mortablunt
Member
I am still a supporter of the big hole school. Bigger bullet = bigger hole = more damage = faster kill.
The hydrostatic shock theory?
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I think that if you run the numbers you will find that this supposed pressure at the point of origin is a little low.
For what it might be worth to anyone interested in the topic, Courtney runs through similar calculations in this paper.
http://arxiv.org/ftp/physics/papers/0701/0701267.pdf
He also discusses situations which would prevent the pressure from falling off as rapidly as one might expect from calculations based on a purely homogenous target medium.
MachIVshooter
Member
Ignoring for the moment that I made disclaimers that I was illustrating concept rather than hard data (hence the nice round numbers and simple, linear math), did you actually look at the chart? (Which, BTW, is flawed at a fundamental level; It is titled "peak pressure wave magnitude", but illustrates only peak pressure, not magnitude [size]. Magnitude and pressure are not synonymous). The chart shows the majority of their test results between 400 and 800 PSI peak pressure, and tapering off dramatically over 1,000 PSI.
From his article, and perhaps what you saw that made you believe my round 1,000 figure was low:
This is so flawed it's not funny:
-100 lbs/.02 ft=5,000 pounds per square foot, which is 35 PSI (5,000/144)
-There is not a simple calculation to turn ft/lbs of kinetic energy into pound of force or pressure; KE variable is found with velocity and mass, force variable is found with distance/ time/acceleration and mass, pressure is force and area:
1 pound dropped from 1 foot generates 1 ft/lb of energy
1 pound force is required to move 1 pound a distance of 1 foot
1 pound per square foot is the pressure generated by a one pound weight distributed over a square foot at earth gravity
-Even if we had the correct force figures for that bullet (nowhere near 5,000), one cannot calculate the pounds per square foot without a variable Courtney ignored; area of application. He does not tell of the contact area of that bullet that is shedding 100 ft/lbs in .2 ft.
-To calculate the pressure of a ballistic projectile in a liquid medium, one needs to incorporate displacement and velocity. Energy doesn't matter; 500 pound weight at 10 FPS generates 777 ft/lbs, a 200 grain bullet at 1320 FPS generates 773 ft/lbs. Do you think that the 10 FPS weight can generate the kind of hydraulic pressure wave the 1,320 FPS bullet can? A static object only dispaces it's own volume, while a fast-moving object will displace a volume far greater than it's own as the medium cannot resume it's static orientation as quickly as it is being displaced. The faster the object, the greater the displacement volume to object volume ratio.
-There is no calculation for the hydraulic pressure a ballistic projectile will generate in a human body; It is wholly dependent on the medium.
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JohnSKa, it seems to me that given the various problems with how Courtney has tried to sell this theory and the way he has shied away from doing experiments that can outright prove it one way or another, there is room for a more sincere investigation into this.
Are you keen to do it (and include your extensions to his theory)?
Are you keen to do it (and include your extensions to his theory)?
MachIVshooter
Member
You're gonna spend a lot of time running around this forum and others correcting every post that has denoted it with a slash instead of a dash. Have fun with that.
As long as you're being persnickety, you might look for all of these as well:
ft.lb
lb.ft
ft/lbf
ft-lbf
lb/ft
lb-ft
foot/pound
pound/foot
pound-foot
When you're done with that, there are always the other divisions of imperial and the whole of SI with similar variations in denotation
In denoting units of measurement, the slash DOES NOT indicate division. PSI is often denoted with lbs/in2 (sorry, can't get the superscript). Inches or millimeters of mercury is denoted in/hg or mm/hg. The list goes on. And on.
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I don't even need to read this entire thread... This thread should be shut down for so much misinformation about physics....
If you really think that a bullet has more energy imparted onto it than onto the gun firing it, please watch this Video of a man standing on ONE foot taking a .308 to the Torso.
If you really think that a bullet has more energy imparted onto it than onto the gun firing it, please watch this Video of a man standing on ONE foot taking a .308 to the Torso.
Mike1234567
member
pound-feet 
MachIVshooter
Member
Again, the energy vs. force thing.
The force driving the bullet forward is equal to the force driving the gun backward. However, due to the mass, the rate of acceleration is inversely proportionate; The much lighter bullet is accelerated much faster and to a much higher velocity. Because energy is velocity squared times mass, the bullet produces much more energy.
Force is mass and acceleration; 1 pound force (~32 poundals) is what is required to accelerate a 1 pound mass at a rate ~32 feet per second squared (or 32 pounds at 1 FPS squared). The pound-force is based on the rate of acceleration for standard gravity at 32.174049 feet per second per second, which is why one can state that a pound dropped from a height of 1 foot will exert one foot pound, because gravity is acting on that mass with 1 pound-force.
However, a pound will not have accelerated to 32 FPS when dropped from a height of one foot, because it began at 0 FPS and is accelerating at the rate of 32 FPS per second, not 32 FPS per foot. It will have attained a velocity of ~8 FPS after 1 foot of travel and covered that distance in a quarter of a second.
Force exists in the absence of motion; Kinetic energy cannot; Force can be exerted on an object but,be insufficient to move it. The force is still there, still acting.
Kinetic energy can be calculated if mass, force and distance are known, because the force vector is a constant; Over a specific distance at a given rate of acceleration, the velocity can be determined, and so can the kinetic energy.
Then there's momentum, which is mass times velocity.
Momentum and kinetic energy can be used to calculate the force a bullet exerts on a target, but this is not the same as the energy it impacts with; A bullet that impacts with 500 ft/lbs does not exert 500 pounds of force, That bullet cannot accelerate a 500 pound object it hits at the rate of 32 feet per second per second. In fact, it can barely budge something that large.
My math isn't good enough to calculate the force a 230 gr. bullet at 1,000 FPS (510 ft/lbs) can exert without spending more time on it that I have to spare at present, but I'm sure someone can. I'm not sure how one figures the force of an object that rapidly decelerates from a high velocity on impact. I know it's momentum (pound-feet-second) over time, more or less.
This should clear up some of the unit confusion:
http://www.racquetresearch.com/units.htm
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Loosedhorse
member
Well, this is progress. Finally an admission that you have been presenting errors as fact.MachIVshooter said:
There is no ambiguity: the midbrain, pons, and medulla (oblongata) comprise the brainstem.
Claimed by whom? Where? My quote was this:
So, just like you have been making errors in anatomy, you are now apparently claiming I said things that I did not. You're really helping your credibility.Loosedhorse said:
Who's this "we"? YOU can make whatever blanket statement you want.MachIVshooter said:
Let's see: we have one poster, you, who repeatedly and unapologetically makes obviously false statements about human anatomy; and when that fails, makes false statements about what another poster "claimed"; and then, when that fails, resorts to ad hominem name-calling (usually taken to be an admission that the person has no logical, cogent response to the points made).
And we have another poster, me, tirelessly correcting those errors. And you award the title of jerk to me? Okay.
I was trying to spare you yet more embarrassment (if you in fact can ever be embarrassed). Tell you what: you explain to us the difference between these waves, okay? (For a bonus, you can explain why any such differences are important to our discussion.) After all, you've been doing so well with anatomy...!
Well, guess what? We agree. Well said.
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He's done some practical experimentation that provided interesting results. Enough to demonstrate that his basic premise has value. Not necessarily enough to demonstrate that it could be turned into something practical.
I have neither the resources nor the inclination to attempt to verify his results given that I'm not at all certain there's much of a way to turn the results into something practical in terms of handgun stopping power. I saw his experimentation and theory primarily as expanding the understanding of terminal ballistics by providing explanations for effects that are fairly commonly observed but not well-explained by other theories of stopping power.
He's not talking about psi/pressure, he's talking about the fact that kinetic energy divided by the distance required to stop the projectile is proportional to the force applied to the target medium by the projectile.
Certainly there is, and it's VERY basic physics.
If you work out the units, you'll find that Kinetic Energy and Force are very closely related.
Force = mass x acceleration
Acceleration = speed/time
Speed = distance/time
Kinetic energy = mass x speed x speed
Kinetic energy = mass x speed x distance/time
Rearranging terms
Kinetic energy = mass x speed/time x distance
Since speed/time = acceleration...
Kinetic Energy = mass x acceleration x distance
And now it's plain that
Force = Kinetic energy / distance.
Exactly what Courtney's calculations show. Not surprising since he has a doctorate in physics.
By the way, Force can also be related to momentum. Force is proportional to the momentum of the projectile divided by the time it takes to stop the projectile.
Force = Momentum / time
That's apparent. What's confusing is that, knowing that you didn't understand the science, you would think it made sense to try to nitpick basic math and physics calculations in a published and peer-reviewed document written by two persons with doctorates in physics from MIT and Harvard.
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Vern Humphrey
Member
You can settle that by a simple expiment -- when you shoot a deer this fall and saw off the top of the skull with the attached antlers, look to see if there is evidence of ruptured blood vessels in the brain.
Mike1234567
member
My head hurts...
MachIVshooter
Member
First, I was addressing the assertions that my 1,000 PSI POO figure was low, which seems to have come from seeing that 5,000 pound force figure, because the chart reflects 1,000 to be on the high end. That was only to illustrate a point; I know the difference between pound force and pound per square foot, as I've demonstrated.
Second, are we are to believe that if you fire a bullet developing 100 ft/lbs into the back of a 5000 pound vehicle and the bullet stops within .2 ft, that vehicle will accelerate at a rate of 32 FPS per second? Because that's what 5000 pounds force will do.
Reality shows that a bullet certainly cannot move an object that size, no matter how deep or shallow the penetration, even with much more than 100 ft/lbs KE.
So a bullet that is stopped at 0.000 Feet penetration exerts no force, and one that travels 1 foot and expends it's 100 ft/lbs generates 100 pounds of force, but that bullet that only travels 0.02 ft exerts 5000 pounds force? Seems a bit flawed, doesn't it?
Except Courtney didn't incorporate time or momentum, only distance and KE with an unknown velocity and mass, which leaves the time and momentum a mystery.
Momentum cannot be found without either A) bullet weight and velocity, B) Bulet weight and KE or C) velocity and KE. Since all he gives is KE, you cannot find momentum and therefore cannot solve for force. Two projectiles that develop the same KE can have very different momentum:
a 20 gr. bullet at 4,100 FPS generates a KE of 747 with a momentum of 11.71 pounds/FPS.
a 200 gr. bullet at 1,300 FPS generates a KE of 750 with a momentum of 37.14 pounds/FPS
Let's use the momentum of a 230 gr. pill at 1,000 FPS. That gives us a momentum of 33 pounds/FPS. Now we divide 33 by, say, 5 milliseconds (0.005 Seconds) that it would take to stop the bullet. Does 211,200 poundals sound right to you? That a .45 ACP slug can accelerate a 3+ ton object to 22 MPH in one second?
Since you seem to believe that you're so good at math (the numbers are a bit more complicated than these over-simplified equations), why don't you solve for force with the aforementioned 230 gr. slug at 1,000 FPS.
The same we you were referring to when you insulted me.
Vern wisely observes:
Indeed that's what I suggested to Michael Courtney. Shoot one deer with a bullet, one with an arrow, have the heads analysed by a veterinarian.
If that's positive, I'll say right here that the ballistic pressure wave exists and that gunshots to the chest can cause brain haemorrhages (even micro) and the debate will be over.
You have to ask why it hasn't been done...
Indeed that's what I suggested to Michael Courtney. Shoot one deer with a bullet, one with an arrow, have the heads analysed by a veterinarian.
If that's positive, I'll say right here that the ballistic pressure wave exists and that gunshots to the chest can cause brain haemorrhages (even micro) and the debate will be over.
You have to ask why it hasn't been done...
Loosedhorse
member
Mike, I think that theory was long ago discarded. (Knowing that, and knowing what the results would show no rupture, some here have still suggested you try it.)
It might be instructive to shoot animals and look for microscopic differences in the vessels of those animals that collapsed immediately (or within, say, 5 seconds) compared to those animals who fell later. You might need a lot of animals to distinguish consistent patterns from random differences, and now you're gettng expensive...
And the PETA folks will be on you!
And then we'd still have the bugaboo that some humans who suffer concussion have no abnormal imaging findings--so why should we be assured there will be microscopic pathological findings?
Vern Humphrey
Member
If we did that (examine the brain of a gun shot deer for evidence of hydrostatic shock) we'd end all debate and gun forums would wither and die.
Mike1234567
member
Anyone have access to a MRI machine?
Vern Humphrey
Member
No need, since we're not talking about examining a live deer -- we're talking about doing a necropsy. All you need for that is a microscope and some staining fluid.
MachIVshooter
Member
Here, let's take a crack at this using Courtney's 100 ft/lbs imparted in .02 ft. number:
A 45 gr. projectile at 1,000 FPS develops almost exactly 100 ft/lbs (99.9)
We'll use a 5,000 pound mass suspended (no rolling resistance)
At 1,000 FPS, that bullet will travel 1 ft in 1 millisecond, so it will travel .02 ft. in 20 microseconds, and we'll assume that's where it stops for "complete energy transfer". Obviously, the bullet will take slightly longer to make that depth, since it is decelerating during that 1/4" of penetration, but that's a whole nother equation. And, of course, Courtney has proposed projectile that sheds 100 ft/lbs in that .02 ft, but does not tell us what the total KE was, nor did he give us the projectile weight or velocity, so we can't know the time it took to reach that depth.
The math tells us the 5,000 pound mass is ~777,778 times that of the bullet
This means the bullet's velocity, once imbedded in the mass, must be divided by the factor of it's mass against the one it's embedding into plus the bullet's mass (employing conservation of momentum):
5000x7000=35,000,000 grains, plus the bullet
35,000,045/45=777,779 ratio
1,000 FPS/777,779=0.001257 FPS
So, in theory, this bullet will accelerate that mass to that velocity in 0.00002 seconds. Agreed?
Divide the velocity by the time it takes to reach that velocity and we get the rate of acceleration per second:
0.001257/0.00002=64.3 fps per second
Now we divide 64.3 FPS by 32.174 FPS that is the base velocity for pound force
64.3/32.174=1.9985
To find pound force required for that acceleration, we multiply mass by that coefficient
1.9985*5,000=9,992
So, by the math and confimed by This calculator, to accelerate that 5,000 pound mass at that rate requires almost 10,000 pound force.
Now, the problem with trying to calculate the force using simple math of energy and distance over time is that it only tells us the maximum velocity the 5,000 pound object will reach. Simple math suggests that it will accelerate to that velocity in the same amount of time that it takes for the bullet to decelerate from 1,000 FPS to 0 FPS, but that directly conflicts with the 100 ft/lbs in .02 ft.= 5,000 pound force.
Let's try this again, but with a 100 pound mass instead of 5,000
100x7000+45=700,045 grains
700,045/45=15,557 ratio
1000/15,557=0.06428 FPS velocity of the 100 lb mass
0.06428/.00002=3,214 FPS per second rate of acceleration
9,989 pound force is required to accelerate a 100 pound mass at 3,214 FPS per second.
Of course, anyone who's been shooting knows full well that a smaller object can be accelerated by the bullet, while a larger one cannot.
Let's do it one more time with the 100 pound mass, but a 90 gr. bullet at 707 FPS (100 ft/lbs)
100x7000+90=700,090 grains
700,090/90=7,779 ratio
707/7779=0.09089 FPS velocity of the 100 lb mass
0.09089/.00002829=3,213 FPS per second rate of acceleration
3,213/32.174=99.86
99.86*100=9,986 pound force
So, with a slower but heavier bullet making the same KE, but having greater momentum, we get roughly the same force? The 45 gr. bullet has a momentum of 6.43, while the 90 gr. has 9.09 pounds-ft/sec momentum.
Now let's really go nuts with the momentum thing using an automobile:
a 3,000 pound vehicle traveling at 1.4646 MPH generates 100 ft/lbs of energy
since momentum=mass*velocity, it has a momentum of 4,394 pounds-ft/sec
So, since it has the same energy, are we to believe that that slow-moving automobile has exactly the same capability of accelerating another mass as the 45 gr./1,000 FPS bullet and the 90 gr./707 FPS bullet?
Or are we to go with the theory that force=KE/distance, and since the car will not penetrate that object at all and imparts it's energy in 0.000...... seconds, that it has no force? Because that's how Courtney's math works out; No penetration=no force and deeper penetration=less force. Kinda goes against conservation of energy, doesn't it?
Now, since I have a little time, we'll run the numbers again accounting for deceleration, but, since Courtney didn't specify that the bullet stopped in .02 feet, and since according to John that's irrelevant, because it's all about the KE dump and distance, we'll use on that doesn't stop. Let's go with a 9mm 115 gr. at 1,200 FPS, producing 368 ft/lbs. If that bullet dumps 100 ft/lbs in the first .02 feet (1/4"), that means it's velocity has been reduced to 1,043 FPS, so it lost 157 FPS in 1/4". That gives us an average velocity of 1171.5 FPS over the 17 microseconds it took to travel that distance and impart that energy.
We'll use our 100 pound mass again, but because the bullet didn't stop, we'll exclude it's weight from the calculation.
So:
100x7000=700,000 grains object
700,000/115=6,087ratio
Now, because the bullet didn't stop, but imparted 157 FPS of it's velocity into the object, we have to change our math a little:
157/6,087=0.02579 FPS velocity of the 100 lb mass
0.02579/.000017=1,517 FPS per second rate of acceleration
1517/32.174=47.14
47.174*100=4,714 pound force
Hmmm.....still not 5,000, even with calculations accounting for deceleration.
How about a .45 ACP 230 gr. @ 830 FPSfor 352 ft/lbs?
still a 700,000 grain object, so now we have a 3.043 bullet to object weight ratio
For the bullet to penetrate .02 feet and shed 100 ft/lbs, it's velocity is reduced to 671 FPS and it takes .00002665 seconds to traverse that 1/4". That's a velocity differential of 159 FPS imparted on the object, so:
159/3043=0.05225 FPS in .00002665 seconds
0.05225/0.00002665=1,961
1,961/32.174=60.95 FPS per second
60.95*100=6,095 pound force
So how, with both the 9mm load and the .45 load "dumping" 100 ft/lbs in .02 ft, is there such a pound force difference? Because again, according to Courtney and JohnKSA, it should be the same.
Who wants to explain for all of us why John's "Simple equations" don't work here? Because, as I've admitted, I lack the mathematical skills to do so.
A 45 gr. projectile at 1,000 FPS develops almost exactly 100 ft/lbs (99.9)
We'll use a 5,000 pound mass suspended (no rolling resistance)
At 1,000 FPS, that bullet will travel 1 ft in 1 millisecond, so it will travel .02 ft. in 20 microseconds, and we'll assume that's where it stops for "complete energy transfer". Obviously, the bullet will take slightly longer to make that depth, since it is decelerating during that 1/4" of penetration, but that's a whole nother equation. And, of course, Courtney has proposed projectile that sheds 100 ft/lbs in that .02 ft, but does not tell us what the total KE was, nor did he give us the projectile weight or velocity, so we can't know the time it took to reach that depth.
The math tells us the 5,000 pound mass is ~777,778 times that of the bullet
This means the bullet's velocity, once imbedded in the mass, must be divided by the factor of it's mass against the one it's embedding into plus the bullet's mass (employing conservation of momentum):
5000x7000=35,000,000 grains, plus the bullet
35,000,045/45=777,779 ratio
1,000 FPS/777,779=0.001257 FPS
So, in theory, this bullet will accelerate that mass to that velocity in 0.00002 seconds. Agreed?
Divide the velocity by the time it takes to reach that velocity and we get the rate of acceleration per second:
0.001257/0.00002=64.3 fps per second
Now we divide 64.3 FPS by 32.174 FPS that is the base velocity for pound force
64.3/32.174=1.9985
To find pound force required for that acceleration, we multiply mass by that coefficient
1.9985*5,000=9,992
So, by the math and confimed by This calculator, to accelerate that 5,000 pound mass at that rate requires almost 10,000 pound force.
Now, the problem with trying to calculate the force using simple math of energy and distance over time is that it only tells us the maximum velocity the 5,000 pound object will reach. Simple math suggests that it will accelerate to that velocity in the same amount of time that it takes for the bullet to decelerate from 1,000 FPS to 0 FPS, but that directly conflicts with the 100 ft/lbs in .02 ft.= 5,000 pound force.
Let's try this again, but with a 100 pound mass instead of 5,000
100x7000+45=700,045 grains
700,045/45=15,557 ratio
1000/15,557=0.06428 FPS velocity of the 100 lb mass
0.06428/.00002=3,214 FPS per second rate of acceleration
9,989 pound force is required to accelerate a 100 pound mass at 3,214 FPS per second.
Of course, anyone who's been shooting knows full well that a smaller object can be accelerated by the bullet, while a larger one cannot.
Let's do it one more time with the 100 pound mass, but a 90 gr. bullet at 707 FPS (100 ft/lbs)
100x7000+90=700,090 grains
700,090/90=7,779 ratio
707/7779=0.09089 FPS velocity of the 100 lb mass
0.09089/.00002829=3,213 FPS per second rate of acceleration
3,213/32.174=99.86
99.86*100=9,986 pound force
So, with a slower but heavier bullet making the same KE, but having greater momentum, we get roughly the same force? The 45 gr. bullet has a momentum of 6.43, while the 90 gr. has 9.09 pounds-ft/sec momentum.
Now let's really go nuts with the momentum thing using an automobile:
a 3,000 pound vehicle traveling at 1.4646 MPH generates 100 ft/lbs of energy
since momentum=mass*velocity, it has a momentum of 4,394 pounds-ft/sec
So, since it has the same energy, are we to believe that that slow-moving automobile has exactly the same capability of accelerating another mass as the 45 gr./1,000 FPS bullet and the 90 gr./707 FPS bullet?
Or are we to go with the theory that force=KE/distance, and since the car will not penetrate that object at all and imparts it's energy in 0.000...... seconds, that it has no force? Because that's how Courtney's math works out; No penetration=no force and deeper penetration=less force. Kinda goes against conservation of energy, doesn't it?
Now, since I have a little time, we'll run the numbers again accounting for deceleration, but, since Courtney didn't specify that the bullet stopped in .02 feet, and since according to John that's irrelevant, because it's all about the KE dump and distance, we'll use on that doesn't stop. Let's go with a 9mm 115 gr. at 1,200 FPS, producing 368 ft/lbs. If that bullet dumps 100 ft/lbs in the first .02 feet (1/4"), that means it's velocity has been reduced to 1,043 FPS, so it lost 157 FPS in 1/4". That gives us an average velocity of 1171.5 FPS over the 17 microseconds it took to travel that distance and impart that energy.
We'll use our 100 pound mass again, but because the bullet didn't stop, we'll exclude it's weight from the calculation.
So:
100x7000=700,000 grains object
700,000/115=6,087ratio
Now, because the bullet didn't stop, but imparted 157 FPS of it's velocity into the object, we have to change our math a little:
157/6,087=0.02579 FPS velocity of the 100 lb mass
0.02579/.000017=1,517 FPS per second rate of acceleration
1517/32.174=47.14
47.174*100=4,714 pound force
Hmmm.....still not 5,000, even with calculations accounting for deceleration.
How about a .45 ACP 230 gr. @ 830 FPSfor 352 ft/lbs?
still a 700,000 grain object, so now we have a 3.043 bullet to object weight ratio
For the bullet to penetrate .02 feet and shed 100 ft/lbs, it's velocity is reduced to 671 FPS and it takes .00002665 seconds to traverse that 1/4". That's a velocity differential of 159 FPS imparted on the object, so:
159/3043=0.05225 FPS in .00002665 seconds
0.05225/0.00002665=1,961
1,961/32.174=60.95 FPS per second
60.95*100=6,095 pound force
So how, with both the 9mm load and the .45 load "dumping" 100 ft/lbs in .02 ft, is there such a pound force difference? Because again, according to Courtney and JohnKSA, it should be the same.
Who wants to explain for all of us why John's "Simple equations" don't work here? Because, as I've admitted, I lack the mathematical skills to do so.
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Mike1234567
member
Your figure was at the point of origin (POO), the chart does not reflect point of origin energies. Your figure was low the way it was stated.
I used the calculator you found and it indicates that 5000 poundals of force will accelerate a 5000 lb vehicle at 1fps per second, not 32fps per second. Not too surprising given that F=ma, or F/m=a. 5000/5000 = 1.
AND, that assumes that all of the energy is used creating force that moves the object. In reality much of the force is used up deforming the objects involved and or creating heat and noise as opposed to being used entirely to create motion. In case it's not obvious, this discussion is PRECISELY about collisions that result in deformation because we're interested in the damage done by the bullet.
The simple fact that penetration is a factor here means that treating the collisions as no deformation takes place is not valid in the least.
Which means, in reality, that the actual acceleration will be less even than the 1fps the calculator indicated. Most, if not all, of the energy was used to deform the target, not to move it.
No, theoretically it would exert infinite force. If you divide a non-zero quantity by zero you get infinity.
It seems a bit flawed because you don't understand the physics or apparently even the math involved.
I didn't so much as hint that he did. I was trying to point out how badly flawed your assessment of converting kinetic energy to force was and thought it might be worthwhile to throw in the bit about momentum to show how momentum and energy relate.
You can calculate force from KE if you know how much distance it took the bullet to stop. I showed you how. Two persons with doctorates in physics from MIT & Harvard showed you how in their paper.
Here's another source.
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html
"Average impact force x distance traveled = change in kinetic energy"
Rearranging the equation by dividing distance traveled on both sides we get:
Average impact force = change in kinetic energy/distance traveled.
You are laboring under a severe misapprehension. The equation that force = KE/distance is NOT theory in any sense that suggests it might be invalid or might not reflect real-world behavior. It is well-accepted science and has been for a few centuries. It has been carefully verified by experimental methods.
Your lack of understanding of the concepts, science and mathematics does nothing to affect their validity.
You're not going to be able to blame this on Courtney. No penetration equals infinite force.
And it's not "Courtney's math", it's simply math & physics. There's nothing special about his calculations and his understanding and application of physics is just fine.
Look. How about a little bit of reality. You may not like Courtney's conclusions, but there is absolutely nothing wrong with his math, nor his understanding of physics. The paper you're referring to was written by two people with doctorates in physics and has been read and reviewed by other people who understand the math & physics. There's nothing wrong with that aspect of his paper.
It's certainly possible to question his overall conclusions, but the nuts and bolts of his calculations are just fine. If you can't make the math work out, then I can assure you that the problem is on your end.
You're performing calculations that prove nothing other than that you don't understand the basic concepts involved in the physics of moving objects. It's a waste of your time to perform the calculations, a waste of your time to post them and a waste of other people's time to read them.
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MachIVshooter
Member
32.174049 poundals in a pound-force; 5,000 pound force acting on a 5,000 pound mass will accelerate it at 32.174049 FPS per second.
Courtney stated pounds of force.
I understand that, but to say that I'm flawed in using that also implies that Courtney's theory is flawed, because pound-force is a vector, and he submits that his bullet dumping energy turns it into force, and he does not quantify the elastic properties of that impact. We all know that conservation of energy tells us that the energy lost from the bullet is converted in several ways, including pound-force acting on the target, pressure within the target, deformation of the slug, etc.
This is also why I was saying that you can't oversimplify this the way it has been.
I oversimplified to a point using inelastic collisions to demonstrate that calculating pound-force with only energy, distance and time that way will give different results if you change other variables like velocity and bullet weight.
So show us how it's done.
I honestly did it expecting to get the same numbers, and re-did them a few times to make sure that in rounding a bit I was using enough digits to the right of the decimal that it was not the cause of the different results.
If you can rework the 9mm/115/1,200 and .45 ACP/230/830 both dumping 100 ft/lbs in .02 feet examples somehow to get the 5,000 pounds of force out of both whilst still having the rest add up, then I'll post in the largest, reddest font possible that I was erred.
Which brings about a point I noticed earlier, but did not mention: In the world of science and medicine, Imperial units are not common in the least. Why wasn't his paper in SI units?
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