.
Resisting force (English units) = 207 ft-lbs / 0.2625 ft = 789 pounds-force

(Likely the 1 pound-force discrepancy is due to rounding.)

So this shows pretty conclusively that, yes, you can most certainly calculate the penetration depth from kinetic energy as long as you have an idea of the resisting force.

06 March 2011, 08:57

jwp475

quote:
Originally posted by someoldguy:
OK. This isn't the physics problem I was looking for, but it's quite similar, so I'll use it. With it I will prove that kinetic energy, resisting force, and penetration depth definitely have a relationship. (But momentum also, yes indeed.)

Here is the problem itself.
It's located on this page:
http://answers.yahoo.com/quest...0100302215610AAmDFEy

quote:
A 9 g bullet hits a tree with impact speed of 250 m/s. The bullet makes a 8 cm long hole in the wood, then stops. Find the average force resisting bullet motion through the wood.

For my purposes, I need to convert these values to English units because this is what I'm most comfortable using. I have also calculated the kinetic energy, in ft-lbs.

9 g bullet = 139 grains
250 m/s = 820 feet per second
8 cm penetration = 3.15 ins (0.2625 ft) [Must be some pretty strong wood!]
KE = 820^2 x 139 / 7000 / 32.2 / 2 = 207 ft-lbs
(What kind of bullet is this? Maybe a .38 Special handgun bullet?)

In the answer, the poster used momentum to arrive at the force, expressed in Newtons. Of course, the answer is in metric units.

quote:

impulse = change in momentum

FΔt = p1 - p2
F = (p1-p2)/Δt

p1 = mv = (.009 kg)(250 m/s) = 2.25 kg*m/s
p2 = 0

v^2 = v0^2 +2a(x-x0)
a = (v^2 - v0^2) / 2d
a = (0-(250 m/s)^2) / 2(.08 m)
a = -390625 m/s^2

v = v0 + at
t = (v-v0)/a = (0-250 m/s) / -390625 m/s^2
t = .00064 sec

F = (p1-p2)/Δt = (2.25 kg*m/s - 0) / .00064 sec
F = 3515.63 N

So they've arrived at the answer:
Force = 3515.63 Newtons

Converting to English units, this quantity is
3515.63 x 0.2248 = 790 pounds-force

But guess what? If you're using English units, there's a quick and easy shortcut to arrive at this same answer.
It's simply the kinetic energy divided by the penetration depth in feet.
Resisting force (English units) = 207 ft-lbs / 0.2625 ft = 789 pounds-force

(Likely the 1 pound-force discrepancy is due to rounding.)

So this shows pretty conclusively that, yes, you can most certainly calculate the penetration depth from kinetic energy as long as you have an idea of the resisting force.

You took one example that benefited your model. The question is this accurate 100% of the time. Also how do we know that this is what we will find when we fire the bullet. Has any of this been proven? The only proven model that works 100% of the time that I am aware of is MacPhearson's

Where are your proofs? How you shot bullets into media to prove this theory of yours?

I know that for some materials, such as military armor Kinetic energy is realevent to calculating penetration, but not realy in tissue

06 March 2011, 09:55

jwp475

http://www.bordeninstitute.arm...apter4/Pages1-12.pdf

quote:

The matmaticl expression for the drag that occurs when a projectile passes through a fluid is:
where CDis the proportionality constant known as the coefficient of drag, is the density of the target me-dium,A is the projectile’s cross-sectional area, and Vits velocity.
The drag equationindicates if remain the same, changing the density of the medium from that of air (0.001 g/cm3) to that of water (1.00 will result in a thousandfold increase in drag. Similarly, increasing the cross-sectional area of the projectile (for example, as when a deforming bullet such as a dumdum hits a solid target, or when a bullet tumbles) will increase drag. Because velocity is raised to the second power in the equation, doubling the projectile’s velocity increases drag fourfold.
All other factors, such as the coefficient of drag and the shape, being the same, drag is much greater with high-than with low-velocity projectiles. For example, high-velocity rifle bullets slow more rapidly than do similarly shaped pistol bullets.
The drag equation omits a factor for the splitting of the medium, a phenomenon that depends upon the medium’s material strength. Because of this omission, the increase in drag that occurs when a projectile passes from air into human soft tissue will be much greater than a prediction based simply on the change in density (from 0.001 in air to 1.05 in muscle). This factor is unknown and is one of the areas that requires the attention of sophisticated ballistics researchers.
When drag is treated mathematically as the force of retardation, it is usually normalized for projectile mass (that is, and can be expressed:
or, by transposition: A
However, a projectile’s sectional density /A) is an important determinant of drag: the higher the sectional density, the lower the drag.