Vmedvil Posted October 18, 2017 Posted October 18, 2017 (edited) 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 October 18, 2017 by Vmedvil
Vmedvil Posted October 18, 2017 Author Posted October 18, 2017 (edited) 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 October 18, 2017 by Vmedvil
Vmedvil Posted October 18, 2017 Author Posted October 18, 2017 (edited) 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 October 18, 2017 by Vmedvil
Vmedvil Posted October 19, 2017 Author Posted October 19, 2017 (edited) Linear Newton Gravitational acceleration to Displacement Δx = ((G * M2 / r2) * Δt2) Linear Newton Gravitational Force to Displacement Δx = ((G * M12 * M2) / r2 ) * Δt2) Linear Joule Gravitational Energy to Displacement Δx = (((G * M12 * M2) / r )* Δt2)1/2 Linear Galileo Power to Displacement Δx = (((G * M12 * M2) / r) * Δt4)1/2 Linear Willem 's Gravesande and Émilie du Châtelet gravitational Momentum to displacement. Δx = ((G * M2 / r2) * Δt2) Linear Galileo Gravitational Impulse to displacement. Δx = (((G * M12 * M2) / r2 ) * Δt4) Gravitational Angular Velocity change to displacement still the greeks? Δx = ((G0 * M2 / 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 I bet coulomb's law shatters at the same point, so does this mean our laws of charge and gravity bases are broken? Edited October 19, 2017 by Vmedvil
Vmedvil Posted October 19, 2017 Author Posted October 19, 2017 (edited) On 10/18/2017 at 0:20 PM, Vmedvil said: 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. Now lets see if Coulomb's law breaks at the same spot. 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. Edited October 19, 2017 by Vmedvil
Vmedvil Posted October 19, 2017 Author Posted October 19, 2017 (edited) 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) Edited October 19, 2017 by Vmedvil
Vmedvil Posted October 19, 2017 Author Posted October 19, 2017 (edited) 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 October 19, 2017 by Vmedvil
Mordred Posted October 19, 2017 Posted October 19, 2017 We really need to teach you how to latex lmao.
Vmedvil Posted October 22, 2017 Author Posted October 22, 2017 Galileo Gravitational Moment of Interial Power to Displacement Δx = (G * M1 * M2 * I * r * Δt4)1/2 Wrong Solutions discontinuous function. Δx = (((G * M12 * M2) / r2 ) * 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. Then power is replaced by Voltage in the electric field so those classical power equation solutions are also wrong and discontinuous. Then that gravity has not been fixed besides being out of those intertial reference frames in GR + SR making newton discontineous over inertial moment. The Other newton equation solution that is wrong has not been fixed.
Vmedvil Posted October 25, 2017 Author Posted October 25, 2017 (edited) 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 * M12 * M2) / r2 ) * 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. Then power is replaced by Voltage in the electric field so those classical power equation solutions are also wrong and discontinuous. Then that gravity has not been fixed besides being out of those intertial reference frames in GR + SR making newton discontineous over inertial moment. 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. Which makes GR continuous at any point even at V = C in SR. Edited October 25, 2017 by Vmedvil
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