TIMO MOILANEN
Registered Member
So you did not see the advantages of a different ampere (my output ,my fault). Physics and math can be challenging , waiting for other's opinions is content for many people.
A metric current and sequel
- Thread starter TIMO MOILANEN
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Nope. It's just a number. It's like changing the inch to be exactly 25mm. Would that have huge advantages? No. It would have some minor advantages; easier length conversions. But not worth it.
TIMO MOILANEN
Registered Member
There is still an error of(+-) 0.13 to 0.09% because of the not exact ampere. One thing that is indisputable with current amps. is that since force of charge is Q*E = (Q1*Q2/r^2*16pi/3*F(Mp))^0.5*Q1/r^2 16pi/3*F(Mp), it give that Q*E =(16pi/3)^1.5*F(Q1,Q2,Mp). This mean that fine structure coefficient "for now" is exact (alpha)=(3/(16*pi))^1.5/2= 0.007290328592=1/137.16803
I'm still working on this. There is a couple of places to put in matching metric amps. Sqrt(3/4)*SI for ex. but then all values would be unrecognizable (and a very different amp)
TIMO MOILANEN
Registered Member
Since the my whole theory is hard to discus "comprehend", I will continue work on it (till I have "nonmathematical" arguments, or maybe even use an other approach to knit amp to (m.s,kg).
One "achievement" however came out of the process. The geometry of Coulomb const.k is 4/3 *1/(F(Mp,r))^2 and combined with definition of electric const. = 1/(4pi *k) give that Q is proportional (16pi/3) compared to electric const. And electric field being similar give E =1/16pi/3*F(Mp,r^2) . This give e^2*(16pi/3)^1.5 for energy of e^2 per e compared to product of constants. And this is the (geometrical) cause and numerical value of fine structure coefficient. (alpha)=(3/16pi)^1.5/2=0.00729033 =1/137.168
Till later
Timo Moilanen
Have you caught up with the redefinition of the SI system that occurred in 2019?
1 Ampere is now a derived unit, equal to the passage of an exact number of elementary charges past a point in one second. The fundamental units on which it is based are the charge on the electron (which now has a defined value), and the second (whose value is defined with reference to a certain atomic transition in caesium atoms).
As long as the definition is used consistently, it cannot be anything other than "compatible" with the rest of the SI system.
TIMO MOILANEN
Registered Member
They have never been able to use this definition in a practical "experiment", and according to my calculations they have to "improvise" noticeably to make numbers fit. Maybe this is the problem of getting the electron count done. As I understand the technology is there (SET-transistors). The method I use to calc. these is dependent on mass of the proton, and I would use a even more exaggerated mass than new measures yield. Therefor I have derived a calculation of Avogadro's number from the Cavendish experiment. The "last" quantification of this constant is done ,I have read. Well, it is at least the costliest, I think. Here the link to exel. calc. sheet. https://1drv.ms/x/s!ArSE2R4ReZrzijhQRlzAbbuZla27?e=RGSOAx
I would be forever thankful for getting genuine experiment data. (G,R1,R2,r)
TIMO MOILANEN
Registered Member
Ones more calculation of Avogadro's number from experiment for G (universal gravitational constant)
https://1drv.ms/x/s!ArSE2R4ReZrzijhQRlzAbbuZla27?e=a6oVCf
https://1drv.ms/x/s!ArSE2R4ReZrzijhQRlzAbbuZla27?e=a6oVCf
Now put All of your Thinking into words in your Native Tongue .
Then have it translated into English .
TIMO MOILANEN
Registered Member
Logic: ex. 1dz*1dz =12dz not 1sqrdz , 6pairs*6pairs =72pairs and 1kg*1kg=NA*1kg
This is not physics, 1 proton/neutron just happens to be about the "size" of the smallest factor in any weight unit. Since G=F*r^2/(M1*M2) ,it contain kg*kg. Changing that and "dividing" with c^2 to get rid of acceleration. The 1/2 is half of total potential in gravity (=work) and 1000g/1kg. What comes to the k values, 1/k is a volumetric efficiency (for a sphere using this formula), calculated from a ratio of two 3-D integrals over a sphere. Somewhat challenging math. but ads up perfectly in gravitation calcs. so why not for Avogadro's number , to get both even better specified, since I'm not sure Avogadro's number is very exact. Well I have quite high criteria.
The Cavendish type experiments are done for hundreds of years and it is almost impossible to find relevant data. I simply ask for help
https://1drv.ms/x/s!ArSE2R4ReZrzijhQRlzAbbuZla27?e=nUIF1z
Last edited:
Logic: ex. 1dz*1dz =12dz not 1sqrdz , 6pairs*6pairs =72pairs and 1kg*1kg=NA*1kg
This is not physics, 1 proton/neutron just happens to be about the "size" of the smallest factor in any weight unit. Since G=F*r^2/(M1*M2) ,it contain kg*kg. Changing that and "dividing" with c^2 to get rid of acceleration. The 1/2 is half of total potential in gravity (=work) and 1000g/1kg. What comes to the k values, 1/k is a volumetric efficiency (for a sphere using this formula), calculated from a ratio of two 3-D integrals over a sphere. Somewhat challenging math. but ads up perfectly in gravitation calcs. so why not for Avogadro's number , to get both even better specified, since I'm not sure Avogadro's number is very exact. Well I have quite high criteria.
The Cavendish type experiments are done for hundreds of years and it is almost impossible to find relevant data. I simply ask for help
In words TIMO .
TIMO MOILANEN
Registered Member
NA=c^2/(2*1000*G*k1*k2) , NA Avogadro's number, c speed of light, G measured gravitation constant in each separate experiment, averages can be any combination (will not do). Any unit in square is Unit*Unit/smallest unit. For kg this make X[kg]*Y[kg]=X*Y/(1000*NA/kg). This "take care of" 1kg unit and the other kg (N) divided by c^2 leaves n kg. From F=G*m1*m2/r^2 <=> G=F*r^2/(m1*m2) the real kg:s(the numbers) are already divided away.
Divided 2 is as business as usual work =pot energy/2 or an integral 1/2 "it just appear ".And 1000g/1kg since mol is in grams (new def. thousands of kg's 0.012kg). k1 and k2 have no units and are calculated from distance /radius (same units) =p. The parameter for "volumetric efficiency or reduced volume 1/k" k=3/2*p^2-3/4*(p^3-p)*ln((p+1)/(p-1)) is integral over a sphere dV/s^2 dx,dy,dz divided by dVcosb/s^2dx,dy,dz , the total V/r^2 divided by the vector of V/r^2 . The later is how mainstream calc. gravity and try to prove shell theory (there is no such thing) and the former total gravitational pressure ( Einstein equality), should be used for ex. for emission of photons in gravity field (mainstream have no idea).
The second one is officially impossible to calc. I did it 4 years ago along whit the first that is probably mainstream useless. Probably not much better
https://1drv.ms/x/s!ArSE2R4ReZrzijhQRlzAbbuZla27?e=Zk4wmVNA=c^2/(2*1000*G*k1*k2) , NA Avogadro's number, c speed of light, G measured gravitation constant in each separate experiment, averages can be any combination (will not do). Any unit in square is Unit*Unit/smallest unit. For kg this make X[kg]*Y[kg]=X*Y/(1000*NA/kg). This "take care of" 1kg unit and the other kg (N) divided by c^2 leaves n kg. From F=G*m1*m2/r^2 <=> G=F*r^2/(m1*m2) the real kg:s(the numbers) are already divided away.
Divided 2 is as business as usual work =pot energy/2 or an integral 1/2 "it just appear ".And 1000g/1kg since mol is in grams (new def. thousands of kg's 0.012kg). k1 and k2 have no units and are calculated from distance /radius (same units) =p. The parameter for "volumetric efficiency or reduced volume 1/k" k=3/2*p^2-3/4*(p^3-p)*ln((p+1)/(p-1)) is integral over a sphere dV/s^2 dx,dy,dz divided by dVcosb/s^2dx,dy,dz , the total V/r^2 divided by the vector of V/r^2 . The later is how mainstream calc. gravity and try to prove shell theory (there is no such thing) and the former total gravitational pressure ( Einstein equality), should be used for ex. for emission of photons in gravity field (mainstream have no idea).
The second one is officially impossible to calc. I did it 4 years ago along whit the first that is probably mainstream useless. Probably not much better
The Big Picture Timo .
TIMO MOILANEN
Registered Member
Gravitation constant do not exist on "distances", but get a value 7.462*10^-11 Nm2/kg2 at infinity. This is why they never will find a stable enough value. I would need Avogadro's number to pin down this transformation coefficient Ti=c^2/(2*1000*NA/kg) . G can be calculated via form dependent k values for even density bodies , but in the end G=Ti and the now vanishing G*dM*(1-cosb) (mainstream) must just be dealt with. This require quite new formulas for gravity calculations , but in the end satellites and planets stay up there even better than before. It's a pity they have to crash so many probes on Mars. I have tried to contact NASA with a "gravity/satellite speed per altitude" chart. Everything is very different than assumed 200 years ago. I hope you're American and might help me connect to NASA (rather than east Asia) because there do not seem to be anything such as non-stable satellite orbits (they are just "not mainstream"). I have most in pencil only. By the way Xkg*Ykg is Xkg*Ykg/(1000*NA), just as 5dozen times 7dozen is 420 dozen (60*84/12=420/12). Times c^2 make the Newton (kgm/s2) , 1000 is because mol is in grams (kg/1000). Half the "potential" energy is g free fall energy (bodies are falling towards each other despite their speed and direction).
My big picture: I have a new gravitation theory (with math. even for galaxies), I need to upload safely ,soon (I'm over 60 and in quite bad health).https://1drv.ms/x/s!ArSE2R4ReZrzijUSNapsoEeJ0EV4?e=RmExXU
Your not ready yet Timo . When you are ready , you won't need mathematics to describe your understanding ; you get the picture ...
TIMO MOILANEN
Registered Member
The mainstream science is not ready for my gravity, though almost a consensus that there is too much missing.
I can only contribute by showing gravity is 12% stronger than believed, but on the other hand orbits need 27% more potential energy (mass) than assumed, and every formula "work".
About the ampere, only useful "news" I had is that in the formula 2*fine structure constant" is (3/16pi)^(3/2) and I'm not sure it is "the" constant, since it is only a geometry calculation (but fit perfectly to mainstream science). This mean it fit in and explain formula e^2=2a*E0*h*c , but it would be quite a coincidence if it fit beyond compensation for E0 (electric constant).
Is it like this ? The universal gravitational constant cant be determined accurately, I have a solution but it is mathematical, sorry. You probably could point out where I get gravitation wrong.
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Don't be sorry , but in the end , to be true , this , gravitational constant will need to be physical to have any consequence on anythings movements . Not some mathematical construct , only .
TIMO MOILANEN
Registered Member
It is physical and explain satellites and planets orbits better (more accurately) than Newtonian. But one can not hold on to the idea that we already know it all because this bring quite big changes to 18. century and modern assumptions. It is essentially a advancement from point mass. The two first methods I determined it was directly from Cavendish experiments as is done still today for G, with help of new geometric math (old math novel integrals). The new masses and "old" orbits explain a multitude of oddities (nonstable orbits,near earth orbits, moon difficulties and foremoust the crashes on Mars. This is why I try to get "info" to nasa .
Now that everybody is flying, we will see crashes because gravitation fields on Mars and Moon is very different from assumptions. Earth's satellite orbits are saved by the massive moon. The near satellites fly around the wobbling earth without seeming to take a shorter "rout", unlike what happens on Mars and moon.
TIMO MOILANEN
Registered Member
My theory of gravity and orbits do not depend on my constant, since a mass times any constant do the same product. My main difference in the theory is that the shell theorem do not work and free fall compared to orbit have different potential energies.It is physical and explain satellites and planets orbits better (more accurately) than Newtonian. But one can not hold on to the idea that we already know it all because this bring quite big changes to 18. century and modern assumptions. It is essentially a advancement from point mass. The two first methods I determined it was directly from Cavendish experiments as is done still today for G, with help of new geometric math (old math novel integrals). The new masses and "old" orbits explain a multitude of oddities (nonstable orbits,near earth orbits, moon difficulties and foremoust the crashes on Mars. This is why I try to get "info" to nasa .
Now that everybody is flying, we will see crashes because gravitation fields on Mars and Moon is very different from assumptions. Earth's satellite orbits are saved by the massive moon. The near satellites fly around the wobbling earth without seeming to take a shorter "rout", unlike what happens on Mars and moon.