KE = ½ mv^2 is disproved in a new falling object impact test.
NoEinstein
smoothness, but are of different weights, when they are dropped they
will have the same air resistance as they fall. If such balls are
dropped from an identical height, and both subsequently impact the
same horizontal bed of soft clay, the heavier ball will penetrate into
the clay further than the lighter ball.
Since the 17th century, people have observed the varying depths of
penetration of a given lead shot or canon ball as the height of fall
is increased. Since Galileo, fall distances could be correlated into
the falling objects’ velocities. Observations that a ball which falls
with double the velocity seemed to penetrate four times as far into
soft clay, led many to suppose that kinetic energy was accruing
parabolically with respect to velocity.
In 1830, Coriolis modified the former, purely parabolic increase in
kinetic energy into a semi-parabolic, but still exponential, rate of
increase. His well known equation is: KE = ½ mv^2. But such has
never been verified by any means which meets Scientific Method
standards.
The resistance of clay to impact varies widely. The individual
particles of clay are free to slide over adjacent particles. And the
resistance to such sliding is determined by the angle of internal
friction of the particles. That is analogous to the force required to
slide a given object down an inclined plane. Most know that the
coefficient of SLIDING friction is less than the coefficient of
friction at rest. So, the internal friction of clay that’s already
beginning to slide (internally) is also less than the friction of clay
particles that are at rest. What that means is: The energy required
to cause a given depth of penetration doesn’t vary linearly.
For the above reason, a single round ball falling from varying heights
and impacting soft clay could never accurately correlate to the amount
of KE present at impact. But a most simple new impact test which I
have devised, which uses two same-size, but different weight balls,
CAN correlate to the KE present! Such experiment is described below:
If any heavy ball is dropped into clay, it will impact with a KE which
has been assumed to be given by Coriolis’s equation. If KE = ½ mv^2
is true, then there must be a greater height from which a lighter, but
same size, ball can be dropped so that the lighter ball will impact
with the SAME kinetic energy as the heavier ball.
The amazing thing about having two different density, but same size
balls impacting soft clay with the SAME kinetic energy, is that the
size of the resulting holes in the clay should be IDENTICAL! Under
the latter conditions, the consistency and material characteristics of
the clay have no influence whatsoever on the KE comparison being
made. *** “If two same size balls of different density impact with
the same KE, they will make IDENTICAL holes regardless of the type and
softness of the clay being impacted.” ***
Coriolis’s KE equation allows calculating the velocity of fall needed
to, say, double the impact of a given ball. But his equation also
allows calculating the impact velocity necessary for a lighter ball to
impact with the SAME kinetic energy as a heavier one.
If Coriolis’s equation is correct, the KE of the heavier ball can be
set equal to the KE of the lighter ball. For any two same size balls
of known weight, the lighter ball will be some percentage of the
heavier ball‘s weight. By substituting the lesser weight (as a
percentage, if you assume that the heavier ball is ‘unity’ weight)
into Coriolis’s equation, it is easy to solve the equation for the ‘v’
needed to cause the lighter ball to have the same KE as the heavier
ball. That velocity can be converted to the distance of fall—using
accepted equations.
If Coriolis’s equation is “a law of nature”, the size of the holes in
the clay should be identical. However, as I have long suspected,
Coriolis’s equation is wrong. It assumes that the KE is accruing semi-
parabolically. But the UNIFORM force of gravity can only be imparting
KE at a LINEAR rate, not at a semi-parabolic one. Coriolis’s equation
violates the Law of the Conservation of Energy. KE must correlate
exactly to the amount of force that gravity can impart. And the
accruing force from gravity increases LINEARLY.
Today, I ran a simple KE test. I dropped a ¾” dia. chrome steel ball
from a height of 3.3684 feet into a small flower pot full of just-
mixed art clay. That ball sank in close to its ‘equator’. I
immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
heavy fluoroplastic ball, weighing .2807 times as much as the chrome
steel ball), from an exact height of 12 feet. The KE value should be .
10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
16.087 feet, the distance of fall in one second. The time of fall is .
86368 seconds for the lighter ball.
The PTFE ball landed 1” from the chrome steel ball. It sank into the
clay only about .75 as deep. If Coriolis’s equation was correct, both
balls would be imbedded equally. Those two balls are stuck in the
clay. I will let everything air dry to serve to document my
experiment.
The above simple experiment can be run with any two equal size, but
different weight balls. (Ping Pong balls excluded.) Use Coriolis’s
equation to make the KE values for each ball weight equal. If you
have access to a tall building where drops can be made from two
heights as required to satisfy the equations, the results, still,
won’t cause equal size holes in the clay. A semi-parabolic equation,
like Coriolis’s, can never predict impact results when the Law of
Nature is a LINEAR increase in KE with respect to velocity! My
correct equation is: KE = a/g (m) + v/32.174 (m).
The above described experiment is another of my conclusive disproofs
of Coriolis. Since Coriolis’s equation gave Einstein the mistaken
notion that the energy progression in traveling to velocity ‘c’ is
parabolic, BOTH of Einstein’s theories of relativity are disproved,
yet again, by yours truly!
Respectfully submitted,
— NoEinstein —
Eric Gisse
hhc...@yahoo.com
> If two perfectly spherical balls have the same diameter and surface
> smoothness, but are of different weights, when they are dropped they
> will have the same air resistance as they fall. If such balls are
> dropped from an identical height, and both subsequently impact the
> same horizontal bed of soft clay, the heavier ball will penetrate into
> the clay further than the lighter ball.
So far, no great surprise, although you appear to be leaving out most
of the experimental parameters. So you are definitely "No Einstein".
Not even a 1st year physics student would make such a gross error.
For example, were the balls of equal volume and of the same equivalent
density and air resistance, or was one made out of cork and the other
lead?
> The resistance of clay to impact varies widely. The individual
> particles of clay are free to slide over adjacent particles. And the
> resistance to such sliding is determined by the angle of internal
> friction of the particles. That is analogous to the force required to
> slide a given object down an inclined plane. Most know that the
> coefficient of SLIDING friction is less than the coefficient of
> friction at rest. So, the internal friction of clay that’s already
> beginning to slide (internally) is also less than the friction of clay
> particles that are at rest. What that means is: The energy required
> to cause a given depth of penetration doesn’t vary linearly.
No one has claimed the penetration depth varies linearly, simply that
it is identical for all spherical objects with the same dimensions and
of the same density. It you try the same experiement with spheres of
lead and spheres of cork, you will obviously obtain very different
experimental resluts because their kinetic energy, even though dropped
from the same height will be quite different.
>
> For the above reason, a single round ball falling from varying heights
> and impacting soft clay could never accurately correlate to the amount
> of KE present at impact.
Oddly enough, it does bcause Kinetic Energy = 1/2 mv^2.
What you are not taking into account is that it the two balls being
dropped, the less dense one will have not only less mass and kinetic
energy on striking the clay, and less because its decreased mass has
less force to overcome atmospheric drag, so it will impact the clay
with less mass and less velocitiy that the denser sphere.
Most highschool students are taught these simple facts and
computations during their junior year physics courses.
> If any heavy ball is dropped into clay, it will impact with a KE which
> has been assumed to be given by Coriolis’s equation.
WHOA RIGHT HERE -- If you don't know the difference between Coriolis
and basic Newtonian Mechanic, you either ride the "short bus" to
school, or are trolling and pretending to be stupid for some purpose.
This type of physics material is so basic that a highschool student
majoring in physical education, or a college student majoring in law
or business should be able to comprehend it...else they should pursue
a vocation such as digging holes in the ground or flipping burgers.
And if you are of the "Meow" mentality, then you should begin to pay
your own way, and stop expecting good old "mom and dad" to do it for
you perpetualy. Get yourself a local job, and pay your own way. It's
character building.
Harry C.
Sanny
When 2 balls are thrown from height "h". Then K.E is zero initially.
later increases as the ball goes down.
The Total energy that the ball has at height "h" is m*g*h That is the
potential energy at height "h".
When the ball touvhes the ground the K.E == Potential Energy So
Kinetic Energy of the ball when touching the ground is m*g*h
Where m is mass of ball
g is acc due to gravity
h is height from which ball was thrown.
And how deep that ball will peneterate depends on the presure/ unit
area by the ball. Sinch area is same, large mass has larger force So
the massive ball will peneterate more deeper.
Bye
Sanny
Get intelligent: http://www.GetClub.com
hhc...@yahoo.com
Eric, he doen't need data, because his observations are that obviously
flawed. He is simply playing "troll" games, and attempting to make
anyone who know anything about physics appear to be a fool. In
reality, he does precisely the opposite.He has a motive for posting
this crap, but I'm not sure what that motive is, unless his past
semester physics graded are at risk of fluking him out of college (or
prep school), and I suspect that he is worried about "good old mom and
dad" pulling the financial plug supporting him.
I have a reason for posting what I did,which largely relates to my
"Meow" comment...I'll leave you to figure that out,
Take a look at his posting history on Google. If you know how to read
it, that much of the story. Take a look at when his/her posting bursts
take place. Correlate this with the student year. Correlate the two.
Harry C.
Dirk Van de moortel
36da8e96-218b-4757...@j33g2000pri.googlegroups.com
> You are making wrong observation.
>
> When 2 balls are thrown from height "h". Then K.E is zero initially.
> later increases as the ball goes down.
>
> That is the potential energy at height "h".
Yes, but it safer to say it like this:
The Potential Energy that the ball has at height "h" is m g h.
So that is the Total Energy as well, since the Kinetic Energy
is zero.
>
> When the ball touvhes the ground the K.E == Potential Energy
When the ball touches the ground, the Potential Energy is zero.
And the Total Energy is conserved, so...
> So
> Kinetic Energy of the ball when touching the ground is m*g*h
Wrong reasoning, correct conclusion.
You were lucky to have reached that conclusion because you
chose a zero initial velocity and zero height for the ground.
Dirk Vdm
NoEinstein
Dear Eric: The needed numbers and formulas are given in my post. Do
the math, if you can. But since I know you to be a complainer, not a
doer, I don't expect you to do anything but complain. — NoEinstein —
NoEinstein
> On Aug 16, 9:36 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > If two perfectly spherical balls have the same diameter and surface
> > smoothness, but are of different weights, when they are dropped they
> > will have the same air resistance as they fall. If such balls are
> > dropped from an identical height, and both subsequently impact the
> > same horizontal bed of soft clay, the heavier ball will penetrate into
> > the clay further than the lighter ball.
>
> So far, no great surprise, although you appear to be leaving out most
> of the experimental parameters.
>
mv^2 (sub chrome steel) = KE = 1/2 mv^2 (sub PTFE fluoroplastic).
There are no "weights" given, just the factor lighter of the PTFE than
the chrome steel. I determined such factor using a steel yard stick.
I rested the 18" mark of the yard stick on a triangular engineer's
scale acting as a fulcrum. I 'set' the steel ball on the beam at the
4" (from the fulcrum) mark and held such in that spot using a 1/8" I.
D. stamped washer. I placed a similar washer under the PTFE ball. To
counteract the asymmetry of the washers, and the 'key-hole-slot' for
hanging up the yardstick, I approximated the location of the PTFE
ball; removed such, and the steel ball; and used a copper penny to
balance the yardstick, evenly. Then, I repositioned both balls and
repeated the adjustment of "level" by removing both balls, and using
the penny to balance just the two washers and the 'key-hole-slot'. By
the third repeat of this process, I determined that the PTFE ball
needed to be exactly 14.25 inches from the fulcrum to balance the
steel ball, which remained at the 4" mark, throughout. Using simple
moment equivalency, I could exactly determine the relative weight of
the PTFE ball compared to chrome steel. The accuracy of this type
scale will rival any two-beam balance as is used in most university
lab work.
> So you are definitely "No Einstein".
> Not even a 1st year physics student would make such a gross error.
>
Which error is that?
>
> For example, were the balls of equal volume and of the same equivalent
> density and air resistance, or was one made out of cork and the other
> lead?
>
You can't 'read', can you, Harry C.?
>
> > The resistance of clay to impact varies widely. The individual
> > particles of clay are free to slide over adjacent particles. And the
> > resistance to such sliding is determined by the angle of internal
> > friction of the particles. That is analogous to the force required to
> > slide a given object down an inclined plane. Most know that the
> > coefficient of SLIDING friction is less than the coefficient of
> > friction at rest. So, the internal friction of clay that’s already
> > beginning to slide (internally) is also less than the friction of clay
> > particles that are at rest. What that means is: The energy required
> > to cause a given depth of penetration doesn’t vary linearly.
>
> No one has claimed the penetration depth varies linearly, simply that
> it is identical for all spherical objects with the same dimensions and
> of the same density. It you try the same experiement with spheres of
> lead and spheres of cork, you will obviously obtain very different
> experimental resluts because their kinetic energy, even though dropped
> from the same height will be quite different.
>
Harry C.: Balls of cork, Styrofoam, or Ping Pong are RULED by their
air resistance. As comedian Benny Hill liked to say: "Use a bit of
common." If you were doing an impact test, would you use a ball so
light that it would be blown off of course by the slightest breeze?
If such was dropped in a vacuum, those would travel at the same rate
of fall as heavy balls. But since the purpose of the test is to
indent the clay, using light balls makes no sense.
>
> > For the above reason, a single round ball falling from varying heights
> > and impacting soft clay could never accurately correlate to the amount
> > of KE present at impact.
>
> Oddly enough, it does bcause Kinetic Energy = 1/2 mv^2.
>
Harry C.: You are a status-quo-only "scientist". You’re gullible
enough, and air brained enough to accept everything you were ever told
in college. For the record, I have disproved enough of what used to
pass for science to justify FIRING 75% of the supposed physics
professors who have been "teaching" that which they haven't understood
well enough to question.
>
> What you are not taking into account is that it the two balls being
> dropped, the less dense one will have not only less mass and kinetic
> energy on striking the clay, and less because its decreased mass has
> less force to overcome atmospheric drag, so it will impact the clay
> with less mass and less velocitiy that the denser sphere.
>
> Most highschool students are taught these simple facts and
> computations during their junior year physics courses.
>
The impacts are forced to be equal by causing the lighter ball to drop
from the height from which Coriolis's equation "says" that they are
(should be) equal. The total air drag on both balls is identical, so
such effect cancels out on of both sides of the equation.
>
> > If any heavy ball is dropped into clay, it will impact with a KE which
> > has been assumed to be given by Coriolis’s equation.
>
> WHOA RIGHT HERE -- If you don't know the difference between Coriolis
> and basic Newtonian Mechanic, you either ride the "short bus" to
> school, or are trolling and pretending to be stupid for some purpose.
>
Dear Harry C.: I am "honoring" you by replying to your shallow
comments. Your stupidity should make any thinking person go WHOA!
>
> This type of physics material is so basic that a highschool student
> majoring in physical education, or a college student majoring in law
> or business should be able to comprehend it...else they should pursue
> a vocation such as digging holes in the ground or flipping burgers.
> And if you are of the "Meow" mentality, then you should begin to pay
> your own way, and stop expecting good old "mom and dad" to do it for
> you perpetualy. Get yourself a local job, and pay your own way. It's
> character building.
>
You’ve never done a lick of science that I know of. I have asked you,
before, to give links to any post or article that shows that you have
any credentials, whatsoever. In this news group, you are a known
dunce! — NoEinstein —
>
> Harry C.
NoEinstein
Dear Sanny: You know a dusting of physics. But your ideas are
unsupported by either the "traditional wisdom" or by a thinking man's
logic. Get off of the sauce, or give it up, fellow. — NoEinstein —
NoEinstein
Dear Harry C.: You and Eric Gisse are cut from the same mold. You
two deserve each other. — NoEinstein —
NoEinstein
SperM.hotmail.com> wrote:
> Sanny <softta...@hotmail.com> wrote in message
>
> 36da8e96-218b-4757-b337-a597eb50f...@j33g2000pri.googlegroups.com
Dear Dirk: Potential energy is given by my correct equation: "PE = w
+ (h/16.087 ft.) w." The potential energy of all objects is just
their static weight, unless there is a fall distance (waiting for a
pending drop of the object). — NoEinstein —
Dirk Van de moortel
ec420fa5-d7dc-4b1b...@8g2000hse.googlegroups.com
> On Aug 17, 5:12 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> SperM.hotmail.com> wrote:
>> Sanny <softta...@hotmail.com> wrote in message
>>
>> 36da8e96-218b-4757-b337-a597eb50f...@j33g2000pri.googlegroups.com
>>
>>> You are making wrong observation.
>>
>>> When 2 balls are thrown from height "h". Then K.E is zero initially.
>>> later increases as the ball goes down.
>>
>>> The Total energy that the ball has at height "h" is m*g*h.
>>> That is the potential energy at height "h".
>>
>> Yes, but it safer to say it like this:
>> The Potential Energy that the ball has at height "h" is m g h.
>> So that is the Total Energy as well, since the Kinetic Energy
>> is zero.
>>
>>
>>
>>> When the ball touvhes the ground the K.E == Potential Energy
>>
>> When the ball touches the ground, the Potential Energy is zero.
>> And the Total Energy is conserved, so...
>>
>>> So
>>> Kinetic Energy of the ball when touching the ground is m*g*h
>>
>> Wrong reasoning, correct conclusion.
>>
>> You were lucky to have reached that conclusion because you
>> chose a zero initial velocity and zero height for the ground.
>>
>> Dirk Vdm
>
> Dear Dirk:
Go away, imbecile.
Dirk Vdm
Eric Gisse
There's no data in your post since you didn't actually do the
experiment.
Spaceman
Poor Eric sticks his head in the sand and yells,
I can't see you so you are not there!
LOL
--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman
Eric Gisse
wrote:
Where did he say he actually did the experiment, spaceshit?
Spaceman
The 11th paragraph and beyond of the original post.
you know.. the part you never read at all Eric.
:)
You should shove some "spaceshit" in your empty brain cavity.
You might become smarter.
:)
Eric Gisse
I don't consider that data relevant to testing the kinetic energy
formula, as materials of differing composition will have differing
penetration depths. Even you - with your boundless depths of stupidity
- should know that.
This is SIMPLE SHIT that is testable with an air hockey table. Or a
pendulum with photogates. Or a bowling ball, stopwatch, and a small
paved hill.
He is an idiot for writing thousands of words about a simple subject
he doesn't understand, and you are an idiot for....everything you
write.
Spaceman
I partly agree, but it does prove that KE fails
when it comes to "absorption" impacts.
> This is SIMPLE SHIT that is testable with an air hockey table. Or a
> pendulum with photogates. Or a bowling ball, stopwatch, and a small
> paved hill.
If it is simple shit to you, why didn't you say the problem
with the experiment to begin with and tell NoEinstein that KE
is only good for "non elastic" collisions and it never has been
good at "absorbtion" problems.?
> He is an idiot for writing thousands of words about a simple subject
> he doesn't understand, and you are an idiot for....everything you
> write.
Poor Eric,
He could have simply said something like this.
Dear No Einstein,
Your experiment is proving KE fails for elastic impacts.
But that was proven before.
(In fact bullet proof vests prove such each day)
(energy is spread out to different directions and direct impact
energy is lost due to the diversion)
KE does not actually fail for "non elastic" conditions though
So maybe you would like to do such for "bounce" testing.
the bounce will tell how much energy was given back
as KE for each object.
So you may want to re-think the experiment and maybe
you can find a flaw in KE with "non elastic" conditions.
:)
But of course.
Eric is too much of an ignorant arrogant asshole to be polite
or have any actual clue about how to show someone
their errors at all, so instead you pretend an experiment
was not even done yet one was done and it proves very
well that KE fails for elastic impacts due to the
absorbtion/deflection characteristics of the material.
Igor
> If two perfectly spherical balls have the same diameter and surface
> smoothness, but are of different weights, when they are dropped they
> will have the same air resistance as they fall. If such balls are
> dropped from an identical height, and both subsequently impact the
> same horizontal bed of soft clay, the heavier ball will penetrate into
> the clay further than the lighter ball.
How long have you been doing this experiment? Where is your data? It
also looks like you may have made an inaccurate assumption about air
resistance.
http://en.wikipedia.org/wiki/Drag_(physics)
> Since the 17th century, people have observed the varying depths of
> penetration of a given lead shot or canon ball as the height of fall
> is increased. Since Galileo, fall distances could be correlated into
> the falling objects’ velocities. Observations that a ball which falls
> with double the velocity seemed to penetrate four times as far into
> soft clay, led many to suppose that kinetic energy was accruing
> parabolically with respect to velocity.
>
> In 1830, Coriolis modified the former, purely parabolic increase in
> kinetic energy into a semi-parabolic, but still exponential, rate of
> increase. His well known equation is: KE = ½ mv^2. But such has
> never been verified by any means which meets Scientific Method
> standards.
Why would it have to be verified when it is the DEFINITION of kinetic
energy in classical mechanics?
> The resistance of clay to impact varies widely. The individual
> particles of clay are free to slide over adjacent particles. And the
> resistance to such sliding is determined by the angle of internal
> friction of the particles. That is analogous to the force required to
> slide a given object down an inclined plane. Most know that the
> coefficient of SLIDING friction is less than the coefficient of
> friction at rest. So, the internal friction of clay that’s already
> beginning to slide (internally) is also less than the friction of clay
> particles that are at rest. What that means is: The energy required
> to cause a given depth of penetration doesn’t vary linearly.
Maybe you need to explain in a little more depth about how you came to
that conclusion.
> For the above reason, a single round ball falling from varying heights
> and impacting soft clay could never accurately correlate to the amount
> of KE present at impact. But a most simple new impact test which I
> have devised, which uses two same-size, but different weight balls,
> CAN correlate to the KE present! Such experiment is described below:
>
> If any heavy ball is dropped into clay, it will impact with a KE which
> has been assumed to be given by Coriolis’s equation. If KE = ½ mv^2
> is true, then there must be a greater height from which a lighter, but
> same size, ball can be dropped so that the lighter ball will impact
> with the SAME kinetic energy as the heavier ball.
>
> The amazing thing about having two different density, but same size
> balls impacting soft clay with the SAME kinetic energy, is that the
> size of the resulting holes in the clay should be IDENTICAL! Under
> the latter conditions, the consistency and material characteristics of
> the clay have no influence whatsoever on the KE comparison being
> made. *** “If two same size balls of different density impact with
> the same KE, they will make IDENTICAL holes regardless of the type and
> softness of the clay being impacted.” ***
You would probably also need to control for local variations in the
way the clay absorbs the energy upon impact before you can jump to any
conclusions.
> Coriolis’s KE equation allows calculating the velocity of fall needed
> to, say, double the impact of a given ball. But his equation also
> allows calculating the impact velocity necessary for a lighter ball to
> impact with the SAME kinetic energy as a heavier one.
>
> If Coriolis’s equation is correct, the KE of the heavier ball can be
> set equal to the KE of the lighter ball. For any two same size balls
> of known weight, the lighter ball will be some percentage of the
> heavier ball‘s weight. By substituting the lesser weight (as a
> percentage, if you assume that the heavier ball is ‘unity’ weight)
> into Coriolis’s equation, it is easy to solve the equation for the ‘v’
> needed to cause the lighter ball to have the same KE as the heavier
> ball. That velocity can be converted to the distance of fall—using
> accepted equations.
>
> If Coriolis’s equation is “a law of nature”, the size of the holes in
> the clay should be identical. However, as I have long suspected,
> Coriolis’s equation is wrong. It assumes that the KE is accruing semi-
> parabolically. But the UNIFORM force of gravity can only be imparting
> KE at a LINEAR rate, not at a semi-parabolic one.
What do you mean by that? In time or in displacement? Only the
latter would be true.
>Coriolis’s equation
> violates the Law of the Conservation of Energy.
How so? Total energy is conserved, not just kinetic.
>KE must correlate
> exactly to the amount of force that gravity can impart. And the
> accruing force from gravity increases LINEARLY.
What?!!! Gravity force near Earth's surface is a constant. I think
you're confused with the terminology.
> Today, I ran a simple KE test. I dropped a ¾” dia. chrome steel ball
> from a height of 3.3684 feet into a small flower pot full of just-
> mixed art clay. That ball sank in close to its ‘equator’. I
> immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> heavy fluoroplastic ball, weighing .2807 times as much as the chrome
> steel ball), from an exact height of 12 feet. The KE value should be .
> 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> 16.087 feet, the distance of fall in one second. The time of fall is .
> 86368 seconds for the lighter ball.
>
> The PTFE ball landed 1” from the chrome steel ball. It sank into the
> clay only about .75 as deep. If Coriolis’s equation was correct, both
> balls would be imbedded equally.
You keep saying that. Please explain why you think that would be
true? Maybe you need to be a good scientist and start looking for
alternative explanations for your results.
>Those two balls are stuck in the
> clay. I will let everything air dry to serve to document my
> experiment.
> The above simple experiment can be run with any two equal size, but
> different weight balls. (Ping Pong balls excluded.) Use Coriolis’s
> equation to make the KE values for each ball weight equal. If you
> have access to a tall building where drops can be made from two
> heights as required to satisfy the equations, the results, still,
> won’t cause equal size holes in the clay. A semi-parabolic equation,
> like Coriolis’s, can never predict impact results when the Law of
> Nature is a LINEAR increase in KE with respect to velocity! My
> correct equation is: KE = a/g (m) + v/32.174 (m).
Nobody understands that. Define your terms.
> The above described experiment is another of my conclusive disproofs
> of Coriolis.
Hardly. Maybe you need to learn about experimental protocols.
>Since Coriolis’s equation gave Einstein the mistaken
> notion that the energy progression in traveling to velocity ‘c’ is
> parabolic, BOTH of Einstein’s theories of relativity are disproved,
> yet again, by yours truly!
And now, as you take a swift turn to the left and drive off of the
road completely, I guess I will bid you goodbye.
Ray Vickson
I don't believe you. Did you really measure the height (above
something) to approximately half the diameter of a human hair? (See
http://hypertextbook.com/facts/1999/BrianLey.shtml ) How did you
manage it? Are you measuring the height above a pot of art clay? Was
the surface of the clay so smooth that there were no bumps or
imperfections of the size of a human hair's diameter?
> That ball sank in close to its ‘equator’. I
> immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> heavy fluoroplastic ball, weighing .2807
What measuring instruments did you use to attain such accuracy? In a
well-equipped lab it can be done, but it is not easily done in a
typical household.
> times as much as the chrome
> steel ball), from an exact height of 12 feet. The KE value should be .
> 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> 16.087 feet, the distance of fall in one second.
The value of g varies from place to place on earth (partly due to
rotational effects, altitude and the like); the variations can be
larger than the so-called accuracy in your figures; see
http://en.wikipedia.org/wiki/Earth's_gravity .
> The time of fall is .
> 86368 seconds for the lighter ball.
Did you really time the fall to that kind of accuracy? How did you
manage?
My suspicion is that you did not really perform an experiment to
anything like the accuracy you claim. Even if you did manage to do so,
the experiment would not be showing what you claim, as others have
already explained to you.
R.G. Vickson
NoEinstein
SperM.hotmail.com> wrote:
> NoEinstein <noeinst...@bellsouth.net> wrote in message
>
> ec420fa5-d7dc-4b1b-bbba-31940ccee...@8g2000hse.googlegroups.com
>
> - Show quoted text -
Dear Dirk: Learning that one is wrong isn't kind to the ego. Take it
like a man. You have some ideas which are right on. — NoEinstein —
NoEinstein
Dear Eric: You are like a child who is "it" playing hide-and-seek.
But instead of running and hiding, all you do is to put your hands
over your own eyes. Because you can't SEE anything, you assume that
no one can see you. You're as deluded as they come.
My experiment invites anyone with two equal sized, but different
weight balls to drop those into soft clay from the distances required
by Coriolis's formula: KE = 1/2 mv^2. In no case will those "same KE"
balls make the same hole or dent in the clay. Coriolis's own formula
disproves his own formula. Is that 'poetic justice' or what! —
NoEinstein —
NoEinstein
wrote:
Dear Spaceman: "Great minds run in the same channels!" :-) —
NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear Eric: I have photographed my experiment. AND I have the clay-
filled flower pot with the chrome steel ball (dropped from 3'-4")
embedded nearly to its 'equator'; and the PTFE ball (dropped from 12')
embedded about 1/8" less. Both balls are 3/4" in diameter and
perfectly spherical. Claim that I haven't done the experiment, and
you only show yourself to be an ignorant fool, Eric. — NoEinstein —
Eric Gisse
How in the hell do you think dropping stuff into _clay_ is a relevant
test of kinetic energy? You didn't run _any_ experiment that is an
actual test of kinetic energy. Open up a freshman college physics
textbook and see how its' *really* done.
Yea, I'll keep claiming you haven't done the experiment cuz you aren't
doing anything relevant to your whining. There isn't a "distance" in
the kinetic energy formula, its' just a measure of _energy_ - it will
NOT tell you how far materials will penetrate.
How do you think that the kinetic
Eric Gisse
Really, where is the "distance required" ???
Regardless the masses were different, stupid. They will penetrate
differently.
PD
[an unbelievable amount of clap trap]
Coriolis made his measurements this way because he did not have access
to a strobe lamp and a camera. Had he had these available, he would
not have needed to measure dents in the ground at all. He could simply
have noted that balls dropped from heights in the ratio 2 acquire
velocities that are in the ratio of square-root of 2, by direct
measurement of distance covered between strobe flashes. Furthermore,
he could have noted that balls with masses in the ratio of 2 landing
on a teeter-totter send a projectile on the other end of the teeter-
totter to heights with a ratio of 2.
In so doing, he would have demonstrated that the kinetic energy
acquired in the fall is proportional to the mass and proportional to
the square of the velocity, by simple and direct measurement.
Now, the fact that Coriolis did not have these measurement tools
available to him and he was *still* able to ascertain the right answer
is remarkable. However, critiquing his cruder methods does not change
the validity of his conclusion at all, especially since the more
precise methods just described DO confirm the conclusion spectacularly
and with much higher precision than Coriolis was able to achieve.
NoEinstein, a basic rule of thumb that may not have occurred to you.
Every seminal experimental is later corroborated by a string of
independent experimental findings, most of which are improvements on
the original and which provide even stronger evidence than the seminal
experiment did. It is thus not sufficient to reproduce the initial
experiment and show that it is flawed. You have to reproduce *all the
successive experiments* and *also* show that they are flawed.
PD
Ray Vickson
You obviously don't get it. Anyone can claim anything; supplying proof
is another matter entirely. Eric is a fool because you choose not to
present proof that you have done the experiment? I don't think so.
After you present the proof you can expect people to, maybe, believe
you, but not until then. Anyway, as has been explained to you many
times, your experiment does not prove what you think it does
(assuming, of course, that you actually DID the experiment).
R.G. Vickson
PD
>
> Today, I ran a simple KE test. I dropped a ¾” dia. chrome steel ball
> from a height of 3.3684 feet into a small flower pot full of just-
> mixed art clay. That ball sank in close to its ‘equator’. I
> immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> heavy fluoroplastic ball, weighing .2807 times as much as the chrome
> steel ball), from an exact height of 12 feet. The KE value should be .
> 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> 16.087 feet, the distance of fall in one second. The time of fall is .
> 86368 seconds for the lighter ball.
>
> The PTFE ball landed 1” from the chrome steel ball. It sank into the
> clay only about .75 as deep. If Coriolis’s equation was correct, both
> balls would be imbedded equally. Those two balls are stuck in the
> clay. I will let everything air dry to serve to document my
> experiment.
>
> The above simple experiment can be run with any two equal size, but
> different weight balls. (Ping Pong balls excluded.) Use Coriolis’s
> equation to make the KE values for each ball weight equal. If you
> have access to a tall building where drops can be made from two
> heights as required to satisfy the equations, the results, still,
> won’t cause equal size holes in the clay. A semi-parabolic equation,
> like Coriolis’s, can never predict impact results when the Law of
> Nature is a LINEAR increase in KE with respect to velocity! My
> correct equation is: KE = a/g (m) + v/32.174 (m).
>
Now, perhaps, you are getting a small dose of the critical review that
publishing experimental results will have to endure, NoEinstein. As
others here have pointed out, there are a number of things that you
document here that are obviously questionable. A published
experimental finding has to be polished and more carefully prepared so
that it will endure this kind of examination.
If you don't want to do that, then you have the wrong hobby.
PD
Androcles
"Ray Vickson" <RGVi...@shaw.ca> wrote in message
news:8537cad2-e357-4c9c...@u6g2000prc.googlegroups.com...
============================================
Good observation, Vickson.
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
Anyone, even the Lord God Einstein, can claim anything; supplying proof
is another matter entirely.
You are a really hypocritical fuckhead, aren't you?
NoEinstein
>
> - Show quoted text -
Dear Eric: If you had been as fired up in college, perhaps you would
have graduated. The "distance" in the KE is a simple conversion of
the velocity numbers from Coriolis's formula. At a specific distance
of fall the velocity will match that required by his equation. Can
you find such a formula in one of your now useless physics texts?
Probably not. For you, bitching is easier than using your limited
brain power. — NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear PD: You are more impressed by the shine on the print paper than
you are by new scientific truths. You should try chasing your own
tail. Evolution never caused yours to disappear. — NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear Ray: My post explains a $30.00 experiment which can be verified
by anyone. Claiming that I haven't done the experiment borders of
libel. Tell me, fellow, why is it that you come out of nowhere and
get your ego so hurt by my accomplishments that you must resort to
libel? — NoEinstein —
NoEinstein
wrote:
>
Dear Spaceman: My background is structural engineering. I understand
issues of elasticity very well. The modulus of elasticity of PTFE,
obviously, is less than that of chrome steel. But both of those balls
STUCK into the clay. The PTFE ball didn’t “rebound” out. The
effective kinetic energy of both balls got converted to force of
impact, which, according to Coriolis’s own formula, should have been
exactly equal for both balls. And the entire energy (except for the
frictional heating of the balls) got transferred into the clay.
Teflon has a very low coefficient of friction—less than steel. Also,
the thermal conductance of Teflon is less than that of steel. So,
Teflon will transfer a higher percentage of its impact force into the
clay than will steel. Even with that higher energy input, the Teflon
ball didn’t penetrate as far. The reason isn’t ‘petty’ issues of
material elasticity, it’s the fact that Coriolis’s formula is semi-
parabolic, when the ‘gravity’ force input (KE input) is accruing
LINEARLY. Arguing over minutia should be replaced by the simple
acknowledgement that Coriolis’s formula violates the Law of the
Conservation of energy. A semi-parabolic equation can NEVER be
correct when the force-KE that’s available accrues LINEARLY. —
NoEinstein —
> Spaceman- Hide quoted text -
NoEinstein
>
“How long have you been doing this experiment? Where is your data? It
also looks like you
may have made an inaccurate assumption about air resistance.”
Dear Igor2: You are a persona non grata. But because you ask
questions that others might like to hear answered, I will respond. I
couldn’t care less whether you understand, or pass on what I teach.
In the past two and a half years I have investigated about six
different methods of measuring KE effects. One of those involved
dropping a small clevis pin head-to-head with the next size larger
pin. There is a height of drop, correctly predicted by my: KE = a/g
(m) + v/32.174 (m), from which the KE of the smaller pin will exactly
match the INERTIA of the larger pin. When that happens the “ringing”
sound of the smaller pin gets dampened before the pin has time to
bounce away. By using voice analysis software, I was able to show the
sound wave patterns (visible in audio spectrographs) corresponding to
drop heights in 2” increments.
If you would just read (rather than skim) my present post, every
parameter for duplicating my experiment is given.
The accepted method for determining air resistance is to mount the
test object on a small, stiff wire, in a wind tunnel. An electronic
scale measures the drag that the object has. Another object on the
same size wire can be compared for air resistance over the speed range
desired. Since both my chrome steel and PTFE balls are the same size,
the air resistance will be identical. The steel ball will deliver
more force, after subtracting the drag, but since both balls subtract
identical amounts, then the effect of air drag can be disregarded.
“Why would it (KE = ½ mv^2) have to be verified when it is the
DEFINITION of kinetic
energy in classical mechanics?”
About 25% of what is shown in physics texts is wrong. Many have known
that things are wrong. But because colleagues would be embarrassed to
be shown-up, the errors are just left in. The textbook publishers
(Jews) love it, because the books keep getting thicker, and more
expensive. And the trusting students are none the wiser.
“Maybe you need to explain in a little more depth about how you came
to that conclusion.”
Lazy Igor2 needs to take the time to read, and learn to comprehend
what he reads.
“You would probably also need to control for local variations in the
way the clay absorbs the energy upon impact before you can jump to any
conclusions.”
The clay is made (squeezed and re squeezed) to be as homogeneous as
possible. Also, the art clay I used is more uniform in texture than
was probably available in the 17th century, when the first round balls
were dropped into clay. While ’local’ variations are possible, I’m
not so much “selling” my results as the last word, as I am inviting
other scientists, anywhere, to do their own similar dropped ball
experiments. The materials cost less than $30.00, and the experiment
took just an afternoon. But it’s still conclusive!
“What do you mean by that? In time or in displacement? Only the latter
would be true.”
KE is imparted LINEARLY. And KE increases LINEARLY with respect to
the TIME that the object falls. The “new physics” which I have added,
is to realize that all objects at rest have a KE = w. My equation: KE
= a/g (m) + v/32.174 (m), is identical to the traditional equation for
MOMENTUM, except for the addition of static weight. And the equation
is apt only for “continuous thrust” accelerations. If an object is
traveling at a uniform velocity, the momentum equation of old would be
the correct one to use. In no case is KE a function of the distance
of fall! The reason that is so is because the great majority of fall
distance is the COASTING carryover distance from the previous second.
Deduct COASTING, and all falling objects will add 16.087 feet more
fall distance than in the previous second. Such is a LINEAR increase
in velocity, and thus in KE.
“How so? Total energy is conserved, not just kinetic.”
The energy that’s THERE must match, exactly, the energy that is, or
can be, imparted. KE accrues in direct proportion to the falling
object’s velocity. The velocity is the SLOPE of the inverted “free
drop” parabola. And such slope varies LINEARLY.
“What?!!! Gravity force near Earth's surface is a constant. I think
you're confused with the terminology.”
Fellow, a constant “force” of gravity (equal to every falling object’s
static weight) near the Earth’s surface is what causes both the
velocity and the KE to be increasing LINERALY. It is YOU who are
confusing terminology!
“You keep saying that. Please explain why you think that would be
true? Maybe you need to be a good scientist and start looking for
alternative explanations for your results.”
Just READ what I wrote. Until you can read, and take the time to do
so, you won’t understand very much. If you are a “good
scientist” (ha!) do one of three things: conduct your own dropped ball
experiment, such as I have described; give a definitive alternative
explanation for the results of my experiment; or… just shut up. The
latter seems most apt, because your mouth tends to precede your brain.
“Nobody understands that. [KE = a/g (m) + v/32.174 (m)] Define your
terms.”
Anyone who has ever taken a course in physics knows what everything
means. The “nobody”… who can’t understand is you.
“Hardly. Maybe you need to learn about experimental protocols.”
If you were as good at thinking as you are with verbiage, you wouldn’t
be so consistently CLUELESS.
“And now, as you take a swift turn to the left and drive off of the
road completely, I guess I will bid you goodbye.”
Goodbye, and good riddance, Igor2! — NoEinstein —
> road completely, I guess I will bid you goodbye.- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
Eric Gisse
You are confused. Your clay experiments are not a concrete model - get
an air hockey table and some pucks, then start playing around. You see
exactly what you want to see because shooting into a material
substance is a messy process, which you do not understand. Hell, I
don't even know how to model it.
Pick up a goddamn pendulum with a weight on the end, and do simple
verifications of the work-energy theorem, or ask someone who does
crash reconstruction. A lot of very, very simple devices use the
formula you think is wrong. You are arguing about high school physics
- poorly. Your experiment is irrelevant to the discussion since you
don't know what you are doing.
Eric Gisse
What are you doing to make these "new scientific truths" known?
PD
It would do you well to answer some of the questions that have been
directed to you about your purported precision on some of the numbers
you claimed in your experimental write-up. Rather than desperately
trying to deflect the questions by calling people names.
You told me once that you are very good at explaining things. Yet when
asked questions that give you an opportunity to explain, you demur.
Why?
PD
Puppet_Sock
[snip]
> Today, I ran a simple KE test. I dropped a ¾” dia. chrome steel ball
> from a height of 3.3684 feet into a small flower pot full of just-
> mixed artclay.
You expect people to believe you measured the height you
dropped this ball from to five significant digits?
Clearly you are some mixture of stupid, a liar, a troll, and insane.
> That ball sank in close to its ‘equator’.
And you then expect people to accept "close to" as a measure
of penetration.
Clearly you are some mixture of stupid, a liar, a troll, and insane.
> I immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> heavy fluoroplastic ball, weighing .2807 times as much as the chrome
> steel ball), from an exact height of 12 feet. The KE value should be .
> 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> 16.087 feet, the distance of fall in one second. The time of fall is .
> 86368 seconds for the lighter ball.
It would appear to be all of the above. You have reported 8
significant
digits for KE, but with no units. And you expect people to believe you
have *measured* the time of fall of this ball to five significant
digits.
At some point you should consider returning to public school
and completing your arithmetic lessons. You are a clue free zone.
Socks
NoEinstein
>
> - Show quoted text -
Dear Eric: Much of your compensation for your huge inferiority
complex hinges on selecting the person (me) with the most going for
them, and then trying to elevate your low self-esteem by berating
everything that I do. Long ago that scenario became very clear to any
reader of sci.physics. If such wasn't the case, you wouldn't give a
tinker's damn about anything I say or do. You are a clueless groupie
of mine, who has never made a positive contribution in the world.
The sad thing is that you have been exposed to enough of the
vernacular of science to "bluff" naive readers that you might actually
know something. But I don't have to pass anything I say off of you
for verification. Your high and mighty pretenses are just symptoms of
a severe mental illness. Unless you can get your parents to have you
committed for treatment, your future prospects are dim, indeed. I
don't hate you, I just feel very sorry for you. — NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear Eric: If you could read and comprehend, you wouldn't have to
ask. Sad... very sad. — NoEinstein —
NoEinstein
Dear PD: You have the prerogative of doing the six grade math
necessary to solve for KE. Unless you do, I'll not to be badgered to
keep providing you with more proofs of this or that. Your mental
illness, like Eric Gisse's, is that you have an over-compensated
inferiority complex. Since you don't like, or agree with anything I
say, just go away. But since "shooting me down" is your primary
recreation, I won't expect a CHILD like you to give up your play. —
NoEinstein —
NoEinstein
> On Aug 16, 9:36 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
> [snip]
>
> > Today, I ran a simple KE test. I dropped a ¾” dia. chrome steel ball
> > from a height of 3.3684 feet into a small flower pot full of just-
> > mixed artclay.
>
> You expect people to believe you measured the height you
> dropped this ball from to five significant digits?
>
"solution" to setting 1/2 mv^2 (sub chrome steel) = KE = 1/2 mv^2 (sub
PTFE fluoroplastic). The height of drop is found by determining the
TIME necessary for the falling objects to reach the "solution"
velocity. Then, the distance of fall necessary to reach such velocity
can be found by squaring the TIME (in seconds) and multiplying the
result by 16.087 feet of fall (in one second). The accuracy of my
fall distances was 1/16" plus or minus for the steel ball, and 1/4"
plus for the PTFE ball. I intentionally made the fall distance of the
lighter ball greater to prevent any nit pickers from claining that
that 1/4" (less) had caused the greatly reduced penetrastion into the
clay.
For the record, the smaller ball would have had to be dropped from a
height of about 250 feet in order to make the same dent in the clay.
12 feet of fall, as had satisfied Coriolis's equation doesn't even
come close to causing the KE needed.
>
> Clearly you are some mixture of stupid, a liar, a troll, and insane.
>
I'll let history decide whether you are I come closer to your
"mixture".
>
> > That ball sank in close to its ‘equator’.
>
> And you then expect people to accept "close to" as a measure
> of penetration.
>
The clay is still soft. When it has dried hard, I will measure the
penetration. I wouldn't want anyone to accuse me of pressing that
steel ball deeper when I made the measurements.
>
> Clearly you are some mixture of stupid, a liar, a troll, and insane.
>
Clearly, you are in an echo chamber!
>
> > I immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> > heavy fluoroplastic ball, weighing .2807 times as much as the chrome
> > steel ball), from an exact height of 12 feet. The KE value should be .
> > 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> > 16.087 feet, the distance of fall in one second. The time of fall is .
> > 86368 seconds for the lighter ball.
>
> It would appear to be all of the above. You have reported 8
> significant
> digits for KE, but with no units. And you expect people to believe you
> have *measured* the time of fall of this ball to five significant
> digits.
>
Units are unnecessary when one is comparing the relative penetration.
My "weights" were determined using a balance beam. I could have taken
the specific gravities of the balls and used that proportion. But
different materials have ranges of specific gravities. My balance
beam solution is very accurate, probably to 100th of a gram.
PD
That's not what I'm asking about. You said you did the following:
====================================
I dropped a ¾” dia. chrome steel ball
from a height of 3.3684 feet into a small flower pot full of just-
mixed art clay. ... The time of fall is .86368 seconds for the
lighter ball.
====================================
I want to know how you placed a ball to be at a height of 3.3684 feet
above the clay to a precision of 1 part in 33000.
I want to know how you measured the time of fall to a precision of 1
part in 85000.
PD
> On Aug 19, 2:53 pm, Puppet_Sock <puppet_s...@hotmail.com> wrote:> On Aug 16, 9:36 pm,
NoEinstein <noeinst...@bellsouth.net> wrote:
> > [snip]
>
> > > Today, I ran a simple KE test. I dropped a ¾” dia. chrome steel ball
> > > from a height of 3.3684 feet into a small flower pot full of just-
> > > mixed artclay.
>
> > You expect people to believe you measured the height you
> > dropped this ball from to five significant digits?
>
> Dear Puppet S.: The five significant digits is the rounded-off
> "solution" to setting 1/2 mv^2 (sub chrome steel) = KE = 1/2 mv^2 (sub
> PTFE fluoroplastic). The height of drop is found by determining the
> TIME necessary for the falling objects to reach the "solution"
> velocity. Then, the distance of fall necessary to reach such velocity
> can be found by squaring the TIME (in seconds) and multiplying the
> result by 16.087 feet of fall (in one second). The accuracy of my
> fall distances was 1/16" plus or minus for the steel ball, and 1/4"
> plus for the PTFE ball.
Ah, so you didn't drop the steel ball from a height of 3.3684 feet,
because your precision
was only 1 part in 500 and not 1 part in 33000.
You DO know how to quote measured values to appropriate precision with
an appropriate
uncertainty, don't you? Misquoting numbers is one reason you'll get
comments for revision
on an experimental paper.
So given your experimental uncertainty in the position of the drops
(and in the masses of
the ball, by the way), what is your percentage uncertainty in knowing
that the kinetic
energies are the same? Is it 0.2% or is it 5% or is it 25%? You DO
know how to calculate
that, don't you?
> I intentionally made the fall distance of the
> lighter ball greater to prevent any nit pickers from claining that
> that 1/4" (less) had caused the greatly reduced penetrastion into the
> clay.
>
> For the record, the smaller ball would have had to be dropped from a
> height of about 250 feet in order to make the same dent in the clay.
What's your experimental evidence for that?
> 12 feet of fall, as had satisfied Coriolis's equation doesn't even
> come close to causing the KE needed.
>
> > Clearly you are some mixture of stupid, a liar, a troll, and insane.
>
> I'll let history decide whether you are I come closer to your
> "mixture".
>
> > > That ball sank in close to its ‘equator’.
>
> > And you then expect people to accept "close to" as a measure
> > of penetration.
>
> The clay is still soft. When it has dried hard, I will measure the
> penetration. I wouldn't want anyone to accuse me of pressing that
> steel ball deeper when I made the measurements.
Doesn't help document the precision with which you placed the balls'
heights initially,
does it?
Doesn't help document the masses of the balls, does it?
>
> > Clearly you are some mixture of stupid, a liar, a troll, and insane.
>
> Clearly, you are in an echo chamber!
>
> > > I immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> > > heavy fluoroplastic ball, weighing .2807 times as much as the chrome
> > > steel ball), from an exact height of 12 feet. The KE value should be .
> > > 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> > > 16.087 feet, the distance of fall in one second. The time of fall is .
> > > 86368 seconds for the lighter ball.
>
> > It would appear to be all of the above. You have reported 8
> > significant
> > digits for KE, but with no units. And you expect people to believe you
> > have *measured* the time of fall of this ball to five significant
> > digits.
>
> Units are unnecessary when one is comparing the relative penetration.
> My "weights" were determined using a balance beam.
With what precision? What were the *measured* masses and mass
uncertainties for the two
balls?
All we're trying to do is get you to report your results the way a 7th
grader is taught to
do it.
> I could have taken
> the specific gravities of the balls and used that proportion. But
> different materials have ranges of specific gravities. My balance
> beam solution is very accurate, probably to 100th of a gram.
Probably? You need to either calibrate to be sure (and describe your
calibration
procedure) or determine the manufacturer's spec.
Darwin123
Don't bother looking at it. He starts off with absurd physical
assumptions.
1) He actually thinks that the formula for kinetic energy is derived
from deformation of materials, not from the conservation of energy. As
most of us know, the derivation of conservation of energy using
Newtonian physics automatically brings about:
KE=0.5mv^2
The v^2 factor comes from an integration
2) He ignores everything about material properties discovered after
Isaac Newton. For example: He ignores the fact that his clay has a
plastic deformation limit.
3) etc.
NoEinstein
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Dear PD: You are a persona non grata. Get help, fellow. —
NoEinstein —
NoEinstein
NoEinstein
Dear Darwin 123: Who the hell are you? The premise of COE is: Energy
out = Energy in. 1/2 mv^2 is an exponential increase in energy. But
the uniform "thrust'' of gravity—equal to the static weight of every
object—can only cause KE to accrue LINEARLY. Therefore, since Energy
out (according to Coriolis) is greater than the energy being imparted
by the force of gravity, then Coriolis's formula, and Einstein's SR,
which used such as it's basic, both violate the Law of the
Conservation of energy, and are thus disproved by yours truly.
Scientific truths trump 'smoke and mirrors' shots-from-the-hip every
time. — NoEinstein —
NoEinstein
Dear PD: You are a persona non grata. Get help, fellow. —
NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear PD: You are a persona non grata. Get help, fellow. —
NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear Eric: You are a persona non grata. Get help, fellow. —
NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear Eric: If you could READ, you would know. The likes of you and
PD are "the static" to getting my scientific truths known. —
NoEinstein —
NoEinstein
Dear PD: You are a persona non grata. Get help, fellow. —
NoEinstein —
NoEinstein
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Dear PD: You are a persona non grata. Get help, fellow. —
NoEinstein —
NoEinstein
NoEinstein
Dear Darwin123: I'm making you a persona non grata, too. Examine
your motives. Science is more important than personal attacks not
grounded in science truths. — NoEinstein —
Jerry
> If two perfectly spherical balls have the same diameter and surface
> smoothness, but are of different weights, when they are dropped they
> will have the same air resistance as they fall. If such balls are
> dropped from an identical height, and both subsequently impact the
> same horizontal bed of soft clay, the heavier ball will penetrate into
> the clay further than the lighter ball.
>
> penetration of a given lead shot or canon ball as the height of fall
> is increased. Since Galileo, fall distances could be correlated into
> the falling objects’ velocities. Observations that a ball which falls
> with double the velocity seemed to penetrate four times as far into
> soft clay, led many to suppose that kinetic energy was accruing
> parabolically with respect to velocity.
>
> In 1830, Coriolis modified the former, purely parabolic increase in
> kinetic energy into a semi-parabolic, but still exponential, rate of
> increase. His well known equation is: KE = ½ mv^2. But such has
> never been verified by any means which meets Scientific Method
> standards.
>
> particles of clay are free to slide over adjacent particles. And the
> resistance to such sliding is determined by the angle of internal
> friction of the particles. That is analogous to the force required to
> slide a given object down an inclined plane. Most know that the
> coefficient of SLIDING friction is less than the coefficient of
> friction at rest. So, the internal friction of clay that’s already
> beginning to slide (internally) is also less than the friction of clay
> particles that are at rest. What that means is: The energy required
> to cause a given depth of penetration doesn’t vary linearly.
>
> and impacting soft clay could never accurately correlate to the amount
> of KE present at impact. But a most simple new impact test which I
> have devised, which uses two same-size, but different weight balls,
> CAN correlate to the KE present! Such experiment is described below:
>
> If any heavy ball is dropped into clay, it will impact with a KE which
> has been assumed to be given by Coriolis’s equation. If KE = ½ mv^2
> is true, then there must be a greater height from which a lighter, but
> same size, ball can be dropped so that the lighter ball will impact
> with the SAME kinetic energy as the heavier ball.
>
> The amazing thing about having two different density, but same size
> balls impacting soft clay with the SAME kinetic energy, is that the
> size of the resulting holes in the clay should be IDENTICAL! Under
> the latter conditions, the consistency and material characteristics of
> the clay have no influence whatsoever on the KE comparison being
> made. *** “If two same size balls of different density impact with
> the same KE, they will make IDENTICAL holes regardless of the type and
> softness of the clay being impacted.” ***
>
> to, say, double the impact of a given ball. But his equation also
> allows calculating the impact velocity necessary for a lighter ball to
> impact with the SAME kinetic energy as a heavier one.
>
> If Coriolis’s equation is correct, the KE of the heavier ball can be
> set equal to the KE of the lighter ball. For any two same size balls
> of known weight, the lighter ball will be some percentage of the
> heavier ball‘s weight. By substituting the lesser weight (as a
> percentage, if you assume that the heavier ball is ‘unity’ weight)
> into Coriolis’s equation, it is easy to solve the equation for the ‘v’
> needed to cause the lighter ball to have the same KE as the heavier
> ball. That velocity can be converted to the distance of fall—using
> accepted equations.
>
> If Coriolis’s equation is “a law of nature”, the size of the holes in
> the clay should be identical. However, as I have long suspected,
> Coriolis’s equation is wrong. It assumes that the KE is accruing semi-
> parabolically. But the UNIFORM force of gravity can only be imparting
> violates the Law of the Conservation of Energy. KE must correlate
> exactly to the amount of force that gravity can impart. And the
> accruing force from gravity increases LINEARLY.
>
> from a height of 3.3684 feet into a small flower pot full of just-
> immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> heavy fluoroplastic ball, weighing .2807 times as much as the chrome
> steel ball), from an exact height of 12 feet. The KE value should be .
> 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> 16.087 feet, the distance of fall in one second. The time of fall is .
> 86368 seconds for the lighter ball.
>
> clay only about .75 as deep. If Coriolis’s equation was correct, both
> balls would be imbedded equally. Those two balls are stuck in the
> clay. I will let everything air dry to serve to document my
> experiment.
>
> The above simple experiment can be run with any two equal size, but
> different weight balls. (Ping Pong balls excluded.) Use Coriolis’s
> equation to make the KE values for each ball weight equal. If you
> have access to a tall building where drops can be made from two
> heights as required to satisfy the equations, the results, still,
> won’t cause equal size holes in the clay. A semi-parabolic equation,
> like Coriolis’s, can never predict impact results when the Law of
> Nature is a LINEAR increase in KE with respect to velocity! My
> correct equation is: KE = a/g (m) + v/32.174 (m).
>
> notion that the energy progression in traveling to velocity ‘c’ is
> parabolic, BOTH of Einstein’s theories of relativity are disproved,
> yet again, by yours truly!
>
>
> — NoEinstein —
The relationship K.E. = 1/2 m v^2 is NOT an empirical formula,
and is thus NOT subject to empirical test under "ordinary" (i.e.
non-relativistic) conditions. It is an inevitable mathematical
consequence when the DEFINITION of force is combined with the
mechanical DEFINITION of work. In other words, it is a simple
consequence of the following two expressions taken together:
F = dp/dt
dW = F dot ds
Unfortunately, NoEinstein, you are obviously lacking in basic
calculus, so rather than present a general proof of the kinetic
energy formula, I will need to present to you a somewhat less
than general heuristic demonstration.
-----------------------------------------------------------------
Heuristic Demonstration of the Kinetic Energy formula
-----------------------------------------------------------------
We need to answer the question, "What is the relationship between
the velocity achieved by a mass m that has fallen a distance h
under constant acceleration g, starting from zero velocity?" and
will combine the answer to this question with the conventional
definition of Potential Energy and the principle of Conservation
of Energy, assuming conditions where relativistic corrections are
not necessary.
By definition, P.E. = mgh (assuming constant g)
Assuming a start from zero velocity, the kinetic energy after
falling the distance h must equal the potential energy of the
mass before falling. In other words,
K.E.final = P.E.initial
The distance traveled under constant acceleration g is given by
h = 1/2 g t^2
Rearranging,
t = sqrt(2h/g)
The velocity reached after time t is
v = gt = g sqrt(2h/g)
Square both sides
v^2 = 2gh
Divide both sides by 2
1/2 v^2 = gh
Multiply both sides by m
1/2 m v^2 = mgh
K.E.final = P.E.initial = 1/2 m v^2
Quod Erat Demonstratum
-------------------------------------------------------------
You wrote, "the Law of Nature is a LINEAR increase in KE with
respect to velocity!" A quantity that increases linearly with
velocity is is already well known as "momentum", and your
experimental results indicate nothing more than than momentum
correlates fairly well with penetration depth.
-------------------------------------------------------------
Jerry
Jerry
TYPO:
> We need to answer the question, "What is the relationship between
> the velocity achieved by a mass m that has fallen a distance h
> under constant acceleration g, starting from zero velocity?" and
> will combine the answer to this question with the conventional
> definition of Potential Energy and the principle of Conservation
> of Energy, assuming conditions where relativistic corrections are
> not necessary.
SHOULD READ:
We need to answer the question, "What is
the velocity achieved by a mass m that has fallen a distance h
under constant acceleration g, starting from zero velocity?" and
will combine the answer to this question with the conventional
definition of Potential Energy and the principle of Conservation
of Energy, assuming conditions where relativistic corrections are
not necessary.
TYPO:
> You wrote, "the Law of Nature is a LINEAR increase in KE with
> respect to velocity!" A quantity that increases linearly with
> velocity is is already well known as "momentum", and your
> experimental results indicate nothing more than than momentum
> correlates fairly well with penetration depth.
SHOULD READ:
You wrote, "the Law of Nature is a LINEAR increase in KE with
respect to velocity!" A quantity that increases linearly with
velocity is is already well known as "momentum", and your
experimental results indicate nothing more than that momentum
correlates fairly well with penetration depth.
Jerry
PD
I get the feeling that NoEinstein has spent most of his tortuous life
making people non grata.
PD
PD
This is the coward's way.
The coward purports to put forward an idea or an experimental result
and then hides from the examination of the work. It is even more
cowardly to put the onus of that behavior on others.
PD
PD
Do you imagine that posting on a usenet group is the way to get your
"scientific truths" known?
You'd have better luck printing out leaflets and passing them out on
street corners.
PD
PD
> NoEinstein —
How shocking that you will not answer questions about your work.
PD
Darwin123
> On Aug 17, 2:38 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Aug 17, 8:42 am, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > > On Aug 16, 10:42 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > > On Aug 16, 5:36 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
> > > > [snip]
>
> > > > Where's the data?
>
> > > the math, if you can. But since I know you to be a complainer, not a
> > > doer, I don't expect you to do anything but complain. — NoEinstein —
>
> > There's no data in your post since you didn't actually do the
> > experiment.
>
> But instead of running and hiding, all you do is to put your hands
> over your own eyes. Because you can't SEE anything, you assume that
> no one can see you. You're as deluded as they come.
>
> My experiment invites anyone with two equal sized, but different
> weight balls to drop those into soft clay from the distances required
> by Coriolis's formula: KE = 1/2 mv^2. In no case will those "same KE"
> balls make the same hole or dent in the clay. Coriolis's own formula
> disproves his own formula. Is that 'poetic justice' or what! —
> NoEinstein —
I acknowledge the results, I just don't believe your
interpretation of them. Your conclusions are based on erroneous
assumptions.
Kinetic energy is not defined in terms of how deep and impression
it makes in clay. That is not how kinetic energy is defined in any
twentieth century textbooks. Given what I know about physics, I
wouldn't expect the impressions to be the precisely the same.
Basically, your argument is one for Aristotlean physics. Aristotle
made the claim that the heavier the object, the faster is falls. The
experiment "proving" this is the case is dropping a feather and a rock
through air. Because of air resistance, the rock falls faster.
However, if you drop two rocks of different weights and densities, as
long as both densities are high, the rocks will drop at the same rate.
Your observation is analogous to making the observation that a feather
falls slower than a hard rock.
Your argument fails the same way Aristotle's arguments fail, only
with clay replacing the rock. The type of linearity you talk about,
where depth of penetration is proportional to kinetic energy, only
applies over a narrow range of conditions. Your straw man physics
theory predicts that a feather pillow will make the same impression as
a rock of the same density, as long as they were dropped in a vacuum.
Not so. The properties of clay are complex.
Anyone can do the experiment you describe. However, your
conclusions are wrong even if the observations are correct. It is not
about people "closing their eyes." Asking people over and over again
to repeat the experiment represents closing your eyes. They aren't
questioning your observations, they question your reasoning ability.
You refuse to acknowledge your lack of knowledge, it isn't that other
people refuse to acknowledge the observations.
There are books and articles that describe the physical
properties of clay. There are also books and articles concerning the
physics of impacting bodies, especially with regards to bullet impact.
You can look them up. Why don't you read those articles, find the
parts relevant to your experiment, and then defend your conclusions.
Surely you can cite a reference saying, "the depth of impression in
clay is always proportional to the amount of kinetic energy in the
impacting body." Providing such a reference will at least show us that
you have thought about the problem.
NoEinstein
>
Dear Jerry: The majority of the errors spread throughout physics are
the result of those who don’t understand what they are doing using
math and units to “derive” that which is wrong. Too often the repeat
of derivations is done so matter-of-factly, that the students don’t
have time to notice or call-out the errors. And when they do, they
aren’t brave enough to point out mistakes to the same “professor” who
will be grading their papers in the course.
Physicists are one gullible and timid group of people! Since “the
illogical” is more difficult to “comprehend” than the logical, there
is a pseudo intellectual superiority in being among those claiming to
comprehend stuff that is difficult or counter-intuitive. That same
psychology is a huge part of the reason you are so offended that I am
showing up your pseudo expertise in science, with my simple science
truths.
An error made throughout science is to wrongly assume that if an
equation uses various units to calculate a result, that the result
must always “do the same math” on the units as was done on the various
integers. For example: A continuously applied force will cause a
uniform acceleration of an object. An accelerating object will cause
an increase in the object’s kinetic energy. Since an object having KE
will cause an impacted object to move’, then, the “units” of KE is…
foot-pounds. (sic!) NOTE: By whatever means that KE is “derived”,
measured, or calculated, the units of KE is just the same lowly FORCE
that caused the acceleration to begin with!
Another grievous example: It was observed by Galileo and Sir Isaac
Newton that falling objects accelerate. Both the distance of fall,
and the velocity are increasing with respect to time. The plot of the
distance of fall vs. time is an inverted parabola. So, the observed
acceleration due to gravity must be a second power increase, too—
expressed: ‘g’ = 32.174 feet per second, per second… or 32.174 feet
per second SQUARE. (sic!) There is nothing “exponential” about
acceleration! The only correct way to write the acceleration due to
gravity is: *** ‘g’ = 32.174 feet per second EACH second! ***
In MATH, two “pers” mean two divides; and two divides can be expressed
two inverts and multiplies, or … “square”. The above 300-plus-year-
old mistake has continued to this day, because the ones doing… “the
math” didn’t understand this fact: A parabolic increase in the
DISTANCE of fall with respect to time doesn’t correlate into a
parabolic increase in ACCELERATION! The latter would be an
acceleration of the acceleration.
When Einstein took his imagined “acceleration” journey to velocity
‘c’, he reached this erroneous conclusion: “There isn’t enough energy
in the entire Universe to cause even a speck of matter to reach
velocity ‘c’.” Einstein’s mistake was due to his blind acceptance of
the 1830 KE equation of Coriolis: KE = ½ mv^2. And his blind
acceptance of the most common acceleration of all, that due to
gravity, as: ‘g’ = 32.174 feet per second SQUARE. (sic!) Albert
Einstein had no earthly idea what acceleration is! The only thing he
knew about was acceleration of the acceleration! The latter, most-
massive mistake in the history of science, caused Einstein to write is
infamous E = mc^2, which yours truly has repeatedly disproved, up,
down, and sideways!
Any formula—if it’s a true law of nature—should be derivable (via
reverse ‘engineering’) from the subsequent empirical data. The
“predictions” of KE = ½ mv^2 are NOT consistent with either the
experimental results, nor with sound reasoning. So such equation is
invalidated!
… [snip Jerry's cook-booked crap]
Jerry said:
“You wrote, "the Law of Nature is a LINEAR increase in KE with respect
to velocity!"
A quantity that increases linearly with velocity is already well known
as "momentum", and your experimental results indicate nothing more
than that momentum correlates fairly well with penetration depth.
Jerry
Dear Jerry: Your understanding indicates that you are as confused
about science as Einstein was. In the ninth grade I invented calculus
to solve a for-extra-credit, circle area in a square problem. Nothing
about KE requires the use of calculus to “derive” the equation. But
if calculus had been required, the resulting equation would still be
wrong!
There is an expression: “Give someone enough rope, and they will hang
themselves.” For physicists, most were given too much rope before
they even graduated with a BS degree. The latter stands for bull-shit
of science degree. — NoEinstein —
> Jerry- Hide quoted text -
NoEinstein
Dear Jerry: Your corrections need to go further than typos. They
need to start from scratch, without any notions that what you were
ever taught was correct. — NoEinstein —
NoEinstein
>
> - Show quoted text -
Dear PD: You are one of about five, so far. Truths in science aren't
about who has the biggest ego, or who says the last word. They're
simply truths that even the meek and lowly to understand. However,
the high and mighty don't like the processes very much. — NoEinstein —
NoEinstein
>
> - Show quoted text -
:-} - ~ - - -
NoEinstein
:-} - ~ - - -
NoEinstein
:-} - ~ - - -
Matthew Johnson
NoEinstein says...
[snip]
>Dear PD: You are one of about five, so far. Truths in science aren't
>about who has the biggest ego, or who says the last word.
He knows that. It is you who repeatedly displays ignorance of this principle.
>They're
>simply truths that even the meek and lowly to understand.
Now why would that be the case? Why would the "truths in science" be accessible
to ALL the "meek and humble"? Was Dalton's theory of the atom "accessible to the
meek and humble"? Was Newton's theory? If it was, then why did to many of the
"meek and numble" miss it completely, until Newton invented calculus?
It was not meekness and humility that made Newton's great ideas accessible, it
was his invention of calculus. Without that, there was simply no way to explain
the motion of the planets based on 1) Newton's 3 Laws of Motion" and 2) the Law
of Universal Gravitation.
This is why Halley was so shocked when Newton simply declared that an inverse
square law attractive force causes elliptical motion. Nobody else knew how to
prove that at that time. See
http://hua.umf.maine.edu/Reading_Revolutions/Newton.html for details.
>However,
>the high and mighty don't like the processes very much.
This, of course, is just a slur on those who know better than you. BTW:
resorting to such slurs if a PERFECT proof that you are NOT one of the "meek and
lowly", so by your own word, you cannot understand the truths in science!
Matthew Johnson
NoEinstein says...
>
>On Aug 20, 12:01=A0am, Jerry <Cephalobus_alie...@comcast.net> wrote:
>>
>Dear Jerry: The majority of the errors spread throughout physics are
>math and units to =93derive=94 that which is wrong.
Nonsense. This is a completely fictitious version of the history of physics.
> Too often the repeat
>of derivations is done so matter-of-factly, that the students don=92t
>have time to notice or call-out the errors. And when they do, they
>aren=92t brave enough to point out mistakes to the same =93professor=94 who
>will be grading their papers in the course.
Well, I suppose I might have fallen for your claim here -- if I had never gone
to college! But even at the lowly junior college I briefly attended, professors
mad few such errors, and student were quick to point them out.
Have you ever even -been- to college, 'NoEinstein'?
>Physicists are one gullible and timid group of people!
I'll bet I am far from the only reader who nearly fell out of his char laughing
at this one!
[snip]
>An error made throughout science is to wrongly assume that if an
>equation uses various units to calculate a result, that the result
>must always =93do the same math=94 on the units as was done on the various
>integers.
You can't correct an 'error', if you can't even state the principle correctly!
What 'integers'?
>For example: A continuously applied force will cause a
>uniform acceleration of an object.
Wrong. It has to be a constant force, too.
> An accelerating object will cause
>an increase in the object=92s kinetic energy.
Wrong again. It is the force that is the cause, NOT the accelerating object.
> Since an object having KE
>will cause an impacted object to move=92, then, the =93units=94 of KE is=85
>foot-pounds. (sic!)
That does not follow, and it is NOT the reasoning physicists apply.
> NOTE: By whatever means that KE is =93derived=94,
>measured, or calculated, the units of KE is just the same lowly FORCE
>that caused the acceleration to begin with!
Wrong. There is a REASON that one of the most common systems of units is called
MKS -- you have to take into account TIME (seconds) in your units. But since
force IS the derivative of momentum, it must has a second factor of time in the
denominator.
NoEinstein
>
Dear Darwin123: What I teach is that the KE values for two balls of
the same size, but of different weights, won’t produce equal effects
(like holes in clay) when each ball’s KE value is equated to that of
the other, and the required fall velocities are calculated. Knowing
the velocity of fall for each ball to cause equal KE effects allows
calculating the height from which each ball must be dropped to
(supposedly) cause the same size holes in the clay. Even in a vacuum,
Coriolis’s formula won’t allow predicting “equality of effects”,
because his equation isn’t a Law of Nature.
D.123 said: “The type of linearity you talk about, where depth of
penetration is proportional to kinetic energy, only applies over a
narrow range of conditions…”
You are ‘almost’ thinking. Clay is a highly variable material. I
took soil mechanics in college, and we did lab tests on the many
variations. The purpose of my present experiment is to cause the clay
variations to become a non-issue! Here’s how: If you dropped numbers
of the same size steel balls into the same consistency bed of clay,
and all of the balls were dropped from the same height, the dents (or
holes) in the clay should be very close to the same. If you repeated
that experiment with a different, say, softer clay, the dents (or
holes) in the clay would be deeper, but all of them would be the same
size. In every case, those balls impact with the same KE.
From the clay’s “point of view” the clay can’t tell what the ball’s
material is that hits it. If a lighter metal, or other type ball, of
the same size and surface smoothness impacts the clay with the same
“calculated” KE, the latter ball “should” make the same size hole in
the clay. And the clay wouldn’t “know” the difference. My requiring
that both balls impact with the same calculated KE is the premise of a
very easy test that doesn’t require ANY information on the clay, other
than that such is uniform for all balls hitting it.
I didn’t want the clay that I used to be so firm that the dents would
be tiny. So, as I mixed the clay (softened the too firm, store-
bought, art clay by adding more water), I kept noting how easily my
thumb could force an indention. When the latter became easy, I took
the gob of wet clay and put it into a plastic bag. I spent the next
ten minutes kneeding the clay to be a uniform consistency. Then, I
started putting such, a tablespoon at a time, into a nominal 4” x 4” x
5” plastic flower pot. I used my knuckles to force the clay inside,
and to force out any loose water or air bubbles.
As comedian Benny Hill liked to say: “Use a bit of common (sense).”
You keep pointing out the ‘extremes’ of the possible tests (balls of
feathers) solely to try to discredit what I have accomplished. But
all that you’ve succeeded in doing is to make yourself look like an a.
h.
Initially, I tried to locate ¾” diameter glass marbles. I ordered a
bunch, but they were all “nominal” size, rather than actual size. At
www.smallparts.com I found PTFE balls which are almost as heavy as
glass. And such are machined much more perfectly than the pitted
marbles I bought.
Darwin123, typical of many who reply on sci.physics, you play being
superior. But your background (whatever it is) comes up short.
Fortunately, I don’t have to CLEAR my results with you. The “real”
Darwin was a great scientist. It’s sad, but you’re just a bluff. —
NoEinstein —
> you have thought about the problem.- Hide quoted text -
NoEinstein
wrote:
> In article <ad4df875-d745-4868-b520-dbb3fd0a2...@m44g2000hsc.googlegroups.com>,
> NoEinstein says...
>
> [snip]
>
> >Dear PD: You are one of about five, so far. Truths in science aren't
> >about who has the biggest ego, or who says the last word.
>
> He knows that. It is you who repeatedly displays ignorance of this principle.
>
> >They're
> >simply truths that even the meek and lowly to understand.
>
> Now why would that be the case? Why would the "truths in science" be accessible
> to ALL the "meek and humble"? Was Dalton's theory of the atom "accessible to the
> meek and humble"? Was Newton's theory? If it was, then why did to many of the
> "meek and numble" miss it completely, until Newton invented calculus?
>
> It was not meekness and humility that made Newton's great ideas accessible, it
> was his invention of calculus. Without that, there was simply no way to explain
> the motion of the planets based on 1) Newton's 3 Laws of Motion" and 2) the Law
> of Universal Gravitation.
>
> This is why Halley was so shocked when Newton simply declared that an inverse
> square law attractive force causes elliptical motion. Nobody else knew how to
>
> >However,
> >the high and mighty don't like the processes very much.
>
> This, of course, is just a slur on those who know better than you. BTW:
> resorting to such slurs if a PERFECT proof that you are NOT one of the "meek and
> lowly", so by your own word, you cannot understand the truths in science!
Dear Matt: The high and the mighty don't like to be shown up. It is
said: "The bigger they are, the harder they fall." You should learn
to be meek, before you go... BOOM! — NoEinstein —
NoEinstein
wrote:
> In article <c64a0635-3731-41ee-a1b8-c5f002892...@c65g2000hsa.googlegroups.com>,
Dear Matt: I have a raspberry for you. :-} - ~ - - - — NoEinstein —
Darwin123
> On Aug 18, 5:37 pm, Ray Vickson <RGVick...@shaw.ca> wrote:
>
>
>
> > On Aug 18, 10:00 am, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > > On Aug 17, 4:14 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > > On Aug 17, 10:41 am, "Spaceman" <space...@yourclockmalfunctioned.duh>
> > > > wrote:
> > times, your experiment does not prove what you think it does
> > (assuming, of course, that you actually DID the experiment).
>
> > R.G. Vickson- Hide quoted text -
>
> > - Show quoted text -
>
> by anyone. Claiming that I haven't done the experiment borders of
> libel. Tell me, fellow, why is it that you come out of nowhere and
> get your ego so hurt by my accomplishments that you must resort to
> libel? — NoEinstein —
He didn't claim that you didn't do the experiment. He implied a
small doubt, but it is clear he thinks you really did the experiment
and actually got the results you describe.
He did say, as I have said many times, that your experimental
results do not logically lead to your final conclusions. The main
issue is not whether you did the experiment. The main issue is as to
what those results mean.
This is obvious. I (mostly) believe your experimental results. I
don't believe your initial assumptions. I said so many times. The fact
that you choose to be defensive about the experimental results, rather
than the initial assumptions under discussion, leads me to believe
that you are being evasive. It is barely possible that we (I and other
skeptical types) haven't communicated to you the real issue. So let me
spell it out.
I don't believe the depth of penetration into clay is always and
forever proportional to the kinetic energy of the impacting body. This
type of linearity is associated with materials under an extremely
narrow set of conditions. If this type of linearity is not present in
the clay that you used, then your conclusions do not logically follow
from your results.
Coriolus may have done an experiment and found that type of
linearity in the particular clay that he used for the impacting bodies
that he was using. Few since then have measured kinetic energy in that
way.
In fact, you have not even told us anything about the shape of
those holes. Coriolus may have observed a certain shape which had a
definite relation to penetration depth. It could be the real rule for
that type of clay is that the volume of displaced material is
proportional to the kinetic energy. Coriolus may have been looking at
cylindrical pits, while you may be looking at crater like hemispheres.
Each shape indicates a different type of interaction with the clay.
If you are serious about this study, you would do the following.
Find a way to calibrate the response of the clay you are using. Find
some other way to impact the clay, not using falling bodies, and
determine whether the penetration depth really is proportional to
kinetic energy. Unless you check the linearity of your system, there
is no way you can claim it tells you anything about gravity. A
calibration curve is the bare minimum you need for a study.
Without calibrating your instrument, you are just splattering mud
all over the place.
Ray Vickson
> On Aug 18, 5:37 pm, Ray Vickson <RGVick...@shaw.ca> wrote:
>
>
>
> > On Aug 18, 10:00 am, NoEinstein <noeinst...@bellsouth.net> wrote:
>
> > > On Aug 17, 4:14 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > > On Aug 17, 10:41 am, "Spaceman" <space...@yourclockmalfunctioned.duh>
> > > > wrote:
>
> > > > > > On Aug 17, 8:42 am, NoEinstein <noeinst...@bellsouth.net> wrote:
> > > > > >> On Aug 16, 10:42 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > > > >>> On Aug 16, 5:36 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
> > > > > >>> [snip]
>
> > > > > >>> Where's the data?
>
> > > > > >> Dear Eric: The needed numbers and formulas are given in my post. Do
> > > > > >> the math, if you can. But since I know you to be a complainer, not a
> > > > > >> doer, I don't expect you to do anything but complain. — NoEinstein —
>
> > > > > > There's no data in your post since you didn't actually do the
> > > > > > experiment.
>
> > > > > I can't see you so you are not there!
> > > > > LOL
>
> > > > > --
> > > > > James M Driscoll Jr
> > > > > Creator of the Clock Malfunction Theory
> > > > > Spaceman
>
>
> > > > - Show quoted text -
>
> > > filled flower pot with the chrome steel ball (dropped from 3'-4")
> > > embedded nearly to its 'equator'; and the PTFE ball (dropped from 12')
> > > embedded about 1/8" less. Both balls are 3/4" in diameter and
> > > perfectly spherical. Claim that I haven't done the experiment, and
> > > you only show yourself to be an ignorant fool, Eric. — NoEinstein —
>
> > is another matter entirely. Eric is a fool because you choose not to
> > present proof that you have done the experiment? I don't think so.
> > After you present the proof you can expect people to, maybe, believe
> > you, but not until then. Anyway, as has been explained to you many
> > times, your experiment does not prove what you think it does
> > (assuming, of course, that you actually DID the experiment).
>
> > R.G. Vickson- Hide quoted text -
>
> > - Show quoted text -
>
> Dear Ray: My post explains a $30.00 experiment which can be verified
> by anyone. Claiming that I haven't done the experiment borders of
> libel. Tell me, fellow, why is it that you come out of nowhere and
> get your ego so hurt by my accomplishments that you must resort to
> libel? — NoEinstein —
If you think I have libelled you, just go ahead and sue me. I claim
you cannot possibly have performed the experiment to the type of
accuracy your figures present, and claiming that I say that because my
ego is hurt is just scurrilous nonsense. Others have also explained
over and over and over again why your experiment may not be measuring
what you think it does. They have asked you for justification, but you
have given none. You seem to have no appreciation for the way that
scientists present their results, including limitations on significant
figures, estimates of measurement errors and the like. Anyone who has
submitted a paper to a refereed journal, or faced a PhD oral
examination committee, or given a seminar has likely faced much more
serious criticism than anything you have faced in this forum.
Professionals have to develop a bit of a thick skin, because criticism
and the need to defend ideas and results come with the territory. Get
over yourself.
R.G. Vickson
Eric Gisse
[snip]
> Dear Matt: I have a raspberry for you. :-} - ~ - - - — NoEinstein —
If this is your level of discourse, I hope you are comfortable with
spending years and years of being ignored.
Darwin123
example, the clay "knows" the inertial mass of the ball. Even if the
falling ball had the exact same kinetic energy and exact same shape,
its response upon hitting the clay would depend on its mass. If the
floor were hard instead of clay, a lighter ball could bounce higher.
Just the air resistance alone would slow a lighter ball more than a
heavy ball. If the falling ball was impacting another ball, instead
of clay, the amount of linear momentum transfer would vary with the
ratio of masses between the two balls. These examples should be
sufficient to tell you that the system can't be as simple as you
thought it would. Your assumption that the inertial mass of the ball
doesn't effect the balls response to the clay is wrong.
In addition, the clay also knows the weight of the ball. Let us
say the ball wasn't dropped, and it was just laid down on the clay.
The degree to which the ball "sinks" in the clay would very much
depend on weight. Even if the static ball wasn't heavy enough to sink,
the pressure caused by the weight plus the linear momentum transfer is
obviously causing "sinking." The shear limit of the clay has been
exceeded. So the weight of the ball does matter, independent of
kinetic energy.
Although you say the ball has the same amount of smoothness,
there are differences in the adhesiveness of different materials. The
clay is going to stick to the surface, so the friction between
different metals will be different. When you pull the ball out of the
depression it makes in the clay, the ball sticks (?) Its not like
every ball was teflon coated (?) An aluminum ball will have different
friction coefficients with the clay than an aluminum ball, and really
different coefficients from a wood ball. The surface tension with
water changes.
I don't know if any of your balls spin. You may be adding an
extra spin on your ball. If you roll them, the spin can be significant
and vary with the ball.
You say you keep the shape the same. I will assume that the shape
is the same. I will assume the same for elasticity.
The clay to the inertial mass of the ball is perhaps the most
obvious error. The transfer of linear momentum from ball to clay is
going to vary due to the differences in inertial mass. There is no way
in the experiment you describe to separately control this linear
momentum transfer. Therefore, you still need a calibration including
both kinetic energy and linear momentum, separately.
> If a lighter metal, or other type ball, of
> the same size and surface smoothness impacts the clay with the same
> “calculated” KE, the latter ball “should” make the same size hole in
> the clay. And the clay wouldn’t “know” the difference.
You are wrong. Look up "momentum transfer." Momentum transfer
depends on the ratio of masses of the system.
> My requiring
> that both balls impact with the same calculated KE is the premise of a
> very easy test that doesn’t require ANY information on the clay, other
> than that such is uniform for all balls hitting it.
You are assuming a symmetry that isn't there. In real mechanics
problems, the amount of momentum transfer has to be calculated knowing
both the kinetic energy and the mass of the colliding bodies. The
ratio of masses is important even in the most simple cases. You can't
ignore mass.
>
> I didn’t want the clay that I used to be so firm that the dents would
> be tiny. So, as I mixed the clay (softened the too firm, store-
> bought, art clay by adding more water), I kept noting how easily my
> thumb could force an indention. When the latter became easy, I took
> the gob of wet clay and put it into a plastic bag. I spent the next
> ten minutes kneeding the clay to be a uniform consistency. Then, I
> started putting such, a tablespoon at a time, into a nominal 4” x 4” x
> 5” plastic flower pot. I used my knuckles to force the clay inside,
> and to force out any loose water or air bubbles.
The shear forces will depend on how rapidly the ball slows down
once it has penetrated the surface. A lighter ball would be slowed
down faster once in the clay than a heavier ball. I suspect that
Coriolus, or whoever your source is, arranged for balls of almost the
same mass to hit the clay.
I don't know how that translates in terms of material science. You
took soil courses. However, I don't think the soil scientists
calculate momentum transfer very often.
NoEinstein
>
Dear Darwin123: You are so determined to have a discourse with me
that you reply to my reply to “Ray”. Why not let Ray reply for
himself, rather than assume that you know what he was thinking? I
sense that you skim what I write, rather than try to comprehend what I
write. All of your “show-off” comments about “clay” and trying to
tell me how experiments should be run it just your ruse to try to be
superior to someone who is your superior—me.
I recommend that you re read my explanation of why the consistency of
the clay will never influence the results of my $30.00 experiment—
which can be duplicated by anyone. I am a person who doesn’t mind
getting my hands dirty to do an experiment. Nor do I mind having
experiments dead-end, because I have the work ethic to make changes
and keep on testing.
If there is a teeter-totter in dynamics labs which can (supposedly)
measure KE, I suggest this simple experiment: Drop a steel ball and
measure how high such will toss a lighter ball. Any falling object
impacting with the same KE should toss that ball an identical height.
Now, use Coriolis’s KE = ½ mv^2 to calculate the velocity necessary to
have the KE of a PTFE ball match the KE of the steel ball. Use d =
t^2 (where d = 16.087 ft.) to find the height of drop necessary to do
such. Finally, drop the PTFE ball. Since the KE at impact (according
to Coriolis’s equation) is identical to the KE of the same size steel
ball, the apparatus ball should be tossed to the same height as
before. But it will be tossed much lower. In fact, the ceiling
height in the lab will likely be less that 10% as high as would be
necessary to get and PTFE ball to hit with enough KE. Do the
experiment that way, and there is no “clay” involved.
If you would like to have me reply to you in the future, don’t try to
snow me with wordiness. You know a smidgeon about science; though
mostly can’t see the forest for the trees. Try to be “the old dog”
who CAN learn new tricks. There are very few dogs out there willing
to even try to do so. — NoEinstein —
NoEinstein
Dear Ray: Nowhere did I claim that my experiment was so accurate!
The numbers given in my initial post are those necessary to make the
KE of the chrome steel ball = that of the PTFE ball. I showed the
numbers to five decimal places so others can confirm that I did the
math.
I don't recall having had you reply to me before. Please explain your
background. You seem to be in the majority who will attack the King
of the Hill, but know very little about scientific logic. The latter
game is childish, really. — NoEinstein —
NoEinstein
Dear Eric: Being ignored by you would be a blessing!!!!! —
NoEinstein —
NoEinstein
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Dear Darwin123: You have already 'used up' your quota of replies from
me. Try again, latter! — NoEinstein —
Jonathan Thiessen
> from a height of 3.3684 feet into a small flower pot full of just-
> mixed art clay. That ball sank in close to its ‘equator’. I
> immediately went up my outdoor staircase and dropped a ¾” dia. PTFE (a
> heavy fluoroplastic ball, weighing .2807 times as much as the chrome
> steel ball), from an exact height of 12 feet. The KE value should be .
> 10469323 for each ball. Note: 12 feet of drop = .745944d, where d =
> 16.087 feet, the distance of fall in one second. The time of fall is .
> 86368 seconds for the lighter ball.
>
> The PTFE ball landed 1” from the chrome steel ball. It sank into the
> clay only about .75 as deep. If Coriolis’s equation was correct, both
> balls would be imbedded equally. Those two balls are stuck in the
> clay. I will let everything air dry to serve to document my
> experiment.
>
> The above simple experiment can be run with any two equal size, but
> different weight balls. (Ping Pong balls excluded.) Use Coriolis’s
> equation to make the KE values for each ball weight equal. If you
> have access to a tall building where drops can be made from two
> heights as required to satisfy the equations, the results, still,
> won’t cause equal size holes in the clay. A semi-parabolic equation,
> like Coriolis’s, can never predict impact results when the Law of
> Nature is a LINEAR increase in KE with respect to velocity! My
> correct equation is: KE = a/g (m) + v/32.174 (m).
>
> of Coriolis. Since Coriolis’s equation gave Einstein the mistaken
> notion that the energy progression in traveling to velocity ‘c’ is
> parabolic, BOTH of Einstein’s theories of relativity are disproved,
> yet again, by yours truly!
>
> Respectfully submitted,
>
> — NoEinstein —
>
Good day,
I'm sorry if someone has already posted something of like manner. I
haven't been as thorough as is due.
Have you taken into account that the two balls won't have the same
net force acting on them [and thus won't accelerate at the same rate]? I
do believe that drag force is proportional to v^2. If you assume the
acceleration is g, your experiment won't work; if you calculate the
approx true acceleration of the first ball and assume it will be the
same for the second, your experiment won't work [it has the same
acceleration _function_, but the domain over which you integrate is
different]. In order to get anywhere, you need to look at all of the
forces involved [and realize that they are different for the two objects].
Have a good one ;)
Jonathan Thiessen
Jonathan Thiessen
Now that I think of it more, I'm not really sure if the former sentence
really belongs. The drag force function is the same, and the
acceleration due to gravity negating drag force is the same, however,
F=ma => a_1 != a_2 [since drag force is the same, but masses differ]. In
any case, it still holds that the net force acting on the two balls is
different.
Jonathan Thiessen
I suppose it would be good to say that the net force functions are
differing and that the net force ratio [between balls] is not equal to
the mass ratio [ie the acceleration functions are different], but I
suppose that I already said that the acceleration functions are different.
So as to not risk obfuscating my point further, I will try to stop
reforming my post.
PD
The drag force function is the same, but one is acting over a larger
distance, since it is dropped from a larger height. Since the energy
removed by the drag (and lost to heat in the air) is the drag force
multiplied by the distance of the fall, the ball dropped from the
larger height is *expected* to lose more energy and therefore have
less energy at the landing and therefore leave a smaller dent.
But of course, NoEinstein is an idiot and does not really have the
first clue how to either apply energy conservation principles in
practice, nor how to run an experiment properly. But to tell him in no
uncertain terms where he is going wrong is, in NoEinstein's
estimation, the act of a nay-sayer, a party-pooper, an arrogant
freedom-snuffer.
PD
Jonathan Thiessen
This was my thinking before, and although it works, differing heights
aren't actually required.
If one were to drop two otherwise identical balls with differing masses
over an equal distance you'd get something like this [down is positive]:
F_1 = g*m_1 - a(v_1)^2
F_2 = g*m_2 - a(v_2)^2
Suppose m_1 > m_2; v_1_0 = v_2_0,
[I really should work out the DE here in full, but alas, I don't want to :P]
=> a_1 > a_2
=> v_1 >= v_2 [only equal initially]
ie When not in a vacuum, heavier things really do fall faster [assuming
they have the same drag coefficient and cross sectional area].
Jonathan Thiessen
'a' was a bad choice of a constant... Substitute with your personal
favourite -- maybe alpha, or k.
> Suppose m_1 > m_2; v_1_0 = v_2_0,
>
> [I really should work out the DE here in full, but alas, I don't want to
> :P]
>
> => a_1 > a_2
These were actually meant to be acceleration.
PD
Right, but equal distances is not what we have here.
Jonathan Thiessen
I suppose it was somewhat irrelevant to the particular problem, however,
it's stronger. My thinking was that it would lead to a better conceptual
understanding in that it re-enforces the idea that the balls'
accelerations are immediately divergent since they also depend on their
masses [rather than the divergence occurring strictly after the moment
of impact of the first ball [or after falling a certain non-zero distance]].
Darwin123
> On Aug 19, 6:50 pm, Darwin123 <drosen0...@yahoo.com> wrote:
>
>
>
> > On Aug 16, 10:42 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > On Aug 16, 5:36 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
> > > [snip]
>
> > > Where's the data?
>
> > assumptions.
> > 1) He actually thinks that the formula for kinetic energy is derived
> > from deformation of materials, not from the conservation of energy. As
> > most of us know, the derivation of conservation of energy using
> > Newtonian physics automatically brings about:
> > KE=0.5mv^2
> > The v^2 factor comes from an integration
> > 2) He ignores everything about material properties discovered after
> > Isaac Newton. For example: He ignores the fact that his clay has a
> > plastic deformation limit.
> > 3) etc.
>
> out = Energy in. 1/2 mv^2 is an exponential increase in energy. But
> the uniform "thrust'' of gravity—equal to the static weight of every
> object—can only cause KE to accrue LINEARLY. Therefore, since Energy
> out (according to Coriolis) is greater than the energy being imparted
> by the force of gravity, then Coriolis's formula, and Einstein's SR,
> which used such as it's basic, both violate the Law of the
> Conservation of energy, and are thus disproved by yours truly.
> Scientific truths trump 'smoke and mirrors' shots-from-the-hip every
> time. — NoEinstein —
Let me put it this way. The way that Joule measured potential energy
was by the temperature increase due to friction. The temperature
increase of the clay just after being hit by the ball should be
approximately linear. There are some problems there with dissipation.
However, the equivalence of energy, kinetic and heat energy, is well
established. This is why we measure energy in Joules. You chose not to
use temperature.
They call this type of measurement for energy a calorimetric
measurement. Calorimetric measurements are routinely done for all
sorts of energy. Scientists do not used penetration depth for kinetic
energy. The temperature changes in your clay may very well be linear
in kinetic energy. The plasticity measurements you describe are too
sensitive to all sorts of parameters. Your results would vary greatly
with the microscopic structure of the clay, as well as kinetic energy.
There is no fundamental theory connecting kinetic energy with impact
depth. Thermodynamics connects temperature rise with kinetic energy.
Maybe you can do a calorimetric series of measurements to support
your penetration depth studies. It would cost a little more than $30
of course. You need a real sensitive thermometer. Put water or
ethylene glycol (to reduce evaporation) in a thermos. Drop your balls
into the water. Measure the rise in temperature with a very sensitive
thermometer (yes, you can buy them). See if you can repeat your
results using calorimetry rather than penetration.
If someone were to check the kinetic energy that you determined
using temperature instead of penetration depth, they would probably
find a different number. They would probably find the same temperature
rise for the same kinetic energy. Suppose someone got interested
enough to check your results with temperature, a more established way.
That would prove conclusively that your penetration depth varied with
something more than kinetic energy.
The method you use to measure kinetic energy has no theoretical
or experimental justification. If you do prove such a correlation,
that would be a breakthrough in itself.
Ray Vickson
Well, I have a PhD in physics from a little place called MIT in
Cambridge, Massachussetts; you may have heard of it. I did a couple of
years post-doctoral work before switching to operations research,
which I have done for all my professional life, teaching and doing
research at the University of Waterloo. I am now retired. I maintain
an interest in Physics, but am not actively engaged in research.
> You seem to be in the majority who will attack the King
> of the Hill
So, are you claiming to be King of the Hill? What hill? Who made you
king?
>, but know very little about scientific logic.
How can you make this claim? I have mostly ignored you postings over
the years because they are not worth bothering about, but in this one
thread I piped up, against my better judgment.
> The latter
> game is childish, really.
As I said before: get over yourself. If you cannot take criticism you
should forget about being taken seriously. Of course, this is the
internet, and anybody has the right to post anything they want. The
question is whether you want your contributions to be recognized as
worthwhile.
R.G. Vickson
> — NoEinstein —
Androcles
"Ray Vickson" <RGVi...@shaw.ca> wrote in message
news:c21fbe33-42bd-4173...@j1g2000prb.googlegroups.com...
R.G. Vickson
=============================================
Hey Vickson! If you cannot take criticism you should forget about
being taken seriously, you stupid bastard.
Ray Vickson
> "Ray Vickson" <RGVick...@shaw.ca> wrote in message
As the John Cleese character said to the Kevin Kline character in "A
fish Called Wanda": you are a true vulgarian.
R.G. Vickson
Androcles
=========================================
Yes, and you are a true sanctimonious hypocritical fool.
Nobody but you gives a shit that you went to MIT, everyone
knows about American sports scholarships and how fucking
worthless they are.
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
Your answer goes here:
________________________________________________________
Other answers have been:
According to Ian Parker:
"We are not talking about the speed of light here we are talking
classical stability theory." -- Idiot Ian Parker.
______________________________________________________
According to cretin harald.vanlin...@epfl.ch
"Easy: he did NOT say that."
According to moron van lintel, Einstein did not write the equation he wrote.
______________________________________________________
According to xxein:
It is an artefactual/superficially imposed yin-yang of sorts.
______________________________________________________
According to Lamenting Shubert:
Why do you want to know?
______________________________________________________
According to Imbecile Jimmy Black:
" In neither system (meaning frame of reference in modern-day terminology)
is the speed of light c-v or c+v. In both systems the speed of light is c."
According to the imbecile Jimmy Black, Einstein did not write the equation
he wrote.
______________________________________________________
According to Dork Bruere
"I don't give a damn what Einstein wrote."
______________________________________________________
According to Spirit of Truth:
that math is correct but WRONG
______________________________________________________
According to constipated Eric Gisse
"I don't give a shit (fill in the blank ____________)."
______________________________________________________
'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Rabbi Albert Einstein
Ray Vickson
Well, Noeinstein DID ask me about my background, so I answered him.
> everyone
> knows about American sports scholarships and how fucking
> worthless they are.
What does that have to do with anything? Anyway, for the record, I am
not an American and am not athletic. Of course, you will say "who give
a shit?" or something similar. Have a nice day.
R.G. Vickson
>
> Why did Einstein say
> the speed of light from A to B is c-v,
> the speed of light from B to A is c+v,
> the "time" each way is the same?
>
> Your answer goes here:
>
> ________________________________________________________
>
> Other answers have been:
>
> According to Ian Parker:
>
> "We are not talking about the speed of light here we are talking
> classical stability theory." -- Idiot Ian Parker.
> ______________________________________________________
>
> According to cretin harald.vanlintelButNotT...@epfl.ch
Jerry
> On Aug 18, 10:00 am, NoEinstein <noeinst...@bellsouth.net> wrote:
> > Dear Eric: I have photographed my experiment. AND I have the clay-
> > filled flower pot with the chrome steel ball (dropped from 3'-4")
> > embedded nearly to its 'equator'; and the PTFE ball (dropped from 12')
> > embedded about 1/8" less. Both balls are 3/4" in diameter and
> > perfectly spherical. Claim that I haven't done the experiment, and
> > you only show yourself to be an ignorant fool, Eric. — NoEinstein —
>
> You obviously don't get it. Anyone can claim anything; supplying proof
> is another matter entirely. Eric is a fool because you choose not to
> present proof that you have done the experiment? I don't think so.
> After you present the proof you can expect people to, maybe, believe
> you, but not until then. Anyway, as has been explained to you many
> times, your experiment does not prove what you think it does
> (assuming, of course, that you actually DID the experiment).
I do not understand all of this focus on whether or not NoEinstein
performed the experiments that he claims to have performed. The
kinetic energy formula is a direct consequence of the definition
of mechanical work.
The mechanical work performed on a mass is given by the product
of the applied force times the distance over which the force acts.
If F is the applied force and s is the distance,
W = Fs
In the absence of friction, a force F acting on a mass m over the
distance s will accelerate the mass to a velocity given by
v = sqrt(2Fs/m)
If can assume, applying the conservation of energy principle, that
all of the work performed on the mass manifests itself in kinetic
energy, then the standard formula for kinetic energy arises
naturally as
K.E. = 1/2 m v^2 = Fs = W
NoEinstein proposes an alternative formula for kinetic energy,
namely
K.E. = mv
For this to be so, the definition of mechanical work must be
altered. In particular, the following bizarre formula arises:
W = sqrt(2mFs) = m sqrt(2as)
The NoEinstein work that can be extracted from a falling mass is
directly proportional to the mass, so that seems OK. But to
extract twice the NoEinstein work from a falling mass, we need to
let it fall four times the distance. That doesn't make sense at
all.
Jerry
Androcles
On Aug 21, 4:55 pm, "Androcles" <Headmas...@Hogwarts.physics> wrote:
> Yes, and you are a true sanctimonious hypocritical fool.
> Nobody but you gives a shit that you went to MIT,
Well, Noeinstein DID ask me about my background, so I answered him.
> everyone
> knows about American sports scholarships and how fucking
> worthless they are.
What does that have to do with anything? Anyway, for the record, I am
not an American and am not athletic. Of course, you will say "who give
a shit?" or something similar. Have a nice day.
R.G. Vickson
===================================================
If you cannot take criticism you should forget about being taken seriously.
Of course, this is the internet, and anybody has the right to post anything
they want. The question is whether you want your contributions to be
recognized as worthwhile, and obviously you do not.
NoEinstein
>
Dear Jonathan:
The force (of gravity) acting on the falling balls is exactly equal to
their static weight. So, balls of different weight don’t have the
same gross force acting upon them. By “net” force, I’m assuming that
you mean… AFTER deducting the drag caused by air resistance. Since
two balls of identical size have the same air resistance, such will
cancel out for most short fall-distance experiments. The exception (a
matter of degree) is due solely to the fact that the ball falling
further is exposed to air resistance longer. Also, because a lighter
ball must fall faster to have the same KE as a heavier ball, the extra
velocity through air tends to reduce the effective acceleration of the
lighter ball.
Issues of air resistance is 1/3rd of an entire branch of science.
Such is part of aeronautics. If air resistance “correctness” was the
sole determinant whether or not my new “clay” experiment is valid, the
air resistance issues could be argued about for a very long time.
The primary issue that should be considered regarding the correctness
of KE = ½ mv^2 is whether or not such equation violates the Law of the
Conservation of Energy. (It does… BIG TIME!) A Styrofoam ball won’t
accelerate as fast as a steel ball. But the force of gravity causing
each to fall, is still just that ball’s static weight.
I intentionally chose two fairly dense balls so air resistance issues
would be close to nil. Somewhere in this country there must be a lab
with a vacuum drop chamber. I invite such lab to do my same test.
Coriolis’s semi-parabolic equation EXAGERATES the KE that is
accruing. My correct KE = a/g (m) + v/32.174 (m) has been confirmed
in a short drop of a small clevis pin head-to-head with the next size
larger clevis pin. When the KE of the smaller dropped pin is equal to
the INERTIA of the larger pin (hanging from light tension springs) the
small pin will remain in contact with the larger pin long enough to
DAMPEN the ringing tone caused by the impact. The loss of ringing
tone occurred at exactly the drop height predicted by my equation.
Coriolis’s equation would have required that the lighter pin be
dropped from about FIFTY FEET further!
My “clay” experiment dropped the lighter PTFE ball with a velocity of
about .87, (where ‘g’ velocity = 1). The steel ball dropped at like .
18 v. Note: The actual velocities are shown in my original post. If
you would like to apply some air resistance equations that you know to
find the effect on my experiment, by all means do so. But I have no
interest in arguing the minutia of air resistance. My main, easy to
understand refutation of Coriolis’s equation is this: “KE OUT must be
equal to the velocity x time that gravity is acting on all (unit
weight) falling objects.” Since gravity is a uniform force (a
downward thrust cause by flowing ether), then no exponential equation
could ever be correct. Why? Because velocity increases LINEARLY for
all dropped objects. So, the velocity x the time has to be increasing
LINEARLY, too—just like my KE equation says!
— NoEinstein —
> Jonathan Thiessen- Hide quoted text -