The fixed grey tube increase height of water, so increase pressure at bottom. External pressure don't change lateral force because grey tube is at right and at left. Inside container, at bottom there is more right force than left but this depend of the diameter of the tube, like tube has more volume inside water, the height of water is greater. Weight=f(h) but Fr-Fl=f(h'). Grey shape is a helix, the length of helix is greater than a triangle so the volume is greater too and the weight too. In the contrary, the right force is the same than a triangle.
Look at h2.png image, s1 and s2 sections are the same and the volume is the same too, so the same weight for container. But with a helix shape, the section is the same than the triangle but the volume is higher, it depend of 2pi. The limit is 0.63 (4R/6piR). Helix don't turn, don't move, it is there only for adding volume inside container, like that the weight is higher and the right force is always equal to the triangle, so the container give more energy than its own weight.
I compute the works for additional weight and F1-Fr force. With:
N: number of turns
S: section of helix (the section of the cable)
a: angle of axis of helix
P: pitch of helix
g: acceleration of gravity
rho: density
R: radius of helix
Or I'am stupid a lot,or you are a fucking GENIUS!
You have to make the container turn to gain energy... :)
Seems genial...too genial....
the container must be like that for have angle constant
why turn ?
Nope
Can be?
P.S: Are you surethe fall of the container doesn't balance energetically the energy gained from volume rised from revolving?
the volume is always the same with last image (parallelogram)
If the helix is in a parallelogram, I compute the volume less forces Fr-Fl but the weight come from the full volume of helix inside water, but there is less surface due to the helix, so the weight is lower than in the first parameter. The total is 0, but if the first parameter is not exactly that, the sum is not 0. In my calculation, I don't compute f1 and f2 forces of the weight, I consider the full weight, but it is less, it's Weight - (f1 -f2)*surface, so the sum is not to 0.
let a part of top helix outside water, like that the volume is lower so the weight too. With several number of turns, more than one turn can be outside water; And if the helix turn (not move, just turn like worm drive), the position of helix in water is always the same, need to choose the good position for have no torque on helix.
You should use an helix like that
If the section of the helix is ovoidal,you can use a rectangle containter insted a parallelogram
It is clear in this way you have to follow during the fall the rotations of the helix
Using an helix like in figure you can mutiply the turns without loosing Fr-Fl
with a double helix, I have a double entry, so double Fl-Fr, why you think you don't have Fl-Fr ?
With this single or double helix you can mutiply the numbers of turns without increasing difference of force from right side to left side... the area covered by the helix in the lateral wall is the same with few or many turns... consequentially you can do same work with less liquid... this aren't my word but yours! and with that helix i think it is possible to achieve overunity
yeah a simple helix but a helix that turn 2 times.
http://footage.shutterstock.com/clip-1079053-stock-footage-red-double-helix-dna-strand-spirals-down-into-frame.html?src=rel/1079020:1
Not from gravity but from temperature. Take this circular helix. The helix is closed to itself, no end. The helix has gas inside at pressure P. Put in a container with pressure at 0, this give energy e1. Increase diameter of the helix without change the volume, the pitch decrease and this give energy e2. Move out helix, this need energy e1. The sum is e2.
Second image: increase d, sure c must change too.
Why from temperature? Energy come from pression....if in the container the helix will change shape due to torque...
imagine a helix with 2 ends (not close to itself), if diameter increase the energy needed come from these 2 ends, but without 2 ends, if diameter 'd' increase, I don't lost energy from these ends, sure if the number of turn is the same this would say the diameter 'c' decrease in the same time but it's not a problem I think
Talkin' about your first idea.. I'm not sure volume increase using an helix insted a common linear hose....
no the first idea don't work, the volume is the integral of the section along the lenght
for the second idea with pressure, I think it's possible to change the number of turn and increase diameter in the same time without change the volume.
In this case,witch value has the pression inside the container,considering the empty flexyble hose inside. F/1 or F/2?
the section increased in the same time, pressure is higher due to the larger volume but the section (surface) is larger in the same proportion. If you want to have the small section you need to basculate (slope) the container in the same time you move up or down.
had you try to understand second idea with pressure ? Entry a circular helix with gas inside, the lenght of helix is N*sqrt(r²+(p/2pi)²) with N the number of turn, r the radius and p the pitch. When the helix is inside container where there is 0 pressure, I transform the pitch to radius (volume = constant).
If there is a torque that change the shape of the helix,it would be better put she in a container ful of water,and varing the pressure by a weight on the surface...maybe...
if the diameter increase, this give energy from a torque (think about a helix with 2 ends, you will see that is ends that lost energy), sure I can't increase diameter from nothing all along the helix, but like helix has pitch length, I can reduce pitch so I can increase diameter. I don't think pitch lost energy because it's linear not circular. This need to be compute for verify.
Another idea with 2 spirals. One spiral turn clockwise, another turn anticlockwise. The container move down with a velocity relative of the angular velocity of spirals. The container move down with the additional volume from the black cable. Black cable is empty and has no weight in theory, in practice imagine it with very thin walls, and it is full of of air inside. Black cable give a up force, but this force is not give to the container, spirals control this force. Container move down with additional weight from the volume of black cable. Spirals turn only, they don't move down. Cable is rool up around red cylinder like a helix.
Second image: the cable slide only, cones don't turn like that no torque from up force inside water.
http://www.youtube.com/watch?v=1dd9NIlhvlI
Watch this, quite an interesting and practical idea to convert gravity into light.
Best,
Other side view. Cable has a F force to up (air inside cable). Cone stop F force from its axis, this don't cost nothing. The cable slide on cone 1, turn like a helix inside water and slide on cone 2. The container has more weight when it move down and the potential energy is always the same.
Or maybe with 4 pulleys like that. 2 up pulleys are conic like that they take the additional length of the cable
Maybe with 2 densities ? The black stem is fixed. Force F give and extra force to the left. 2 containers have always the same position.
The stem can move like the system because of force is perpendicular to the surface. I think it must be exist one position for the stem where the lost of surface of the container 1 is not equal to the lost of liquid at a slope giving.
Maybe like that ?
Triangle has weight and it is stable in water without black stem (it's like a boat). There are fixed black stems inside water. Triangle move from right to left. It's possible to see there is always a left force on triangle except at final where this need very few energy.
Seems logic. But how you return to initial situation? you have to give the same energy you have gained... or not? Perhaps if the black hose is weightless you can re-put inside the triangle gaining energy
Imagine the container like a disk (top view), triangle turn in the container. All black stems are fixed. The left position is like right position (return to initial position). Triangle is always inside water. Triangle pass through stems from right to left.
At final step, it's possible to move down grey rectangle because there is less up force, when all rectangle is there, move to the left and move up and repeat cycle.
Doesn't work.. Cause right covered surface is bigger than left...
in the case of a triangle and a rectangle too...
And the last shape too.. >:(
An idea with half torus inside water. All the system is under gravity. I compute forces F1 and F2 and they don't depend of radius (R), only of height (h) in the integrale. So this would say I can suppress bottom surfaces (grey color) so the weight is reduced too. Normally, surfaces must be compensated by volume of air, but here it's lower volume. I can move up or down all the system (water+air) the energy is not the weight by distance. Like F1=F2 all the time I can let at bottom air volume without force, and this suppress surfaces at bottom.
The result is always 2h, for any R.
small error, it's more that: 2h/pi, don't depend of R
So you are saying F1=F2? Are you sure of that?
arf, this don't work with homogenous liquid, maybe if density change
New idea with container of water with an object with air inside. It's like a weighing scale, but here I use the difference of distance for change the torque. If container can be like drawings, the air object give a difference of torque. It's possible to change the position of air object because container of water is always at horizontal. Red line is a rope that attach the air object to the stem.
doesn't work. The air object must not t be linked with the lever.But to the container.And perhaps it will not work
Quote from: Gabriele on March 28, 2014, 07:49:11 PM
doesn't work
Why, could you explain please ?
Like water is at ALL the bottom surface, the container is in equilibrium position, with or without air container. If I attach the air container to the lever, the distance is not the same and the torque too, no ?
You are right,the torque is not the same. But when each air container rises or fall,lose potential energy respect the water tank
If there is a torque, energy can be receover when lever basculate, no ? For move right to left (or left to right) air container I don't need to give energy (without count friction), no ?
No,no,no... i'm interested insted in the drawing you made about a circular spring that compressed change his shape
ok, but for the #44 post ? air container can be attached directly to the water container, I think it's the same.
Each cycle you move the lever the air container lose his potential energy untill he is completly out of the water
if I use only one container and turn it around like drawing shows, in half part I don't have more energy ?
Maybe with electric like that ? V can be multiplied with external iron. At secondary, reluctance is divided by 2 because there are 2 iron in parallel. If reluctance is divided by 2, there are 2 fluxes created by I, so this multiplied voltage by 2, no ?
With number of turns at primary = number of turns at secondary:
Input: give V*I
Output: recover 2V*I
V is divided by 2, so at primary I need VI/2 and at secondary I recover VI/4 ?
So if I place source in the middle with 2 charges, I give VI/2 and recover VI ?
This case destroy energy ? Same number of turns N for each coil. Reluctance is the same in each iron circuit.
a test with NI SIM : there is a little difference of power
first and second : win energy
third: lost energy
more output than input
I changed value of Kx, because it must lower than sqrt(Lx*Ly), but the system is very stable like that (even voltages are high !)
if K < 0.95 the system don't give energy.
works with Kx=0.9 but lower power
with possible values of L and K. You can test, NI sim is free for 15 days. This circuit need superconductivity material for coils.
interesting
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.
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 ?
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.
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.
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.
nobody tested this circuit ?
I don't have tecnical instruments to try this circuit...why don't you try by yourself?
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 ?
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.
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.
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
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
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).
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
Step1/ Accelerate and recover energy from red tube
Step2/ Decelerate and recover energy from acceleration of step 1
Repeat
This suppose it's possible to recover 100% of energy from acceleration, it's possible in theory, in practise this increase temperature and temperature is energy, so the system increase its own energy. The only energy lost is the blank part inside container, but it's a very low energy compared to red tube.
If the velocity to introduce green tube is high, the axis have a force F that cancel themself on one turn but not in each time. In top view, the green tube is fixed, move only in front view. The force F move the object alone. GREEN TUBE HAS NO MASS.
a new test with Multisim (first image) and LTSpice (second). With LTSpice I added a RL for transformer.
For the message # 76, I tested with Algodoo, it's ok the disk move alone. I give you the file for test. You can put all velocity at 0 except rotationnal speed. play, you'll see the disk move alone :)
The disk don't accelerate, it's ok, because force cancel themself in a turn.
I tested again with Algodoo and the CG move. You can put velocity at 0 for all the scene except rotationnal velocity and play. The velocity of CG is not very high because force from pressure cancel themself on one turn but it move. If you test with ice, you'll see the CG don't move with same scene.
Even CG move when object is inside liquid, the CG is compute with linear law. The law of force of pressure has a cosine function in it.
even with frequency at 10000 the volume moves alone. It's evidence with second case that velocity change like the position of block inside water.
With a macroscopic model with compressible balls it's easy to see additional forces F from balls to solid.
File R1: Works with only 1 ball and 1 spring. If you turn clockwise the system lost energy in the contrary the system win energy. Without the spring the system lost energy clockwise or anticlockwise.
I think this append because circle is not perfect. I try with this new file, it's the same result (file angle1). Sum of forces on ball is 0, the ball don't move. Sum of forces on system (except ball) is 0, but radius on apply forces of ball are different so there is a torque => energy move up or down depend of the direction of rotation.
Sum of torque is (R)*m1-(R+d)*m2, d is constant, so the relation between radius is (R+d)/R, R can be like I want. The relation between m1/m2 is 1.362/1.292 = 0.07 (look at figures). I can choose R like want sot there is a difference of torque.
The système turn clockwise. The sum of forces WITH friction give a torque. Pieces 1 are link together. Piece 2 is alone. Friction is at 0.5, restituion 0.5.
Your compressible balls, thats air, or any other gas. Isn't it?
And water is not compressible, it will not get more dense and will not lift your mass. It will leave the container, trying to stay in place.
There is no water. Take last Algodoo scene. No friction from air. No gravity.
The ball can be not compressible in fact. The system move alone (no axis) and sum of energy increase, without never stop. Even there is friction at 0.5, the friction must lost energy but in fact with friction the system win energy. Friction change the sum of forces. I take frequency at 12000 Hz, results don't change from 4800 to 12000 Hz. At start I added only an angular velocity.
I think friction act on center of gravity of small ball (2), and act on piece (1), there are not same center of gravity and radius are not the same. So the torque is not the same.
If you change mass of one part of the system this change the efficiency.
Without linear velocity, the system lost energy.
I tested with bigger dimensions. The difference of energy is bigger. I choose parameters for have small forces in the same range than previous test.
I think I understood how energy is won.
Image 1/ Average power measure by Algodoo is 10.2 W
Image 2/ Average angular velocity is 0.7044 rd/s
Image 3/ Average force is 0.0647 N (only a dy for compute only axis Y, relative to system at start)
Image 4/ A torque exist on Object2, the power from torque is 11.2 W
Image 5/ In the contrary Object1 moves only 0.686 m, so the power is 0.686 * 0.0647 = 0.04 W
Image6/ The system, scale
Like I measure average of force from -15° to 15°, I need to correct value of 11.2 W, this is (1+cos(85))/2*11.2 = 11 W it's very close to 10.2 W.
I study this new case. More stable and I think it's the different trajectories that give energy. Object2 (red and orange colors) turn, but not Object1 that is more in translation trajectory like image showing.
Object 2 receive a torque, the radius come into equation, it's 0.026 m. In the contrary, Object1 move only in translation for a big part, and the radius of rotation is at 0.0001 m. Forces from torque are sum of T and N. For compute, take Power = Torque * angular velocity = 22 N * 0.026 * 18.4 = 10.5 W, it's near Algodoo find (Algodoo don't compute friction).
For the sum of forces on each center of gravity I need to compute the direction and for Object 1 it's logical the force works.
On Object1, forces from Object2 pass by its center of gravity and like it is less than 1/10 of the trajectory of Object2, the works it's lower (1/10).
Wow, this is so CLEAR & Easy to make, I'm sure everybody here will make one RIGHT AWAY. LOL
.
It's only 3 objects in free rotation. Not easy to build in practise the free rotation, but like Algodoo don't compute friction, if someone can test with another simulator it can be confirm (or not) these results, the test is very short to do. Nobody has another simulator ? With an axis for Object 1 it seems the system increase energy too. P = 18.5 rd/s * 0.026 m /2 * 88 * cos(60) = 10.5 W
edit: cvf2
In some cases, F1 don't work around its trajectory, F2 too. But F1/F2 give a torque, no ?
I understood and now I can optimised the efficiency. Sum of force 93.45 N act on Object1, like the center of gravity move in linear trajectory, this increase the linear velocity of the system (when you are at 10 m/s and add a rotational force, the system accelerate because the rotationnal velocity increase in the same time). For object2, the trajectory is circular and linear, this increase the rotational velocity. I added the scene. I added a possibility for torque that explain a difference.
30 W with good parameters for masses.
I added a gif for show where is the torque that create energy. The torque exist between Object1 and 2.
I understood too in the system where there is not direct torque from F(cg1)/F(cg2). In the system with trajectory like first image showing, forces are like second image shows. Center of gravity of Object2 moves like red circle. F1 and F2 apply a torque, this torque is exactly the same for Object1. For Object2, F1 is separated by d from F2, there a torque but F1 works more than F2. The difference of power is F*w*d.
If I change angular velocity to 6 rd/s, forces are divided by 8 and angular velocity by 3, the power is divided by 24, it's linear. I change frequency, this don't change the result.
I tested with several frequencies. I thought 4800 Hz was near the best but if I let turn I have energy like second image shows. BUT only when forces are like third image, I compute with spreadsheet all datas, and energy come from red/orange object not green object. Maybe torque come only from centrifugal forces red/orange object because they must keep trajectory give by green object.
I compute all torques, it's only the difference of trajectories that give energy. It's logical. When I find 22*3 = 66 W Algodoo find less: 15 W. It's possible to move red and orange objects and look at energy after one object touch green object, if forces are like I show before, the energy increase. Red object give 66 W and orange object give 171 W with calculation (without friction). Algodoo give 15 W for red and 35 W for orange object. Maybe the difference is somewhere else than friction. Note that orange and red object must be like image shows for have best result (if red and orange objects touch green object in the same side the energy is lower). Maybe the difference come from my method of compute torque. I think Algodoo compute with mesh 2D and I compute with center of gravity maybe the result is not the same because each object has its own trajectory. Like Algodoo don't give force of a part of object I can't compute torque like that.
I added the scene with unstable forces, energy increase only when forces are in a special position.
I understood where come from the problem, it's only when red object has a torque (around itself) from green object. Last scenes, red object is not perfectly perpendicular because Algodoo change the position of red object (need to zoom a lot for show it). It's possible with red object to give at green object 0 torque (2 torques cancel themselves), the same is done at red object but like position of gravity (and trajectory) are not the same the red object has a torque if forces are not perpendicular. This torque create energy. If you look at slope of trajectories of center of gravity, it's logical: there is a phase angle between object red and object green.
bad scene
With this scene it's possible to have an oscillation for energy. This oscillation decrease with frequency but never diseapear.
In this last scene I can recover 1.3 % at frequency 12000 Hz. If it's an error of Algodoo it's 0.25 J of 18 J at this frequency and with stable forces I don't think it's possible. Someone else can test with another software ?
New scene. Explanation where the energy come from.
If blue object turn at constant rotational velocity around axis "axis" (external system not drawn limit its rotational velocity). Orange object is forced to be like image because there is Fc force. If I apply forces F1 and F2 on orange object, it will turn and I increase potential energy. But in the same time blue object has a torque from -F1/-F2 because distances d1 and d2 are not equal. The energy from blue object can be recover by external system and limit rotational velocity of blue object. With a rotational velocity clockwise, orange object can receive a standard torque. F1/F2 on orange object will slow down orange object and decrease potential energy but this energy can be recover in the same time.
With Algodoo it's works, until torque is apply to small disk, the big disk receive a torque too.
If you change the direction of rotational velocity of big disk this change the sum of energy. If I compute energy from motor, this is less than the additional energy giving to the system.
I tested with Algodoo. The motor give 0.15 Nm * 9 rd/s = 1.35 J each second (in fact less because I take the last rotational velocity not the mean). Algodoo give 2 J, look at the scene please.
Another test:
rotational velocity at start of big object = -6.28 rd/s
rotational velocity at start of small object = 6.28 rd/s
rotational velocity at end of small object = 7.77 rd/s
energy at start for all system = 114.56 J
energy at end for all system = 116.408 J
Motor torque = 0.15 Nm
Loss without motor for all system at 6.28 rd/s = 0.1 W
Power from motor = (7.77+6.28)/2*0.15 = 1.05 W
Power from Algodoo = (116.408-114.56)/1 + 0.1 = 1.95 W
Near the double because when torque is giving to small object the big receive the same free. Why not exactly the double ? because small object increase its rotational velocity at 7.77 rd/s but the big increase its rotational velocity to 6.34 rs/s only. The energy is 0.946 W from the torque 0.15 Nm. So the result find by Algodoo is 1.95W and I must find 2 W !!! very close.
In the last post, maybe you can think the energy give by motor to big disk it's an energy from the motor, but not, the stator like the rotor turn around central axis and the motor don't turn really around central axis, it's not a true rotation because the rotor like the stator turn together.
I think it's important to take w1=-w2 like that motor "see" a real rotation for disk2. And it's necessary to recover torque give from motor to disk1 like that angular velocity around central axis is constant.
If I take stem and object at end like image shows. I give rotational velocity to stem around axis "x", object rotates around "x" not around "y". The system has less energy than if object turns around "y" too. Stem turns at w around "x", object turns around "x" at w, but don't turn around "y". Object and stem don't have same rotational velocity (w compare to 0). Now I can brake from stem to object, object add a torque (and I can recover energy from brake) to stem and add its rotational velocity, the system add energy.
I apply F3 to object, stem receive F1, axis "y" receive F2 and F4 ? so, all these torque cancel themselves but the system has more energy at end.
The system in this case seems to lost energy, but it's possible to give rotational velocity to object around axis "y" and accelerate the object around "y". At start all turn at w around "x" and object turns at w around "y". After, I accelerate rotational velocity of object around "y".
I think it's works only if moment of inertia are different for object and stem (around their axis), and it is I think. Energy giving by a torque is torque*angle, but like moment of inertia are different, angle will be different even torque are the same in value.
I tested with a spring, it's easy to sum the energy. First scene without spring, the system lost energy. With a spring the system won energy. A torque is apply to circle, the same torque is apply to stem but moment of inertia are not the same and like energy win/lost is torque by angle of rotation, the system can win energy. I added force on circle because frequency don't must to be high, in this case there are a lot of vibration.
With specific velocity (x,y) the system with 2 arms and an axis give energy. Arms are attrack with gravitational forces.
Maybe I can use this last technic with this example. The idea is to accelerate an inertial mass with a motor. All energy consume by motor is in mass in rotation. The motor accelerate only rear bottom green pole from magenta pole ant front top green pole from another magenta pole, the rotor has always Fr/-Fr and turn around its axis of rotation and the stator has Fs/-Fs and turn around axis X. Radius r2 > r1 so the torque on stator is not the same. Magenta pole is fixed to stator.
For the stator it's not more difficult to repuls green pole even it rotates in the contrary direction because the rotor rotates perpendiculary to axis X.
For the last case, angular velocity around axis X can be fixed at w rd/s if an external system (not drawn) recover energy in the same time. No rolling element in the last idea, let rotor change its axis. Or with a rooling element in the center of rotor. With rolling element in the center of the rotor, all forces cancel themselves, at final there are only 2Fr that increase more and more angular velocity of rotor and 2Fs that give torque to external system for recover energy.
3 last images: slope the motor and give forces from magenta to green poles. Fr forces rotates only rotor around its axis and Fs forces rotates around axis red (X).
With a motor parallel to the axis X, with an inertia very high for the rotor, I apply a torque small, small enough for don't change a lot the angular velocity of rotor around its axis. If the angular velocity around axis X is high, the energy win is Torque by angle, but the energy needed is very low.
270 W
Frequency don't change the final result, even the time to reach 14000 J is the double.
Yes! Because precession provides more energy than it takes to run the gyro.
If this was not true then this device would not work: http://www.ebay.com/itm/GAVIC-Sports-Gyroscope-LED-Power-Gyro-Wrist-Ball-Birthday-Gifts-For-Men-Him-/141012768553?pt=LH_DefaultDomain_0&hash=item20d5041729 (http://www.ebay.com/itm/GAVIC-Sports-Gyroscope-LED-Power-Gyro-Wrist-Ball-Birthday-Gifts-For-Men-Him-/141012768553?pt=LH_DefaultDomain_0&hash=item20d5041729)
where there is precession, my simulation it's a 2d object, could you explain please ?
I reduced force from spring for control perturbations. Without perturbations slopes of 2400 Hz and 4800 Hz are the same, with 6000 Hz there are less perturbation and the slope is near 2400 Hz.
With 5 axes the system change the sum of energy. Result don't change from 2400 to 12000 Hz.
range of difference of velocity is very different, logical energy can't be the same from the same torque no ?