Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: Recover energy from gravity  (Read 71793 times)

Gabriele

  • Full Member
  • ***
  • Posts: 248
    • Formerelax
Re: Recover energy from gravity
« Reply #60 on: April 06, 2014, 07:21:27 PM »
interesting

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #61 on: April 07, 2014, 08:30:27 AM »
I simulated another case more stable, but here the sum of energy seems to be to 0.

Do you try to simulate ?

It's only when the system increase current and voltage very quickly that the energy recover is higher than energy giving.
« Last Edit: April 07, 2014, 12:54:46 PM by rc4 »

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #62 on: April 07, 2014, 05:53:23 PM »
First image show high voltage and high current for R1 and R2. Voltage of source is always 120 Vrms. So there is a big difference of energy. Look voltage at Oscilloscope not instantaneous probes !

Second image show how is the current and voltage at start when it is low, there are sinusoidal curves but with a high positive or negative value.

If frequency of source increase, the voltage/current at secondaries increase more slowly. Try at 20000 Hz or more. At 50000Hz, the voltage is only 1000 V after 0.5s.

This state not works with all values of L. In the contrary works with severals values of voltage/freq of source. Don't forget to respect K < sqrt(LxLy). You can change resistance too.

The thing very important is the coef K2, they must be close to 1, in this case L can be at any value, sure, the coef K must be lower than sqrt(LxLy). K1 must be nearto 1 but can be at 0.7, it's works. If K1 is too lower this effect don't works.

Someone can test with another software ? or in reality ?
« Last Edit: April 07, 2014, 11:43:24 PM by rc4 »

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #63 on: April 08, 2014, 06:25:50 AM »
First image: here the system give high voltage/current after 1 s. I show you only first 0.1 s. It's just for look at signals,for me angle phase of voltages and currents are like I think they must be in theory. This is why it's works. If return from K2 is lower the system can't loop (no enough current) and if K1 is lower the current from source is not enough to loop.

Second image: here K1 = 0.1, so the circuit is stable. In fact, flux create by right coil destroy flux in central coil, so no voltage at central coil.

In the first image, there is a positive slope for input current (blue) and positive current slope from right coil (red) each slope give -Ndphi/dt voltage to central coil, this is why central voltage coil increase more and more.

For good physics reality, I need to put L2 at 2L1 or 2 L3, but this don't change the result.
« Last Edit: April 08, 2014, 10:11:39 AM by rc4 »

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #64 on: April 08, 2014, 10:53:52 PM »
Even with no current in R1 and R2 and with low inductances like 0.001 H, the voltage increase very quickly. Software find like me Vr1 = V + Vr2 and not V - Vr2. Like Vr2 is a part of Vr1, this is a positive loop. It's the same with current.

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #65 on: April 09, 2014, 08:03:04 AM »
I simulated with Spiceopus, it's the same result:

Transformer

vV1 0 3 dc 0 ac 1 0
+      sin(0 {12*1.414213562} 500 0 0 0)

rR4 1 0 1e-05 vresR4 
.model vresR4 r(  )


rR3 5 0 1e-05 vresR3 
.model vresR3 r(  )


rR5 8 0 1e-05 vresR5 
.model vresR5 r(  )

K1 LL1 LL2 0.9
K2 LL2 LL3 0.9
K3 LL1 LL3 0.0001

rR2 0 7 6 vresR2 
.model vresR2 r(  )


rR1 0 2 9 vresR1 
.model vresR1 r(  )


lL3 7 8 1

lL2 2 5 1

lL1 3 1 1

.end

Sure, iron will saturate but even in transitory the sum of energy seems to be not at 0.

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #66 on: April 11, 2014, 07:08:37 AM »
nobody tested this circuit ?

Gabriele

  • Full Member
  • ***
  • Posts: 248
    • Formerelax
Re: Recover energy from gravity
« Reply #67 on: April 11, 2014, 09:24:46 AM »
I don't have tecnical instruments to try this circuit...why don't you try by yourself?

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #68 on: April 11, 2014, 09:46:29 AM »
I tried it but fuse from alimentation is destroyed each time, the current increase very quickly in the primary ! Why others people don't simulate and watch what's happened ?
« Last Edit: April 11, 2014, 05:13:57 PM by rc4 »

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #69 on: April 12, 2014, 09:15:40 PM »
1 image/ Another idea with capacitors, use N plate capacitors (or others shapes, cylinder for example). Charge in step 1 with plates in front of them. Discharge in resistor with plate far away, this increase energy (not the charge but the field), so the voltage.  Use plate - from N capacitor and plate + from N+1 capacitor for example. It's not a linear law and I think it's better to user small larger of plates. Or do the reverse STEP2->STEP1. Maybe discharge 1,3,5,7, etc. and after discharge 2,4,6,8,etc. like that first capacitors give same energy than alone but after the voltage increase.

Something must change when I use 2 differents plates, because the cycle could be:
s1/ charge capacitors
s2/ move one plate, need energy
s3/ discharge, if I discharge same energy at s1, where goes energy at s2 ?

2 image/ use N/N+2 capacitors

3 image/ charge all capacitors, and after increase red diameter, it's easierto increase because red surfaces repuls themselves.

c9.png: when capacitors move away, there are attractive forces (magenta) and repulsives forces (black). It's easier to separate capacitors than alone.
« Last Edit: April 13, 2014, 09:01:02 AM by rc4 »

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #70 on: April 13, 2014, 11:03:41 AM »
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

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #71 on: April 14, 2014, 08:29:57 PM »
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

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #72 on: April 16, 2014, 10:58:41 PM »
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

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #73 on: April 17, 2014, 10:24:28 AM »
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).
« Last Edit: April 17, 2014, 01:05:24 PM by rc4 »

rc4

  • Full Member
  • ***
  • Posts: 147
Re: Recover energy from gravity
« Reply #74 on: April 18, 2014, 03:18:59 PM »
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

« Last Edit: April 18, 2014, 05:20:52 PM by rc4 »