# Free Energy | searching for free energy and discussing free energy

## Discussion board help and admin topics => Half Baked Ideas => Topic started by: EOW on October 12, 2014, 11:36:02 AM

Title: Sum of torque
Post by: EOW on October 12, 2014, 11:36:02 AM
Blue liquid can turn around red point. I suppose red disk like a mass M, this mass attrack liquid like gravity can do in water on Earth. I suppose pressure linear with distance like on Earth. There is buoyancy in liquid. I compute sum of torque :

With:

R1 = 1
R2 = 2

In this case L1 = 1 and L2 = sqrt( 2²-1² ) = sqrt(3)

T1:

integrate from 0 to 1 of (1+x)*(1-x) this give 0.666

T2:

integrate from 0 to sqrt(3) of x*(sqrt(3)-x)/sqrt(3) this give 1/2

Sum of torque is not 0, the system turn alone and give energy I think. Gravity can be change by electrostatic or magnetism forces, small balls can replace liquid.

Title: Re: Sum of torque
Post by: EOW on October 14, 2014, 04:23:23 PM
Oops, there is an error, the integration for T2, the sum is zero.

It's possible to use 3 disks like second image showing, they turn at different rotationnal velocity X and Y but in the same direction without friction. Red disks don't turn around itself. At time t, friction is on, this accelerate one disk and decellerate others but if inertia are differents work from torques are different, so the sum of energy is not constant (take in account friction = energy = temperature).

The Work for torque = Torque * angle of rotation ( W=T*Theta )

Theta = 1/2 * a * t² with a the rotationnal acceleration

Second law of Newton : Torque = I * a with I the inertia

Angle of rotation = 1/2 * T/I * t²

Work of torque = 1/2 * T²/I * t²

W1 = 1/2 * T²/I1 * t² with I1 the inertia of grey disk
W2 = -1/2 * T²/I2 * t² with I2 the inertia of red disks

It's logical that W1 is not equal to W2 (in value) if I1 is not equal to I2, no ?
Title: Re: Sum of torque
Post by: EOW on October 15, 2014, 04:51:59 PM
Even with linear velocity there is a problem. E=1/2mV². Two objects move at +V1 and +V2. At a moment Object1 give +F to Object2, and Object2 give -F to Object1 without friction. Object1 move faster than Object2. The sum of energy is:

S=1/2m1(V1+F/m1*t)²+1/2m2(V2-F/m2*t)² - (1/2m1V1²+1/2m2V2² )

With data:

m1=10
m2=10
F=100
V1=10
V2=15
t=1

S=500 J
the sum of energy is positive

Title: Re: Sum of torque
Post by: EOW on October 20, 2014, 06:58:00 PM
3 disks (no gear just friction). Before start, blue disk, red disk and grey disks turn without friction. Note, red disk turn at w2 around itself and at w1 around blue point. A t=0, friction1 and friction2 are ON. This give a torque to grey and blue disks, energy of blue and grey disks increase. Red disk increase its rotational velocity around blue point (w1 increase). Rotationnal velocity of red disk around its center of gravity slow down SO the energy increase too (energy is 1/2I(w1-w2). All energies increase here. And there is energy from friction too. Friction 1 and 2 are choose for have a torque on red disk and only a torque. For this, friction must change all along the distance. It's just a technical problem. Note this works only fex second, until w2 is stable.
Title: Re: Sum of torque
Post by: EOW on October 21, 2014, 07:54:23 AM
Note about w2, w3 and w1. absolute value of w2 must be greater than 0. Absolute value of w2 must be greater than (abs( ((R+r)/r)*w1) +abs(w3)  ). R=radius of big disk. r=radius of small disk. This works only for a transitional regime due to these conditions.

I think big disk (disk that turn at w1) must be lower than small disk (disk that have w2) for have a difference of energy (w1-w2) increasing when w2 increase like image shows

better with 2 diameters for  disk 2 (second image). Remember energy come from red/yellow disk, it gives energy to grey disk and to blue disk. The goal is to increase energy of red/yellow disk in the same time.
Title: Re: Sum of torque
Post by: EOW on October 21, 2014, 09:56:41 PM
Magenta disk has a velocity around red point and around blue point too at start, this energy is gave. At t=0, 2 disks are like figure, with friction between 2 disks. Magenta disk increase rotationnal velocity of grey disk. Grey disk turn around blue point only. Magenta disk decrease its velocity around blue point but decrease its velocity around red point too. The sum of energy is not 0. Black arm is force to turn like grey disk.
Title: Re: Sum of torque
Post by: EOW on October 22, 2014, 09:14:38 AM
A disk with radius R turn at w1 around red axis of rotation. A cylinder inside the disk of radius r turn at w2 around black axis of rotation. Black axis of rotation is supported by disk. At start, disk and cylinder are turning. Friction exist between cylinder and disk, this friction give torque F1/F2 and -F1/-F2. Like radius of disk is different of radius of cylinder torques not works the same value, because it's possible to have the same angle when choose good value w1 and w2. The works of torques can be positive with good value and friction add energy too.

F1/F2 works with a radius r on cylinder. -F1/-F2 works with a radius R on disk.
Title: Re: Sum of torque
Post by: EOW on October 22, 2014, 03:31:40 PM
2 disks are turning together. Blue disk turn at w1 and yellow disk turn at w2. With w2<w1. A support not drawn aplly forces like image shows. torque on blue disk don't change energy because forces pass to axis of rotation. In the contrary, forces on yellow disk increase energy like (w1-w2)², like w1>w2, when w2 decrease, the energy increase. Friction add energy too. For maintain yellow disk at contact of black walls, this need to add a spring or something like that.
Title: Re: Sum of torque
Post by: EOW on October 23, 2014, 08:24:37 AM
Maybe like that ? N disks all around disk1. Each disk (except disk1) receive a torque that slow down rotationnal velocity of diskX, so the energy increase is W1 > Wx with X from 2 to N. And N the number of disks (except disk1). W1 is clockwise, Wx are anticlockwise. W1 don't decrease because forces pass through axis of rotation (blue point in center of grey disk1)

Friction create forces. Each disk except disk1 receive a torque, this torque decrease rotationnal velocity and increase energy because energy is a function of (W1-Wx)². W1 > Wx. Disk1 turn faster than others disks.
Title: Re: Sum of torque
Post by: EOW on October 23, 2014, 10:43:08 PM
At start the system is turning freely (no axis of rotation), the system turn around its center of gravity. w2 > w1 like that when w2 slow down, the energy increase, energy if a function of (w1-w2)².
Title: Re: Sum of torque
Post by: EOW on October 24, 2014, 10:27:52 PM
All red disks will slow down, so like energy is (w1-w2)², energy increase. Except for 4 corners. But N can be great and there are always 4 corners. Before start, give energy for turn red disks around themselves and around green axis. After, friction is ON, this increase energy of the system.
Title: Re: Sum of torque
Post by: EOW on October 25, 2014, 08:16:36 AM
With severals layers.

With one layer only: 4 corners give 4 torques but 4 disks at corners don't decrease their rotationnal velocity around their center of gravity.
With 2 layers only: the same for 4 next corners, previous torque are reported at this layer but 4 disks at layers one decrease rotationnal velocity =>energy is bigger
Wit 3 layers: 4 disks at corners of the second layer win energy

Etc.

R=Radius of first square layer (a side of inside square = 2R)
F=basic force from friction
t=time

1/ With first layer only, there are:
- 20 torques work 20*T*(w1-w2)t = 2*20*Fr*w2t = 40Fr(w1-w2)t
- Friction 1 = 40Frw2t
- 8 torques back (R-r)Fw1t, with R=6r for example, this give -48Frw1t+8Frw1t
Sum = 40Fr(w1-w2)t + 40Frw2t -48Frw1t+8Frw1t = 0 so no energy from first layer if alone

2/ With second layer only, there are:
- 28 torques work 28*T*(w1-w2)t
- Friction 2
- 8 torques back (R+2r-r)Fw1t
Sum is 0 too like first layer

3/ With first AND second layers only, there are:
- 20+28+8 torques work 52*T*(w1-w2)t = 2*56Fr(w1-w2)t so that add 16Frw1t-16Frw2t
- Friction 1+Friction2 + 8Frw2t
- 8 torques back (R-r)Fw1t + 8 torques back (R+2r-r)Fw1t

The difference is 16Frw1t-8Frw2t, the sum is not 0, like w1>w2, the energy is positive.

Forces on corners' s disk will receive more torque (4F), so rotationnal velocity at start must be higher for 4 corners's disk for have better efficiency.

Title: Re: Sum of torque
Post by: EOW on October 25, 2014, 03:58:37 PM
The sum is 0:

1/ With first layer only, there are:
- 24 torques work 24*T*(w1-w2)t = 2*24*Fr*(w1-w2)t = 48Fr(w1-w2)t
- Friction 1 = 48Frw2t
- 8 torques back (R)Fw1t, with R=6r for example, this give -48Frw1t

2/ With second layer only, there are:
- 32 torques work
- Friction 2 =  64Frw2t
- 8 torques back (R+2r)Fw1t
Sum is 0 too like first layer

3/ With first AND second layers only, there are:
- 24+32+8 torques work 64*T*(w1-w2)t = 2*64Fr(w1-w2)t add 16Fr(w1-w2)t
- Friction 1+Friction2 + 8Frw2t add 16Frw2t
- 8 torques back (R+2r)Fw1t  + 8 torques back (R+2r)Fw1t add 16Frw1t

Sum = 0

------------------------------------ IF W2 LAYER_2 IS DIFFERENT OF W2 LAYER_1---------------------------------

Maybe if w2 layer 2 noted w2' is lower than w2 of layer 1 the sum is not 0. I will compute it. I suppose frictions are different for have the same force.

- 24+32+8 torques, Frt( 24(w1-w2)+64(w1-w2')+16(w1-w2)), don't change if w2'=w2 because force is the same and torque is apply on first layer not the second
- Friction 1+Friction2 + 2*8Frw2t add 8Fr(w2+w2')t here friction change because it depend of w2 and w2'
- 8 torques back (R+2r)Fw1t + 8 torques back (R+2r)Fw1t , don't change if w2' not equal to w2 because force is the same

here the difference exist it is 8Fr(w2+w2')t that can be different of 8Fr(2w2)t

Difficult to build in practice because friction must change in dymamics, corners's disks must have a friction that change for have the same force even velocity is not the same.
Title: Re: Sum of torque
Post by: EOW on October 26, 2014, 02:44:30 AM
I compute work from torques:

+24+32+8 torques on red disks = Frt(48(w1−w2)+64(w1−w′2)+16(w1−w2))
Friction : +24Fr(w2+w2)t+32Fr(w′2+w′2)t+8Fr(w2+w′2)t
-8 torques layer1 - 8 torque layer2 =−8(R+2r)Fw1t−8(R+2r)Fw1t=−8(6r+2r)Fw1t−64Frw1t=−128Frw1t

The sum of works from torque = 128Frtw1−64Frtw2−64Frtw′2+56Frtw2+72Frtw′2−128Frw1t

The sum = −8Frtw2+8Frtw′2=8Frt(w′2−w2) it's different of 0
Title: Re: Sum of torque
Post by: EOW on October 26, 2014, 10:12:31 AM
How disks can be for have different friction.

Before t=0, an external system give energy for rotate disks around themselves and around green axis. At t=0, the external system is OFF and friction is ON. The energy must be conserved but not.
Title: Re: Sum of torque
Post by: EOW on October 27, 2014, 08:12:54 PM
Maybe this idea ?

Before t=0, turn at w1 and w2 disk. We need to give energy for that. W1>W2. Ar t=0, walls of disk has friction with air only at red point, this give forces F1 and F2 and reduce w2, like w1>w2, the energy increase, no ?
Title: Re: Sum of torque
Post by: EOW on October 27, 2014, 11:47:05 PM
It's possible to give a torque to disk with external green object. Each object is accelerating, its energy is increasing. w2 slow down, not w1. The energy is :

1/2md²w1²+1/2mr²(w1−w2)²

with :

d = lenght of black arm
m = mass of disk

The energy of disk increase. The system increas its energy.

Edit: use a ring instead of disk for have (w1-w2) term.
Title: Re: Sum of torque
Post by: EOW on October 28, 2014, 08:29:38 PM
nobody for help me ?
Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 12:27:06 AM
There is no motor after t=0, only before for launch disk. At t=0, apply friction ONLY at red point, not all around the disk. Red points change all the time their positions, this need special wall if you use friction (w1>w2). I don't know if it's clear enough. I think the formula is good only for a ring it's a little more complex for a disk. The disk win energy because w2 decrease. It's logical if you think with some points around the disk, at external the velocity is dw1-rw2 but at internal it's dw1+rw2, like the energy is a square function the external disk "cancel" more energy than it add at internal. I tested with Algodoo, you can test if you want it's very easy, fixe w1 and change w2 like you want and look at the sum of energy you will see the energy increase when w2 decrease (w1>w2). It's strange to say the energy increase if you slow down something but it is. And you can slow down the disk only at red point where w1 is the same for 2 points.

Quote
What is it that you would like help with?
maybe if you understand it allow me to explain the system, after if you can test for watch the energy increase with Algodoo, if you agree with that it's a first step for me. Test in reality must be difficult due the dynamic wall needed.
Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 07:16:18 AM
Yeah, I think we're ok. When I said it's difficult to built, it's for a pure 100% mechanical system.

What's number is wrong, the formula ? I think it's ok for a ring and maybe for a disk, a french guy on a forum said the formula is ok for a disk but I'm not sure, another guy on another forum said it's ok for a ring not for a disk. I looked for 10 hours on Internet and I don't find the formula.
Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 12:36:51 PM
Quote
I kind of find it funny that a ring or a disc might be able to use different formulas just because it is either a ring or a disc.
me too... I don't find the formula in Internet but anyway it's works with a ring, and this idea can be use with a ring.

you tested on what software ?

You tested on a real machine ? where you speak about it ? you must apply torque only a red point, ideally the torque must be applied a very short of time (function of the rotationnal velocity), this need sensors, precise electronic control, etc. You built a mortor like that ?
Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 02:55:51 PM
Trajectories can be like first image or like second image if w1 and w2 are choosen correctly (depends of the radius).

Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 04:49:47 PM
No, it's not the same principle. My idea is to use external objects (or something external like air), it's very important because your link shows a system "close" nothing external. Or something fixed to the ground that can recover energy. Here all the system turn no ? And like I said, it's not possible to give a torque like that. You must apply the torque when w1=w2, this in a short time, you need sensors and electronic command.
Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 07:12:01 PM
I don't understand your message, maybe if you drawn an image I can. Where are results of your tests and the model to test, have you image and link ?

Quote
Where is the torque in your system if you apply an equal force in opposite directions on opposite sides of the disc?
look at image of post #16, there is a torque due to external object (that can be done with fixed and controlled coils). And look at trajectories of points with w1 and w2, it's not a point, it's a line, your coil must recover energy perpendiculary to the trajectory, like the shock with balls for example. It's not easy to build. I don't know your system and if I can watch what you do I can say if it's the same or not.

For my study, the trajectory must be good for have a torque, like image shows. Maybe it's not the best. Choose point at P1/P2 and adjust the mass of object to schock for have the same force (in value).
Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 08:49:03 PM
Like that, there is a torque ?
Title: Re: Sum of torque
Post by: EOW on October 29, 2014, 09:19:15 PM
If it's possible to create a torque with external objects on disk, the energy is won. So, I think it's possible to change the disk, one part with mass, another without mass, and to place point at the good position for have only a torque on it.
Title: Re: Sum of torque
Post by: EOW on October 30, 2014, 08:30:10 AM
If I take the disk in a double rotation w1 and w2, with w1>w2. It's possible to add a translation to all the system for have to point P1 that move in one direction. Other point P2 friction with black stem, this add forces F1 to disk and F2 to stem. Shock with external purple object for the point P1. P2 friction only, not shock.
Title: Re: Sum of torque
Post by: EOW on October 30, 2014, 09:06:33 AM
With velocities like that :

Title: Re: Sum of torque
Post by: EOW on October 30, 2014, 10:05:59 AM
With this ? this need a translation in addition for have forces like that.
Title: Re: Sum of torque
Post by: EOW on October 30, 2014, 10:59:55 AM
Vleft=R1w2−(R1−d)w1
Vright=R2w2−(R2−d)w1 With:
R1=10
R2=4
w1=10
w2=2
d=4
I find:
Vleft=-40
Vright=8

Use ring to be sure the formula is good: 1/2md²w1²+1/2mr²(w1−w2)²
Title: Re: Sum of torque
Post by: EOW on October 30, 2014, 01:33:36 PM
I made a mistake R1 must be < R2, so value can be:

Vleft=-R1w2+(R1−d)w1
Vright=-R2w2+(R2−d)w1 With:
R1=0.5
R2=15
w1=10
w2=8
d=4
I find:
Vleft=-49
Vright=-20

Add a translation for have +30:

Vleft= -19
Vright=+10

CQFD
Title: Re: Sum of torque
Post by: EOW on October 30, 2014, 02:09:42 PM
I think like that it's correct, someone can verify ?
Title: Re: Sum of torque
Post by: EOW on October 30, 2014, 04:30:33 PM
If I take a simple example. Trajectories are in doted lines. F1/F2 are created with a shock from external object. F11/F21 create a torque on ring, this increase the energy of disk. F12/F22 create a force to blue axis, this axis is free to move in space, so forces can't give a torque on axis but only move it, this give energy too. For simplify the problem, think with mass of blue axis like very high compared to the grey ring. I think like that the energy is not constant.
Title: Re: Sum of torque
Post by: EOW on October 31, 2014, 08:55:37 AM
I forgot a 2 forces, so there is a torque on black arm, this decrease energy. But the force (F12+F22) on blue axis move all the system, this must increase a little its energy, no ? What's the difference between fixed blue axis and free ? Fixed blue axis F12+F22 don't works and free F12+F22 works.
Title: Re: Sum of torque
Post by: EOW on November 02, 2014, 07:57:29 AM
It's seems it's not possible to have trajectories like I want. But it's possible to add N system in parallel. Each system give two energy E1 in heating and energy from disk E2. Sum for each disk is 2E1+E2. With N system this will give N(2E1+E2). I need to add energy at 2 last systems, need 2E1, so the sum of energy is N(2E1+E2)-2E1.

edit: for have friction, it's possible to place blue axis of rotation further (or closer), and change the angle, like angles can be shows between black and magenta lines. Add 3° at each new system. It's only for have "constact" during a time for have friction.

edit: in a last message I said it's not possible to find 2 points where horizontal velocity is negative, but it's possible with r > d (the mass is in the green point only, not at all the surface of the disk). Like that I can think like that: hits only one ball, and for have a torque I give the energy for this, like that I recover energy from ball but I lost energy I gave, BUT th energy of the disk increase !
Title: Re: Sum of torque
Post by: EOW on November 02, 2014, 07:36:08 PM
A force F is needed for generate friction, so energy is needed for have friction. Force F is apply from an angle 'a1' to another angle 'a2'. It's not possible to have friction from 0 to 360° but only a part of 360°, it depend of the configuration.This force give energy to disks because F increase w1 of each disk. Friction don't decrease F, so all energy from F is giving to w1. Green arrows are forces from friction. Magenta arrows are forces that I need to give energy.

Edit: it's possible to add forces F at bottom and at top, like I'm sure I don't lost energy from forces F.
Title: Re: Sum of torque
Post by: rc4 on November 04, 2014, 05:52:53 AM
Works of forces

F is the value of green or magenta force

wdisks=+N1/2mr²((w1−w2i)²−(w1−w2f)²)

with w2f<w2i, w2f=w2 at final, w2i=w2 initial

Wfriction=2(N−1)Fdw3t with w3 the mean of w2

WF1=2dF−2dF=0
WF2=2dF−2dF=0

Wmagentaforce=−2Fdw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−1)Fdw3t−2Fdw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Fdw3t

Sum of energy:

Before t=0, the system (N disks) has the energy N(1/2md²w1²+1/4mr²(w1−w2)²)

At final, the system has the energy:

N(1/2md²w1²+1/4mr²(w1−w2)²)+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Fdw3t
Title: Re: Sum of torque
Post by: rc4 on November 05, 2014, 07:28:58 AM
I made a mistake in my calculations:

Works of forces

F is the value of green or magenta force

wdisks=+N1/2mr²((w1−w2i)²−(w1−w2f)²)

with w2f<w2i, w2f=w2 at final, w2i=w2 initial

Wfriction=2(N−1)Frw3t with w3 the mean of w2 **************************** I noted 'd' but it is 'r'

WF1=2dF−2dF=0
WF2=2dF−2dF=0

Wmagentaforces=−2Frw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−1)Frw3t−2Frw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Frw3t

Sum of energy:

Before t=0, the system (N disks) has the energy N(1/2md²w1²+1/4mr²(w1−w2)²)

At final, the system has the energy:

N(1/2md²w1²+1/4mr²(w1−w2)²)+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Frw3t
Title: Re: Sum of torque
Post by: rc4 on November 06, 2014, 09:22:43 AM
I posted here :

http://physics.stackexchange.com/questions/143377/one-disk-ring-in-double-rotation-and-sum-of-energy

for have a reply, maybe if you up (+1) the question, more people will be interested about it

Title: Re: Sum of torque
Post by: EOW on November 09, 2014, 11:17:10 AM
The problem with my last idea are the trajectories. I can't have green forces in this direction because points don't move like I thought. For that, I need to have a rotationnal velocity higher than w1. I use for this gears, that will increase rotationnal velocity. These addionnal gears has no mass (in theory) like that they don't lost energy.

For the cycle: give rotationnal velocity w1 and w2 (and kw2), this is for launch the system. And after let "live" the system like it is, no external motor. There is only friction between magenta/magenta disks. Forces are like that because kw2 > w1. Here I can take w1=10 clockwise, w2=-7 counterclockwise and kw2=21 clockwise.

First image: At start: all disks (or rings) are turning around blue axis at closckwise w1. All bigger green disks are turning at w2 counterclockwise around green axis. All smaller magenta disks are turning at kw2 clockwise with k =3 (in this example but k can be higher). Green disks have mass. Magenta disks haven't mass. Note there is no friction between magenta/green disks because it's gears (and in theroy I consider no friction here).

Second image: Look at forces. Friction generate F1 and F2 forces. Note there is heating => energy. F1 and F2 create F3, F4, F5, F6, F7, F8, F9, F10 like image shows and 2 additionnal -F3 -F4 not drawn. Each magenta disk receive a counterclockwise torque. They reduce their rotationnal velocity but note they have no mass. Each green disk receive a clockwise torque and increase their rotationnal velocity, and like they have mass, they increase kinetic energy. In the lab frame reference, the kinetic energy is 1/2md²w1²+1/4r²w2² with m the mass of the disk, d the lenght of the arm and r the radius of the disk. Fx-1 and Fx+1 show the forces come from another basis system when I repeat them.

Third image: repeat N systems

I define H the energy from one magenta/magenta friction. I define K the additionnal kinetic energy of one green disk. The sum of energy increases, we have (N-1)*H energy from heating . Add N*K kinetic energy from each green disk. Remove H for two last system, I need to give energy for give Fx-1 and Fx+1 at 2 terminal system. The sum of energy is (N-2)*H+NK.

Edit: like before, blue axes are fixed to the ground. Think with k very high (radius of magenta disks are very small), it's easy to see forces can be like I drawn but not all around the circle just like I drawn at start with arms horizontal.
Title: Re: Sum of torque
Post by: EOW on November 11, 2014, 09:56:02 AM
The idea is to increase rotationnal velocity (in the lab frame reference) of disks with friction forces. Friction is energy and kinetics energy too. I try to choose good velocities, but event there is a problem, it's possible to chose one disk without mass if its rotationnal velocity decrease. For those disks that increase their rotationnal velocitiy set mass not at 0. I hope it's clear enough. Look at images, please.

Before $t=0$: I give rotationnal velocity $w_1$, $w_{2a}$, $w_{2b}$ and $w_3$, this is for launch the system, during this step I need to give energy. And after I let "live" the system like it is, no external motor. The sum of energy must be constant. I count all energies in the system, heating too. Green disk is turning around blue axis clockwise and around itself counterclockwise (green axis). Magenta disk is turning around blue axis clockwise and around itself clockwise (magenta axis) for the upper and counterclockwise around itself for the lower. If I want to guess no sliding between green and magenta disks, I need to have $w_3$ different of $w_3'$. There is only friction between magenta/magenta disks, but before $t=0$, I set $friction=0$ for launch the system. I need to set $|w_1| > |w_2|$. I can take, for example, $w_1=-10$, $w_{2a}=-7$, $w_{2b}=-5$ and $w_3=+9$ and $w_3'=-29$ (all angular velocities are labo frame reference). There is a relation between $w_{2x}$ and $w_3$ because there is no sliding between magenta and green disks.

Blue axes are fixed to the ground. Arms must turn together and gears (not drawn and without friction) force all arms to turn together at the same rotationnal velocity. If a torque is present on an arm it will be apply on others through gears. Green disks have mass. Magenta disks haven't mass.

I need to have friction between magenta disks AND in the good direction, for that I need to set $w_{2x}$ different for each green disk with $|w_1| > |w_2|$. Look the description and the image where I calculate velocities.

Note I study the sum of energy only in a transcient analysis. From $time=0$ to $time=t_x$ with $t_x$ very small.

At start:  Note there is no sliding between magenta/green disks because green disk force to turn magenta disk like gear can do, it's not a gear but consider contact magenta/green disks are like gears without friction (no heating dissipiation) and no sliding. Sure, magenta disk force green disk to turn like gear can do. But I guess no friction between magenta/green disks.

Look image N°1

Friction generate $F1$ and $F2$ forces. I noted all others forces I see. Note there is heating => energy between magenta/magenta disks. $F1$ and $F2$ create others forces. Each magenta disk receive a counterclockwise torque. They reduce their rotationnal velocity but they have no mass, so they don't decrease their kinetic energy. Each green disk receive a clockwise torque and increase their rotationnal velocity, and like they have mass, they increase kinetic energy. Image shows $Fx-1$ and $Fx+1$ forces, it comes from another basis system when I repeat them. Look image N°2

Another position: Look image N°3

To be sure trajectories are correct, I calculate linear velocities in this position: look image N°4

$P1$ and $P2$ are on each magenta disk, where there is friction. With $r'$ the radius of magenta disk. The linear velocity of $P1$ is $(d-r')(10)-r'(15)=-25r'$ and the linear velocity of $P2$ is $(d+r')(10)-r'(19)=-9r$, I counted positive the right direction. So, $P1$ move faster at left than $P2$ and forces can be like I drawn.

So, with: $|w_{2a}| > |w_{2b}| > |w_{2c}| > |w_{2d}|$ it's ok.

I need to give $Fc1$ and $Fc2$ forces for have friction between magenta disks. These forces don't work. Look Image N°5

I repeat $N$ systems like that last image

I define $H$ the energy from one magenta/magenta friction. I define $K$ the additionnal kinetic energy of one green disk. I define N the number of basic system {magenta disk + green disk}. The sum of energy increases, we have $(N-1)*H$ energy from heating . Add N*K kinetic energy from green disks. Remove $H$ for two last system: I need to give energy for give $Fx-1$ and $Fx+1$ at 2 terminal disks. The additionnal energy is $(N-2)*H+NK$, it's not 0.

Edit: with friction between magenta and green disks it could be easier to find good velocities. Maybe with no friction between magenta/magenta disks. I will calculate this. If necessary change radius of disks, magenta disks can be with different radius; The same for green disks.

Title: Re: Sum of torque
Post by: EOW on November 11, 2014, 02:12:01 PM
Like image shows. F5 and F6 forces (and others) increase w1. There is heating. 4 red disks decrease kinetics energy but 3 black disks increase theirs and there is heating. |w1|>|w2|
Title: Re: Sum of torque
Post by: EOW on November 12, 2014, 12:19:54 AM
With 'R' radius of purple disk.  With 'r' the radius of grey disk. 'F' the force from friction. 't' the time. A grey disk is rotating at w1 around red axis and at w2. Gears are turning too. I need to give kinetic energy for that. After, I let "live" the disk and gears. There is on friction between purple disk and grey disk, no between gears. No external motor. I let the device like it is and I count energy. The device works only few seconds. Purple disk is fixed to the ground. Friction generate forces F1 and F2, energy goes to heating. w2>(R+r)w1/r, with labo frame reference.

Gears are turning clockwise and counterclockwise. I guess between gear/grey_disk it's like a gear: no sliding, no heating dissipation.

Cycle:

1/ Friction is OFF. Launch grey disk and gears at w1 and w2
2/ Set friction ON
3/ Measure the sum of energy, heating too (H) !

H= | F(Rw1-r(w2-w1))t |

The sum of energy must be constant, but it decreases in this example. So with gears I can have good rotationnal velocities and forces. Like gears has mass (like disk), when grey disk decelerates, gears want to keep their rotationnal velocity due to inertia, this give forces. I can set forces like I want, for example F/2 for first gear and F/4 for second gear because forces depends of the inertia of each gear and it can be like I want. Here the delta energy is H+Ft(-rw2+Rw1+1/2rw1). I need to have w2>(R+r)w1/r for have F1 and F2 in these directions. So with the limit case w2=(R+r)w1/r the sum of energy is -1/2Frtw1+H. In the limit case H=0, the sum is not 0, it is < 0. If w2=(R+r)w1/r+x, the sum of energy is -Ftx-1/2Frtw1+ Ft ((Rw1-r(w1(R+r)/r+x)-w1)t)=-1/2Frtw1, it's not 0.

Calculations :

Interface PurpleDisk/GreyDisk: ( -1/2rw2+3/2(R+r)w1 )Ft
Interface GreyDisk/FirstGear: ( -1/4rw2-3/4(R+3r)w1) Ft
Interface FirstGear/SecondGear: ( -1/4rw2+1/4(R+5r)w1) Ft

Sum is (-rw2+Rw1+1/2rw1) Ft

Title: Re: Sum of torque
Post by: EOW on November 12, 2014, 01:07:00 PM
My last case, the energy increases of -rw2-1/2rw1, because I forgot one torque. Like H is FRtw2 it's not possible to have 0.

Now, grey disk rotate around itself counterclockwise. Forces are like image shows. The energy from torques is Ft ( rw2-1/2rw2-3/2(R+r)w1+1/2(R+3r)w1-1/2rw2 ) = -FRtw1

This result don't depend of w1, but for friction w2 is a parameter, H = Ft((R+r)w1-rw2), look below I verified this calculation

The additionnal energy is Ft((R+r)w1-rw2)-FRtw1= Ftr(w1-w2) and w2 < w1, the energy is positive
Title: Re: Sum of torque
Post by: EOW on November 12, 2014, 08:09:05 PM
I'm not sure about my forces and my sum of energy but in the first image, grey disk don't around itself. The energy from heating H=+FRtw1, and torques give the energy +Frtw1-F(R+r)rw1, the sum is 0.

In the second image, grey disk turns around itself counterclockwise. The energy from heating is Ft((R+r)w1-rw2). The energy from torque is +Frtw2-F(R+r)rw1, here the sum is 0, like that I'm sure of the calculation of the heating H for the last message where I find a sum different of 0.
Title: Re: Sum of torque
Post by: EOW on November 14, 2014, 05:44:24 PM
To simplify the study of this device. Think with an external theoretical device that add the energy FRtw1, like that all rotationnal velocitiy w1is  constant. I give FRtw1 and I receive Ft((R+r)w1-rw2), the difference is: Frt(w1-w2) with w1>w2, the energy is positive. I can't keep w2 constant, because I need the force F4 (and F5), if w2 constant => w3 constant and F4=0. So, w2 must change with time. Maybe a little sliding between grey disk / brown disk can be adjust the kinetic energy.

Heating gives H=Ft((R+r)w1-rw2)
Torques need Ft ( rw2-1/2rw2-3/2(R+r)w1+1/2(R+3r)w1-1/2rw2 ) = -FRtw1

Here F4=1/2F but it's possible to have another value it depends of the inertia.
Title: Re: Sum of torque
Post by: EOW on November 16, 2014, 12:16:31 AM
I'm not sure when I drawn direction of rotation, so I give an example:

w1 = 10 rd/s
w2 = 8rd/s
w3 = 12rd/s

w1 = 10 rd/s, the arm is turning clockwise
w2 = 8rd/s, the grey disk is turning clockwise at w1 around red axis (labo reference) and grey disk is turnin around itself counterclockwise at 2 rd/s (arm reference)
w3 = 12rd/s, the brown disk is turning clockwise at w1 around red axis (labo reference) and brown disk is turning around itself clockwise at 2 rd/s (arm reference)

I guess |w2| < |w1| < |w3| . There is a relation between w2 and w3 at start because there is no slinding between grey disk and brown disk. In the arm frame reference w2 = -w3 at start.
Title: Re: Sum of torque
Post by: EOW on November 17, 2014, 01:32:35 AM
In the case with the arm fixed and the purple disk is turning, I have a work from torque at -FRtw1+Frtw2+kFrtw2-kFrtw3, heating don't change, the sum is at kFrtw2-kFrtw3, it is 0 too. 'k' is kF force from brown disk. If there is a little sliding between the grey disk and the brown disk, the sum is always kFrtw2-kFrtw3. So there is more heating but +kFrtw2-kFrtw3 seems to be higher because w2 seems to increases more than w3.
Title: Re: Sum of torque
Post by: EOW on November 17, 2014, 11:23:37 AM
First image:

Another idea. A motor is fixed to the arm accelerates the red disk, more and more. The slave disk (grey) is accelerated in the same time, no sliding between the red disk and the grey disk. All the system is turning at w1 around the magenta axis. I drawn all forces. The motor receive a negative  torque, it must give energy E1. The slave disk and the red disk receive its energy E1 and increase their rotationnal velocity. But the grey disk has the force F6 that incresase w1. In the contrary, F5 don't work. The sum of work seems to be positive.Like the grey disk is turning in the other direction that w1, choose the rotationnal velocity of the grey disk > w1 at start.

Second image: no motor after t=0

w3 > w1, and w2=w3 (or not exactly). All rotationnal velocities are counterclockwise. Before start, launch the device with w1, w2, w3 and set friction between disks at 0. At t=0, set friction ON, there are forces like I drawn. F6 can give energy to the arm via w1. Disks slow down but it is compensated by heating from friction.
Title: Re: Sum of torque
Post by: EOW on November 18, 2014, 08:46:47 AM
Like this there is a torque on arm.
Title: Re: Sum of torque
Post by: EOW on November 18, 2014, 08:41:13 PM
An arm of length d is turning clockwise at w1. A red disk of radius r is turning at w2. w1 and w2 velocities are in labo frame reference. I define w2′ the rotationnal velocity in the arm frame reference. I set w2′ for have the velocity at the origin of F1 at 0: w1(d+r)−rw2′=0, w2′=w1(d+r)/r. An external force F1 is applied during a very short time, like there is no movement, the work needed by the force F1 is 0. I define F the value of F1 or F2. Like I give F1 on disk, F2 appears on arm. The work from F2 on arm is Fdtw1. The work from F1 on disk is −Frtw2′=−Ftw1(d+r). The sum of energy is not 0.
Title: Re: Sum of torque
Post by: EOW on November 20, 2014, 11:01:41 PM
The black arm is turning clockwise at w1. The blue axis is fixed to the ground. The grey cruz (grey arms) is turning at w2, with w2 < w1, so the grey cruz is turning around itself counterclockwise in the black arm reference. Each  red disk is turning at w3, with w3 < w2. For example, w1=10, w2=9 and w3=1. Friction from disk/disk gives black forces. The rotationnal velocity of each red disk on grey arm reference is w3' with w3'= w2-w3 = 8, so each red disk is turning counterclockwise around itself on the grey arm reference. So w3' can give forces like I drawn. These black forces give a torque on each disk and give a torque on the grey cruz. Each red disk increases its kinetic energy, because the torque is clockwise and w3 < w1. The black arm don't have a torque, so w1 is constant. Green forces reduces w2. BUT, the inertia of grey cruz can be like I want and don't depend only of the disks, I can add a big mass in the center of the cruz for example. Remember, torque=inertia*acceleration and kinetic energy is 1/2*inertia*w², so kinetic energy is 1/2w²/inertia. If inertia of cruz is very high the rotationnal velocity of the cruz don't decrease (imagine a mass very high), or very few in practise. If I increase the inertia of the cruz, I change w1 ? no. I change w3 ? no. I change the work from friction ? no. It is an independant parameter.

a/ w1 is constant
b/ I can choose the inertia of the grey cruz as high I want (in the center of the cruz)
c/ The heating is a positive energy
d/ The kinetic energy of each red disk increases
e/ I can reduce the lost of kinetic energy from grey cruz by adding a mass in the center of the cruz

Cycle:

1/ Friction is OFF. Launch the device with w1, w2, w3 with external motor
2/ No external motor or force
3/ Set friction ON
4/ Measure the sum of energy, heating too, for me the energy increases, the heating and the kinetic energy.

What do you think about that ?
Title: Re: Sum of torque
Post by: EOW on November 22, 2014, 12:58:01 PM
For recover energy (if the heating is not the goal), just add 4 generators between red disks. Generator must be with few mass. Interface between red disk / rotor =  gear. I think I need to adjust forces from friction (or torque from generator) at each time for have no torque on w1. At inner radius, set friction higher and at outer set friction lower.
Title: Re: Sum of torque
Post by: EOW on November 25, 2014, 02:28:40 PM
With one arm and one disk, the energy is not constant. Friction between disk/arm gives F1 and F2.
Title: Re: Sum of torque
Post by: EOW on December 01, 2014, 08:41:32 AM
No friction, no gravity.

Case A
1/ turn counterclockwise the ring
2/ turn the ring with arm clockwise
3/ eject small part of ring when point are like green points, recover kinetic energy

Case B
1/ turn clockwise the ring
2/ turn the ring with arm clockwise
3/ eject small part of ring when point are like green points, recover kinetic energy

The energy needed for turn the ring around itself and with the arm is the same in case A and case B. In the contrary, the energy recover in case A is not the same than in case B.

Title: Re: Sum of torque
Post by: EOW on December 03, 2014, 05:45:37 PM
If the ring is turning at w2 and the arm is turning at w1 with w2 < w1, I take 2 external blue objects that are turning at w1, like the ring turns counterclockwise in the arm reference, it's possible to eject these 2 objects and apply a clockwise torque to the ring in the same time. The two external objects don't lost the energy from the rotation w1 and win energy from the ring. Like this:
Title: Re: Sum of torque
Post by: EOW on December 05, 2014, 04:34:53 PM
The arm is turning clockwise at w1 and the ring is turning at w2, with w2 < w1. With one external object, free to move in space, that move in a straight line at the velocity V.

There is a shock at time = 1, IN HORIZONTAL AXIS (the shock is not in vertical axis), between the blue object and the ring.

The trajectory of the ring is more at left than at right (look at image)

For example, with:

r=1
d=3
w2=0
w1=10
t=1e-5

the result is :

9.9999e-5 at left
9.9985e-5 at right

So the shock move at left the free object and increases the kinetic energy of the ring. The arm don't receive a torque because the force is in the axis.

I did a video:

Title: Re: Sum of torque
Post by: EOW on December 06, 2014, 06:14:26 PM
The black arm is turning clockwise at w1. The red ring is turning clockwise at w2 with w2 < w1. There are 2 magenta disks that are turning at w3 around themselves. The grey arm is turning at w1 at start. The grey arm can turn around itself. Magenta disks are perpendicular at the ring. There is friction between the red ring and magenta disks, this gives a force F (I guess F constant even rotationnal velocity is changing with time).
Title: Re: Sum of torque
Post by: EOW on December 07, 2014, 02:36:03 AM
Edit: Consider the grey arm like a gyroscope, like that the torque on the grey arm don't move the grey arm around the blue axis but around a perpendiculary axis. And like this the grey arm don't lost energy. There is only friction, w2 is increasing and w3 is increasing too.

Title: Re: Sum of torque
Post by: EOW on December 07, 2014, 01:19:39 PM
With only the black arm, the red ring and the gyroscope. The gyroscope must turn at w1 at start.

Title: Re: Sum of torque
Post by: EOW on December 07, 2014, 02:52:20 PM
No friction, no gravity, no external motor. The black arm is turning clockwise at w1. The gyroscope is turning at w3=0 like that it keeps its position in the labo frame reference. Springs give a clockwise torque on the black arm, the kinetic energy of the black is increasing. The gyroscope don't lost energy because it turns in a perpendiculary plan.
Title: Re: Sum of torque
Post by: EOW on December 07, 2014, 06:27:36 PM
No gravity. No heating dissipation. The black arm is turning clockwise at w1. A motor accelerate more and more a slave disk (inertial disk). I guess w1 constant because I recover the energy from the black arm for keep constant w1. The belt don't receive a torque, all forces Fa, Fb, Fc, and Fd cancel themselves. The motor receives -F but this force is in the axis, so no torque from -F. The slave receives F, and the black arm receives a clockwise torque +FR . The energy won is +FRw1t. The energy gives by the motor is in the kinetic energy of the slave, so this energy is not lost, it can be recover later. The belt gives only forces where I drawn them (special belt), look at second image.
Title: Re: Sum of torque
Post by: EOW on December 09, 2014, 09:35:00 AM
It's difficult to have forces like I drawn with a belt. With an hydraulic system it would be easier. An external hydraulic system (not drawn) gives a pressure in the hydraulic pipes. This pressure accelerates 2 inertial disks. The sum of forces on each hydraulic pipe is 0, so there is no torque. The black arm is turning at w1 clockwise. The disk1 is turning clockwise at w2, with w2=0 at start. The disk2 is turning counterclockwise at -w3, with w3=0 at start.

The black arm receives two forces F and -F, this gives a clockwise torque FR on the arm. The energy recovered is FRtw1. An external system (not drawn) recover the energy from the black arm and keep constant w1. If w1 is very high the work from FRtw1 is very high too.

The hydraulic system must give an energy E, but all this energy is in the kinetic energy of 2 disks. So this energy E can be recover later. At final, the energy gave by the system is FRtw1.

All the system is turning like the second image.

It's possible to compare with the system in the third image. The energy is conserved (I added a ring on the disk1, the mass is in te grey inner part). The only difference if the system in the first image

I think the hydraulic pump must turn at w1 like that all the device is turning at w1 except the disk1 and the disk2.
Title: Re: Sum of torque
Post by: EOW on December 09, 2014, 06:26:18 PM
A big green plate has N small grey disks. Disks can turn around themselves. The plate turns at w clockwise like disks at start. I have on the plate 2*N devices (not drawn) that can ejects small black masses. These devices don't give a force or a torque on the green plate because masses are the same and I eject them with the same velocity. These masses shock grey disks and increase the rotationnal velocity of disks. Sure at final I have a mass, alone, that MUST give a torque FR (if F is the force that a mass give) during a time 't'. But I can change the radius of the mass because I have time, the energy for change the radius is 0 because the force is always perpendiculary to the trajectory, but if the radius decreases the rotationnal velocity is increasing because kinetic energy is constant=1/2mr²w². So I can shock the plate with a clockwise torque. I need the same energy for accelerate disks when w=0 or w=100, but in the case where w=100, the kinetics energy of each disk is increasing more than when w=0:

Second image, I can use springs.

Thrid image: use electromagnetic system for increase rotationnal velocity of each grey disk. Prepare a lot of masses M (black) for have a continous movement.

Don't forget, masses rotate at w so their velocity is V clockwise even the force reduce the velocity, the direction can give a clockwise torque.

The  5° image is a basic system, with only a spring for understand the main idea !
Title: Re: Sum of torque
Post by: EOW on December 11, 2014, 09:21:43 AM
No gravity. No friction. Only one motor. An external device not drawn recover energy from the black arm for have w1 constant.

First image: I guess the energy is conserved in this case. It is the basic system that I replicate. At start, all the device is turning clockwise at w1. The motor need energy E1 for turn the inertial disk. There is a torque FR on the black arm but I guess this is the motor that give this energy.

Second image: I use 2 basic devices. The torque on the black arm is 2FR and the energy from the motor is the same than in the case 1/. The inertial disk has te same kinetic energy than in the case 1.

Third image: I can use N basic devices. The torque is NFR.

Fifth image: the small disk is turning at R/rw but I want the big is turning at w, so I can use magnet and like there is a sliding so the external ring can turn more slowly than the small ring
Title: Re: Sum of torque
Post by: EOW on December 13, 2014, 04:25:24 PM
Look at images please. The motor accelerates more and more the disk with N intermediate devices. All the device is turning at w1. The motor gives a counterclockwise torque to the main axis (red color) -300 J, the disk win +100 J and with 3 intermediates devices I recover 3*200 J, the sum is 400 J in my example.

The first image shows a basic device motor+belt+disk, I compare the energy when w1=0 and w1=100 at start. I can find the conservation of the energy.

The goal is to accelerate more and more a disk with a motor and have a lot of devices that give the torque from (F5,F6) * (R2-R1) on the main support. An external device recover energy from the support for keep constant w1.

Like I drawn the reductor, it is divided by 3.3, but the rapport motor/disk is 1.45, so this must say with N=100, the disk accelerate very few, so I can guess the energy won by te disk is 0. The motor lost 300 J in my example but I can set N like I want, and each intermediate device win 200 J.
Title: Re: Sum of torque
Post by: EOW on December 14, 2014, 10:35:18 AM
w1 is constant, an external device recover energy from the red axis. I guees no friction with pulleys, belts and gears. I guess a mass as lower as possible with pulleys, belts and gears. The motor gives only energy for the brake. The support receives a clockwise torque from (N-1)F(R1-R2) with N the number of intermediate devices. I think I can take w=2w1. w and w1 are constant.
Title: Re: Sum of torque
Post by: EOW on December 15, 2014, 08:24:33 PM
I resumed the device in the image.
Title: Re: Sum of torque
Post by: EOW on December 18, 2014, 11:42:35 AM
The mass of the belt give Fc1 and Fc2, with Fc1 > Fc2, like that the support has a net torque on it. An external device recover energy from the main axis, like that w1 is constant. A motor on the support drive the Pulley1 but with a mass = 0 for the belt the torque is 0 on the support. But if the mass is not 0, there is a net torque on the support.

(http://imagizer.imageshack.us/v2/xq90/673/sM8cws.png) (https://imageshack.com/i/ipsM8cwsp)

(http://imagizer.imageshack.us/v2/xq90/540/0D83oF.png) (https://imageshack.com/i/f00D83oFp)
Title: Re: Sum of torque
Post by: EOW on December 19, 2014, 12:35:30 AM
All this device is on a support not drawn. The support is turning clockwise at w1. A method for cancel the torque of the counterclockwise blue disk. There is no torque on 3 disks. But the motor gives a torque to the support. If the support is turning at w1 very high, the energy can be recover.
Title: Re: Sum of torque
Post by: EOW on December 19, 2014, 07:10:20 PM
It's only a motor, 2 disks on a support. The motor give a counterclockwise torque on 2 disks, so the support receives a clockwise torque. There is friction between 2 disks at point A and B. If 'd' is very small all the counterclockwise torque on the support can be canceled. Like the support is turning clockwise at w1, it's possible to recover energy from the support because w1 can be very high, in the contrary the energy needed for give friction is very small (local angular velocity not dependant of w1).

The motor gives: -F1 to the red disk and -F2 to the magenta disk. The motor receives F1 and F2.

The red disk gives to the magenta disk F3, the red disk receives -F3

The magenta disk gives to the red disk F4, the magenta disk receives -F4

Axis of rotation of the magenta disk is on the support. Axis of rotation of the motor is on the support. When all disks are turning they are like the image all the time.
Title: Re: Sum of torque
Post by: EOW on December 20, 2014, 10:24:25 AM
Look at the image. The support receives a clockwise torque Fd. Each disk receives a torque equal at 0. The motor give energy to friction but if the support is turning very quickly the energy won is high.
Title: Re: Sum of torque
Post by: EOW on December 21, 2014, 01:07:38 AM
With one motor it could be ok.
Title: Re: Sum of torque
Post by: EOW on December 22, 2014, 08:53:43 AM
Maybe like this ?

The support is turning clockwise at w1, it receives a clockwise torque from 3 motors. There is no counterclockwise torque from pulleys or belts on the support. w1 can be very high. The energy gave by motors is lost but the energy gave to the electromagnetic devices can be recover.

I need only special belts that can push only, or push only or push and pull. I drawn 6 pulleys but it can be 100.
Title: Re: Sum of torque
Post by: EOW on December 22, 2014, 07:51:20 PM
With an acceleration.
Title: Re: Sum of torque
Post by: EOW on December 24, 2014, 06:22:00 PM
Look at the images.

The green disk is very small for cancel all torque on the support but it can be attached to a bigger disk with a mass. The green disk I drawn is a pulley attache to a bigger disk. If the diameter of the blue ring is 0.1m and the diamter of the green disk is 0.001m, if w=1 rpm, then the angular velocity of the green disk must be at 100 rpm. If I applie a force F=20 N (a torque of 1Nm), the support receive a torque = 0.1/2*5 - (0.1-0.001)/2*20 = 0.01 Nm and it can be very low. The work on the green disk is 100Cwt=0.001/2*20*100 = 1*t Nm. The work on the support is 0.01*w*t=0.01*t. So it's possible to compare these 2 different energies.
Title: Re: Sum of torque
Post by: EOW on December 25, 2014, 10:05:11 AM
At start, all velocities are at 0. I accelerate the support clockwise, I let pulleys (all pulleys have mass) like they want to turn, all the pulleys are free to turn like they want. The green ring want to turn at -w (in the support frame reference) like the blue pulleys. But the difference of radius force the blue pulleys to follow the green ring. The support receives a clockwise torque, value = 1/10Fd. When I turn the support, I give the energy for turn only the support with all pulleys but I don't give the energy for turn around themselves pulleys. So I receive a clockwise torque and in the same time pulleys turn around themselves, I don't care about the value of this angular velocity, positive or negative, because a value is necessary greater than 0.

For find the final angular velocity of the pulley I can write:

1/2J1w1²+3*1/2J2w2²=C, C is calculated with a final value of w
and
w2=Kw1

There is a difference of 1/10 torque.
Title: Re: Sum of torque
Post by: EOW on December 25, 2014, 10:48:26 PM
Pulleys can't have the same angular velocity, one must accelerate, other need to decelerate. Each pulley receives a torque FR, but the support receives the torque -F(R1-R2). And if I set the belt ON when the support is at w, 2 pulleys are turning at -w. So the work won by 2 pulleys is 2FRwt and the support lost F(R1-R2)wt.

Cycle:

1/ Turn the support with the belt OFF at w, in the labo frame reference disks don't turn, if a torque is appied on a disk its kinetic energy is increasing
2/ Set the belt ON
3/ The support receives -F(R1-R2), one disk +FR1 the other disk -FR2. FR1 increases the kinetic energy and FR2 too.  It's like -FR2 must decrease the energy of the smaller disk but not, in reality (labo frame reference) the smaller disk don't turn and the torque increase the kinetic energy of the smaller disk.
Title: Re: Sum of torque
Post by: EOW on December 26, 2014, 04:49:32 PM
No motor ! it's only a transcient analysis. All velocities are in labo frame reference.

Case 1/ At left, the support don't turn, the small disk turns at -6.2rd/s, the big disk turns at -2rd/s. The big disk will drive the smaller, I need a spring for adjust angular velocities. I noted R=R2. The big disk win 6wFRt, the small disk lost 6.2wF3Rt and the spring win the rest. The sum of energy won/lost is well 0.

Case 2/ At right, it's the same case except that the support is turning clockwise at w. The sum of forces are the same. Except one thing the support receive a clockwise torque +2wFRt and like it turns it can produce an energy. Here, the sum of energy won is 2wFRt.

Forces can be like I drawn because in the support frame reference: the bigger disk turns at -w and the smaller at -5.2w, the smaller turns more quickly even the radius is not the same.

I use a spring for take the difference of angular velocities. The spring has no potential energy at start, but it could if necessary.

In the second case the device win energy.
Title: Re: Sum of torque
Post by: EOW on December 28, 2014, 07:22:19 AM
The power is transmit like that:

With 2 basic devices (not drawn) :

I can reverse the angular velocity without give a torque on the support if the diameter of P2 and P3 are very small. The motor gives -FRwt to the support but receives 2FRwt from pulleys. The motor needs the energy -2Frwt, P4 gives Frwt. P4 has a mass no others pulleys. The motor accelerates only P4. Forces F1/F2 gives 2FRwt. P2 drives P3 like a gear (not drawn). The sum is well at 0. So, I need to place at least 3 basic devices in serial.

With 4 basic devices (the image):

Motor is P1
P2 is driven from P1 with the belt
P3 is driven from P2 with the gear (each pulley has a gear fixed to it)
P4 is driven from P3 with the belt
P5 is driven from P4 with the gear (each pulley has a gear fixed to it)
P6 is driven from P5 with the belt
P7 is driven from P6 with the gear (each pulley has a gear fixed to it)
P8 is driven from P7 with the belt

And for reverse the velocity of bigger disk, I can use gears with small diameter in the center of each pulley P4 and P5, like that the torque on the support is near 0. The device can win 2FRwt

The power transmit from P4 to P5 is higher if the angular velocity is high: 101 to -99 for example. If I take 3 to -1 I transmit 3FRwt to -FRwt and I think the power is reduced of 1/3. With 101/99 the ration is 99%.
Title: Re: Sum of torque
Post by: dieter on December 29, 2014, 04:36:01 AM
Most of this is over my head  :-[ , but I higly respect the amount of work you put into it. The lack of responses is rather cold, but see it that way: Many people follow the creed "when I can't say anything negative about it then I'd say rather nothing at all.", so in that sense this is good ^^

The most fascinating demonstration of a sum of torque that I have seen is gyroscopical precession, where a rotating wheel that is supported only on one side of the axle does not tip over and fall down, because the torque of the tipping sums up with the rotation and the torque chases the angular momentum as long as the rotation continues. (Or something like that)

Peace

Title: Re: Sum of torque
Post by: EOW on December 29, 2014, 07:24:12 AM
I can reduce the angular velocity without give a torque on the support. It's possible if I multiply the number of pulleys. 3 pulleys receive the energy from one with sliding. The small pulley receives 4FRwt, but in the support frame reference it turns at 3w, I can transmit the power at 3 bigger pulleys without lost energy and with an angular velocity of w. The small pulley lost 4FRwt but each bigger pulley win 2FRwt because in the labo frame reference its angular velocity is 2w not w.

For that I need to have a sequential device. I can't give energy from the motor and recover in the last disk all the time. I need to give power 1/3 time to each pulley P3, P4, P5. The rest of the time the pulley is not driven.
Title: Re: Sum of torque
Post by: EOW on December 29, 2014, 07:50:56 PM
Or with a gyroscope like that. The goal is to decelerate (in the support frame reference) all purple disks because they turn at -w in the support frame reference. In the labo frame reference disks don't turn (w=0), so if they decelerate in the support frame reference, they accelerate in the labo frame reference. The gyroscope don't turn, it can cancel near all torque from 2 ending disks, can't cancel all the torque, there are forces fx more and more bigger with the angle. Need to add a lot of gyroscope all around the support with different angles because the gyroscope don't turn.

There is friction between disks and between the arm of the gyroscope and 2 ending disks. Friction can be replace by any system for recover energy like electromagnetic.

The gyroscope receives a torque around the axis z but it turns around the axis x.

I can place the disks in the 'center' like the gyroscope:
Title: Re: Sum of torque
Post by: EOW on December 30, 2014, 09:33:42 AM
With this cycle of water. Imagine the size of the container (fluid) very large. The mean density of object to move is 1.1. Density=0 in half part and density=2.2 in the other half part.
Title: Re: Sum of torque
Post by: dieter on December 30, 2014, 06:40:06 PM
Hi EOW,
I am trying to understand ...  :o ... :-[ am I stupid?  ??? What is density 0 ?!? Vacuum? And how is the object rotated?

Peace
Title: Re: Sum of torque
Post by: EOW on December 30, 2014, 11:18:48 PM
Hi dieter, I will explain later my last message, density 0 is vaccum. I come back with my friction. The goal is to accelerate 2 sprockets in the labo frame reference without energy. The support turns clockwise at w so sprockets turns counterclockwise in the support reference. If I can reduce the angular velocity of sprockets, I win energy because they accelerate in the labo frame reference. I use a bad quality roller chain for that. There is friction in the roller chain, the wheel lost angular velocity and the support receives a clockwise torque.  I can use a special roller chain, friction exist only when I curve it, no when I straigth it.
Title: Re: Sum of torque
Post by: EOW on December 31, 2014, 08:32:53 AM
The support receives don't receive a torque. Each sprocket receives a clockwise torque. There is friction when I curve the chain but no friction when the chain become a straight line. Forces I drawn don't come from an external device it's the link that force the chain to be like that. So when the support don't turn, the chain lost energy in friction (it's an energy) and each sprocket has a clockwise torque on it (it lost kinetic energy or apply a negative torque on the motor). But here, the clockwise torque increase the kinetic energy.
Title: Re: Sum of torque
Post by: EOW on December 31, 2014, 10:48:44 AM
With gravity and an asymetric container:

Title: Re: Sum of torque
Post by: dieter on December 31, 2014, 09:09:35 PM
Hi EOW,

Ok, I  think I understand the last one. Centrifugal force may be insignificant at low RPM.

Happy New Year!
Title: Re: Sum of torque
Post by: EOW on January 01, 2015, 11:41:32 AM
Hi Dieter,

Happy new year you too !

I think the last case with water, the device don't work because while the support turns the red/white container don't work.

For 2 sprockets + roller chain (post #92), if I guess no friction except between links of the chain, and I guess the chain is tensioned very high, the friction between links gives heating (this is an energy), this heating  come from the lost of angular velocity of each sprocket. But sprocket turn only in the support frame reference, not in the labo frame reference. So the friction increases the angular velocity of sprockets. The only thing to verify is the support musn't receive a torque like that the support don't lost a kinetic energy.

Maybe it's possible to place magnets on each sprocket, like that magnets attrack parts of chain, the sprocket lost kinetic energy (in labo frame reference, the sprocket win kinetic energy) and there is heating. For "eject" the part of chain that is magneted when the chain becomes straight, I can use the centrifugal force, the sprocket will lost angular velocity and the part of the chain becomes straight ? Like that I don't need a chain, just one disk and small stems around the disk:

Title: Re: Sum of torque
Post by: dieter on January 01, 2015, 12:31:25 PM
Hi EOW,

I watched #92 10 minutes, I still do not understand it, sorry.  :-[

BR

Title: Re: Sum of torque
Post by: dieter on January 01, 2015, 12:41:31 PM
Still studying it... wait a minute, do you mean the friction / no friction axles are bearings with no friction in clockwise rotation and much friction in counterclockwise rotation?

BR
Title: Re: Sum of torque
Post by: EOW on January 01, 2015, 12:44:27 PM
Quote
Still studying it... wait a minute, do you mean the friction / no friction axles are bearings with no friction in clockwise rotation and much friction in counterclockwise rotation?
I'm not sure to understand. My goal is to slow down the disk (or sprocket) in the support frame reference WITHOUT return a torque on the support. Slow down the disk in the support frame reference = accelerate the disk in the labo frame reference.

1/ Forget the chain, it's possible to use only stems and a disk. The disk is on a support. If a disk don't turn in the labo frame reference, its angular velocity is -w if the support turns at w, ok ?

2/ Now, if I slow down the disk in the support frame reference, the angular velocity goes from -w to -0 in the support frame reference, the disk accelerates in the labo frame reference, ok ?

3/ The disk has magnets and can attrack the stem (steel), but there is friction in the axis between the stem and the disk. The friction gives heating (it's an energy). This energy come from the kinetic energy of the disk: the disk slow down in the support frame reference.

4/ I use the centrifugal force for eject the stem. I need to give an energy for eject the stem, the energy can come only from the angular velocity of the disk, so the angular velocity of the disk increases in the labo frame reference.

The only thing to verify is: the support musn't receives a torque, and if the axis of rotation of the disk has no friction, it's not possible to return a torque I think

Tell me if you don't understand, I can explain more
Title: Re: Sum of torque
Post by: EOW on January 01, 2015, 01:04:01 PM
All could be fine with one disk, but in the labo frame reference the disk don't turn, so there is no centrifugal force, it's for that I would like to use 2 sprockets and a roller chain. But without return a torque on the support.
Title: Re: Sum of torque
Post by: dieter on January 01, 2015, 01:34:45 PM
I even don't know what "Labo" means, I tried babelfish translation, not found.
I made some gears, like planetary gears, but basicly this is not really my field of expertise, so I hope you don't mind if I seem rather stupid  ???
I will read it again after some sleep. I didn't sleep since last year ^_^
BR
Title: Re: Sum of torque
Post by: EOW on January 01, 2015, 07:41:05 PM
labo frame reference: when you are in the laboratory and you look at the device turn, http://en.wikipedia.org/wiki/Laboratory_frame_of_reference

support frame reference: when you are in the support and you look at the disk

maybe I can use a massless roller chain like that it's easy to straigth the chain.
Title: Re: Sum of torque
Post by: EOW on January 02, 2015, 10:21:56 AM
If I use a massless roller chain with friction, I can roll up the chain around the red disk, friction will slow down the red disk (in the support frame reference) = accelerate the red disk in the lab frame reference. The red disk has a mass.

Cycle:

1/ The red disk don't turn in the lab frame reference, launch the support at w. The roller chain is all around the support. The energy needed is the energy for turn the support at w
2/ Roll up the chain around the red disk, at final the roller chain is all around the red disk, not on the support
3/ Slow down all the support. The energy recover is the energy from the support and the energy from the red disk.
Repeat
Title: Re: Sum of torque
Post by: dieter on January 02, 2015, 10:28:21 AM
I think I am beginning to understand slooowly  :) .  In the next weeks I won't have much time to hang around in forums  :-\ . This was interesting. Maybe you should build one of these devices?
What I usually do is:
1 have the idea
2 draw it in 2D
3 simulate it in 3D in the Computer (here I often discover flaws in my thinking!)
I'm using Blitz3D, but any flexible 3D engine will do. Making use of an attached physics engine may be useful, maybe not always.
4 make a crude model, eg. of wood.

Kind Regards
Title: Re: Sum of torque
Post by: EOW on January 02, 2015, 10:42:07 AM
Thanks for the software, I will simulate. It's difficult to start with Blitz3d ?

Regards
Title: Re: Sum of torque
Post by: EOW on January 02, 2015, 11:18:19 AM
I think it's better with a straigth roller chain to roll up on the red disk (the chain turns at w). Like (in theory) the roller chain has no mass there is only the friction that need a force. Like there is only one disk maybe it gives a force to the support, imagine the forces is like the first image, the support receives a counterclockwise torque but in this case I can place the red disk like in the second image. So, I think there is no force on the support. But there is friction and the red disk is accelerating.
Title: Re: Sum of torque
Post by: dieter on January 02, 2015, 11:19:06 AM
Hi EOW,
Blitz3D is a basic compiler with an embedded 3D engine for DirectX. It is rather easy when you know a bit about programming. But it is not for free. Some years ago there was a free version with a 1000 lines of code limitation, but now I guess there is only a 30 days trial version.

There is also freebasic with some 3D extensions, but I  not sure if it has all the important features, like parenting one object to an other.

There may be other solutions, including Blender, although Blender is very unintuitive and hard to learn.

BR
Title: Re: Sum of torque
Post by: dieter on January 02, 2015, 11:23:54 AM
CORRECTION!!! BLITZ3D is now Freeware!!!

www.blitzbasic.com/ (http://www.blitzbasic.com/)

Totally reccommended! I paid 100$and never regreted a thing! BR Edit: BTW, here you should find several physic engine extensions, like Physix, Bullet and others: www.blitzbasic.com/Community/topics.php?forum=94 The blitz3D compiler allows to use those 3rd party DLLs with a "wrapper". Basicly it may be useful to search the forum using the word "physic" Title: Re: Sum of torque Post by: EOW on January 02, 2015, 12:08:04 PM With Blitz it's possible to have the sum of forces, the energy, etc ? Title: Re: Sum of torque Post by: dieter on January 02, 2015, 12:16:33 PM You have to add one of the physic engines (see my last posting), and then yes, to some degree they try to simulate physics. Some are really good, probably even with friction parameters. Ok, this is not the best, because the best physics simulations usually cost$. But it is interesting and may be very useful.

BR

Title: Re: Sum of torque
Post by: EOW on January 02, 2015, 03:01:08 PM
With a stem with mass, the left half fixed circle can stop the stem. With a fixed spiral, the stem can have a massless, the spiral gives always a prependicular force to the stem. There is friction between the stem and the red disks. Forces F1 and F2 are forces from friction between the red disk and the purple stem. Forces F3 and F4 force the purple stem to be on the disk, F3 and F4 cancel themselves, the spring never move, it don't lost energy.

It's ok because S<pi*R
Title: Re: Sum of torque
Post by: EOW on January 03, 2015, 12:25:50 AM
I tested on Algodoo for look at all angular velocities, it seems ok. I done a video :

The stem move a little, the mass of the stem is very small, there is friction between the purple stem and red disks. The support has only an angular velocity at start, not disks.

Need the spiral for test the sum of the energy. The spiral is fixed and I think forces can be like I drawn.

The gif:

http://imageshack.com/a/img903/8053/6oVzC4.gif

And a bigger video:

http://youtu.be/TVv2GTwHhY0

The purple stem must block at the point B not A, because the distance move by the stem is not very big.
Title: Re: Sum of torque
Post by: EOW on January 03, 2015, 12:48:41 PM
I think the length travelled by the purple stem is not enough for reach the spiral. But I can fixe one part of the stem to the spiral, like forces F1 are always reported to the center of the support like I drawn, this could say the red disk accelerate counterclockwise. But at start, the red disk don't turn around itself, so this is an additionnal energy. The purple stem has no mass.

2 Cases:

1/ the length travelled by the stem is greater than the "step" of the spiral, in this case I can block the stem and red disk increase their angular velocity clockwise
2/ the lenght travelled by the stem is lower than the step of the spiral, in this case I can attach the stem to the spiral, red disks increase their angular velocity counterclockwise.

The support don't receive a torque in case 1 nor in case 2.

Cycle:

Accelerate more and more the support, but at start angular velocity of the red disk is 0. The purple stem will accelerate the red disk in the counterclockwise direction.
Title: Re: Sum of torque
Post by: EOW on January 04, 2015, 01:45:31 AM
Maybe instead of use a spiral I can use another circle with the center just aside just for a transcient analysis.
Title: Re: Sum of torque
Post by: EOW on January 04, 2015, 02:21:17 PM
I can unroll a roller chain from one disk to roll up to another disk. Red disks don't turns at start in the labo frame reference = turn counterclockwise in the support frame reference. Like I unroll and roll up the roller chain, the support don't receive a torque. But there is friction and red disks decelerate in the support frame reference = accelerate in the lab frame reference. Think with the same diameter for disks (think in 3D, disk can have depth). The roller chain has no mass (or as lower as possible). Disks can be very close from one to other like that the torque on the support is very small.

Cycle:

1/ Turn clockwise at w the support
2/ Unroll and roll up the roller chain from one disk to the other => win energy
3/ Decelerate the support and recover energy

I can repeat the cycle because the roller is roll up to the left disk. I need to change the position of the disk left <-> right only.

The friction can be asymmetric, no friction when I unroll and friction when I roll up or the other.

----------------------------------------------------------------------------------------

Or the idea with a pipe and a container of water, the idea is to move down the container with a higher weight from the volume of the pipe. There is air inside the pipe. For the shape of the reel : http://en.wikipedia.org/wiki/Spiral_of_Theodorus, like that the up force from the water to the pipe is always perpendicular to the trajectory, there is no torque on the reel from the force up.
Title: Re: Sum of torque
Post by: EOW on January 05, 2015, 10:28:40 AM
With 2 spiral of Theodore of Syrene, the force Fp from the water is gave to the spirals but this sort of spiral has Fp/2 at the center of rotation. I can full spirals with water inside, like that Fp = 0. I need energy for move up the water but that energy can be recover later.  I move up the water from the bottom spiral to the upper spiral and I move down the container full of water. Or move up the water only, but need to change the step of the spiral more and more lower when the diameter increase. With d1< d2 < d3 < d4 < d5 < d6.
Title: Re: Sum of torque
Post by: dieter on January 05, 2015, 12:48:02 PM
I think your creativity as well as your ability to imagine the devices in operation is remarkable. Genius at work I'd say.  :)
BR
Title: Re: Sum of torque
Post by: EOW on January 05, 2015, 04:23:29 PM
Thanks, but all my ideas are bad :(

I try with this new one, there is a big difference of radius of the left pulley and the right pulley, I try to move down a pipe with air inside. The bigger pulley decrease its diameter of x, and the small pulley increase its diamter of y with y >> x. If there is a torque this could say in the seond image the belt turn alone.
Title: Re: Sum of torque
Post by: dieter on January 05, 2015, 05:09:22 PM
You know, Edison used to say "I didn't fail, I have just found one more way the lightbulb does not work". So, work in progress is not bad only because it is not finished.

BR
Title: Re: Sum of torque
Post by: EOW on January 05, 2015, 07:03:14 PM
Thanks for your positive feedback !

The problem with the spiral is the torque on the pulley.

Maybe here, move to the right 3 objects in the same time, there is a force F to the triangle. The black is fully in water, it don't receive a left force.
Title: Re: Sum of torque
Post by: EOW on January 07, 2015, 11:26:18 PM
The support is turning clockwise at w. The disk don't turn in the lab frame referenceI => the disk is turning counterclockwise in the lab frame reference. I give the force F3 and I give the energy FRwt but I receive the energy 2FRwt if the diameter 'x' of the disk is small.  Like that, this work until the green disk is not at w.

It's possible to prevent the disk to turn: put inside the green disk a gyroscope, like that it's a continous movement.
Title: Re: Sum of torque
Post by: dieter on January 08, 2015, 10:58:27 AM
Instead of a gyroscope you could use a gear system. I did that once, with wooden gears  :) , it really worked, the non-rotation I mean. Like, when the green radius is 1 and the grey radius is 3 then there needs to be a second gear on the axle of the gray one, radius 1 and speed 2 times that of the grey.

BR
Title: Re: Sum of torque
Post by: EOW on January 08, 2015, 05:59:57 PM
You try to add gears to the device ?

I come back to the system with the spiral Maybe I can let a slope for the pipe and have Fp/2 on each reel with no torque.
Title: Re: Sum of torque
Post by: EOW on January 11, 2015, 03:58:02 PM
Wheels don't turn at start but like the support turns at w clockwise => wheels turn counterclowkwise at w. There is friction between wheels => forces F. Each wheel give a counterclockwise torque to the support, but instead to give that torque to the support I give it to each gyroscope: magenta forces. The gyroscope can cancel a part of this torque. It's necessary to take one force from other wheel and distribute to each gyrocopes. I think each wheel can have a gyroscope, like that, wheels don't turn even there there is a torque on it, like that each gyroscope keep its position like I drawn. I drawn 6 wheels but it can be more. Each gyroscope turn in a perpendicular axis when it receives a torque.

Friction give heating but it's possible to use an electromagnetic device for recover the energy.
Title: Re: Sum of torque
Post by: dieter on January 11, 2015, 06:09:54 PM
How do you handle the friction?

BR

Title: Re: Sum of torque
Post by: EOW on January 11, 2015, 07:53:48 PM
Friction is an energy => heating. When I calculated the sum of work, the energy won by the heating is lost by the support: it receives a counterclockwise torque, but here with gyroscopes, the torque to the support can be reduced. The friction can be replaced by an electromagnetic device and recover the energy to the electric energy instead of use heating. Like gyroscope are always in that position, they never rotate (they rotate in a perpendicular plan), it must be possible to recover more energy even while a short time.

Maybe I don't understand the word "handle", all wheels don't turn in the lab frame reference => turn counterclockwise in the support reference. Each wheel touch 2 wheels => forces F.
Title: Re: Sum of torque
Post by: dieter on January 12, 2015, 08:59:55 PM
By "handle" I meant what do you do with friction. Electromagnetic device is a bit vague, but ok. I guess the wheels do not really have to touch eachother.

The gyroscope will resist rotation, but when a torque is applied (as the big wheel rotates), then I guess the sum of torque 1+2 will cause "recession", eg. in the direction into the picture (or against the surface of the big wheel), see also youtube videos about "Eric Laithwaite".

BR

Title: Re: Sum of torque
Post by: EOW on January 12, 2015, 11:35:21 PM
Yes, the gyroscope cause a precession. The support turns around the axis Z. The gyroscope cause a precession around the axis Y.  In the lab frame reference the wheel and the gyroscope don't turn around the axis Z, it's a translation, its angular velocity around the axis Z is 0. The problem is to give the rest of the force that the gyroscope can't cancel to the support. The gyro can't be fixed to the support with an axis because the axis will absord all forces and the gyroscope will receive nothing. It's possible to imagine the wheel with an axis to the support and the gyroscope is "floating" in the wheel. Forces are gave to the gyroscope not the wheel, like that the force the gyroscope can't cancel will give to the wheel.
Title: Re: Sum of torque
Post by: dieter on January 13, 2015, 10:18:39 PM
"Floating" ?  :)

BTW. aren't the arrows of F in the wrong direction?  :o

BR

Title: Re: Sum of torque
Post by: EOW on January 18, 2015, 05:39:46 PM
I think forces are good but it's the trajectory the problem, the distance change when the gyro turns.

A new idea with 2 supports:

Support1 and Support 2 are turning clockwise at w around the white axis
On the Support1 there is the Pulley1 with the radius 3R, the Pulley1 is turning at 2w clockwise (lab frame reference), the Pulley1 is turning at w in the lab frame reference
On the Support2 there is the Pulley2 with the radius R, the Pulley2 is turning clockwise at 4w (lab frame reference), the Pulley2 is turning at 3w in the lab frame reference
The Pulley2 is a brake, it transforms the mechanical energy to heating, the energy from heating is 4FRwt, because the Pulley2 turns at 4w and the radius is R
The Pulley1 receives the energy from a motor fixed on the Support1 (the motor is not drawn). The motor needs the energy -3FRwt and the Support1 receives a counterclockwise torque that lost the energy -3FRwt
The sum is not 0, it's -2FRwt. If the Pulley1 and the Pulley2 are fixed to the same support, the support receive a clockwise torque that give the energy 2FRwt, here with 2 supports, the force F and -F works at the same value in one round and cancel themselves.

I drawn different positions.
Title: Re: Sum of torque
Post by: EOW on January 19, 2015, 09:40:30 AM
I corrected my message

Support1 and Support 2 are turning clockwise at w around the white axis
On the Support1 there is the Pulley1 with the radius 3R, the Pulley1 is turning at 2w clockwise (lab frame reference), the Pulley1 is turning at w on the Support1 frame reference
On the Support2 there is the Pulley2 with the radius R, the Pulley2 is turning clockwise at 4w (lab frame reference), the Pulley2 is turning at 3w on the Support2 frame reference
The belt turns because the radius goes from 3R to R and the angular velocity goes from w to 3w on the pulley
The Pulley2 is a brake, it transforms the mechanical energy to heating, the energy from heating is 4FRwt, because the Pulley2 turns at 4w and the radius is R
The Pulley1 receives the energy from a motor fixed on the Support1 (the motor is not drawn). The motor needs the energy -3FRwt and the Support1 receives a counterclockwise torque that lost the energy -3FRwt
The sum is not 0, it's -2FRwt. If the Pulley1 and the Pulley2 are fixed to the same support, the support receive a clockwise torque that give the energy 2FRwt, here with 2 supports, the force F and -F works at the same value in one round and cancel themselves in one round.

It's possible to use something else than the brake. It's just for have the force F. Here the energy is destroyed. For created energy, turns Support 1 and Support2, turn Pulley1 and Pulley2 and brake the Pulley2, it gives 4FRwt from the brake and it cancel -2FRwt from the Pulley1 (if the pulley has a mass).

I added 2 points A and B for look the angular velocity. A is fixed on Pulley1, B is fixed on Pulley2.
Title: Re: Sum of torque
Post by: dieter on January 20, 2015, 12:47:13 AM
This is interesting, and in my special case: mind bending  :o

BR

Title: Re: Sum of torque
Post by: EOW on January 20, 2015, 08:54:37 AM
Hi Dieter,

Do you understand my idea ?

For create energy, the motor can be at the Pulley2. The motor need to give -3FRwt to the Pulley2, the Support2 lost -FRwt. The Pulley1 recover +6FRwt because the Pulley1 turns at 2w in the lab frame reference and the radius is 3R.

Have a good day
Title: Re: Sum of torque
Post by: EOW on January 21, 2015, 04:15:07 PM
The Support1 receives the force F and the Support2 receives the force -F. One support is helped to turn clockwise when the other support receives a counterclockwise torque. With gears (1, 2 and 3) it's possible to cancel the work of forces F and -F. Like that supports don't accelerate or decelerate. If the motor is put on the Pulley1 the energy is destroyed and if the motor is put on the Pulley2 the energy is created, no ?
Title: Re: Sum of torque
Post by: dieter on January 21, 2015, 10:38:52 PM
Hi EOW,

I'm referring to #130. Example: when you rotate the left purple wheel, the big pulley (fixed) will rotate the small pulley 3x. So the small pulley may not be fixed to the right purple wheel, otherwise the belt will be blocked.

I have to confess, I do not see where the excess energy comes in or out.

BR
Title: Re: Sum of torque
Post by: dieter on January 21, 2015, 10:55:56 PM
Hi EOW,

referring to #133, well, some people get upset by the terms "create + destroy energy "  ;)

Let me see if I understand it:  The pulleys will give F and -F to the supports. But, one support gets 3 lengths of 1/3 force and the other gets -1 length of 1 force. Any rotation of the pulleys would make the supports rotate in opposite direction, or with a speed ratio if 1:3 (And then the belt would fall offl). But the gears will force the supports to rotate at the same speed, in the same direction?
So it cannot rotate at all?
I guess I didn't understand it.  :)

BR

Title: Re: Sum of torque
Post by: EOW on January 21, 2015, 10:59:53 PM
Hi, Dieter,

In the post #130, the Pulley2 is not fixed. The Pulley2 turn of 360°, one round, because its angular velocity is 4w (lab reference), the Pulley1 turns at 2w (half round). I will draw another example.

For understand, put the Pulley1 and the Pulley2 in the same support, if the Pulley1 is the motor, the motor need to give -3FRwt and the support lost -3FRwt, the Pulley2 receives 4FRwt and the support receive 2FRwt the sum is at 0. Now, with 2 supports, the sum of the torque on Support1 and Support2 is 0. If the motor is the Pulley1, the support lack 2FRwt. If the motor is the Pulley2, the support don't lost -2FRwt.

Quote
But the gears will force the supports to rotate at the same speed, in the same direction?
Yes, it's not necessary to add gears for understand. But like the Support1 receives a torque T and the Support2 receives the torque -T, the sum is at 0, if I use gears +T-T = 0, like that supports turn always at w. For one support the torque change its sign in a round.

Quote
So it cannot rotate at all?
Support1 and Support2 turn at w. Pulley1 turns at 2w (lab reference), Pulley2 turns at 4w (lab reference).

Quote
The pulleys will give F and -F to the supports.
correct

I added an image with an angle of 45°, for look at A and B. The support turns of 45°(w), so A turns of 90° (2w), B turns of 180° (4w).
Title: Re: Sum of torque
Post by: dieter on January 21, 2015, 11:46:09 PM
Hi EOW,

ok, that makes sense now.

BR

Title: Re: Sum of torque
Post by: EOW on January 21, 2015, 11:59:25 PM
Are you ok with a torque at 0 for Support1 and Support2 if I use gears ?
Title: Re: Sum of torque
Post by: EOW on January 22, 2015, 09:43:23 AM
Hi Dieter,

I think d1=d2, always. The force F on the Support1 gives a clockwise torque T but the force -F gives a counterclockwise torque -T. |T|=|-T|. I drawn forces with the Pulley1 a motor.

Have a good day
Title: Re: Sum of torque
Post by: dieter on January 22, 2015, 08:47:44 PM
Hi EOW,

But the torque of the pulleys will affect the supports only when there is, example given, friction...?

Have a nice day too!  8)

BR
Title: Re: Sum of torque
Post by: EOW on January 22, 2015, 09:22:37 PM
Now, I drawn forces with one support like I spoke before (the motor is on the Pulley1, it's an example), look at the support it works at 2FRwt. It's possible to compare these 2 cases, one with one support and other with 2 supports. Pulleys1 is at 2w in 2 cases, Pulleys2 is at 4w in 2 cases, support is at w in 2 cases.

In the case with one support, the sum of energy is:

-3FRwt for the motor on the Pulley1
-3FRwt for the support, the motor is on the support is apply a counter torque
+2FRwt for the support, torque due to pulleys
+4FRwt for the Pulley2

The sum is at 0

In the case with 2 supports, the sum of energy is:

-3FRwt for the motor on the Pulley1
-3FRwt for the support, the motor is on the support is apply a counter torque
+4FRwt for the Pulley2

The sum is at -2FRwt

-----------------------------------------------------------------------------------------------------------

If the motor is at the Pulley2, the motor need to give -3FRwt, true ?
The Support2 receives -FRwt because the motor apply a torque, ok ?
The Pulley1 turns at 2w and there is a radius 3R, so the energy recover is 6FRwt
Title: Re: Sum of torque
Post by: dieter on January 22, 2015, 10:37:14 PM
I was asking about friction because, as fas as I see, when both pulleys can rotate freely, then the support won't receive any torque or force at all. (?)

Please excuse me, I may seem really stupid here.

BR

Title: Re: Sum of torque
Post by: EOW on January 22, 2015, 10:45:41 PM
You don't see the torque with 2R ? F1 and F2 give a torque 2FR, no ? this is not a torque from friction in the axis. Forces come from the belt and are reported to the axis of each pulley.

And you're not seem stupid, it's not easy to understand ideas from others, and you, you try, it's friendly :)
Title: Re: Sum of torque
Post by: dieter on January 23, 2015, 06:05:52 PM
I really don't see it, no matter how hard I try..,  :o

Maybe I should just build it. Maybe as a side project.

BR

Title: Re: Sum of torque
Post by: EOW on January 23, 2015, 06:19:06 PM
Hi Dieter,

With one support: F1 and F2 in the image, no ? The big pulley force the belt to turn clockwise, and drive the small pulley, there is the force Fb on the belt, this force is put on the axis : F2. The belt receive the force Fa, and this force is reported to the axis of the big pulley: F1. F1 and F2 gives a torque with the value 2FR (with |F| = |F1| = |F2|). I think with one support it's logical to have 2FRwt on the support for have the sum of energy at 0, but with 2 supports I don't find the torque.

With 2 supports, I tested today with a simulation for look at the angular velocities and all seems correct.

If you have gears and pulleys, you can test it (I order a package of gears for test).
Title: Re: Sum of torque
Post by: dieter on January 23, 2015, 07:16:24 PM
I would have to make them, with wood and saw, and a couple of inline-skate ballbearings, probably I go and get some plywood this evening :)

Can your simulation software export animations? That would help a lot. Also, looking foward to see your build!

BR

Title: Re: Sum of torque
Post by: EOW on January 23, 2015, 11:08:29 PM
Like I saw my simulation, the angular velocities must not be w, 2w and 4w but w, 2w and 6w. With 2 supports, the motor and the Support1 need -6FRwt, the Pulley2 recover 6FRwt. But with one support, the motor and the support need -6FRwt and the Pulley2 recover 6FRwt, but there is the torque on the support, that torque gives the energy 2FRwt. It's necessary to check the sliding of the belt with w (support),2w(Pulley1) and 6w (Pulley2), if the sliding is 0 this could say the case with one support must give an extra energy to the support.
Title: Re: Sum of torque
Post by: EOW on January 23, 2015, 11:53:34 PM
For explain the forces F1 and F2 in my last example. Take another simple example, I renamed forces, with one pulley. Apply a force F1 to the rope, that force is transmitted to F2 along the rope. F2 want to move the pulley to the left, so F2 apply a force F3 to the axis, the fixed axis reply and give the force F4. F1/F4 is a torque that turn the pulley but the support receive the force F3. With 2 pulleys there are 2 forces, and if the size of the pulley is not the same there is a torque. Do you understand ?
Title: Re: Sum of torque
Post by: EOW on January 24, 2015, 10:53:29 AM
I made a mistake with the force with one support. If I brake the Pulley2 on the support, you're right there is no torque (first image shows all forces, grey color -> to the pulley, magenta color -> to the support and black colo -> to the belt).

Now, second image, I don't brake the pulley2, imagine it with a mass. I accelerate more and more the Pulley2, the energy is in the kinetic energy in the pulley. The sum of forces are like I drawn, no ? And the support decelerate because it receive a counterclockwise torque from force F1 / F2, no ?
Title: Re: Sum of torque
Post by: dieter on January 25, 2015, 01:24:47 AM
Hi EOW,

I would say yes, as long as the pulley is accellerating, the inertia of the mass will give an opposite torque to the support. I guess this happens due to an imbalance of centrifugal force on the pulley shaft. Not sure tho.

BR

Title: Re: Sum of torque
Post by: EOW on January 25, 2015, 01:26:54 PM
I think I forgot the force -F6 on the big pulley. So:

At start, the support turns clockwise at w, the big pulley turns clockwise at 2w and the small pulley turns clockwise at 6w. The belt don't slide. Pulleys accelerate more and more (no brake), the energy is recover after (kinetic energy). An external motor keep constant the angular velocity of the support (need energy because the support receives a torque F3/F4).

At a time:

The motor needs -3FRwt, the motor gives F1 and F2 to the big pulley, the motor gives F3 and F4 to the support
The support receives -3FRwt (F3 and F4)
The belt gives the force F5 to the big pulley, this gives the force F6 to the support and -F6 to the big pulley. The belt gives the force F7 to the small pulley
The support receives the torque F6/F8, and the energy 2FRwt

The sum is 2FRwt, no ?
Title: Re: Sum of torque
Post by: dieter on January 25, 2015, 11:31:56 PM
Hi EOW,

Again, this has knocked out my brain  :o ... I guess you understand this much better than I. Personally, I would rather construct it in 2 hours than to (fruitlessly) think about it for 4 hours. In fact I brought home some plywood, so maybe I will. I just don't know how to provide the pulleys with power when the support rotates.

BR

Title: Re: Sum of torque
Post by: EOW on January 26, 2015, 02:21:29 PM
Maybe it's possible to test the torque on the support without a motor. The big pulley has a mass and it's possible to turn it at 2w before to turn the support, the big pulley can drive the small pulley with the inertia. With no friction to the axes of rotation => no torque must be appear to the support, if a torque appear that could say the system can give en extra energy. The small pulley is braking on the support at the radius R, for me the sum of energy is:

The big pulley lost -2w*3R*F*t = -6FRwt
The friction win 5w*F*R*t = 5FRwt
The support win w*F*3R*t = 3FRwt

The sum is at 2FRwt. The small pulley has a mass as lower as possible.

I think the key is the torque on the support.
Title: Re: Sum of torque
Post by: dieter on January 27, 2015, 03:36:36 AM
Maybe you're right. But then there is still an other problem in a practical implementation: There will be friction on the pulley shafts. The skate bearings are made for about 20 kg, they have a lot of friction. It can he reduced by cooking out the grease, but still... so we wouldn't know if the torque is caused by the friction.

Bearings from CD players would be better, but I got only one right now... (those with the "clickable" cd shaft, not the normal PC cdrom drives).

Maybe I'll find one more in my pile if stuff :)

BR

Title: Re: Sum of torque
Post by: EOW on February 06, 2015, 02:50:44 PM
For test easily, it's possible to put all things on the support and filmed for watch the energy input/output. The DC motor drives the big pulley, the small pulley drive a DC generator, the generator can give energy to a resistor, all instruments are on the support, need only 4 instruments. If the generator is fixed on the support, there is no torque on the support. I calculate the energy from the system and it's not a factor of the angular velocity of the motor but factor of the angular velocity of the support:

E=2FRwt

The generator must recover energy from the motor: the force F

If the support turn slowly the energy recover is low

With :

R=0.1 m
F=0.5 N
w=200 RPM
The power is 1W

Title: Re: Sum of torque
Post by: dieter on February 06, 2015, 09:06:44 PM
Hi EOW,

did I understand this correctly:
the motor turns the big pulley, the small pulley recovers the energy, and the support rotates due to inertia?

Did you get the gears yet?

BR

Title: Re: Sum of torque
Post by: EOW on February 07, 2015, 07:51:32 AM
Quote
the motor turns the big pulley, the small pulley recovers the energy, and the support rotates due to inertia?
yes, there is no torque on the support IF pulleys don't accelerate.

Quote
Did you get the gears yet?
yes but the energy recover is lower (there is friction and efficiency of generator is not 1 maybe 0.8 or lower) beause w is small and the motor turn very quickly => the force is small, I need a low speed motor and a support that turn at w high enough. I gave an example, the power recover is smal with a small force.
Title: Re: Sum of torque
Post by: dieter on February 07, 2015, 08:22:09 PM
There may be some low RPM dc motors using built in gears, like eg. a grill motor. I also got one from the mentioned laserprinter, although more of a motor with attached gear set. However defective laserprinters are a great source for gears, I got mine for 5 bucks, was even still working, but I wasn't interested in printing...

Other sources may be Microwave oven plate motor (dc?) or scanner motor, they all need to be low RPM for what they do.

BR

Title: Re: Sum of torque
Post by: EOW on February 08, 2015, 11:07:59 AM
Quote
There may be some low RPM dc motors using built in gears
with gears inside, there is a counterclockwise torque on the support, the motor needs to rotates slowly or the angular velocity of the support must be high.

With:

w (support) = 157 rd/s (1500RPM)
w (motor) = 157*2rd/s and P=300 W, the torque is 1.91 Nm => the force is 19 N if 3R = 0.1 m, R=0.1/3

The generator turns at 6w but the stator of the generator turns at w, so the difference is 5w. The generator can give 5FRwt.

The power won is 2FRw = 200 W

If the efficiency of the motor and the generator is 0.8, the global efficiency is 5/3*0.8*0.8 = 1.06 so it is necessary to use brushless motor (some can have an effciency of 0.9) and without take in account of the efficiency of the electronic card and the belt... maybe 0.95 for the belt and for the electronic card, the global efficiency become 5/3*0.8*0.8*0.95*0.95=0.96. With a motor/generator at 0.9 this could be: 1.21

Maybe the best is to test a basic dc motor/generator and look the efficiency without rotation of the support and turn more and more the support and look at the efficiency, the efficiency must increase more and more like the angular velocity of the support. The small pulley can be smaller, 3R for the bigger and 0.2R for the smaller, the efficiency will be at 29/15*0.8*0.8*0.95*0.95=1.11
Title: Re: Sum of torque
Post by: dieter on February 08, 2015, 05:30:49 PM
There are also losses in getting the required current to the pulley's motor. As the support is rotating, you maybe need to use brushes. Or you could try inductive coupling if the support has near zero mechanical losses: an additional disc on the support's shaft, maybe 1-2 foot away to prevent magnetic interferences with the dc motor etc. Although the Lorentz force is the only cost in frictionless inductive coupling (other than friction etc.), the implementation would complicate things.

Furthermore, you also got to extract the power from the generator to measure it ...

If the COP is over 1.0 then it may be possible to have everything on the support...

BR

Title: Re: Sum of torque
Post by: EOW on February 08, 2015, 06:04:04 PM
Quote
Furthermore, you also got to extract the power from the generator to measure it ...
It's easy with brushes and a plate with paths of copper, it's possible to use 2 brushes for input and 2 brushes for output. The main problem is to find a motor with a good efficiency.
Title: Re: Sum of torque
Post by: dieter on February 08, 2015, 09:28:06 PM
Hard enough to find high efficiency dc motors, let alone low RPM without gears... ebay has only high RPM, down to 3600 RPM, as far as I see.

BR

Title: Re: Sum of torque
Post by: EOW on February 09, 2015, 04:20:31 PM
Hi Dieter,

Maybe I can use a stepping motor with a big difference of radius like that I can use a dc motor for the generator. I will look for the efficiency of the stepping motor.

Title: Re: Sum of torque
Post by: DreamThinkBuild on February 09, 2015, 07:10:50 PM
Hi EOW,

Here is a patent that you may find interesting, it's in French but he gives examples in the drawings on how to match input motor, to gear reduction, to generator rating. He's using servo motors.
Title: Re: Sum of torque
Post by: EOW on February 09, 2015, 11:13:05 PM
Hi DreamThinkBuild,

I speak french, so it's easy for me. Thanks but there is an error in the patent and there is no rotationnal support.
Title: Re: Sum of torque
Post by: dieter on February 10, 2015, 04:14:10 AM
EOW,

Steppermotors is a good idea to get a very low RPM. But usually they are driven by synthesized Sine waves / Phases. If you can control these shapes PWM wise, then it may be possible to reduce the power consumption.

Otherwise, my experience with stepper motors tells me, the slower they are rotating, the hotter they get (because of unneccessary  DC currents), and heat means loss here.

Also, the rotation is rather vibrating, not very smooth. Not sure if this is a problem.

The SMC800 control card does in fact allow the shaping of the 2 phases, but only in a 2 bit resolution. Like:  0,1,2,3,2,1,0.. not really a smooth sine wave. And that is used to control -12 to +12 volts. I don't know if this card is still manufactured. There may be other, similar ones.

BR

Title: Re: Sum of torque
Post by: EOW on February 10, 2015, 01:51:36 PM
I watched a lot of specifications of stepper motors and I think it's not a good solution. The easiest it's to build a small device and turn it at 750 RPM or more. It's not necessary to have an efficiency greater than 1, if there is no torque on the support (it can be test with w=0 for the support) and if the efficiency is greater with the support in rotation I think it could be ok. For example, if I have an efficiency of 0.2 for the device when the support don't turn, the efficiency must increase of 5/3*0.2=0.33 (if ratio of bigpulley/smallpulley = 3/1 and if the motor turns at 750RPM). Like that a simple test can be done with a drill and 2 dc motor (any efficiency could be ok). The efficiency must increase like the angular velocity of the support (linear law).

The general formula for the extra power is :

Power = F(R1w - R2w)

with:

R1 the radius of the big pulley
R2 the radius of the small pulley
F the force on the belt, it's Pm/(R1*wm), with wm the angular velocity of the motor and Pm the power used by the motor
w the angular velocity of the support

So the extra power is:

Power = Pm*(R1w - R2w) / (R1 * wm)

The efficiency of the device is :

Eff = (Power * ηg*ηb + Pm*ηg*ηb*ηm) / Pm = (Pm*(R1w - R2w) / (R1 * wm) * ηg*ηb + Pg*ηg*ηb*ηm) / Pm = (R1w - R2w) / (R1 * wm) *ηg*ηb + ηg*ηb*ηm

With:

Pm the power consumed by the motor
Pg the power recover by the generator
ηb the efficiency of the belt
ηm the efficiency of the motor
ηg the efficiency of the generator

I don't take in account friction and lost by Joule effect in electric circuit.

For example, with Pm=20W, w=750RPM, ηm=0.6, ηg=0.6, ηb=0.95, R1=3, R2=0.5, wm=4000 RPM, the efficiency is 0.43, without the support the efficiency is 0.34. With a motor with 750 RPM, the efficiency move up to 0.81
If ηm = ηg = ηb = 1, then the efficiency is 1.15 with 4000 RPM and 1.83 with 750 RPM for the motor
Title: Re: Sum of torque
Post by: EOW on February 11, 2015, 09:59:23 AM
With the same support, the small pulley don't turn at 6w but at 4w, so the sum of energy is 0.
Title: Re: Sum of torque
Post by: dieter on February 11, 2015, 06:37:19 PM
Sum of energy is zero? You mean energy gain is zero, llike in efficiency = 1.0 - losses?

BR

Title: Re: Sum of torque
Post by: EOW on February 12, 2015, 01:16:56 PM
Yes, efficiency=1 (I recover that I lost).

I study this new idea. Without the support3, the efficiency is 1. I will calculate all energies for find the global efficiency.

The motor needs -9FRwt
Torque F6/F9 gives 2R*2w*Ft=4FRwt
Torque F5/F8 gives 4R*2w*Ft=8FRwt
The friction (brake from the pulley2 to support3) gives 3FRwt

The sum is +6FRwt

Title: Re: Sum of torque
Post by: dieter on February 12, 2015, 11:52:52 PM
Looks good, as far as I understand it.

BTW. I'm off for a break

BR
Title: Re: Sum of torque
Post by: EOW on February 13, 2015, 09:21:06 AM
Ok. I drawn anothers views at different positions. The torque from F7/F8 is not constant it depends of the position, so forth image shows the system lost energy but it's possible to change the position of the friction and have a clockwise torque to the support3 (fifth image). The sum of torque from F3 and F4 to support1 and support2 is 0 like I show before.

with

angular velocities (lab ref):
Support1 = 0
Support2 = 0
Support3 = +2w
Pulley1 = +4w
Pulley2 = +12w

The motor need -12FRwt
The friction gives +2FRwt
The Support3 gives +12FRwt
The Support2 need 0
The axes of pulleys gives +4FRwt

The sum is at +6FRwt
Title: Re: Sum of torque
Post by: EOW on February 14, 2015, 09:57:46 AM
With one motor and one brake on a support. The motor drives the brakes with a crossed belt. The motor turns clockwise at +10w around itself. The brake turns counterclockwise around itself at -10w. The support turns clockwise at +w. So, the motor turns at +11w in lab ref and the brake turns at -9w in lab ref. The motor needs -11FRwt, the brake gives +11FRwt (the brake turns at -10w and the support turns at +w, the difference is 11w), torque F1/F2 gives +2FRwt to the support, torque F3/F5 needs -FRwt to the support, the sum is at +FRwt. All angular velocities are constant. I guess no mass for pulleys and belt.
Title: Re: Sum of torque
Post by: EOW on February 15, 2015, 01:06:12 PM
I can replace the friction with an electric generator for example. I guess the efficiency of the motor, generator, belt, etc is 1 (no losses). The motor gives F1 and F2 to the Pulley1. The motor gives F3 and F4 to the support. The belt gives the force F5 to the Pulley2. The belt gives the force F6 to the Pulley1. F5 gives F7 to the axis of the Pulley2. F6 gives F8 to the axis of the Pulley1.The generator gives F9 and F10 to the support.

I noted R the radius of the Pulley1 and the Pulley2.

With |2F1|=|2F2|=|2F3|=|2F4|=|F5|=|F6|=|F7|=|F8|=|2F9|=|2F10|=|F|

The motor need to give -10FRwt
The support reveices -FRwt from F2 and F3
The support receives -FRwt from F9 and F10
The support receives +2FRwt from F7 and F8
The generator receives +11FRwt because the difference between the rotor and the stator is 11w

The sum of energy is +FRwt

All angular velocities are constant. The sum of torque on the support is 0, so its angular velocity is constant. The energy lost from the motor is -10FRwt and the generator can recover +11FRwt.
Title: Re: Sum of torque
Post by: EOW on February 16, 2015, 01:59:48 PM
The stator of the motor is fixed on the support so it turns clockwise at +w. The stator of the generator is fixed on the support so it turns clockwise at +w.

The rotor of the motor turns clockwise at +10w relatively from the stator, so the rotor turns clockwise at +11w (lab ref), the power needed for the motor is -10FRw and the support needs -FRw. The rotor of the generator turns counterclockwise at -11w but the stator turns clockwise at +w so the difference is 12w, but the support receives a torque -FR and need the power FRw.

Forces on axes of the pulleys give a power 2FRw to the support and with -2FR from 2 startors, the sum of torque on the support is 0.

So, the motor needs -10FRw and the generator gives 12FRw.
Title: Re: Sum of torque
Post by: EOW on February 19, 2015, 03:21:20 PM
A closed container has a pipe enrolled on a grey disk. For enter the pipe inside the container I need to give the energy E1. All the device is at a linear velocity V, with V = constant. When the pipe move out the container I recover the energy E1. The container has 2 forces F2 and F4 on it, this forces give an energy E2 = (F2+F4)*V*t. All concentric forces on the pipe cancel themselves because the pipe has a mass and with rotation there is centrifugal forces. P2 > P1, and imagine the device with P1=0 in theory. There is no gravity.
Title: Re: Sum of torque
Post by: EOW on February 20, 2015, 07:40:00 AM
In fact, all forces from pressure*surfaces cancel themselve => the container has no net force on it. The force come from the centrifugal forces only. The gas can't give a contrary force, I think the gas give a force that add the net force of the centrifugal forces.
Title: Re: Sum of torque
Post by: EOW on May 31, 2015, 03:01:52 PM
The idea is to use the device I drawn in the message #40. But here I use pressure. No gravity. No modification of the volumes. I suppose no friction. It's a 2D but it's possible to imagine in 3D replace circles with spheres. Outside the pressure is constant at P (atmospheric pressure for example at 1 bar). A basic device is composed by an arm and a circle. The black arm can turn around the blue axis. The circle is fixed to the arm. The circle don't turn around itself. I drawn 2 basic devices but it possible to have N devices, in the third image I drawn 16 basic devices. Circles are smart, walls between circles can be removed : red in the drawing. I suppress wall like that a part of each circle receives a torque on it. Like all circles don't turn around the same center, the torque is not the same.

First drawing: the device turns from time=0 to time=2.
Second drawing: the device turns from time=0 to time=0.5

It's possible to look at torque on each arm, like the distance d1 is greater than d2, the torque is higher counterclockwise from time=0 to time=1.

At time=1, the pressure inside circles must changed. This don't need energy. It possible to think with a device with N basic devices (arm+circle) and X circles are time=0, X circles are at time=1, X circles are at time=2, etc. For change the pressure inside circle it's easy: take the pressure inside one and replace from another cirche that is in another time (or position). Like olumes are constant, change pressure don't need energy in theory, a little in practise.

The third drawing shows a device with 16 circles, 2 circles at each time, all 16 amrs turn at the same angular velocity. Two circles work together. At the exact position: time=1 time=4 time=6 and time=8 there is no work. At time=1it's necessary to put P/2 inside circles so change with the circles at time=4.

Inside circles the pressure is 2P from time=8 to time=1, from time=4 to time=6.
Inside circles the pressure is P/2 from time=1 to time=4, from time=6 to time=8.
Title: Re: Sum of torque
Post by: EOW on June 01, 2015, 09:08:39 AM
First image: for look the sum of torque from time=0 to time=1.  There is no torque if radius are the same ! need one small circle and one bigger. Look at third image.

Second image:  look of the rotation of the gaskets. Even from time=0 to time=1, the device have a net torque on it and gaskets can't compensate it.

For change the pressure inside the circles it's possible :

a/ to exchange physically circles with a device with 16 circles (low angular velocity of the device)
b/ to use an external device, this device will need an energy but this energy will be transform in heating of gas

I explain with a gas but it's ok with a liquid too (less losses for change the pressure inside circles).
Title: Re: Sum of torque
Post by: EOW on June 01, 2015, 04:27:30 PM
With gas: there is no torque if circles have the same diameter. The torque is not the same with 2 diameters but the angular velocity must be different for one circle. So the sum of energy is conserved.

Now, if I put liquid inside circles, with same diameter, centrifugal forces are not the same inside one circle to another because the centers of rotation are not the same. The torque must be different. Maybe the liquid move between 2 circles. The velocity in a fluid change like the speed of sound in the fluid, the pressure can't be faster than this. So if the device turn at high velocity, and like the red wall change all the time with the angular position, the pressure cannot be the same and the sum of torque is not 0.
Title: Re: Sum of torque
Post by: lumen on June 01, 2015, 07:40:57 PM
I think the design could be made less complicated by reducing the working principal to just two working components.
1: centrifugal force
2: Roberval action
I have done some testing on the principal and have not found any cause for the device to fail so I am now building a test model.
There must be a condition to cause the device to fail but it's simply not apparent.

Title: Re: Sum of torque
Post by: EOW on June 01, 2015, 08:35:25 PM
What is Roberval action ? The balance ? there is no gravity here.

Yes, sure for a non continous machine it's possible to test with just 2 circles. Better efficiency with several circles. But do you tested with liquid or gas ? It can be tested with a "velocity" higher than the speed of sound. The "velocity" is the velocity of the red line. Where are you discussed about that ? The speed of sound in water is 1500m/s and for test, it's not easy. I drawn lines of equal pressure from another circle when the door is open. The force F from the red line don't give the same torque, it's logical. But, this additionnal torque don't work the same in one turn (of 2 circles) and it works a small time, "waves" (curved lines black, red, magenta and blue) move at 1500 m/s
Title: Re: Sum of torque
Post by: lumen on June 01, 2015, 09:31:21 PM
Roberval action is simply the mechanics that maintains the orientation as the chambers rotate around the circle.
If they were aligned to north before starting, they would remain pointing north while rotating.
In your drawing 1-1 is pointing the same direction as 2-2 and 3-3 then 4-4 , 5-5, 6-6. They point the same direction as the disk rotates.

This is what forces the gas (or liquid) to move from one chamber to another by centrifugal force.
There is always an apparent gain in energy even at slow speeds.

Title: Re: Sum of torque
Post by: EOW on June 01, 2015, 09:40:11 PM

Sure, it's work at any velocity but:

1/ for test with sensors, you need to detect something that run at 1500m/s
2/ with 2 circles, the device like I drawn don't work, in one turn the sum of work is 0, the door must be closed some steps and open others steps

With a high velocity, the waves  don't have time to stabilize them, so it's better. For me it's the delay that can give energy and it's logical because the angular velocity of circles can be very high. If the device lost energy when the liquid moves (water is compressible so a few liquid will move), this energy is near constant, but the angular velocity can be 10 times or 100 times higher.

The centrifugal force is mv²/R, so with a big v and a big R, the centrifugal force is not high, but the distance that circle run is high, so it must be ok. Tell me more about your tests, on a software ?

Title: Re: Sum of torque
Post by: lumen on June 02, 2015, 12:15:10 AM
Your plan to use a Gas and extract energy is not exactly what I was doing but there are many ways to extract energy from the two basic systems combined.

I plan to generate electrical current directly from centrifugal force using a roberval system on a rotating disk.
The tests I have been doing are to find out if the torque from the centrifugal force can in some unknown way apply back against the rotation of the disk generating the centrifugal force.

Torque on a roberval device cannot apply in any way to cause rotation of the disk but centrifugal force is radial and some things change because of the vector of the force has vector differences over the working area involved.

Regardless of these facts there is a way to correct for that problem so again I'm left searching for the cause of failure in such a device.

It will be interesting to know the exact cause of failure for such a device because it appears so elusive.

Title: Re: Sum of torque
Post by: EOW on June 02, 2015, 01:16:48 AM
I prefer a liquid it must be more efficiency. Let the door open and like the pressure moves all the time this will give a torque on the device, like that ? Yes, but I'm not sure of:

1/ water will move because it is compressible, even a small volume of water, this amount can't be ignored, but the energy seems very low
2/ an object rotates => no torque, so this is the delay of pressure to be stabilized that can give a torque, here with the door always open, the pressure change all the time. Look at positions: it's necessary to close door and reopened because the torque is counterclockwise and after clockwise, are you agree ?

I drawn the device, and I think there is one wave inside the upper circle and another wave in the lower circle. In the upper the pressure increases, in the lower the pressure decreases, a part of time, and this must give forces F1 to F4. Length of arm can be higher. The material must absord the wave for prevent round trips of waves.

have you tested on a software ? Maybe if you draw something for understand why you want to recover energy from this device I could understand.

Title: Re: Sum of torque
Post by: lumen on June 02, 2015, 02:31:28 AM
Ok, you are saying that the release of the pressure into the second chamber causes movement and then the centrifugal force would cause the next compression cycle and the release would be delayed until a time when the release would again act to increase the speed. The delay of the gas may cause a shift in weight that works against the rotation like shifting your weight on a swing to slow yourself down.

Would this be easier if the centrifugal force was used instead to rotate a generator and some of the power was used to increase the speed of the disk.
At some point there would be excess power that could be directly used for other purposes and the design would be achieving the same goal using the same forces but would be easier to design and could operate at lower speeds.

Title: Re: Sum of torque
Post by: EOW on June 02, 2015, 09:48:08 AM
There is no gravity, so no weight. For recover energy, maybe turn the device at the max angular velocity possible to do, and recover energy at constant velocity is better. With a constant velocity the device must be more stable.

Maybe it's possible to use a piezoelectric vibrator for create the waves and test with only one circle. Step2: the valve at left, move out the valve quickly. Step3: return of 180° the device around itself, now the valve is at right. Step4: move in the valve. Step1: return the device around itself of 180°, now the valve is at left. Repeat the cycle.

And you, how do you want to recover energy from this device ? I don't understood your Roberval system
Title: Re: Sum of torque
Post by: EOW on July 19, 2015, 05:00:57 PM
I don't use gravity here. All volumes are constant. I use only the red shape, I drawn black circle for show the center where I put the springs.

The pressure come from small balls inside the red 'S' shape. The balls are attrack to the center or repuls from it. This gives a pressure p, 2p, 3p on the side where there are the forces F1 and F2. The balls are vey small.

I give the force F1, this need the work W1 and I recover the work W2 from F2. The move like a snake. But for have a continious mouvement, it's necessary to add another part of circle and give the pressure. Put a pressure inside a shape don't need energy in theory.

Title: Re: Sum of torque
Post by: sm0ky2 on July 20, 2015, 03:12:07 AM
the balls do not behave exactly as a fluid, but much slower. pressure does not have to be constant through the length of the tube.
When pressure varies in one part from the springs, more balls move out of the "higher pressure" area quickly.
this lowers the pressure by a proportionate amount in the part of the tube just before the springs.
meaning, to maintain the original pressure in the rest of the tube, you have to increase the input of balls
to bring the pressure back up by an amount of balls squeezed out by the springs.
the torque comes at a cost.

Title: Re: Sum of torque
Post by: EOW on July 20, 2015, 07:41:39 AM
Right the number of balls of 3p must increase compared to the number of balls of p. Springs are used only for balls that are inside the curved shapes, not in the straight red shape. So the pressure adjust itself in the part where there is no spring: red straigth shape and especially when the curve shape becomes straight, or straight becomes curved. When a space increases between balls of 3p why spring necessary gives the work ? Why it's not a closer ball that can move ?

Title: Re: Sum of torque
Post by: EOW on July 22, 2015, 09:53:38 AM
The idea is to increase the potential energy of this device. I increase the angular velocity of the torus more and more like that I can keep the relative position of the Object O always. I gave the position after 45°. Inside the Object O there is a lot of small balls, the pressure come from spring attached directly on the Object O. I create pressure on 2 sides of the Object O. Outside there is no pressure  (or near 0). The volume of the Object O is constant, so I don't lost the potential energy of the springs. The torus don't receive a torque from the Object O because it is composed of circle lines. And the sum of forces F1+F2 (in vector) at the center C2 is zero BECAUSE the pressure inside the Object O is adjust for have F1=F2 in value. The center C2 is fixed to the torus. The center C1 is fixed on the ground.

It is very important to have partial circle for 2 ends of Object O, like that there is only the torque from F1 and F2.

It is very important too, to have |F1|=|F2| in value, for that the pressure inside the object must be adjust.

I give the energy for rotate more and more the torus and the Object O on it but the Object O has more and more kinetic energy because it turns more and more too around the center C2. Give for receive :)

Like it's very important to have F1=F2, it's possible to use gravity but the device will be more complex. Pressure from balls are easier to do. The balls are smaller in reality.

Anf if possible to have F1>F2 like that there is a clockwise torque on the Object O and on the torus too.

The Object O keep the same angle relatively to the torus, so if I'm on the torus I see always the same angle from the torus and the Object O.
Title: Re: Sum of torque
Post by: EOW on July 22, 2015, 10:15:36 PM
I give an image for show 8 positions of the device. It's possible to show the angle between the Oject O and the torus. Noted that the angular velocity of the torus increases like the angular velocity of the Oject O in the same time. It's very important to keep these two angular velocities exactly the same. The angular velocities I gave are in the laboratory reference (not in the torus for example).

I drawn big balls but they are really small in reality, it's important for prevent another torque on the Object O.

I drawn the device with balls above the torus, but it's a 2D drawing so I added another image with the torus inside the Object O, the torus pass throught the Object O. (S4.png)

I shows on s4.png the difference of the distances d1 and d2. Like the distances are not the same I need to adjust the pressure for have F1=F2 with balls.

I don't drawn the springs that attrack the balls but the springs are fixed on the Object O.

The energy is recover after. This device increases the potential energy. The Object O turns more and more quickly for free around its center of rotation C2. But I need to give an energy for rotates the torus, this energy can be recover later. If F1 can be higher than F2, in this case the torus has a clockwise torque too and it's possible to recover the energy in the same time. Noted that the angular velocity MUST increase more and more.

Title: Re: Sum of torque
Post by: EOW on July 23, 2015, 10:57:13 AM
Pressure with balls is like water and gravity. The pressure is more and more higher at bottom, here the pressure at point p1 is greater than at point p0, so even d2>d1 it's possible to have F1=F2.
Balls are very small compared to the shape, in the contrary another small torque appear on the Object O. If the Object is 1 meter of size balls can be like 1cm. I drawn springs but it can be another technologie like magnets.

Note that even I can have F1>F2 the energy from the torus can't be recover in the same time, it's necessary to increase more and more the angular velocity. And at a time, recover all the energy from the torus and from the rotation of the Object O around its center of rotation C2.
Title: Re: Sum of torque
Post by: EOW on July 24, 2015, 07:47:08 AM
Or for have F1=F2 in value I can use centrifugal force. I put a liquid inside the Object O. Like the Object O keep its relative position the centrifugal forces are always the same. Or use 2 liquids with different density for have F1=F2 and no other force on the center C2.

Or in one part of the Object 1 put a gas and in other part small balls (or a liquid). The goal is to have F1=F2 or F1>F2.

If I attrack balls from the torus I can have F1=F2 if I use the mean of the pressure. At bottom I will have the same pressure 10 but I can play with the start of the pressure, in one part I start with 0 to 10 and in other part I start from 8 to 10, like that I can have maybe F1>F2. The sum of forces on C2 from the springs is 0, I change just the mean not the final pressure. Balls must be compressible a littlefor transmit pressure in all directions.
Title: Re: Sum of torque
Post by: EOW on July 25, 2015, 01:43:34 AM
Maybe with the attraction from springs like that. I can change the shape for find the good forces: the force on C2 is like the yellow force. Here I have 2 areas with 2 different gravities.

F1, F2, F3 come from the springs, note that like there is a curve (torus) the forces from springs are lower than the forces from pressure so I need to add F7, F8 and F9
F4, F5, F6 come from the forces of pressure
F7, F8, F9 come from the forces of pressure, gravity2 is small compared to gravity1

Dotted arrows are the reported forces on C2 or for construct the sum of forces. The torus turns counterclockwise. Don't forget, the torus accelerates more and more for follow the acceleration of the Object. The angular position between the Object and the torus is ALWAYS the same.
Title: Re: Sum of torque
Post by: EOW on July 26, 2015, 05:54:02 AM
The torus and the square object turn counterclockwise. The torus turns around C1. C1 is fixed to the ground. The square object turns around C2. C2 is fixed on the torus. The square object if free to turn on the torus. Consider the mass of the springs and the mass of the balls like zero, like that there is no problem with centrifugal forces, or like the position of each ball don't move in reference of the torus it's possible to cancel with an external arm for each ball the centrifugal force.

First image: the square object without the torus for understand how forces are on it. The are two areas, first Attraction1 where the pressure is higher than the other area Attraction2. The springs attrack more and more and give more and more pressure at "bottom" (bottom of the image). The forces F1, F2, F3 com from the springs. The forces F4, F5, F6 come from the pressure of the first area. The forces F7, F8, F9 come from the pressure of the second area. The sum of forces F1+F2+F3 is lower than the sum F4+F5+F6 because the shape of the torus is curved. I drawn small red forces on springs to look how springs attrack, I drawn only 3 springs but there is one spring for one ball. And the balls are very small.

Second image: it's the same but now with the torus. Look at the sum of forces F1+F2+F3+F4+F5+F6+F7+F8+F9 on the center C2: the yellow force. The yellow force don't give a torque on the torus. So I can accelerate like I want the torus  and after recover all this energy. Even, I can set F7+F8+F9 higher for have a counterclockwise torque on it, I lost a part of a torque on the square object but I win a bigger torque because the distance with C1 is higher.

Third image: I drawn the device for 4 positions. The angle between the square object and the torus is always the same. The square object want to turn counterclockwise, the square object accelerate more and more BUT I accelerate with an external motor the torus like that the square object keep its position. The springs don't move inside the square object so the potential energy of the springs is contant. I noted the letter 'a' for look at the position.

I can't recover the energy in the same time, I need to accelerate the torus and recover the energy after. I recover the potential energy from the square object only, it turns more and more.

The device is unstable, it's necessary to control very well the angular acceleration of the torus. And the center C2 must be with friction as lower as possible.

Title: Re: Sum of torque
Post by: EOW on July 28, 2015, 06:58:19 AM
I try to explain like I drawn the forces:

Image1: Water with gravity, there are force from pressure at left and at right and forces at bottom (the weight). I did not drawn all forces, just the first, the middle and the last, because there are a lot of forces from pressure.
Image2: I replace gravity+water with springs+balls, like that I have the same forces from pressure at right and at left, sure there is no "weight" F1+F2+F3 cancelled by -F1-F2-F3. I use compressible balls or I change the arrangment like that I have a pressure on the side walls. Springs attrack exactly like gravity or with another law. I did not drawn all springs, but there is one spring for each ball.
Image3: The forces on the square object only (not forces on the torus). There are 2 differents areas where springs don't attrack with the same force, like 2 different gravities. I have forces from pressure at left and at right. I have only the up forces from the springs. I don't have the down forces because there are on the torus.
Image4: Now I drawn the forces on the torus too. There only a point of connection between the torus and the square object, it is the center C2. So, the sum of forces on the square object will be on the center C2. The sum of forces is drawn by the yellow force. The goal is to cancel the torque on the torus, and have the yellow force like I drawn. F4+F5+F6>F1+F2+F3 because there is a curve so it's necessary to have the forces F7+F8+F9.
Image5: Details of forces
Image6: The device
Image7: The device at 4 positions, look at the point 'a' it rotates like the torus. The angular velocity of the torus is exactly the same than the square object.

There is a torque on the square object so it will accelerate more and more, if you let the square object alone on C2 the device will turn a little and springs will lost their potential energy. But if the square object is place on the torus, and it the torus is accelerate more and more EXACTLY like the square object in this case the angle between the square object and the torus is always the same. There is no torque on the torus, look at the yellow force so I can accelerate it without lost an energy, I can recover the energy I give to the torus later.

Cycle: just accelerate the torus for have the same angular velocity of the square object. It necessary to accelerate more and more the torus and never stop until you can.
Recover energy from the rotation of the square object.

I think it's possible to have a better efficiency with a torque on the torus from F7+F8+F9 because when F7+F8+F9 increase it don't decreases the forces F1+F2+F3 but only F4+F5+F6, I lost a part of torque on the square object but I win the torque on the torus and like the radius is higher the efficiency is higher. Or reduce the distance between C1 and C2 like that it could be easier to turn the torus.
Title: Re: Sum of torque
Post by: EOW on July 30, 2015, 01:56:45 PM
I noted:

'c' the lenght of the square
'R' the external radius of the torus
'd' the distance c1c2 x or y axis
Values: c=3 and R=6.5, d=4.34

I consider the pressure like the height of the balls like I can choose with springs.

**************************************************************************CALCULATION*********************************************************************

******************************************Springs**************************************
Start integration x:
s1=d-c/2=2.84

Middle integration x:
sm=d=4.34

End integration x:
s2=d+c/2=5.84

Height:
H=d+c/2=5.84

Integration1:
$$\frac{1}{c/2}\int_{s1}^{sm} (H-\sqrt(R^2-x^2))(d-x) dx$$

Integration2:
$$\frac{1}{c/2}\int_{sm}^{s2} (H-\sqrt(R^2-x^2))(d-x) dx$$

Values: -0.20+1.63

Sum of forces:
$$\frac{1}{c}\int_s1^s2 (H-\sqrt(R^2-x^2)) dx$$

Value: 1.15

******************************************Right side**************************************

$$\frac{1}{c/2}\int_{0}^{c/2} x(-x+\frac{c}{2}) dx$$

$$\frac{1}{c/2}\int_{c/2}^{c} x(x-\frac{c}{2}) dx$$

Values: -0.37+1.87

Sum of forces:
Value: c/2=1.5

******************************************Left side**************************************

I want cancel sum of forces on C2, so I need Fx: 1.5-1.15=0.35

Torque:

Value: -0.35*4.34=-1.5

*******************************************Results**************************************

Torque on the object: -0.20+1.63-0.37+1.87-1.5=1.42

Torque on the support: 0
Title: Re: Sum of torque
Post by: EOW on August 02, 2015, 12:50:40 AM
The red part is full with small compressible balls. All springs are attached on C2 and each spring attrack a ball. The force F1 come from the springs. The support (black) turns around C1, C1 is fixed to the ground. The red object turns around C2, C2is fixed on the support.

The support and the object turn at the same angular velocity.

The force F1 don't give a torque to the support. The force F2 don't give a torque to the support. The force F2 give a torque to the object.
Title: Re: Sum of torque
Post by: EOW on August 02, 2015, 04:11:13 PM
The goal is always the same: have a torque on the red object and no torque on the support. The support turns around C1. C1 is fixed to the ground. The red object turns around C2. C2 is fixed to the support. I attrack balls from C2, all balls are attracked from C2. I want to have F1=F2 and the sum of forces of attraction from part1 =  the sum of forces of attraction from part2. For that I use the left part with a higher radius than the right part (ratio 9/7). But I need to have F1=F2, so it's important to change the pressure in the same time. When the radius at left increases, the volume of balls increases too, so I can cancel the forces of attraction left/right. If I take the good value of the pressure I can cancel in the same time F1+F2 on C2. One end of each spring is attached on C2, the other end of the spring is attached on a ball. There is one spring for each ball.

Like that the red object has a torque on it. It's unstable like before, but I accelerate more and more the support in the same time. I win the energy of the rotation of the red object. The center of gravity of the red object is on C2.
Title: Re: Sum of torque
Post by: EOW on August 03, 2015, 09:53:59 AM
I changed the radius of the left part for have a counterclockwise torque on the support.

With the radius at 5.5 at left and the radius 3.5 at right I have :

The vector at left :
x=-6.27
y=1.49

The vector at right:
x=5.21
y=-1.9

The sum is :
x=-1.06
y=-0.41

So the force give a counterclockwise torque on the support.

The force on C2 from F1 and F2 is 0 because I take the pressure at 3.5 for the right and 2.22 for the left.

Title: Re: Sum of torque
Post by: EOW on August 04, 2015, 01:16:46 AM
If I attrack balls at outer circle I can have F1=F2 and a counterclockwise torque on the support, no ?
Title: Re: Sum of torque
Post by: EOW on August 04, 2015, 06:21:04 PM
Here I have a counterclockwise torque on the support and each red object receives a counterclockwise torque too.
Title: Re: Sum of torque
Post by: EOW on August 11, 2015, 02:13:53 PM
Black lines = to the center C1. Red line = to the center C2.
Attraction: height = pressure.

Torque1=2.66
Torque2=-4
Torque3=-1
Torque4=+1
Torque5=+0.166
Torque6=-0.833

Sum of torques=-2

Force3+Force4=4
Force5+Force6=2

Sum of F3+F4+F5+F6 (vectors)=sqrt(4²+2²)=4.47

equation of the line of the sum of forces F3+F4+F5+F6 on C2: y=-2x+1; Equation of perpendicular: y=0.5x; Intersection: x=0.4,y=0.2; Distance =sqrt(0.4²+0.2²)=0.447

Torque from forces on C2 from F3+F4+F5+F6 = 4.47 * 0.447 =2

Title: Re: Sum of torque
Post by: EOW on August 12, 2015, 11:52:58 AM
Hello,

A wheel is turning and is moving in translation exactly like a wheel of a bike on the road. I put a gas under pressure P inside the black shape. The black shape is moving in translation (no rotation). Outside the black shape there is no pressure (a theoretical problem to simplify the problem). The angle $$\alpha$$ is very small, it can be at 0.00000001 rd for example.

I drawn several positions of the wheel+black_shape. A point "w" fixed on the wheel turns and moves in translation:

All the volumes are constant. I don't drawn the gaskets between the wheel and the black shape but there are (no gas escapes). I suppose the friction at 0 to study the sum of energy.

I don't need an energy to move in translation the black shape. So, the sum of energy must be at 0 too for the wheel, but:

If I'm looking at points A and B, one is moving more at right than one another. I calculated this with a big precision, like that it's possible to decrease the angle $$\alpha$$:

The radius of the wheel is at 1.

Quote
from mpmath import *

mp.dps=100; mp.pretty=True

da=0.000000001
a=pi/2

x1=a-sin(a)
y1=1-cos(a)

x2=a+da-sin(a+da)
y2=1-cos(a+da)

x3=a+pi-sin(a+pi)
y3=1-cos(a+pi)

x4=a+pi+da-sin(a+pi+da)
y4=1-cos(a+pi+da)

print "dx=", x2-x1-(x4-x3)
print "dy=", y2-y1+(y4-y3)

print "dx1=", x2-x1
print "dx2=", x4-x3
print "dy1=", y2-y1
print "dy2=", y4-y3

The result is that the point B moves more at right than the point A, so with the pressure of the gas this would say the point B gives more and energy than the point A need.

dx= 0.000000000000000001000000000000000124479849582226383362699685857886561868869889953568598575440719416999541559130733522
dy= -7.143671195514218638848647176908380996574437467642491632594536520172703974417113722971550136626075185  e-102

I calculated for A and B like I drawn but it's possible to change the angle of the black shape.

Title: Re: Sum of torque
Post by: EOW on August 13, 2015, 10:04:56 PM
The numerical solution:

from mpmath import *

mp.dps=150; mp.pretty=True

D=0.1
S1=pi/2
S2=pi

w=0.0000000000000001

def rectangles(f,a,b,n) :
h=(b-a)/n/1.0
z=0.0
for i in range(n) :
z=z+f(a+i*h)
return h*z

def sx1(t):
y1= -1/( sin(t) / (1-cos(t)) )
if t<=pi/2.0:
a1=atan(y1)+pi
a2=-t-pi/2.0
a3=abs(abs(a1)-abs(a2))
if t>pi/2.0 and t<=pi:
a1=atan(y1)+pi
a2=-t+pi/2.0
a3=abs(abs(a1)-abs(a2))
if t>pi and t<=3.0*pi/2.0:
a1=atan(y1)+pi
a2=-t+pi/2.0
a3=abs(abs(a1)-abs(a2))
if t>3.0*pi/2.0 and t<=2.0*pi:
a1=atan(y1)+pi
a2=-t+pi/2.0
a3=abs(abs(a1)-abs(a2))

return sqrt( sin(t)*sin(t)+(1-cos(t))*(1-cos(t)) ) * cos(a3)

m1=rectangles(sx1,S1+w,S1+D+w,40000)
m2=rectangles(sx1,S2+w,S2+D+w,40000)

print m1
print m2
print abs(m1)-abs(m2)

Title: Re: Sum of torque
Post by: EOW on August 13, 2015, 10:06:16 PM
The maths solution

All the volumes are constant. I didn't draw the gaskets between the wheel and the black shape but there are (no gas escapes). I suppose the friction at 0 to study the sum of energy. The radius of the wheel is at 1.

If I'm looking at points A and B, one is moving more at right than one another. I calculated this with a big precision, like that it's possible to decrease the angle $$\alpha$$

I don't need an energy to move in translation the black shape. So, the sum of energy must be at 0 too for the wheel, but for find the work of the circle I must integrate

$$\int_{pi/2}^{pi/2+da}(x-sin(x)) sin(x) dx$$
$$\int_{pi/2}^{pi/2+da} (1-cos(x)) cos(x) dx$$
$$\int_{3pi/2}^{3pi/2+da} (x-sin(x)) sin(x) dx$$
$$\int_{3pi/2}^{3pi/2+da} (1 - cos(x)) cos(x) dx$$

$$\int_{0}^{x}(x-sin(x))sin(x)dx=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))$$
$$\int_{0}^{x}(1-cos(x))cos(x)dx=-x/2.0+sin(x)-1/4.0*sin(2.0*x)$$

The result is 0.5 for an angle of 0.1

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxNUMERICAL SOLUTIONxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

[quote]from mpmath import *

mp.dps=200; mp.pretty=True

da=0.000000001
a=pi/2

x1=a-sin(a)
y1=1-cos(a)

x2=a+da-sin(a+da)
y2=1-cos(a+da)

x3=a+pi-sin(a+pi)
y3=1-cos(a+pi)

x4=a+pi+da-sin(a+pi+da)
y4=1-cos(a+pi+da)

w1=(x2-x1-(x4-x3))*sin(a+da)
w2=(y2-y1-(y4-y3))*cos(a+da)/2

print w1
print w2

print "dx=", x2-x1-(x4-x3)
print "dy=", y2-y1-(y4-y3)

There is a difference of distance:

dx=1.00000000000000012447984e-18
dy=2.00000000000000012422984e-9

But, I calculated the mean works with the worst case:

w1=(x2-x1-(x4-x3))*sin(a+da)
w2=(y2-y1+(y4-y3))*cos(a+da)/2  ***** I need to calculate the integrale

print w1
print w2

And the results of the works are:

w1= 0.00000000000000000100000000000000012397984
w2=-0.00000000000000000100000000000000012422984

I can reduce the angle there is always a difference in the axis x.

I calculated for A and B like I drawn but it's possible to change the angle of the black shape, this could change the sign of the result.

from mpmath import *

mp.dps=200; mp.pretty=True

da=0.000000001
a=pi/2

def rectangles(f,a,b,n) :
h=(b-a)/float(n)
z=0
for i in range(n) :
z=z+f(a+i*h)
return h*z

def sx(x):
return (x-sin(x)) *sin(x)

def cx(x):
return (1-cos(x)) *cos(x)

def sxp(x):
return (x+pi - sin(x+pi)) *sin(x+pi)

def cxp(x):
return (1 - cos(x+pi)) *cos(x+pi)

m1=rectangles(sx,a,a+da,10000)
m2=rectangles(cx,a,a+da,10000)
m3=rectangles(sxp,a,a+da,10000)
m4=rectangles(cxp,a,a+da,10000)

print m1
print m2
print m3
print m4

print m1-m2+m3-m4

The result is:

Sum of works = 0.00000000514

But I'm not sure about my integration

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxMATH INTEGRATIONxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Here all math integrations:

x=pi/2.0
p1=1/4*(-2*x+4*sin(x)+sin(2*x)-4*x*cos(x))
p2=-x/2+sin(x)-1/4*sin(2*x)
x=pi/2.0+da
p3=1/4*(-2*x+4*sin(x)+sin(2*x)-4*x*cos(x))
p4=-x/2+sin(x)-1/4*sin(2*x)

x=pi/2.0
p5=-x/2-sin(x)+1/4*sin(2*x)+(x+pi)*cos(x)
p6=1/2*(-x-2*sin(x)-sin(x)*cos(x))
x=pi/2.0+da
p7=-x/2-sin(x)+1/4*sin(2*x)+(x+pi)*cos(x)
p8=1/2*(-x-2*sin(x)-sin(x)*cos(x))

The result is 0.00000000514

The difference is small because the angle is small, if I take $$\alpha$$ at 0.1 rd the result is 0.5

I change the math integrals and I obtain this :

x=pi/2.0
p1=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p2=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
x=pi/2.0+da
p3=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p4=-x/2.0+sin(x)-1/4.0*sin(2.0*x)

x=3.0*pi/2.0
p5=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p6=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
x=3.0*pi/2.0+da
p7=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p8=-x/2.0+sin(x)-1/4.0*sin(2.0*x)

print p1
print p2
print p3
print p4
print p5
print p6
print p7
print p8

z1=-(p3-p1)+p4-p2
z2=p6-p5+p7-p8
print abs(z1)-abs(z2)

It is the same result. And if I take the angle à pi the sum is 0.

Title: Re: Sum of torque
Post by: EOW on August 26, 2015, 07:58:16 AM
The idea is to use small balls and springs to have a pressure inside a container like a fluid do with the gravity. The black container is full of small balls and springs, the pressure is from 0 to pmax in the top view (not the side view). One spring for one ball, I didn't draw all springs, only three. Balls are compressible or there are in random position (not like I drawn), I need the lateral force from the pressure. I drawn big balls but there are smaller in reality. In the top view I drawn the disk like an ellipse because there is an angle on the side view.

I attrack the disk with the springs like the balls. IT IS THE DIFFERENCE with older ideas. The red disk has 2 axis of rotation, one around the black axis and the other around itself. It is very important to have this axis of rotation around itself too, because the energy come from this axis.

The container is an object. The disk is an object.

Look at the forces on the red disk. I attrack the red disk like the  blue balls. This gives a torque on the red disk, it's logical. Like it's a disk there is no other torque on it.

Look at the forces on the red arm. Like the disk is attracked like the balls it's like an object with the same density in water with gravity, the force on the red arm is F2. F2 is only located on the arm, it is the sum of forces on the disk. F2 can't give a torque on the red arm.

Look at the forces on the container: only F1. There are lot of F1 because they are forces from the pressure but each F1 is parallel to the axis of rotation, so no torque.

Like I attrack the balls AND the red disk the force on bottom (top view) from pressure is cancelled by the force from springs, so there is no force on the side where there are green points.

The sum of torques is not 0, but the device is unstable. If I let the device like that, the springs will move and lost their potential energy. So it is necessary to accelerate more and more to increase the kinetic energy of the device. The red disk has a mass so like there is a torque on it, it will accelerate but I accelerate the black arm (the container) at the same acceleration EXACTLY the same acceleration. Like that the red disk keeps its position inside the black container. I need an energy to accelerate the container but this energy can be recover later. I win the energy from the acceleration of the red disk.

I can put several disks inside the container and no balls between disks like that I don't have torque on the red arm.
Title: Re: Sum of torque
Post by: EOW on August 26, 2015, 09:54:11 AM
To look of 2 axis of rotation of the red disk. Like before each spring attrack.

I noted all surfaces Sx:

The torque on the  container:

S1 and S2 cancel themselves the torques on the recipient
S3 and S4 give the forces F1: F1 are parallel to the axis of rotation: no torque
S5: no pressure (top)
S6: the force from pressure is cancelled by the force from springs BECAUSE I attrack the disk too

The torque on the disk around itself S7 to S9:

S7: it's a sum of segments, at final the torque from springs is always there
S8 and S9 give forces parallel to the axis of rotation (red), there is a difference of surface between S8 and S9

The sum of forces on the disk:

There are some forces like F2, but these forces can't give a torque on the red arm because they are parallel on it.

Torque on the red arm:

With severals disks inside the container I can suppress balls between disks. And if I reduce the size of the container I can suppress the difference from S8/S9

Title: Re: Sum of torque
Post by: EOW on August 26, 2015, 10:29:57 PM
The problem on the red arm come from the difference of surface S8/S9, so if I put balls inside the part of the red disk (only the part that is inside the container) I can cancel the pressure from S8/S9 (I attrack the balls inside the red disk from springs from the container). The force on the curve part (circle) don't change the torque on the disk. The red disk has always its torque around itself.
Title: Re: Sum of torque
Post by: EOW on August 27, 2015, 04:01:38 PM
First image: if I take this simple device the force F will lost an altitude so a potential energy will be lost. But the disk will turn more and more. The sum of energy is constant or not... ?

Second image: I put balls inside the part of disk that is inside the container. I attrack balls with springs. The disk will have the force F so a torque. But the container don't have any torque. The container has balls +springs.

Title: Re: Sum of torque
Post by: EOW on August 30, 2015, 10:03:58 AM
If I take this device:

(http://s5.postimg.org/kzk1nvd8z/a70.jpg) (http://postimg.org/image/kzk1nvd8z/)

The support turns at w0 and the purple disk can turn around itself, at start the angular velocity of the purple disk is 0.
I can adjust the angle alpha like I want before start but the angle is constant after start.

If alpha=0° a point A fixed on the purple disk will be at differents position with time:

(http://s5.postimg.org/91n2npyvn/a73.jpg) (http://postimg.org/image/91n2npyvn/)

If alpha=2 or 3° it's the same but the point moves up/down like that:

(http://s5.postimg.org/rej2kjjyr/a71.jpg) (http://postimg.org/image/rej2kjjyr/)

And like that:

(http://s5.postimg.org/67t41rbgz/a72.jpg) (http://postimg.org/image/67t41rbgz/)

The angle can be increased more at 20° for example.

So if the point A moves down or up, it's possible to attach a mass, this mass will move up (under gravity) and the purple will turn a little. I win the energy of the rotation of the purple disk and the mass that moves up. THERE IS NO TORQUE ON THE SUPPORT.

Title: Re: Sum of torque
Post by: EOW on September 03, 2015, 03:17:18 PM
The video of the device, I simulated with Ansys, 10 seconds, it's possible to look at the rotation of the disk, at start the disk has no rotation around itself, a small black circle is fixed on the disk to look at the rotation :

https://youtu.be/Pjc4dIf1aWI

If with this device the sum of energy is constant. I can use the last device with the balls+springs but WITHOUT MASS, or a mass as lower as possible. Like that I don't have these problems of inertia. I can recover the torque on the red axis from the pressure and accelerate more and more the device.
Title: Re: Sum of torque
Post by: EOW on September 04, 2015, 08:04:40 PM
I can place the disks like that:

http://s5.postimg.org/i21birqon/image.png

to increase the angular velocity of each disk
Title: Re: Sum of torque
Post by: EOW on September 05, 2015, 10:27:06 PM
With a unique disk +arm there is a problem with the linear momentum too. If I explode the disk after it rotates around itself the linear momentum is not conserved => antigravity device
Title: Re: Sum of torque
Post by: EOW on September 10, 2015, 06:53:54 PM
The spring like that.

The spring increases its length.

The red arm receives a positive torque.

The disk receives a positives torque.
Title: Re: Sum of torque
Post by: EOW on September 11, 2015, 05:48:56 PM
First idea:

I come back with the container full with springs and small balls. The pressure like a fluid inside the container. The part of the red solid is attrack with spring like that there is no force at "bottom". I need to take a special shape of the red solid, it is not a disk, it is an addition of circles in translation. With a circle I don't have a torque on the red solid from the pressure. With the shape I drawn I don't have a torque on the arm. The disk is always in the position I drawn because I accelerate the arm more and more.

The problem is I need to have an infinity of arms !

Second idea: start the devices each disk will rotates at w*cos(a) (not the perpendicular disk):

I can take 3 disks, 2disks and one disk in a perpendicular plane. I start the device, the 2 disks turns slower than the support so it's possible with friction between disks to accelerate the perpendicular disk, and in the same time 2 disks accelerates in the laboratory reference.
Title: Re: Sum of torque
Post by: EOW on September 12, 2015, 10:17:48 AM
With 2 plates free in space to move up or down. The plates are in rotation. There is friction between the plates and the orange disk.

Or with 2 grey circles fixed to the ground. The orange torus increases its angular velocity in the labo reference. Don't forget friction gives an energy of heating. the best is to have the friction in one direction higher than in other direction. I want to decelerate the torus in the arm reference with friction but in this case the friction will slow down the red arm. If I can have the friction of rotation around the magenta axis equal to zero it's better. I would like to have the forces F1 and F2 on the torus and no force from friction on the axis of the torus like that the red arm don't decelerate. Even the red arm decelerate due to friction the energy is not lost it is in heating.
Title: Re: Sum of torque
Post by: EOW on September 12, 2015, 07:08:59 PM
With the device in the same plane like I studied before. It's possible to use a belt. A unique belt. A belt without a mass. The belt has a friction. I use the friction of the belt to decrease the angular velocity of the disk (arm reference) like that I recover the energy from friction and the disk turns more in the garage reference.

I can take my idea of the message #74

http://overunity.com/15011/sum-of-torque/msg428104/#msg428104

The image case1.png

The belt is like a "8" shape.

The friction of the belt is not between the belt and the support nor the belt and the disk but when the belt changes its shape, at each time the belt change its angle.

Title: Re: Sum of torque
Post by: EOW on September 12, 2015, 09:37:15 PM
No gravity here.

I think like that it's a good device:

The torus has a mass.

1/ Turn the red arm at w. Turn the grey circles at 0.999w. Grey circles have mass. The angular velocity of the orange torus is -w*cos(a) in the arm reference with a the angle of the axis from the vertical.
2/ I use the torus like a catapult. The grey circles can be cut at small parts when I want. Like that the torus can have only a torque on it, the red arm don't receive a torque. The small parts of the circles can be eject and the velocity is higher than before the friction with the torus. Need to recover the energy from the small parts of circles. Small parts are free to move in 3d space. A small part has it velocity from the rotation and the velocity up/down from the torus, it's an addition of velocities (vectors).

It's not really a friction between the torus and the circle it's more use the torus like a catapult. The goal is to keep no torque on the grey circles.
Title: Re: Sum of torque
Post by: EOW on September 13, 2015, 11:31:48 AM
I add an image in 3d to look the orange disk between the grey torus.

The energy won is : 2*(-1/2*I*(w*(1-cos(a)))^2) + (1/2*I*w^2)) = 2*1/2*I*(-(-2w^2cos(a)+w^2*cos²(a))) = Iw² (2cos(a)-cos²(a))
I: the inertia of the orange torus around itself.
a: the angle of the axis of the orange torus from the vertical.

In practice the vectors of velocities are not perfectly perpendicular (i8.png), V2 cancel a little of V1 but V3 add a little of V1, the sum is near 0.
Title: Re: Sum of torque
Post by: EOW on September 14, 2015, 12:29:35 PM
With the device of the video :

I use 2 devices like this I turn one red arm clockwise and the another red arm counterclockwise and I exchange the torus ! I lost an energy.

Or I use friction between the disks
Title: Re: Sum of torque
Post by: EOW on September 15, 2015, 12:36:32 PM
Cycle:

1/ I launch the red arm and the green arm at the angular velocity w. The orange disk turn around itself at w*(1-cos(a)) in the laboratory reference. The blue disk don't turn around itself.
2/ I use stems (like a parts of belt) for give the forces F3 and F4 to the blue disk. The blue disk increases its angular velocity around itself. The orange disk increases its angular velocity around itself too. The red arm and the green arm don't receive a torque.

Title: Re: Sum of torque
Post by: EOW on September 16, 2015, 09:11:56 AM
And sure, I can do this with spring(s) no need the blue disk. Replace the stem by spring(s). There is a positive torque on the red arm. The orange disk increases its angular velocity around itself. The device must turn at an infinite angular velocity. The distance of the spring never change because the arm accelerates more and more. The spring is always like I drawn: vertical.
Title: Re: Sum of torque
Post by: EOW on September 16, 2015, 09:59:22 PM
Last is bad, the disk turn in the opposite direction so I can do this :

Title: Re: Sum of torque
Post by: EOW on September 17, 2015, 11:19:45 PM
With a thickness for the disks I can have a difference of local velocity to have the forces F1 and F2. No torque on arms. Like each disk don't turn around itself (without friction), F1 and F2 will change the angular velocity of each disk so the kinetic energy increases.
Title: Re: Sum of torque
Post by: EOW on September 18, 2015, 04:04:16 PM
Like that the cylinder receive a positive torque like the arm from F1. The torque from F2 is cancelled by F3y. F3x don't give a torque. I want to keep constant the length of each spring, for that I need to accelerate the arms like the cylinder accelerate. Even the torque from F1 and F2 are not enough I will accelerate the arms like the cylinder. I recover the energy later.
Title: Re: Sum of torque
Post by: EOW on September 19, 2015, 06:30:32 AM
I added some informations on the image (I didn't draw all springs, but there is one for one ball and some for the sphere), the balls in reality must be smaller than I drawn, it's a theoretical study where I suppose I can obtain the pressure of a fluid under gravity inside the container. The sphere is a full object, there isn't balls between the sphere and the container but I attract the sphere with the springs.

1/ The sphere receives a clockwise torque from the springs
2/ The part of the sphere in contact with balls don't receive a torque because the shape is an addition of part of circles
3/ The red arm receives a force F2 because the springs attract the spehre but balls give a force F3 (composed of F3x, F3y and F3z), and F3y=-F2 (in vectors) so there is no torque from the sum F2 and F3y
4/ F3z and F3x can't give a torque on the red arm
5/ There is a surface with balls at left greater than at right in the container because at right there is the sphere . Like the pressure is higher at "bottom" the force by unit of surface is greater at "bottom" so there is a clockwise torque on the black arm, I resumed this torque by the force F1
6/ At "bottom" there is no force because springs attract all balls and the volume of the sphere

There is a last problem, the sphere will receive a torque on it and like it must turn at the angular velocity w too it seems the sphere lost the energy I won with the black arm. Now, imagine all the mass of the sphere is in the center (like particule in physics, you know, some people say: it's a point particle with mass), so if the mass is in the center I don't lost any energy from the sphere because the energy of rotation is 1/2Iw² and I is at 0 with a point particle. I need the sphere with a radius but I want the mass in the center. In our reality, the mass can't be at a point but can be near the center, so the energy lost by the sphere can be small.

The black arm will receive a torque. The red arm don't receive any torque. Two arms turn at w. There is a last problem, the sphere will receive a torque on it and like it must turn at the angular velocity w too it seems the sphere lost the energy I won with the black arm. Now, imagine all the mass of the sphere is in the center (like particule in physics, you know, some people say: it's a point particle with mass maybe it's not really a point), so if the mass is in the center (or near the center) I don't lost any energy from the sphere because the energy of rotation is 1/2Iw² and I is at 0 with a point particle. I need the sphere with a radius but I want the mass in the center. In our reality, the mass can't be at a point but can be near the center, so the energy lost by the sphere can be small.

Maybe like this the inertia is not very high for you. So, in this case, with the mass in the wall of the sphere (or in the center) when I apply a torque on it, I apply a torque on a gyroscope if 'w' is high so the sphere turns in the perpendicular axis like a gyroscope can do and don't lost its energy.

If you don't want a high 'w' you place the mass in the center of the sphere and turn it very quickly around this axis you want like that when the springs want to turn the sphere it turns a gyroscope. Sure I use springs for show the device but imagine an electrostatic force for example.

And to be ok with my theory with gravitation, I need 'w' very high, it's great because the precession of the gyroscope is a function of 'w', higher is 'w' lower is the precession.
Title: Re: Sum of torque
Post by: EOW on September 19, 2015, 06:45:05 PM
The sphere will works like a gyroscope. Put the mass in the center and turn it around itself at a very high angular velocity w'. The black arm receives a torque and can increase the energy of the device. The sphere receive a torque on it too, this could cancel the energy won by the arm but if the sphere is like a gyroscope, the gyroscope precession.

The precession of the gyroscope is very low (low inertia but a very high angular velocity) and in one turn the mass return to its initial position because the recipient keeps it position.
Title: Re: Sum of torque
Post by: EOW on September 20, 2015, 02:01:40 PM
Maybe I don't need the arms, just the sphere and its mass in the center. With the container + balls + springs. The main axis of rotation is the blue axis.
Title: Re: Sum of torque
Post by: EOW on September 21, 2015, 09:12:07 AM
Like that with a gyroscope. Or without a gyrsocope it must work.
Title: Re: Sum of torque
Post by: EOW on September 22, 2015, 07:29:06 PM
At start the disks don't turn around themselves. I launch the arms at w. I can put inside the orange disk a gyroscope because in the laboratory reference the orange disk don't turn.

2 springs: one attracts the other repuls.

There is one gyroscope inside the orange disk. The blue disk don't have a gyroscope. The blue arm receives a positive torque. The orange arm don't receive any torque because the gyroscope take the torque.

For 180° Spring1 repuls, the Spring2 attracks. For the next 180° Spring1 attracks, the spring2 repuls.
.
Title: Re: Sum of torque
Post by: EOW on September 24, 2015, 09:17:49 PM
Details of the disks
Title: Re: Sum of torque
Post by: EOW on September 25, 2015, 03:55:53 PM
The orange disk is not necessary. Onle the blue disk and the gyrsocope. The blue arm receives a torque. And the gyroscope receive the negative torque but it does a precession. I drawn several position of the gyroscope and the blue disk to show the electrostatic charge (or it can magnetism or springs or hydraulic device). The arms turn clockwise but the blue disk turn counterclockwise in the arm reference (the disk like the gyroscope don't turn in the laboratory reference).
Title: Re: Sum of torque
Post by: EOW on September 26, 2015, 10:42:24 AM
No gravity !

The gyroscope must have only one torque on it to precession correctly. So, the angular velocity of the arms must be constant. I add an external device that recover the energy from the arm like that w is constant. The gyroscope don't turn around itself in the laboratory reference without a torque on it. But it turns like that :

I add a torque from springs (or something else) so the gyroscope does a precession, this is only its angular velocity. The torque from the blue arm is not constant in one turn so the device that recovers the energy must be precise. The blue disk must turn at the precession.

When I commute the torque (each 180° of the blue disk in the arm reference) on the gyroscope, the gyroscope will do a nutation. The angular velocity of the precesion is small and it's good. In the arm reference, the angular velocity of the gyroscope will be at -w.cos(a)-d or -w.cos(a)+d with 'd' the angular velocity of the precession. I can choose a gyroscope with a spinning wheel very high, so high that the nutation and the precession will be low if the inertia Is of the gyroscope is very high. Sure, I need to choose a torque on the gyroscope in relation with Is.
Title: Re: Sum of torque
Post by: EOW on September 27, 2015, 04:30:13 PM
For start the device I use 2 gyroscope with their wheel turning in the opposite direction. After, I use 2 separated gyroscopes.
Title: Re: Sum of torque
Post by: EOW on September 28, 2015, 11:22:52 AM
w is the angular velocity of the arm in the lab reference
d is the angular velocity of the precession of the gyroscope when it has a torque.

I'm sure that the gyrsocope don't turn in the lab reference when the gyroscope turns at -w.cos(a) in the arm reference and when the arm turns at w. If I give a torque on the gyroscope, the gyroscope will do a precession small but it exists. And maybe when the gyroscope is not exactly at -w.cos(a) in the arm reference and the arm is at w, the gyroscope don't turn in one direction but in two directions: I'm not sure about that ! So, I will change the angular velocity of the arm in the same time I will add 2 torques on the gyroscope so I can't.

Maybe like d << w and like I change the sign of the torque each 180° so each pi/w second, this is not a problem, the gyroscope can have a mean position good enough to work like I want. The modification on the angular velocity from d to w is small.

Example, with w=10 rd/s and d=0.01 rd/s with an angle of 45°, the angular velocity of the gyroscope in the arm is -7+0.01 and 7-0.01 the precession don't change the position of the gyroscope. In 1.4 turns of the arm, the gyroscope must return to its initial position.

---------------------------------------------------------------------------------------------------------------------------------------------------------

Second method: I friction the gyroscope and the orange disk. The orange disk increases its angular velocity in the lab reference. The gyroscope precession.
Title: Re: Sum of torque
Post by: EOW on September 29, 2015, 02:02:06 PM
The arm moves the center of gravity of the gyroscope and the torque from the friction change the orientation of the gyroscope.

There is an energy from friction, an energy from the rotation of the orange torus. The arm don't receive a torque.
Title: Re: Sum of torque
Post by: EOW on September 29, 2015, 03:27:05 PM
And without a gyroscope. There are vertical disks (green). All others disks are with a slope (like before). All the device turns at w around the black center C.

I drawn ellipses for disks because it's a top view. I drawn one blue disk alone, one orange disk alone and one green disk alone to show the side view.
Title: Re: Sum of torque
Post by: EOW on September 30, 2015, 01:29:03 PM
And this idea with the balls and springs to create a pressure like a fluid under gravity. There is a torque on the red center. Use a law in 1/d^2 for the attraction. For example at 1 meter the pressure is 1 and at 2 meters the pressure is 4 not 2. Better with d^3 and don't start at p=0 at the top of the half disk => the disk in at 6 meter at proof for example.
Title: Re: Sum of torque
Post by: EOW on October 01, 2015, 10:38:42 PM
With the device and with gravity.  :

I place at the top of the orange disk a small mass $$m$$ after the device is launched. With an angle of 60°. This mass will move down of $$2R.sin(60°)$$ so the energy lost is $$2R.sin(60°)mg$$. With $$g$$ the gravity. The disk is turning in the arm reference at $$-0.5w$$ and the arm turns at $$w$$. The mean torque on the red arm is $$\frac{2}{\pi}Rmg.sin(60°)$$ so the erergy is a the torque by the angle of the arm, if the disk rotates of $$\pi$$ the arm rotates of $$2\pi$$ so the energy won is $$\frac{2}{\pi}Rmg.sin(60°).2.\pi$$ it's twice than the energy lost by the mass when it moves down.
Title: Re: Sum of torque
Post by: EOW on October 03, 2015, 08:00:58 PM
There is gravity. A small mass 'm' is put at the point 'A' at top of the disk with the linear velocity V+wRcos(a) with 'w' the angular velocity of the arm, 'a' is the angle from the vertical, 'R' is the radius of the disk.  The mass 'm' is ejected at the point 'B' with the linear velocity V+wRcos(a). 'V' depends of the length of the arm. The mass will move down and lost an energy, the arm accelerates and will win an energy. In the image or in the video, the angle 'a' is at 60°.

The mass 'm' lost the energy 2mgRsin(a), with 'g' the gravity.

The mean torque on the arm is 2/pi*mg*cos(a). Like a=60°, when the disk turn of pi (from the top to the bottom, the disk must turn of pi), the arm turns of 2pi. So the energy win by the arm is 2pi*2/pi*mgR*cos(a)=2mgR

There is a difference of energy of 2mgR(1-sin(a)). And at start the disk don't turn around itself, when the mass 'm' is at the point 'B' the disk turns a little.

Title: Re: Sum of torque
Post by: EOW on October 05, 2015, 02:38:23 PM
I changed the direction of the gravity like that the mass don't lost any energy. The arm receives a torque when it rotates. Gravity is always like I drawn even the arm rotates.

Or I can use 2 springs if F1>F2, the goal is to have F1+F2 (in vector) like the slope of the disk. F1=1.73N and F2=1N for example when a=60°. The arm receives a negative torque that cancel the energy 2RFsin(a). One spring win the energy RF another spring win 1.73*1.73RF+RF=3RF
Title: Re: Sum of torque
Post by: EOW on October 06, 2015, 10:54:32 PM
If I use a spring, the arm will receive a negative torque but the spring will receive a positive torque. But if I turn the arm with 180° not 360° the point A turns around the center C of little less than 180°, it's maybe something like 175° in the example. The difference of the energy come from the fact the arm turns of 180° but not the point A, so the spring will receive a difference.
Title: Re: Sum of torque
Post by: EOW on October 08, 2015, 10:03:44 AM
I move in translation all the device. Like that I can have -X and +X for the velocity on the green part. The green part receive a torque. The arm don't receive a torque and the disk accelerates the rotation around itself.
Title: Re: Sum of torque
Post by: EOW on October 09, 2015, 08:56:22 PM
With the pressure with the law 1/d². The torque from the SideA is not the same than the torque from the SideB. I drawn only 2 springs but there are in all the blue surface.
Title: Re: Sum of torque
Post by: EOW on October 10, 2015, 09:28:23 AM
In the last image, all the black lines are link together. All the device I drawn turn around the red axis. Black lines "separate" the pressure from the top.

On the SideB the mean pressure is 0 if the number of black line is infinite. But with 100 lines for example the force is decrease by 100.

On the SideA, the torque is integrale(x²(x-R)dx) from 0 to R = R^4/4-R^3*R/3

With R=5, the torque on the SideB is hum maybe 5 and it is 52.0833 for the SideA.

There is no torque from the curved shape because the device turns around the red center and this is the center of the half circle.

There is no torque from straight horizontal lines and springs because they cancel themselves all torques IF the number of horizontal lines inside the half disk is high.

The attraction can be in another law than 1/d²

I added an image, N is the number of black lines. Here N=9 for example. At top in each slice the pressure is 0, at bottom the pressure is R/N.

I drawn the device after 30°, to look how the device rotates, all the device rotates.
Title: Re: Sum of torque
Post by: EOW on October 10, 2015, 11:01:38 PM
Image 1/ With half disk the sum of torque is 0

Image 2/ With a full disk. Springs are only in the right half disk. Springs push balls. The law of repulsion for springs is like d². The force is higher when the cos(angle) is lower so there is a difference.

Image 3/ With balls inside in PartB and balls outside in partA1/PartA2

Image4/ With water. F1>F2 so the disk want to turn clockwise. F3 gives a clockwise torque to the device around the red center. All the device is always in that position because I synchronized 2 angular velocities.
Title: Re: Sum of torque
Post by: EOW on October 11, 2015, 11:51:21 AM
I came back with this device. If I take a half disk, the centrifugal forces turns the half disk and the arm receives a torque. The arm turns at w (labo reference) and the disk turn around itself at w.cos(a) (labo reference) like that the disk is always in the position I draww.
Title: Re: Sum of torque
Post by: EOW on October 12, 2015, 08:24:41 PM
The calculations for the torques in the first image. Eq1 is the torque from the centrifugal force from the half disk itself around itself to the arm. Eq2 is the torque on the arm from the centrifugal force around the support. Eq3 is the torque on the disk. With:

d: the distance of the arm
alpha: the angle
R: the radius of the disk
w: the angular velocity of the arm

I can take 2 half disks like the image shows, like that Eq1=0 for 2 half disks. And like d1 different of d2 the torque on the arm + torque on the half disks are not at 0.

Title: Re: Sum of torque
Post by: EOW on October 16, 2015, 07:03:46 PM
The sum of torques is not 0.
Title: Re: Sum of torque
Post by: EOW on October 17, 2015, 10:09:27 AM
Better, if the center of the quarter circle of the white shape is the red center not the black center.

All the last device I explained are mounted on a device with an arm with an angle and a disk, I drawn only the disk, look at the third image.
Title: Re: Sum of torque
Post by: EOW on October 18, 2015, 07:51:33 PM
With the spring. The arm lost the energy FRwtsin(a)cos(a) and spring win the energy FRsin(wt.cos(a))sin(a), the sum of energy is not constant. It's like to compare:

b.cos(a) and sin(b*cos(a))

and the result is not the same, an example:

a=2.09
b=0.1

The result is -0.05 and -0.049979

Another example:

a=0.001
b=60*pi/180

The result is 1.74e-5 and 3.04e.7, the relation is 57.29 = 180/pi, it's the maximum of the relation

Title: Re: Sum of torque
Post by: EOW on October 21, 2015, 04:00:49 PM
At start, the disks don't turn around themselve in the lab reference. The arm turns at w. The spring gives the forces F1 and F2. Like the disks don't turn around themselves the lenght to the spring changes very few. But the forces F1b and F2b give a torque to the arm, this torque depends of w.
Title: Re: Sum of torque
Post by: EOW on October 24, 2015, 08:35:48 PM
With 2 different radius and a belt.

Or 2 disk with the same radius and a circle with friction to accelerate the disks
Title: Re: Sum of torque
Post by: EOW on October 25, 2015, 10:58:33 AM
The radius of the grey disk is lower than the red disk but the angle alpha of the grey disk is lower than the red disk, so the grey disk turns quickly than the red disk, the arm receives a clockwise torque.
Title: Re: Sum of torque
Post by: EOW on October 26, 2015, 08:54:55 AM
Like that the spring increases its potential energy and the arm receive a positive torque. The horizontal torus turns at -w in the arm reference. The disk turns at -0.707w in the arm reference.
Title: Re: Sum of torque
Post by: EOW on October 27, 2015, 07:37:02 PM
2 springs and 2 disks.
Title: Re: Sum of torque
Post by: EOW on October 28, 2015, 01:14:27 PM
On the red arm I can put 2 spirals. The spiral1 will accelerate counterclockwise in the arm reference and it is correct because R1 becomes smaller more and more and R2 becomes higher more and more. So, the spiral1 accelerates in the laboratory reference.The spiral2 decelerates in the arm reference but it accelerates in the laboratory reference. There is a pressure from a gas outside the device.
Title: Re: Sum of torque
Post by: EOW on October 29, 2015, 10:23:15 AM
At start, I want the same angular velocity of the spiral1 and the spiral2, but like the diameter is not the same I need to increase the diameter of the tube. Increase the diameter cost nothing in theory. Like that the arm has a torque on it. The angular velocity of each spiral at start is -w.cos(a). The Spiral1 has a counterclockwise torque on it, so it will accelerate in the laboratory reference counterclockwise. The Spiral2 has a clocckwise torque on it so it will accelerate in the lab reference clockwise. In the arm reference, the Spiral1 accelerates from -w.cos(a) to X and the Spiral2 deccelerates from -w.cos(a) to Y, but radius of each spiral change in the same time and it's possible to adjust the thickness of the tube when it goes from the Spiral1 to the Spiral2.

This device must works with the angle a=0: axis of spiral are vertical.

I can use a full material for the tube with a low mass, it's easier to see the volume can be constant from Spiral1 to Spiral2.
Title: Re: Sum of torque
Post by: EOW on October 31, 2015, 09:52:40 PM
Take the example in the image. There is no arm, just 2 pulleys and one tube. I want to have the same angular velocity for the Pulley1 and Pulley2 even the radius of the Pulley1 is twice than the Pulley2. The pulleys turn counterclockwise. There is pressure from a gas outside at P=100000Pa and inside the closed tube there is p=1000 Pa. The Pulley1 receives a counterclockse torque from F1, the Pulley2 receives a clockwise torque from the force F2. Like I want the same angular velocity if the section (surface) of the tube when it is on the Pulley1 is 2mm * 2 mm. I need to keep constant the surface of the material, it is 4 surfaces by 2*2 = 16 mm². Like the Pulley1 has a radius twice than the Pulley1 I need to move 16 in surface but the length of the arc is 2 times lower it's not 2 mm but 1 mm, so if I change only 2 surfaces, I need to have 2 surfaces at 2*1 and 2 surfaces at 6*1 = 16. So the front surface from F1 is 2*2=4 and from F2 it's 1*6=6. So the Force F2=1.5*F1. Or if I change 4 surfaces I will have 4 surfaces at 4mm*4mm.

The sum of the energy:

1/ The Pulley1 has a counterclockwise torque from F1 the energy is 2*R*F1*wt=8(P-p)Rwt, where (P-p) is the difference of the pressure
2/ The Pulley2 has a clockwise torque from F2 the energy is -R*F2*wt=-16(P-p)Rwt
3/ The volume of the tube increases like 8(P-p)Rwt
Note: move the wall don't need any energy because the force from pressure is perpendicular to the movement except where there are F1 and F2.

Even the sum of energy is constant do that:

Now place all this device on an arm that turns at w. The Pulley1 and the Pulley2 turn at -w in the reference arm. But look at the torque on the arm, it is positive. And the Pulley2 receives a clockwise torque but it turns at -w in the arm reference not in the lab reference ! so it decelerates in the arm reference (accelerates in the lab reference).

The sum can't be constant.
Title: Re: Sum of torque
Post by: EOW on November 02, 2015, 10:12:32 PM
In the last example, the volume of the tube increase but it's possible to let it constant, for that I need to enter the tube inside itself. Like that repeat the cycle is easier.

I place the center of rotation of the device in the red center like that the force from pressure due to the conic section can't give a torque.

The pulley1 can support the tube in the upper side of the magenta line (third image) and the pulley2 can support the lower side of the magenta line. Like that there is a torque on the arm.
Title: Re: Sum of torque
Post by: EOW on November 03, 2015, 07:20:18 PM
I change the direction of the force, like that the arm receives a clockwise torque. The volume of the tube is constant so I don't lost any potential energy. Each pulley will accelerate in the lab reference (decelerate in the arm reference). Like that the difference is the bigger pulley lost more energy more quickly (in the arm reference) but I don't gave any energy in fact :)

Example:

Radius of Pulley1 = 1000 mm
Radius of Pulley2 = 500 mm

Surface of tube in the Pulley1 = 2 mm * 2 mm = 4 mm ²
Surface of tube in the Pulley2 = 7.4641 mm * 0.535898 mm = 4 mm² (resolv 2x+2y=4 and xy=4)

The tube don't lost any potential energy. The arm don't receive a torque.

Example2:

Radius of Pulley1 = 1000 mm

Surface of tube in the Pulley1 = 2 mm * 2 mm = 4 mm ²
Surface of tube in the Pulley2 = 0.472136 mm * 16.9492 mm = 8 mm² (resolv 2x+2y=8 and xy=8)

Here the volume of the tube increases and the arm receives a clockwise torque.
Title: Re: Sum of torque
Post by: EOW on November 04, 2015, 10:56:38 AM
At start the squares don't turn around the blue center. The arm turns at w. The blue walls are supported by the squares so each square accelerates counterclockwise. The red center supports the red walls, in one turn of the squares, the torque could be at 0 on the arm, but not because the squares increase their angular velocities and there are less and less time for the torque. So the arm accelerates clockwise and the squares accelerate counterclockwise. Imagine with the thickness T near 0.

Outside the device there is a small pressure.
Inside the blue/red shape there is a bigger pressure from a gas for example.

The pressure at 100000Pa is only in the shape composed by 2 blue walls and 2 red walls (small volume because T is near 0). This shape gives the forces F1 to F4.

I give the Python code to test. With k=0 (no acceleration) there is no torque on the arm. But with k=0.2 for example there is a torque of 58 in one turn of the squares.

The squares turn together with the same angular velocity, they keep constant their relative position from each other.
Title: Re: Sum of torque
Post by: EOW on November 04, 2015, 11:07:14 PM
Another position,the arm turned of 45° the squares too
Title: Re: Sum of torque
Post by: EOW on November 05, 2015, 04:55:25 PM
I change the gas every 90°.
Title: Re: Sum of torque
Post by: EOW on November 06, 2015, 07:01:47 PM
This device don't create energy, just use the energy from the temperature of a gas (pressure). The arm turns at w clockwise. The disk don't turn around itself. The pneumatic cylinder has a spring inside and has the pressure P everywhere except between the cylinder and the disk. The disk decreases its angular velocity in the arm reference and the cylinder compress the spring. The temperature of the gas decreases. I drawn the device at the time the pneumatic cylinder must work, exactly at this time, it's a transcient action.

Second image: all the device turns clockwise around the red center. The square can turn around the blue center. The device don't receive a torque but the square receives one. The square receives the forces Fa and Fb and Fc, these forces give F2 on the blue center. Inside the red circle there is no pressure.

The torque on the red circle from F1 is:

C1=(R1+R2/2)*R2*P  (I suppose the depth is 1)

The torque on the red circle from F2 from the blue center is:

C2=R1*R2*P+(R1+(R2²-2R1²)/2/R1)R1+ (R1-(R1+(R2²-2R1²)/2/R1))*R1

The square in the contrary receives a torque of:

C3=(R1+(R2²-2R1²)/2/R1)R2*P

Numerical application:

R1=10
R2=1
C1=10.5
C2=10.50012
C3=0.05

If the red circle accelerates more and more, the square keeps its relative position with the red circle, so the square is always like I drawn. I need to give an energy to turn the red circle (the device) but I win the energy of the rotation of the square around itself.
Title: Re: Sum of torque
Post by: EOW on November 07, 2015, 09:21:30 PM
I can use one or 2 half torus. Outside the pressure is near 0 and inside the grey container there is 1bar. The arm turns clockwise at w. At start the half torus don't turn around itself so it turn at -w in the arm reference.
Title: Re: Sum of torque
Post by: EOW on November 08, 2015, 07:15:45 PM
Shows the image
Title: Re: Sum of torque
Post by: EOW on November 09, 2015, 10:30:28 PM
The torque on the half disk is 2 times the torque on the device.
Title: Re: Sum of torque
Post by: EOW on November 10, 2015, 07:13:15 PM
The device is always like I drawn. F1/F1' gives counterclockwise torque. F2/F2' gives a counterclockwise torque. The only problem is springs rotates the torus in the clockwise direction, so I will attract the torus 1 with the container 2 and attract the torus2 with the container1. I can't attract all the part of the torus but the bigger part so the torque is counterclockwise.
Title: Re: Sum of torque
Post by: EOW on November 11, 2015, 08:36:05 AM
I hatching the area where I exchange the attraction (transparency mode):

Container 1 attract the torus 2
Container 2 attract the torus 1

Before the green line attract the green torus and the red line attract the red torus. But now, the green line attract the red torus where the hatched area is (don't forget there are 2 torus) and the red line attract the green torus where the area is hatched. The green line attract the rest of the green torus (not hatched area), and the red line attract the rest of the red torus like that there is no force on the green line and on the red line. Like that I don't have any force on the red line and on the green line.

I drawn container1+torus1 on one image and the same for the container2+torus2.

With one device the sum of energy is keeped. F1 and F2 want to turn counterclockwise but the attraction from the springs on the torus want to turn clockwise, so the sum is constant.

With 2 devices, I attract from the green line a part of the torus1 and a part of the torus2. The torque on the torus1 is near 0 or maybe it's counterclowise when I see the surface of the hatching compared of the all surface inside the container. It's the same for the torus2.

The device turns around the red center, it is always like I drawn: it's important to keep constant the length of the springs. The springs don't lost any energy.
Title: Re: Sum of torque
Post by: EOW on November 12, 2015, 02:26:02 PM
The green line is fixed to the Container1, the red line is fixed to the Container2.
I don't attract all the torus, only the part that is inside the Container1 or the Container2.
The springs keep their length constant, the device turns around the red axis but all turns at the same angular velocity.
Title: Re: Sum of torque
Post by: EOW on November 13, 2015, 07:56:33 PM
The device can be limited to one torus and one container. But the walls of the torus are separate of the interior torus. The inner wall and the outer wall of the torus is in contact with pressure, this pressure can't give any torque to the torus but gives the force F1 and F1 gives a counterclockwise torque on the device. Now, when I attract the interior of the torus it will rotates clockwise, but it's not the same energy I lost. With inner radius at 7 and outer radius at 8.72 the torque on the interior torus is 11.8 but the force F1 gives a torque of 17 and the force on the walls of the container gives a counterclockwise torque too.

At start, all the device turns counterclockwise.
Title: Re: Sum of torque
Post by: EOW on November 14, 2015, 11:16:09 PM
If the green (interior) torus is fixed to the red walls the solution doesn't works. I need to get the torque from F1 in the center of the torus, for that I need to have the force from pressure from the red walls. So when I attract the green part of the torus from the green line, I give only one clockwise torque ,  but if the green torus and the red walls are the same piece the torque on the torus is twice.

With the green torus and red walls the same piece:

- torque from F1 =  17
- torque from torus = -11.6*2
- torque from the container = 6

So the sum is at 0

With the green torus not the same piece that red walls:

- torque from F1 =  17
- torque from torus = -11.6
- torque from the container = 6

The sum is not 0

To keep constant the length of the springs I can use a motor on the blue axis, this motor needs an energy but less than the device can recover. I adjust the torque of the motor to have the length of the springs constant.
Title: Re: Sum of torque
Post by: EOW on November 15, 2015, 04:46:34 PM
Like that the red center don't have any torque from torus.
Title: Re: Sum of torque
Post by: EOW on May 23, 2016, 07:56:22 PM
I calculated the sum of forces on this device and I don't find 0. I used small balls attracted by springs like before.  Friction is low. The law of attraction is in 1/d² not 1/d. There are a lot of balls inside the area (or the volume). Balls give pressure like water inside a recipient under gravity but the law need to be in 1/d² (like gravity) or another law 1/d³ for example.

The device is in a unstable position but like I drawn the sum of forces is not 0. I don't try to rotate the main black arm, just have the sum of forces on the gray device, and I don't find 0.

Datas:

Radius of the circle : 0.5
Length d = 1
Pressure at top : 0
Pressure at bot : 0.5

RED WALL:
Force on axis x from red wall:
integrate( 0.5-1/(2-x) dx from 0 to 1 = 0.193

Torque on the black arm from the red wall:
integrate( (0.5-1/(2-x))*(x-2) dx from 0 to 1 = 0.25

BLUE WALL
Force on axis x from the black axis:
0.5*integrate( 0.5-1 / ( sqrt( (0.5*cosx)² +(1.5+0.5*sinx)² ) ) * cosx dx from -pi/2 to pi/2 = 0.1666
Force on axis y from the black axis:
0.5*integrate( 0.5-1 / ( sqrt( (0.5*cosx)² +(1.5+0.5*sinx)² ) ) * sinx dx from -pi/2 to pi/2 = 0.364
Torque on the black axis from the force:
1.5*0.1666 = 0.25

The torques cancel themselves so the half circle don't give any energy.

The force from the springs on the red axis:
In x : 0.0674
In y : 0.129

The sum of forces on x: -0.2+0.1666+0.0674 = 0.034 MAYBE if I divide 0.0674 by 2, I have 0.034 but not exactly, it is 0.03371 so the sum is not 0 too, the result 0.2 and 1.6666666 is known, the result 0.03371 is very precise, the sum is 0.00037
The sum of forces on y: 0.364-0.129 = 0.235, here if I divide by 2 or 4 it's worse..., the sum is not 0

I can't have 0 in x and y axis in the same time. I'll integrate to find the best result.

Below, an example for calculate the sum of springs in the red center, I changed the result because I tried to find 0 :

#include <stdio.h>
#include <math.h>

int main()
{
double N=2000;
int i, j, c=0;
double s1=0, s2=0 ,x,y1,y2;
for(i=0;i<N;i++)
{
x=(double)i/N*0.5;
for(j=0;j<N;j++)
{
y1=+(double)j/N*0.5;
y2=-(double)j/N*0.5;
if ((x*x+y1*y1)<=0.25)
{
s1+=fabs(cos(atan((y1+1.5)/x))/(x*x+(y1+1.5)*(y1+1.5)));
s1+=fabs(cos(atan((y2+1.5)/x))/(x*x+(y2+1.5)*(y2+1.5)));
s2+=fabs(sin(atan((y1+1.5)/x))/(x*x+(y1+1.5)*(y1+1.5)));
s2-=fabs(sin(atan((y2+1.5)/x))/(x*x+(y2+1.5)*(y2+1.5)));
c++;
}
}
}
printf("\nc=%i\n",c);
printf("s1=%lf\ns2=%lf\n",s1/c,s2/c);
return 0;
}

In the i7.png image, I drawn the red line side by side the black arm, the red wall don't have a link with the black center, and the red forces is reported to the red axis, sure.

The sum of forces in x and in y axis is not 0. It's possible to move the device in any direction.  I drawn spring but it's possible to use any device for have the attraction like magnets, electrostatic of hydraulic.

At start, I calculated for the torus device, in this last device, there is nothing in the half disk. The balls are all around the torus, attracted by springs from the red center, center of the torus. The torus is stable, it will move in straight line with a force  and turn in the same time.

I drawn half disk but it's possible to use another shapes.
Title: Re: Sum of torque
Post by: EOW on May 24, 2016, 08:07:30 AM
The forces from red and blue walls are correct. Note, the half blue disk can turn around the black center, sure the half disk don't turn around the black center. The black arm don't turn because the sum of torque is 0 but it is in a unstable position.

I need to divide by 4 for my program because I integrate a square 0.5 by 0.5

My program:

x: 0.0337
y: 0.232

Double integration:

x:0.0308 or 0.0249 if I divide only by pi
y:0.282 or 0.228 if I divide only by pi

I think I need to divide by pi because the length of the arc is 0.5*pi but I divide by 0.5 due to the first integral, so the result seems near from the program:

integration:

x: 0.0249 near 0.033
y: 0.2288 near 0.232

But the sum of forces is not 0

For cancel the forces in x, need to have 0.0264
For cancel the forces in y, need to have 0.182, I'm far with 0.23

With the lib MPFR and a precision of 200 bits (62 digits in base 10), the sum at x is -0.193+0.166666+0.0337 =0.007 and in the y axis, the sum is 0.182 - 0.232 = 0.05 the result converge, even I change NN, the result don't change for 7 first digits.

Double integration with double integration gives 0.078 in x axis, I think I need to multiply by the area of the disk (mean) so it is 0.392*0.0785 = 0.0308 and the result is not 0.193 but 0.197, note that the program and Wolfram find near the same result

With the y axis the difference is 0.282 - 0.182 = 0.1

I think Wolfram don't integrate with 200 bits precision so I think my program is more accurate.

Centrifugal forces can do the same.

Title: Re: Sum of torque
Post by: EOW on May 25, 2016, 08:33:28 AM
I think I need to divide by 4 the integrals not by pi, like that the sum of forces are at 0.

Now, for increase the torque, I can change the position of the center for the black arm, the length of the black arm must be variable (telescopic). I take the green center at 1.5 in horizontal, like that I have a radius of 2.121.

X: the force in axis x, it is 1/6
Y: the force in axis y, it is 0.182
A: the small angle to rotate, like 0.1°
L: the length of the black arm, it is 2.12

The energy from the black arm is sqrt( X^2 + Y^2 ) * A / 2 * L * sin(45°+atan( X/Y )= 0.2617*A and I need to take in account the length of the black arm change, so the length increase, I lost energy. This energy is dL*F, dL is the difference from start to end:

dL = L - sqr( ( L/sqrt(2)*cos(A) ) ^2 + ( (1+sin(A)*L/sqrt(2) ) ^2 )
F = sqrt( X^2 + Y^2 ) * cos(45+atan(X/Y))

If A = 0.1, dL = 1.85e-3 and F=0.011 so the energy lost is 20e-6 it is very small compare to 0.02617 (A=0.1)

sqrt( X^2 + Y^2 ) * A / 2 * L * sin(45°+atan( X/Y ) - ( L - sqr( ( L*sqrt(2)*cos(A) ) ^2 + ( (1+sin(A)*L*sqrt(2) ) ^2 ) ) * (  sqrt( X^2 + Y^2 ) * cos(45+atan(X/Y)) ) - 0.25 * A

At final, the energy is 0.02617-20e-6-0.025 = 0.001053 J for an angle of 0.1°
Like the angle can be very small, I don't take the effect of the direction at final, I consider the direction of force and arm is the same than at start.

I added true value of the integral because all value depend of the angle and it changes all the time

Note the result is low but the force on the black center is 0.182 N it is small. With an angle of 5° and a force of 200 N the energy recover is 58 J and the device can turn at a very high angular velocity. Note I don't need mass, it's only shapes with attraction, like that I don't have any problem from the centrifugal force, it's possible to have a high angular velocity.
Title: Re: Sum of torque
Post by: EOW on May 26, 2016, 06:30:06 AM
1/ I'm sure of the values of the forces because I found the sum of forces at 0. I need only the forces on the red wall and on the black center because the device turns around the red center
2/ The device don't move in straight direction, it's only a rotation around the red center. The black center turns too around the red center even the  black arm turns around the green center because the length of the black arm increases during the rotation
3/ The springs never lost any potential energy because all the springs are attached in the red center and the device turns around the red center.
4/ I'm sure about the torque on the red wall, it is 0.25
5/ I'm sure about the forces on the black center: x: 1/6 y:0.182
6/ Like I drawn the device can turn because I turned it with an angle of 30°, I need only to adjust the length of the black arm
7/ The last thing to be sure is the energy won by the black arm:
I'm sure of:
7.1/ If the device turns of X degrees then the black arm turns of X/2 degrees, I calculated and verify with the geometry
I need to take in account:
7.2/ The torque on the black arm, and take care about the direction of the force. The value of the force is always : sqrt( 0.166² + 0.182² ) because the springs don't change their lengths
7.3/ I need to increase the length of the black arm, so I need an energy for that.

So, my calculation is:

---------------------- ok from 0 to pi/2 ---------------------------
Fx=1/6
Fy=integrate( 0.5 - 1/sqrt( 0.5*cos(x) + (1.5 + 0.5*sin(x) ) )*sin(x)*0.5 from -pi/2 to pi/2 = 0.182355
L=sqrt( (1.5*cos(2*x))^2+(1.5+1.5*sin(2*x))^2) // I need to have 2x inside trigo functions because when I integrate I do from 0 to A/2
F=sqrt( Fx^2 + Fy^2 )
B=atan( Fx / Fy )
Energy recover from black arm = Integrate( F * L * sin( pi/4 -x + B ) from 0 to A / 2
Energy needed to increase the length of the black arm = Integrate( F * L ) )* cos( pi/4 -x + B ) from 0 to A / 2
Energy lost from red arm = A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1 = 0.25 * A

At final the energy won is :

Integrate( F * L * ( sin( atan( C ) + B ) -  cos( atan(C) + B ) ) from 0 to A/2 - A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1

Example, A=0.1 rad sum at : 0.005829 J

With the force 1000 times higher (182 N) the device gives 0.1J and if it turn at 1000 tr/s it will give 100 W

I will verify today again but I think my calculations are correct.
Title: Re: Sum of torque
Post by: EOW on May 26, 2016, 08:00:37 PM
I checked my equations, all seems fine, there is a difference of torque. I compute with 500 digits and I find the same value than Wolfram, so the values of the integrals are correct and from 0 to 0.1 rd the value 0.0494171 it is not 0.5.

Datas:

A: the angle of rotation of the red arm, the black arm turns of A/2
Fx=1/6
Fy=integrate( 0.5 - 1/sqrt( 0.5*cos(x) + (1.5 + 0.5*sin(x) ) )*sin(x)*0.5 from -pi/2 to pi/2 = 0.182355
L=sqrt( (1.5*cos(2*x))^2+(1.5+1.5*sin(2*x))^2) // I need to have 2x inside trigo functions because when I integrate I do from 0 to A/2
F=sqrt( Fx^2 + Fy^2 )
B=atan( Fx / Fy )
Energy recover from black arm = Integrate( F * L * sin( pi/4 -x + B ) from 0 to A / 2
Energy needed to increase the length of the black arm = Integrate( F * L ) )* cos( pi/4 -x + B ) from 0 to A / 2
Energy lost from red arm = A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1 = 0.25 * A

At final the energy won is :

Integrate( F * L * ( sin( pi/4 - x + B ) -  cos( pi/4 - x + B ) ) from 0 to A/2 - A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1

I try with an attraction with the law 1/d and another law and it's the same, there is a difference. The final integral is logical. The difference come from sin(x)-cos(x). The integral from 0 to 2pi is 0 but from 0 to 0.1 for example the sum is not at 0.

The max is gave from x=0.7629 to pi+0.7629

Nobody ?
Title: Re: Sum of torque
Post by: EOW on May 27, 2016, 10:22:48 AM
Note the start of the integration, it's the vertical not the horizontal. And the direction is clockwise. Axis x is horizontal and axis y is vertical.

I found where come from the difference, it is :

integrate sqrt( (1/6)^2+0.182355^2 )*(sqrt( (1.5*cos(2*x))^2+(1.5+1.5*sin(2*x))^2)-1.5*sqrt(2))*(sin( pi/4-x+0.7404789456 )-cos(pi/4-x+0.7404789456))  dx from x=0 to 0.05

it's exactly that. This work come from the length of the arm increases, at start it is 1.5*sqrt(2), but the arm increases when the red arm turns so this extra length give an extra energy.

The energy won is the extra length of the black arm * the force (constant) * sin (angle) and the sinus is 0.99 at 0.05 - extra length of the black arm * the force (constant) * cos (angle)  and the cos is 0.094 the difference is here

The energy recovered is 0.2 J for a turn, so with a force 1000 times higher (180N) and a rotation of 1000 tr/s the power is 200 kW. The rotation can be high because there is no mass.
Title: Re: Sum of torque
Post by: EOW on May 28, 2016, 08:50:21 PM
For me it's logical but maybe not for all: the device is not composed of one body. There are 2 parts:

1/ The red wall with its red arm connected to the red center.
2/ The half circle and its black center of rotation. The black arm is connected to the black center.

If I let the device like that, the red wall rotates to the counterclockwise and the half circle clockwise, and sure, the balls move closer to the springs, and the springs lost their potential energy. But it's not like that I want to use the device. I have a motor on the red center that force the red wall to rotate and follow in the clockwise direction the half circle. Like balls are very small (in theory imagine balls like molecules of water) I need to add gaskets. So, I calculated the energy I need to give to the motor and I calculated the energy I recover from the black arm. The black arm needs to change its length because the are 2 centers of rotation.

It's possible to replace the springs by magnets or electrostatic force or any hydraulic device.  It's possible to use a law of attraction different of 1/d², like (1/d^x) with x any real number. It's possible to use another shape than a circle like an ellipse. Here, it's a 2D device, but it's possible to have a 3D device.

Here the force of pressure is very small (0.5) but it's easy to have a high pressure to increase the power of the device.

Title: Re: Sum of torque
Post by: EOW on May 29, 2016, 01:09:34 PM
Maybe I made a mistake about the force F2y, it is (maybe) 0.14709. My program at start was good for the forces F3x and F3y.The result change but the sum of torque is not 0. In fact, this is not very important because it is possible to change the law of attraction and to find a vector with a good value.

With 500 binary digits and 3e6 steps the result is 0.025034 and even I change a little the values this don't change the difference enough to have the sum of energy at 0.

I got it, it's 0.14780 not 0.182 for F2x. It's nothing the result it's the same, there is a difference to have the sum of energy not at 0.

And don't forget, the force is small and I integrated for a small angle, if I integrate from 0 to 0.5 rd I have 0.219, it's far from 0.25

Someone can confirm ?
Title: Re: Sum of torque
Post by: EOW on May 30, 2016, 11:33:50 AM
I think the integrals are good, I had a problem of coef in my program and it is ok now. Like that I have the sum of forces at 0 and without the green center I have the sum of torques at 0 too.

But with the green center I can have a difference in the sum of energy.

I have the definite integral so it can't be a problem of accuracy

Title: Re: Sum of torque
Post by: EOW on May 31, 2016, 07:33:48 AM
I drawn an image to show how is the black axis, without the balls and the red wall to watch details. Spoke are not in the same plan like that they don't interact with balls. The semicircle can only turn around the black axis and the black axis is fixed to the black arm. The semicircle can only rotate around itself but there is no torque on it (because it is a part of a circle and the forces of pressure are perpendicular to the surface). The semicircle is in an unstable position but in theory there is no torque on it and in practice it's possible to correct with an external device the position.

I don't wrote but I show an image with a rotation of the device, but note:

The red wall is free to rotate around the fixed red axis, a motor force the red wall to rotate clockwise and follow the half disk
The black arm (telescopic) is free to rotate around the fixed green axis, it has the force F2 at end (black dot). The black arm rotates around the green center and increases its length
the semicircle is free to rotate around the black dot but it don't rotate around it because there is no torque
The trajectory of the black dot is a circle with a radius of 1.5 and the center is the red axis.
I used balls inside the half circle but it's possible to use another shape like ellipse.

The black axis is not connected with the red arm.
Title: Re: Sum of torque
Post by: dieter on May 31, 2016, 01:59:34 PM
I see you're very creative. As usual this is way over my head ^^ nevertheless: keep up the good work!
Title: Re: Sum of torque
Post by: EOW on May 31, 2016, 05:35:44 PM
Thanks. Maybe someone could understand what I'm trying to explain :)
Title: Re: Sum of torque
Post by: LibreEnergia on June 01, 2016, 10:07:44 AM
Thanks. Maybe someone could understand what I'm trying to explain :)

Noether's Theorem tells us that for every differential symmetry there is a corresponding conservation law.  In my opinion if you are finding anything other than zero change in energy while integrating over a path that starts and ends in the same physical positions then your analysis is wrong.

In this case 1, The motion is continuously differentiable, ie it has no inflection points, and it does represent a mathematical  'symmetry' , in this case a rotation,  so it meets the conditions for which Noether's theorem applies.

In my opinion you will find the error in analyzing the way you are calculating the integrals numerically, in that small errors introduced when summing numbers containing large differences in numerator and denominator give rise to an error in the totals.
Title: Re: Sum of torque
Post by: EOW on June 01, 2016, 01:26:19 PM
In my opinion you will find the error in analyzing the way you are calculating the integrals numerically, in that small errors introduced when summing numbers containing large differences in numerator and denominator give rise to an error in the totals.

I had this problem with the F3x and F3y forces because it is a double  integral. Even Wolfram is not perfect for the double integral I wrote a program with a high accuracy, and like that I found the sum of forces at 0 with the number of digit I want (Wolfram not, my program yes). But I don't need F3x and F3y in my use of the device because  the red axis is fixed. And I have the indefinite integrals for F2x, F2y, F1x and for the torques.

Do you understand the device ?

For Noether, don't forget there is an hypothesis: the potential energy don't change with time.

I corrected my integrals for the F3x and F3y, I divided by 4 but the true integral take x inside. The results are the same but the integrals are easier to calculate
Title: Re: Sum of torque
Post by: EOW on June 02, 2016, 04:23:16 PM
The indefinite integral for the force F2. I don't think this is a problem of accuracy.

And the difference of energy is more than 20 % for an angle of 1rd, with the force I use, the difference of energy is 0.25-0.203=0.046 J it is near 20%. If the force is 1000 times higher (F2=247 N) the difference is 1000*0.046=46 J. It's easy to increase the force because the springs like the balls are always in the same relative position inside the half disk. If the device turns at 10 tr/s, the power is 460 W.

@LibreEnergia: Do you understand how the device works ?
Title: Re: Sum of torque
Post by: EOW on June 03, 2016, 11:23:02 PM
With an extreme accuracy and with a very small angle it's possible to have directly the result without an integral, there is always a difference:

#include <stdio.h>

#include <gmp.h>
#include <mpfr.h>

#define DG 2000

int main()
{
mpfr_t   x,l,angle,c,s,r,r2,sum,dl,fx,fy,f,temp1,temp2,temp3,temp4,temp5,temp6,temp7,temp8,temp9;
mpfr_init2 (x, DG);
mpfr_init2 (l, DG);
mpfr_init2 (angle, DG);
mpfr_init2 (c, DG);
mpfr_init2 (s, DG);
mpfr_init2 (r, DG);
mpfr_init2 (r2, DG);
mpfr_init2 (dl, DG);
mpfr_init2 (sum, DG);
mpfr_init2 (fx, DG);
mpfr_init2 (fy, DG);
mpfr_init2 (f, DG);
mpfr_init2 (temp1, DG);
mpfr_init2 (temp2, DG);
mpfr_init2 (temp3, DG);
mpfr_init2 (temp4, DG);
mpfr_init2 (temp5, DG);
mpfr_init2 (temp6, DG);
mpfr_init2 (temp7, DG);
mpfr_init2 (temp8, DG);
mpfr_init2 (temp9, DG);

mpfr_set_d(x, 1e-20, MPFR_RNDD);
mpfr_set_d(l, 0.0, MPFR_RNDD);
mpfr_set_d(fx, 1.0, MPFR_RNDD);
mpfr_div_d(fx, fx, 6.0, MPFR_RNDD);
mpfr_set_d(fy, 0.182355, MPFR_RNDD);
mpfr_div(angle, fx, fy, MPFR_RNDD);
mpfr_atan(angle, angle, MPFR_RNDD);

mpfr_mul(temp2, fx, fx, MPFR_RNDD);
mpfr_mul(temp3, fy, fy, MPFR_RNDD);
mpfr_sqrt(f, temp2, MPFR_RNDD);

mpfr_mul_d(temp2, x, 2.0, MPFR_RNDD);
mpfr_sin(temp1, temp2, MPFR_RNDD);
mpfr_mul_d(temp1, temp1, 4.5, MPFR_RNDD);
mpfr_sqrt(l, temp1, MPFR_RNDD);

mpfr_set_d(temp2,3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196,MPFR_RNDD);
mpfr_div_d(temp2, temp2, 4.0, MPFR_RNDD);
mpfr_sub(angle, temp2, x, MPFR_RNDD);

mpfr_sin(s, angle, MPFR_RNDD);

mpfr_mul(temp5, l, s, MPFR_RNDD);
mpfr_mul(r2, temp5, x, MPFR_RNDD);

mpfr_sqrt_ui(temp7, 2.0, MPFR_RNDD);
mpfr_mul_d(temp7, temp7, 1.5, MPFR_RNDD);
mpfr_sub(dl, temp7, l, MPFR_RNDD);
mpfr_cos(c, angle, MPFR_RNDD);
mpfr_mul(r, dl, c, MPFR_RNDD);

mpfr_mul(sum, sum, f, MPFR_RNDD);

printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, l, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, angle, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, s, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, c, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, r, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, r2, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, sum, MPFR_RNDD);
printf("\n");
printf("\n");

//mpfr_clear (s1);

return 0;
}

Title: Re: Sum of torque
Post by: Gabriele on June 04, 2016, 04:28:40 PM
Hello. Why don't you model your idea with WorkingModel2D... should be simple...
Title: Re: Sum of torque
Post by: EOW on June 04, 2016, 05:42:55 PM
Thanks for the name of the software but it's not possible to simulate N balls I tried with Comsol.

I found my error it was in the integral about the length of the arm.

Another idea: take 3 semicircles. One like before, radius of 0.5 and 2 smaller with a radius of 0.25. There is no straight wall like before. I do the same with balls and springs with the law 1/d^2 (or 1/d^3). I place the device on an object in rotation.  Each semicircle can turn with its arm (color).

The torque from the 3 semicircles device is 0 on the green dot but the object turns and it can receive a difference.

I calculated the 3 forces (radial), I have:

Tangential forces :
F1=1/28 (black)
F2=1/6 (red)
F3=15/100 (green)

The value are exact it is indefinite integrals.

The torque at 1.5 is 1/28*1.75-1/6*1.5+15/100*1.25 = 0
But at 10 it is 1/28*10.75-1/6*10.5+15/100*10.25 = 0.1904

Don't remember the semicircles don't receive any torque around themselves from pressure of balls because the pressure is perpendicular to the surface (center of the circle). I want the device keeps its shape because walls are 3 bodies, so I need to apply a torque on the center, for example I give a torque on the red center to compensate the 1/6 torque. I need an energy but I can recover more from 2 others.

Note the main object will receive a torque on it.
Title: Re: Sum of torque
Post by: EOW on June 05, 2016, 02:47:32 PM
I verified my tangential forces, they are correct.

Note, it's very important do not block the semicircle on the main object. Each semicircle must give it's force on the gray axis to have the sum of forces at 0 on the gray axis. So the device must accelerate more and more, it wins only a potential energy. This energy can be recover later not in same time.

And it's pretty good with the Neother's theorem :)

I added an image with the forces:

F1: force from pressure on the black semicircle to the black center
F2: force from pressure on the red semicircle to the red center
F3: force from pressure on the green semicircle to the green center

Like each semicircle can turn around the gray axis, forces on the red/green/black axis are reported to the gray axis
F1' : from F1
F2' : from F2
F3' : from F3

The springs gives a force on the gray axis:
Fs1 : for attract all balls of the bigger semidisk
Fs2: for attract all balls of the smaller semidisks

I added the external force Fext because I want the red semicircle follows the others semicircles.

The sum of forces on the gray axis is 0 (x or y)

At d=0 the torque is 0 but with d different of 0 there is a counterclockwise torque.

I don't need radial forces, but tangential forces are:

F1t = 1/28
F2t = 1/6
F3t = 3/20
Fs1t = 0.02648
Fs2t = 0.007433
Fext = 1/6

Tangential force on the gray axis  = 0

3 images to show each axis alone.

It's not only a sum of torque different of 0 but when the forces F1, F2, F3 apply their force on axis, these forces are applied to the gray axis, so the sum of force is at start (tangential forces):

F1=+1/28
F2=-1/6
F3=+3/20
On gray axis=-0.019

The sum is at 0 but the forces F1-F2+F3 go to the gray center so the sum is:
On gray axis: -0.019 +1/28-1/6+3/20 = 0
But the forces F1-F2+F3 is not 0 it is 0.019

----------------------------------------------------------------------------------------------------------------------------

If I come back with only one semicircle with straight red line (look before) it's strange too, the sum of force on the gray axis is 0 but the forces are not the same so at distance the sum of torque is not 0. If d=10 then the red torque is integrate((0.5-1/(2-x))*(x-10) dx from x=0 to 1) = 1.79 but the black dot torque is 9.5 * 1/6 = 1.583 there is a difference. There is no force on the red axis.

---------------------------------------------------------------------------------------------------------------------------

The last device, the torque is -0.25+0.0568 +1/6*1.5-1/6*0.5 different of 0
Title: Re: Sum of torque
Post by: EOW on June 06, 2016, 12:01:45 PM
Note: axes don't interact with balls ! Each basic heart shape is composed with 3 bodies (3 semicircles). Axes are in another plane.

I take the basic shape I drawn before (like a heart shape). But there is nothing inside the heart shape. I imagine I have a lot of sizes, bigger, smaller, like that I can full all the blue disk with basics heart shapes. Each basic heart shape gives a clockwise torque. It's possible to reduce the size of the heart shape so small that it can full all the blue disk. So at final the springs attracted balls from the white center with the law 1/d^2 (or another law) and there are very few balls if balls are very small. In fact, I need only one thickness of ball and if the size of ball is near 0, the volume attracted by springs is near 0, so I can have no torque on the white center from springs.

In fact, I can dot like before and place balls inside the heart shapes IF I take a lot of heart shapes with different sizes, like that I attract all around the device balls everywhere. And like that I'm sure of my results. The torque is counterclockwise in this  case.

I plot the difference of torque, it's logical that it is always in the same direction but like that I'm sure all basic heart shapes give a clockwise torque (nothing in it).
Title: Re: Sum of torque
Post by: EOW on June 07, 2016, 06:21:01 AM
I drawn the side view: balls don't touch the axes.

Note that each semicircle receives the forces only from the balls not from another semicircle.

I need to give to the red semicircle a torque in the good direction but I recover more from 2 others semicircles.

My integral at start was good. For example, the big circle (main device) is at 20, so I used 1/20 in the equation. I plot the result of the integral with plot3d, function of 'd' and function of 'r'. 'd' is the distance from the white center (main device) and 'r' is the radius of the semicircle. A basic heart shape is composed of 3 semicircles, 2 with the radius 'r' and one with the radius '2r' like you can see in the equation of the integral.
Title: Re: Sum of torque
Post by: dieter on June 07, 2016, 07:31:48 AM
It's just amazing how productive you are. Maybe you should really try it in some physics simulation software, as Gabrielle suggested.
Title: Re: Sum of torque
Post by: EOW on June 07, 2016, 08:28:45 AM
I don't have a simulator for now. Maybe it's possible to simulate only one heart shape to show if I'm right with the torque, someone ?

I think it's not a problem of sum of torque, but sum of forces. The sum of torques is always at 0 but the sum of tangential forces of 3 dots (green, red, black) is not 0, with one shape the springs compensate this difference but if I place a lot of sizes I don't have anything to attract (very few balls) so the main device can be a sector of a disk full with heart shapes. The forces from pressure on the walls 1 and 2 are canceled by the buoyancy force that each heart shape has on it. But the sum of tangential forces from the difference of green/red/black forces can't be canceled.
Title: Re: Sum of torque
Post by: Gabriele on June 07, 2016, 12:01:51 PM
I think any solid on which we apply a an equal force for each point of its surface,will obtain only equilibrium
Title: Re: Sum of torque
Post by: Gabriele on June 07, 2016, 12:04:10 PM
i dont know
Title: Re: Sum of torque
Post by: EOW on June 07, 2016, 01:09:54 PM
My forces are correct for one device, I found the sum of forces at 0 and the sum of torques at 0 too. The difference of tangential force from the green/red/black force is canceled by the springs because the shape is asymmetrical. But I can take a sector of disk full with heart shapes with different sizes, in theory I can full the device and like that there is very few balls inside the sector and the springs can't compensate the sum of tangential forces. The buoyancy force is compensated by the forces on the straight lines (1) and (2). I calculated the indefinite integral and I plotted in 3d with the parameter 'd' and 'r' to show there is always the difference in the same direction. Have you any idea what can cancel the tangential forces ?

And I can take any shape, like half disk (straight line and center of the semicircle), or another shape, I need only to fill the sector with all the basic shape with different sizes. The sum of torque is 0 but not the sum of forces.

I need to take small semidisks, very small to prevent the asymmetrical forces from fa/fb. For example, if the sector has a radius of 10 I take for the semidisks a radius from 1e-10 to 1e-6 not more.

I calculated the difference of force when I reduce the size of the semidisk, if I divide by 2 the radius of the disk, the force is divided by 4 but like the surface is multiply by 4 I don't lost anything in the force:

#include <stdio.h>
#include <math.h>

#define N 100

int main()
{

double sum=0, sum2=0,r,d=2,l;
int i;

l=1.0;
r=l/(double)N/2.0;

for(i=0;i<N;i++)
{
l=1+1/(double)N*(double)i;
sum+=r*(-sqrt(l*l+4.0*r*l+4.0*r*r)/(r*l+r*r)+l/(r*l+r*r)+2.0/d);
//   printf("\nl=%f",l);

sum2+=-log(l+2*r)+log(l)+2*r/d;
}

printf("\nr=%f\ns=%f\ns2=%f\ndiff=%f\n",r,sum,sum2,sum-sum2);

return 0;
}

And for a big precision, it's possible to see the force is divided always by 4 when the radius is divided by 2 but the surface is 4 times higher:
#include <stdio.h>
#include <math.h>

#include <gmp.h>
#include <mpfr.h>

#define DG 400
#define N 1000

int main()
{

mpfr_t   s1,s2,r,d,l,t1,t2,t3,t4,diff;
mpfr_init2 (s1, DG);
mpfr_init2 (s2, DG);
mpfr_init2 (r, DG);
mpfr_init2 (d, DG);
mpfr_init2 (l, DG);
mpfr_init2 (t1, DG);
mpfr_init2 (t2, DG);
mpfr_init2 (t3, DG);
mpfr_init2 (t4, DG);
mpfr_init2 (diff, DG);

mpfr_set_d(s1, 0.0, MPFR_RNDD);
mpfr_set_d(s2, 0.0, MPFR_RNDD);
mpfr_set_d(d, 2.0, MPFR_RNDD);
mpfr_set_d(l, 1.0, MPFR_RNDD);

mpfr_div_d(t1, l, N, MPFR_RNDD);
mpfr_div_d(r, t1, 2.0, MPFR_RNDD);

//long double sum=0.0, sum2=0.0,r,d=2.0,l;
int i;

//l=1.0;
//r=l/(long double)N/2.0;

for(i=0;i<N;i++)
{
//l=1+1.0/(long double)N*(long double)i;
mpfr_set_d(t1, 1.0, MPFR_RNDD);
mpfr_div_d(t1, t1, N, MPFR_RNDD);
mpfr_mul_d(t1, t1, i, MPFR_RNDD);

//sum+=r*(-sqrtl(l*l+4.0*r*l+4.0*r*r)/(r*l+r*r)+l/(r*l+r*r)+2.0/d);
//   printf("\nl=%f",l);
mpfr_mul(t2, r, r, MPFR_RNDD);
mpfr_mul(t1, r, l, MPFR_RNDD);

mpfr_mul_d(t1, r, 4.0, MPFR_RNDD);
mpfr_mul(t3, t1, r, MPFR_RNDD);

mpfr_mul_d(t1, r, 4.0, MPFR_RNDD);
mpfr_mul(t4, t1, l, MPFR_RNDD);

mpfr_mul(t1, l, l, MPFR_RNDD);

mpfr_sqrt(t1, t1, MPFR_RNDD);
mpfr_div(t1, t1, t2, MPFR_RNDD);

mpfr_div(t3, l, t2, MPFR_RNDD);

mpfr_d_div(t4, 2.0, d, MPFR_RNDD);

mpfr_sub(t1, t2, t1, MPFR_RNDD);
mpfr_mul(t1, t1, r, MPFR_RNDD);

//sum2+=-logl(l+2*r)+logl(l)+2*r/d;
mpfr_mul_d(t1, r, 2.0, MPFR_RNDD);
mpfr_log(t1, t1, MPFR_RNDD);
mpfr_sub(s2, s2, t1, MPFR_RNDD);

mpfr_log(t1, l, MPFR_RNDD);

mpfr_div(t1, r, d, MPFR_RNDD);
mpfr_mul_d(t1, t1, 2.0, MPFR_RNDD);

}

mpfr_sub(diff, s1, s2, MPFR_RNDD);

//printf("\nr=%Lf\ns=%Lf\ns2=%Lf\ndiff=%Lf\n",r,sum,sum2,sum-sum2);
printf("\n");
mpfr_out_str (stdout, 10, 0, s1, MPFR_RNDD);
printf("\n");
mpfr_out_str (stdout, 10, 0, s2, MPFR_RNDD);
printf("\n");
mpfr_out_str (stdout, 10, 0, diff, MPFR_RNDD);
printf("\n");

return 0;
}
Title: Re: Sum of torque
Post by: EOW on June 08, 2016, 10:40:55 AM
Maybe like that it's easier to the sum of forces: there are the forces from the springs and there are the forces from the axis, all at left.
Title: Re: Sum of torque
Post by: dieter on June 13, 2016, 10:52:33 PM
Have you found a physics simulation software yet?

Do you believe a physics simulation software could accurately and correctly simulate it?

Title: Re: Sum of torque
Post by: EOW on June 14, 2016, 12:25:25 PM
I found an error in my device, if I reduce the size of the semi disk the force decreases in the same time. I tried to find a good simulator on Linux but they are difficult to use.

I'm trying to find where is my error in this case with 2 wheels.

1/ The two wheels move to the right
2/ One wheel rotates clockwise, the other counterclockwise
3/ The pressure is at 0.01 P everywhere except in the square shape delimited by the green/red walls
4/ The pressure is at P inside the shape delimited by the green/red walls
5/ Altitude of the wheels don't change
6/ There is no gravity
7/ The green V shape is free to rotate around the orange axis
8/ The orange arm is free to rotate around the black axis
9/ The red V shape is free to rotate around the blue axis
10/ The black arm is free to rotate around the pink axis
11/ Like there is a difference of pressure, the green V shape has a torque on the orange axis, this torque is 1
12/ Like there is a difference of pressure, the red V shape has a torque on the blue axis, this torque is 1
13/ I add a spring between the green V shape and the red V shape (not at the axes), to cancel the torque on each V shape, I need a force of 1/sqrt(2) because the distance is sqrt(2) and the torque is at 1
14/ So, now with the spring, each V shape don't want to rotate, they keep their relative position
15/ The force from the difference of pressure on the V shape on the orange axis is f11, this force =1
16/ The force from the difference of pressure on the V shape on the orange axis is f10, this force =1
17/ Each wheel receives a torque of 1-sqrt(2)*R,  like each wheel turn they can give an energy
18/ All volumes are constant, the gas don't lost any potential energy (I hope...)

I have only a problem with the wall of the circle between the V shapes but it's possible to take wheels just side by side at start, like the straight velocities are the same, the wheels are always side by side, an example:

The equation of the cycloid with a radius of the circle of 1:

x=a-sin(a)
y=1-cos(a)

But be careful that the reference is not the same for the left wheel and the right wheel.

For the spring at start:
Left wheel, the point at start is (0,0)
Right wheel, the point at start is (2,0)
For the spring at final, after 0.01 rd for each wheel:
Left wheel, the point at final is (0.02,50e-6)
Right wheel, the point at final is (2+1.666e-7,50e-6)

The difference of length is 0.02

The force needed to compensate the torque is 1/sqrt(2)

The energy for the spring is 1.414 e-2

The force from the V shape is 2 so the force on each axis is 2-1/sqrt(2). Note I take a big V shape, the radius of the wheel is 1 and a side of the V shape is sqrt(2).

The energy from the wheels is 2*1*(2-1/2)*0.01 = 2.58 e-2

There is a difference

The forces:

F1 and F4 come from the spring
F1 gives F2 to the orange axis and F2 gives F3 to the center of the left wheel
F4 gives F5 to the blue axis and F5 gives F6 to the center of the right wheel
F10 come from the pressure on the red V shape
F11 come from the pressure on the green V shape
Title: Re: Sum of torque
Post by: dieter on June 14, 2016, 07:38:33 PM
Wikipedia has a long list of open source simulations:

https://en.m.wikipedia.org/wiki/Physics_engine

Sorry, I don't see where the error is.
Title: Re: Sum of torque
Post by: EOW on June 14, 2016, 09:41:24 PM
Thanks for the link, if I don't find my error I will simulate the device.

With the following example and an angle of 0.001 rd:

spring lost 0.0010005
wheels win 0.001

In this example I can find near the sum of energy at 0. But not others examples.

It seems before pi/4 the wheels win more energy than the springs lost and after it is the contrary. I calculated for pi/8 (near like the first case I drawn in a last message). I found 2.7e-4 for the springs and 7.8e-7 for the wheels (the force on the axis is near 0)
Title: Re: Sum of torque
Post by: EOW on June 15, 2016, 01:18:33 PM
I took another example:

Radius of the wheels = R = 1
β = pi/6 rd
δ = 0.01 rd (the small angle I calculate the energy)

The length of one V shape's wall is R/2*√2, so the force is  R/2*√2 = 1/√2 for each wall (I consider the pressure outside at 0 to simplify), like forces are at 90° the force from the difference of pressure is 1 N

The torque is 2*R/4*√2 = 1/√2 Nm

The force Fs for the spring is 1/√2 /(2*sin(β)) = 1/√2 N

The force on the orange or the blue axis is 1-1/√2 = 0.2929 N
The energy for the spring is 2*(cos(β)-cos(β+δ))*Fs = 7.13 e-3 J
The energy for the wheels is δ*2*Fa*Rf = 2.93 e-3 J

Maybe I can take Fb in another position and the red object could be not half square (an angle different of 90°)

I can replace the spring by 2 stems (telescopic stems like hydraulic cylinders) like that I don't lost any energy.

If I cancel the torque from the center I lost sqrt(2)*angle but I win angle/2
Title: Re: Sum of torque
Post by: EOW on June 16, 2016, 09:23:39 AM
I was wrong about the torque with a constant pressure.

If I use small balls attracted from the center or use a fluid (liquid is better, centrifugal forces are higher), the force and the torques are not the same.

Title: Re: Sum of torque
Post by: EOW on June 17, 2016, 08:31:21 AM
Maybe if I transmit the torque from one arm to another.

The yellow arm can turn around the pink axis. The black arm can turn around the white axis (center of the circle). Each arm has a different torque, higher to the yellow.
Title: Re: Sum of torque
Post by: EOW on June 18, 2016, 06:04:53 AM
I drawn all forces and details. I supposed the torque can be at 2/3 for the yellow arm and 1/3 for the black arm, it's an example. Like I can adjust the law of attraction I can have this example. But maybe my forces are wrong.

With gears it's logical.

I can rotates the device around the pink axis to have a continuous movement. Look at the epicycloïde. But I need to have N device with a phase angle at 360/N.

If the wheel moves to the right the energy is destroyed, if it moves to the left the energy is created.

Title: Re: Sum of torque
Post by: EOW on June 19, 2016, 09:02:05 AM
I had a problem with the force from the gears, like the radius of the gear is 2 times lower, the force is 2 times higher so the sum of energy is at 0.

Now, I used a torus (like a wheel of a bike that moves in translation and rotates). There are small balls in a part of the torus.

Title: Re: Sum of torque
Post by: dieter on June 19, 2016, 09:30:29 PM
I am currently using the open dynamics engine ODE with an Api wrapper for Blitz3D.

So now I can do this in a BASIC programming language / compiler. This is good not only due to simplicity, but also due to extremly fast compilation/run times, compared to bulky, complicatified Cpp paths.

But the accuracy / precision of ODE is low, so maybe it's too low for you. Anyway, I could send you the wrapper. ODE is open source, Blitz3D too, on github I think.
Title: Re: Sum of torque
Post by: EOW on June 20, 2016, 07:15:03 AM
Thanks for the name of the software  :) do you use it to simulate some devices ?

I can give the torque from the ground and give the force at the bottom from the ground in the same time.

Title: Re: Sum of torque
Post by: EOW on June 21, 2016, 12:13:59 AM
This case is better, I found the torque at 0.74 at top, 0.37 at the center and 0.26 from the torque of the blue wall. 0.74-0.34-0.26=0.11
Title: Re: Sum of torque
Post by: EOW on June 22, 2016, 07:21:46 AM
I added all informations for verify my calculations, if someone has a little time to verify and tell me if it is correct ?

Title: Re: Sum of torque
Post by: dieter on June 24, 2016, 05:36:12 AM
Sorry, I was offline for a while.

I had in mind to simulate magnets together with normal newtonian physics, but it turned out to be rather complicated, so I started programming a magnet simulation entirely in Blitz (without ODE).

It works, a bit too well tho: my permanent magnet motor works in almost every configuration :-) very much unlike in real live.

ODE has all kinds of links, also friction and things. I did some basic tests, like a curtain or chains, a rotor with hinge, stuff like that.

If one takes some time to get into ODE, I think such devices as you designed can be simulated quite effectively.

At least this is a package that works out of the box and you don't have to compile the ODE Api by your own.  (that is for "DC"'s api distro)
Title: Re: Sum of torque
Post by: EOW on June 25, 2016, 08:59:06 AM
It could be useful to simulate.

I drawn the device in many positions. Note if I don't use the ground for the force F1, the sum of energy is always at 0. If I use the ground for F1, I can win an energy in a few positions like the first I drawn, the angle is pi/2 I think. In other cases when I can't win an energy, I don't use the ground for the force F1. I wait the next good position and I use the ground.
Title: Re: Sum of torque
Post by: dieter on June 27, 2016, 02:58:29 AM
To me equations are the secret code of the academic class ... invented to disguise simple terms in a dense fog of mysterious haze and archaic greek letters . <:^)

Title: Re: Sum of torque
Post by: EOW on June 27, 2016, 07:23:16 PM
Thanks Dieter, I will try to simulate it because I don't find the error in this device. I'm not sure the device is well explained...
If someone knows maths I will appreciate comments or questions about the device.

When I wrote the wheel moves and rotates, it is at the time I studied the sum of energy. Like the device can accelerate or decelerate it is not easy to count the energy, so it is easier to think with an external device ED that force the wheel to move to the right and rotate in the same time, that device ED recovers an energy for some forces or moments and needs an energy for others forces/moments. I count these energies.

I corrected some errors in my drawings. I gave the sum of energy with formulas.

Axes:
The green axis is fixed on the wheel

Fixed bodies:
The ground
The device that gives the force Fq

Discreet mobile bodies:
All the device moves in translation and rotates clockwise like a wheel of a bike

The red wall
The outer circle of the wheel and the blue wall (blue color = one body)
N balls (small like molecules of water, just for simplify the calculations), without mass (simplify)
N springs, without mass (simplify calculations)

Connections and constraints:
The red wall can only turn around the green axis, the red wall is in contact with balls only, the red wall receives the pressure from the balls
Each ball is attracted by a spring from the center C
The device like that is unstable but I give the force Fq from an external device fixed on the ground.

Constraints on motion :
When the wheel moves and rotates, all the device turns and moves, the springs never lost their potential energy because the balls are always in the same relative position

Physical properties of all bodies:

All bodies are rigid except the springs
The springs and the balls are without mass
The wheel has a mass.
Each ball is attracted with the law 1/d² with 'd' the distance between the center of the wheel and the ball

Title: Re: Sum of torque
Post by: dieter on June 29, 2016, 04:42:19 AM
I was still waiting for you to answer my personal message, so I can mail you the physics sim files, if you want it.

I'm having some access provider troubles here, so I may be offline for a while.
Title: Re: Sum of torque
Post by: EOW on July 01, 2016, 01:01:07 PM
Sorry, I was busy these last days. I sent a pm.

I find the error for the last device but I try with a triangle. I changed only the shape and the point of attraction. I sent equations later.

Have a good day :)
Title: Re: Sum of torque
Post by: EOW on July 02, 2016, 10:07:06 AM
Maybe like that.
Title: Re: Sum of torque
Post by: EOW on July 02, 2016, 11:51:07 PM
Maybe directly on the circle (no ground)
Title: Re: Sum of torque
Post by: EOW on July 03, 2016, 08:07:17 PM
With the wheel in rotation the energy is :

Energy lost by the moment: √3*R*δ*F = 0.001*√3*1*0.45 = 7.79e-4 J
Energy win by the center: √3*R*δ*F = 7.79e-4 J
Energy needed for the force F2: ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) *2*F = 4.27e-4 * 2 *F = 3.84 e-4 J
Title: Re: Sum of torque
Post by: EOW on July 05, 2016, 09:38:15 PM
With a gas under pressure inside the triangle, the sum of energy is not 0:

L = 1 m =  thickness of the device
P = 1 Pa = pressure of the gas
R = 1 m = radius of the wheel

The force on the red arm is √3 N so F2 = √3 N
The torque on the red arm around the red axis is 1.5 Nm

Energy studied for an angle of δ = 0.001 rd

I apply the force F2 from the ground, I lost an energy because the length increases. The force F2 come from a spring for example. I consider there is no acceleration, I can use a big mass for the wheel of use an external device to keep constant the velocity.

Energy lost by the moment on the red axis: √3*R*δ  = √3e-3 J
Energy win by the center in translation: √3*R*δ = √3e-3 J
Energy needed for the force F2: ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) *2*F = 4.27e-4 * √3 = 7.39 e-4 J
Title: Re: Sum of torque
Post by: EOW on July 11, 2016, 01:47:58 PM
I don't need the balls and the springs. I can use the pressure of a gas.

The wheel rotates and turns before I studied the sum of energy.

There are 4 bodies:
A: the wheel + 2 red walls, the center of the wheel receives the black force, the outer circle of the wheel receives the red forces
B: the black wall, can turn around the red axis (i), receives the gray forces
C: the spring, gives F2 to the black arm and -F2 to the ground
D: the ground, it is fixed, receives the green force

Radius of the wheel = 1 m
All forces at sqrt(3) N
P = 1 Pa

F1 come from the difference of pressure of the gas on the black wall
F2 come from the spring to the black arm
-F2 come from the spring to the wall
Fi come from the black arm on the axis
-Fi come from the axis to the black arm
Fo come from Fi to the center of the wheel
-Fo come from the center to the outer circle
F4 come from the difference of pressure of the gas on the red walls

I studied the sum of energy for a small angle δ like  0.0001 rd for example

3 walls (2 red+1black) shape a equilateral triangle, there is a gas under pressure at 2P inside the triangle. The surface of the triangle is always constant, the gas never lost its potential energy. Outside the triangle there is the pressure P.

The wheel moves to the right and rotates clockwise like a wheel of a bike, BUT THERE IS NO FRICTION BETWEEN THE GROUND AND THE WHEEL.

Like I want the surface of the triangle constant, I need the spring and the force F2.

There is a gasket between the black arm and the red wheel but I don't drawn it.

The mass is in the center of the wheel and in the outer circle of the wheel.

The center of the wheel (move to the right) win the energy the moment on the wheel lost (the moment is counterclockwise and the wheel rotates clockwise) : +1.5Rδ-1.5Rδ = 1.5 e-3 -1.5e-3 = 0 J
The spring lost 7.4e-4 J
The force F4 works very few compared to others, it is  √3Rδ²/2 = 8.6e-7 J

The sum of energy is not at 0, in this case the device destroy an energy

Title: Re: Sum of torque
Post by: EOW on July 12, 2016, 09:41:40 AM
I drawn forces with less length .

First: destroy an energy
Second: create an energy

I don't drawn them before but there are spokes on the wheel
Title: Re: Sum of torque
Post by: EOW on July 13, 2016, 02:45:35 PM
In fact, I don't need pressure nor the black and red walls, just a bike and a spring. I drawn only the wheel not the bike. There is NO friction between the ground and the wheel.

The force Fo gives the same energy than the torque -Fo/Fw on the wheel lost. But the spring lost an energy.

The energy won by the force Fo to the bike is lost by the torque -Fo/Fw on the wheel but the spring lost an energy. The sum of energy is not at 0. The center win FRδ. The wheel lost FRδ. If F=1 N and R=1m, the energy lost/win is δ. For δ=0.001 rd the energy win/lost is 0.001 J.

The length increases: dl = ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) with θ at 300° The energy for the force F2 is at dl*F2 = 4.27e-4 * 1 = 4.27 e-4 J
The spring lost 4.27e-4 J

Maybe I can attach the spring with a wheel bigger (trochoid)  like that the spring can win/lost more.
Title: Re: Sum of torque
Post by: EOW on July 14, 2016, 10:36:15 PM
The spring is attached to the blue object and the black arm. The ground receives nothing. The blue device lost the energy that the wheel win in translation but the wheel lost an energy in rotation. The spring lost near nothing.
Title: Re: Sum of torque
Post by: EOW on July 30, 2016, 04:27:36 PM
The spring directly on the wheel, no force from the ground or to the ground. The trajectories of the points X and Y are not the same.
Title: Re: Sum of torque
Post by: Gabriele on July 31, 2016, 11:12:12 AM
No my friend
Title: Re: Sum of torque
Post by: Gabriele on July 31, 2016, 11:14:19 AM
Better...how do you link the spring to the wheel?
Title: Re: Sum of torque
Post by: EOW on July 31, 2016, 11:39:43 AM
The springs is attached on the wheel, two ends of the spring is attached on the wheel. Note, there is no axis of rotation like a wheel of a bike, the mass is only at the outer circle.

Title: Re: Sum of torque
Post by: EOW on August 01, 2016, 01:03:45 PM
I use a continuous track and a road wheel. I suppose the road wheel can receive a torque from Fg (there is an axis of rotation of the wheel). The energy lost by the spring is won by the track but the energy win by the torque on the road wheel is greater than the energy lost by the center of the wheel.

Lg3 is higher than 1/sqrt(2) so the sum of energy is not 0 because the center of the wheel receives always F/sqrt(2) because the direction of the force is always the same: pi/4.
Title: Re: Sum of torque
Post by: EOW on August 02, 2016, 11:27:45 AM
I suppose the force from the spring constant even the length increases.

The center of the wheel is (0,0)
With an angle of rotation of 0.01 rd

Point P1:
At start (2 , 1)
At final (2.02 , 1)

Point P2:
At start (1 , 0)
At final (1.00995000043 , -0.009999833)

At start the angle of the forces from the spring is atan( (1-0) / (2-1)) = pi/4 rd
At final the angle of the forces from the spring is atan( (1+0.009999833) / (2.02-1.00995000043) ) = 0.7853733291 rd

cos(pi/4) = √2/2
cos(0.7853486586) = 0.7071243415

The difference is 1.75e-5 so the force on the center of the wheel change from √2/2 to √2/2-1.75e-5
And the torque is exactly √2/2+(0.01-1.75e-5)/√2 the sum of energy can't be at 0

Title: Re: Sum of torque
Post by: EOW on August 03, 2016, 02:44:43 PM
I calculated with a program and I find a difference. With quadmath it's the same difference.
Title: Re: Sum of torque
Post by: EOW on August 04, 2016, 03:54:49 PM
I found my error it was in the equation of the distance from the center.

Maybe with friction and forces from the ground with 2 wheels
Title: Re: Sum of torque
Post by: EOW on August 05, 2016, 03:24:26 PM
Maybe a wheel and a spike full of water under gravity. I can adjust the contact wall/water like I want. The trajectory of the point is in red. I named the wall of the wheel elastic but maybe it is more dynamic, it is possible to imagine a wall theoretical  that can move like the wheel. I just want to benefit the pressure on the wall A for example.

Image m4p7: it's possible to rotate the spike in the same time the wheel rotates and moves. The spike needs a lower energy to rotate than the wheel won.
Title: Re: Sum of torque
Post by: EOW on August 10, 2016, 04:12:52 PM
The torque from the liquid is not the same the torque from the solid, the wheel rotates and moves alone:

The law of gravity is 1/d² not 1/d.
Title: Re: Sum of torque
Post by: EOW on August 11, 2016, 09:00:18 AM
Radius of the wheel = 0.5 m
Direction of gravity : vertical
Origin of gravitation: 0.5 m below the ground
Law of attraction: 1/d²

Torque on the solid:

integrate from 0 to 0.5 of integrate from -pi/2 to pi/2 of x^2*cos(y)/(1+x*sin(y))² dx = 0.0986123 Nm

Torque on the walls from the liquid:

integrate from 1.5 to 1 of (x-1)*(1/1.5-1/x) - integrate from 1 to 0.5 of (x-1)*(1/1.5-1/x) dx = 0.0112016 - 0.1098 = - 0.0986123 Nm

There is no torque on the wheel.

Now, if I attract from a point below the red dot at 0.5 m, the pressure on the walls from the liquid don't change but the torque on the liquid change

Like it is the same density for the liquid and the solid, the potential energy of gravity don't lost any energy when the wheel rotates.

Title: Re: Sum of torque
Post by: EOW on August 11, 2016, 11:43:54 PM
Like the attraction come from a point (the green point), I need to give an energy in the center of the wheel but that energy is a potential energy that the wheel wins. The torque 0.0986123 - 0.083333 = 0.0152 Nm is the torque that can create an energy when the wheel rotates counterclockwise. Maybe it's possible to use a sphere.
Title: Re: Sum of torque
Post by: EOW on August 12, 2016, 02:28:41 PM
Like that the cg of the semi disk solid don't work. There is negative torque on the wheel from the liquid on the walls but the work is :

( 0.1098 - 0.0112 ) x = 0.0986 x with x a small angle of rotation

There is an energy from the center of the wheel:

(0.1098 + 0.0112 ) = 0.121 x

The vertical force don't work and the cg of the semi solid disk can't work
Title: Re: Sum of torque
Post by: EOW on August 13, 2016, 09:03:20 AM
I have the good values for the forces with the integrals I have the sum of forces at 0 (not with my programs).

The torque from the solid part is -0.252879+0.0675103+1.36114 = 1.5465087 Nm

The torque from the liquid is 1.38147-0.0277572 = 1.3537128 Nm

There is a difference.
Title: Re: Sum of torque
Post by: EOW on August 13, 2016, 02:39:21 PM
The wheel lost an energy because the force on the white object is perpendicular of the trajectory, it is possible to controlled the velocity by an external device like that there is no forces on the balls from the white object. When the wheel rotates and moves the white object lost a potential energy.
Title: Re: Sum of torque
Post by: EOW on August 13, 2016, 07:25:39 PM
In fact, like I drawn before the spring, the poitn don't work but the spring increases its length !!

The spring is attached between the wheel and the ground. Its length pass from 1.414 R to 1.57 R

Cost nothing for the wheel because the positive clockwise torque from the force is cancelled by the force on the center of the wheel.

The wheel rotates and moves before I attached the spring. There is an inertia, so the wheel continue to move and rotate.
Title: Re: Sum of torque
Post by: EOW on August 14, 2016, 11:20:05 AM
I can use a water+gravity device. There is no force on the green point so I don't need any energy to rotate the black arm. There is a torque on the pink arm and like the angle between the black arm and the pink arm change I can recover an energy not with a motor because in this case the motor will give a torque on the black torque, but simply rotate a disk, the disk is on the green axis without friction, it can't give any torque on the black arm. I win the rotation of the disk, potential energy in the rotation of the disk.
Title: Re: Sum of torque
Post by: EOW on August 14, 2016, 11:50:39 PM
Use a local gravity, replace water by small balls and gravity by springs attached on the yellow point. Turn all the device clockwise. The yellow point turn, it is in the device, the springs to have gravity don't lost any energy.
Title: Re: Sum of torque
Post by: EOW on August 15, 2016, 04:57:32 PM
The law of attraction need to be linear in the contrary there is a torque on the container I think.

It's easier to think with a line of attraction like that it's possible to use the formulas of buoyancy/pressure in water with gravity.

I need to adjust the law of attraction in the same time the radius of the black arm change, like that I can have no torque on the container and have one to the pink arm. Change the law of attraction cost nothing in energy because the balls are always in the same position in the container so the springs don't lost any energy.

An example to calculate the law of attraction:
law in 1/d^0.2 with big length for the black arm
Title: Re: Sum of torque
Post by: EOW on August 16, 2016, 01:03:31 PM
I need a negative pressure, it is possible to do with an attraction from the left side.

All the device turns, even green line and yellow line.

Title: Re: Sum of torque
Post by: EOW on August 17, 2016, 08:09:45 AM
laws 1/x^0.5 and 1/x^0.3
Title: Re: Sum of torque
Post by: EOW on August 17, 2016, 09:09:20 PM
The idea is to use balls1 attract with a law of 1/d². In the container I put a lot of balls2 (empty) but not attract, if I was inside water under gravity I could say these balls2 are bubble with gas inside. I calculate the torque on the black arm around the green center, like balls2 give the complementary pressure I have no torque on the black arm around the green center. Now, the force in the green center is 1.5555 N and if I place the green axis at 11 meter from the center of rotation of the device I have 11*1.5555 = 17.11 Nm. But the torque on the left side is not 17.11 Nm it is 18.197 Nm. Look at the integrals.

The function is (1/3-2*(1/3^2-1/x^2))
Title: Re: Sum of torque
Post by: EOW on August 18, 2016, 09:36:24 AM
Or maybe like that with disks or spheres, I think disks are better

The volume of the blue ball is:

H*H-((H/-2*R)²/sqrt(3)*2*pi*R²) = constant = 0.0931 H² with R the radius of the white ball, the result don't depend of the radius of the white ball.

There is a up force on the object. The disk (2d) 7.519 % than a square (don't forget white balls are quinconx) so I could say the up force is 7.519% of the up force without white balls.

I need a force of 7.5 % and the blue balls move down with 9.31 %

If I move the object down, don't forget each ball win an energy.
Title: Re: Sum of torque
Post by: EOW on August 19, 2016, 07:27:02 AM
Forces change because blue balls can't access all the part between a circle and a line, it depends of the radius of the blue ball
Title: Re: Sum of torque
Post by: EOW on August 20, 2016, 12:00:35 AM
I think the pressure on the white balls is higher because there is a part (J) where there is no balls between each white balls. So the result could be more negative.
Title: Re: Sum of torque
Post by: EOW on August 20, 2016, 12:27:20 PM
I suppose the sum of forces at 0, so I need the parameter 's' with the container alone so I added the parameter 's' but now, with the object I can't have the sum of forces at 0...
Title: Re: Sum of torque
Post by: EOW on August 21, 2016, 07:47:16 AM
I found the sum of forces at 0 in the containers without the Object X.

But not with the Object X.
Title: Re: Sum of torque
Post by: EOW on August 22, 2016, 07:38:20 AM
I verified my calculations and I corrected a little the values. I take in account the area the green line don't attract. All in the device is static, nothing moves, it is the sum of forces that is not 0, so all the device can create or destroy an energy and move the CG alone.
Title: Re: Sum of torque
Post by: EOW on August 22, 2016, 10:29:21 PM
I need to add a thin tube for each white disk to allow the pressure  of blue disks to be transmitted below. This lost a little.

If that can work like that, it could say it is possible to have a up force of buoyancy higher than the volume of water (under gravity with a container and water), and I don't need disks just a shape like I drawn. Block the parts between disks with a wall.
Title: Re: Sum of torque
Post by: EOW on August 23, 2016, 11:31:19 AM
Rotate the Object X ?
Title: Re: Sum of torque
Post by: EOW on August 24, 2016, 01:51:35 AM
If I compare the energy recover from a disk, half is a solid, the other half is air, I lost more energy in the water.
Title: Re: Sum of torque
Post by: EOW on August 25, 2016, 12:16:19 PM
With the shape I described before.
Title: Re: Sum of torque
Post by: EOW on August 26, 2016, 10:51:44 AM
like that
Title: Re: Sum of torque
Post by: EOW on August 28, 2016, 11:15:33 AM
Maybe like that
Title: Re: Sum of torque
Post by: EOW on August 29, 2016, 12:18:30 AM
The torque from white depends of 2 pressures
Title: Re: Sum of torque
Post by: EOW on September 02, 2016, 12:12:59 AM
With 2 pressures
Title: Re: Sum of torque
Post by: EOW on September 04, 2016, 10:58:44 AM
With my method I found the sum of energy at 0 with a gas or with attracted disks (or spheres) but I don't find the sum of energy at 0 with not attracted disks (or spheres).
The sum of forces are at 0 too.

With gas:
from the torque and the force:  0.5*δ*1.1-0.5=0.05
from the volume: 0.005*10*δ=0.05

Only attracted disks:
from the torque and the force: 0.5*δ*1.1-0.516666=0.03333
from the volume: 0.005*10*δ*0.666=0.03333   , I need to take in account the fact that more disks with pressure at 10 moves out than 0 so I used the calculation of the center of gravity of a square triangle

With disks, attracted disks and not attracted disks
from the torque and the force: 0.9535*1.1*δ-1.045*δ=0.0385δ
from the volume: 0.005δ*9.689=0.04844δ

With spheres, attracted spheres and not attracted spheres
from the torque and the force: 0.87*1.1*δ-1.0361333*δ=0.079δ
from the volume: 0.913*0.005δ*10=0.0456δ

Title: Re: Sum of torque
Post by: EOW on September 07, 2016, 08:35:05 PM
The tube turns clockwise around the red dot and counterclockwise around the black dot.

The difference of volume are the same.

To find the same volume resolve the equation:

(1.1²-1²)/2*pi-(3.1²-3²)/2*pi/x     this gives x at 2.904761    like that the volume is constant and the difference of energy is 2.90895-1.0004*2.904761=0.003
Title: Re: Sum of torque
Post by: EOW on September 08, 2016, 11:25:47 AM
Like that

Title: Re: Sum of torque
Post by: EOW on September 09, 2016, 08:19:11 AM
Like that
Title: Re: Sum of torque
Post by: EOW on October 01, 2016, 02:32:02 PM
--
Title: Re: Sum of torque
Post by: EOW on October 04, 2016, 09:15:29 PM
I take a last idea (first image) but I change a little the device (second image). I place the tube inside water and the tube is full of water, the density of the tube is like water.
Title: Re: Sum of torque
Post by: EOW on December 17, 2017, 04:56:15 PM
torque vs translation