Recover energy from gravity

[rc4]

With a cilynder capacitor, the external plate want to move outside, so it's possible to charge the capacitor, after, move away 4 external curved shapes and after voltage is greater. I tested with FEMM. I tried with +/-10V source voltage for charge. After I keep constant charge and move away 4 external curved shapes. I win 2V for a moving of +0.1 in each direction.

[rc4]

another ide, tested with Multisim



*## Multisim Instrument XWM3 ##*
XWM3 15 3 XXWM3_98073480

*## Multisim Instrument XWM2 ##*
XWM2 14 1 XXWM2_98073480

*## Multisim Instrument XWM1 ##*
XWM1 13 8 XXWM1_98073480

*## Multisim Component V1 ##*
vV1 13 0 dc 0 ac 1 0
+      distof1 0 0
+      distof2 0 0
+      sin(0 {120*1.414213562} 600 0 0 0)

*## Multisim Component R3 ##*
rR3 3 0 10 vresR3 
.model vresR3 r(  )

*## Multisim Component C2 ##*
cC2 2 15 1e-006

*## Multisim Component R2 ##*
rR2 1 0 1000 vresR2 
.model vresR2 r(  )

*## Multisim Component L3 ##*
lL3 10 14 0.001

*## Multisim Component T1 ##*
xT1 6 7 10 0 2 0 Tran_T1
.subckt Tran_T1 p1pos p1neg s1pos s1neg s2pos s2neg
***Primary coil 1
G1 p1pos p1neg value={-1/10*(5*I(Es1)+5*I(Es2))}

***Secondary coil 1
Es1 s1pos s1neg value={V(p1pos,p1neg)*5/10}

***Secondary coil 2
Es2 s2pos s2neg value={V(p1pos,p1neg)*5/10}

.ends

*## Multisim Component L2 ##*
lL2 0 7 0.001

*## Multisim Component C1 ##*
cC1 5 6 1e-006

*## Multisim Component R1 ##*
rR1 4 5 0 vresR1 
.model vresR1 r(  )

*## Multisim Component L1 ##*
lL1 8 4 0.001


.subckt XXWM1_98073480 3 4
Vamp 3 4 0
.ends


.subckt XXWM3_98073480 3 4
Vamp 3 4 0
.ends
.subckt XXWM2_98073480 3 4
Vamp 3 4 0
.ends

[rc4]

This circuit works with K1=K2=K3=0.9 or lower values. Works with Multisim and LTSpice. There is a differenceof energy 8 kJ in one second. I let the circuit turn 100 s and the energy is always 8kJ. I suppose the shape of the current = signe sign give energy near 0 at third secondary. The difference of energy is only in fist second, after current in third secondary goes to 0.

The code LTSpice:


vV3 14 9 dc 0 ac 1 0
+      sin(0 {90*1.414213562} 60 0 0 90)
vV2 12 6 dc 0 ac 1 0
+      sin(0 {90*1.414213562} 60 0 0 90)
rR5 12 0 1e-009
rR4 11 4 1
rR1 1 5 1
rR3 2 0 10
vV1 7 3 dc 0 ac 1 0
+      sin(0 {120*1.414213562} 60 0 0 0)
K1 LL1 LL2 0.5
K2 LL2 LL3 0.5
K3 LL1 LL3 0.5
rR2 10 9 1
lL3 13 0 1
lL2 5 0 1
lL1 0 8 1
.ends

[rc4]

Works only is K<1, it's ok with 0.9 or 0.5 not 1. Works if angle phase is not at 0, let it at 60° for example. At start, the energy inside selfs are 2*0.5*L*I^2 but circuit recover only 10 % of this energy.

I simplify the circuit:

vV3 14 9 dc 0 ac 1 0
+ sin(0 {90*1.414213562} 60 0 0 60)

vV2 0 6 dc 0 ac 1 0
+ sin(0 {90*1.414213562} 60 0 0 60)

rR4 11 4 1

K2 LL2 LL3 0.9

rR2 10 9 1

lL3 13 0 1

lL2 5 0 1


With :

vV3 14 9 dc 0 ac 1 0
+      sin(0 {90*1.414213562} 60 0 0 60)

vV2 0 6 dc 0 ac 1 0
+      sin(0 {90*1.414213562} 60 0 0 240)

The output energy is 2 times input (even take in account energy stock in selfs before start).

[rc4]

If sources of voltage are in 180° phase (flux at right destroy flux at left), this could say with a linkeage of 0.95, energy needed for have current (at start) in inductances could be near 0. Like sources don't use energy during 10 s, resistors recover energy from nothing.

Step1: disconnect inductance from circuit and put it to special circuit that give DC current inside inductances, no need energy
Step2: disconnect ciruit of step1 and connect to sinus alim (Circuit Source-R-L)
Step3: recover energy in first seconds, until there is current
Step4: goto step1