Perhaps so, but he kept saying he thinks he has fusion, so I felt he could become more definitive with the measurement of helium and  performance of mass balance analysis. It was just a suggestion

-------- Original Message --------
Subject: Re: [All-E] Re: micro ac tukomak
From: Prune <prune.etna@ntlworld.com>
Date: Mon, February 09, 2009 10:16 am
To: All-Energy@yahoogroups.com

Ionisation is not really relevant to the fusion reaction. Stripping away the electron requires only a fraction of the energy needed to force the nuclei together.

I'm not sure how he would measure helium, unless he has access to a mass spectrometer. I reckon that his proposal to measure energy output is a better bet.

editor@gbanalysts.com wrote:

Hello Chris,

I believe that by fusion, you mean that the hydrogen is ionizing and then fusing into helium such as occurs in the Sun - correct me if I am wrong. Can you therefore carry out a quantitative analysis of the helium gas you produce at the end of the each run?

EditorGbanalysts

-------- Original Message --------
Subject: [All-E] Re: micro ac tukomak
From: "chris64strev" <cjrs@chrisspages.co.uk>
Date: Sun, February 08, 2009 3:36 pm
To: All-Energy@yahoogroups.com

I think I may have fusion in the tube after all, here is my
reasoning based on circuit theory and the measurents I made.

In fact the ionised gas shows a series resistance that decreases the
dynamic impedance that creates a small mismatch. If the gas conducts
better this reduces the extra series resistance and will increase
the dynamic resistance, making this smaller or zero will not change
the dynamic resistance as the effective series resistance approaches
its value that it was without the tube. The only way the dynamic
resistance can increase is if the effective series resistance
decreases and this can only occur if the plasma resistance goes
negative meaning that power is being produced and this can only come
from fusion in the gas.

The mismatch was a VSWR of about 1.2 with Forward Power=100 and
reverse power = 2
so RP/FP= 0.2
so reflection coefficient = 0.2 = (ZL-ZS)/(ZL+ZS) = 0.2= (50-Z)/
(50+Z)

0.2 *50+ 0.2 * Z = 50-Z so 10-50 = -Z-0.2*Z 
so -40=-Z(1.2) so Z = 33.3 ohms.
L/Cr=RD = 11250 and RD = 7492.5
L=2.5 x 10^-6 C=7.39x10^-10
2.5E-6/r=11250 = 3383/r r=11250/3383=3.3 ohms
r with 2 watt reverse power =  7492.5/3383=2.2 ohms
so the plasma showed a negative resistance of 1.1 ohms.
Since Power = I^2*R then the extra current this power of 2 watt
generated by the 1.1 ohm was 1.3 amp.

Chris.