# 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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2. ### I'm stubborn

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1. Now explain with no mathematics in your explanation .

3. ### TIMO MOILANENRegistered Member

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Since ammaters are tuned to SI "standard" (assuming e=1.602*10^-19) their tuning is over compensated compared to ei=1.585*10-19. Shortly they are set to show 1A at an expected count 6.2415*10^18 . A "electron counting chip" will count the real n=6.3748 *10^18 . The electron "packages" are too small to give SI reading and it is also expected a smaller amount n to do, twice compensated bigger A-reading with smaller charges. Metric ampere has less than SI A "juice" and the real charges ei are smaller. So only one need to "adjusted" to get metric n=(6.3748*10^-19/6.2415*10^-19)*(1/2) =6.2746*10^-19 counts , ei=n^-1 =1.5853*10^-19.
The SI ammeters are tuned 1 notch to compensate for a smaller than expected elementary charge , and equally much (second notch)to up amperage to SI. Since metric ampere is smaller (less juice) and elementary charge is what it is, only the scale need adjustment (one notch) to show metric Ai . (x1*x2)^0.5 is exact compared to (x1+x2)/2 this time. The electron counter-chip tell how many charges and have not been used (nor invented ) when SI ampere was "set". And again since V*A=W is accurate for any pair of units U=RI , P=I*U there has been no technical problem . When using amperes and coulombs there is a 1% difference to metric units , too small to cause technical issues.

5. ### HipparchiaRegistered Senior Member

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You need to fix your quote function.

My ill mannered story is that I comment on what appears like gibberish when I see it. I recognise that there are three possible explanations for something seeming like gibberish:
• It is gibberish
• It is way over my head
• It is not gibberish, but has been presented in a chaotic and confusing manner
I have read a lot of material that is over my head that I am still able to recognise is not gibberish. Your posts don't fit that category.

The most hopeful explanation is the third one. If correct, you need to revist your presentation style.

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Gibberish.

8. ### TIMO MOILANENRegistered Member

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I am sorry for my substandard presentation , sadly I can probably not improve it, and I will not try to explain math without math in future. I have a simpler version with mass of a proton separated .
Version 17.10-20
Using CODATA mass for proton Mp=1.67262E-27 From integral and E=m*c^2 =>
Coulombs Ki=4/(3*Mp*c^2) = 8869508257 Nm^2/Ci^2
Electric c. E0i=3/(16pi)*Mp*c^2 = 8.97202745E-12 Fi/m
Magnetic c. µ0i=16pi/(3*Mp*c^4) = 1.2401322E-6 N/Ai^2
Making the metric: Coulomb F=Ki*Q^2/r^2 Faraday e=1/(Q*NA) =>
Metric elementary charge ei=1/((F*r^2/Ki)^0.5*NA) , putting in F=1N, r=1m and NA=6.02214067*10^23
ei=1.5638636E-19 Ci
Fine structure constant I use ai=2*(3/16pi)^2 =0.0071241
From formula ei^2=2*ai*E0i*hi*c I get hi =6.3815129 Js
Finally the metric ampere Ai =6.394412E18/s*1.5638636E-19 Ci or 1Ai*ei/e=0.9760869 (SI)A
Without measurement (count of elementary charges /(A s) these results are best I can accomplish.
Other indifferent (metric ) constants accordingly.
Timo Moilanen 17.10-20

9. ### TIMO MOILANENRegistered Member

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One example from mainstream officially CODATA proven math. From coulomb's and Faraday's formulas comes that:
e=1/((F*r^2/K)^0.5*NA) =1/((1N*1m^2/898755179Nm^2/C^2)^0.5*6.02214076*10^23mol^-1) =1.574235858*10^-19 C
CODATA e=1.602176634*10^-19C. delta 1.7%
I may be the worst writer on cite but I can calculate like anyone and do math compared to anyone. Please try follow my "add up" thoughts. There are loads of mainstream scientists approving wider error margins. I can not write math without math, so I don't.

10. ### exchemistValued Senior Member

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If you had any idea what you were doing, you would at least define what physical quantities these various terms: e (do you mean the charge on the electron?), F, r, K, N, A and C (do you mean c, the speed of light?), are supposed to represent and you would explain what your calculation is supposed to be doing. Also, you would not be quoting numbers to unnecessary numbers of significant figures that add nothing to the point you are trying to make.

Without any of that, it is, well, gibberish.

If you are not schizophrenic, my next best guess would be that you are on the autistic spectrum, like Reiku. Whatever your problem is, nobody will be able to follow what you write if you make no attempt to explain what you are doing.

Hipparchia likes this.
11. ### billvonValued Senior Member

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So can handheld calculators.

Doing math and posting it isn't all that insightful or interesting. For example, I could post things like 2.337438e-12 j / 7.43839e6 m3 = 3.14239775005e-19 j/m3 all day - but that means nothing. It's just random math.

12. ### DaveC426913Valued Senior Member

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This is worth repeating.

If the OP is serious about his/her proposal, there are places s/he could submit it. Rather than a place to get his ideas reviewed, the OP might be better off using SciFo to hone his communications skills to make a proposal that has a chance of getting looked at.

13. ### TIMO MOILANENRegistered Member

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The "constants" do not ad up , and never will with todays ampere. By renewing the amp. to fit (m,s,kg) SI units constants not only fit , they get predicted values from multiples of pi and speed of light.

14. ### exchemistValued Senior Member

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So why not explain, in words, how you propose to calculate the ampere from various physical constants. Then we can all see what you are doing and we may be able to comment on it. Start by listing the constants you use in the calculation and give the symbol you intend to use for each.

15. ### TIMO MOILANENRegistered Member

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I have no intention nor time to hone any "skills" . I hope I manage getting a note applied here

16. ### DaveC426913Valued Senior Member

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You can't post pictures from sites that are behind account walls such as Facebook.
You see that pic but everyone else sees a busted link.​

17. ### TIMO MOILANENRegistered Member

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The constants compared to todays ampere definition are ei=elementary charge and 1/ei . But this do not tell anything about the SI ampere nor Ai . as products of both pairs is 1.000 Cs Easiest I can do is comparing to old definition . Two infinite long "wires" 1m apart with 1 amp current in each induces a force of 2*10^-7N between them . "My" Ai induces 1.987172*10^-7N and compared to that Ai is 0.9967877 A SI . Then all electromagnetic constants get set values and the "scientists " can finally go do something productive (after repeating themselves in vane for up to 200 years)
Got it

?

18. ### exchemistValued Senior Member

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Yes, I understand the modern definition is that an ampere is a charge of 1 Coulomb flowing for 1 second, the Coulomb being defined in terms of the charge on the electron e.

And the old definition was a current producing a force of 2 x 10⁻⁷ N, in a pair of infinitely long parallel wires. Agreed.

But why do you think there is anything wrong with either of these? Please answer in words.

Last edited: Oct 21, 2020
19. ### billvonValued Senior Member

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Joules and meters cubed are indeed SI units in the MKS system.
Scientists (and engineers) have not been hampered by not using your units. They make some pretty good motors using today's units.

20. ### TIMO MOILANENRegistered Member

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They do not ad up mathematically . That for words.
Ex . Coulomb and Faraday (school math) give e=1/(1N*1m^2/k)^0.5'NA) .CODATA coulomb const.=8.9875518*10^9[xxxunits] but the eq. give e=1.574235*10^-19C and all "low end" constants are directly coupled to k and "highend to e. The tower topples . The only things on proving the definition directly is excuses on small currencies or slow counting chips . A chip of 4 GHz speed Would give a 1µA current . By my understanding the speed is modest and current measurable (if you insist finishing the experiment in 1s). My guess the results are discarded as impossible or at best inconvenient. By the way all constants need to work in all equations at the same time . That is a lot of math. but the result surprisingly simple. In the end I have changed the constants quite little (average 0.1%) by adjusting amperage by 0.3%

21. ### TIMO MOILANENRegistered Member

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How can I send a pic. directly from computer ? Help me !

22. ### TIMO MOILANENRegistered Member

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The ampere as it is define define voltage and V*A=W so power and energy ad up better than you can compare them to thermodynamics and could cause well under a percent error in traditional electric components .I'm sure motors can be engineered mush more "exact" than the result (power) or supply voltage can be measure/provided. Scientists have adjusted the constants for decades declaring some officially "right" till they next decade change it again. I'm also sure a 0.3 % adjust to ampere could not hamper anything , and neither is not needed. In science they have no "multimeters" only extreme precision so the ampere is always a calculation anyway.

23. ### DaveC426913Valued Senior Member

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These two posts indicate some counter-productivity in your intentions. You are more interested in posting a Facebook picture than you are in being able to communicate with your audience.