The ABSOLUTE N^0 orders the General Relativity. N^0 = E = m (C^2 kg) = m (9×10^16 m^2 s^-2) kg C is the speed of the electromagnetic VOLUME, in mass. We obtain it through this syllogism:
1) 10^3 × Log 10^3 m s^-1 = 3×10^3 m s^1 is the complex volume’s speed, as its advancing mass in kg. 2) m^0.5 = 10^5 is the TOTAL electric quantity in kg. 3) Therefore 10^5 × 3×10^3 m s^-1 = 3 × 10^8 m s^-1 is the electromagnetic volume’s speed as its advancing mass in kg. Exactly 3×10^8 m s^-1 kg. The calculus m^0.5 considers the electric advancement of the mass on the base of the magnetic one, and vice versa. We have these two opposite vectors, having the same direction and quantity, and opposite advancements: (+3 × 10^8 m s^-1) kg, mass electrically advancing, on the active base of: (-3 × 10^8 m s^-1) kg, mass magnetically advancing. The product of the interaction, between these two vectors, stops the advancement in the unity of the space-time squared m^2 s^-2, and the quantity of mass is contained in the transversal area of the electromagnetic flow, having one side electric, and a magnetic one. (9×10^16 m^2 s^-2) kg It is C^2, area of electromagnetic mass in kg, as the Energy of the flow, which, to can be activated, needs TIME. In fact E s = mass in movement. Romano Amodeo affirms that is a BIG MISTAKE to extrapolate the “c light speed” IN THE TIME, because E = m c^2 are “energies” also in “m” and “c^2”. The c speed isn’t a E “without time” but a E “having real time in action”. To can extrapolate the E of the light, the electric flow in speed must become the pure quantity of the volume’s mass. Therefore c = 299 792 458 m s^-1 (a line of flow in m) must multiply the unitary front big m^2 (300000000 : 299792458), in way to become 3×10^8 m^3. To can obtain the “mass” of these m^3 (that are “expansion” 10^3 of 1 kg), we have to transform 3×10^8 m^3 in kg, multiplying by 10^3, and are 3×10^11 kg of expansion 10^3. Afterwards, we have to reduce at 10^-3 its “mass”. The result is again 3 ×10^8, but now in kg of “mass” (as 10^-3 m^3 of H2O at 4° C). So acting, in the transversal plane m^2 s^-2, we have 9×10^16 kg, and we CAN have. in E= m c^2, the product between the mass <m> (in kg) and the <9×10^16 kg> of the area quantified in m^2 s^-2. Romano Amodeo affirms that the C speed ISN’T an “optional”, but a SIMPLE and OBLIGED DERIVATION by the first ASSUMPTION of m, kg and s, by the Earth mass. 10^10 dm was the length ¼, of the Earth Meridian as a full circle of 4×10^10 dm^3 of water at 4° C and so 4×10^10 kg. dm^3 (10^10 : 10^3), (those in 1 m^3) became ¼ of circle in m^3, having 1 m^2 of section, long 10^7 m, and having always the same 10^10 kg as ¼ of the Meridian made by cubes in line. In this way ¼ of the Earth Meridian was 10^7 m long, having 10^10 kg, and was the <TIME ¼> of all the Meridian. In the same time, ¾ of the Earth Meridian was 3×10^7 m long, having 3×10^10 kg, as the <SPACE ¾> of all the Meridian. This 1 mass has it most expansion in line becoming 10 in line, since all the SI units were referred to the 10 cycle of the unit. So 10 × 3×10^7 m = 3×10^8 m s^-1 is ALL the cycle 10 in 10 s, of 3×10^7 m s^-1. Therefore 3×10^8 m s^-1 is THE ONLY POSSIBLE “C speed”, as Volume’s flow. Its mass is of 3×10^8 kg since 3×10^10 kg have 10^2 as the unitary front, and kg 3×10^10 : 10^2 (that is the dm^2 in 1 m^2) are 3×10^8 kg in the line of the flow having 1 dm^2 of section. The area 3×10^8 m^2 s^-2 contains 9×10^16 m^2 s^2 in kg of mass. The E is only in kg, or in atomic weight. The ratio E/m is that existing in the WATER, because the Champion in Platinum Iridium is the weight of 1 dm^3 of H2O at 4° C. In the H2O molecule the H has 1 atomic weight, and there are two ones: 2 a.w. The bond energy – as mass – is (2 + 16) a.w. (being 16 a.w.the O mass). 18 a.w. / 2 a.w. = 9/1 pure number ! Romano Amodeo affirms that E/m is A PURE NUMBER !!! Romano Amodeo affirms that, since all the SI units have the 10 cycle, the only possible invariant PURE NUMBER is the number 9 !! --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Epistemology" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/epistemology?hl=en -~----------~----~----~----~------~----~------~--~---
