Consider this,
ln(cosh(G√2mc2aψc))=14
G√2mc2aψc=cosh−1(e14)
and,
ln(cosh(G√2mc2aψπ))=1
G√2mc2aψπ=cosh−1(e)
So,
aψπaψc=cosh−1(e)cosh−1(e14)
aψπaψc=2.24921
From the post "Sizing Them Up" dated 13 Dec 2014,
aψ=19.34nm
aψ=16.32nm
aψ=15.48nm
aψ=14.77nm
of all the particles proposed none of them pair up to give a ratio of 2.24921. The extreme pair,
19.3414.77=1.30941
Maybe aψc readily coalesce into big aψπ. Furthermore, if these are the positive particles and the last an electron, where are the other two conjugate negative particles; the negative temperature particle and the negative gravity particle. It could be that at higher temperature, the negative particles are drive off, which implies that at lower temperature, these particles will emit a faint spectra. Negative gravity particles may be more prominent under micro gravity; a new spectra line could be seen when the experiment is conducted in outer space. Under Earth's gravity, negative gravity particles will fall to the ground not affected by the electric or temperature gradients.
Still, the difference in aψ can be due the prevailing fields, temperature, electric and gravity, that the experiment is subjected to. The bigger the field the bigger the particle. Theoretically all particles with the same number of basic particles n, should be of the same aψπ.
But in consolation,
15.4814.77≈7874=1.05
which is just ψ going around in circle...
For that matter, as we go boldly maybe four experimental values of aψ from the post "Sizing Them Up" suggests the presence of another type of field and its associated particle pair. The last value for aψ=14.77 is not an electron, but a positive particle of a field relatively least in strength during the course of the experiment. The weakest field and so a particle of smallest aψ.
What is this possible field?