You have to remember that my need for help becomes more and more obvious the further I go without it. I know there is a conventional wisdom or attitude that says we ATMers don’t listen, but if you go back through my threads you will find that I do listen. I am not asking for someone to jump on board believing in the ISU. I am asking for collaboration to get the best presentation of my ATM idea to you, and you are the ones that I am asking to help me do that. Without help, this is what you get (would you please offer to help? Send me email or PM if you want.):

Relationship #1 is the surface vs. the volume of the proton. Simple math comparing the formula for the surface of a sphere and the formula for the volume of a sphere with these formulas: Surface Area = 4 pi r2; Volume = 4/3 pi r3.

To determine the specific numbers we need the radius.

For relationship #2, we know the relationship between the mass of the electron and the mass of the proton, both consisting entirely of EEPs, albeit at drastically different densities of EEPs; 1:1,836.

So we solve for the radius that gives us the same relationship between the surface and the volume of the proton as we get between the mass of the electron and the mass of the proton. That way, since my idea is that the number of EEPs in the electron at rest is equal to the number of EEPs on the surface of the proton, and a proton is a sphere in the ISU, then I have the number of EEPs in an electron at rest when I have the number of EEPs on the surface of a proton.

I did this using the trial and error method, this website, and an Excel spreadsheet and about five minutes. It turns out that a radius of 5,508 units gives us the right relationship, i.e. 1:1,836.

The result of 5,508 occurs at a surface of 381,239,356 units, and a volume of 699,955,457,517 units, right at 1:1,836. The electron at rest equals the number of surface EEPs and so the electron contains 381,239,356 EEPs.

Some concern exists about using the average diameter of the EEP as a unit of measure since clearly the EEP at this average diameter has volume itself, so you may want to suggest a better way to go about this. Anyway, the two relationships working together should allow us to quantify the EEP even though the next round of calculations based on your input may change the numbers. Until then I will go with the first round numbers.

According to this site the diameter of a proton = 10^-15 m. The radius I obtained for the proton in average EEP diameters times two equals a diameter of 11,016 average EEP diameters. Dividing the diameter of the proton in meters by the number of average EEP diameters, it turns out that the average diameter of an EEP is 9.08 x 10^-20 meters.

((10^(-15)) m) / 11 016 = 9.07770516 × 10^-20 meters is the diameter of EEP.

The energy of an EEP is the electron ground state energy of -13.6 eV divided by 381,239,356 EEPs in the electron cloud at the ground state (equivalent to the quantum state n=1 in QM?), and the energy of an individual EEP in joules is that energy in eV times 1.6 × 10^–19.

(13.6 eV / 381,239,356) x (1.6 x 10^-19) = 5.71 x 10^-27 joules, the joules of one EEP.

Simple math, right?

Note that the EEP always has positive energy. The electric charge is the energy density differential between two energy environments. The negative charge assigned to an electron is an energy differential between the EEPs that make up the electron and the low energy density space surrounding the proton.

It would be understandable if you had questions about some of this. I will defend it (or cave! ).