[LB7 (OP)]

I think it is better to take green disks and blue spheres. Like that I can have only one layer of blue spheres for the thickness (perpendicularly to the screen):


syh.png (42.45 kB . 1035x689 - viewed 4243 times)

A disk with spheres all around:

sbbv.png (43.9 kB . 526x476 - viewed 4212 times)

An example with the blue spheres bigger than I drew:


iij.png (225.76 kB . 1103x685 - viewed 4214 times)

Again: each sphere or disk is taken by an arm and the length of the arms is constant, then I need to move out the green disks and even the blue spheres I move in just before. Move out/in don't need/give an energy.

Note: I need the contact between the disks/disks, spheres/spheres, or disks/spheres, so I need a little pressure inside the container, it is easy to do because I move out/in a volume inside the closed container. The pressure could be realized by something else.

I don't think I have a problem with the sum of forces for the blue spheres in contact with the disk because it is symmetrical. But there is a problem with some spheres inside the areas I move out/in because it is very difficult to have the sum of forces always at 0N. But the number of blue spheres that have 0N is high in comparaison. 

I drew the arms for the disk and the spheres:


plm.png (77.82 kB . 723x571 - viewed 4190 times)


Each disk or sphere can rotate around itself (orange axis in the center of the disk of the sphere). At start they don't turn around themselves but like the arms rotate counterclockwise, then the disks and the spheres see the others rotate clockwise. There is contact between the spheres and the disks so there is friction and energy from heating (a lot). The sum of forces for near all the spheres and the disks is 0N. Even there is problem with the last layer at bottom (where I move out) I have 6 contacts for near all spheres so the energy created could be (6-1)*k/6 with k the sum of energy from heating.

[LB7 (OP)]

The sum of energy is (6-2)*k/6 with k the sum of energy from heating.

The only problem arrives when there is no a number integer of blue spheres in a space given, but if I reduce the size of the blue spheres that number is constant but I increase the number of contact of blue spheres: each time I have a blue sphere I win 6 because I have 6 contact and I lost 2 because at the bottom (where I move out) I don't have a contact. So, in theory the energy won by friction increase like 1/R and the energy needed because the number is not an integer is constant. The energy lost because there is no an integer doesn't mean I lost all the energy, it is only a part.

The expression of the energy could be:

N1*(6-2)/6 - N2*f
with f a factor 0 ≤ f ≤ 1 and N2<√N1, f depends of the efficiency to cancel all forces on all blue spheres I move out/in an around them
with N1=0.9*(1-S1)/(πR²), 0.9 because there is a circle packing, S1 is the surface occupied by the green disks, the container is 1m² with a thickness equal at R

The energy from friction is a square law, the problem of the integer is a linear law.

The sum of energy is (6-2)*k/6 with k the sum of energy from heating.

The only problem arrives when there is no a number integer of blue spheres in a space given, but if I reduce the size of the blue spheres that number is constant but I increase the number of contact of blue spheres: each time I have a blue sphere I win 6 because I have 6 contact and I lost 2 because at the bottom (where I move out) I don't have a contact. So, in theory the energy won by friction increase like 1/R and the energy needed because the number is not an integer is constant. The energy lost because there is no an integer doesn't mean I lost all the energy, it is only a part.

The expression of the energy could be:

N1*(6-2)/6 - N2*f
with f a factor 0 ≤ f ≤ 1 and N2<√N1, f depends of the efficiency to cancel all forces on all blue spheres I move out/in an around them
with N1=0.9*(1-S1)/(πR²), 0.9 because there is a circle packing, S1 is the surface occupied by the green disks, the container is 1m² with a thickness equal at R

The energy from friction is a square law, the problem of the integer is a linear law.

To have a continuous movement with the move out of the green disks I replace the green disk by half green spheres, when I move out/in the blue spheres I move up/down (perpendicularly to the screen) :


str.png (50.73 kB . 899x728 - viewed 4168 times)

But I move out the green half spheres up:


sybb.png (41.04 kB . 640x734 - viewed 4160 times)

To have f=0, maybe I can use arms with a fixed but different length in rotation too. In the following image, the sphere has the red arm but it gives the red trajectory (red dotted line), but I need the green trajectory (even for a small movement) so I can replace and use the green arm, move out/in cost nothing. The green arm rotates too, so I the sphere can give force from friction to others. It is difficult in practice but in theory, if I can have f=0, the job is done.


ryg.png (26.79 kB . 637x448 - viewed 4174 times)

If the sum of forces on each blue sphere is 0 I can without any energy change the arm. I recover energy from the friction not from the move out/in.

Even when the blue spheres rotate, I will have space between them, so I need to add not one but  two smaller spheres to have 6 contacts for the friction:


sdx.png (125.52 kB . 795x434 - viewed 4128 times)

[opportunity]

LB7, I really admire the detail you have exercised with your idea.


You're using the idea of circles, proposing the idea of spin, as a one-dimensional idea, lines, circles, rotations, projecting I assume to a 3-d reality.


You're then assuming the annexation of points in space as though such points represent "mass", something that can confer a reality. Am I right?

[LB7 (OP)]

You're then assuming the annexation of points in space as though such points represent "mass", something that can confer a reality. Am I right?

Yes, here I need mass, because I use friction between disks or spheres. Is that your question ?

I drew that example:


avf.png (73.47 kB . 1120x609 - viewed 4070 times)

after a little rotation it is:

rer.png (87.25 kB . 1145x665 - viewed 4084 times)

like it is drew it is like there is a perspective, I don't know if you see it. So my goal is to have in that example 4 contacts to cancel the force on each sphere or disk. I have the lateral contact but I have more and more space between the elements in the height. So, if I use part of cone like that:


cone.gif (2.1 kB . 168x138 - viewed 4080 times)

I can set a horizontal layer like that:


sdt.png (12.81 kB . 454x253 - viewed 4085 times)

And to resolve the friction for the upper layer, I can use another plane:


sedd.png (19.51 kB . 386x250 - viewed 4087 times)

Ok, maybe I need to set the mass at the inner of the cone and change a little the diameter for one cone:


ghh.png (24.82 kB . 472x279 - viewed 4079 times)


rf.png (115.9 kB . 1202x733 - viewed 4087 times)

Like that I don't need to change the shape:


rb.png (51.03 kB . 909x517 - viewed 4063 times)

[opportunity]

Yeeesss….that's what I was asking.

I think what you've done is marvellous.

However, your description of spheres assumedly as mass-points "do not" wash with actual reality (and any theoretical physicists in the forum, go easy).

Your whole matrix of thought is a giant circle itself, and so well constructed at that, and of course as a giant circle it is balanced with energy. "Yet".....do that with known facts of the atom, and "then" you'll have your theory.

Take a look at mine, I should know. I've done pretty much what you have done, yet I acknowledge particle physics:

http://www.equusspace.com/index-2.htm

[LB7 (OP)]

as mass-points

You mean a dot = mass ? I use mass at the inner of the cones if I need to change the shape if not, the mass can be all in the shape.

I drew 2 positions for the layer 2 in 2 times to show how I can have friction always:


ess.png (61.36 kB . 925x530 - viewed 4048 times)

And for the planes if I watch from the left in the front view:


swxx.png (26.85 kB . 251x778 - viewed 4061 times)

It is important to note that the friction between the cones is not all along the cones but only at ONE point (not the segment in contact between the cones).

[opportunity]

Yeah, of course, but friction of "theoretical what"?

[LB7 (OP)]

Yeah, of course, but friction of "theoretical what"?

I don't understand

[opportunity]

You're using circles, making the suggestion that reality on the atomic level operates in that manner.

On a gross level, you have a "circular" idea, a butter-zone for the layman of physics, yet there is an entire absence of all the terms you use, such as "friction", "energy", "mass", and so on...….that for a scientist are terms defined and held according to the atomic level, and must be, and the intricacies of those mechanics and inter-relationship......defined as "absolutely necessary", and thus "ultimate".

[LB7 (OP)]

yet there is an entire absence of all the terms you use, such as "friction", "energy", "mass", and so on

If I understood, physics don't recognize friction, energy and mass ? I don't know if you read my theory about gravity, for me the mass doesn't exist really. But the energy, the friction, why not ?

[LB7 (OP)]

I resumed the movements there:


ruj.png (238.81 kB . 1228x765 - viewed 3993 times)

The goal is to have the sum of forces on each cone equal to 0 N. Like in the front view for the lateral friction I can have only a friction for one point (in rotation) I do the same the vertical friction up/down (I do not use the line of the cone only a point).

With one device I don't need to cancel the up/down forces but like that I can do that:


drbv.png (292.12 kB . 1088x676 - viewed 4065 times)

Or at least, sure, use my parallelogram continue it and form a circular shape like a rectangular torus:


tore.jpg (11.51 kB . 194x112 - viewed 3974 times)

All the external surface has cones or disks

Yes, the vertical forces from friction (if the cones turn around themselves at start) give the heating from friction, the cones lost the same energy in kinetics but there is a net sum of torque for the last layer at bottom:


yhy.png (108.67 kB . 975x776 - viewed 4025 times)

There is a small distance between disks

[LB7 (OP)]

The distances are not the 2 times:


ubf.png (90.98 kB . 1076x750 - viewed 3967 times)


hyk.png (68.63 kB . 1094x477 - viewed 3950 times)

I would have 2 times the distance of the rotation of the basic circle but I have more. So even the disks rotate and lost a kinetic energy they will have more friction than the disks lost. The difference is 2*0.76-1.78 = 0.26 for 5° of rotation in my example.

[LB7 (OP)]

In the following image, I use a rigid connection (like a belt but rigid):


F12.png (115.21 kB . 806x735 - viewed 3943 times)

The disks don't rotate around themselves before start. I deform the device (the arms rotate), the disks rotate clockwise and there is the force F1 and F2 from the belt. Like I use cones it is easy to adjust. I use the "belt" to have friction. Maybe it is possible to use a timing belt and a "brake" is one each cone to recover heating but the friction must be a the points:


trfg.png (116.73 kB . 786x750 - viewed 3944 times)

Mechanically it is better to give the friction to the next arm (the left cone give a force to the next arm):


dyhs.png (111.22 kB . 801x658 - viewed 3926 times)

[LB7 (OP)]

No the next arm increases more than the disk rotates, so it helps, maybe it can increase the angular velocity of the disk when it turns before start around itself:


tuu.png (102.35 kB . 1002x661 - viewed 3938 times)

[LB7 (OP)]

Yes, the arm (d1) increases more than the rotation of the disk (d2):


stg.png (87.81 kB . 1029x809 - viewed 3903 times)


dun.png (77.18 kB . 1120x692 - viewed 3924 times)

There a difference so I think I can use it to change the sum of energy.

In fact, the important thing is d1-d2 ≠ 0

Or I fixed a spring between the dot A (orange disk, at R, not to the arm on the disk) and B (center of the red disk):


detv.png (72.41 kB . 866x606 - viewed 3858 times)

If the disks don't turn before start and if they have a big inertia, it gives:


duj.png (82.68 kB . 995x604 - viewed 3879 times)

d1 ≠ d2 the springs increases less but the arm lost the same.

[LB7 (OP)]

I tried to simplify but the only solution could be:


sxt.png (183.19 kB . 1149x704 - viewed 3860 times)


iko.png (149.22 kB . 1097x618 - viewed 3821 times)


rtg.png (164.25 kB . 1006x724 - viewed 3866 times)

Or with a belt for vertical cones:


gui.png (154.58 kB . 967x686 - viewed 3868 times)

The yellow lines show if there is a belt between 2 vertical cones, the torque on the arms are not the same.

I can place the axis of rotation of the black arm far away like I drew or at vertical:


DTEE.png (121.73 kB . 930x663 - viewed 3860 times)

[LB7 (OP)]

If the bottom line is forced to move to the right and give a torque and the friction to the last cone. The energy needed by to move the bottom line is recovered by friction and for the arm. But I win the friction of all cones.

[LB7 (OP)]

Or maybe rotate counterclockwise around itself a big disk inside a circular container before start, with small cones inside the container. The sum of forces on the arm of the big disk will be 0 N. The small cones don't rotate before start around themselves. The big disk lost an energy because it receives a torque from the friction from the coned. The arms of the cones lost the same energy. I recover the same energy from heating. The sum is - 1-1+1=1

[LB7 (OP)]

The  blue cones don't rorate around themselves, but i rotate the big disk around itself at the same rotational velocity than the arm. The disk lost the same energy the friction give. But the arms of the blue cones lost another energy

[LB7 (OP)]

Only friction between the disk and the cones, not between the cones. The disk wins δFRN with N=πR/r with F the force of friction, R the radius of the disk, r the radius of the cone (at the contact). The energy from friction is δFNr. The energy needed for the black arms of the cones is δFNR/2. The energy to rotate the disk is 0.