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Cold fusion paper
Host Element Fusion
Unpublished Work
Copyright © 1990
Earl Laurence Lovings
1. Proton (1 Hydrogen 1) Energy: 938.3 Mev = 1.007825 amu
2. Neutron (1 Neutron 0) Energy: 939.6 Mev = 1.008665 amu
3. Deuterium (2 Hydrogen 1) = 2.014102 amu
4. [(1 Neutron 0) + (1 Hydrogen 1) - electron] =
(939.6 Mev + 938.3 Mev - .511 Mev) = 1877.389 Mev =
2.015447128 amu
5. [(1 Neutron 0) - (1 Hydrogen 1) + electron] =
(939.6 Mev - 938.3 Mev + .511 Mev) = 1.811 Mev =
1.9441763 x 10 - 03 amu
6. (105 Palladium 46) = 104.905064 amu
7. (103 Rhodium 45) = 102.905511 amu
The host element fusion experiment begins with a Palladium
electrode submersed in Deuterium. A energy source is supplied,
which enables the fusion process to begin. My theory on this
subject is explained below:
The Deuterium atoms are allowed inside the Palladium electrode due
to the electric field on the electrode. Once the Deuterium atoms
are inside, the Deuterium causes the Palladium to become unstable.
This is done by this process:
[ (105 Pd 46 - (1 Neutron 0 + 1 Hydrogen 1 - electron) +
(1 Neutron 0 - 1 Hydrogen 1 + electron)] or,
[ 104.905064 amu - 2.015447128 amu + 1.9441763 x 10-03 amu] =
102.8916 amu.
The closest element Palladium can try to become stable is
(103 Rhodium 45).
Take Palladium's new mass and subtract it with Rhodium's mass.
(103 Rhodium 45) - 102.8916 amu, or
102.905511 amu - 102.8916 amu = 1.394653 x 10-02 amu.
To find out how many electrons that is equivalent to:
(1.394653 x 10-02 amu x 931.5 Mev/amu)/(.511 Mev/electrons) =
25.42309 electrons
This is the amount of electrons required to be ionized to enable
host element fusion with Deuterium.
That is the first process of host element fusion.
The second process begins when the ionized electrons from the
palladium atom shields the deuterium atoms to allow host element
fusion.
The Equation:
Q = [(2 Hydrogen 1) + (2 Hydrogen 1) + (25.423 e) -
(1877.389 Mev) + (1.811 Mev) - (2 Hydrogen 1)] x 931.5 Mev or,
Q = [(2.014102 amu + 2.014102 amu + .01394653 amu - 2.015447 amu
+ 1.944176 x 10-03 amu - 2.014102 amu)] x 931.5 Mev
Q = 13.55 Mev
You must realize for this process to work for host element fusion,
you have to have a host element before Deuterium will fuse.
My equation also theoretically works for known Deuterium fusion
processes.
Known Equation:
1. [(2 Hydrogen 1) + (2 Hydrogen 1)] -> (3 Helium 2) +
(1 Neutron 0)] = or,
[(2.014102 amu + 2.014102 amu - 3.016030 amu - 1.008665 amu) x
(931.5 Mev/amu)] = Q = 3.27 Mev
My Equation:
Host Element = (3 Helium 2) + (1 Neutron 0)
[(3 Helium 2) + (1 Neutron 0) - 1877.389 Mev + 1.811 Mev] =
[3.016030 amu + 1.008665 amu - 2.015447 amu + 1.944176x10-03 amu]
= 2.011192 amu
The element whose mass is closest to the new unstable "element" is
(2 Hydrogen 1)
(2.014102 amu) - 2.011192 amu = 2.909899 x 10-03 amu excess mass
convert to electrons
(2.909899 x 10-03 amu x 931.5 Mev/amu) /(.511 Mev/electrons) =
5.304445 electrons
Now the fusion of Deuterium atoms
[(2 Hydrogen 1) + (2 Hydrogen 1) + 5.30444e - 1877.389 Mev
+ 1.811 Mev - (2 Hydrogen 1)] x 931.5 Mev =
[(2.014102 amu + 2.014102 amu + 2.909899x10-03 amu - 2.015447 amu
+ 1.944176x10-03 amu - 2.014102 amu)] x (931.5 Mev/amu)
Q = 3.27 Mev
Known Equation:
2. [(2 Hydrogen 1) + (2 Hydrogen 1) -> (4 Helium 2)] = or,
[(2.014102 amu + 2.014102 amu - 4.002603 amu)] x
(931.5 Mev/amu) = Q = 23.85 Mev
My Equation:
Host Element = (4 Helium 2)
[(4 Helium 2) - 1877.389 Mev + 1.811 Mev] =
[(4.002603 amu - 2.015447 amu + 1.944176x10-03 amu)] = 1.9891 amu
The element whose mass is closest to the new unstable "element" is
(2 Hydrogen 1)
(2.014102 amu) - 1.9891 amu = 2.500188 x 10-02 amu excess mass
convert to electrons
(2.500188 x 10-02 amu x 931.5 Mev/amu) /(.511 Mev/electrons) =
45.57585 electrons
Now the fusion of Deuterium atoms
[(2 Hydrogen 1) + (2 Hydrogen 1) + 45.58 electrons - 1877.389 Mev
+ 1.811 Mev - (2 Hydrogen 1)] x 931.5 Mev =
[(2.014102 amu + 2.014102 amu + 2.500188x10-02 amu - 2.015447 amu
+ 1.944176x10-03 amu - 2.014102 amu)] x (931.5 Mev/amu) =
Q = 23.85 Mev
Known Equation:
3. [(2 Hydrogen 1) + (2 Hydrogen 1) -> (3 Hydrogen 1) +
(1 Hydrogen 1) = or,
[(2.014102 amu + 2.014102 amu - 3.016050 amu - 1.007825 amu)] x
(931.5 Mev/amu) = Q = 4.03 Mev
My Equation:
Host Element = (3 Hydrogen 1) + (1 Hydrogen 1)
[(3 Hydrogen 1) + ( 1 Hydrogen 1) - 1877.389 Mev + 1.811 Mev] =
[3.016050 amu + 1.007825 amu - 2.015447 amu + 1.944176x10-03 amu]
= 2.010372 amu
The element whose mass is closest to the new unstable "element" is
(2 Hydrogen 1)
(2.014102 amu - 2.010372 amu) = 3.729582x10-03 amu excess mass
convert to electrons
(3.729582x10-03 amu x 931.5 Mev/amu)/(.511 Mev/electrons) =
6.798641 electrons
Now the fusion of Deuterium atoms
[(2 Hydrogen 1) + (2 Hydrogen 1) + 6.799 electrons - 1877.389 Mev
+ 1.811 Mev - (2 Hydrogen 1)] x 931.5 Mev =
[(2.014102 amu + 2.014102 amu + 3.7296x10-03 amu - 2.015447 amu +
1.944176x10-03 amu - 2.014102 amu)] x (931.5 Mev/amu) =
Q = 4.03 Mev
Known Equation:
4. [(1 Hydrogen 1) + (1 Hydrogen) -> (2 Hydrogen 1) +(electron)=
[(1.007825 amu + 1.007825 amu - 2 electrons - 2.014102 amu)]
x (931.5 Mev/amu) = Q = .42 Mev
My Equation:
Host Element = (2 Hydrogen 1)
[(2 Hydrogen 1) - 1877.389 Mev + 1.811 Mev] =
[(2.014102 amu - 2.015447128 amu + 1.9441763x10-03 amu)] =
5.990302x10-04 amu
The element whose mass is closest to the new unstable "element"
is (1 Hydrogen 1)
(1.007825 amu - 5.990302x10-03 amu) = 1.007226 amu excess mass
convert to neutrinos
(1.007226 amu x 931.5 Mev/amu) / (.42 Mev/neutrinos) =
2,233.884 neutrinos
Now the fusion of 1 Hydrogen 1 atoms
[(1 Hydrogen 1) + (1 Hydrogen 1) + 2,234 neutrinos - 2 electrons
- 1877.389 Mev + 1.811 Mev - (1 Hydrogen)] x 931.5 Mev/amu =
[(1.007825 amu + 1.007825 amu + 1.007226 amu - .001097 amu
-2.015447128 amu + 1.9441763x10-03 amu - 1.007825 amu)]
x 931.5 Mev/amu = Q = .42 Mev
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