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Author Topic: Sum of torque  (Read 90551 times)

Offline EOW

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Re: Sum of torque
« Reply #45 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.

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Re: Sum of torque
« Reply #45 on: November 09, 2014, 11:17:10 AM »

Offline EOW

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Re: Sum of torque
« Reply #46 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.


« Last Edit: November 11, 2014, 12:01:51 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #47 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|

Offline EOW

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Re: Sum of torque
« Reply #48 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

« Last Edit: November 12, 2014, 11:56:04 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #49 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
« Last Edit: November 13, 2014, 12:20:44 AM by EOW »

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Re: Sum of torque
« Reply #49 on: November 12, 2014, 01:07:00 PM »
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Offline EOW

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Re: Sum of torque
« Reply #50 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.
« Last Edit: November 12, 2014, 11:51:24 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #51 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.

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Re: Sum of torque
« Reply #51 on: November 14, 2014, 05:44:24 PM »
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Offline EOW

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Re: Sum of torque
« Reply #52 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.
« Last Edit: November 16, 2014, 08:23:09 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #53 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.

Offline EOW

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Re: Sum of torque
« Reply #54 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.

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Re: Sum of torque
« Reply #54 on: November 17, 2014, 11:23:37 AM »
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Offline EOW

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Re: Sum of torque
« Reply #55 on: November 18, 2014, 08:46:47 AM »
Like this there is a torque on arm.

Offline EOW

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Re: Sum of torque
« Reply #56 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.

Offline EOW

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Re: Sum of torque
« Reply #57 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 ?

Offline EOW

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Re: Sum of torque
« Reply #58 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.

Offline EOW

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Re: Sum of torque
« Reply #59 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.

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Re: Sum of torque
« Reply #59 on: November 25, 2014, 02:28:40 PM »

 

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