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Wikipedia also shows the constant vibration in Cavendish Experiment #157 New Physics #266 ATOM TOTALITY 5th ed
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Archimedes Plutonium
Feb 1, 2012, 3:06:22 PM2/1/12
to
Below I have quoted the precision or dare I say "lack of precision"
involved
with the Cavendish Experiment and that lack of precision is in all
replications of the Cavendish Experiment. In subsequent posts, I will
show that the Cavendish Experiment is EM-gravity and that the
Cavendish constant vibration also disproves the Casimir Effect
Experiments in that they are both experiments of parallel plate
capacitors in which there is not only a attraction but a repulsion.
The constant vibration in the Cavendish Experiment as what Wikipedia
notes below, is that lead balls form a parallel plate capacitor and to
see those vibrations even better, if Cavendish had designed the balls
to be rectangular in shape as my experiment uses 50lb
steel weights and use a side of one weight next to the side of another
weight, forming a better plate capacitor. My experiment highlights
that Cavendish could never eliminate the vibration because the balls
were fluctuating between attraction and repulsion of the electricity
and magnetism set up by the lead balls-- in mine the steel 50lb
weights. In the Casimir Effect, it alternates between the plates
attracting and repulsing.
I am happy to learn that in the Cavendish Experiment they required
telescopes to read the twist motion.
--- quoting Wikipedia on the precision involved in Cavendish
Experiment ---
http://en.wikipedia.org/wiki/Cavendish_experiment
To find the wire's torsion coefficient, the torque exerted by the wire
for a given angle of twist, Cavendish timed the natural oscillation
period of the balance rod as it rotated slowly clockwise and
counterclockwise against the twisting of the wire. The period was
about 20 minutes. The torsion coefficient could be calculated from
this and the mass and dimensions of the balance. Actually, the rod was
never at rest; Cavendish had to measure the deflection angle of the
rod while it was oscillating.[10]
Cavendish's equipment was remarkably sensitive for its time.[9] The
force involved in twisting the torsion balance was very small, 1.74 x
10–7 N,[11] about 1/50,000,000 of the weight of the small balls[12] or
roughly the weight of a large grain of sand.[13] To prevent air
currents and temperature changes from interfering with the
measurements, Cavendish placed the entire apparatus in a wooden box
about 2 feet (0.61 m) thick, 10 feet (3.0 m) tall, and 10 feet (3.0 m)
wide, all in a closed shed on his estate. Through two holes in the
walls of the shed, Cavendish used telescopes to observe the movement
of the torsion balance's horizontal rod. The motion of the rod was
only about 0.16 inches (4.1 mm).[14] Cavendish was able to measure
this small deflection to an accuracy of better than one hundredth of
an inch using vernier scales on the ends of the rod.[15]
Cavendish's experiment was repeated by Reich (1838), Baily (1843),
Cornu & Baille (1878), and many others. Its accuracy was not exceeded
for 97 years, until C. V. Boys' 1895 experiment. In time, Michell's
torsion balance became the dominant technique for measuring the
gravitational constant (G), and most contemporary measurements still
use variations of it. This is why Cavendish's experiment became the
Cavendish experiment.[16]
--- end quoting ---
I wonder if Michell's torsion balance required a telescope in order to
read the twist results?
I have to redesign my experiment because my steel 50 lb weights that I
used yesterday are rectangular in shape and I was measuring the gap
between the two rectangles on its edge of one weight and the edge of
the
second weight. I have to do the experiment so that the gap is for the
full side of one weight and the full side of the second weight as to
increase the
exposure as a parallel plate capacitors.
Both the Cavendish and Plutonium experiments involve a constant
vibration.
The constancy of the vibration is not a Newtonian gravity nor a
General Relativity, but is due to EM of the materials and the
materials involved are not just the weights but the planet Earth
itself-- the EM of Earth and the EM of the weights.
If Newtonian gravity and General Relativity were correct, then there
should be a gravity force in this equation:
Separation of two 50 lb weights = EM force + gravity force
Because that separation always has an oscillation between attraction
and repulsion indicates there is no gravity force involved and that
the only force involved due to the constant oscillation is the EM
force.
So what we are seeing in the Cavendish Experiment is an elaborate set
up that is in fact a deformed Parallel Plate Capacitor experiment.
Now I propose that using magnets will be better than using 50 lb steel
weights or the lead balls that Cavendish used. And using Galileo
inclined planes for more accurate precision measure of magnets. So we
start with magnets at 90 degrees to the ground stuck onto a
rectangular inclined plane
and we decrease the strength of that magnet until it can no longer
stick to the surface. We then use that magnet on inclined planes to
check for accuracy.
What we are doing is showing that the mass of object A does not
attract mass of object B, but what is going is that the EM inside of
object A has
an attraction force for the EM of object B. That Newtonian gravity
and General Relativity are both fake forces. The size of an object
usually tells us it has a lot of mass, and thus has a lot of EM within
and that EM is attracted to other objects of EM, and so gravity is EM-
gravity. EM-gravity is the uneven distribution of its net positive and
negative charges. The Sun holds all the planets together in the Solar
System because the surface of the Sun can be analyzed as all electron
negative charge surface and the planets surface can be analyzed as all
positive charge.
Now I have my experiment in the garage but am going to try in the next
several days to set it up here at home so that at all hours I can
tinker and fiddle with it. The success of many experiments are
accomplished in the
tinkering and fiddling stages.
Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
involved
with the Cavendish Experiment and that lack of precision is in all
replications of the Cavendish Experiment. In subsequent posts, I will
show that the Cavendish Experiment is EM-gravity and that the
Cavendish constant vibration also disproves the Casimir Effect
Experiments in that they are both experiments of parallel plate
capacitors in which there is not only a attraction but a repulsion.
The constant vibration in the Cavendish Experiment as what Wikipedia
notes below, is that lead balls form a parallel plate capacitor and to
see those vibrations even better, if Cavendish had designed the balls
to be rectangular in shape as my experiment uses 50lb
steel weights and use a side of one weight next to the side of another
weight, forming a better plate capacitor. My experiment highlights
that Cavendish could never eliminate the vibration because the balls
were fluctuating between attraction and repulsion of the electricity
and magnetism set up by the lead balls-- in mine the steel 50lb
weights. In the Casimir Effect, it alternates between the plates
attracting and repulsing.
I am happy to learn that in the Cavendish Experiment they required
telescopes to read the twist motion.
--- quoting Wikipedia on the precision involved in Cavendish
Experiment ---
http://en.wikipedia.org/wiki/Cavendish_experiment
To find the wire's torsion coefficient, the torque exerted by the wire
for a given angle of twist, Cavendish timed the natural oscillation
period of the balance rod as it rotated slowly clockwise and
counterclockwise against the twisting of the wire. The period was
about 20 minutes. The torsion coefficient could be calculated from
this and the mass and dimensions of the balance. Actually, the rod was
never at rest; Cavendish had to measure the deflection angle of the
rod while it was oscillating.[10]
Cavendish's equipment was remarkably sensitive for its time.[9] The
force involved in twisting the torsion balance was very small, 1.74 x
10–7 N,[11] about 1/50,000,000 of the weight of the small balls[12] or
roughly the weight of a large grain of sand.[13] To prevent air
currents and temperature changes from interfering with the
measurements, Cavendish placed the entire apparatus in a wooden box
about 2 feet (0.61 m) thick, 10 feet (3.0 m) tall, and 10 feet (3.0 m)
wide, all in a closed shed on his estate. Through two holes in the
walls of the shed, Cavendish used telescopes to observe the movement
of the torsion balance's horizontal rod. The motion of the rod was
only about 0.16 inches (4.1 mm).[14] Cavendish was able to measure
this small deflection to an accuracy of better than one hundredth of
an inch using vernier scales on the ends of the rod.[15]
Cavendish's experiment was repeated by Reich (1838), Baily (1843),
Cornu & Baille (1878), and many others. Its accuracy was not exceeded
for 97 years, until C. V. Boys' 1895 experiment. In time, Michell's
torsion balance became the dominant technique for measuring the
gravitational constant (G), and most contemporary measurements still
use variations of it. This is why Cavendish's experiment became the
Cavendish experiment.[16]
--- end quoting ---
I wonder if Michell's torsion balance required a telescope in order to
read the twist results?
I have to redesign my experiment because my steel 50 lb weights that I
used yesterday are rectangular in shape and I was measuring the gap
between the two rectangles on its edge of one weight and the edge of
the
second weight. I have to do the experiment so that the gap is for the
full side of one weight and the full side of the second weight as to
increase the
exposure as a parallel plate capacitors.
Both the Cavendish and Plutonium experiments involve a constant
vibration.
The constancy of the vibration is not a Newtonian gravity nor a
General Relativity, but is due to EM of the materials and the
materials involved are not just the weights but the planet Earth
itself-- the EM of Earth and the EM of the weights.
If Newtonian gravity and General Relativity were correct, then there
should be a gravity force in this equation:
Separation of two 50 lb weights = EM force + gravity force
Because that separation always has an oscillation between attraction
and repulsion indicates there is no gravity force involved and that
the only force involved due to the constant oscillation is the EM
force.
So what we are seeing in the Cavendish Experiment is an elaborate set
up that is in fact a deformed Parallel Plate Capacitor experiment.
Now I propose that using magnets will be better than using 50 lb steel
weights or the lead balls that Cavendish used. And using Galileo
inclined planes for more accurate precision measure of magnets. So we
start with magnets at 90 degrees to the ground stuck onto a
rectangular inclined plane
and we decrease the strength of that magnet until it can no longer
stick to the surface. We then use that magnet on inclined planes to
check for accuracy.
What we are doing is showing that the mass of object A does not
attract mass of object B, but what is going is that the EM inside of
object A has
an attraction force for the EM of object B. That Newtonian gravity
and General Relativity are both fake forces. The size of an object
usually tells us it has a lot of mass, and thus has a lot of EM within
and that EM is attracted to other objects of EM, and so gravity is EM-
gravity. EM-gravity is the uneven distribution of its net positive and
negative charges. The Sun holds all the planets together in the Solar
System because the surface of the Sun can be analyzed as all electron
negative charge surface and the planets surface can be analyzed as all
positive charge.
Now I have my experiment in the garage but am going to try in the next
several days to set it up here at home so that at all hours I can
tinker and fiddle with it. The success of many experiments are
accomplished in the
tinkering and fiddling stages.
Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
Archimedes Plutonium
Feb 1, 2012, 3:43:17 PM2/1/12
to
On Feb 1, 2:06 pm, Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
Newtonian gravity and General Relativity are fakes.
We have a refrigerator door and a magnet stuck to it. Now no planet
sticks to the Sun as a magnet sticks to the door. But we weaken the
magnet to the point where it just barely hangs on. Still that is far
greater of a force than the Sun on any planet. Now, we weaken that
magnet so it no longer hangs on the door at 90 degrees to the ground
and thus it falls from the door.
But now we take that weak magnet and place it on a metal inclined
plane like Galileo once
performed. And the magnet fell off the 90 degree inclined plane-- the
door. Now the magnet is
on a 89degree inclined plane and the question is does it fall off? If
it does then we place it on an inclined plane to the point where it
does not roll down but stays put.
We finally get to the point where we have an inclined plane of 1
degrees and a weak magnet that stays put. Now we can get to a magnet
that stays put in fractions of the 1 degree inclined plane. Now we
recall that the planets are attracted to the Sun in orbit, and a
magnet of enough force to stay put on a 1 degree inclined plane. We
weigh the magnet that stayed put at 1 degree inclined plane.
We find that the mass of that magnet should not have stayed put on the
inclined plane.
So logically, what we have shown is that gravity as a Newtonian
gravity or as General Relativity is nonexistent. And that gravity is
all about EM and the uneven distribution of charge
in all objects of mass.
Mercury, Venus, Earth and Jupiter are not magnets stuck to the Sun at
90 degrees, but are magnets stuck to the Sun at perhaps a 5 degree
inclined plane.
<plutonium.archime...@gmail.com> wrote:
(snipped)
>
> Now I propose that using magnets will be better than using 50 lb steel
> weights or the lead balls that Cavendish used. And using Galileo
> inclined planes for more accurate precision measure of magnets. So we
> start with magnets at 90 degrees to the ground stuck onto a
> rectangular inclined plane
> and we decrease the strength of that magnet until it can no longer
> stick to the surface. We then use that magnet on inclined planes to
> check for accuracy.
Alright, let outline the logic of why EM-gravity exists and that
> Now I propose that using magnets will be better than using 50 lb steel
> weights or the lead balls that Cavendish used. And using Galileo
> inclined planes for more accurate precision measure of magnets. So we
> start with magnets at 90 degrees to the ground stuck onto a
> rectangular inclined plane
> and we decrease the strength of that magnet until it can no longer
> stick to the surface. We then use that magnet on inclined planes to
> check for accuracy.
Newtonian gravity and General Relativity are fakes.
We have a refrigerator door and a magnet stuck to it. Now no planet
sticks to the Sun as a magnet sticks to the door. But we weaken the
magnet to the point where it just barely hangs on. Still that is far
greater of a force than the Sun on any planet. Now, we weaken that
magnet so it no longer hangs on the door at 90 degrees to the ground
and thus it falls from the door.
But now we take that weak magnet and place it on a metal inclined
plane like Galileo once
performed. And the magnet fell off the 90 degree inclined plane-- the
door. Now the magnet is
on a 89degree inclined plane and the question is does it fall off? If
it does then we place it on an inclined plane to the point where it
does not roll down but stays put.
We finally get to the point where we have an inclined plane of 1
degrees and a weak magnet that stays put. Now we can get to a magnet
that stays put in fractions of the 1 degree inclined plane. Now we
recall that the planets are attracted to the Sun in orbit, and a
magnet of enough force to stay put on a 1 degree inclined plane. We
weigh the magnet that stayed put at 1 degree inclined plane.
We find that the mass of that magnet should not have stayed put on the
inclined plane.
So logically, what we have shown is that gravity as a Newtonian
gravity or as General Relativity is nonexistent. And that gravity is
all about EM and the uneven distribution of charge
in all objects of mass.
Mercury, Venus, Earth and Jupiter are not magnets stuck to the Sun at
90 degrees, but are magnets stuck to the Sun at perhaps a 5 degree
inclined plane.
Archimedes Plutonium
Feb 1, 2012, 3:53:56 PM2/1/12
to
> Below I have quoted the precision or dare I say "lack of precision"
> involved
> with the Cavendish Experiment and that lack of precision is in all
> replications of the Cavendish Experiment. In subsequent posts, I will
> show that the Cavendish Experiment is EM-gravity and that the
> Cavendish constant vibration also disproves the Casimir Effect
> Experiments in that they are both experiments of parallel plate
> capacitors in which there is not only a attraction but a repulsion.
> The constant vibration in the Cavendish Experiment as what Wikipedia
> notes below, is that lead balls form a parallel plate capacitor and to
> see those vibrations even better, if Cavendish had designed the balls
> to be rectangular in shape as my experiment uses 50lb
> steel weights and use a side of one weight next to the side of another
> weight, forming a better plate capacitor. My experiment highlights
> that Cavendish could never eliminate the vibration because the balls
> were fluctuating between attraction and repulsion of the electricity
> and magnetism set up by the lead balls-- in mine the steel 50lb
> weights. In the Casimir Effect, it alternates between the plates
> attracting and repulsing.
>
> I am happy to learn that in the Cavendish Experiment they required
> telescopes to read the twist motion.
>
> --- quoting Wikipedia on the precision involved in Cavendish
> Experiment ---http://en.wikipedia.org/wiki/Cavendish_experiment
> involved
> with the Cavendish Experiment and that lack of precision is in all
> replications of the Cavendish Experiment. In subsequent posts, I will
> show that the Cavendish Experiment is EM-gravity and that the
> Cavendish constant vibration also disproves the Casimir Effect
> Experiments in that they are both experiments of parallel plate
> capacitors in which there is not only a attraction but a repulsion.
> The constant vibration in the Cavendish Experiment as what Wikipedia
> notes below, is that lead balls form a parallel plate capacitor and to
> see those vibrations even better, if Cavendish had designed the balls
> to be rectangular in shape as my experiment uses 50lb
> steel weights and use a side of one weight next to the side of another
> weight, forming a better plate capacitor. My experiment highlights
> that Cavendish could never eliminate the vibration because the balls
> were fluctuating between attraction and repulsion of the electricity
> and magnetism set up by the lead balls-- in mine the steel 50lb
> weights. In the Casimir Effect, it alternates between the plates
> attracting and repulsing.
>
> I am happy to learn that in the Cavendish Experiment they required
> telescopes to read the twist motion.
>
> --- quoting Wikipedia on the precision involved in Cavendish
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
testing for tampering
> where dots of the electron-dot-cloud are galaxies
Jim Burns
Feb 1, 2012, 4:37:59 PM2/1/12
to
[Re: Wikipedia also shows the constant vibration in
Cavendish Experiment #157 New Physics #266 ATOM TOTALITY 5th ed]
On 2/1/2012 3:06 PM, Archimedes Plutonium wrote:
> I am happy to learn that in the Cavendish Experiment they required
> telescopes to read the twist motion. [...]
If the vibration to which you refer is the same vibration I am
familiar with (from conducting Cavendish's experiment myself,
for a physics class) then that vibration is needed for the
analysis of the experiment. The frequency of vibration and
the damping rate will tell you how much torque is required to deflect
the torsion pendulum by a given angle. Without that information,
you can measure the deflection, but then you can't turn that into
a measurement of the gravitational force between the lead balls.
The force on the two suspended lead balls does indeed vary.
The wire that suspends the lead balls has a restoring force
that increases as its twist increases, in a way similar to
the usual kind of spring increasing its pull as it gets
stretched.
Cavendish Experiment #157 New Physics #266 ATOM TOTALITY 5th ed]
On 2/1/2012 3:06 PM, Archimedes Plutonium wrote:
> I am happy to learn that in the Cavendish Experiment they required
If the vibration to which you refer is the same vibration I am
familiar with (from conducting Cavendish's experiment myself,
for a physics class) then that vibration is needed for the
analysis of the experiment. The frequency of vibration and
the damping rate will tell you how much torque is required to deflect
the torsion pendulum by a given angle. Without that information,
you can measure the deflection, but then you can't turn that into
a measurement of the gravitational force between the lead balls.
The force on the two suspended lead balls does indeed vary.
The wire that suspends the lead balls has a restoring force
that increases as its twist increases, in a way similar to
the usual kind of spring increasing its pull as it gets
stretched.
Archimedes Plutonium
Feb 1, 2012, 5:41:53 PM2/1/12
to
not have the Cavendish Experiment, hands on in front of me to be able
to list the shortcomings. I do have my own mock Cavendish which
demonstrates what we all know, that when you put a magnet near another
magnet you sense and perceive a force of attraction which can be
measured. When you put one mass up close to another mass, there is
never that sense or perception or measurement of a attraction,
regardless as to how massive they are. In other words, there is no
Newtonian gravity nor GR.
What there surely is, is a uneven distribution of electric charge in
two bodies close to one another, and because the Coulomb force is
inverse square, what we end up measuring in a Cavendish Experiment is
not gravity but the Coulomb inverse square on those unevenly
distributed charges inside both objects.
Now I wish I could fabricate spherical magnets to cut down on surface
friction. And I may have to be satisfied with disc shaped magnets and
use the edge to roll the magnets down steel inclined planes. I do have
2.5 cm disc magnets which are strong enough magnetically to stay put
on a 90 degree inclined plane. So I reduce the magnetism and keep the
mass the same to find at what inclined plane such a magnet stays put.
What I achieve with the magnets and inclined plane is that at some
point the magnetic strength keeps the magnet stayed put, yet the mass
of that magnet should have forced it to roll down the inclined plane.
Thus I would have achieved, by logic, that there is no force of
Newtonian gravity and no General Relativity.
Do you see the logic of that Jim, or is it unclear?
What I have devised is an Experiment that shows there is no force of
mass gravity, but that all of gravity is due to EM, the charge
distribution between two objects.
Your lucky Jim that you have the actual Cavendish Experiment materials
before you, and the best I have is a mock up version of 50lb steel
weights. The Cavendish model is so full of illogic and imprecision
that the only valid take away from the Cavendish Experiment is that
there is this constant vibration or oscillation. Now in the magnet
stuck to the refrigerator door, there is no constant vibration to
measure, but when the magnet is very weak and cannot stay stuck, then
the magnet presents us with a constant vibration and oscillation also.
So the only true feature of the Cavendish Experiment is that we are
dealing with a force that has a constant oscillation and that force
could only be EM.
Question Jim, can you set up inclined steel planes with spherical
magnets and have an instrument that can measure the amount of
magnetism in each sphere. And then roll those magnets of various
strengths down the inclined plane. What I need to find out is what
magnet strength and mass of magnet for which gravity mass should force
the ball down the plane, yet instead, it is stuck fast to the plane
regardless of the mass. You see, that point is where gravity is
nonexistent and that the EM of the magnet, the plane and Earth yield
that situation.
Jim Burns
Feb 1, 2012, 7:37:11 PM2/1/12
to
Not only are there the usual arguments for a gravitational force,
arguments that Newton used when he published _Principia_ many
years before Cavendish, but we ourselves can measure directly the very
small force between two less-than-planetary objects.
A mock experiment will not do it for you. The forces involved are
very small. I had a working apparatus, and I needed to record
the position of the slowly swinging structure every minute or so
for some hours, as the non-swinging lead balls pulled it first
one way and then another. (Okay, maybe I overdid the data-taking,
but I had nowhere else to go while the experiment ran.) The
"torsion pendulum" was a piece of wire exactly strong enough
to support the swinging balls, to the point that setting the apparatus
down too hard would snap the wire. (Sigh.)
I mentioned that I have measured it myself. I hope I didn't sound like
I was bragging. It's a common experiment for budding young physicists
to repeat. It is a very interesting experiment, not only because
Newton's gravity opened the flood gates of scientific progress,
and put the realms of the gods and of the streetsweepers under the
same laws, but also because it is a technically challenging experiment,
yet not too challenging, that shows the student (ahem, me, that is)
that it is possible to measure suprisingly small effects with the
application of sufficient quantities of cleverness.
My rather long-winded point is that you do not have to take anyone's
word for it. With the proper apparatus -- your own NON-mock apparatus
-- you too can see the direct effect of gravity between two rather
small balls of lead. It would be possible, I suppose, for you to
construct your own working apparatus, but I wouldn't recommend that
route. Whatever your own skills may be, it would be a lot of trouble,
and probably cost more, when very nice apparati (sp?) can probably be
bought at Edmund Scientific. (I would bet so. I haven't looked.) I
would not be surprised if they cost much less than 50 lb
spheres of steel.
(Please note that I do not have the apparatus any more. I used
it in a class some years ago. The thing you really want to do is
see for yourself, though, or so it seems to me.)
Archimedes Plutonium
Feb 2, 2012, 1:24:33 AM2/2/12
to
On Feb 1, 6:37 pm, Jim Burns <burns...@osu.edu> wrote:
> On 2/1/2012 5:41 PM, Archimedes Plutonium wrote:
>
(snipped)
> On 2/1/2012 5:41 PM, Archimedes Plutonium wrote:
>
x 15cm x20cm
with a steel handle in it. I do not know what they were made for.
Anyway, can you tell me the measure of the lead balls used and what
their weight is in the
Cavendish Experiment?
Can you provide some calculations of those lead balls? We have
Newton's gravity as
F= G* m_1 * m_2/ d^2, can you provide F for Cavendish's lead balls?
Next, we have for Coulomb's law F = k* q_1*q_2/d^2
Jim, I used to know a Stanford physicist that could do this in his
head. I am not that good at calculation and better at the logic.
So now we have the Force of gravity for the lead balls, and I do not
know their weight, and computed is ??
Now we translate that gravity force into EM and plug it into Coulomb's
law and we know k and
d but we do not know q_1 or q_2. Jim, can you compute how many
negative charges and how many positive charges are involved in the
Cavendish Experiment? Can you compute the number of electrons and
protons for q_1 and q_2 as to give the force of gravity for the
Cavendish Experiment? Wish that Stanford physicist were still around
but deceased almost a decade ago.
Szczepan Bialek
Feb 2, 2012, 4:41:32 AM2/2/12
to
"Jim Burns" <burn...@osu.edu> napisal w wiadomosci
news:4F29DAB7...@osu.edu...
> >
> I mentioned that I have measured it myself. I hope I didn't sound like
> I was bragging. It's a common experiment for budding young physicists
> to repeat. It is a very interesting experiment, not only because
> Newton's gravity opened the flood gates of scientific progress,
> and put the realms of the gods and of the streetsweepers under the
> same laws, but also because it is a technically challenging experiment,
> yet not too challenging, that shows the student (ahem, me, that is)
> that it is possible to measure suprisingly small effects with the
> application of sufficient quantities of cleverness.
In this way was possible to measure the "G".
> I mentioned that I have measured it myself. I hope I didn't sound like
> I was bragging. It's a common experiment for budding young physicists
> to repeat. It is a very interesting experiment, not only because
> Newton's gravity opened the flood gates of scientific progress,
> and put the realms of the gods and of the streetsweepers under the
> same laws, but also because it is a technically challenging experiment,
> yet not too challenging, that shows the student (ahem, me, that is)
> that it is possible to measure suprisingly small effects with the
> application of sufficient quantities of cleverness.
Now people were on the Moon and they did the direct measurement of the "G"
(gravity constant).
Does anybody know the result?
S*
>
Jim Burns
Feb 2, 2012, 7:59:22 AM2/2/12
to
the force, but you could estimate the force, too.
I really think it would be a valuable experience for you, not
just estimating the force, I mean, but repeating Cavendish's
classic experiment.
It would take a bit of effort on your part, but you already seem
to have invested quite a bit of effort and not really gotten very
far. If you go just a bit further than you have, you could get
some results at last.
The gravitational force between two planets or between a planet
and a middle-sized object are all very well, but I find it hard
to _feel_ their reality without an effort of imagination. Entire
planets are outside my everyday experience, really. All I really
see is the little bit of one planet that I drive around on.
Seeing the gravitational attraction between four balls of lead
that I can hold in my hand (two swing, two don't) brings home
the reality of the gravitational force. Perhaps you remember
in _The Wizard of Oz_ where Toto pulls back the curtain and
reveals the Wizard? It feels something like that.
>
> Can you provide some calculations of those lead balls? We have
> Newton's gravity as
> F= G* m_1 * m_2/ d^2, can you provide F for Cavendish's lead balls?
>
> Next, we have for Coulomb's law F = k* q_1*q_2/d^2
I don't need to know much about the situation to be able to say
that.
However, there are both positive and negative electric charges
that balance each other -- balance each other _exactly_ very
often, and balance each other _nearly_ exactly when we find
what we think of as "electromagnetic forces". When we produce
a magnetic field by running current through a wire, for example,
there is exactly the same positive and negative charge.
A fraction of the negative charges are moving, though, and
that is enough to produce forces many times the gravitational
forces involved.
The reason we do not notice electromagnetic forces more often
than we do is -- ironically -- because they are so strong.
Positive and negative charges have a very strong tendency
to combine and neutralize each other. Since we only have
_positive_ masses, this is not an option for gravitational
forces.
Archimedes Plutonium
Feb 2, 2012, 4:00:47 PM2/2/12
to
On Feb 2, 6:59 am, Jim Burns <burns...@osu.edu> wrote:
> On 2/2/2012 1:24 AM, Archimedes Plutonium wrote:
>
(snipped in parts)
> On 2/2/2012 1:24 AM, Archimedes Plutonium wrote:
>
Actually
this is the first time any physicist computed what it takes for EM
charges
to replace all of mass-gravity with EM-gravity. No-one before had the
bravery
of saying, "oh, let us not have charges hinder us from thinking that
the Sun holds Earth due to EM charge and not due to mass".
The physicist that is not brilliant, looks at Saturn Rings or barred
spiral galaxies
and sees rigid body rotation and then due to his lack of physics
skills invokes
dark matter and dark energy. The brilliant and true physicist seeing
rigid body rotation
says "gravity can no longer be mass-gravity but must be EM-gravity."
The Cavendish Experiment is a fool's experiment.
The mass of the Sun is 2x10^30 kg and the surface area is 6x10^12 km^2
The mass of Earth is 6x10^24 kg and its surface area is 5x10^8 km^2
The mass of a single proton is 1.6x10^-27 kg
The force coupling strength of EM versus mass-gravity is 10^39 to
10^40
The number of protons that make up the Sun is 10^57 and for Earth it
is 10^51
Since the coupling strength of EM to gravity-mass is 10^40 then the
number of
protons needed for Earth is 10^51/10^40 is 10^11
The 10^11 protons is about the number of protons in a iron nail
The number of electrons needed for the sun replaced of its mass-
gravity with
EM gravity is 10^57/10^40 = 10^17 which is about the number of protons
in a box of iron nails.
The diameter of the hydrogen atom is about 10^-13 km
The surface area of Sun is 6x10^12 km^2 so that for the surface of the
sun
would have 6x10^25 protons.
So all it takes for the Sun to hold Earth in a bonded condition of EM
force is a box of iron nails
of ions as a net overall charge and for Earth to have a net overall
charge of the protons in one nail.
It is remarkable that we all know the Sun has a huge number of ions
that do not cancel out but leave a net charge. And the number of
cosmic electron rays that hit Earth every minute, is remarkable to
think that Earth would not have a net charge the amount of protons in
a single nail.
The concept of Net Overall Charge is perhaps to new, too revolutionary
for all those old minds in physics who have been brainwashed into
thinking that mass, just plain mass attracts other mass.
Archimedes Plutonium
Feb 2, 2012, 4:28:58 PM2/2/12
to
On Feb 2, 3:00 pm, Archimedes Plutonium
Sorry that should have read 6x10^12/10^-26 for there to be 6x10^38
protons or electrons
that compose the surface of the Sun, and that the Sun only needs a net
imbalance of
10^17.
Now many are going to complain as to why and how the Sun is charged
opposite to the charge of a planet to have them attract. And my answer
is that in Faraday's law, not Coulomb law
that the Faraday's law needs a bar magnet, not isolated charges and
that a net charge differential make the entire Sun and a planet be a
magnet.
protons or electrons
that compose the surface of the Sun, and that the Sun only needs a net
imbalance of
10^17.
Now many are going to complain as to why and how the Sun is charged
opposite to the charge of a planet to have them attract. And my answer
is that in Faraday's law, not Coulomb law
that the Faraday's law needs a bar magnet, not isolated charges and
that a net charge differential make the entire Sun and a planet be a
magnet.
Szczepan Bialek
Feb 3, 2012, 3:32:12 AM2/3/12
to
news:4F2A88AA...@osu.edu...
> When we produce
> a magnetic field by running current through a wire, for example,
> there is exactly the same positive and negative charge.
No. To make the flow of electrons there must be the gradient of electron
density. So in the wire is always the excess/deficit of electrons.
> A fraction of the negative charges are moving, though, and
> that is enough to produce forces many times the gravitational
> forces involved.
>
> The reason we do not notice electromagnetic forces more often
> than we do is -- ironically -- because they are so strong.
> Positive and negative charges have a very strong tendency
> to combine and neutralize each other. Since we only have
> _positive_ masses, this is not an option for gravitational
> forces.
The "G" (and the Earth's mass) measured by Cavendish is without any doubts
wrong.
S*
>>
> >
> Electromagnetic forces are much larger than gravitational forces.
> I don't need to know much about the situation to be able to say
> that.
>
> However, there are both positive and negative electric charges
> that balance each other -- balance each other _exactly_ very
> often, and balance each other _nearly_ exactly when we find
> what we think of as "electromagnetic forces".
The Earth has the excess of electrons. The Cavendish balls also.
> Electromagnetic forces are much larger than gravitational forces.
> I don't need to know much about the situation to be able to say
> that.
>
> However, there are both positive and negative electric charges
> that balance each other -- balance each other _exactly_ very
> often, and balance each other _nearly_ exactly when we find
> what we think of as "electromagnetic forces".
> When we produce
> a magnetic field by running current through a wire, for example,
> there is exactly the same positive and negative charge.
density. So in the wire is always the excess/deficit of electrons.
> A fraction of the negative charges are moving, though, and
> that is enough to produce forces many times the gravitational
> forces involved.
>
> The reason we do not notice electromagnetic forces more often
> than we do is -- ironically -- because they are so strong.
> Positive and negative charges have a very strong tendency
> to combine and neutralize each other. Since we only have
> _positive_ masses, this is not an option for gravitational
> forces.
wrong.
S*
>>
Archimedes Plutonium
Feb 3, 2012, 11:41:05 PM2/3/12
to
On Feb 3, 2:32 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
> "Jim Burns" <burns...@osu.edu> napisal w wiadomoscinews:4F2A88AA...@osu.edu...
exploration and ask yourself, since EM is so much stronger, then is
there
some way, any way that we can drop gravity altogether and replace it
with EM. You see the person that asks such a question is not a parrot
of science but rather a thinker of science.
> > However, there are both positive and negative electric charges
> > that balance each other -- balance each other _exactly_ very
> > often, and balance each other _nearly_ exactly when we find
> > what we think of as "electromagnetic forces".
>
Yea,yea yea, I heard the balancing act by everyone in science
classrooms.
But we all know that electric charge is unevenly distributed over
Earth, for example
lightening bolt strikes, static electric, and even in rocks such as
lodestone or in
water such as the polar-bond. And our planet is bombarded every day by
cosmic
protons and cosmic electrons. As I computed that a needle of protons
for earth and a
few needles of protons for the Sun is all it takes to cover all of
gravity of Sun to Earth.
So it is time for physicists to stop being parrots and go beyond the
"charge is all cancelled
because it is not all cancelled" and that only a little differential
of charge is all it takes to
compose all of gravity.
> The Earth has the excess of electrons. The Cavendish balls also.
>
Szczepan, I could not get the vibration to stop in my mock-Cavendish
Experiment.
Scientists have not eliminated the fact of seismography that the
ground is constantly
shaking and which could have some influence on the Cavendish reading.
Also, the
Cavendish set up is a large antennae and can easily pick up the radio
frequency of the
Jupiter or Sun and how much does those radio waves have on the result.
But I think
more important is that the materials of lead balls, wood planks, steel
wires used, all of them
are held together by EM charges, and that only a tiny differential of
EM charges would account
for the entire torque that Cavendish alleged in 1797. So I am saying
that instead of the Cavendish Experiment being a great experiment, I
am saying it was a flop for it never separated out EM from mass
gravity. And it is a crying shame that modern day physicists
are blind to the crudeness and lack of precision and illogic of that
experiment.
If we did the experiment using Galileo inclined planes with balls that
are magnetic and find where the magnet ball due to mass should roll
off the inclined plane from mass alone but is
stuck due to magnetic force, we thence prove that Newtonian gravity
and General Relativity are fake forces.
> > When we produce
> > a magnetic field by running current through a wire, for example,
> > there is exactly the same positive and negative charge.
>
> No. To make the flow of electrons there must be the gradient of electron
> density. So in the wire is always the excess/deficit of electrons.
>
Well said.
> > A fraction of the negative charges are moving, though, and
> > that is enough to produce forces many times the gravitational
> > forces involved.
>
> > The reason we do not notice electromagnetic forces more often
> > than we do is -- ironically -- because they are so strong.
> > Positive and negative charges have a very strong tendency
> > to combine and neutralize each other. Since we only have
> > _positive_ masses, this is not an option for gravitational
> > forces.
>
> The "G" (and the Earth's mass) measured by Cavendish is without any doubts
> wrong.
> S*
>
Question. How do we convert the magnet force of a bar magnet into
terms of how many electrons that bar magnet represents? So I have two
bar magnets and their strength is approx 10^-2 Tesla, and so I want to
know how many individual electrons is a B field of 10^-2 Tesla. The
surface of Earth is said to be 10^-4 Tesla and so how many negative
charges of an electron does that represent? How do we convert Tesla to
number of unit charges?
> "Jim Burns" <burns...@osu.edu> napisal w wiadomoscinews:4F2A88AA...@osu.edu...
>
>
>
> > Electromagnetic forces are much larger than gravitational forces.
> > I don't need to know much about the situation to be able to say
> > that.
>
But then, as a scientist, why did you never take the next best turn of
>
>
> > Electromagnetic forces are much larger than gravitational forces.
> > I don't need to know much about the situation to be able to say
> > that.
>
exploration and ask yourself, since EM is so much stronger, then is
there
some way, any way that we can drop gravity altogether and replace it
with EM. You see the person that asks such a question is not a parrot
of science but rather a thinker of science.
> > However, there are both positive and negative electric charges
> > that balance each other -- balance each other _exactly_ very
> > often, and balance each other _nearly_ exactly when we find
> > what we think of as "electromagnetic forces".
>
classrooms.
But we all know that electric charge is unevenly distributed over
Earth, for example
lightening bolt strikes, static electric, and even in rocks such as
lodestone or in
water such as the polar-bond. And our planet is bombarded every day by
cosmic
protons and cosmic electrons. As I computed that a needle of protons
for earth and a
few needles of protons for the Sun is all it takes to cover all of
gravity of Sun to Earth.
So it is time for physicists to stop being parrots and go beyond the
"charge is all cancelled
because it is not all cancelled" and that only a little differential
of charge is all it takes to
compose all of gravity.
> The Earth has the excess of electrons. The Cavendish balls also.
>
Experiment.
Scientists have not eliminated the fact of seismography that the
ground is constantly
shaking and which could have some influence on the Cavendish reading.
Also, the
Cavendish set up is a large antennae and can easily pick up the radio
frequency of the
Jupiter or Sun and how much does those radio waves have on the result.
But I think
more important is that the materials of lead balls, wood planks, steel
wires used, all of them
are held together by EM charges, and that only a tiny differential of
EM charges would account
for the entire torque that Cavendish alleged in 1797. So I am saying
that instead of the Cavendish Experiment being a great experiment, I
am saying it was a flop for it never separated out EM from mass
gravity. And it is a crying shame that modern day physicists
are blind to the crudeness and lack of precision and illogic of that
experiment.
If we did the experiment using Galileo inclined planes with balls that
are magnetic and find where the magnet ball due to mass should roll
off the inclined plane from mass alone but is
stuck due to magnetic force, we thence prove that Newtonian gravity
and General Relativity are fake forces.
> > When we produce
> > a magnetic field by running current through a wire, for example,
> > there is exactly the same positive and negative charge.
>
> No. To make the flow of electrons there must be the gradient of electron
> density. So in the wire is always the excess/deficit of electrons.
>
> > A fraction of the negative charges are moving, though, and
> > that is enough to produce forces many times the gravitational
> > forces involved.
>
> > The reason we do not notice electromagnetic forces more often
> > than we do is -- ironically -- because they are so strong.
> > Positive and negative charges have a very strong tendency
> > to combine and neutralize each other. Since we only have
> > _positive_ masses, this is not an option for gravitational
> > forces.
>
> The "G" (and the Earth's mass) measured by Cavendish is without any doubts
> wrong.
> S*
>
terms of how many electrons that bar magnet represents? So I have two
bar magnets and their strength is approx 10^-2 Tesla, and so I want to
know how many individual electrons is a B field of 10^-2 Tesla. The
surface of Earth is said to be 10^-4 Tesla and so how many negative
charges of an electron does that represent? How do we convert Tesla to
number of unit charges?
Szczepan Bialek
Feb 7, 2012, 12:57:15 PM2/7/12
to
Uzytkownik "Archimedes Plutonium" <plutonium....@gmail.com> napisal w
wiadomosci
news:b8dc8b99-cf1c-450c...@p13g2000yqd.googlegroups.com...
On Feb 3, 2:32 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>
>> The Earth has the excess of electrons. The Cavendish balls also.
>
>Szczepan, I could not get the vibration to stop in my mock-Cavendish
Experiment.
Scientists have not eliminated the fact of seismography that the
ground is constantly
shaking and which could have some influence on the Cavendish reading.
Also, the
Cavendish set up is a large antennae and can easily pick up the radio
frequency of the
Jupiter or Sun and how much does those radio waves have on the result.
But I think
more important is that the materials of lead balls, wood planks, steel
wires used, all of them
are held together by EM charges, and that only a tiny differential of
EM charges would account
for the entire torque that Cavendish alleged in 1797. So I am saying
that instead of the Cavendish Experiment being a great experiment, I
am saying it was a flop for it never separated out EM from mass
gravity.
You are right.
>
>> The Earth has the excess of electrons. The Cavendish balls also.
>
>Szczepan, I could not get the vibration to stop in my mock-Cavendish
Experiment.
Scientists have not eliminated the fact of seismography that the
ground is constantly
shaking and which could have some influence on the Cavendish reading.
Also, the
Cavendish set up is a large antennae and can easily pick up the radio
frequency of the
Jupiter or Sun and how much does those radio waves have on the result.
But I think
more important is that the materials of lead balls, wood planks, steel
wires used, all of them
are held together by EM charges, and that only a tiny differential of
EM charges would account
for the entire torque that Cavendish alleged in 1797. So I am saying
that instead of the Cavendish Experiment being a great experiment, I
am saying it was a flop for it never separated out EM from mass
gravity.
>And it is a crying shame that modern day physicists
are blind to the crudeness and lack of precision and illogic of that
experiment.
"G".
I do not know when it will not be a secret.
>
>> No. To make the flow of electrons there must be the gradient of electron
>> density. So in the wire is always the excess/deficit of electrons.
>
>Well said.
>Question. How do we convert the magnet force of a bar magnet into
terms of how many electrons that bar magnet represents? So I have two
bar magnets and their strength is approx 10^-2 Tesla, and so I want to
know how many individual electrons is a B field of 10^-2 Tesla. The
surface of Earth is said to be 10^-4 Tesla and so how many negative
charges of an electron does that represent? How do we convert Tesla to
number of unit charges?
The excess of the electrons rotate with the Earth (convectional current) and
terms of how many electrons that bar magnet represents? So I have two
bar magnets and their strength is approx 10^-2 Tesla, and so I want to
know how many individual electrons is a B field of 10^-2 Tesla. The
surface of Earth is said to be 10^-4 Tesla and so how many negative
charges of an electron does that represent? How do we convert Tesla to
number of unit charges?
creates this 10^-4 Tesla.
Somebody can calculate the current.
S*
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