good visual representation of power factor.
in this video,
http://www.youtube.com/watch?v=4MPSDjJQIv4
hickok45 shoots a swinging steel plate with different calibers to see what happens.
doesn't draw conclusions, doesn't offer to explain. just see what happens.
the target swings more in this order:
22 LR
.38 SPL
9mm
.357 MAG
.40 S&W
.45 ACP
10mm
.44 MAG
well here's what is happening and what we can make of it.
notice the .357MAG in the middle of the list. it has more kinetic energy than .40SW and way more than .45ACP yet it doesn't swing the target as much as those.
similarly .40SW and 9mm have more KE than .45ACP, so why don't they swing the target more?
because the list is in order of power factor. that is how much momentum is imparted to the steel plate. exactly like one billiard ball hitting another.
shooting these steel plates is a good visual representation of power factor.
http://www.youtube.com/watch?v=4MPSDjJQIv4hickok45 shoots a swinging steel plate with different calibers to see what happens.
doesn't draw conclusions, doesn't offer to explain. just see what happens.
the target swings more in this order:
22 LR
.38 SPL
9mm
.357 MAG
.40 S&W
.45 ACP
10mm
.44 MAG
well here's what is happening and what we can make of it.
notice the .357MAG in the middle of the list. it has more kinetic energy than .40SW and way more than .45ACP yet it doesn't swing the target as much as those.
similarly .40SW and 9mm have more KE than .45ACP, so why don't they swing the target more?
because the list is in order of power factor. that is how much momentum is imparted to the steel plate. exactly like one billiard ball hitting another.
shooting these steel plates is a good visual representation of power factor.
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"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'. 0 -
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.0 -
Ifn ya hits the cue ball hard enough against a steel plate it would splatter too
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.0 -
Being attacked by steel plates is a scenario I do not anticipate at Wal-Mart. [}:)] 0 -
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.0 -
It would appear that a lot of folks just discovered momentum. 0 -
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
Although the correct term is "inelastic" you are correct.0 -
quote:Originally posted by Mr. Perfect
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
Although the correct term is "inelastic" you are correct.
That's it, couldn't think of the correct term. Always sucked at names and nomenclature.0 -
bigger bullet takes more energy to crush. results are still proportional. 0 -
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
I thought the subject was the force applied to the plate, not a discussion of what happens to the bullet or cue ball.
The transfer of momentum from the projectile to the plate is affected very little by what happens to the cue ball or projectile at and after impact. The fact that the pieces of the projectile retain energy in a different manner than does the cue ball is not significant.
We could get into a discussion about the rotational momentum of the projectile as compared to backs or side spin applied to the cue ball, but this too would be outside the intent of the video and the conversation.0 -
quote:Originally posted by Mr. Perfect
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
Although the correct term is "inelastic" you are correct.
Elastic is the correct term. A pure elastic collision results in no energy being absorbed by the deformation of either body. A billiard ball collision is very close. When energy is lost due to deformation, it becomes an inelastic collision were inelastic (or plastic) deformation occurs.
The OP was discussing momentum, however, not kinetic energy.0 -
quote:Originally posted by Don McManus
quote:Originally posted by Mr. Perfect
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
Although the correct term is "inelastic" you are correct.
Elastic is the correct term. A pure elastic collision results in no energy being absorbed by the deformation of either body. A billiard ball collision is very close. When energy is lost due to deformation, it becomes an inelastic collision were inelastic (or plastic) deformation occurs.
The OP was discussing momentum, however, not kinetic energy.
Don, the correct term for "non-elastic" is "inelastic". I see little resemblance between a bullet that gets permanently deformed and breaks apart to that of a billiard collision. I suppose it's just degree where we disagree, however.0 -
quote:Originally posted by Don McManus
quote:Originally posted by Mr. Perfect
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
Although the correct term is "inelastic" you are correct.
Elastic is the correct term. A pure elastic collision results in no energy being absorbed by the deformation of either body. A billiard ball collision is very close. When energy is lost due to deformation, it becomes an inelastic collision were inelastic (or plastic) deformation occurs.
The OP was discussing momentum, however, not kinetic energy.
Momentum and kinetic energy are related. Given that some of the kinetic energy is converted into the deformation of the bullet, the momentum transfer gets reduced also. The height that the plate rises is directly proportional to the kinetic energy imparted to the plate at impact, which is equal to the kinetic energy of the bullet minus the amount of energy it took to fragment or deform the bullet.0 -
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by Mr. Perfect
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
Although the correct term is "inelastic" you are correct.
Elastic is the correct term. A pure elastic collision results in no energy being absorbed by the deformation of either body. A billiard ball collision is very close. When energy is lost due to deformation, it becomes an inelastic collision were inelastic (or plastic) deformation occurs.
The OP was discussing momentum, however, not kinetic energy.
Momentum and kinetic energy are related. Given that some of the kinetic energy is converted into the deformation of the bullet, the momentum transfer gets reduced also. The height that the plate rises is directly proportional to the kinetic energy imparted to the plate at impact, which is equal to the kinetic energy of the bullet minus the amount of energy it took to fragment or deform the bullet.
You are correct, but you seem to be ignoring the kinetic energy retained by the cue ball as it either rebounds or caroms off to the side. The conservation of momentum deals with the transfer of energy from one body to the other based upon the total mass and velocity of the two bodies. I don't see how you can state that the kinetic energy that is retained by the bullet through deformation and fragmentation differs significantly from the kinetic energy retained by the cue ball.
You may be right at the margins, but given that the total momentum of the system is the same prior to and after impact, it would be my assumption that motion of the plate is going to be the same whether the impacting body retains its shape or is fragmented.0 -
For all you super-geniuses discussing this, what effect does the material used in the bullet have on displacement of the plate? I would theorize that a softer bullet would not displace as much as a more solid bullet. I.e., while different results would be anticipated by larger calibers, does 38 cal soft-lead ball displace as much as 38 cal FMJ? 0 -
quote:Originally posted by Rocky4winds
For all you super-geniuses discussing this, what effect does the material used in the bullet have on displacement of the plate? I would theorize that a softer bullet would not displace as much as a more solid bullet. I.e., while different results would be anticipated by larger calibers, does 38 cal soft-lead ball displace as much as 38 cal FMJ?
Softer materials make the scenario even more inelastic.0 -
Spoiler alert... 44 mag made plate swing the most! 0 -
And this is definitive proof that the .22 caliber is worl class! 0 -
quote:Originally posted by Mr. Perfect
quote:Originally posted by Don McManus
quote:Originally posted by Mr. Perfect
quote:Originally posted by AzAfshin
quote:Originally posted by Don McManus
quote:Originally posted by MIKE WISKEY
"exactly like one billiard ball hitting another.".........no, not even close. when a bullet hits a steel plate the bullet shatters with lots of energy 'lost'(sideways) due to bullet 'splatter'.
I would suggest that it is, while not 'exactly like', it is very close.
The projectile is moving in one direction, and the energy is directed in that one direction. After hitting the steel plate, all of that energy is re-directed. Just as a cue ball changes direction at impact, so does the projectile. The fact that the cue ball retains its shape and is re-directed as a unit rather than fragmenting does not change the resultant transfer of energy significantly.
Mike is absolutely correct. The two extremes of collisions are elastic and non-elastic collisions. Billiard balls are the closest example of elastic collisions where almost none of the kinetic energy is lost into the colliding objects, whereas a bullet hitting a plate is a great example of non-elastic collision where a large amount of kinetic energy is lost due to the destruction of the bullet.
Although the correct term is "inelastic" you are correct.
Elastic is the correct term. A pure elastic collision results in no energy being absorbed by the deformation of either body. A billiard ball collision is very close. When energy is lost due to deformation, it becomes an inelastic collision were inelastic (or plastic) deformation occurs.
The OP was discussing momentum, however, not kinetic energy.
Don, the correct term for "non-elastic" is "inelastic". I see little resemblance between a bullet that gets permanently deformed and breaks apart to that of a billiard collision. I suppose it's just degree where we disagree, however.
A pure inelastic collision is one where the the bodies completely fuse during the collision. I used 'non elastic collision' because a bullet hitting a steel plate with bits and pieces spattering about is obviously not, nor does it approach, an inelastic collision.
I would suggest that a bullet hitting a steel plate; fragmenting and dispersing is much closer to an elastic collision (conservation of all kinetic energy) that an inelastic collision (maximum loss of kinetic energy). I would suggest that two billiard balls falls into the same category.0 -
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Don, I'm making certain assumptions that allow us to simplify the problem. If you'd like to include all the splatter effects then by all means, I'll go that route. After all, I used to write Monte Carlo simulations of nuclear fragmentation reactions (one of less than 100 nuclear physicists in the world qualified to do so at that time).
The collision of a bullet and a plate is far more inelastic than elastic. The simplest picture is that of the bullet being pancaked and then falling to the ground (at bullet velocities we don't get electromagnetic Coulomb barrier penetration leading to adhesion of bullet fragments to the target plate). But this scenario doesn't lead to any bounceback, hence:
KE (Plate) = KE (Bullet) - E (bullet deformation)
Now let's throw some splatter in there. The most reasonable assumption of splatter is a symmetric pattern around the point of impact. As such, conservation of momentum says that there is zero net momentum to the sides, hence 100% of the KE due to motion parallel to the plate is wasted and not imparted to the plate. Therefore:
KE (Plate) = KE (Bullet) - E (bullet deformation) - E (fragmentation leading to parallel motion) + KE (fragmentation perpendicular to plate)
Now, the argument can be properly focused to how big an effect each of the 4 terms play.0 -
quote:Originally posted by buschmaster
in this video,http://www.youtube.com/watch?v=4MPSDjJQIv4
hickok45 shoots swinging steel targets with different calibers to see what happens.
doesn't draw conclusions, doesn't offer to explain. just see what happens.
well here's what is happening and what we can make of it.
the target swings more in this order:
22 LR
.38 ACP
9mm
.357 MAG
.40 S&W
.45 ACP
10mm
.44 MAG
notice the .357MAG in the middle of the list. it has more kinetic energy than .40SW and way more than .45ACP yet it doesn't swing the target as much as those.
similarly .40SW and 9mm have more KE than .45ACP, so why don't they swing the target more?
because the list is in order of power factor. that is how much momentum is imparted to the steel plate. exactly like one billiard ball hitting another.
shooting these steel plates is a good visual representation of power factor.
Good post and excellent description using the billiard balls.0 -
quote:Originally posted by AzAfshin
Don, I'm making certain assumptions that allow us to simplify the problem. If you'd like to include all the splatter effects then by all means, I'll go that route. After all, I used to write Monte Carlo simulations of nuclear fragmentation reactions (one of less than 100 nuclear physicists in the world qualified to do so at that time).
The collision of a bullet and a plate is far more inelastic than elastic. The simplest picture is that of the bullet being pancaked and then falling to the ground (at bullet velocities we don't get electromagnetic Coulomb barrier penetration leading to adhesion of bullet fragments to the target plate). But this scenario doesn't lead to any bounceback, hence:
KE (Plate) = KE (Bullet) - E (bullet deformation)
Now let's throw some splatter in there. The most reasonable assumption of splatter is a symmetric pattern around the point of impact. As such, conservation of momentum says that there is zero net momentum to the sides, hence 100% of the KE due to motion parallel to the plate is wasted and not imparted to the plate. Therefore:
KE (Plate) = KE (Bullet) - E (bullet deformation) - E (fragmentation leading to parallel motion) + KE (fragmentation perpendicular to plate)
Now, the argument can be properly focused to how big an effect each of the 4 terms play.
If any of the impacts in the video has something that resembled a bullet 'being pancaked and falling to the ground', I would agree with your assessment. The fact is, however, that significant amount of the kinetic energy of the bullet is retained by the fragments, resulting in a more elastic than inelastic collision.0 -
quote:Originally posted by Don McManus
quote:Originally posted by AzAfshin
Don, I'm making certain assumptions that allow us to simplify the problem. If you'd like to include all the splatter effects then by all means, I'll go that route. After all, I used to write Monte Carlo simulations of nuclear fragmentation reactions (one of less than 100 nuclear physicists in the world qualified to do so at that time).
The collision of a bullet and a plate is far more inelastic than elastic. The simplest picture is that of the bullet being pancaked and then falling to the ground (at bullet velocities we don't get electromagnetic Coulomb barrier penetration leading to adhesion of bullet fragments to the target plate). But this scenario doesn't lead to any bounceback, hence:
KE (Plate) = KE (Bullet) - E (bullet deformation)
Now let's throw some splatter in there. The most reasonable assumption of splatter is a symmetric pattern around the point of impact. As such, conservation of momentum says that there is zero net momentum to the sides, hence 100% of the KE due to motion parallel to the plate is wasted and not imparted to the plate. Therefore:
KE (Plate) = KE (Bullet) - E (bullet deformation) - E (fragmentation leading to parallel motion) + KE (fragmentation perpendicular to plate)
Now, the argument can be properly focused to how big an effect each of the 4 terms play.
If any of the impacts in the video has something that resembled a bullet 'being pancaked and falling to the ground', I would agree with your assessment. The fact is, however, that significant amount of the kinetic energy of the bullet is retained by the fragments, resulting in a more elastic than inelastic collision.
I think I'll put my time to a more fruitful task and go teach liberals about the joys of gun ownership.0 -
some energy is certainly used when the bullet gets smashed.
it's my understanding that an elastic collision is when the two objects bounce off of each other like billiard balls, and an inelastic collision is when some or all of the first object sticks to the other and its mass is now part of the second object. like if you shoot a billiard ball with a wad of chewing gum.
I think it's interesting to point out that when the first object strikes the second object, whatever deformation (chewing gum) or splintering-apart (wood block) or bullet-smashing is accounted for by a loss of kinetic energy, while the remaining velocity of its parts (chewing gum stuck on the 2nd object, chunks of wood flying apart, first billiard ball going off on another direction) is accounted for by momentum.
so we have heard that "momentum is conserved" but really? it's not, when some of the energy is translated into kinetic energy in an inelastic collision, and that would be how to define it as such.0
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