I have had some complaints (justifiable in my opinion) on how I derived the operating voltage for a given fusion reactor to within a few percent. I dont't take into account Einstein since the particle energies we are talking about for a proton are on the order of .6 Mev maximum and a proton "weighs" around 900 Mev. It might matter in the fifth place decimal (I estimate about a 1 part in 300,000 difference) but we don't really need to know the voltage that close since creation of the polywell will cost us around 10% of the drive voltage, indicating that any reasonable supply should be capable of about the calculated voltage +20% or maybe +25% to give a little more margin.
Well off to the races. First let me define the terms.
m1 = mass of the lowest mass particle amu (protons)
m2 = the other mass in amu
c1 = the charge of m1 in electron charge units
c2 = the charge of m2 in electron charge units
AV = accelerator voltage in KV
TV = the voltage from a chart or graph where one particle is stationary
V1 = the velocity of m1 when accelerated by AV
V2 = the velocity of m2 when accelerated by AV
V1tv = the velocity of m1 when accelerated by TV
V1 + V2 = V1tv
When deriving the equations the = sign from now on will mean exactly proportional to.
m1 * V12 = c1 * AV
m2 * V22 = c2 * AV
m1 * V1tv2 = c1 * TV
Since we are using Newtonian mechanics velocities add:
V12 = c1 * AV / m1
V1 = √( c1 * AV / m1 )
V22 = c2 * AV / m2
V2 = √( c2 * AV / m2 )
V1tv = √( c1 * TV / m1 )
I'm going to skip a few obvious steps here for brevity.
√( c1 * TV / m1 ) =
√ AV * ( √( c1 / m1 ) + √( c2 / m2 ))
√ AV = (√( c1 * TV / m1 )) / ( √( c1 / m1 ) + √( c2 / m2 ))
Which is in a spread sheet availabe at the IEC Fusion Newsgroup as an attachment.