Misinformationists at Work
Herbert Blenner
Recently Mythbusters showed that bullets do not send victims flying
through the air and used the wrong ideas to explain their case.
Mythbusters argued that a bullet transfers all of its kinetic energy
to a target that stops it. This idea is physically incorrect and
contrary to their conclusion.
Suppose a 160-pound victim stopped a 1/43 pound MC bullet that struck
at a speed of 2100 foot per second. The kinetic energy of the striking
bullet equals one half the 1/43-pound weight of the bullet divided
by the acceleration of gravity, 32 foot per second squared, multiplied
by the square of the 2100-foot per second striking speed. Doing the
arithmetic gives the kinetic energy as 1602 foot-pound. If the 160-
pound victim acquired this kinetic energy then they would fly through
the air with a speed of 25.3 foot per second.
Using the proper idea that a striking bullet transfers its lost
momentum to the target gives a reasonable result. The momentum of the
bullet at striking equals the mass of the bullet, its weight divided
by the acceleration of gravity, multiplied by its striking speed. When
the bullet stops within the target, its momentum equals its weight
divided by the acceleration of gravity multiplied by the common speed
of the target and the bullet, Vc. Hence the loss of momentum of the
bullet becomes 1/43 pound divided by 32 foot per second squared
multiplied by the difference between the striking and common speeds of
the bullet. Equating this loss of momentum by the bullet to the
momentum gained by the target gives an equation for the common speed
Vc.
[(1/43) / 32 ] ( 2100 - Vc ) = 160 /32 Vc
So conservation of momentum gives the speed of the target as Vc =
0.305 foot per second. In general when the bullet stops within the
target, the common speed equals the striking speed multiplied by the
weight of the bullet divided by the sum of the weights of the target
and the bullet. Hence Vc = (2100) ( 1/43) / ( 160 + 1/43) = 0.305.
The target and bullet moving at 0.305 foot per second has a kinetic
energy of
0.5 ( 160 + 1/43) / 32 ( 0.305)^2 = 0.233 foot-pound that is a tiny
fraction of the 1602 foot-pound kinetic energy of the striking bullet.
Continuing with this analysis shows that this tiny fraction equals the
weight of the bullet divided by the sum of the weights of the target
and the bullet.
Mythbusters used another faulty idea that the distance moved by the
target is a measure of the force from the bullet. They observed that a
struck target suspended by a rod slid a small distance and incorrectly
concluded meekness of the force from the striking bullet. This
argument neglected the work done by sliding friction that consumed the
merger kinetic energy transferred from the bullet. In general the
stopping distance S equals the initial kinetic energy, T, divided by
the force of friction, f = uW, where u, the coefficient of sliding
friction is approximately 0.3 and W is the weight of the target.
For the case of the MC bullet, the initial kinetic energy T = 0.233
foot-pound and weight of 160.023 pound gives a stopping distance of
0.0049 foot or 0.058 inch. On the other hand if the idea that a
bullet transfers its kinetic energy to the target were correct then
the stopping distance would have been 1602 foot-pound divided by ( 0.3
X 160.023 pound) or 33.4 foot. Frankly I am surprised that these
"experimenters" did not shoot themselves.
This demonstration by Mythbusters does not represent the shooting of a
standing victim where the force of static friction at the feet and the
force from the bullet generate a torque that unbalances and perhaps
knocks down the victim.
Mitch Todd
"Herbert Blenner" <a1e...@verizon.net> wrote
> Recently Mythbusters showed that bullets do not send victims flying
> through the air and used the wrong ideas to explain their case.
>
> Mythbusters argued that a bullet transfers all of its kinetic energy
> to a target that stops it. This idea is physically incorrect and
> contrary to their conclusion.
It's perfectly correct. The KE of the bullet is completely
absorbed by the combined boady and bullet. However,
the bulk of the KE is transferred through a handful of
mechanisms, most notably frictional heating.
Kenneth A. Rahn
I have always felt that the strictest way to deal with this question is
to solve the equations for conservation of momentum and total energy (not
kinetic energy) simultaneously, for they both must hold. For total energy I
use kinetic plus "potential," and separate the terms. The colliding objects
are bullet and body.
When you do this, you quickly see that most of the original kinetic
energy disappears after the collision, and goes into the "potential energy"
term. So kinetic energy is not even close to being conserved (another way of
saying that the collision is "inelastic.") The missing KE has been degraded
to lower forms, like "heat." (I know this is fuzzy, but it makes the point.)
This whole thing is treated in more detail on my Web site.
Best wishes,
Ken Rahn
"Herbert Blenner" <a1e...@verizon.net> wrote in message
news:eda2d082-da59-4e02...@r27g2000yqb.googlegroups.com...
BobR
I think that if the bullet is brought to a complete stop within the
body then it can be said the bullet transferred all of its momentum to
the body. If the bullet goes through the body then it did not.
BobR
Let me try this again.
After reading your post more closely I agree with what you are saying.
You are saying that the bullet and body have a common velocity after
the body stops the bullet. So , only in those instances where the
body stops the bullet AND the body remains at rest did the bullet give
up all its momentum to the target.
John McAdams
wrote:
I'm afraid you need to reconcile your calculations with the actual
experimental results.
If the calculations don't jibe with the experimental results, you need
to reconsider the calculations.
.John
--------------
http://mcadams.posc.mu.edu/home.htm
Herbert Blenner
> Herbert,
>
> I have always felt that the strictest way to deal with this question is
> to solve the equations for conservation of momentum and total energy (not
> kinetic energy) simultaneously, for they both must hold. For total energy I
> use kinetic plus "potential," and separate the terms. The colliding objects
> are bullet and body.
> When you do this, you quickly see that most of the original kinetic
> energy disappears after the collision, and goes into the "potential energy"
> term. So kinetic energy is not even close to being conserved (another way of
> saying that the collision is "inelastic.") The missing KE has been degraded
> to lower forms, like "heat." (I know this is fuzzy, but it makes the point.)
> This whole thing is treated in more detail on my Web site.
>
> Best wishes,
> Ken Rahn
>
>
The total energy is constant when the change in kinetic energy is equal
and opposite the change in potential energy. This requires that the
conversions between kinetic and potential energies are lossless and
completely reversible.
The second law of thermodynamics explicitly excludes the conversion of
thermal to kinetic energies as a reversible process.
Herbert
Herbert Blenner
> On 23 Jul 2010 00:03:28 -0400, Herbert Blenner <a1ea...@verizon.net>
>
> - Show quoted text -
I calculated that a MC bullet would impart a low speed to the target
and friction would stop the target after traveling a negligible
distance. How do these results not "jibe" with the experiments?
Herbert
Herbert Blenner
>
You have recognized the logical flaw in the idea that a bullet transfers
all of its kinetic energy or all of its momentum to a target that stops
it. Good for you.
Mythbusters used other logically flawed ideas to describe their
experiments. I suspect that you would prefer to recognize these flaws for
yourself so I will withhold my criticisms of these ideas.
Herbert
WhiskyJoe
> This demonstration by Mythbusters does not represent
> the shooting of a standing victim where the force of
> static friction at the feet and the force from the
> bullet generate a torque that unbalances and perhaps
> knocks down the victim.
The bottom line is that bullets do not contain a lot
of momentum. A Carcano bullet weights about 1/43th of
a pound, goes roughly 1360 MPH, so it can, at most,
deliver enough momentum to push a one pound weight at
1 MPH. Or a 32 pound weight at 1 MPH. Or a 100 pound
weight at 1/3th of a MPH. Or rotate one end of a 100
pound weight at 2/3rds of a MPH. Or push JFK's head
and torso back at 2/3rds of a MPH.
Since one would expect only half the momentum of the
bullet to be transferred to JFK, the other half carried
away by the bullet or it's fragments, JFK's head and
torso should be pushed at 1/3th of a MPH.
The bottom line is that the Mythbusters did not
misrepresent the facts. Just as they showed a body
will not go flying away when hit by a bullet. Although,
yes, it's possible an inert body might topple over and
fall when hit by a bullet, as it might if a butterfly
lands on it. A bullet contains more momentum than a
butterfly, but still less than most people realize.
The Mythbusters simply tried to correct this mistaken
belief and did so quite honestly.
There is no experiment that will satisfy you.
If an experiment is made, shooting a dummy wearing
a bullet proof bullet, so 100% of the momentum is
transferred which does not cause the dummy to fall,
you will claim the dummy was too stable. Only if the
dummy was unstable, easily toppled in any direction,
or at least away from the rifle, will you be satisfied.
And it should be noted that such a demonstration would
be highly misleading because most of the momentum of
the dummy would come from gravity, not from the bullet.
John McAdams
wrote:
>On Jul 23, 8:36 pm, John McAdams <john.mcad...@marquette.edu> wrote:
So you are admitting that a bullet hitting Kennedy from the right
front would not throw him back and to the left?
.John
--------------
http://mcadams.posc.mu.edu/home.htm
Herbert Blenner
> On 23 Jul 2010 23:18:58 -0400, Herbert Blenner <a1ea...@verizon.net>
>
>
> >I calculated that a MC bullet would impart a low speed to the target
> >and friction would stop the target after traveling a negligible
> >distance. How do these results not "jibe" with the experiments?
>
> So you are admitting that a bullet hitting Kennedy from the right
> front would not throw him back and to the left?
>
> .John
>
You claimed that my calculations do not agree with the experiments, I
challenged the assertion and you folded by attempting to change the
subject.
I'm afraid that it now appears you made up the claim to divert attention
from my exposure of the conceptual errors made by Mythbusters.
Herbert
Kenneth A. Rahn
Eh? (1) Total energy is always constant during a collision. (2) I was
talking about a one-way process, and said nothing about reversibility. (3)
Recall that I used the term "potential energy" but said that I wasn't
satisfied with it.
Ken Rahn
HB: The total energy is constant when the change in kinetic energy is
equal and opposite the change in potential energy. This requires that the
conversions between kinetic and potential energies are lossless and
completely reversible.
HB: The second law of thermodynamics explicitly excludes the conversion of
John McAdams
wrote:
>On Jul 24, 10:19 am, John McAdams <john.mcad...@marquette.edu> wrote:
>> On 23 Jul 2010 23:18:58 -0400, Herbert Blenner <a1ea...@verizon.net>
>> wrote:
>>
>>
>> >I calculated that a MC bullet would impart a low speed to the target
>> >and friction would stop the target after traveling a negligible
>> >distance. How do these results not "jibe" with the experiments?
>>
>> So you are admitting that a bullet hitting Kennedy from the right
>> front would not throw him back and to the left?
>>
>>
>
>You claimed that my calculations do not agree with the experiments, I
>challenged the assertion and you folded by attempting to change the
>subject.
>
>I'm afraid that it now appears you made up the claim to divert attention
>from my exposure of the conceptual errors made by Mythbusters.
>
I don't follow you. You were assuming that the bullet transferred all
it's energy to the person hit, and claiming that it would throw the
person (Kennedy) backwards.
But MYTHBUSTERS tried that -- and with a .50 cal. sniper rifle, and it
did not happen.
Are you aware that the range of bullet weights for plausible sniper
rifles in 1963 was 120-18 grains, and a .50 slug is 600 grains plus?
.John
--------------
http://mcadams.posc.mu.edu/home.htm
Anthony Marsh
> On 24 Jul 2010 12:22:52 -0400, Herbert Blenner<a1e...@verizon.net>
> wrote:
>
>> On Jul 24, 10:19 am, John McAdams<john.mcad...@marquette.edu> wrote:
>>> On 23 Jul 2010 23:18:58 -0400, Herbert Blenner<a1ea...@verizon.net>
>>> wrote:
>>>
>>>
>>>> I calculated that a MC bullet would impart a low speed to the target
>>>> and friction would stop the target after traveling a negligible
>>>> distance. How do these results not "jibe" with the experiments?
>>>
>>> So you are admitting that a bullet hitting Kennedy from the right
>>> front would not throw him back and to the left?
>>>
>>>
>>
>> You claimed that my calculations do not agree with the experiments, I
>> challenged the assertion and you folded by attempting to change the
>> subject.
>>
>> I'm afraid that it now appears you made up the claim to divert attention
>>from my exposure of the conceptual errors made by Mythbusters.
>>
>
> I don't follow you. You were assuming that the bullet transferred all
> it's energy to the person hit, and claiming that it would throw the
> person (Kennedy) backwards.
>
> But MYTHBUSTERS tried that -- and with a .50 cal. sniper rifle, and it
> did not happen.
>
But Mythbusers DID try that -- and it DID happen. It knocked the dummy
back several inches. Didn't you even watch it?
> Are you aware that the range of bullet weights for plausible sniper
> rifles in 1963 was 120-18 grains, and a .50 slug is 600 grains plus?
>
We are not postulating an ARMY sniper on the knoll. We are considering a
second assassin using the same type of weapon and similar ammo. There
were and are M-C spire point bullets in the range of 156 grains.
> .John
> --------------
> http://mcadams.posc.mu.edu/home.htm
Anthony Marsh
Friction? JFK's head? What friction? From what source? Air friction?
Herbert Blenner
> On 24 Jul 2010 12:22:52 -0400, Herbert Blenner <a1ea...@verizon.net>
>
The first and second paragraphs of my post makes it clear that I employed
Mythbusters assumption. I wrote.
"Recently Mythbusters showed that bullets do not send victims flying
through the air and used the wrong ideas to explain their case."
"Mythbusters argued that a bullet transfers all of its kinetic energy to a
target that stops it. This idea is physically incorrect and contrary to
their conclusion."
In the third paragraph, I used Mybusters' argument to calculate a speed
comparable to a flying victim.
How you equation my actions with "assuming that the bullet transferred all
it's energy to the person hit" boggles any rational mind especially one
that read the first sentence of my next paragraph.
"Using the proper idea that a striking bullet transfers its lost momentum
to the target gives a reasonable result."
So which portions of the post did you read?
Herbert
Herbert Blenner
Show us where Mythbusters said, implied or even hinted that momentum
transfer from the bullet is the measure of the motion of the shot
victim. For your convenience I post a link to their video.
http://www.youtube.com/watch?v=QCzD5uhSViY
They mentioned transference of kinetic energy several times while
referring to momentum once in the wrong context of "lethal momentum"
absorbed by their dummy.
>
> There is no experiment that will satisfy you.
> If an experiment is made, shooting a dummy wearing
> a bullet proof bullet, so 100% of the momentum is
> transferred which does not cause the dummy to fall,
> you will claim the dummy was too stable. Only if the
> dummy was unstable, easily toppled in any direction,
> or at least away from the rifle, will you be satisfied.
> And it should be noted that such a demonstration would
> be highly misleading because most of the momentum of
> the dummy would come from gravity, not from the bullet.
Experiments neither satisfy nor dissatisfy me. Instead they are
objects for a complete analysis of the underlying physics. They call
this the physical model upon which science draws their conclusions.
In case, I concluded that the distance moved by a target that remained
on the rack equaled the kinetic energy acquired by the target divided
by the force of friction between the rack and the bar that supported
the target.
By contrast, the showmen of Mythbusters related the force from the
bullet to the distance moved by the target. I invite you to defend
Mythbusters' use of this discredited Aristotelian idea.
Herbert
Bob
Speaking of frictional heating, you can see the white hot impact point
on jfks skull when the second headshot hits him at zapruder frame 317
http://www.youtube.com/watch?v=uvtpjHyhEPg
Compare that white hot spot on jfk's skull to this xray of jfk's
skull. That perfectly round hole is an entry wound
http://hogtuner.net/gauges/jfk_skull.png