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Δx physics equations needed, Physics Renormalization


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Ya, This is not really a question but more of a request, I would like a list of every equation or summation with Δx = F(anything) that is confirmed to be correct in physics please list who made them originally and what they describe thanks if you would like to post on this thread doesn't matter what just who wrote them and what they apply to.

Edited by Vmedvil
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Well, since no one has taken the initiative, I will begin. 

Displacement from two positions, Some greeks

Δx = (x1-x2)

Velocity  to displacement still the greeks?

Δx = (V* Δt)

Newton acceleration to Displacement

Δx = (a * Δt2)

Newton Force to Displacement

Δx = ((F/M) * Δt2)

Joule Energy to Displacement

Δx = ((E/M) * Δt2)1/2

Galileo  Power to Displacement

Δx = ((P/M) * Δt3)1/2

Willem 's Gravesande and Émilie du Châtelet  Momentum to displacement.

Δx = ((p/M) * Δt)

Galileo Impulse to displacement.

Δx = ((J/M) * Δt3)

 

Just the beginning, chime in if you want to add to this.

 

 

Edited by Vmedvil
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277px-Moment_of_inertia_solid_sphere_svg.png.84475ae7bab63b33997c19fb409f7217.png

Angular Velocity change  to displacement still the greeks?

Δx = (V0Δt + (V2/r)Δt2)

Newton angular velocity to Displacement

Δx = ((V2/r) * Δt2)

Newton Moment of Inertial Force to Displacement

Δx = ((F/I)* r2 * Δt2)

Joule Moment of Interial Energy to Displacement

Δx = ((E/I)* r2  * Δt2)1/2

Galileo  Moment of Interial Power to Displacement

Δx = ((P/I)* r2  * Δt3)1/2

Willem 's Gravesande and Émilie du Châtelet  moment of Inertial Momentum to displacement.

Δx = ((p/I)* r2  * Δt)

Galileo Moment of Interial Impulse to displacement.

Δx = ((J/I)* r2  * Δt3)

Edited by Vmedvil
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200px-NewtonsLawOfUniversalGravitation_svg.png.a2ca087c00900820fa6dd132aef7cfa9.png

Linear Newton Gravitational acceleration to Displacement

Δx = ((G * M/ r2) * Δt2)

Linear Newton Gravitational Force to Displacement

Δx = ((G * M1* M2) / r) * Δt2)

Linear Joule Gravitational Energy to Displacement

Δx = (((G * M1* M2) / r )* Δt2)1/2

Linear Galileo  Power to Displacement

Δx = (((G * M1* M2) / r) * Δt4)1/2

Linear Willem 's Gravesande and Émilie du Châtelet  gravitational Momentum to displacement.

Δx = ((G * M/ r2) * Δt2)

Linear Galileo Gravitational Impulse to displacement.

Δx = (((G * M1* M2) / r) * Δt4)

 

 

binorbit.gif

Gravitational Angular Velocity change  to displacement still the greeks?

Δx = ((G0 * M/ r2)Δt + ((G2 * M22/r3)Δt2)

Newton  Gravitational angular velocity to Displacement

Δx = ((G2 * M22/r3) * Δt2)

Newton Gravitational Moment of Inertial Force to Displacement

Δx = (G * M1 * M2 * I * Δt2)

Joule Gravitational Moment of Interial Energy to Displacement

Δx = (G * M1 * M2 * I * r  * Δt2)1/2

Galileo Gravitational Moment of Interial Power to Displacement

Δx = (G * M1 * M2 * I * r * Δt4)1/2

Willem 's Gravesande and Émilie du Châtelet  Gravitational moment of Inertial Momentum to displacement.

Δx = (M1 * G * M2 * I * Δt)

Galileo Gravitational Moment of Interial Impulse to displacement.

Δx = (G * M1 * M2 * I * Δt4)

 

This is where Newton's Gravity shatters into pieces, how again can that be correct, its not.

Δx = (G * M1 * M2 * I * Δt4) = (M1 * G * M2 * I * Δt)

and 

Δx = (G * M1 * M2 * I * r * Δt4)1/2 = (G * M1 * M2 * I * r  * Δt2)1/2

IC725763.png.3d681a3e2370632df0460c89962944bb.png

I bet coulomb's law shatters at the same point, so does this mean our laws of charge and gravity bases are broken?

Coulombs_Law_Diagram.png.e12b852ccdd6786ba73cd9e1562144fd.png

Edited by Vmedvil
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On 10/18/2017 at 0:20 PM, Vmedvil said:

277px-Moment_of_inertia_solid_sphere_svg.png.84475ae7bab63b33997c19fb409f7217.png

Angular Velocity change  to displacement still the greeks?

Δx = (V0Δt + (V2/r)Δt2)

Newton angular velocity to Displacement

Δx = ((V2/r) * Δt2)

Newton Moment of Inertial Force to Displacement

Δx = ((F/I)* r2 * Δt2)

Joule Moment of Interial Energy to Displacement

Δx = ((E/I)* r2  * Δt2)1/2

Galileo  Moment of Interial Power to Displacement

Δx = ((P/I)* r2  * Δt3)1/2

Willem 's Gravesande and Émilie du Châtelet  moment of Inertial Momentum to displacement.

Δx = ((p/I)* r2  * Δt)

Galileo Moment of Interial Impulse to displacement.

Δx = ((J/I)* r2  * Δt3)

The Picture on this broke, whatever just as long as everyone knows what this geometrically describes.

220px-Moment_of_inertia_solid_sphere_svg.png.0f3853f166b62ff1555ac390bc681f8f.png

Now lets see if Coulomb's law breaks at the same spot.

Coulombs_Law_Diagram.png.e12b852ccdd6786ba73cd9e1562144fd.png

Linear Electric Velocity  to displacement still the Coulomb-greeks?

Δx = (((K*q1*q2) / r2) * Δt2)

Linear Coulomb-Newton Electric acceleration to Displacement

Δx = (((K*q1*q2) / r2) * Δt2)

Linear Coulomb-Newton Electric Force to Displacement

Δx = (((K*q1*q2* M) /r2) * Δt2)

Linear Coulomb-Joule Electric Energy to Displacement

Δx = (((K*q1*q2*M) /r) * Δt2)1/2

Linear Coulomb-Galileo  Electric Power to Displacement

Δx = (((K*q1*q2*M) /r) * Δt4)1/2

Linear Coulomb-Willem 's Gravesande-Émilie du Châtelet  Electric Momentum to displacement.

Δx = (((K*q1*q2) / r2) * Δt2)

Linear Coulomb-Galileo Electric Impulse to displacement.

Δx = (((K*q1*q2* M2) / r2) * Δt4)

Electric Angular Velocity change  to displacement still the Coulomb-greeks?

Δx = (((K0*q1*q2) / r2)Δt2 + ((K2*q12*q22) / r3)Δt3)

 Coulomb-Newton Electric angular velocity to Displacement

Δx = (((K*q12*q22) / r3) * Δt2)

Coulomb-Newton Electric Moment of Inertial Force to Displacement

Δx = (K*q1*q2*I*Δt2)

Coulomb-Joule Electric Moment of Interial Energy to Displacement

Δx = (K*q1*q2*I*r*Δt2)1/2

Coulomb-Galileo  Electric Moment of Interial Power to Displacement

Δx = (K*q1*q2*I*r*Δt2)1/2

Coulomb-Willem 's Gravesande and Émilie du Châtelet  Electric moment of Inertial Momentum to displacement.

Δx = (M*K*q1*q2*I*Δt2)

Coulomb-Galileo Electric Moment of Interial Impulse to displacement.

Δx = (K*q1*q2*I*M*Δt4)

Actually Coulomb's law doesn't error out in any situation, I think it is because of how we define force as "Mass * Acceleration = Force" that errors out newton's gravitational law.

units1.jpg

Edited by Vmedvil
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On second look yes, it does still error out at, I think I doing something wrong at this point or coulomb's law  and Newton's Law Both Error at the same point.

Δx = (K*q1*q2*I*M*Δt4) = (M*K*q1*q2*I*Δt2)

and

Δx = (G * M1 * M2 * I * Δt4) = (M1 * G * M2 * I * Δt)

IC725763.png.3d681a3e2370632df0460c89962944bb.png

Edited by Vmedvil
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This makes this wrong after checking or at-least not continuous, Later we will see if Maxwell's equations and GR fix it, but for now, Newton's law and Coulomb's law are not entirely correct on Impulse and Momentum breaking them against themselves on Δx. If this is a recursive error, I am going to want to go back in time and severely remand these scientist for being sloppy, unless I am at fault if anyone wants to check this go for it.

 

 

Edited by Vmedvil
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Galileo Gravitational Moment of Interial Power to Displacement

Δx = (G * M1 * M2 * I * r * Δt4)1/2

Wrong Solutions discontinuous function. 

Δx = (((G * M1* M2) / r) * Δt4)

 

Coulomb-Galileo Electric Moment of Interial Impulse to displacement.

Δx = (K*q1*q2*I*M*Δt4)

 

Linear Coulomb-Galileo  Electric Power to Displacement

Δx = (((K*q1*q2*M) /r) * Δt4)1/2

These equations are incorrect solutions to these equations due to J being solved for in maxwell's equations these classical solution being in error being replaced by Maxwell ampere equation.

fig-1b-maxwell-equations2.thumb.gif.d696ff3f4bd7fdc375a7c0fcbc7061b4.gif

 

Then power is replaced by Voltage in the electric field so those classical power equation solutions are also wrong and discontinuous. 

11001.png

Then that gravity has not been fixed besides being out of those intertial reference frames in GR + SR making newton discontineous over inertial moment.

198829.image0.jpg

 

The Other newton equation solution that is wrong has not been fixed.

 

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On ‎10‎/‎21‎/‎2017 at 10:10 PM, Vmedvil said:

Galileo Gravitational Moment of Interial Power to Displacement

Δx = (G * M1 * M2 * I * r * Δt4)1/2

Wrong Solutions discontinuous function. 

Δx = (((G * M1* M2) / r) * Δt4)

 

Coulomb-Galileo Electric Moment of Interial Impulse to displacement.

Δx = (K*q1*q2*I*M*Δt4)

 

Linear Coulomb-Galileo  Electric Power to Displacement

Δx = (((K*q1*q2*M) /r) * Δt4)1/2

These equations are incorrect solutions to these equations due to J being solved for in maxwell's equations these classical solution being in error being replaced by Maxwell ampere equation.

fig-1b-maxwell-equations2.thumb.gif.d696ff3f4bd7fdc375a7c0fcbc7061b4.gif

 

Then power is replaced by Voltage in the electric field so those classical power equation solutions are also wrong and discontinuous. 

11001.png

Then that gravity has not been fixed besides being out of those intertial reference frames in GR + SR making newton discontineous over inertial moment.

198829.image0.jpg

The Other newton equation solution that is wrong has not been fixed.

 

Actually that other discontinuous part is also fixed by SR + GR, so all of the problems with Coulomb's and Newton's discontinuity has been fixed in Inertial Power with the first equation of SR the second fixes Inertial Momentum, which were both wrong via classical predictions. SR is discontinuous at  V = C and neither Maxwell's equations nor Ohm's are discontinuous. So many of the problems have been fixed so, this post is done I believe until QM and QFT is visited which I know fixes SR's Discontinuity at V = C, due to Schwarzschild Metric and etc.

 

the%20schwarzschild%20metric,%20inverse%

 

eq0033SP.gif

img73.gif

Which makes GR continuous at any point even at V = C in SR.

Edited by Vmedvil
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