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The absolute electron mass in kg
           

So the energy-pound of the particle named electron is exactly 9 times the dimension 10^16(existing in the transversal area if 1 m^2 existing every s^2) x 10^16 (the same "charge" counted now how mass in the direction of the mass flow) gets the 10^32 of the 4th power of the 10^8 dimension of the light. These 10^32 are (m/s)^4 that, in its 1/4 as "the quantity of the unitary potential presence", are m/s.       All the fourth dimension of the dimension 10^8 of the electromagnetism is so 10^32 m/s, where the unitary mass in line is of 1 dm = m/10. So we must divide for 10 as m, as s. The "time" 1/10 of the second becomes the quantity 10^32 x 1/10, while m/10 gets 1 kg. So 9 x 10^31 electron energy masses are 1 kg.            
This confirms the 9 exact energies as an unit, also of the unitary electron particle, and this at dimension of 10^-31 kg.

We obtain the electron mass through this  numerical calculus, referred to 1 dm^3 of water equal to 1 kg:
900/100 x 10^-30 x 10^-1 kg  = 900/100  (the energies of 10^-2 a.u.m.) having 10^-30 x 10^-1 kg of mass.   
900/100 are the 10^3 particles of matter in 1 a.u.m. divided in the 10^-6 of the matter-antimatter complex.
When these 10^3 are referred to 1 a.u.m., we have 540 particles (how the unitary intensity of the Candle, 540 x 10^12 cycles/s) how the "real matter" 500 (as 1/2 of 10^3) +40 (the real cycle), summed to 500 (the antimatter) -40 (that negative reality of the negative time). In this reference to the a.u.m., the number of the mass precises the 10^-3 unit of the positive volume containing 10^3 units how the 540/10^6 complexes particle of 1 a.u.m., that is 0,000540 a.u.m.

But here we have to refer to the a.u.m. that contains 10^6 complex particles, expanded in the 6 coordinate lines +x +y +z -x -y -z, while the kg gravitation happens in only one. So 0,000540, divided by the 6 directions, is 0,00009 in one, while the 10^2 of its section are 0,009. 
In this way we have the 9 "third minutes" corresponding to the time of 54/100 of second minutes that goes on simultaneously in the 6 positive-negative directions of the complex space-time-mass. Where the time advances in 6 directions in the 0,54 s of the time minimum of the Sidereal Year (365,25 days, 9 prime, 9 second and 0,54 second), the "third time" is 9 too, because 9 prime and 9 second (with a dimension 6x10 of unitary relative difference) already both respect the expansion 6 in the cycle 10. On the contrary, 0,54 second minutes, to respect the same cycle and to count in "third minutes", must divide itself by 6/100, referring the 6 quantities +0,09x +0,09y +0,09z -0,09x -0,09y -0,09z = 54/100, to the total 100 "red colored" quantities of the area 10x10 of the flow. So 54/6 x 10^-2 x 10^2 = 9 "third units" = "third minutes", and 10^-3 = 0,001 how 0,000999 times 0,0000001  is presented in 9 "prime, second and third" minutes... in which 0,001 (the 10^-3 unit of the mass 10^3) is all divided in third minutes how 1  +9 +90 +900 units 10^-3 equal to 10^3 third minutes.

The same happens passing from the 0,00054 a.u.m, to the 0,009 energies of the cycle of ten 0,0009.  
A ideal balance measures as 100 units of pound this mass. The other 900/1.000 of the 1.000 thousandth masses of the electron are green colored and are positioned on the first red level of the mass, and add all how "pound added in gravitational energy". In fact the ideal balance doesn't touch this 900 cubes, but accuses its gravitational energy added to the only 100 known by the balance, because, caused by this energy added in pound, the balance measures all the 1.000/1.000 of the particle 10^-1 of the u.a.m.
So, when in the electron mass, we put in 900/100 of energy, we consider only the green colored quantity of the movement of the red presence of the masses. In fact this 100 red colored masses are in gravitational free fall. The situation is this other, in which I show the 100 masses in the high position.
This free fall is acted by the electron masses always collocated at distance 10, how a body in free circular orbit that is in eternal fall but is always collocated to the same distance (1/10 of the a.u.m. cube) because the tangential escape is always equal to the simultaneous fall, and the result of this eternal fall is the curvature of the tangent line in a circular form indicating the free eternal fall. Every turn of the orbit is how the unitary cycle of this fall that we have divided in 10 units. So the electron (red) mass, that is always at 1/10 of the 10 distance of the unitary side of the cubed u.a.m., in its real movement is only the green part that doesn't appear, but is.
The division 900/100 = 9 reduces to one alone the 100 masses of the red transversal plane, and the situation is this sequent, that represents a pure line, because dividing a volume by its transversal plane we have the only vertical line of the ideal fall of one only particle of the u.a.m. equal in space at 10^-6 u.a.m. cube, having its side 1 equal to 10 decimals.

I here show the green 9 decimals of the energy of the electron eternal fall, since it is rotating around its nucleus.
We have these 9 decimal in space equal to 9 unitary energies, and use the space to count the energy, because 1 energy is the movement of 1 decimal of the space 10 of all the line 1 of the side of the u.a.m. cube.
Using the space to count the energy of the movement, and using the same space to count the stopped position 1/10 of the unit of the u.a.m. cube, this space is a pure line of 9/10 of the u.a.m. side, where the unitary mass in line is collocated exactly at 10 decimals of distance, and it is "pound" expressed in kg, the unit of the pound-mass.

To calculate the quantity in kg, we must consider 10^10 u.a.m. collocated one upon the other, in way that the distance was in space dimension the same of 1 dm, the space reference of the kg cube.
10 u.a.m. cubes in line form a length of 1 Å (Angstrom, unit of the atomic space) and 10^10 Å are 1 m. So the 1/10 of the Å is the space reference of the u.a.m. and the same 1/10 of the m is the space reference of the kg cube.
So it's evident that 10^10 a.u.m. cube in full linear sequence are the 1 dm of the kg cube.

This kg cube is a volume whose calculus is made in line: 1 dm x 1 dm x 1 dm.
If we put 10^10 u.a.m. sides, at the place of 1 dm we precise the same length of 1 dm in the a.u.m. sides corresponding.
So the dm^3 (equal to 1 kg when is 1 dm^3 of H2O) is 10^10 x 10^10 x 10/10 = 10^30 a.u.m. sides of 10/10 = 1 kg.
To precise the electron quality of 9 energies of the particle that is 1/10 in line of 10/10 of a.u.m., we have to introduce the 9 energies ad the 10 positions of the side of the 10^3 cube containing the 1.000 particles of the a.u.m., that are 10 in line.
In this way, 10^30 u.a.m. in line, referred to the 10 lines (energy + mass) of the electron eternal gravitational fall, becomes the quantity of 9 energies of 10^30 x 10 electrons (1 red + 9 green) = 1 kg.    If 10^31 electrons are equal to 1 kg, 10^-31 kg is 1 electron having 9 energies. So - since the energy and the mass is the same, because it is the energy of the mass 1, we don't precise the quality of the 9, and count 9 masses in movement, only the green colored. and finally 9 x 10^-31 kg is the energy-mass of 1 electron.

WE CAN ALSO REASON IN THIS OTHER WAY, to have exactly 9 times 10^31 kg.
The mass 1 is the cycle 10 that has lost 9 unitary spaces, to can be amassed in 1 only space. 
So 1 mass, in space, is worthed a -9 quantity of space, lost in the amassing. 
When we measure this -9 of space we use 0,109389754 quantities kept away (so subtracted) from -9, so they are negative ones, are -0,109. The practical result of this subtraction to -9, a negative quantity is: -9 -0,109 = -9,109. 
This mass, for us that consider positive the space, is negative one because it is an amassing, but we, for the 3th Dynamic principle "action-reaction", perceive +9,109 (as reaction) the action -9,109 of the unit mass counted. In fact we measure the mass through the opposition to the same mass going down, really acted in up by a balance.  
On the contrary the expansion is positive, is +9 times 1 and the subtraction of the TIE to c subtracts a quantity to the positive 9, that is reduced to 8,98755...

How is it possible to unify this electron mass 9,109389754 with its 8,98755... expansion, when in a case the tie is added and in the other it is subtracted, in way that afterwards the numbers are different ones?

The lack in UNIFICATION is evident one!

E (energy in power) can be equal to two opposite -9 and +9 that summed annihilate these two opposite works of the amassing and of the expanding, or can present the Expansion as 9/1 (a 1 expansed 9 times) and the amassing as the inverse 1/9 (of the 9 amassed in 1). In this way mc^2 is in number 1/9 x 9/1 = 1 E.
The two oppositions -9 and +9 and 9 and 1/9, are the same thing.
10^+9 and 10^-9 is the absolute case of the index +9 and -9, and are the two inverse 1.000.000.000/1 and 1/1.000.000.000.

Why today the electron appears to have the mass of 9,109389754 x10^-31 kg? This happens how a consequence of our unitary measurement, that put in unitary geometrical TIES that are extraneous to the quantity measured. Here are all this TIES at every dimension referred to the unitary meter:

  1. 10^-1 is the unit of 10^1. It is the action 1/10 that shows the reaction 10/1 of the unitary cycle of the space. The tie is 1/10, so it is 0,1.
  2. 10^-2 is the unit of 10^2. It is the transversal area xy = 10x10, of the flow of the mass that flows only in z direction. So we ave  the tie 0 at dimension 10^-2, that is 0,00.
  3. 10^-3 is the unit of 10^3. The time 1/10 is entered in joke and the energy 3x3 of the transversal area xy, now can work in direction z, as the tie 9 that, at the dimension 10^-3 is worthed  0,009.
  4. 10^-4 is the unit of 10^4. The volume 10^3 is in full real movement and shows the tie 3 on the base of its numerical cycle. At the dimension 10^-4 the tie 3 is  0,0003.
  5. 10^-5 is the unit of 10^5. This, referred to all the 10^10 Angstrom that form 1 m, is the dimension (10^10)^1/2 of the electric flow. 10^3 x 10^5 counts all the masses electro-magnetics, and are 10^8, whose tie is the 8 on the base of its numerical cycle 10. At the dimension 10^-5 the tie 8 is 0,00008.
  6. 10^-6 is the unit of 10^6 (all the complex space 10^3 x 10^3). The unit of the mass volume 10^3, multiplied by all the line 10^6, gets the tie 10^9, of the 9 on its decimal base. At the dimension 10^-6 the tie 9 is 0,000009.
  7. 10^-7 is the unit of 10^7. All the space in 10^10, divided bu the 10^3 unitary masses get the number 10^7, whose tie is the 7 on the base of its numerical cycle 10. At the dimension 10^-7 the tie 7 is  0,0000007.
  8. 10^-8 is the unit of 10^8, that is all the length electromagnetic having the absolute area 10^2 in 10^10. This precise calculus  10^8 : 10^3 = 10^5 quantifies all the masses electrical of the electron. 5 is the tie on the base of its numerical cycle 10. At the dimension 10^-8 the tie 5 is 0,00000005.
  9. 10^-9 is the unit of 10^9,  =that is all the space runned by the cycle 10, in 10^10 (the usual Angstroms = 1 m). The calculus  10^9 : 10^5 = 10^4, numbers the electrical masses. Th tie is 4 on the base 10 of its numerical cycle. At the dimension 10^-9 the tie 4 is 0,000000004.
The sum of all the ties ad every dimension is:
0,1
0,00
0,009
0,0003
0,00008
0,000009
0,0000007
0,00000005
0,000000004
==========
0,109389754

The Physicians cannot have more numbers than 9,109389754, because by the cycle 10 can be obtained till 10 decimal numbers. A eleventh decimal number exists, but "isn't yet present" for the reasons told by Planck.