# A metric current and sequel

Discussion in 'Alternative Theories' started by TIMO MOILANEN, Oct 1, 2020.

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## Should they redefine the kg

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1. ### TIMO MOILANENRegistered Member

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The thought : spikes on a sphere representing charges dq . The spikes don't bend , they represent potential .
Amount of dq integral -RtoR for r=-1to1 , a=arccosR ,dq =2pi R*sina The force F =Sigma dq |Rcosa| /(r+Rcosa)^2 ,r/R=p , dividing with pi to obtain average and so on. Shortly I get INT -1 to1 of 2|R|(1+R^2)^0.5 /(p+R)^2 . Is=4(((p^2-1)^.5*(2P^2-1)*arcsin(1/p) )-2p^2+2)/(P^2-1) . This multiplied with p^2 goes to 4/3 for p=>infinity. For "Coloumbs force law I get F=4/3Q^2/r^2*const. For constant I put mass of proton Mp*c^2 So I have F=K*Q^2/r^2 , where K=4/(3Mp*c^2) . Force per mass(in nuc. units) of proton. I use index i to separate "invented" from official . Ki=8.9*10^9 Nm^2/Ai^2 and electric constant E0i =3*Mp*c^2/(16pi)= 8.9*10^-12 Fi/m , magnetic constant (µ0i=16pi/(3Mp*c^4) =1.25*10^-6 N/Ai^2 . Planck constant is not per mass and removing 4pi to , but electron fall in (kin+pot energy and photon take both energies with it =>*4). hi=16/3c^4 =6.6*10^-34 Js Fitting these into formula e^2=2aE0hc give fine structure coefficient get value of and "look like a fit" to be ai=2*(3/16pi)^2 =7.12*10^-3 =1/140 . Elementary charge become ei= 3/8*(Mp/pi^3*c)^.5 =1.58*10^-19 Ci . But ai should also apply for ei so ei = 2ai/c^2 =4/c^2*(3/16pi)^2 .

As a sum up putting in all "assumptions" ei^2 = 2*ai*E0i*hi*c become : (2*2*(3/(16pi))^2/c^2)^2 = 2*2(3/16pi)^2*3Mp*c^2/(16pi)*16/(3*c^4)*c and from this Mp=9/(64pi*c^3)=1.66*10^-27 kg. I read that a physical constant can not be anything even or such , and by my opinion it can't if you calculate and measure with a random unit , in this case the SI ampere. Since codata e =1.602176634*10^-19 C and I get ei=1.584968575*10^-19 Ci , the metric ampere Ai=ei/e =0.989259574 A(SI)
As the mass of a proton is a result of calculations the only input are 4/3 from integral pi and c .
Some constants: Ki =4/3*Mp*c^2= 8929903726 Ci
E0i=3*Mp*c^2/(16pi)= 8.9113470857 *10^-12 Fi/m
µ0i=1/(e0i*c^2)= 1.2485767251*10^-6 N/Ai
ei=2ai/c^2 = 1.585336228 *10^-19 Ci
hi=16/(3c^4)= 6.602614118592*10^-34 Js
Rki=hi/ei^2= 26270.7909 (ohm)i and so on
A value for mass of proton 9/(64pi*c^3)= 1.661309521*10^-27kg
Of course this work only with the metric values , Vi= 1/ 0.989259572 V(SI) and so on.
Now this looks easy but these affiliations was "not easy" to find and the process would fill a little book.
There are many particulars on a iron lever 1.10-20 Timo Moilanen

3. ### TIMO MOILANENRegistered Member

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I am sorry for using wrong numbers for the ampere Ai=0.98948904532648 A(SI) and Vi=1/0.98948904532648 V(SI)

5. ### DaveC426913Valued Senior Member

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What is this post about? What are you trying to tell us?

7. ### TIMO MOILANENRegistered Member

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The Ai ampere is comparable to the (m,kg,s) system directly Ex. F=K*Q1Q2/r^2 and Fm/L=µ0i*I^2/(2pi*r)
The F/L was definition of ampere =1.25663706 *1A^2/(2pi*1m)=2*10^-7N (new definition is very close to this) ,but I derived the constants directly from speed of light c and pi by solving an integral for the force between two charges putting in mass of a proton arranging all the formulas together till they summed up perfectly, getting out the elementary charge (my version ei) and a mass of the proton that fit to it. With these inputs all constants apply to m,s and kg in a known way. The F/L I get 1.9871716*10^-7N (smaller current).
1V*1A is still 1W because Vi*Ai even out ( by definition) but the (ohm)i =0.9786345 (ohm)(SI) . And all other units are off too. Other energies like electron volt is also little of . I is never said they are comparable (they do it anyhow) but they were aiming for it originally . That why there is only some percent difference . My inspiration is from the sad accident that the kg was defined to planck constant , that is measured and calculated with a mere "politically" defined ampere. My version of planck const. hi =16/(3*c^4)=6.602614119 Js that coincide with m,s and kg . plus consist of known constants itself. So by my meaning there is no reason to use technical units in science , since I made this kind of break through in what was "big" science some 50 years ago , but probably not "sexy" anymore. My hope is to get "big boots" scientists to have a look. One of the first of whom is you

The new tihng is that I combine energy with mass E=mc^2

8. ### DaveC426913Valued Senior Member

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The first thing you need to do is provide a synopsis explaining briefly, the problem as you see it, and then your proposal of a solution. No numbers, just a description.

9. ### TIMO MOILANENRegistered Member

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The technic (SI)ampere is not coherent with (m,s,kg) system and there fore measuring the scientific constants always give values multiplicated divided and often *(delta)^2 and ^-2 (also .5 ,-.5) This is why some scientist say constants can not be exact (like pi or n) . This is because using a close but random definition of ampere never is going to fit all the constants at same time ,and mass of proton have been "useless in these" calcs. Furthermore comparing classical physics values with electrical values is some 1%of , and atom physics use electro volt (officially approximately 1.602 *10^-19J
I have derived the exact converging electron volt =9/(64pi^2*c^2) = 1.585336228*10^-19 J . So using i (indifferent I call them) constants and units would make the results of the separate domains exactly comparable .
Scientists will disclaim this because it is "unfair" that a unnamed wannabe scientist claim this , and furthermore set the values on several constants they have worked on for decades . What comes to the math I just scratched the surface, and to really understand math is a necessity

What comes to me I have a degree in chemistry bsc , and I'm used to pull of miracles (you can read it from the pharmaceutical industry publications NEVER)

10. ### originHeading towards oblivionValued Senior Member

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If you use LaTex I would read your posts, but the equations in their current from are indecipherable.

11. ### river

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So now tell me in 3D what is going . Physically , an object(s) doing what , in plain English . If you can't do this then you really don't understand what your thinking . Take your time , hour , day , week , month etc. From now is fine , just let us know .

12. ### TIMO MOILANENRegistered Member

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Very simply when using metric ampere (Ai) all electric and electromagnetic constants "become" derivable from n (natural numbers) pi and c(speed of light) . The difference to CODATA values is a multiplicand of metric elementary charge /codata e =1.585336*10^-19/1.602177*10^-19=0.989498
Ex Boltzman constant kBi (metric)=1024*pi^2/(27*c^3 )=1.389229*10^-23 J/K . CODATA value 1.380649 *10^-23 J/K
kB (coda)/0.989489=1.395315*10^-23 J/K . this is 0.4 % of metric value calculated by me.
In the integral there is two electrically charged spheres ( The laboratory setup for measuring Coloumb's constant) and as math Coloumb's formula . The 3D is that the spheres are 3D . I move one of the spheres toward infinite dist. This give me 4/3 /r^2 (the integral at "long" distance)then I put in per mass of a proton in nuclear units Mp*c^2 and get metric Coloumb's constant Ki =4/(3*Mp*c^2)=8929903727Nm^2/Ci^2 .[ After a ton of easy math Ki is also =256/27*pi*c ]. Proceeding trough related constants math and mathematical relations I end up getting metric elementary ei the coupling to ampere . From there the formula e^2=2*(alpha)*(epsilon)0*h*c leave the mass of proton Mp =9/(64*pi*c^3) , and "everything" boils down to n ,pi and c. So far 10 constants, all /,*^(-2to2) 0.9895 make very good match (+- 0.2 %)with CODATA.
Essentially I say "all" constants can be derived with natural numbers ,pi and speed of light when using a different value for current Ai= 0.989489 A(SI) . (You understand ?)

13. ### TIMO MOILANENRegistered Member

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Ok I try a translator. Beginning integral: divided pi and p^2 readily put in :
\dfrac{2p^2\sqrt{1-r^2}\left|r\right|}{\left(r+p\right)^2} is=
\dfrac{4p^2\left(\sqrt{p^2-1}\left(2\arcsin\left(\frac{1}{p}\right)p^2-\arcsin\left(\frac{1}{p}\right)\right)-2p^2+2\right)}{p^2-1} by putting in bigger distances p integral goes to area |4/3|

14. ### TIMO MOILANENRegistered Member

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One use full formula : ei^2=2*E0i*h*c and results of constants . I don't translate literature to LaTex .Here my(metric) examples
Coloumbs const Ki =
\dfrac{256{\pi}c}{27}
electric cons. E0i=
\dfrac{27}{1024{\pi}^2c}
magnetic c. µ0i=
\dfrac{1024{\pi}^2}{27c}
fine structure const. ai=
\dfrac{9}{128{\pi}^2}
Planck constant hi=
\dfrac{16}{3c^4}
elementary charge ei=
\dfrac{9}{64{\pi}^2c^2}
Klitzing const. RKi=
\dfrac{65536{\pi}^4}{243}
Boltzman c. kBi=
\dfrac{1024{\pi}^2}{27c^3}
Josepson const. KJ=
\dfrac{27c^2}{512{\pi}^2}
I have put in the mass of proton . Here what I use Mpi=
\dfrac{9}{64{\pi}c^3} where it is applicable so here is only the shortest versions . The intermediates are quite beautiful and very compatible , but I have most math in pencil .

Hope this LaTex is comparabel.

15. ### billvonValued Senior Member

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In general, one of the better signs that you are reading woo is improbable numbers of significant digits, like 1.585336228 *10^-19 Ci, 8.9113470857 *10^-12 Fi/m or 1.2485767251*10^-6 N/Ai.

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16. ### originHeading towards oblivionValued Senior Member

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11,713
Yeah, looks like numerology to to me. Apparently in his world a joule has the units of s^2/m^2.

17. ### TIMO MOILANENRegistered Member

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There is the input of kg via mass of proton *c^2 =9/(64pi*c^3) that is divided away in some constants , and 9/(64pi^2*c^2) is ei (elementary charge) in coulombs C. It is math and I have been lucky (this time) not to get out trigonometry or logarithms (ln) and what it do not show is the input of 1Ci , 1Ai ,1m,1s and 1kg. Pi and c are numeric values only and mathematically reduced to shortest form.

18. ### HipparchiaRegistered Senior Member

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648
One wonders if you may have been too intimately involved with some products of that industry. Or not.

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19. ### exchemistValued Senior Member

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11,636
This is mathematical word salad. Are you being treated for schizophrenia?

20. ### river

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Why , What is the problem . Is it the understanding of what he is saying , is above your head , or is it actually mathematical non-sense ?

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22. ### TIMO MOILANENRegistered Member

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My idea is that the unit for electric current is not compatible with other SI units. It is close because at one time A was tried to adapt. Now the ampere is a little too big and by definition volt is a little too small. As a result I*U=P watt is spot on 1W=1J/s=1Nm/s but all other technical electric units are off by about 1% and 2%. The main tool I used is the formula e^2=2*a*E0*h*c and between constant the connections are mathematically simpler. So now the above equation in numbers is as follows :
(2*2*(3/16pi)^2/c^2)^2=2*2*(3/16pi)^2*27/(1024*pi^2*c)*16/(3c^3)*c this is eq. to 1=1 . I can not calculate the ampere Ai from this (like not m or kg either) since all units is 1. This is only (almost) about logics and math. and a huge (to my surprise ) simplification of physics . These expands and reduces do not give a value on the metric ampere Ai and comparing to constants is +- 0. 12% depending on const.

23. ### TIMO MOILANENRegistered Member

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I found a quite easy way of measuring ei (the real value ) in Ci , and compare the "energies" of e and ei. Since the (SET) single electron transistors have evolved in accuracy and speed one could count the elementary charges (not only in theory) in a current. The SI ampere is defined 1/e (units) * e =1A. Well this is always true (e/e=1) , but since the CODATA e is 1% of this proof have not been published , they would prove me right while other struggle with a impossible 1% error.
So a measurement as follows: A current source , a electron counter(SET) and ammeter in series. Im =measured amps ,nm =count during time tm=measured time. The measured elementary charge is em=Im*tm/nm . Simple but when e (1%)is wrong it do not give CODATA e . The ammeter is calibrated to CODATA and have a "inbuilt" e value et= e tuning , and I ad the true value ei . We get nm=Im*tm /ei *(et/ei) T he parenthesis to correct ammeter "adjustments" => nm=Im*tm*et/ei^2 We know et = CODATA e and from these 3 e's follows that ei^2=em*et I have proved my theory right since the measured em will be about 1.568845*10^-19C and (1.60217*10^-19*1.568845*10^-19)^(1/2)=1.585 *10^-19C
Every amperemeter in the world is adjusted to prove me right (they have other uses too