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New AP-Maxwell Equations, under construction
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Archimedes Plutonium
Aug 16, 2017, 1:56:19 AM8/16/17
to
New AP-Maxwell Equations, under construction
taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
This is taking my time, because I have to straighten out these equations;;
Ampere
V' = i'*B*L + i*B'*L + i*B*L'
= i*B*L + i*V/m^2*L + i*B*L'
Faraday law
(V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
= (V'iL - Vi'L -ViL')/ i^2L^2
=
New Coulomb law and Gravity EM
(V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
= (V'BL -VB'L - VBL') / B^2L^2
Spin law
(V/i*B)' = (V'*i*B - V*i' B - V*i*B') / (i*B)^2
Justification
(i) derivative with respect to time s, 1/s velocity, 1/s^2 acceleration
current i = dq/ds so current is 1/s
Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3,
Current 1/s, Conductance A^2*s^3, Capacitance A^2*s^4,
(ii) Resistance R = B*L in case of electrical wire and substance matter
AP
taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
This is taking my time, because I have to straighten out these equations;;
Ampere
V' = i'*B*L + i*B'*L + i*B*L'
= i*B*L + i*V/m^2*L + i*B*L'
Faraday law
(V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
= (V'iL - Vi'L -ViL')/ i^2L^2
=
New Coulomb law and Gravity EM
(V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
= (V'BL -VB'L - VBL') / B^2L^2
Spin law
(V/i*B)' = (V'*i*B - V*i' B - V*i*B') / (i*B)^2
Justification
(i) derivative with respect to time s, 1/s velocity, 1/s^2 acceleration
current i = dq/ds so current is 1/s
Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3,
Current 1/s, Conductance A^2*s^3, Capacitance A^2*s^4,
(ii) Resistance R = B*L in case of electrical wire and substance matter
AP
Archimedes Plutonium
Aug 16, 2017, 11:35:37 AM8/16/17
to
On Wednesday, August 16, 2017 at 12:56:19 AM UTC-5, Archimedes Plutonium wrote:
> New AP-Maxwell Equations, under construction
>
> taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
>
> This is taking my time, because I have to straighten out these equations;;
>
>
> Ampere
>
> V' = i'*B*L + i*B'*L + i*B*L'
>
> = i*B*L + i*V/m^2*L + i*B*L'
>
> Faraday law
>
> (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
>
> = (V'iL - Vi'L -ViL')/ i^2L^2
>
> =
>
> New Coulomb law and Gravity EM
>
> (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
>
> = (V'BL -VB'L - VBL') / B^2L^2
>
Now somewhere in these equations, I need to get speeds varying from proportional to R, to 1/R, to 1/R^2 as the Coulomb force. A force that is variable in range, from Solid Body Rotation to 1/R to that of gravity as 1/R^2. I believe it is the above. And involves the term B^2L^2 which normally is Resistance in conduction wires.
> New AP-Maxwell Equations, under construction
>
> taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
>
> This is taking my time, because I have to straighten out these equations;;
>
>
> Ampere
>
> V' = i'*B*L + i*B'*L + i*B*L'
>
> = i*B*L + i*V/m^2*L + i*B*L'
>
> Faraday law
>
> (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
>
> = (V'iL - Vi'L -ViL')/ i^2L^2
>
> =
>
> New Coulomb law and Gravity EM
>
> (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
>
> = (V'BL -VB'L - VBL') / B^2L^2
>
Archimedes Plutonium
Aug 16, 2017, 8:31:00 PM8/16/17
to
The single most important concept that the New AP Maxwell Equations brings to light, is the idea that Coulomb Force is a variable force, ranging from R to 1/R to 1/R^2. This truth then trashcans the Big Bang, Dark Matter, Dark Energy, General Relativity, and even into particle physics trashcans the Higgs boson.
AP
AP
burs...@gmail.com
Aug 16, 2017, 8:39:52 PM8/16/17
to
AP brain farto, producing internet trash,
already for 30 years. Not a single line of
math or physics, only nonsense.
already for 30 years. Not a single line of
math or physics, only nonsense.
benj
Aug 16, 2017, 9:55:26 PM8/16/17
to
On 8/16/2017 8:30 PM, Archimedes Plutonium wrote:
> The single most important concept that the New AP Maxwell Equations brings to light, is the idea that Coulomb Force is a variable force, ranging from R to 1/R to 1/R^2. This truth then trashcans the Big Bang, Dark Matter, Dark Energy, General Relativity, and even into particle physics trashcans the Higgs boson.
>
> AP
>
I smell Nobel! (Of course Goldbach didn't work out so well)
> The single most important concept that the New AP Maxwell Equations brings to light, is the idea that Coulomb Force is a variable force, ranging from R to 1/R to 1/R^2. This truth then trashcans the Big Bang, Dark Matter, Dark Energy, General Relativity, and even into particle physics trashcans the Higgs boson.
>
> AP
>
Archimedes Plutonium
Aug 17, 2017, 2:26:45 AM8/17/17
to
On Wednesday, August 16, 2017 at 12:56:19 AM UTC-5, Archimedes Plutonium wrote:
> New AP-Maxwell Equations, under construction
>
> taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
>
> This is taking my time, because I have to straighten out these equations;;
>
>
> Ampere
>
> V' = i'*B*L + i*B'*L + i*B*L'
>
> = i*B*L + i*V/m^2*L + i*B*L'
>
> Faraday law
>
> (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
>
> = (V'iL - Vi'L -ViL')/ i^2L^2
>
> =
>
> New Coulomb law and Gravity EM
>
> (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
>
> = (V'BL -VB'L - VBL') / B^2L^2
>
i' = (V/B*L)'
>
> taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
>
> This is taking my time, because I have to straighten out these equations;;
>
>
> Ampere
>
> V' = i'*B*L + i*B'*L + i*B*L'
>
> = i*B*L + i*V/m^2*L + i*B*L'
>
> Faraday law
>
> (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
>
> = (V'iL - Vi'L -ViL')/ i^2L^2
>
> =
>
> New Coulomb law and Gravity EM
>
> (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
>
> = (V'BL -VB'L - VBL') / B^2L^2
>
(V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
= (V'BL -VB'L - VBL') / B^2L^2
Now how do we get R to 1/R to 1/R^2 out of the derivative of current i?
= (V'BL -VB'L - VBL') / B^2L^2
So we list the derivatives with respect to time of EM parameters
derivative with respect to time s, to 1/s velocity, to 1/s^2 acceleration
current i = dq/ds so current is 1/s what is derivative of current, is 1/s^2 and what is that?
Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3
Derivative of B would be 1/s^2 to 1/s^3, so that derivative of B is either Volt or Resistance and the clear choice here is Volt
Derivative of L and here we have L as 1/s so the derivative is 1/s^2 and the clear choice here is a force, a torque, and now, if we have a torque times magnetic field B we end up with Voltage.
Derivative of V, voltage, and here we have 1/s^3, and the only s^4 I know of is Capacitance current Capacitance A^2*s^4, even though it is in the numerator. Let me denote it by i_C
Now, one more justification is that Resistance is B*L for some materials such as copper electric wires, or crystals. And we are free to substitute R for B*L
i' = ((i_C)*V -VVL - VV) / R^2
Now keep in mind Resistance R is not the radius R in R to 1/R to 1/R^2, but, in a serendipity fluke of luck, we can see Resistance as a distance parameter. So what does that leave us?
If we use the Old Ohm's law also of V/R = i
Then we have
i' = ((i_C)*V -VVL - VV) / R^2
i' = (i_C)*i / R - i*L - i^2
Does that look like a R to 1/R to 1/R^2 where R is radius (not resistance)
Not to me,,,,, more later
Arindam Banerjee
Aug 17, 2017, 2:47:41 AM8/17/17
to
On Thursday, August 17, 2017 at 4:26:45 PM UTC+10, Archimedes Plutonium wrote:
> On Wednesday, August 16, 2017 at 12:56:19 AM UTC-5, Archimedes Plutonium wrote:
> > New AP-Maxwell Equations, under construction
> >
> > taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
> >
> > This is taking my time, because I have to straighten out these equations;;
> >
> >
> > Ampere
> >
> > V' = i'*B*L + i*B'*L + i*B*L'
> >
> > = i*B*L + i*V/m^2*L + i*B*L'
> >
> > Faraday law
> >
> > (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
> >
> > = (V'iL - Vi'L -ViL')/ i^2L^2
> >
> > =
> >
> > New Coulomb law and Gravity EM
> >
> > (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
> >
> > = (V'BL -VB'L - VBL') / B^2L^2
> >
>
>
> i' = (V/B*L)'
> (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
>
> = (V'BL -VB'L - VBL') / B^2L^2
>
> Now how do we get R to 1/R to 1/R^2 out of the derivative of current i?
>
> So we list the derivatives with respect to time of EM parameters
>
> derivative with respect to time s, to 1/s velocity, to 1/s^2 acceleration
>
> current i = dq/ds so current is 1/s what is derivative of current, is 1/s^2 and what is that?
>
> Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3
>
> Apparently, derivative of current i is Magnetic Field B
actually, derivative of current i with respect to time is the back voltage
> On Wednesday, August 16, 2017 at 12:56:19 AM UTC-5, Archimedes Plutonium wrote:
> > New AP-Maxwell Equations, under construction
> >
> > taking my time Re: PreliminaryPage22, 3-7, AP-Maxwell Equations of New Physics
> >
> > This is taking my time, because I have to straighten out these equations;;
> >
> >
> > Ampere
> >
> > V' = i'*B*L + i*B'*L + i*B*L'
> >
> > = i*B*L + i*V/m^2*L + i*B*L'
> >
> > Faraday law
> >
> > (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
> >
> > = (V'iL - Vi'L -ViL')/ i^2L^2
> >
> > =
> >
> > New Coulomb law and Gravity EM
> >
> > (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
> >
> > = (V'BL -VB'L - VBL') / B^2L^2
> >
>
>
> i' = (V/B*L)'
> (V/B*L)' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
>
> = (V'BL -VB'L - VBL') / B^2L^2
>
> Now how do we get R to 1/R to 1/R^2 out of the derivative of current i?
>
> So we list the derivatives with respect to time of EM parameters
>
> derivative with respect to time s, to 1/s velocity, to 1/s^2 acceleration
>
> current i = dq/ds so current is 1/s what is derivative of current, is 1/s^2 and what is that?
>
> Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3
>
> Apparently, derivative of current i is Magnetic Field B
as per
v = L d(i(t)/dt
where L is the inductance assumed not to change with time.
A magnetic field is always associated with a current following the Biot-Savart
empirical relation. The creation of this
magnetic field requires energy which is stored in the circuit's inductance.
Because of this inductance, the voltage across any resistive load does not
rise instantaneously. It takes a while to push the current through till the
maximum value is reached when the circuit gets fully magnetised.
Putrid-Poo, stop making an ass of yourself. Read a basic book on electrical
circuits. I doubt if much will penetrate your fat head, though.
Cheers,
Arindam Banerjee
Archimedes Plutonium
Aug 17, 2017, 3:20:49 AM8/17/17
to
Now if we take Magnetic Field times torque what we get is another i_C, a second capacitor current
> Derivative of V, voltage, and here we have 1/s^3, and the only s^4 I know of is Capacitance current Capacitance A^2*s^4, even though it is in the numerator. Let me denote it by i_C
>
> Now, one more justification is that Resistance is B*L for some materials such as copper electric wires, or crystals. And we are free to substitute R for B*L
>
> i' = ((i_C)*V -VVL - VV) / R^2
i' = ((i_C)*V -VVL - V(i_C) / R^2
That reduces instantly to
i' = -VVL/R^2
And what that reduces to, is
i' = - (i*L)
Now, let me reflect over night, as to whether a Coulomb variable range Force from Radius to 1/Radius to 1/Radius^2 is embodied by a
- (i*L)
Does a multiplication of current by angular momentum, provide us with Solid Body Rotation, also 1/Radius rotation and finally the familiar 1/Radius^2 rotation
Or back to the drawing board
pora...@gmail.com
Aug 17, 2017, 3:25:34 AM8/17/17
to
in short
he is a cureless imbecile moron psychopath
that is doing his best to steal
more and more as possible
but what can he do ??
while he is a blockhead psychopath spammer?
Y.Porat
==========================
pora...@gmail.com
Aug 17, 2017, 3:29:41 AM8/17/17
to
and by that
I agree with Arindam
==
Y.Porat
=======================
Archimedes Plutonium
Aug 17, 2017, 6:02:27 AM8/17/17
to
Beautiful, beautiful, i have my variable range Coulomb force for it lies not in i' but rather in L'. The i' turns out to be the spin law and the L' angular momentum is the New Coulomb law that includes gravity EM
L' = (i^2B - B^3- V^2i) / i^2B^2
With the B term we see it is reflected as 1/B then B then 1/B^2
On a commonsense view the Coulomb force is the motion of a test charge and what describes that motion is derivative of angular momentum
What describes the Spin law is derivative of current which results in current divided by angular momentum.
AP
L' = (i^2B - B^3- V^2i) / i^2B^2
With the B term we see it is reflected as 1/B then B then 1/B^2
On a commonsense view the Coulomb force is the motion of a test charge and what describes that motion is derivative of angular momentum
What describes the Spin law is derivative of current which results in current divided by angular momentum.
AP
Archimedes Plutonium
Aug 17, 2017, 6:17:45 AM8/17/17
to
My mistake the spin law turns out to be i' = - i*L, a current multiplied by angular momentum.
Now it is time to sleep on all of that.
I should mention what a glorious day yesterday was in 2 inches of rain breaking the severe drought and freeing me up from watering the plants-- can't bear to see hazels and lilac suffer. So can start to put the hoses away and get to tree pruning next project.
AP
Now it is time to sleep on all of that.
I should mention what a glorious day yesterday was in 2 inches of rain breaking the severe drought and freeing me up from watering the plants-- can't bear to see hazels and lilac suffer. So can start to put the hoses away and get to tree pruning next project.
AP
burs...@gmail.com
Aug 17, 2017, 6:24:34 AM8/17/17
to
Even with flowers in your garden,
you are the village idiot AP.
Cant even identify a conic section
as an ellipse. What a complete moron.
you are the village idiot AP.
Cant even identify a conic section
as an ellipse. What a complete moron.
Archimedes Plutonium
Aug 17, 2017, 4:58:33 PM8/17/17
to
Alright, somewhat of a review so far, for the new reader, just joining in.
We develop the New Maxwell Equations by differentiating the Ohm's law V= iR, except, we revise Ohm's law itself where we see R resistance is B*L, magnetic field times angular momentum L.
So differentiating the New Ohm's law gives us not only Faraday law and Ampere Maxwell law, but gives us two brand new laws that James Maxwell never saw. These two new laws I call the Coulomb-gravity law and the Spin law.
Now at first, I thought the spin law was the differentiation of L, angular momentum and the Coulomb gravity law was the differentiation of current i. Turns out, I had it backwards, for the differentiation of L was the Coulomb gravity law and the differentiation of current i was the spin law. How did I arrive at that conclusion? Because the Coulomb law has to allow for a VARIABLE FORCE RANGE, ranging from proportional force strength of Radius (solid body rotation) to 1/Radius (intermediate strength) to 1/Radius^2 (our usual perceived strength of gravity).
So, what happened yesterday as seen below, is that I started out thinking differentiation of i as i' was the Coulomb gravity law and differentiation of L was the spin law
B = kg /A*s^2 Law of Magnetism
V = i*B*L Law of Electricity
V' = (i*B*L)' Ampere-Maxwell law
(V/(i*L))' = B' Faraday law
(V/(B*L))' = i' the new law, the EM-Gravity law
(V/(i*B))' = L' the new law of spin
Let me correct that as to this::
B = kg /A*s^2 Law of Magnetism
V = i*B*L Law of Electricity
V' = (i*B*L)' Ampere-Maxwell law
(V/(i*L))' = B' Faraday law
(V/(B*L))' = i' the new law of spin
(V/(i*B))' = L' the new law of Coulomb force with EM gravity force
Below are yesterday's posts showing how I came to this realization::
getting R to 1/R to 1/R^2 Re: New AP-Maxwell Equations, under construction
On Thursday, August 17, 2017 at 5:17:45 AM UTC-5, Archimedes Plutonium wrote:
getting R to 1/R to 1/R^2 Re: New AP-Maxwell Equations, under construction
We develop the New Maxwell Equations by differentiating the Ohm's law V= iR, except, we revise Ohm's law itself where we see R resistance is B*L, magnetic field times angular momentum L.
So differentiating the New Ohm's law gives us not only Faraday law and Ampere Maxwell law, but gives us two brand new laws that James Maxwell never saw. These two new laws I call the Coulomb-gravity law and the Spin law.
Now at first, I thought the spin law was the differentiation of L, angular momentum and the Coulomb gravity law was the differentiation of current i. Turns out, I had it backwards, for the differentiation of L was the Coulomb gravity law and the differentiation of current i was the spin law. How did I arrive at that conclusion? Because the Coulomb law has to allow for a VARIABLE FORCE RANGE, ranging from proportional force strength of Radius (solid body rotation) to 1/Radius (intermediate strength) to 1/Radius^2 (our usual perceived strength of gravity).
So, what happened yesterday as seen below, is that I started out thinking differentiation of i as i' was the Coulomb gravity law and differentiation of L was the spin law
B = kg /A*s^2 Law of Magnetism
V = i*B*L Law of Electricity
V' = (i*B*L)' Ampere-Maxwell law
(V/(i*L))' = B' Faraday law
(V/(B*L))' = i' the new law, the EM-Gravity law
(V/(i*B))' = L' the new law of spin
Let me correct that as to this::
B = kg /A*s^2 Law of Magnetism
V = i*B*L Law of Electricity
V' = (i*B*L)' Ampere-Maxwell law
(V/(i*L))' = B' Faraday law
(V/(B*L))' = i' the new law of spin
(V/(i*B))' = L' the new law of Coulomb force with EM gravity force
Below are yesterday's posts showing how I came to this realization::
getting R to 1/R to 1/R^2 Re: New AP-Maxwell Equations, under construction
getting R to 1/R to 1/R^2 Re: New AP-Maxwell Equations, under construction
burs...@gmail.com
Aug 17, 2017, 5:48:46 PM8/17/17
to
You suck at physics.
pora...@gmail.com
Aug 18, 2017, 12:02:01 AM8/18/17
to
Archi will not go on with his mission
without my
Y.Circlon
and my Y circlon mechanism !!
ATB
Y.Porat
======================
Archimedes Plutonium
Aug 18, 2017, 12:53:22 AM8/18/17
to
On Thursday, August 17, 2017 at 3:58:33 PM UTC-5, Archimedes Plutonium wrote:
(snipped)
where the BB is like distance squared and where the other two preceding terms fall to 0 in value, then the last third term is all we see, such as the planets gravity with Sun. But in the case of Saturn and its rings, the first two terms predominate and the last third term falls to near zero.
> On a commonsense view the Coulomb force is the motion of a test charge and what describes that motion is derivative of angular momentum
>
> What describes the Spin law is derivative of current which results in current divided by angular momentum.
>
> On Thursday, August 17, 2017 at 5:17:45 AM UTC-5, Archimedes Plutonium wrote:
> getting R to 1/R to 1/R^2 Re: New AP-Maxwell Equations, under construction
>
> My mistake the spin law turns out to be i' = - ii*L, a current multiplied by current multiplied by angular momentum.
>
So in that spin law
i' = ((i_C)*V -VVL - V(i_C) / R^2
That reduces instantly to
i' = -VVL/R^2
And what that reduces to, is
i' = - (ii*L)
Earlier I missed the current term i as a single i but it is a i^2
Now here, in Physics history, we have really only one famous experiment dealing with Faraday's law, where a thrusting bar magnet through a closed loop wire produces a current i.
Let me add to that famous experiment the Eddy Current Experiment where you simply drop a magnet through a copper tube, recreating the Faraday law, only what we want to observe is produced spin on the magnet as it falls through the tube. We start the magnet with no spin and observe where it lands, whether it had picked up spin.
My Experiment is a copper tube 21cm long, 1.4 cm diameter. My magnet is a neodymium magnet 1 cm long with .7 cm diameter, and weighs 3.5 grams. Now if I drop a plug of the same weight through the tube, it falls 3 times faster than the magnet. So that if I had a tube of plastic 3 times the length 3x21cm = 63cm of the copper tube, that the magnet falling in the plastic and copper tube would reach the floor at the same time, for it is Lenz law in the magnet copper tube that is slowing the fall down. And that is the fusion barrier principle but here I am concerned only with spin.
So, I start the marked magnet at a marked point on tube and examine the magnet when it reaches the floor (a stack of magazines). Repeatedly, I find the magnet has revolved 1/2 turn, or 180 degree revolution.
Here, I am asking if the formula - iiL can compute the 1/2 turn.
AP
(snipped)
>
> Let me correct that as to this::
>
> B = kg /A*s^2 Law of Magnetism
> V = i*B*L Law of Electricity
> V' = (i*B*L)' Ampere-Maxwell law
> (V/(i*L))' = B' Faraday law
> (V/(B*L))' = i' the new law of spin
> (V/(i*B))' = L' the new law of Coulomb force with EM gravity force
>
>
> Let me correct that as to this::
>
> B = kg /A*s^2 Law of Magnetism
> V = i*B*L Law of Electricity
> V' = (i*B*L)' Ampere-Maxwell law
> (V/(i*L))' = B' Faraday law
> (V/(B*L))' = i' the new law of spin
> (V/(i*B))' = L' the new law of Coulomb force with EM gravity force
>
>
> L' = (i^2B - B^3- V^2i) / i^2B^2
>
> With the B term we see it is reflected as 1/B then B then 1/B^2
>
Now notice the Newton gravity or General Relativity or gravity is all in the last term of - VVi/iiBB
>
> With the B term we see it is reflected as 1/B then B then 1/B^2
>
where the BB is like distance squared and where the other two preceding terms fall to 0 in value, then the last third term is all we see, such as the planets gravity with Sun. But in the case of Saturn and its rings, the first two terms predominate and the last third term falls to near zero.
> On a commonsense view the Coulomb force is the motion of a test charge and what describes that motion is derivative of angular momentum
>
> What describes the Spin law is derivative of current which results in current divided by angular momentum.
>
> On Thursday, August 17, 2017 at 5:17:45 AM UTC-5, Archimedes Plutonium wrote:
> getting R to 1/R to 1/R^2 Re: New AP-Maxwell Equations, under construction
>
>
So in that spin law
i' = ((i_C)*V -VVL - V(i_C) / R^2
That reduces instantly to
i' = -VVL/R^2
And what that reduces to, is
Earlier I missed the current term i as a single i but it is a i^2
Now here, in Physics history, we have really only one famous experiment dealing with Faraday's law, where a thrusting bar magnet through a closed loop wire produces a current i.
Let me add to that famous experiment the Eddy Current Experiment where you simply drop a magnet through a copper tube, recreating the Faraday law, only what we want to observe is produced spin on the magnet as it falls through the tube. We start the magnet with no spin and observe where it lands, whether it had picked up spin.
My Experiment is a copper tube 21cm long, 1.4 cm diameter. My magnet is a neodymium magnet 1 cm long with .7 cm diameter, and weighs 3.5 grams. Now if I drop a plug of the same weight through the tube, it falls 3 times faster than the magnet. So that if I had a tube of plastic 3 times the length 3x21cm = 63cm of the copper tube, that the magnet falling in the plastic and copper tube would reach the floor at the same time, for it is Lenz law in the magnet copper tube that is slowing the fall down. And that is the fusion barrier principle but here I am concerned only with spin.
So, I start the marked magnet at a marked point on tube and examine the magnet when it reaches the floor (a stack of magazines). Repeatedly, I find the magnet has revolved 1/2 turn, or 180 degree revolution.
Here, I am asking if the formula - iiL can compute the 1/2 turn.
AP
Archimedes Plutonium
Aug 18, 2017, 1:21:15 AM8/18/17
to
On Thursday, August 17, 2017 at 11:53:22 PM UTC-5, Archimedes Plutonium wrote:
(snipped)
>
(snipped)
>
> So in that spin law
>
> i' = ((i_C)*V -VVL - V(i_C) / R^2
>
> That reduces instantly to
>
> i' = -VVL/R^2
>
> And what that reduces to, is
>
> i' = - (ii*L)
>
> Earlier I missed the current term i as a single i but it is a i^2
>
> Now here, in Physics history, we have really only one famous experiment dealing with Faraday's law, where a thrusting bar magnet through a closed loop wire produces a current i.
>
> Let me add to that famous experiment the Eddy Current Experiment where you simply drop a magnet through a copper tube, recreating the Faraday law, only what we want to observe is produced spin on the magnet as it falls through the tube. We start the magnet with no spin and observe where it lands, whether it had picked up spin.
>
> My Experiment is a copper tube 21cm long, 1.4 cm diameter. My magnet is a neodymium magnet 1 cm long with .7 cm diameter, and weighs 3.5 grams. Now if I drop a plug of the same weight through the tube, it falls 3 times faster than the magnet. So that if I had a tube of plastic 3 times the length 3x21cm = 63cm of the copper tube, that the magnet falling in the plastic and copper tube would reach the floor at the same time, for it is Lenz law in the magnet copper tube that is slowing the fall down. And that is the fusion barrier principle but here I am concerned only with spin.
>
> So, I start the marked magnet at a marked point on tube and examine the magnet when it reaches the floor (a stack of magazines). Repeatedly, I find the magnet has revolved 1/2 turn, or 180 degree revolution.
>
> Here, I am asking if the formula - iiL can compute the 1/2 turn.
>
>
Now, if that law is correct, there is one astronomical body to put the formula to test, a excellent test.
>
> i' = ((i_C)*V -VVL - V(i_C) / R^2
>
> That reduces instantly to
>
> i' = -VVL/R^2
>
> And what that reduces to, is
>
> i' = - (ii*L)
>
> Earlier I missed the current term i as a single i but it is a i^2
>
> Now here, in Physics history, we have really only one famous experiment dealing with Faraday's law, where a thrusting bar magnet through a closed loop wire produces a current i.
>
> Let me add to that famous experiment the Eddy Current Experiment where you simply drop a magnet through a copper tube, recreating the Faraday law, only what we want to observe is produced spin on the magnet as it falls through the tube. We start the magnet with no spin and observe where it lands, whether it had picked up spin.
>
> My Experiment is a copper tube 21cm long, 1.4 cm diameter. My magnet is a neodymium magnet 1 cm long with .7 cm diameter, and weighs 3.5 grams. Now if I drop a plug of the same weight through the tube, it falls 3 times faster than the magnet. So that if I had a tube of plastic 3 times the length 3x21cm = 63cm of the copper tube, that the magnet falling in the plastic and copper tube would reach the floor at the same time, for it is Lenz law in the magnet copper tube that is slowing the fall down. And that is the fusion barrier principle but here I am concerned only with spin.
>
> So, I start the marked magnet at a marked point on tube and examine the magnet when it reaches the floor (a stack of magazines). Repeatedly, I find the magnet has revolved 1/2 turn, or 180 degree revolution.
>
> Here, I am asking if the formula - iiL can compute the 1/2 turn.
>
>
The planet Uranus has a unique rotation of the planets in that it rotates on its side. Its axis of rotation is not perpendicular to the path of its motion around the Sun, but is parallel. Earth's axis if we discount its tilt of 23 degrees is perpendicular to its path around the Sun. Jupiter's axis of rotation is perpendicular to its path around the Sun. Uranus, however, has a axis of rotation parallel to its motion path around the Sun.
So, can the formula of spin as -ii*L give a axis that is parallel instead of perpendicular using a right hand rule. And the justification would be the EM forces of Jupiter with Saturn versus Sun affecting Uranus.
AP
Archimedes Plutonium
Aug 18, 2017, 6:49:17 PM8/18/17
to
So if spin of planets, their daily rotation is caused by electricity magnetism, then it seems likely that the spin of Uranus on its side, is caused by being so close to Saturn and Jupiter, so the pull of these two giants on Uranus with the pull of the Sun can turn the axis of rotation parallel to path of orbit.
AP
AP
Archimedes Plutonium
Aug 19, 2017, 4:16:37 AM8/19/17
to
Justification schemata
---------------------------------
So we list the derivatives with respect to time of EM parameters
Derivative with respect to time s, to 1/s velocity, to 1/s^2 acceleration
Current i = dq/ds so current is 1/s what is derivative of current, is 1/s^2 and what is that?
Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3
Derivative of B would be 1/s^2 to 1/s^3, so that derivative of B is either Volt or Resistance and the clear choice here is Volt
Derivative of L and here we have L as 1/s so the derivative is 1/s^2 and the clear choice here is a force, a torque, and now, if we have a torque times magnetic field B we end up with capacitor current i_C
----------------justification schemata ----------------------
Ampere Law
V' = (iBL)' = i'*B*L + i*B'*L + i*B*L'
= B*B*L + i*V*L + i*B*(i_C)
Now, does that make sense to what we know as the Ampere-Maxwell law?
It makes a lot of sense in that we have a input current and get out a magnetic field in the first term of B^2(L). Then we have a displacement current in the third term as - i*B*(i_C). But finally we have a mid-term of - iVL which is some spin term, unknown in Old Maxwell Equations.
Faraday law
B' = (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
> = ((i_C)iL - VBL -Vi(i_C))/ i^2L^2
Now if we assume currents are the same, where i and i_C are the same we reduce Faraday's law to this
= 1/L - VB/i^2L - V/L^2
Now if we take Resistance = V/i we get this reduction
= 1/L - RB/iL - V/L^2
Now, does that look like resembling the Old Maxwell Equation of Faraday law? Remember Maxwell had just one term (Heaviside attrition), nothing for a Lenz law. We can see a current produced in the middle term - RB/iL and we see a Lenz law of third term - V/L^2 but we see a new term of 1/L as some sort of spin. What could spin in Faraday's law? I suspect the coil closed loop wire spins, but then that is probably so securely fastened down in generators and motors, that the spin becomes heat.
AP
---------------------------------
So we list the derivatives with respect to time of EM parameters
Current i = dq/ds so current is 1/s what is derivative of current, is 1/s^2 and what is that?
Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3
Derivative of V, voltage, and here we have 1/s^3, and the only s^4 I know of is Capacitance current Capacitance A^2*s^4, even though it is in the numerator. Let me denote it by i_C
Derivative of current i is Magnetic Field B
Derivative of B would be 1/s^2 to 1/s^3, so that derivative of B is either Volt or Resistance and the clear choice here is Volt
----------------justification schemata ----------------------
Ampere Law
V' = (iBL)' = i'*B*L + i*B'*L + i*B*L'
= B*B*L + i*V*L + i*B*(i_C)
Now, does that make sense to what we know as the Ampere-Maxwell law?
It makes a lot of sense in that we have a input current and get out a magnetic field in the first term of B^2(L). Then we have a displacement current in the third term as - i*B*(i_C). But finally we have a mid-term of - iVL which is some spin term, unknown in Old Maxwell Equations.
Faraday law
B' = (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
> = ((i_C)iL - VBL -Vi(i_C))/ i^2L^2
Now if we assume currents are the same, where i and i_C are the same we reduce Faraday's law to this
= 1/L - VB/i^2L - V/L^2
Now if we take Resistance = V/i we get this reduction
= 1/L - RB/iL - V/L^2
Now, does that look like resembling the Old Maxwell Equation of Faraday law? Remember Maxwell had just one term (Heaviside attrition), nothing for a Lenz law. We can see a current produced in the middle term - RB/iL and we see a Lenz law of third term - V/L^2 but we see a new term of 1/L as some sort of spin. What could spin in Faraday's law? I suspect the coil closed loop wire spins, but then that is probably so securely fastened down in generators and motors, that the spin becomes heat.
AP
Archimedes Plutonium
Aug 19, 2017, 7:49:37 PM8/19/17
to
On Saturday, August 19, 2017 at 3:16:37 AM UTC-5, Archimedes Plutonium wrote:
> Justification schemata
> ---------------------------------
>
> So we list the derivatives with respect to time of EM parameters
>
> Derivative with respect to time s, to 1/s velocity, to 1/s^2 acceleration
>
> Current i = dq/ds so current is 1/s what is derivative of current, is 1/s^2 and what is that?
>
> Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3
>
> Derivative of V, voltage, and here we have 1/s^3, and the only s^4 I know of is Capacitance current Capacitance A^2*s^4, even though it is in the numerator. Let me denote it by i_C
>
> Derivative of current i is Magnetic Field B
>
> Derivative of B would be 1/s^2 to 1/s^3, so that derivative of B is either Volt or Resistance and the clear choice here is Volt
>
> Derivative of L and here we have L as 1/s so the derivative is 1/s^2 and the clear choice here is a force, a torque, and now, if we have a torque times magnetic field B we end up with capacitor current i_C
>
> ----------------justification schemata ----------------------
>
>
> Ampere Law
>
> V' = (iBL)' = i'*B*L + i*B'*L + i*B*L'
>
> = B*B*L + i*V*L + i*B*(i_C)
>
> Now, does that make sense to what we know as the Ampere-Maxwell law?
>
> It makes a lot of sense in that we have a input current and get out a magnetic field in the first term of B^2(L). Then we have a displacement current in the third term as - i*B*(i_C). But finally we have a mid-term of - iVL which is some spin term, unknown in Old Maxwell Equations.
>
>
> Faraday law
>
> B' = (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
>
> > = ((i_C)iL - VBL -Vi(i_C))/ i^2L^2
>
> Now if we assume currents are the same, where i and i_C are the same we reduce Faraday's law to this
>
> = 1/L - VB/i^2L - V/L^2
>
Here I am going to have to go back into the laboratory, to see where this spin term is located. This 1/L.
> Justification schemata
> ---------------------------------
>
> So we list the derivatives with respect to time of EM parameters
>
> Derivative with respect to time s, to 1/s velocity, to 1/s^2 acceleration
>
> Current i = dq/ds so current is 1/s what is derivative of current, is 1/s^2 and what is that?
>
> Magnetic field 1/A*s^2, Volt 1/A*s^3, Resistance 1/A^2*s^3
>
> Derivative of V, voltage, and here we have 1/s^3, and the only s^4 I know of is Capacitance current Capacitance A^2*s^4, even though it is in the numerator. Let me denote it by i_C
>
> Derivative of current i is Magnetic Field B
>
> Derivative of B would be 1/s^2 to 1/s^3, so that derivative of B is either Volt or Resistance and the clear choice here is Volt
>
> Derivative of L and here we have L as 1/s so the derivative is 1/s^2 and the clear choice here is a force, a torque, and now, if we have a torque times magnetic field B we end up with capacitor current i_C
>
> ----------------justification schemata ----------------------
>
>
> Ampere Law
>
> V' = (iBL)' = i'*B*L + i*B'*L + i*B*L'
>
> = B*B*L + i*V*L + i*B*(i_C)
>
> Now, does that make sense to what we know as the Ampere-Maxwell law?
>
> It makes a lot of sense in that we have a input current and get out a magnetic field in the first term of B^2(L). Then we have a displacement current in the third term as - i*B*(i_C). But finally we have a mid-term of - iVL which is some spin term, unknown in Old Maxwell Equations.
>
>
> Faraday law
>
> B' = (V/i*L)' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
>
> > = ((i_C)iL - VBL -Vi(i_C))/ i^2L^2
>
> Now if we assume currents are the same, where i and i_C are the same we reduce Faraday's law to this
>
> = 1/L - VB/i^2L - V/L^2
>
I suspect it is a spin of the closed loop of wire.
But that spin term is going to be some tiny that almost immeasurable in my crude lab.
So I think what I need to do is use a electric motor and find a "spin term of 1/L"
I think those in engineering have a term for this already. Something having to do with the fact that the electric equipment-- motor-- generator tend to spin, and so they have to anchor the apparatus down securely.
This spin term of 1/L would go out as heat, waste in energy.
pora...@gmail.com
Aug 20, 2017, 12:10:11 AM8/20/17
to
in the ***existing physics***
SPIN IS NOT ROTATION !!!!!
may be in my Y Circlon !!!!
----
ATB
Y.Porat
==============================
Archimedes Plutonium
Aug 20, 2017, 3:45:56 AM8/20/17
to
Haste makes waste, and surely I have not be hasty on this Maxwell revision. In fact I have dawdled and dawdled for months now at this juncture. So let me review that all the 6 Equations of AP-Maxwell theory are correct, before I get too excited about a 1/L term in Faraday law.
AP
AP
burs...@gmail.com
Aug 20, 2017, 9:07:56 AM8/20/17
to
AP brain farto, spamming
nonsense on sci.physics
altready for 30 years, not a single line
of physics, only profound confusion.
nonsense on sci.physics
altready for 30 years, not a single line
of physics, only profound confusion.
Serg io
Aug 20, 2017, 10:24:43 AM8/20/17
to
On 8/20/2017 8:07 AM, burs...@gmail.com wrote:
no need to redo Maxwells Equns, they works just fine, since 1884.
https://en.wikipedia.org/wiki/History_of_Maxwell%27s_equations
no need to redo Maxwells Equns, they works just fine, since 1884.
https://en.wikipedia.org/wiki/History_of_Maxwell%27s_equations
Archimedes Plutonium
Aug 20, 2017, 6:45:02 PM8/20/17
to
On Sunday, August 20, 2017 at 9:24:43 AM UTC-5, Serg io wrote:
>
> no need to redo Maxwells Equns, they works just fine, since 1884.
>
> https://en.wikipedia.org/wiki/History_of_Maxwell%27s_equations
>
totally wrong on that, the mistakes are everywhere, starting with the idea there is no repulsion, there is no electric field, Faraday's law has no Lenz law in it, Coulomb law is a force that ranges from R to 1/R to 1/R^2 and the list of mistakes goes on and on
>
> no need to redo Maxwells Equns, they works just fine, since 1884.
>
> https://en.wikipedia.org/wiki/History_of_Maxwell%27s_equations
>
Serg io
Aug 21, 2017, 3:00:20 PM8/21/17
to
Michael Moroney
Aug 21, 2017, 6:35:26 PM8/21/17
to
since 1884.
Dingus Dirtbag McGee
Aug 23, 2017, 10:32:36 PM8/23/17
to
Why you posted this?
Did your brain fell out?
Did your brain fell out?
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