[lbmii]

Let us say we have two bullets that are made of strong non-expanding materials and both of these bullets are the same diameter and have the identical octave or shape in every way possible. Both bullets are shot into a perfectly identical and uniform media and neither bullet yaws or wobbles or expands.

IF:
Bullet #1 has twice the energy (due to either having a different mass and/or a different velocity) of bullet #2. Will bullet number 1 penetrate twice as far as bullet #2?

IF:
Bullet #1 has twice the momentum (due to either having a different mass and/or a different velocity) of bullet #2. Will bullet number 1 penetrate twice as far as bullet #2?

IF:
Bullet #1 has twice the mass but the same velocity of bullet #2. Will bullet number 1 penetrate twice as far as bullet #2?

IF:
Bullet #1 has twice the velocity but the same mass of bullet #2. Will bullet number 1 penetrate twice as far as bullet #2?

I do not know the above answers but would like to know the answers.

[EghtySx]

Well, no. Just basic physics. The farther into the mediam the projectile gets the more rapid it's deceleration. Just like when you stop your car. The closer you get to the red light the more quickly you decrease your speed (i.e. it takes you longer to get from 60 to 50 than it does from 30 to 20 and much faster than from 10 to 0). The rate of deceleration increases exponentially. This is in part your increasing effort on the brake pedal and partly due to the physics of an onbect in motion wanting to stay in motion until acted upon by an outside force. Through the first inch of the medium it was going, lets say 2000 fps, it hits the second inch going, lets say 1600 fps. Now, one inch of the identical media is going to have more of an effect on an identical projectile moving 400fps slower than another. Never mind that the slow one is starting to mushroom and tumble. =)

[RJ357]

That is probably a difficult question because there are so many factors. One in particular is that the material being impacted is going to pile up in front of the projectile and add to the mass that has to move through the medium.

Decelerating from 60 mph to 50 mph is no different than decellerating from 10 mph to zero mph. Both require the same amount of energy.
The exact same laws hold true for all objects in uniform, nonrotating motion (the principle of relativity).
An object travelling at 60 mph is perfectly justified in claiming it standing still, for example.
There could be a difference in the efficiency of braking system at different speeds, though.

[RJ357]

One thing you could do is try to find some penetration tests that were done in ballistic gelatin or water. If you have the bullet weights and velocities, you can estimate the relationship between momentum and penetration (and energy and penetration)

Something that might be relavent also, is the volume of the temporary wound cavity as a function of velocity or momentum.

[Molon Labe]

Decelerating from 60 mph to 50 mph is no different than decellerating from 10 mph to zero mph. Both require the same amount of energy.This is not true. It takes 11 times more energy to decelerate from 60 mph to 50 mph vs. decelerating from 10 mph to 0 mph.

[McCall911]

It's rather "geeky" but bullet penetration is addressed in this Italian website:

http://www.earmi.it/balistica/baltermi.htm



(If you right click, you can use the Windows translator into English, but you will get several nonsensical and comical translations probably because of the lack of technical Italian vocabulary. For example, the title is "Terminal Ballistics" but the translation gives "Ballistics finishes them!" LOL :D )
The formula presented is under the heading Penetration in the soft woven ones of the human body (which should have been "Penetration in soft tissue of the human body.")
How accurate this formula is I do not know, but I think it's interesting nonetheless.

[dfaugh]

since you've eliminated variables like bullet expansion, I would think the equations are simple. So my answers swould be:

Yes
Yes
Yes
Not so sure about this one

Now, they probably should be more like

Yes(roughly)
Yes(roughly)
Yes(roughly)
I still don't know

But, in short, with twice the energy (all else being equal) would think you'd get twice the penetration.

Just and opinion, based on my fuzzy recolection of college physics. :p

[TimRB]

Since your projectile needs to do the work of penetration, energy is required. If the force that resists penetration is constant, double the energy would give double the penetration.

E=(1/2)mv^2=f(dot)d

Tim

[Hawkmoon]

Acceleration/deceleration is an exponential function. Eliminating as many external variables as possible, it requires four times as far to decelerate to zero when the speed is doubled. That means double velocity equals four times as much energey.

In this case, the hypothesis is that the ENERGY is double, not the velocity. I believe the amount of penetration would be increased by a factor of the square root of 2.

[RyanM]

http://home.snafu.de/l.moeller/Penetration_Calculator_2.html

Here. You can try the experiments yourself.

300 mps, 10 g, 9mm, 3 kg-sec, 450 J = 73.22 cm penetration

600 mps, 10 g, 9mm, 6 kg-sec, 1800 J = 110.66 cm
300 mps, 20 g, 9mm, 6 kg-sec, 900 J = 146.43 cm
424 mps, 10 g, 9mm, 4.24 kg-sec, 899 J = 91.78 cm
300 mps, 10 g, 4.5mm, 3 kg-sec, 450 J = 292.84 cm
300 mps, 10 g, 18mm, 3 kg-sec, 450 J = 18.32 cm

Doubling mass doubled penetration depth, while doubling velocity increased it about 1.511x. Increasing velocity by 1.41x increased penetration by 1.25x.

Looks like the conclusion is what I've been saying all along, on other threads. You need to consider the factors of bullet weight, velocity, expanded diameter, expanded shape, and retained velocity independently.

[lbmii]

Interesting calculator, I have come upon it before but never sat down and figured it out.

It looks like all other things being equal, mass has a greater influence on penetration than velocity.

Now I need to get some metric conversions and have at it!

[blackrazor]

That calculator program is awesome. The question is very difficult, I wonder where this program came from? I asked several physics professors about the relationship between penetration, momentum, and energy, and noone could give me a straight answer...

[RJ357]

"This is not true. It takes 11 times more energy to decelerate from 60 mph to 50 mph vs. decelerating from 10 mph to 0 mph."

Acceleration and deceleration are equivalent.

So:

A gun with a heavy bullet and weak charge fires the bullet. Initially at rest, 0 fps, the bullet accelerates to 10 fps.
Now we put the gun on a railway car travelling at 50 fps. We fire the gun and the bullet, initially at 50 fps accelerates to 60 fps.
Same energy, same acceleration.

[rlq9thrk]

The expression for kinetic energy was given earlier. The actual experimental data beats all mathematical models.

A complete model would be fairly complex: some of the bullet's energy goes into deforming the target material, some is converted into heat, some into deforming the bullet. Higher velocities do more damage to the bullet. You have probably seen those pictures of an armor-piercing round going through a chunk of steel and only a super-heated gas jetting out the other side.

[Yooper]

More specifically, the Poncelet equation supports sectional density as a parameter for penetration. It should be independent of caliber.

[RJ357]

So does this all show that it really is about energy?

Penetration is proportional to mass.
Energy is proportional to mass.
Penetration is proportional to energy.

Penetration is proportional to velocity squared.
Energy is proportional to velocity squared.
Penetration is proportional to energy.


Penetration is proportional only to the mass component of momentum.
Penetration is not proportional to the velocity component.
Penetration is not necessarily proportional to momentum.

[Yooper]

RJ357

I wrote this in response to another thread:

I believe that penetration is the only practical value to be sought, after all, what do we want the bullet to do when it hits the target other than penetrate? I think it may boil down to sectional density and velocity as determinants in projectile potential, regardless of caliber, and penetration is the net result. For a (theoretically) non-expanding bullet there will be an optimum velocity beyond which velocity increases show a diminishing return, but will always increase (though slightly) as velocities increase. Greater sectional densities will yield higher optimum velocities. For expanding bullets, there also will exist an optimum velocity for proper expansion and when velocities increase beyond this point, penetration will actually decrease due to the accelerated expansion. An expanding bullet is effectively changing its sectional density as it expands. All of this is relative to the density and makeup of the medium the bullet is penetrating. We already have the figures for one medium, air, through ballistic tables. As the density increases, the penetration decreases dramatically. Wet newspapers, ballistic gelatin, etc. all have their own characteristics, but none are as diverse as, say, a game animal. An animal has a very elastic skin, muscle tissue, bone, cartilage, internal organs, various fluids. etc. As far as "knock-down" potential or "killing power" are concerned, I don't think they exist. The mass of a bullet is so minute compared with most living targets that it can't "move" the larger mass effectively.

I hope this helps!

[RJ357]

lbmii seemed to have discovered a correlation between stopping power and momentum, though, not between stopping power and penetration, which would be more in line with energy.

Knock down power is funny; anything that could do that would also knock the shooter on his butt.

Sectional density and velocity would be important to penetration, no doubt.

What about central nervous system shock? What might correlate better to that?

I guess it also depends on whether you are hunting or defending.

[RyanM]

Penetration is proportional to velocity squared.
Energy is proportional to velocity squared.
Penetration is proportional to energy.

Incorrect. If you were paying attention, you'd have noticed a 2x increase in velocity (4x increase in energy) results in a 1.5x increase in penetration. I did some more numbers, with interesting results. There are completely linear correlations between penetration and frontal area, and between penetration and mass. Increased velocity, however, gives diminishing returns. Looks like reducing drag increases the efficiency in using extra velocity.

9mm, 10 g, .57 drag
vel pen. penetration compared to 100 m/s
50, 8.56 0.38
100, 22.48 1.00
200, 52.13 2.32
300, 73.22 3.26
400, 88.63 3.94
500, 100.72 4.48
600, 110.66 4.92
700, 119.09 5.30
800, 126.41 5.62
900, 132.87 5.91
1000,138.66 6.17

[blackrazor]

this means that for M193 vs M855, since M855 has greater energy, greater mass, greater momentum, but only slighly lower velocity, that M855 should get better penetration under all circumstances. Especially when you consider that the M855 is steel core and M193 is lead core (softer). If you get diminishing returns on velocity, then the argument that M193 penetrates more simply because it has a slight velocity edge is completely false... but some people claim M193 penetrates better than M855, especially in real world tests... comments?

[cracked butt]

An easy everyday nonmathimatical world way of looking at it is if you took a toy balloon and threw it across your living room with a moderate throw, then picked it up again and threw it as hard as you could. Your second throw won't go much farther than the first.

[RyanM]

On M16 ammo, M855 is longer and more base-heavy than M193, so it "tumbles" faster on impact with flesh, a wall, etc., at least in theory. In practice, both rounds are all but random in how soon they tumble.

Still trying to generate a reasonably accurate algorithm for penetration. Best I've managed is:

0.00000000093803815542258 X^3 - 0.0000031923097447629 X^2 + .00420116344237 X - .03722554156363

Put X in meters per second in there, and it gives you a penetration factor relative to 300 m/s (if you get 2, then penetration is double the penetration at 300 m/s). That's for bullets with a coefficient of drag of .5, which I think is the equivalent of an FMJ pistol bullet. I think. It's only accurate from 0 to 1250 m/s, though; beyond 1250, the penetration goes up way too fast.

[Molon Labe]

RJ357 said:
A gun with a heavy bullet and weak charge fires the bullet. Initially at rest, 0 fps, the bullet accelerates to 10 fps.
Now we put the gun on a railway car travelling at 50 fps. We fire the gun and the bullet, initially at 50 fps accelerates to 60 fps.
Same energy, same acceleration.No. You are wrong. It takes 11 times more energy to accelerate the bullet from 50 to 60 fps vs. 0 to 10 fps. Even if you fire the bullet from a gun on a train (as long as all energy calculations are refrenced to the earth).

[RJ357]

(as long as all energy calculations are refrenced to the earth).

That's what I was referring to. It depends on your frame of reference. Unfortunately I neglected to specify what reference frames I was using.

Now what does a passenger in a decelerating car observe, about the energies involved, since he is not moving relative to the car?

[taliv]

this means that for M193 vs M855, since M855 has greater energy, greater mass, greater momentum, but only slighly lower velocity, that M855 should get better penetration under all circumstances. Especially when you consider that the M855 is steel core and M193 is lead core (softer). If you get diminishing returns on velocity, then the argument that M193 penetrates more simply because it has a slight velocity edge is completely false... but some people claim M193 penetrates better than M855, especially in real world tests... comments?

blackrazor, IANAPhysicist, but the higher velocity (= higher rotational rate) and different construction of M193 make it prone to fragmentation quickly (relatively) where M855 stays together longer, and thus, penetrates deeper

just a guess

[taliv]

it certainly takes a heckuva lot more energy to go from 299,792,448 m/s to 299,792,458 m/s than it does to go from 50 to 60 mph

:neener:

[RJ357]

"it certainly takes a heckuva lot more energy to go from 299,792,448 m/s to 299,792,458 m/s than it does to go from 50 to 60 mph"

Yes if you're watching it. Not if you're the one who's doing it.

That's what the whole thing was about. Someone braking a car (and obviously moving along with it).

[Mal H]

No. You are wrong. It takes 11 times more energy to accelerate the bullet from 50 to 60 fps vs. 0 to 10 fps. Even if you fire the bullet from a gun on a train (as long as all energy calculations are refrenced to the earth). I'm not sure about that, Molon Labe. When referenced to the gun, which is most likely what RJ357 is intending, the bullet will accelerate from 0 to 10 FPS and will take the same amount of energy to do so whether stationary on the earth or riding along on a train at 50 FPS. However, when referencing to the earth, it does indeed take additional energy to accelerate the bullet, case and gun to the initial 50 FPS. (Actually, that would be a very small amount of energy in comparison to accelerating the train to 50 FPS. But tht's neither here nor there.) Is that the extra energy you are talking about? Let's ignore air resistance for the time being.

Now, if you say the energy will be considerably more to accelerate the bullet on the train (train velocity + 10) than when stationary on the earth no matter what, then why stop at the earth. Let's reference it to the Sun. Now we're going 18.6 miles per second. How much energy now? Or let's take it to the extreme and reference to the Milky Way as it speeds along in the local group. Now we're going roughly 600 km/sec. How much energy now? (I'm going on memory here, the velocities may not be accurate and are for illustration only.)

Yes, it's convenient to reference only to the earth, but I hope you can see what RJ357 is saying. The bullet is leaving the muzzle at 10 FPS relative to the muzzle in both cases and the energy required is the same in both cases.

[Double Maduro]

"Knock down power is funny; anything that could do that would also knock the shooter on his butt."

I have heard this over and over and it just isn't true.

Will a 12 gauge slug knock a person down if it hits him in the chest?

What about a .454?

How about a .5o BMG from a sniper rifle? Or a 2 bore or 4 bore?

All of these rounds have been fired from the off hand position and though they may knock down the target they seldom knock down the shooter.

I believe the difference is in the surface area that the force is applied to. The .50 contacts the target in an area of less than half an inch while the equal and opposite force is applied to the shooter over many square inches of butt plate.

The same holds true for handguns. The difference is that due to the angle of the grip, a lot of energy is directed up in recoil.

Just some ravings.

DM

[blackrazor]

Here's the deal folks. Let's say you have a small charge of powder, that can only accellerate a bullet to a muzzle velocity of 10 m/s. If you fire it on the ground, you get a bullet moving at 10 m/s, if you fire it from the front of a moving train already moving 50 m/s, somehow the same powder charge has accellerated the bullet from 50 m/s to 60 m/s, relative to the ground. This appears at first not to make sense, since the same charge of powder has given the bullet 11x as much additional energy in the moving train case. However, you must remember, that the train in going to slow down when the bullet is fired. As a matter of fact, the train will slow down by whatever velocity is necessary to ensure the the total energy of the bullet+train is the same, whether or not the bullet is fired from a staitionary guy or from a guy on the front of the train. Trust me, it must work out: conservation of energy is a beautiful thing.

[blackrazor]

Yes, it is true, assuming the target is about the same size and build as the shooter, there is no way the target can be hit with more momentum (knockdown potential, so to speak) than hits the shooter. None of those rounds will knock down the target, even the 50 BMG. Sure... he'll fall down after being shot, but that's because he's dead (from the energy, not the momentum).

The difference is not in the grip, or whatever, the difference is that you are getting momentum and energy mixed up. Energy kills, momentum "knocks down". The amount of momentum transfered to the shooter is always higher than the amount of momentum transferred to the target, if anyone is going to be knocked down, it's the shooter. However, the amount of energy delivered to the target is WAY higher than the amount of energy delivered to the shooter, this is because the bullet is heavier than the gun. This is also why it hurts to shoot most "ultralight" guns (more recoil ENERGY).

[Double Maduro]

Blackrazor,

"this is because the bullet is heavier than the gun".

Not many shoulder fired or hand held weapons where the bullet is heavier than the gun.

DM

[blackrazor]

Sorry, meant to say gun is heavier than the bullet. The point is, if the gun and bullet were the same weight, the gun would hit your hand with the same energy as a bullet, effectively destroying your hand. However, even in this extreme case, the momentum transfered to your hand still hasn't changed.

[Molon Labe]

For the third time, it takes 11 times more energy to accelerate the bullet from 50 to 60 fps vs. 0 to 10 fps. It doesn't matter if you're on a train, on a rocket, or on a slow boat to Hong Kong.

Only one person so far has given the right answer: blackrazor.

Let's say I can throw a baseball (0.142 kg) to a maximum speed of 10 mph (4.47 m/s). The kinetic energy of the baseball is thus 1.42 Joules. This also means my arm is only capable of giving the baseball 1.42 J of kinetic energy.

Now let’s say I’m riding a train that’s moving at a constant speed of 50 mph (22.35 m/s). Relative to the earth, the baseball has 35.47 Joules of kinetic energy. I throw the baseball to a speed of 10 mph (relative to the train). Relative to the earth, the speed of the baseball is now 60 mph (26.82 m/s), and the kinetic energy of the baseball is now 51.07 Joules.

Now here's what's puzzling: when I was on the ground, I was only able to give the baseball 1.42 Joules of energy. Likewise, it is assumed I can only give the baseball an additional 1.42 Joules of energy while on the train. Yet on the train, it would seem I was able to give the baseball an additional 15.61 Joules of energy.

So did I give the baseball an additional 15.61 Joules of energy when I threw it on the train?? No. As stated above, my arm is only capable of giving the baseball 1.42 J of kinetic energy. To accelerate the baseball from 50 mph to 60 mph, I contributed 1.42 J and the train contributed 14.18 J.

Now you’re probably asking, “What do you mean the train contributed 14.18 J when you threw the ball? The train is simply moving at 50 mph. I don't see where the train is adding 14.18 J when you throw the ball.”

Ah, but it is. When I threw the ball, the train's speed sensor (whatever that may be) noticed a slight drop in speed. This is because the action of me throwing the ball introduced a small, momentary drag force in the opposite direction the train was traveling. To keep the train at 50 mph, more diesel fuel had to be burned. In other words, when I threw the ball, the train had to burn an additional 14.18 J of fuel in order to maintain a speed of 50 mph.

So there's your physics lesson for the day. ;)

[Molon Labe]

Mal H said,

Yes, it's convenient to reference only to the earth, but I hope you can see what RJ357 is saying. The bullet is leaving the muzzle at 10 FPS relative to the muzzle in both cases and the energy required is the same in both cases.This is true, relative to the train. And if we’re only talking about energy relative to the train, it doesn’t matter what speed the train is traveling. But I don’t think that’s what RJ357 was saying. In his original post, he said, “We fire the gun and the bullet, initially at 50 fps accelerates to 60 fps.” Those numbers are relative to the earth, not the train.

[only1asterisk]

Teach on Dr. Labe!

[blackrazor]

You can also think of it this way, (Kinetic) Energy = Force*Distance. The force your hand exerts on a thrown baseball is the same whether you're on the ground or on the train. However, on the ground, the distance over which you exert force on the ball is much smaller than the distance over which you are exerting the same force on the moving train. The faster the train is going, the greater the distance the force of your arm acts on the ball, and therefore the greater the energy added to the ball. OK, bed time.

[Mal H]

Molon Labe said: ... when I threw the ball, the train had to burn an additional 14.18 J of fuel in order to maintain a speed of 50 mph.True. However, who said the train's velocity had to remain constant? Even if it did, the energy required to maintain that constant velocity is imparted on the train, not on the bullet which left the train/gun system long before the extra fuel started burning. It's energy imparted to the bullet that is the crux of the question. I maintained that no additional energy is required to accelerate the bullet to 60 FPS on a train as opposed to accelerating the bullet to 10 FPS while stationary on the earth. And yes those velocities are relative to the earth.

I will give you the fact that it will take a small, an infinitesimally small, amount of additional energy to accelerate the bullet on the train, but it certainly won't be 11 times. The mass of the train is considerably less than that of the earth (not that that matters), so an instantaneous decrease in the train's velocity (an infinitesimal decrease) will occur over the time the bullet is travelling in the barrel so some additional energy will be required while the bullet is still in the barrel. The decrease in bullet velocity wouldn't be easily measurable if that additional energy is not supplied.

All of the above assumes a rigidly attached gun to a massive freight train on which the laws of physics have not been relaxed, i.e., momentum and inertia do count.

[walking arsenal]

This thread makes my head hurt. :(

[TimRB]

"This thread makes my head hurt."

To each his own. These physics threads usually make me laugh my head off.

Tim

[Molon Labe]

Mal H said:
who said the train's velocity had to remain constant? Even if it did, the energy required to maintain that constant velocity is imparted on the train, not on the bullet which left the train/gun system long before the extra fuel started burning.No. This is wrong. As pointed out by blackrazor, the additional energy imparted to the bullet while it is traveling down the barrel (or the baseball while my arm is swinging forward) is not simply the energy supplied by the gun (or my arm). The train is also giving the projectile additional energy during this time.

Let me go back to the baseball example (since I have all the numbers in a spreadsheet).

Let’s say I am capable of throwing a baseball to a speed of 10 MPH. And when I do it, my hand travels 2 ft. Here is the pertinent data in SI units to three significant digits:

Mass of baseball: 0.142 kg
Distance my hand traveled while it is pushing baseball: 0.610 m
Speed of baseball after it leaves my hand: 4.47 m/s
Kinetic energy of baseball after it leaves my hand: 1.42 J
Constant force applied by hand while arm is pushing baseball: 2.33 N
Constant acceleration of ball while arm is pushing baseball: 16.4 m/s/s
Amount of time arm is pushing baseball: 0.273 s

Now lets say I’m standing on a train that is traveling at a constant speed of 50 mph. The baseball is in my hand. I have not yet thrown the baseball. Here is the pertinent data:

Mass of baseball: 0.142 kg
Speed of train: 22.4 m/s
Speed of baseball before throwing it: 22.4 m/s
Kinetic energy of baseball before throwing it: 35.5 J
Force on baseball: 0 N
Acceleration of baseball: 0 m/s/s

I swing my arm back, and then throw the baseball! The baseball has just left my hand. Before, during, and after throwing the ball, the train continued to chug along at 50 mph. Here is the pertinent data:

Mass of baseball: 0.142 kg
Speed of train: 22.4 m/s
Constant force applied by hand while arm was pushing baseball: 2.33 N
Distance train moved while I was throwing the baseball: 6.10 m
Overall distance the ball moved while I was throwing the baseball: 0.610 + 6.10 = 6.71 m

O.K., now notice that the kinetic energy of the baseball before I threw it was 35.5 J. This was calculated using the familiar 0.5*mass*velocity^2 formula. But to calculate the additional kinetic energy imparted to the baseball after I threw it, I will use a different formula: energy = force * distance. Or more precisely,

Total Energy of Baseball after throwing it =
Energy of Baseball before throwing it + (∆ force * ∆ distance)

Total Energy of Baseball before throwing it = 35.5 J
∆ force = 2.33 N
∆ distance = 6.71 m

Total Energy of Baseball after throwing it = 35.5 + (2.33*6.71) = 51.1 J

Now let’s check this answer by calculating the total energy of baseball after throwing it by using the 0.5*mass*velocity^2 kinetic energy formula. The mass of the baseball is 0.142 kg. The velocity of the baseball after throwing it is 50 + 10 = 60 mph (26.8 m/s). Therefore

Total Energy of Baseball after throwing it = 0.5*0.142*(26.8)^2 = 51.1 J

For the fourth time, it takes 11 times more energy to accelerate a bullet from 50 to 60 fps vs. 0 to 10 fps. It doesn't matter if you're on a plane, on a sled, or jumping up and down on a pogo stick.

[Mal H]

Just to be sure I've got this straight. Your magical train is going to impart energy to the bullet while it is in the barrel equivalent to 11 times the energy required when the gun is stationary on the earth. Yet the bullet's velocity will only increase by 10 FPS over it's resting velocity in either system. The bullet already has an additional amount of kinetic energy when it is "resting" in the chamber at 50 FPS relative to the earth.That energy cannot be taken into account when the cartridge is fired and added to the energy required to accelerate the bullet further.

Anyway, got it.


Oh, BTW, your train isn't named "Blain" by any chance? ;)

[Molon Labe]

Your magical train is going to impart energy to the bullet while it is in the barrel equivalent to 11 times the energy required when the gun is stationary on the earth.Yes.

Yet the bullet's velocity will only increase by 10 FPS over it's resting velocity in either system.This is only true when you fire the gun while standing on the earth, as the bullet starts with v = 0. When you fire the gun on the train, the bullet is already traveling at 50 FPS, and then the velocity changes from 50 FPS to 60 FPS.

Maybe this will help. If you fire a gun on a train, the final kinetic energy of the bullet has three components, and is calculated as follows:

KE of bullet after firing gun =
KE of bullet before firing gun
+ extra KE given to bullet by gun while the bullet is traveling down the barrel
+ extra KE given to bullet by train while the bullet is traveling down the barrel

Let’s look at each coomponent one at a time.

1. The KE of bullet before firing gun is easy to calculate. It’s 0.5*m*v^2, where v = speed of the train.

2. The extra KE given to bullet by the gun while the bullet is traveling down the barrel is just the muzzle energy of the bullet. To calculate this energy, shoot the bullet through a chronograph and use the 0.5*m*v^2 formula, where v = muzzle velocity.

3. The extra KE given to bullet by the train while the bullet is traveling down the barrel is a little more difficult to calculate. To do this, use the formula “energy = force * distance”. The force in this equation is the force on the bullet exerted by the gun as the bullet travels down the barrel. The distance is the total distance traveled by the bullet as the bullet travels down the barrel. This distance is equal to the barrel length + the distance the train travels during the very brief time the bullet is traveling down the barrel.

In the case of a gun, I am guessing #2 is probably much greater than #3, and thus #3 could probably be ignored. But it is there. For my baseball example, all three components (#1, #2, and #3) were significant.

[Mal H]

Yet the bullet's velocity will only increase by 10 FPS over it's resting velocity in either system.
This is only true when you fire the gun while standing on the earth, as the bullet starts with v = 0. When you fire the gun on the train, the bullet is already traveling at 50 FPS, and then the velocity changes from 50 FPS to 60 FPS. No, not true only when you fire the gun while standing on the earth.

As I said the vel will increase by 10 FPS over its resting vel in either system. On the earth, the resting vel is indeed 0 FPS. However, on the train, the resting vel is 50 FPS in the direction of the train's travel. In both instances the velocity increases by 10.

I have a sneaking suspicion that we are both right, but neither of us have stated our propositions in such a manner as to illustrate why we are both right. We are arguing from different viewpoints about the same immutable principles. (IOW, we're both being stubborn. :) )

[entropy]

and have the identical octave If it's Bь, the round will shoot straighter, and if it's G#, don't touch the tip, it could cut you!http://img.photobucket.com/albums/v196/mosinfan/giggle.gif

I can't believe all you physicists missed that; http://img.photobucket.com/albums/v196/mosinfan/Wipe-Glasses.gif the correct term is ogive, the radial curve of the nose of the bullet, which affects sectional density. ;)

Re the M193/M855 debate; having fired more than a few of both, I find the M855 is more accurate, having been designed for stability, whereas the the M193, particularly when fired through higher twist rate barrels, is much less stable in flight. The terminal ballistics in a soft medium are affected accordingly. ;)

[blackrazor]

The M193, when fired through a 1/7 or 1/9 twist barrel, is OVERstabilized in flight. However, both M855 and M193 are so unstable when traveling through a medium like flesh (as opposed to air), it really doesn't matter: they both become unstable.

[Molon Labe]

As I said the vel will increase by 10 FPS over its resting vel in either system. On the earth, the resting vel is indeed 0 FPS. However, on the train, the resting vel is 50 FPS in the direction of the train's travel. In both instances the velocity increases by 10.True. But here’s what started this whole mess: when using the earth as a reference, it requires 11 times more energy to accelerate an object from 50 FPS to 60 FPS than to accelerate from 0 FPS to 10 FPS. This is a simple and undeniable fact. It’s also an undeniable fact that, if you are capable of throwing a ball at 10 MPH, and you do this on a train that is traveling 50 MPH, the ball will leave your hand at 60 MPH (relative to the earth) and that most of energy required to accelerate the ball from 50 MPH to 60 MPH did not come from you; it came from the train.

I have a sneaking suspicion that we are both right, but neither of us have stated our propositions in such a manner as to illustrate why we are both right. We are arguing from different viewpoints about the same immutable principles. (IOW, we're both being stubborn.Perhaps. And yep, I’m a stubborn SOB. ;)

[ExMachina]

To clear up one point :)

Intuitively, it *is* reasonable to think that one would have to apply the same “something” to increase the baseball’s velocity by 10m/s, irrespective of whether the initial velocity was 0 m/s or 100 m/s.

That “something", it turns out, is called the impulse which the ball receives, and is simply a force applied over time. In the baseball example, it would be correct to speak of the thrower’s arm as imparting a constant impulse (ie, change in momentum) to the ball in either frame of reference. Similarly, model rocket engines are rated in terms of their impluse.

FWIW, I believe the most common mistake people make in thinking about classical mechanics is in confusing impulse/momentum with work/energy...

[Deavis]

Labe,

I think you are confusing them because you are being too loose with your calculations in regard to the frame of reference you choose. :)

[RJ357]

My gun example is inaccurate with the reference frames that were implied by it. My point there was that the powder charge was the same.
In other words, the gun did not need to be loaded differently to achieve the 60 fps.

There is no prefered reference frame in physics. All reference frames are equally valid and all phenomena (like observed energy) are real for that observer. They are also arbitrary; you can pick any one you like.

There is no absolute motion. It is meaningless for a driver in a car in uniform motion to say he is traveling at 60mph. He must say 60mph relative what. Of course we don't say that because it is assumed that it is referenced to the earth's surface.

The use of those stated speeds (10 mph, 50 mph etc) were intended to describe the situation. Unfortunately, they also implied the road as the reference frame, whereas I was trying to describe the driver's point of view. Of course, I should have mentioned that.

I had chosen the driver of the braking car as a refernce frame since it seemed a logical choice.

[Mal H]

RJ357 - "In other words, the gun did not need to be loaded differently to achieve the 60 fps."

Exactly my point and the basis of my arguments.

"There is no absolute motion."

Which is why I brought up galactic motion. That's as good a reference as any.


We're done with this I hope.

Now, next discussion - Grand Unification Theory and the inflationary universe starting at 10 ^ -34 seconds forward.

:D