On Apr 28, 8:31 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
On Apr 28, 12:10 pm, PD <TheDraperFam...@gmail.com> wrote:> On Apr 26, 12:51 pm, NoEinstein <noeinst...@bellsouth.net> wrote:

1. Does gravity do continuous work on falling objects?
Yes, though when an object reaches terminal velocity in the
atmosphere, the net work done on the object is zero.
Yes AND no... In the ranges of most free drop experiments (less than
300 feet) of COMPACT objects, the terminal velocity doesn't come into
play.

Sure it does. Skydivers.

But considering air resistance, the total work done by gravity
vs. the total "effect" is still equal! But the movement or frictional
heating of the air and the falling object will account for the energy
not recoverable from the object's "impact effects".

2. If so, would you say that gravity is the source energy for the
mechanical effects exhibited by falling objects? If not, why?
No. When a baseball pitcher throws a fastball, it is falling the
moment it leaves the pitcher's hand (as there is no other force acting
on the ball but gravity after that point). However, the mechanical
effects exhibited by the fastball on the batter's noggin are mostly
NOT due to gravity.
OK... But most free drop experiments relate to the VERTICAL component
of the drop, not the horizontal.

Well, I know that's what you had in mind, but you made a general
statement (which made no restriction about vertical or horizontal
motion), and so I pointed out that the unrestricted statement is
incorrect.

If that falling baseball hits you on
the exact TOP of your head, the "pain" you feel will be due to the sum
of the ball's weight, plus the effective weight increase due to the
velocity a millisecond before the ball hits.

What "effective weight increase"? Is that another, made-up, undefined
term?

How hard the pitcher
threw the ball won't matter in such case.

3. Can any falling object that began at velocity zero manifest a
force delivery potential,
That is a private term. You'll have to define what it means.
OK. If you have a 50 pound steel plate resting on your chest (while
lying on your back), you feel the FORCE, in pounds, on your chest. If
you raise that same plate up 16.087 feet and drop it, the same plate
will hit your chest with double the force, or 100 pounds.

This is completely wrong. I spent years designing impact testing

No, the force will depend on the distance the target moves in slowing
down the plate upon impact. The force the plate exerts on a cement
floor when dropped from 16 ft is different than the force exerted on a
mattress when dropped from the same height. The rule you have that
says that raising an object 16 feet will double the impact force is
simply wrong.

So, the
"force delivery potential" of an object falling 16.087 feet starting
from velocity zero is 2 weight units.

But your rule is wrong. I haven't got the foggiest idea where you
picked up such a crappy idea.

As PD says, there is no direct relationship.

or kinetic energy, greater than the sum of
the forces being imparted to it by gravity as it falls? If not, why?
Well, yes. A bird flapping vigorously in a dive can end up with a
kinetic energy that is larger than if it just folded its wings and
fell.
Wait a minute... We're talking about free drops of compact objects
(dense), NOT powered drops of flying birds!

Well, you are NOW, but you didn't specify it at first. This is the
problem you have. You make general, unrestricted statements when in
fact you have a particular, restricted case in mind.

Try to think with the
"box"!

4. If a force deliver potential, or KE, is imparted to any object,
and in only ONE particular second of its fall, is it OK to recount,
anew, and on an accruing basis, that same numeric value in every
succeeding second,
Is it OK? Depends on whether you want to count the total amount of KE
available or only the change in the KE in the last second. If you have
a savings account that pays 2.5% interest annually, and you started
with $100 in the account, is the amount of money you have in that
account after year two $105.06, or just $2.50? If you needed to buy a
new set of cinderblocks to put under the car in the front yard, I'd
expect you'd be able to tap the whole $105.06.
To pass a test, you must be able to read! Clearly, my "question"
describes "interest" (energy) earned and counted during, say, one
month. The question did NOT say to use that same interest rate from
now on!

OK, is the amount of money you have in the bank after year one
$102.50, or just $2.50?

until the object impacts another, or until its
available kinetic energy can be extracted via mechanical means?
5. The equation for kinetic energy that is shown in most texts is
that of Coriolis, and dates from 1830. It says: KE = 1/2mv^2.
Yes, and that has been tested in thousands of applications after
Coriolis first surmised it, and with much higher precision than what
Coriolis was able to achieve.
The equation was accepted and USED thousands of times, but it was
never TESTED by any means that would pass the SCIENTIFIC METHOD!

This is tested every day by all the labs doing materials testing

Of course it was. Are you that ignorant of experiment?

The
results given by that equation—like earthquake values on the Richter
scale—are not proportional comparisons, but they are still
comparisons! No pilot can withstand the 'g' forces that are claimed.
But the various thresholds of survivability can still be "defined" by
values that aren't linear.

And in fact, in the last 80 years or so, we've been able to measure
that at sufficiently high v, that equation is clearly wrong, though it
does get the right answer to high precision at low v.
Right... compared to what?

Compared to measurement.

Einstein used such equation as the basis for his theories of
relativity.
Actually, no, he didn't. If you think he did, it might be useful for
you to point out where you think he did.
The book name and page number was given to answer that several times
before. I don't have time to chase it down for you.

No, it wasn't. You're lying. A Google search of your posts shows you
never did.

But Einstein had his own unique ideas about what “energy”
is.
Why? What do you think are his unique ideas about what energy is?
That "his" energy is an increase in the mass (HA, ha, ha).

You didn't read what I just wrote about mass. Mass doesn't increase,
in the modern definition of mass. Which Einstein accepted.

Except for
subatomic particles, there has never been any observed mass increase
of ANY object.

What? Subatomic particles don't count?
His view is that if they disagree with what he wants to be true

If there had been, people would be making gold go very
fast to make more gold!

But you can't spend fast-moving gold. And when you stop it, you're
back to the original mass. What's the issue?

Should Coriolis’s equation be evaluated based on issues of
mechanics that were in place during Coriolis’s day?
No. It should be evaluated based on measurements that can be performed
today.
Einstein ACCEPTED the equation of Coriolis.

He accepted the hundreds of experimental verifications that were done
between 1830 and 1905. You seem completely oblivious to them.

But to require that one
inject... relativity into the simple discussion of impact forces is
nothing more than an escape of those who are wrong!

Or should the
above equation be evaluated using Einstein’s much more controversial
ideas about what energy is?
You still haven't established what you think his controversial ideas
about energy are.
I've disproved Einstein up, down, and sideways! So, discussions of
that man's "roles" in understanding the mechanics are thus moot.

You have only shown that you do not understand physics and have no

Depends on whether you think you've disproved something he never said.

6. If a small test car is COASTING on a very smooth surface on a
windless day, and the velocity stays essentially unchanged during a
particular second, does the KE ‘increase’ because the car rolled, say,
20 feet during that second? Or does the KE stay essentially the same
during that 20 feet of roll?
It stays the same. If however, there is work done on the small test
car and its velocity is changing, then the KE is definitely changing.
Moreover, if the car is observed to go 20 feet in one second and 40
feet in the next second, then you KNOW there is work being done on the
small test car and that the KE is definitely changing.
You continue to ESCAPE, by expanding the discussion outside of the
simple mechanics issues stated. I SAID that the car was COASTING!

Yes, I know. And I answered that case, too.

7. If the same small test car has a motor that will increase its
velocity by 32 feet per second EACH second, is the car’s velocity
increasing at a uniform rate?
Yes.
FANTASTIC! Since such velocity increase is what happens to all free-
falling objects, then you are acknowledging that the velocity—caused
by the work done by gravity—increases uniformly!

The velocity increases uniformly, yes.

So... the work done
must be uniform, too!

No, not at all. That would be true if there were a linear relationship
between work and velocity. But there's not. You cannot insist that
there MUST be a linear relationship between work and velocity when it
just doesn't happen.

You have this bonehead idea that if A is related to B in any way, and
B increases linearly, then A must increase linearly too. That is
indeed a bonehead idea. I'll give you a simple example, you bonehead.
Suppose I told you that 3 pints of paint are required to paint a large
room. Now, suppose I want to paint another room that is twice as big
in every dimension. Well, I've doubled the dimensions of the room, and
by your argument, then I should double the paint. But if I send you to
the store to buy the paint, you're going to screw up. Because you're a
bonehead.

And so the KE can only be increasing LINEARLY.

Nope, it doesn't.

The latter deductive reasoning disproves that KE increases
exponentially and thus disproves Einstein's theories of relativity!

8. If the car’s clutch is depressed when the car reaches a velocity
of 64 feet per second, how far will the car COAST during the next
second?
64 feet.
CORRECT!

9. Is the KE of the car being increased during the time that it is
COASTING?
Not if it goes the same distance on subsequent seconds. If, however,
the car goes further in the 2nd second than it does in the 1st second,
then the KE is being increased, yes.
Once again, you are expanding the answers outside of the clearly
stated problem. But the first part of your reply is correct.

Or is the total KE equal only to the accrued force that was
imparted during the time that the clutch, between the motor and the
car’s wheels, was engaged?
Actually, if you have the clutch engaged between the motor and the
car's wheels, and you are going at a constant 20 mph that whole time
(which I can certainly do, either on an icy road or in a school zone)
then the KE is not changing during that whole time either. Perhaps you
are confusing KE and force?
No! KE is a "force delivery potential".

I've already told you your definition of "force delivery potential" is
hogwash.

KE becomes a "force" as soon
as a resistance is in place.

No, it doesn't. Once again, you have this absolutely BONEHEADED idea
that if a collision produces forces, momentum transfers, kinetic
energy transfers, etc., then they all must be the same quantity,
measured in the same units.

So both KE and force are measured in
pounds.

10. Would you say that the ‘gravity and KE’ relationship must conform
to the LAW of the Conservation of Energy?
Yes. But this does not say that gravity can only dole out KE at a
constant rate with time. I don't know where you ever got the notion
that conservation of energy says that gravity can only dole out energy
at a constant rate with time.
It's a constant rate of increase, because the velocity increase is
linear;

See paint example. If A increases linearly, and B is related to A,
this does not demand that B increases linearly.

the work available is uniform (the object's static weight x
the 16.087 ft. of "new" distance increase ABOVE the coasting
distance); and the accepted momentum is increasing linearly.

Or is the latter law to be
disregarded because Coriolis and Einstein didn’t understand such
relationship?
What relationship did you have in mind?
THINK! Or is that too hard for you?

Thinking like you is indeed difficult, as it would require too much
boneheadedness for me to muster.

11. Einstein says that the KE of all objects becomes infinite at
velocity ‘c’.
All massive objects, yes. This is part of where he shows that
(1/2)mv^2 can't be right at all speeds, because... well... at v=c,
this obviously isn't infinite.
That’s because you are considering just set numerical values. Since
the plotted free drop curve is an inverted parabola, ALL parabolas go
to (effective) infinity very fast.

What is "(effective) infinity"? Another made-up term?

He came to that conclusion by realizing that Coriolis’s
equation, as in question 5, above, shows an exponential increase in KE
with respect to velocity.
No, he didn't. First of all, a v^2 dependence is not exponential. You
keep confusing an exponential behavior with a power law behavior.
Secondly, notice that when v=c, then (1/2)mv^2 isn't infinite.
Dear... "One neuron": Exponents are what "powers" are called. The
general term 'exponential' merely requires that the rate of increase
always be greater in one second than in the previous second.

No, it doesn't. Exponential increase means something very specific,
not the very general and coarse meaning you give it. Try googling
"exponential growth".

Are equations “laws of nature” just because
they have been written so many times?
No, because they've been tested in experiment, with the results
published for you to see if you would care to remove your butt from
the chair and go to the library.
So, you use "publishing" as the standard of proof... Get something
published, and that meets your status quo ideas of infallibility.

No, but experiment IS a standard of proof. And in order to find out
what experimental data has been gathered, it is useful to look up
where all that is recorded.

Or are equations subject to on-
going critical examination as to their viability?
Certainly subject to ongoing critical examination. That's why
experimenters continue to publish the results of tests, even to this
day. Do you have something you'd like to publish?
So, why don't you let go of your "defend the status quo with your
life" attitude! The reason: You don't have an analytical neuron in
your brain—just that "one" status quo neuron!

So you don't have anything to publish? Or do you?

12. Einstein also says that as an object’s velocity is increased
toward ‘c’ that its mass will become infinite,
That definition of "mass" is not commonly used any more. This was a
usage of "something like mass" that Einstein used as a teaching
device, but more precise language today says that mass is a constant.
Einstein himself later adopted this more precise meaning of mass.
That’s one of those there "Is... isn't..." smoke screens thrown up by
Einstein.

No, actually, it wasn't Einstein that made that clarification. Do you
need help with the history of the development of relativity?

That's just his way of seeming to be right no matter how
the subject is approached. You, too, answer in a conditional way,
because you need "escapes" if someone disagrees with you. You keep
changing the point of reference.

Just stating a fact, NoEinstein. I realize it comes as a surprise to
you, because you thought something else was true. That's not my
problem, is it?

and the object will be
flattened like a pancake by its own increasing inertia.
It is flattened like a pancake but that isn't an effect if its own
increasing inertia. You just made that up.
WRONG! Inertia is the ONLY resistance that "might be" available to
flatten an object traveling in outer space.

Nope, the flattening has to do with something else entirely. I realize
you cannot imagine what that might be. That is because you're a
bonehead.

And the only inertia felt
by an object traveling at acceleration 'g' is its own UNIFORM weight,
NOT an increasing force!

Can any
object experience a force greater than its static weight by traveling
at a uniform acceleration (or.. x ft./sec. EACH second)?
Certainly. Air Force pilots encounter it all the time, and that's why
they wear "gee" suits to keep themselves from passing out from those
forces during tight turns and other maneuvers.
You're escaping again! The discussion has only mentioned linear
velocity and linear velocity increases. Objects on curving courses
have off-of-discussion-area lateral forces.

OK, then Space Shuttle astronauts encountering linear velocity
increases upon launch of the rocket certainly do experience forces
greater than their static weight.

So do drivers in cars that hit bridge abutments.

If you get shot by a bullet from a gun, the force the bullet exerts on
your body equals the force your body exerts on the bullet (Newton's
3rd law, see your 7th grade science book), and both those forces are
more than the weight of the bullet.
So… 2 is greater than either of the 1 + 1 numbers. How profound! You
should get that one copyrighted!

??

13. Is it possible for an object traveling at a uniform acceleration
to exceed velocity ‘c’ if there is no impedance by impacts with other
matter or with the ether that is in the vacuum of space?
No. And this has been tested.
Ether resistance will increase the mass of some sub-atomic particles.

Already deflated this for you. Mass doesn't increase.

You keep escaping outside of the force of impact discussion area of
mechanics.

14. Einstein’s space-time ideas have been used as ‘escapes’ from the
rational discussions in the present subject areas.
What constitutes a "rational discussion" in your mind?
That's the easiest of all: ANY discussion not with you! HA, ha, ha…!

Oh dear. You seem to have some spittle on your chin now.

Is it OK to still
use those ‘escapes’ even though the Michelson-Morley experiment that
caused Einstein to contrive his space-time postulates
Actually, Einstein didn't know much about the M-M experiment when he
was working on relativity. He was much more concerned about Maxwell's
equations.
Maxwell recommended to Michelson that he "try" detecting velocity
changes with his interferometer. When 'that' experiment failed,
Lorentz, a mentor of Einstein, CONTRIVED the contraction factor Beta—
that is the divisor under the SR equation. So, everything that
Einstein did is the direct result of his inability to understand the
workings of M-M.

As I said, you have a very poor grasp of the history of this
development. What crap books have you been reading?

has been
invalidated as a velocity-detecting device, because it lacked a
CONTROL?
Well, a control is not needed in this case.
You, like Gisse, don't understand that all interferometers COMPARE two
light beams to measure things.

Sorta. Roughly. No, not really. Do you know what the *interference*
part does?

Since velocity detection demands that
one of the light courses be designed so as to change its TIME of
travel due to velocity, M-M failed because all of the optics were
mounted on the same horizontal plane.

He is channelling seto here.

Yes, but the time in one direction is changed and the other is not. Do
you see why?

So, the TIME of travel of BOTH
light beams remains unchanged, even when the apparatus is rotated.
You are an "expert" on particle stuff, not on optics nor mechanics.

He has been unable to calculate average velocity. He still thinks that

Oh, I've done optics and mechanics, too. That's one of the beauties of
being an experimentalist. I've gotten familiar with all sorts of
things:
- analog and digital electronics design
- coding computer software
- optics
- cryogenics
- high vacuum methods
- structural engineering
- steel metallurgy
- EMF shielding
- surveying and alignment
- clean room methodology
- photosensors
- servo motor control and location sensing
- high voltage methods
- gas mixing and flow control
- power conditioning
- calorimetry

Secondly, dozens --
repeat, DOZENS -- of experiments have been conducted since M-M that do
a much better job of testing relativity, and they all confirm
relativity. None of them are dependent on the M-M results.
All they TEST is that Einstein's equations are a close empirical
ANALOGY to the observed effects.
Einstein knew about those effects
BEFORE he wrote his equations.

Nope, many of them were not found until *after* he developed
relativity. And relativity accurately predicted what was *later*
found. Again, your understanding of the history of the development of
relativity is simply pathetic.

Now, when any observation seems to
agree with those equations people say: "See, Einstein predicted
this!" But he predicted nothing—because he knew about unexplained
observations BEFORE he wrote the equations! To his death Einstein
believed that gravity was just: objects accelerating along their world
lines in the space-time continuum. He had no idea of the true cause
of gravity, so he said that his equations were the cause! HA, ha,
ha...!

15. Which is more correct? (A.) Science is a static collective of
ideas. Or (B.) Science is a continuing re evaluation of all that has
been presented in the past.
B. But the latter is done via confrontation with published results of
experiment. Do you have something you'd like to publish?
Already answered. PD, you are still very much a PNG. Don't expect me
to continue talking to someone with zero reasoning ability, like you.

Whatever. I don't expect anything of you. I don't have high
expectations of boneheads.

— NoEinstein —

— NoEinstein —
Those who are interested are urged to read these earlier posts:
Conservation of Energy Law Invalidates Einstein
http://groups.google.com/group/sci.physics/browse_thread/thread/042a8...
Where Angels Fear to Fallhttp://groups.google.com/group/sci.physics/browse_thread/thread/1e3e4...
Cleaning Away Einstein’s Mishmashhttp://groups.google.com/group/sci.physics/browse_thread/thread/5d847...
Dropping Einstein Like a Stonehttp://groups.google.com/group/sci.physics/browse_thread/thread/989e1...

Noeinstein should really listen to and learn from PD. PD is far more

Walking down the Ave, a man with cigarettes ( like me )
gets hit on constantly.. the " rich " are never lonely.

Everyone needs something.. be it a cig, a beer, money,
someone to talk to, a place to sleep, whatever.

On Apr 28, 8:31 pm, NoEinstein <noeinst...@bellsouth.net> wrote:

On Apr 28, 12:10 pm, PD <TheDraperFam...@gmail.com> wrote:> On Apr 26, 12:51 pm, NoEinstein <noeinst...@bellsouth.net> wrote:

1.  Does gravity do continuous work on falling objects?

Yes, though when an object reaches terminal velocity in the
atmosphere, the net work done on the object is zero.

Yes AND no...  In the ranges of most free drop experiments (less than
300 feet) of COMPACT objects, the terminal velocity doesn't come into
play.

Sure it does. Skydivers.





 But considering air resistance, the total work done by gravity
vs. the total "effect" is still equal!  But the movement or frictional
heating of the air and the falling object will account for the energy
not recoverable from the object's "impact effects".

2.  If so, would you say that gravity is the source energy for the
mechanical effects exhibited by falling objects?  If not, why?

No. When a baseball pitcher throws a fastball, it is falling the
moment it leaves the pitcher's hand (as there is no other force acting
on the ball but gravity after that point). However, the mechanical
effects exhibited by the fastball on the batter's noggin are mostly
NOT due to gravity.

OK...  But most free drop experiments relate to the VERTICAL component
of the drop, not the horizontal.

Well, I know that's what you had in mind, but you made a general
statement (which made no restriction about vertical or horizontal
motion), and so I pointed out that the unrestricted statement is
incorrect.

 If that falling baseball hits you on
the exact TOP of your head, the "pain" you feel will be due to the sum
of the ball's weight, plus the effective weight increase due to the
velocity a millisecond before the ball hits.

What "effective weight increase"? Is that another, made-up, undefined
term?

 How hard the pitcher
threw the ball won't matter in such case.

3.  Can any falling object that began at velocity zero manifest a
force delivery potential,

That is a private term. You'll have to define what it means.

OK.  If you have a 50 pound steel plate resting on your chest (while
lying on your back), you feel the FORCE, in pounds, on your chest.  If
you raise that same plate up 16.087 feet and drop it, the same plate
will hit your chest with double the force, or 100 pounds.

No, the force will depend on the distance the target moves in slowing
down the plate upon impact. The force the plate exerts on a cement
floor when dropped from 16 ft is different than the force exerted on a
mattress when dropped from the same height. The rule you have that
says that raising an object 16 feet will double the impact force is
simply wrong.

 So, the
"force delivery potential" of an object falling 16.087 feet starting
from velocity zero is 2 weight units.

But your rule is wrong. I haven't got the foggiest idea where you
picked up such a crappy idea.



or kinetic energy, greater than the sum of
the forces being imparted to it by gravity as it falls?  If not, why?

Well, yes. A bird flapping vigorously in a dive can end up with a
kinetic energy that is larger than if it just folded its wings and
fell.

Wait a minute...  We're talking about free drops of compact objects
(dense), NOT powered drops of flying birds!

Well, you are NOW, but you didn't specify it at first. This is the
problem you have. You make general, unrestricted statements when in
fact you have a particular, restricted case in mind.





 Try to think with the
"box"!

4.  If a force deliver potential, or KE, is imparted to any object,
and in only ONE particular second of its fall, is it OK to recount,
anew, and on an accruing basis, that same numeric value in every
succeeding second,

Is it OK? Depends on whether you want to count the total amount of KE
available or only the change in the KE in the last second. If you have
a savings account that pays 2.5% interest annually, and you started
with $100 in the account, is the amount of money you have in that
account after year two $105.06, or just $2.50? If you needed to buy a
new set of cinderblocks to put under the car in the front yard, I'd
expect you'd be able to tap the whole $105.06.

To pass a test, you must be able to read!  Clearly, my "question"
describes "interest" (energy) earned and counted during, say, one
month.  The question did NOT say to use that same interest rate from
now on!

OK, is the amount of money you have in the bank after year one
$102.50, or just $2.50?



until the object impacts another, or until its
available kinetic energy can be extracted via mechanical means?

5.  The equation for kinetic energy that is shown in most texts is
that of Coriolis, and dates from 1830.  It says: KE = 1/2mv^2.

Yes, and that has been tested in thousands of applications after
Coriolis first surmised it, and with much higher precision than what
Coriolis was able to achieve.

The equation was accepted and USED thousands of times, but it was
never TESTED by any means that would pass the SCIENTIFIC METHOD!

Of course it was. Are you that ignorant of experiment?

 The
results given by that equation—like earthquake values on the Richter
scale—are not proportional comparisons, but they are still
comparisons!  No pilot can withstand the 'g' forces that are claimed.
But the various thresholds of survivability can still be "defined" by
values that aren't linear.

And in fact, in the last 80 years or so, we've been able to measure
that at sufficiently high v, that equation is clearly wrong, though it
does get the right answer to high precision at low v.

Right... compared to what?

Compared to measurement.



Einstein used such equation as the basis for his theories of
relativity.

Actually, no, he didn't. If you think he did, it might be useful for
you to point out where you think he did.

The book name and page number was given to answer that several times
before.  I don't have time to chase it down for you.

No, it wasn't. You're lying. A Google search of your posts shows you
never did.



But Einstein had his own unique ideas about what “energy”
is.

Why? What do you think are his unique ideas about what energy is?

That "his" energy is an increase in the mass (HA, ha, ha).

You didn't read what I just wrote about mass. Mass doesn't increase,
in the modern definition of mass. Which Einstein accepted.

 Except for
subatomic particles, there has never been any observed mass increase
of ANY object.

What? Subatomic particles don't count?

 If there had been, people would be making gold go very
fast to make more gold!

But you can't spend fast-moving gold. And when you stop it, you're
back to the original mass. What's the issue?



Should Coriolis’s equation be evaluated based on issues of
mechanics that were in place during Coriolis’s day?

No. It should be evaluated based on measurements that can be performed
today.

Einstein ACCEPTED the equation of Coriolis.

He accepted the hundreds of experimental verifications that were done
between 1830 and 1905. You seem completely oblivious to them.

 But to require that one
inject... relativity into the simple discussion of impact forces is
nothing more than an escape of those who are wrong!

 Or should the
above equation be evaluated using Einstein’s much more controversial
ideas about what energy is?

You still haven't established what you think his controversial ideas
about energy are.

I've disproved Einstein up, down, and sideways!  So, discussions of
that man's "roles" in understanding the mechanics are thus moot.

Depends on whether you think you've disproved something he never said.



6.  If a small test car is COASTING on a very smooth surface on a
windless day, and the velocity stays essentially unchanged during a
particular second, does the KE ‘increase’ because the car rolled, say,
20 feet during that second?  Or does the KE stay essentially the same
during that 20 feet of roll?

It stays the same. If however, there is work done on the small test
car and its velocity is changing, then the KE is definitely changing.
Moreover, if the car is observed to go 20 feet in one second and 40
feet in the next second, then you KNOW there is work being done on the
small test car and that the KE is definitely changing.

You continue to ESCAPE, by expanding the discussion outside of the
simple mechanics issues stated.  I SAID that the car was COASTING!

Yes, I know. And I answered that case, too.



7.  If the same small test car has a motor that will increase its
velocity by 32 feet per second EACH second, is the car’s velocity
increasing at a uniform rate?

Yes.

FANTASTIC!  Since such velocity increase is what happens to all free-
falling objects, then you are acknowledging that the velocity—caused
by the work done by gravity—increases uniformly!

The velocity increases uniformly, yes.

 So... the work done
must be uniform, too!

No, not at all. That would be true if there were a linear relationship
between work and velocity. But there's not. You cannot insist that
there MUST be a linear relationship between work and velocity when it
just doesn't happen.

You have this bonehead idea that if A is related to B in any way, and
B increases linearly, then A must increase linearly too. That is
indeed a bonehead idea. I'll give you a simple example, you bonehead.
Suppose I told you that 3 pints of paint are required to paint a large
room. Now, suppose I want to paint another room that is twice as big
in every dimension. Well, I've doubled the ...

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PD wrote:
On Apr 28, 8:31 pm, NoEinstein <noeinst...@bellsouth.net> wrote:
On Apr 28, 12:10 pm, PD <TheDraperFam...@gmail.com> wrote:> On Apr 26, 12:51 pm, NoEinstein <noeinst...@bellsouth.net> wrote:

1.  Does gravity do continuous work on falling objects?
Yes, though when an object reaches terminal velocity in the
atmosphere, the net work done on the object is zero.
Yes AND no...  In the ranges of most free drop experiments (less than
300 feet) of COMPACT objects, the terminal velocity doesn't come into
play.

Sure it does. Skydivers.

 But considering air resistance, the total work done by gravity
vs. the total "effect" is still equal!  But the movement or frictional
heating of the air and the falling object will account for the energy
not recoverable from the object's "impact effects".

2.  If so, would you say that gravity is the source energy for the
mechanical effects exhibited by falling objects?  If not, why?
No. When a baseball pitcher throws a fastball, it is falling the
moment it leaves the pitcher's hand (as there is no other force acting
on the ball but gravity after that point). However, the mechanical
effects exhibited by the fastball on the batter's noggin are mostly
NOT due to gravity.
OK...  But most free drop experiments relate to the VERTICAL component
of the drop, not the horizontal.

Well, I know that's what you had in mind, but you made a general
statement (which made no restriction about vertical or horizontal
motion), and so I pointed out that the unrestricted statement is
incorrect.

 If that falling baseball hits you on
the exact TOP of your head, the "pain" you feel will be due to the sum
of the ball's weight, plus the effective weight increase due to the
velocity a millisecond before the ball hits.

What "effective weight increase"? Is that another, made-up, undefined
term?

 How hard the pitcher
threw the ball won't matter in such case.

3.  Can any falling object that began at velocity zero manifest a
force delivery potential,
That is a private term. You'll have to define what it means.
OK.  If you have a 50 pound steel plate resting on your chest (while
lying on your back), you feel the FORCE, in pounds, on your chest.  If
you raise that same plate up 16.087 feet and drop it, the same plate
will hit your chest with double the force, or 100 pounds.

This is completely wrong. I spent years designing impact testing
equipment and it is clear you have not looked at the situation.
The weight lifted over a certain distance has more energy. The force
on a target is a function of the compliance of the target. The
energy absorbed is the integral of the force versus distance curve so
that the larger the displacement, the lower the total force. In general
there is no direct relationship between height and force. You have shown
that you do not understand energy and force but that could be fixed
with some study on your part.

No, the force will depend on the distance the target moves in slowing
down the plate upon impact. The force the plate exerts on a cement
floor when dropped from 16 ft is different than the force exerted on a
mattress when dropped from the same height. The rule you have that
says that raising an object 16 feet will double the impact force is
simply wrong.

 So, the
"force delivery potential" of an object falling 16.087 feet starting
from velocity zero is 2 weight units.

But your rule is wrong. I haven't got the foggiest idea where you
picked up such a crappy idea.

As PD says, there is no direct relationship.





or kinetic energy, greater than the sum of
the forces being imparted to it by gravity as it falls?  If not, why?
Well, yes. A bird flapping vigorously in a dive can end up with a
kinetic energy that is larger than if it just folded its wings and
fell.
Wait a minute...  We're talking about free drops of compact objects
(dense), NOT powered drops of flying birds!

Well, you are NOW, but you didn't specify it at first. This is the
problem you have. You make general, unrestricted statements when in
fact you have a particular, restricted case in mind.

 Try to think with the
"box"!

4.  If a force deliver potential, or KE, is imparted to any object,
and in only ONE particular second of its fall, is it OK to recount,
anew, and on an accruing basis, that same numeric value in every
succeeding second,
Is it OK? Depends on whether you want to count the total amount of KE
available or only the change in the KE in the last second. If you have
a savings account that pays 2.5% interest annually, and you started
with $100 in the account, is the amount of money you have in that
account after year two $105.06, or just $2.50? If you needed to buy a
new set of cinderblocks to put under the car in the front yard, I'd
expect you'd be able to tap the whole $105.06.
To pass a test, you must be able to read!  Clearly, my "question"
describes "interest" (energy) earned and counted during, say, one
month.  The question did NOT say to use that same interest rate from
now on!

OK, is the amount of money you have in the bank after year one
$102.50, or just $2.50?

until the object impacts another, or until its
available kinetic energy can be extracted via mechanical means?
5.  The equation for kinetic energy that is shown in most texts is
that of Coriolis, and dates from 1830.  It says: KE = 1/2mv^2.
Yes, and that has been tested in thousands of applications after
Coriolis first surmised it, and with much higher precision than what
Coriolis was able to achieve.
The equation was accepted and USED thousands of times, but it was
never TESTED by any means that would pass the SCIENTIFIC METHOD!

This is tested every day by all the labs doing materials testing
all over the world. None of their equipment would work if you
were correct. Thus you are not correct.





Of course it was. Are you that ignorant of experiment?

 The
results given by that equation—like earthquake values on the Richter
scale—are not proportional comparisons, but they are still
comparisons!  No pilot can withstand the 'g' forces that are claimed.
But the various thresholds of survivability can still be "defined" by
values that aren't linear.

And in fact, in the last 80 years or so, we've been able to measure
that at sufficiently high v, that equation is clearly wrong, though it
does get the right answer to high precision at low v.
Right... compared to what?

Compared to measurement.

Einstein used such equation as the basis for his theories of
relativity.
Actually, no, he didn't. If you think he did, it might be useful for
you to point out where you think he did.
The book name and page number was given to answer that several times
before.  I don't have time to chase it down for you.

No, it wasn't. You're lying. A Google search of your posts shows you
never did.

But Einstein had his own unique ideas about what “energy”
is.
Why? What do you think are his unique ideas about what energy is?
That "his" energy is an increase in the mass (HA, ha, ha).

You didn't read what I just wrote about mass. Mass doesn't increase,
in the modern definition of mass. Which Einstein accepted.

 Except for
subatomic particles, there has never been any observed mass increase
of ANY object.

What? Subatomic particles don't count?

His view is that if they disagree with what he wants to be true
it must not be true. Very immature children are like that.







 If there had been, people would be making gold go very
fast to make more gold!

But you can't spend fast-moving gold. And when you stop it, you're
back to the original mass. What's the issue?

Should Coriolis’s equation be evaluated based on issues of
mechanics that were in place during Coriolis’s day?
No. It should be evaluated based on measurements that can be performed
today.
Einstein ACCEPTED the equation of Coriolis.

He accepted the hundreds of experimental verifications that were done
between 1830 and 1905. You seem completely oblivious to them.

 But to require that one
inject... relativity into the simple discussion of impact forces is
nothing more than an escape of those who are wrong!

 Or should the
above equation be evaluated using Einstein’s much more controversial
ideas about what energy is?
You still haven't established what you think his controversial ideas
about energy are.
I've disproved Einstein up, down, and sideways!  So, discussions of
that man's "roles" in understanding the mechanics are thus moot.

You have only shown that you do not understand physics and have no
desire to learn it.



Depends on whether you think you've disproved something he never said.

6.  If a small test car is COASTING on a very smooth surface on a
windless day, and the velocity stays essentially unchanged during a
particular second, does the KE ‘increase’ because the car rolled, say,
20 feet during that second?  Or does the KE stay essentially the same
during that 20 feet of roll?
It stays the same. If however, there is work done on the small test
car and its velocity is changing, then the KE is definitely changing.
Moreover, if the car is observed to go 20 feet in one second and 40
feet in the next second, then you KNOW there is work being done on the
small test car and that the KE is definitely changing.
You continue to ESCAPE, by expanding the discussion outside of the
simple mechanics issues stated.  I SAID that the car was COASTING!

Yes, I know. And I answered that case, too.

7.  If

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1.  Does gravity do continuous work on falling objects?

2.  If so, would you say that gravity is the source energy for the
mechanical effects exhibited by falling objects?  If not, why?

3.  Can any falling object that began at velocity zero manifest a
force delivery potential, or kinetic energy, greater than the sum of
the forces being imparted to it by gravity as it falls?  If not, why?

On Apr 26, 9:51 am, NoEinstein <noeinst...@bellsouth.net> wrote:

1.  Does gravity do continuous work on falling objects?

2.  If so, would you say that gravity is the source energy for the
mechanical effects exhibited by falling objects?  If not, why?

Gravity turns the mass of the falling body into motion or kinetic
energy.



3.  Can any falling object that began at velocity zero manifest a
force delivery potential, or kinetic energy, greater than the sum of
the forces being imparted to it by gravity as it falls?  If not, why?

Objects do not start falling at zero. They start at 16 feet in the
first second. Falling is instantaneous speed that happens all at once.
Light is a good example of the phenomnon of jumping to velocity.

The confusion is with inclined planes. But that is not freefall.

Mitch Raemsch

Water is no replacement for the flavor and smell of smoke;
and it's fun to light / nurse / watch the ember.

On Apr 29, 11:02 pm, mitch.nicolas.raem...@gmail.com wrote:
On Apr 26, 9:51 am, NoEinstein <noeinst...@bellsouth.net> wrote:

1. Does gravity do continuous work on falling objects?
2. If so, would you say that gravity is the source energy for the
mechanical effects exhibited by falling objects? If not, why?
Gravity turns the mass of the falling body into motion or kinetic
energy.



3. Can any falling object that began at velocity zero manifest a
force delivery potential, or kinetic energy, greater than the sum of
the forces being imparted to it by gravity as it falls? If not, why?
Objects do not start falling at zero. They start at 16 feet in the
first second. Falling is instantaneous speed that happens all at once.
Light is a good example of the phenomnon of jumping to velocity.

The confusion is with inclined planes. But that is not freefall.

Mitch Raemsch

Dear Mitch: You are like a tone deaf singer—your science ideas don't
"blend" well. Objects have an ACCELERATION of 32.174 ft. per sec.
EACH second even while they are being held before dropping. But the
velocity of that same object IS zero while it is being held. You are
confusing velocity and acceleration. But you’re in "good" company!
Einstein didn't know what acceleration is, either! — NoEinstein —