Grease power

[preoccupied]

There needs to be 3 reversed gears to make the purple weights weightless to shift. The middle reversed gear needs to connect to one of the outer reversed gears. I represented this in the top left of the image. Sam Peppiatt and SHADOW does this give a better explanation of what I share? I posted about it in your thread and you didn't say anything at all about it! I thought I was missing something. This drawing is meant to turn CCW. I'm not perfect and I have possible brain injury. I've got to say though. Progress made if I have this concept correct.

[SHADOW]

Bonjour Preoccupé,
Vous devriez representer le mouvement de votre principe sur plusieurs dessins phase par phase.
Votre représentation est trop surchargée pour être lisible!
Pour l'idée des roues lestées et transfert vers un pignon, il me semble que cela a été testé!
Je fais une Sim pour voir.

Hello Preoccupied,
You should represent the movement of your principle on several drawings phase by phase.
Your representation is too overloaded to be readable!
For the idea of weighted wheels and transfer to a sprocket, it seems to me that this has been tested!
I make a Sim to see.

[SHADOW]

Est ce que cela corespond à votre concept?

Est ce que cela corespond à votre concept?

[SHADOW]

Votre concept après un quart de tour!

Your concept after a quarter turn!

[preoccupied]

I'd like to apologize for my previous drawing which was flawed inherently. This new drawing here tries to give new logic to something else.

Premise of this idea is that I can take the drag on the wheel of the weights off by putting it against a rim of a stationary track in the center. Then it will apply the force of the levers on the distance of the stationary track (the track doesn't move) compared to the wheel. So the distance of the levers applies to the wheel but its connected to the track in the center instead of applying force of gravity on its positions. I think that the track in the center even though its small cancels out the positions of the weights and applies force to the track only. The red right angles are pressing against the track using a reverse gear and the blue right angles are pressing against the track in the direction they are falling. the reverse gear is needed for the red right angles because they are turning CW while the wheel is turning CCW. THE WHEEL IS TURNING CCW. The inner right angles in the drawing is where I was measuring to see when it would be balanced. It should be balanced there if the weights were there.

WEIGHTS ON THE RIGHT hanging on the wheel

The circle in the center holstering the gears is 400 radius. The right angles are 2400x2400. And what's the purpose of using right angles? At this point it's probably structural because the levers are very long. It should probably be a right triangle for structural reasons.

2400+400=2800 right angle to the right and center
diagonals 2800*0.7=1960 AND X2 IS 3920! I'm was also negating the right angles that cancel each other out on the bottom right and top right. Okay never mind so it's actually 400+0.7=280 and (2400*0.7*2+(400*0.7))=3,640 for the total 3,920! which is the same thing.
2400 right angle in red on the bottom
2800+2400+3920=9120
TOTAL IS 9120

weights on the track
2400*4=9600

480 over balanced as it sits in the drawing. I lose some overbalance on the left most right angle if it continues to move along the track as it pulls inwards. I will lose 2400 there. But I gain some overbalance on the wheel to the left side of the wheel from the top most right angle on the right side of the wheel in which I gain 2400 at the end of its turn.

[preoccupied]

Close up on the gears, it's the very center circle that is the stationary rim that the gears push against. It does not move. The gears are connected to the lever and strapped to another gear that connects to the stationary rim. The red lever needs to be a reverse gear because the lever will be tuning CW while the wheel turns CCW.

As you can probably notice right away the gears and the stationary rim in the center are all the same size. What this does is make the lever turn the same angle as the wheel is turning. That is what I intend because the right angle levers turn 90 degrees as the wheel turns 90 degrees also. It also should works in the reverse gear for the red levers.

I'm not yet sure on what the specific mechanics will be for switching to a reversed gear for the red levers after first being on a regular gear for the blue levers. I wonder would it use something similar to what Bicycles use to shift gears? This is in the mechanical engineering realm of thinking. I am stuck in my making drawings as a hobby realm of thinking without proper education in engineering. I got a concept and that was my goal.

[preoccupied]

The rim in the center should be two quarter circles parallel to each other in the z axis direction. The gear connecting to the rim in the center would have a reverse gear in the z-axis direction that would touch the offset second quarter circle rim. This solves the mystery of how the wheel will go in the direction of the wheel on the rim and then reverse it. Does Fletcher have the simulation ability to make my drawing on his software or it is purely 2 dimensional?

[preoccupied]

The premise of this idea is the opposite of my previous idea. I make the assumption that the position of the weights effect the torque even tough its connected to an inner rim. Previous idea, Rim removes weight positions. This idea weights effected by their positions on the wheel and pressure is applied to wheel at the yellow circle from the rim.

If this were Bessler's wheel I think the banging of weights on the falling side of the wheel would be from the weight at the top starting to tilt CW and then fall down CCW again. There is a 90 degree range of motion for the right angle lever in which weights shift 180 degrees sort of because there are two weights. The middle weight looks like it makes it shift 180 degrees when it shifts from the middle to the side. There is some description in the image too. The yellow circle in the center is where the pressure against the wheel from the rim is pulling. The green and red rims separately impact a CW and CCW direction gear movement that allows movement of the right angles at 1:1 ratio with the movement of the wheel. (this is possible by the gear connecting to the inner stationary rims to have a parallel reversed gear that connects to the other rim). The weight possibly offsetting it self on the very top might be able to be managed some by a spring. Because the spring would help in one direction and contradict in the other only to have a null effect by being there but the spring would hold the top most right angle lever a little from falling out of place some which would limit the change that allows the banging on the descending side. The wheel in the drawing is supposed to turn CCW. The overbalance of the whole wheel is shifting the weights by use of stationary inner rims that press against gears.

[preoccupied]

I mentioned in previous post that spring would be needed too. This is kind of how the spring could be positioned in this drawing. And I think seeing the spring there helps you visualize the 90 degree range of motion that I talked about also.

I need help. I think this is Bessler's wheel. It would be embarrassing if I came up with the solution and don't get a working model before the end of the year. I believe my best assumption was that using a stationary rim/catch in the center of the wheel would be better for shifting weights. I think in an actual build I need to split it up into two wheels connected to each other because the gears might not have room in the center to be on the same platform. With everything stuffed in one 2d picture, do I make any sense to anyone? I make sense to me.

I was once a child genius before concussions. I don't have any records that can back that up right now but it's true and my adopting family doesn't believe me. My mother insists she didn't adopt me but it's not something that I agree with. I had a bank that I owned and an early University education. My mother before this mother became a Congresswoman. If I found Bessler's wheel I would be very proud of myself.

[preoccupied]

Tell me if you think my description here is a good representation of where the weights are overbalanced.

All of the weights have a common pivot. So I want to make the pivot point neutral and the position of the weights out of the pivot point plus or minus towards overbalance. I'll compare the right angle to its opposite side right angle. On the very top and bottom in my drawing they cancel out so it's zero. It has got a weight pointing to the right on the top and a weight pointing to the left on the very bottom. On the top left and bottom right, the top left right angle has the same on each side of the pivot so it's like it's all in the pivot or zero. On the bottom right one weight is off to the left and one is neutral at the pivot point. The wheel is trying to turn CCW so that's +1. The middle left and middle right right angles there is two weights, one on the left side of the wheel and one on the right side of the wheel, that are positioned left of the pivot point. That's +2 there. So 3 so far. On the bottom left and top right, on the bottom left two weights are to the left for 1.4 but on the top right one weight is off to the right for minus 1, so the difference of the bottom left and top right is +0.4. 3.4 lever length altogether going CCW, if these positions on the wheel express force then what needs to be overcome is the force necessary to lift four weight sets that are shifting plus the disadvantage of the position of the catch. I moved the catch to the center of the wheel to make it as small as possible. On the bottom a full lever length is needed to shift that right angle and on the bottom right a full lever length is needed to shift that right angle so -2 from those. But on the middle right that right angle is actually moving in the direction it wants to go so that is +1. Then on the top right right angle a full lever distance is needed to shift that right angle so -1. The total in the pictures position is -2+1-1=-2. It costs -2 to shift the weights. The wheel being overbalanced 3.4 and the shifting of the weights being -2 then the loss from the catch in the center can't be more than 1.4. The catch if made close to the axle it is reduced, so as you can see in my drawing it is in the center and very small. Also what can make the difference between the overbalance of 1.4 and the loss from the catch is making the right angles larger. The catch can remain relatively small but the right angles could be larger which would make the total overbalance much more than the loss of using the catch. The loss of using the catch in my drawing is maybe a 1/4th of a lever distance, it is indicated by the yellow circle in the center. The yellow circle is a measuring position and that is where axle of the gear is connected to the wheel. If moving 4 catches it's -1. So my wheel is only theoretically 0.4 overbalanced then. But to make the catch less significant the right angles could be larger. The spring should both resist about equally and then assist so it could be a mute point.

[preoccupied]

In my previous calculation I had 3.4 overbalance minus 2 required to shift the weights and a loss from using the catch in the center of 1. But in this drawing the right angles are larger so the catch is smaller, the yellow circle indicating where the gear is connected to the wheel is less distance from that axle compared to the length of the levers. So I guess it's about 10th of a lever so 0.4 loss from using the catch then. And the wheel would then be 1 lever length overbalanced in the position the wheel is drawn, I mean my hypothesis is that the positions of the weights apply force on the wheel and the effort to shift the weights with also a loss from using the catch equals 1 unit of force over balance CCW in this drawing in the position the drawing is in.

[preoccupied]

What I am doing with this drawing is making the levers larger. This causes the bottom right and middle right levers to extend beyond the axle to the other side of the wheel more. So it should be like 4.4 CCW torque now with the levers this size roughly estimated. The most that shifting the weights can have is 4 resistance plus the catches loss of leverage. Actually it might be a little better than 4.4 because I was measuring the differences from the pivot point of the right angle levers before to find how over balanced it was and now part of the pivot is right over on the other side of the axle of the wheel. The size I made the levers in this drawing is as large as I can have without the levers banging into each other. It could be larger but in an actual build the levers would have to have special positions to be larger than this.

[preoccupied]

I'm trying something here with gears and springs. I sort of have these ratios in mind the 1020.5/1532.32=0.6659 fraction is the smaller red circle connected to the larger gear on the outside perimeter. The 162.5r/25=6.5 is the lever of the red circle connected to the small 1:1 ratio connected to the catch in the center. The 100/244 is the ratio of the size of the lever to the size of the large gear at the perimeter of the wheel. So when the red path is being used by the gears 100/244=0.41 represents the power from the levers and springs there. I would take the power of the lever and springs and go x0.41 and then take that and x0.6659 that, and I think that the remainder 6.5x lever is actually a lever so I want to take the reductions and multiply it by 6.5.

I am making my springs extra strong because I have the second half in red have an extra 45 degree rotation of the wheel in order to try to utilize the strong springs in more places, in exactly one more place. So I have about a 100 pixel lever and I am trying to add 1000 pixels lever worth of spring. I'll go to the end of the turn for the final calculation which would have the worst results. There I will have about 900*0.41*0.6659=245.7171, and that force is applied to the lever for x6.5 equaling 1597.61. I have this in three positions and they are roughly going to be similar so that's 4,792.83.

The blue positions are 1:1 ratios against the wheel and they have 1100 fighting the wheel twice, there is always two positions. 2,200 total. 4792.83-2,200=2592.83. The weights positions around the wheel is about 240 overbalanced and if I have this right, and I have my doubts - the springs lift up their own by arrangement of levers and gears and spring force with a bonus of 2592.83 pixels distance lever force. 2,592.83+240= 2832.83 <---- that should be how much the wheels power is outputting, just about. Now the weights position around the wheel is not as significant but grabbing the wheel by its rim would produce a lot of resistance because the leverage given to the wheel is applied near the center and actions out further would have greater leverage against the spring and gear set up. There is no single weight on the rim of the wheel applying full leverage of the radius of the whole wheel. They are balancing each other out and leaving a small remainder about 100 at a time the basic lever distance for the weight when it overbalances the wheel.

I found this and I don't know if I made my calculations correctly or not, by playing with gears positions and sizes with the idea in mind that I could try using more spring and giving the spring more locations to apply on the second gears catch in red. It starts off reverse flow pulling into the wheel in blue catch in the center then it switches catches and in red pulls in the direction of the wheel with the extra spring force.

Let the bodies hit the floor. Nothing wrong with me.

[preoccupied]

I think that I drew the gear ratio backwards. But what I'm trying to do is move the weights together slower with a spring. I think if I drew the blue circles along the rim with the weights that it would be the correct drawing. The spring pulls in together moving in the same direction as the wheel is turning. When the spring is pulled apart it's supposed to be 1:1 ratio and when the springs pull together with the wheel it's supposed to be 4:1 ratio. I think that the spring should cancel out each other folding and unfolding and the loss from the catch versus the overbalance of the wheel is what's left.

EDIT This ought to be the correct drawing that I intended. The springs should cancel out and the weights should be overbalanced.

Here is another variation in an attempt to be more overbalanced in which the levers swing wider. The gear ratios are 2:1 in and 2:1 out and the springs should cancel out with this and if the wheel is overbalanced in the weight positions it's an overbalanced wheel, minus the loss from the catch. The catch is near the middle of the wheel and very small to reduce the loss from using the catch.

[preoccupied]

tilting right angle27.png

I'm trying something here with gears and springs. I sort of have these ratios in mind the 1020.5/1532.32=0.6659 fraction is the smaller red circle connected to the larger gear on the outside perimeter. The 162.5r/25=6.5 is the lever of the red circle connected to the small 1:1 ratio connected to the catch in the center. The 100/244 is the ratio of the size of the lever to the size of the large gear at the perimeter of the wheel. So when the red path is being used by the gears 100/244=0.41 represents the power from the levers and springs there. I would take the power of the lever and springs and go x0.41 and then take that and x0.6659 that, and I think that the remainder 6.5x lever is actually a lever so I want to take the reductions and multiply it by 6.5.

I am making my springs extra strong because I have the second half in red have an extra 45 degree rotation of the wheel in order to try to utilize the strong springs in more places, in exactly one more place. So I have about a 100 pixel lever and I am trying to add 1000 pixels lever worth of spring. I'll go to the end of the turn for the final calculation which would have the worst results. There I will have about 900*0.41*0.6659=245.7171, and that force is applied to the lever for x6.5 equaling 1597.61. I have this in three positions and they are roughly going to be similar so that's 4,792.83.

The blue positions are 1:1 ratios against the wheel and they have 1100 fighting the wheel twice, there is always two positions. 2,200 total. 4792.83-2,200=2592.83. The weights positions around the wheel is about 240 overbalanced and if I have this right, and I have my doubts - the springs lift up their own by arrangement of levers and gears and spring force with a bonus of 2592.83 pixels distance lever force. 2,592.83+240= 2832.83 <---- that should be how much the wheels power is outputting, just about. Now the weights position around the wheel is not as significant but grabbing the wheel by its rim would produce a lot of resistance because the leverage given to the wheel is applied near the center and actions out further would have greater leverage against the spring and gear set up. There is no single weight on the rim of the wheel applying full leverage of the radius of the whole wheel. They are balancing each other out and leaving a small remainder about 100 at a time the basic lever distance for the weight when it overbalances the wheel.

I found this and I don't know if I made my calculations correctly or not, by playing with gears positions and sizes with the idea in mind that I could try using more spring and giving the spring more locations to apply on the second gears catch in red. It starts off reverse flow pulling into the wheel in blue catch in the center then it switches catches and in red pulls in the direction of the wheel with the extra spring force.

Let the bodies hit the floor. Nothing wrong with me.

When I made the lever smaller than the gear I changed the proportions to the entire equation. As I drew it I would have had to have multiplied 2,200 by 3 to be at the proportion of the length of the lever behind the gear. So as a correction making the levers larger helps the design because it helps the spring and levers effects. Really there was no reason for me to make the levers so small in my example. If the levers were the size of the gear I think that I would have to double 2,200 winding of the spring to 4,400 winding of the spring and then the extra spring use in this scenario would produce 4,400-4792=392 and that would add to the 240 overbalance of the wheel to make 630 overbalance. So my thinking here is that because the lever is behind the gears rim smaller than it I have to add a proportions larger to the rest of the calculation to give value to the lever. Because the value would be less than zero lever length under the gears rim so I have to base it on the proportion of the smaller than gear length as a value and make the rest of the levers larger in value proportionally.