Cutting corners

[Tarsier79]

Lifting the weight on to the inner smaller wheel/radius and on to the outer larger wheel/radius from the smaller inner wheel/radius was solved , getting a weight back to the top of the arm was also solved , that's what i was showing .....

Perhaps I am missing something....

[johannesbender]

Lifting the weight on to the inner smaller wheel/radius and on to the outer larger wheel/radius from the smaller inner wheel/radius was solved , getting a weight back to the top of the arm was also solved , that's what i was showing .....

Perhaps I am missing something....

I will do a quick rehash for you , just remember these are simplified line drawings showing instances of what i am discussing.

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ex1.png

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In the above example drawing , the typical torque design scenario is drawn out , a large diameter and a small diameter ,which is setup with a mass up at A , a mass down at C (bright blue).

Given the large diameter (red) , and the small diameter (green) , the mass at A will torque the system (lets presume its CCW) such that the mass at A (bright blue) ends up at the bottom at D (dark blue), the mass at C (bright blue) on the smaller diameter will be torqued upwards to B (dark blue) , and there the system would stop without a reset.

To reset (as is well known) will require a lift for the mass at D to C (along the white vertical lines) , and a lift for the mass at B to A , for which there is not enough energy to complete the lifts back to reset and has been experienced and shown by almost everyone no matter the weight of the masses or the exact size of the diameters or the location of the diameters or the speed of the masses or the distance of the lifts there is not enough energy to do those lifts.

My proposal for this problem , was to not lift the masses by traversing them along the white vertical lines , but to design a skip somehow , because as you know w=f*d the work required to traverse the mass along the distances of the lifts is to much for what is available to such a system .

To me the only way to have a structure or arm that maintains the force of the mass and can be extended and shaped as needed , no matter the position of the weight is placed along the or on the structure or arm , is the same principle in robervals .

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ex2.png

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The Arms A and the arms B on the robervals , will receive the force of the masses placed on those arms , no matter the positions of the masses placed on those arms.

So the only mechanical principle for a mass to skip traveling the distance of a lift , i could come up with , is the principle employed by robervals , so if i add arms on the outer diameter that functions according to that principle , like drawn in the above image at the top and bottom of the large diameter , which extends up and down past the smaller diameter , which makes up the distance of the required lifts the masses must move across to reset , then i can use the arms to place the masses in the required heights to transition from the large diameter to the small diameter and from the small diameter to the large diameter .
(Edit: correction , when the masses are on the arms/structures the masses forces are acting on the diameter those arms are on , when the masses are not on these arms their forces are not on the diameter the arms are on , in the example i have the arms on the large diameter )
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1ex2.png

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The lift can be skipped by having the masses on the arms which already extends the required distances for free and then transition over from the arms to the diameters and from the diameters to the arms .

However to do this without wasting energy which is low in stock ,the masses also need to transition either free of charge or not using a lot of energy , any movement along a distance of a mass needs energy , and a transition would need energy as we know w=f*d.

So to address this transition energy , i came to the following solution , in the image above i drew out an example scenario of the transition area with the two diameters and arms in the positions to transition across .

The arms(white) on the large diameter(red) , the arms extend further than the lifting distances the masses would have to have traveled (they extend beyond/past the distances between A + B and D + C towards the smaller diameter).

If you remember in the previous image i indicated , the mass at the top ended at the bottom as it torqued around CCW and the mass at the bottom of the smaller diameter ended at the top of the smaller diameter as it torqued upwards , and there was no reset , i continue from this scenario.

The mass which were to be at the bottom of the large diameter at D before i added arms , now sits on the bottom arm at E instead .
The mass which were at the top of the smaller diameter at B is still at B.
The mass at E can transition free of charge by falling down gravity enabled , on to the smaller diameter to C.
The mass at B on the inner diameter can transition free of charge to the arm at F by falling down with gravity on to it.
So the masses can reach the heights they needed to reach to reset free of charge now.
There is also a simplified side view to make this clearer.

But there are other issues with these arms , usually on a roberval these type of arms dont rotate which leads to another problem , so in the following image i am discussing this instance only .

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1ex3.png

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If i singled out the arm/structure (white) with its weight on it (bright blue) , and i drew this arm at certain time frames as it moved CCW from the top to the bottom , then the because the arm is not rotating , the mass ends up at the bottom of the arm and not the top of the arm as i need it to.
There is no energy to lift the weight from the bottom of the arm up to the top of the arm , there is no energy to torque the arm around 180 degrees that the mass ends at the top of the arm , there is no way to cancel the torque with another weight added to the system that does not negatively impact energy.
So , to solve this lifting problem free of charge (trust me i did many many drawings and alterations and ideas), i came to the only solution i could find after about a year (just highlighting how difficult this was) .
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ex4.png

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In this drawing , i added more arms along certain angles , and as the top arms would torque around , the arm from the top would align with another arm at the bottom , then the mass on such top arm can transition free of charge to the top of a bottom arm with gravity , as drawn out for an example to imagine at A and B , so this solves the lift for the mass which would have ended at the bottom of the arms problem free of charge again.

However , this brings about another problem which i haven't described here yet..

[johannesbender]

I actually also want to clear something up , these skipping and getting mass in the required positions is to me not exclusive to the specific design instance i have been drawing as examples , i think these principles are the only way to achieve those problems free of charge for a wheel that is gravity only , however i am not saying the problems i have not discussed yet would be resolved, just that this is the only solution to solve lifting free of charge i can imagine and have theorized about.

[johannesbender]

ex5.png

I mentioned a problem happens because of the way the lifts were solved , this image shows an example of that , if the arm at 12 o'clock were followed along downwards CCW at certain timed intervals and drawn out , and depending at the angles they are spaced at , in this example that angle is 30 degrees.
So if it moved from A to B it would complete 30 degrees , from B to C it would complete 30 degrees , 60 degrees in total from A to C.

if the mass were to transition at D , from an arms at C to an arm at E , it would skip the degrees of travel between C and D which in total is 60 degrees.
(edit: correction: if the mass were to transition at D , from an arms at C to an arm at E , it would skip the degrees of travel between C and E which in total is 60 degrees.)

If it moved from E to F it would complete 30 degrees , from F to G it would complete 30 degrees , 60 degrees in total from E to G.

So 60 from A to C , 60 skipped from C to E , and 60 from E to G ,that totals to 60+60=120 degrees , 60 degrees less than 180 degrees is under torque force , this means a reset would have to happen within less than 360 degrees because of the skipping principle that places the masses at the required positions free of charge.

The upwards torqueing smaller diameter , that moves one of the masses back to the top arm would also have to skip 60 degrees , so if i were to stick to this theme then , 360-(60+60)=240 degrees that would be used to torque from the start and back to start again .

[agor95]

Hello johannesbender

You have spent years on this so if you do not mind then can you help me get a better understanding.

Simple question as I have to go slow so I am not missing anything.

Position A - C as the arm rotates. Can I assume the arm is pivoted on the outer rim?

After the D transition E - G were is the arm pivoted?

They could still be pivoted on the outer rim. But the weight could fall onto the inner rim for lifting.

Would the weight on the inner rim get captured by arm A at the top when it arrives?

Note.
In effect the weights load is on the outer rim on the way down via the arm's pivot point.
But the weights load is on the inner rim on the way up. Until arm A captures the weight where the load is now outer rim pivot point loading.

The arms are either pendulum or inverted pendulum in nature. So no torque produced on them just structural stress in the vertical.

All the Best

[johannesbender]

Hello johannesbender

You have spent years on this so if you do not mind then can you help me get a better understanding.

Simple question as I have to go slow so I am not missing anything.

Position A - C as the arm rotates. Can I assume the arm is pivoted on the outer rim?

After the D transition E - G were is the arm pivoted?

They could still be pivoted on the outer rim. But the weight could fall onto the inner rim for lifting.

Would the weight on the inner rim get captured by arm A at the top when it arrives?

Note.
In effect the weights load is on the outer rim on the way down via the arm's pivot point.
But the weights load is on the inner rim on the way up. Until arm A captures the weight where the load is now outer rim pivot point loading.

The arms are either pendulum or inverted pendulum in nature. So no torque produced on them just structural stress in the vertical.

All the Best

r1.png

Just imagine the arms are fixed on wheels/gears like this ramelli geared type of roberval , and the radius they take when moved in a circle at their attached positions are like the large diameter , i know the center gear must be fixed but i don't really mind that.
you just have to imagine their is more than just 2 arms , think around 12 arms geometrically spread out .
the idea is the arms can capture the weights and the inner radius can capture the weights (think marble machine wheels) because the have to transfer.
But the design im working on has (stil on paper) has arms on the inner radius too , because i need to figure out the internal radius skip too.

[agor95]

Hello johannesbender

Well I have a good imagination supported with 3D modelling.

Fine many arms using a gear trains to keep the arms vertical.
With two arms per weight where you are showing via 30 degree increments.

All the Best

[johannesbender]

Something i have been wondering about was , what kind of geometric shape would have a total internal 240 degrees though.

[Fletcher]

According to the formula the number of interior angles must be a positive integer ..

You'll note from the formula that not all the internal angles of a geometric shape must be the same angle, and will add the same regardless of shape i.e. length of sides can vary ..

3 gives 180 degs
4 gives 360 degs

** Therefore impossible to find the number of interior angles to equal 240 degs IINM ..

[WaltzCee]

Something i have been wondering about was , what kind of geometric shape would have a total internal 240 degrees though.

It would be non-Euclidean, hyperbolic.

Eta
maybe a wobble in rotation.

[johannesbender]

Okay well I'm not sure what to call it ,i was just wondering .

I am going to continue onward by showing the following example which does not have 2 skips but only 1 skip , because i would like to show what the problem would be if there is just one skip.

Its difficult showing whole drawings because if i zoom out you cant see much , and explaining whole drawings would be massively confusing , this is why i did many small drawings that focuses on just certain features .
I feel like i explained enough singled out features , that i can show a better drawing now without having to worry about it.

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ex6.png

ex6zoomed.png

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In this image where i zoomed in and labeled certain things , from A to B is 60 degrees ,at C a mass transfers from a top to a bottom arm, from B to D is a skip of 60 degrees , from D to E is 60 degrees .

A large red diameter has inside it at an offset , a smaller green diameter , such that the angle from G to H is 60 degrees and the angle from H to I is 60 degrees , and the diameter is sufficiently large enough to make up the distance for the arms to meet at G and I.

So if the outer diameter torques from A to B then the inner torques from G to H , and if the outer diameter torques from D to E then the inner torques from H to I , in total that is 60+60+60+60 = 240 (60*4)degrees.

If a mass were at J on the top arm (which already transferred from the inner green diameter at I to the arm at J in purple) at 12 , and a mass were on G on the inner diameter (which already transferred from the bottom arm at F to G in yellow) .

Then a problem would be seen ,if the torque was calculated for the mass at G or I on the inner green diameter , it would be seen that from the green diameter's center line at K to G is 30 degrees , and from L to I is also 30 degrees , so any mass at those locations would have more torque than a mass on the arm at J or on an arm at F because those arms would be at 0 torque when calculated.

So would lead to a torque problem where the inner would have more torque than the outer when the masses were transitioned from the inner to the outer diameter and from the outer to the inner diameter via the arms.

[johannesbender]

At one time i though , well perhaps i should change the torque angles such that a smaller diameter's torque would be less than a larger diameters torque by rotating the whole thing 30 degrees .
That can work but that's not a solution because it would only give torque advantage for a limited degrees as long as the inner diameter's angle is less than the outer diameter's angle.
For instance , the angle from H to A is 30 degrees , and if a mass were at E on the inner diameter , and a mass were at G on the arm of the outer diameter , it would torque from A to around B and the inner would torque from E to around F , but the large diameter would not torque from C to D because from F to G on the inner diameter the torque would be to much , since from I to G is 60 degrees and C to J is 30 degrees and J to D is counter productive torque of 30 degrees.

ex7z.png

So the point is , if a Skip happens on the large diameter , then a skip must happen on the small diameter too , so both maintain their angles of travel such that the outer torque would remain bigger than the inner torque which is established by diameter and not angle.

[johannesbender]

As the bible saying goes , separate the wheat from the chaff , so you better thresh to separate the good from the bad.

[WaltzCee]

As the bible saying goes , separate the wheat from the chaff , so you better thresh to separate the good from the bad.

I'm no scholar, yet I've read it a few times.
Since you brought up the point, I'd like to read it.

Do yo have a link, johannesbender?

[johannesbender]

https://www.bible.com/bible/compare/LUK.3.17
https://www.selbl.org/separate-the-whea ... tthew-312/