Dear Members,
See the attached link.
https://youtu.be/NTKw2Hp6tPw
Now I explain in simple words how it will work.
In the video you can see a seesaw having two ball each side .the red color ball side arm of seesaw is slightly heavier than green color side arm.the weight of each ball is 30 gram and weight of each box is 20 gram.
The red ball is a counterweight and green ball will work to create energy.
The green ball will be free to move but counterweight will be permanently mounted.
Now I explain:
When I lift up the arm of green color ball up to a certain height and then you can see the movement of ball in the video that ball is moving due to a slope .
But when I left the arm then the ball is getting back it's previous position or in other words the seesaw is getting back it's horizontal position due to overunity.
It is totally against physics laws as ,as per physics laws the seesaw shouldn't get back it's horizontal position but in the video you can see it clearly.
Now replace this ball with 10 kg.ball and counterweight 10.200 kg.and calculate input and output using potential energy formula.
In this mechanism don't think about input as input will be very minimal but think about output.
The movement of ball in the box will work as a output so the length of arm and length of box will be also important.
The ball mass=10 kg.
The length of arm is=1 meter
The length of box is=80 cm.
The width will be depend on the diameter of ball.
Now calculations:
Input is very minimal
Output if a 10 kg ball will be rolled down from 10 cm height(though the height will be more) than using potential energy formula
10*10*.10=10 joule
So the output will be 10 joule but input will be very minimal.
I would like to insist on some following points.
(1)the angle of box will be 120 degree downside(see the Sketch)
(2) two piston generator will be mounted on the sidewall of the box.
If still you have any doubts then pls tell me.
(3)there is overunity in this mechanism otherwise seesaw wouldn't get back it's horizontal position.
I request you that if you any doubt regarding to understanding then pls tell me as I am ready to clear your all doubts.
Vikram Kumar Gupta
"as per physics
laws the seesaw shouldn't get back it's horizontal
position but in the video you can see it clearly"
Its getting a free ride?
The most important point is in this mechanism we can increase the potential energy of the ball as per our desire by doing some changes in the design of box.and yes in other words the potential energy is the most important and if it is increasing after applying minimal input then the output will be more than input.
I have already sent you a video.it is not a big hurdle to increase the potential energy in this mechanism.
The green ball will move between piston generators and will hit these generator .when the ball hit the generator then in response generator will work to hit it back and this will be an advantage in this mechanism.
There is nothing to think more and more in this mechanism as we are increasing potential energy by using minimal input.
There are only three points.
(1) the input is very minimal to lift a heavy mass(10kg)
(2) we can increase potential energy of 10 kg.mass using minimal input. (3) the seesaw is getting back it's initial position without using any extra energy. If these points works in a system then no physics can prevent it to get OVERUNITY
See the link
https://youtu.be/9yQ9Ln7jBQ8
In the video the potential energy is being increased of the ball using minimal energy but after getting a certain height it will fall down and get back it's initial position due to the unique design of seesaw without any extra energy.
The design ,as per sketch,of box will work to hold the ball.
Some research scientist from Kharkiv Polytechnic institute has approved it.
I would like to tell everyone that Noether theorem will not work in this mechanism.i have consulted with a very renowned Chartered Physicist and he is very confused as neither he is deny nor accepting the feasibility of this mechanism.
Hello sir,
Overunity achieved.
Overunity achieved
See the link
https://youtu.be/1raDr-qRtrg
The weight of ball is 500 gram and counterweight is 800 gram.the seesaw is tilted towards counterweight side.
The ball is falling down from 80 cm height at the time of tilting and from 70 cm height at the time of reversing of arm.
The counterweight is lifting up only 30 cm.Now calculate input and output.
There is a very interesting point in it which is clearly proving overunity.what is that?
The point is that input will be same if ball doesn't fall down or in rest position when I tilt the seesaw from 70 to 120 degree angle.the seesaw will be also get reversed in the position if ball doesn't fall down.it is not a big matter but main point is if input is same in both condition if ball fall down or doesn't then output??? Or what is about kinetic energy of ball ??
There is clear cut Overunity in this mechanism.
It looks good.
I have not witnessed the experiment in person but I don't want to be on the side of the debunking crowd.
I wish to get a prototype to satisfy myself.
EDIT: To solve the "world problem", you have to make it permanently unstable. Hope you get an investor for this!
Hello,
One more point if I add a 40 cm long box with the top (remember top ,not bottom where ball is located)then the height will be increased and ball will traveled 120 cm and 110 cm.but interestingly the input energy will be same as the height of bottom is remain same.just increased the height of top.
The input will remain same but output will be increased.
Hello Vikram,
As per conversation on Messenger, I have not got time to look into your latest video in detail. But what I suggests, is that the mass in the long tube is falling down because the angle of the tube makes it to.
Further, the experiment needs a hand to repeat the cycle. That means, per definition, that this is not over unity. It might be over unity for the first half cycle, but the device is doing only a half cycle. It need to be self sustained motion. Even acelerate. If you achieve this, you got over unity.
You also lack precicion in your judgment of the experiment. You hand cannot tell the exact energy input. More easily, the energy input is little due to the counterweight. I can see that. However, the experiment is only a half cycle experiment. The "wheel" should continue to spin. But it's not. I do, somewhat, understand how you're thinking. Lifting a mass with little effort, and yet, the mass is falling down with greater energy than the input. That is true for the half cycle. The next half cycle require the rest of the energy, and more, to be accepted as over unity. That's my thoughts.
If you got time, you're welcome to discuss my submerged buoyancy gravity wheel in a new thread here. I also sent you pictures on Messenger. I cannot see where I do the mistake. It MUST conserve energy, but I cannot see where. Not yet.
Vidar
Hello Sir,
There is overunity in it .it doesn't matter whether it is half way or full way.you are forgetting one point that the ball has been rotate at 360 degree completely or completed one cycle.
Can you please tell me such device where half way Overunity is being demonstrate d.?
There are three main reason due to which the device is not getting more acceleration.
(1) the tube is not perfectly bound with arm.
(2) the arm get twisted as nail and hole are not perfect.
(3) the most important the counterweight is fixed .if I use a ball as a counterweight in the box then the ball will get jump when 500 gram of long tube hit the top of box .it means the counterweight will be weight less upto that time when ball is in air in the box so the device will get momentum.
These are main reason.
I want to know if I use two piston generator to get energy from this device and this energy work to lift up the system again then will it be considered as an Overunity as the main purpose is to get more output than input.
Quote from: vikram_gupta11 on December 05, 2017, 02:36:17 AM
I want to know if I use two piston generator to get energy from this device and this energy work to lift up the system again then will it be considered as an Overunity as the main purpose is to get more output than input.
If you attach a generator to the falling mass, the mass will fall slower due to its load. Then the kinetic energy will drop, but in return you get another form of energy. In that energy conversion there is loss. So the total outpup will be less than without the generator.
In the video, the seesaw turns some degrees, around 90 degrees or so. If you attach a generator, some of the kinetic energy which earlier was used to turn the seesaw 90 degrees, will drop, so it will turn less than 90 degrees. If you load it 50%, it will turn somewhere around 50-60 degrees.
Vidar
Hello sir,
I know about it but I am talking about spring based piston generator not coil + magnet.as spring based generator will not create word of the kinetic energy of ball when ball hit it at the time of timing and reversing.
I already mentioned that the kinetic energy can be in increased by adding more box with the top of long tube but input same so the friction of generator can be compensated.
Quote from: vikram_gupta11 on December 05, 2017, 03:24:40 AM
Hello sir,
I know about it but I am talking about spring based piston generator not coil + magnet.as spring based generator will not create word of the kinetic energy of ball when ball hit it at the time of timing and reversing.
I already mentioned that the kinetic energy can be in increased by adding more box with the top of long tube but input same so the friction of generator can be compensated.
An ideal spring does not take or give away kinetic energy. The mass will bounce up and down forever like a resonance system. The spring load itself does not apply or help the generator in any way. As soon as the generator is loaded, the spring will not longer bounce freely, but become slower and damped. Se it this way: A load works just like a shock absorber. A shock absorber picks up the kinetic energy so the car does not bounce out of control when driving on a bumpy road. The shock absorber becomes hot, because its job is to introduce friction to the resonance system.
Vidar
Hello sir,
But in this device this will not happened .spring generator means if I use a wall made with flexible rubber and this rubber wall hit a piston generator mounted outside from the device then I don't thing that it will not bounce back.it will definitely bounce back.
One more thing if I use two piston generator mounted on the both end of long tube then the ball will also bounce back after transferring it's kinetic energy to the piston.
No need to worry about bounce back of ball as the counterweight will work to get the seesaw it's original position as there is no hurdle in it.
I have already mentioned that the system will get back it's initial position if ball fall down or doesn't but interesting point is kinetic energy of ball.
Now you will again raise a question regarding load that the load will create problem but ball is free to move in the tube so ball and generator will work separate ly .there will be no hurdle regarding load.
Hello sir,
If you will see closely in the video that ball has been stopped for a moment after hitting with top and again bounce back so the generator will not work to create hurdle after connecting load .
One more thing I forgot to tell you that when I took counterweight equal to the weight of ball500 gram and tilted the arm then the system get it's initial position also from 120 degree to 360 degree.
Now if I have a spring generator then it will be a mechanical advantage as fall is falling down at the time of tilting and reversing so the ball will bounce back more due to compressed spring .
Quote from: vikram_gupta11 on December 05, 2017, 10:32:19 AM
Hello sir,
But in this device this will not happened .spring generator means if I use a wall made with flexible rubber and this rubber wall hit a piston generator mounted outside from the device then I don't thing that it will not bounce back.it will definitely bounce back.
One more thing if I use two piston generator mounted on the both end of long tube then the ball will also bounce back after transferring it's kinetic energy to the piston.
No need to worry about bounce back of ball as the counterweight will work to get the seesaw it's original position as there is no hurdle in it.
I have already mentioned that the system will get back it's initial position if ball fall down or doesn't but interesting point is kinetic energy of ball.
Now you will again raise a question regarding load that the load will create problem but ball is free to move in the tube so ball and generator will work separate ly .there will be no hurdle regarding load.
How can you get energy out when the "spring" use all its stored kinetic energy to bounce the piston back?
Vidar
Hello sir,
It is very simple if spring has stored all kinetic energy then it will release it also when ball will reversed as ball is free to move.the best example is" Hand power fan" to understand it.
Quote from: vikram_gupta11 on December 05, 2017, 10:55:16 PM
Hello sir,
It is very simple if spring has stored all kinetic energy then it will release it also when ball will reversed as ball is free to move.the best example is" Hand power fan" to understand it.
Ok, but that means you cannot take energy out to run anything useful. The stored kinetic energy in the spring is not more than the kinetic energy applied from the piston, right?
So to achieve over unity, the spring must by magic create energy or pick up additional kinetic energy from some, yet unknown, energy source.
You must first understand that conservation of energy is constant everywhere in the observable universe. Then you must try to understand how to create energy from nothing. I believe that will be a very hard task that takes literally infinite time to figure out. I think you must travel that Sanatana Dharma (सनातन धर्म) to achieve over unity some day - well, actually not one day since that forever lasting road has no end :-)
Make a replica of the "device" below. With some modifications, it might be able to create energy instead of the world.
Vidar
Dear Sir,
I have clearly shown and proven the kinetic energy of falling ball and you say that from where it will come.
If spring based piston generator doesn't work in the device then all Hydropower plant must be shut down as no energy will be generated.
The only difference is between this device and hydro power plants is input energy.in Hydropower the output will be less than input but in this mechanism output is more than input.
The spring can store 99% energy so I preferred the spring based generator.
But I clear your doubts about using coil+ magnet generator.
Ok.this generator can be also used in it but use this generator only halfway of tube.
Suppose if tube length is 120 cm then only 60 cm in middle of tube.
You will found same result.(overunity)
Edit: there will be need of very less kinetic energy in this mechanism to reverse it.you say that the coil+ magnet will work to slow the kinetic energy,yes you are correct but once ball pass the coil then it's kinetic energy will be increased again and this energy will be sufficient to reversed the device.
Dear Sir,
I have already studied very much about "Dharma and Darsan." The image you send me is of "Natraj".a God of energy destroyer not creator but I send you a image see it .you will get surprised that my device is very much similar to that replica.
Lord Vishnu an energy creator.
This is not a particular image.just type "Lord Vishnu in Khseer Sagar". The image shows if Lord Vishnu is sleeping in this position then energy is being created and billions universes are being created.
I would like to insist on one point that mass is important in it .so the device will also work in that case if we use coil + magnet generator but mass must not reduced.as mass is important factor in it.
Quote from: vikram_gupta11 on December 06, 2017, 11:25:27 AM
I would like to insist on one point that mass is important in it .so the device will also work in that case if we use coil + magnet generator but mass must not reduced.as mass is important factor in it.
I guess I have to build one and measure input energy and output energy. Your system is simple, so it should not take long.
I already have a structure with a small flywheel with low loss ball bearings. I got to mount a seesaw on that one, with counterweight and a tube where the mass is falling down into.
Vidar
Sure,test it.there is clear cut Overunity in it without any flaw.
The sketch is clearly proving overunity in this mechanism.
Consider that box are weight less so input will be same in both cases but kinetic energy of ball.?
No doubt kinetic energy will be increased in long tube.
Input less output more.
See the link.i did some changes to prove Overunity.for calculations.
https://youtu.be/XjfA3GqRTXg
If calculate then .the length of tube is 1 meter and it's 30 cm length is mounted with arm.the rest 70 cm is upside so I will have to lift up only 30 cm with 10 kg mass.
If counter weight is 13 kg then the seesaw will be tilted already 70 degree.
The arm length is 2 meter.
Now if I lift up the 10 kg mass then toal mgh=130 Joule but minus 100 Joule of counter weight.
30 joule + 30 Joule of extra counterweight energy so total 60 Joule.
But ouput if ball fall down from 1 meter then 100 Joule and again fall down at the time of reversing then 90 Joule so total output energy will be 190 Joule but input is only 60 Joule.
Even I myself feeling the energy of falling ball in the device but at the time of booking the energy is very less.though hand is not perfect way but I think the difference can be find out by anyone falling ball and input energy.
The arm is getting twisted due to lack of perfection between nail and hole otherwise there would be clear-cut momentum in it.
I did it with 80 cm arm but if I take 2 meter arm then the center of mass can be kept equal to fulcrum.
Noether theorem will not work in this mechanism as it is stopping for a moment in each back and forth step.
E = mgh(initial) - mgh (final)
Quote from: sm0ky2 on December 12, 2017, 12:03:35 PM
E = mgh(initial) - mgh (final)
0 - 0 to be exact.
Vidar
The real issue with See-saws comes into play when you
consider Momentum.
Mass in motion, at whatever velocity, will have its own
Energy equations, that include energy from the gravitational
Field.
A lot of people get hung up trying to "grab" that.
A tiny imbalance can set motion to innumerable mass
Now you have a numerical conservation of energy problem.
And you can spend a lifetime of debate over it.
But none of the debators will take the Archer Quinn challenge
And stand under innumerable mass see-saw with a pebble on your end.
100 lbs can lift 100 lbs.
100 lbs + 1 gram can reverse the situation
You can lift extra any amount >1gram to regain the energy of your gram.
And also you have the momentum of
BOTH halves of the See saw crashing into whatever is under it.
See more at www.howthermodynamicscanblowme.com
Hello sir,
The velocity of mass is equal to velocity of box.but at the time of falling down the velocity of ball will be increased.it means the kinetic will be increased.
At the time of tilting the box velocity will be 0 but ball velocity will be not 0.
You say 100 lbs can lift up 100 lbs and 1. Gram can reverse the situation but 1 gram cannot lift up 100 lbs and reverse the situation.
But in this mechanism the less mass is lifting up heavy mass.
At the time of tilting the box + ball velocity is equal but at the point when ball start to falling down then box velocity is 0 but ball velocity has been increased.
If I drop a 10 kg ball from ,1 meter height .it will never get it's previous height but in this mechanism the ball is getting it's previous height.
If ball would slide in the box then there would be no gain but the is falling down (Free fall)at the time of tilting and reversing despite loss of energy due to impact.
So what is this if it is not overunity?
I have already mentioned that the arm length is only 80 cm but if I take a 2 meter long arm then I can locate center of mass above fulcrum.
And no doubt the arm will get its original position without any external influence.
I am interested in Free fall of ball.
If the height were 80cm
10x0.8/0.1/0.1 =800 Joules
Impact force of ball = 9.8 N
Kinetic energy gained by ball = 96.04 Joules
If height is raised, so is time of fall.
Time is important because it is s^2 in the equation
The longer the fall, the more acceleration.
Greater impact force, and greater kinetic energy change
for the ball.
Also greater input to lift ball.
If lift is leveraged, force x distance of lift decreases.
Thus total change in energy becomes more favorable.
At some height, with some leveraged mass:
There is a mathematical gain in kinetic energy.
This is because the gravitational constant is altered
by the lever, but not the free fall.
Dear sir,
The height is 80 cm
So it is as per mgh= 80 joule
But you forget that the ball is also falling down at the time of reversing from 70 cm
So mgh= 70 Joule
So total output is 150 Joule
But input is only 60 Joule.
Now you say that time is important in it.
Yes time can be increased but input will be same as interesting ly there will be need of increasing very less input.
How input will be same.
I explain it.
In my experiment the box length is 80 cm and arm length is also 80 cm.
But if I take arm length 2 meter then I can take box length 2 meter also.
If I mount 60 cmlength of this 2 meter long boxthen the ball fall down from 2 meter height
So output will be
Mgh=200 Joule
Again reverse output
Mgh=190 Joule
So total output
200+190= 390 Joule
you say that 1 gram can reverse the 100 lbs but I take counterweight 1 kilogram.
Now if 60 cm box length is mounted underfulcrumthen input
Mgh=10*10*.6=60 Joule
And 1 kilogram weight
Mgh=1*10*1=10 Joule
So total input is =70 Joule
Edit: counterweight=11 kg.or 1100 gram
Dear sir,
There is double falling down of ball.inputbis single but output is double.the ball is falling down at the time of tilting and also reversing.
You say that it is not due to free fall but I have already mentioned that the hole and nail is not perfect as arm is getting twisted and working as a brake but if build perfectly then the leverage will be increased due to free fall.
See the image.
The sketch is clearly showing Overunity in this mechanism.
I used a massless box to increase the circumference in the video
If build perfectly then there will be only need of tilting just 30 degree more from 90 degree to decrease force* distance and increase leverage.
There is no flaw in this mechanism.
Do you plan to build it?
Dear Sm0ky2 sir,
I want to build it but I think a team work is needed to build it perfectly as I don't have too much sources.
So now guide me and tell me about your thoughts.
Vikram
If the arm and one half on the tube formed two sides of a right-triangle
Then a hypotenuse drawn from the center of rotation to the outer end
Of one side of the tube
Represents your gravitational radius
This new drawn radius represents the gravitational path of travel
For the counterweight, and partial path for the ball when it is opposite the weight
If you position the device so the imaginary hypotenuse is vertical (straight up)
That is the height the weights must be lifted
When the rotating arm is positioned horizontal, and the tube is vertical
A line drawn vertical through the tube represents the center of gravity of the device
When the arm is at 45-degrees, with the counterweight up, ball down
This is the pivot point, where gravity will decide to pull clockwise, or counterclockwise
Because of the location of the weight, in relationship to the center of gravity
In your setup, you have the tube flipped upside down from my above analysis.
So that as the arm approaches the vertical, center of gravity
The ball falls.
When this happens, the new center of gravity becomes a vertical line drawn
directly below the counterweight and ball, at the bottom end of the tube.
And the ball returns to its end of the tube when the counterweight and ball reach
the 45-degree pivot point.
From this analysis, we can now draw another line, 90-degrees from the pivot-point
To the vertical imaginary hypotenuse
Forming a 45/45/90 triangle
A measurement taken vertically from the pivot to the top of the hypotenuse
And again from the bottom of the tube (on the center of gravity line) to the pivot
Now we see,
the weight being lifted is 2x lifted one half the distance,
During half of its' motion
1x weight lifted during the other half.
The ball falls twice, one half distance
(If you see the intentional exclusion here, give a thumbs up to Chas Campbell
and keep reading)
Adding these together, we have 1x weight lifted 1.5 distance
and 1x weight falling 1 distance
We are still losing 1x weight lifted 0.5 distance
This can be balanced by extending the arm beyond the center of rotation
And placing a counterweight on the opposite end of the arm.
Now the energy of the ball falling twice balances out with no gravitational loss.
Only lost is the small energy required to rotate the device.
-----------------------------------------------------------------------------------------
Suppose the ball in the tube were replaced with a fixed weight, equal to the counterweight
And the rotating arm became a tube with a ball.
Now, when the device reaches the pivot point, the ball will fall down the length of the arm,
Adding to the counterweight we placed on the extension
This imparts rotational force to the arm, bringing it around to the bottom
Where it will remain. However, analysis of this shows that we lift the ball 1/2 diameter (r)
Then it is balanced, and requires little energy to continue rotation to the pivot point.
Our ball falls 1/2 D (or the radial length of the arm from the rotational center)
And we ragain our input.
Only lost is the small energy required to rotate the device.
This is the moment of inertia of the mass of the device.
-------------------------------------------------------------------------------------------
In a complete circle of rotation, both devices, as described above, are balanced.
There is no net energy gain around a complete circle.
By breaking each step of the rotation down into pieces,
We see that the potential and kinetic energy changes drastically during different steps.
If the device is only allowed to move during certain parts of rotation
As shown in your demonstration
The question remains whether or not the changes in potential and kinetic energy
are equal to a non-zero value across the range of restricted rotation.
-----------------------------------------------------------------------------------------------
Back to your original set-up:
(In this analysis the transitional momentum of the ball cancels out)
With the arm horizontal, and tube vertical at the center of gravity line.
Both the ball, and the counterweight are at the bottom of the tube.
(This is below the center of rotation)
The height of "lift" in your set-up, is the vertical distance between this low-stage
And the height of the pivot-point.
This height is 1.5r
The ball falls 2r
And is balanced on the return path (opposite direction)
We see here, that 2x the weight are being lifted 1.5r
But only 1x weight (the ball) is falling.
The missing quantity (1x weigh, height of 0.5r) is regained below the pivot-point
Where 2x weight is falling 1.5 r
Now we have an extra quantity of 1x weight falling 0.5r
In the complete circle analysis, we see the opposite value subtracted on the opposite
quadrant of rotation.
However, we restrict the rotation, so this is not allowed to occur.
We have now a situation, where each oscillation results in a net energy gain of:
[1m x 0.5r x 9.8m/s^2 ] - [friction, moment of inertia, wind resistance, etc.]
This is because for 1/2 r the lifted ball mass is cancelled out by the balancing counterweight.
Up to this quantity of energy could possibly be extracted at the axis of rotation
During the clockwise (down) transition until the arm reaches horizontal, with the tube
vertical on the center of gravity line.
The details are in how we translate this back into oscillating motion.
3x the gained energy could be extracted, the. 2/3 of it put back in?
Congratulations Mr. Gupta
Your device has survived round 1 of my gravitational analysis.
Of course, I do miss obvious factors sometimes, so I could be mistaken about some detail.
But so far, this looks legit.
If anyone reading this thread has some thoughts, input, test suggestions, or comments of any kind,
Please add to my analysis, or make corrections.
We would want to restrict the rotation to 10-degrees left of vertical,
And on the lower end, to the horizontal line. (Tube vertical).
Any more than this will decrease from the gain.
And we probably want to lock it less than completely horizontal.
As suggested by the author.
Dear Sm0ky2 sir,
See the sketch.
As per sketch if we mount the long tube at 70 degree angle and curved the arm of counterweig ht towards upside then the tilting phase will be reduced .
There will be need of only just 10 degree more to tilt the device from 90 degree angle.
Hope these two small changes will help you to analyze the device more perfectly.
The main reason to curved the arm of counterweight is center of mass.
If apply perfectly then there is no doubt regarding the feasibility of this device.
Sketch
It appears to be mathematically coherent.
We should build some kind of test model
Dear Sm0ky2 sir,
If you could build it then try it and as per my video everything is fine in this mechanism.
Hello Webby Sir,
See the sketches to clear your doubts.
Dear Webby Sir,
The length of tube is 2 meter and this tube is mounted 60 cm.under the fulcrum and it's rest part 1.6 meter will be mounted above fulcrum.
The counterweight arm will be designed in this way (curved upside 60 cmt)that center of mass will be equal to fulcrum.
The device will be in tilted position if you take 18 kg counterweig ht.but if possible then try to take it 50 degree in rest position.
There will be no problem up to 90 degree to tilt it but after 90 degree there will be need of only 10 degree more to tilt it so that the ball could fall down.
Now calculate it .as far as I think this mechanism is completely an overunity mechanism.
Hello Webby Sir,
The right one is correct as ball is positioned at lower part of tube.
But in third one just change the position of lower and upper side of tube.the orange circle is lower part of tube and blue circle is upper part.in this position the ball is falling down from lower to towards upper and after hitting upper side the arm is reversing and ball is again falling down from upper to lower side of tube.
In this mechanism the counterweights arm will be curved upside to locate the center of mass equal to fulcrum.
Hello Webby Sir,
What's your thoughts?
All mathematics is favour in this mechanism as well as technical.
There is a need of collaboration to build it completely.
Hello Sm0ky2 sir,
your and other members guidance is important now.
Dear Webby Sir,
Did you mount the counterweight on the" Flip" side of arm?
As when I mounted both mass on a single side of arm the input was not good but when I mounted the counterweight on the flip side of arm then input was good.
Dear Webby Sir,
The device will work as per your expectation if we use a lock mechanism to hold counterweight after tilting the device.this lock mechanism will work in this way that it will be locked at the time of tilting the device after tilting it from 90 degree( as there is no problem up to 90 degree due to counterweight).
So that the counterweight will be lifted upside due to falling mass when the falling mas hit the upper side of tube.in this way the device will get momentum.
The lock system will be designed in this way that it will be unlocked due to energy falling counterweight as the counterweight will get momentum so as the whole device.
Now your main problem is solved.
As far as I think there will be need of maximum 5 joule energy to this lock mechanism.
Dear Webby Sir,
You are correct that lock doesn't need any energy but I want to tell you that it will work to prevent the falling of counterweight after tilting the device from 90 degree so that the counterweight could move up to provide momentum to the device and decreasing input and increasing leverage.
But you are expert so I think you better than me.
Thank you sir.
But you can also play better than me.
Try rotating your degrees of freedom more into the vertical domain.
At no point should the center of gravity drop below the axis.
If the arm is 90-degrees facing the right side:
Draw a vertical line where the arm is.
Now raise the arm until the center of mass
Crosses that line near the 45-deg or so
The weight will fall left.
Balancing the arm to either side of this line
Should be the operation of the device.
The ball falls twice.
Left then right
Dear Sir,
As per sketch there will be need of tilting the device just few degree more so that the ball could fall down.
There will be a lock mechanism which will work to prevent the counterweight up to that time when ball hits the upper part of tube .
The lock mechanism will work in this way that it will be unlocked due to impact of counterweight.
(bumped for upcoming experiments)
Dear Sm0ky2 sir,
As per my calculations it must work.if there is something wrong then please clarify it.
The scale length is 2 meter.
There is a 2 or 4 meter long tube.
Let's take tube length 2 meter.
This 2 meter long tube is mounted on left arm of seesaw in middle so it's 1 meter part is above fulcrum and 1 meter is below fulcrum.
A 10 kg.ball is resting at the bottom of the tube.
Now counterweight.
After calculating distance from fulcrum the counterweight will be 14.4 kg.to balance the seesaw as the distance of ball is 1.44 meter from fulcrum as the ball is in rest position in the tube.
Now the seesaw is in balanced position.
But I will take counterweight 20 kg.due to this excess counterweight the seesaw will be in tilting position at 90 degree angle.
Now when I lift up the seesaw then I will not have to calculate the input energy of the 1 meter distance of tube which is mounted below fulcrum as I have added mass in counterweight.
Now calculate input using mgh formula
Mgh=10*10*1=100 Joule
But calculate output as ball will fall down from 2 meter height
Mgh= 10*10*2=200 Joule
Ball will again fall down from 2 meter height at the time of reversing
Mgh=10*10*2=200 Joule
So total output is 400 Joule but input is 100 Joule.
The very interesting point is that it is not important that the device must complete one cycle as ball has completed one cycle.
See the link.you can see momentum in the video.
https://youtu.be/TQr-MR2yj_U
The latch-pin is a nice addition, as it release the ball when you need it to
Dear Sm0ky2 Sir,
See the sketch.as per sketch the tube length is 2 meter.
It is mounted in middle with left arm.
A10 kg.ball is in rest position in this long tube below 1 meter from fulcrum.
The counterweight is 20 kg.
I have taken a ball as a counterweight in a 1 meter long tube.
The long tube rest 1 meter part is above fulcrum.
Now as per my calculations input and output.
If seesaw is in horizontal position and ball is in rest position below 1 meter from fulcrum then counterweight will be 14.4 kg.
But I have taken 20 kg counterweight.so seesaw will be in Vertical position.
Now input will be 100 Joule
As ball will be lifted up only 1 meter.
But output.
At first ball will fall down from 2 meter height
So output=200 Joule
Ball will again fall down from 2 meter height
So output=200 Joule.
So total output is=400 Joule
But there is more output from counterweight.
When ball will hit the top of tube at the time of tilting then the counterweight ball will also jump in the tube and it will again fall down.
If 20 kg.ball jump 75 cm in the ball then the output will be 150 Joule.
So total output will be 550 Joule.
There will be a great advantage if counterweight ball jump in the tube as device will get great momentum.
The device get great momentum if counterweight ball jump in the air in the tube as the counterweight will become massless up to that time when ball is in air in the tube.
In this way the counterweight will also work as a output.
I did an experiment to know the feasibility of device .when I drop the ball then the counterweight ball is bouncing in the tube to provide leverage.
This experiment is proving difference between input and output.
Only as an oscillatory device.
When turned in a complete circle,
The energy gained in the top half is lost at the bottom.
also the balancing seems important, it needs to be heavier on the
counterweight side, so at rest, the ball is above the axis.
otherwise, the effect is mostly lost.
But when it's too heavy, the down-motion of the ball is fighting the
counterweight.
Still working on a demo model, it looks like the arm with the weight
should be angled, opposite to the angle formed by the end of tube to axis.
this way the machine can balance with both weights at the axis.
Dear Sir,
As per sketch if the long tube is mounted very near to fulcrum with 10 kg weight and counterweight(13kg) is mounted 1 meter from fulcrum then I think input energy will be less.
There is a spring based brake system to store the energy of counterweight as well as to provide it momentum.
So input energy will be only 30 Joule if counterweight is 13 kg.
I am having some issues with my test rigs
First: when the ball falls to the right, most of the weight is balanced
by the counterweight.
So the force it creates is very small.
Next:
When the ball falls to the left, it has more force but
We spend that by either compressing a spring or lifting the weight.
To get it back the other way.
Because the weight has moved, it becomes no longer balanced.
If we make the ball heavy on the right side,
and balanced on the left, so it resets itself
We spend energy to lift it from right to left.
Perhaps the standard "wheel analysis" is not valid here,
because it does not make a complete circle.
we might need to look at this from a perspective of 'absolute height'
i.e.: ball initial height and ball final height
This would give a more accurate energy diagram
Quote from: sm0ky2 on January 31, 2018, 05:20:39 PM
I am having some issues with my test rigs
First: when the ball falls to the right, most of the weight is balanced
by the counterweight.
So the force it creates is very small.
Next:
When the ball falls to the left, it has more force but
We spend that by either compressing a spring or lifting the weight.
To get it back the other way.
Because the weight has moved, it becomes no longer balanced.
If we make the ball heavy on the right side,
and balanced on the left, so it resets itself
We spend energy to lift it from right to left.
Dear Sm0ky2 Sir,
See the sketch.if we use a lock mechanism under counterweight to prevent the falling of counterweight then the force will be increased very much but there will be no need of extra energy in the lock mechanism.
Second if we setup the device as per the sketch then there will be need of very less input .for example,if the entire mass of device is 25 kg then we will have to tilt it just 10 degree more so that the ball could fall down.
Please consider these two points.
First lock mechanism and second rest position of device in this particular angle as per sketch.
If counterweight is also a ball in a tube then it will work to provide momentum due to impact of weight.
A spring will work to bounce back the device after hitting with counterweight side so that the device could get back it's initial position.
Thank you Sir.
Dear Sir,
If the device is set at this angle in initial position then the input will be very minimal even if we calculate it's entire mass as there will be need of just 10 degree more to tilt it.
If the entire mass is 25 kg then input is only 25 joule to tilt it 10 degree more.
I get what you are saying, and can see/feel the momentum
But won't this be absorbed by compressing the spring?
falling ball compresses spring, spring pushes the ball back up
our 25 J is (partially) returned to us when it resets
But we have to put 25 J more do go another cycle.
So we lose part of the 25 J each time.
Like in your video, when you use your hand to lift it up.
It is a very strange way to transfer gravitational potential into momentum
But in experiments I am not able to achieve excess power, greater than the input.
I'm really curious now how a Chaos-Pendulum would perform,
if the weights were in tubes......
Dear Sir,
We will have to use the absorbed energy in the spring to tilt the device.if we could use this absorbed energy then there will be no need of extra energy to tilt it and device will sustain it's momentum.
Quote from: vikram_gupta11 on February 01, 2018, 11:11:41 AM
Dear Sir,
We will have to use the absorbed energy in the spring to tilt the device.if we could use this absorbed energy then there will be no need of extra energy to tilt it and device will sustain it's momentum.
Even if we perfectly balance momentum and springs
There is still friction and wind resistance.
even though the total change in (mgh) =0
I think it is the same as just spinning a wheel
It just seems mathematically confusing.
I was incorrect by using the wheel analysis
To use that, the weights must revolve a full circle
around the axis.
The lever analysis gives the correct answer.
Change in E is -
Each time the ball moves, the center of gravity changes
Dear Sir,
Each time center of gravity will be not changed due to counterweight.
Seethe sketch.as per sketch the arm length is 2 meter.a 2 meter long tube is mounted on the right side arm in middle.
A 10 kg ball is in rest position and a " latch pin" will work to hold the ball.the latch pin will work in this way that the ball will fall down as per our desire.counterweight is 16 kg.so now total weight of device is 26 kg.
There will be two lock system which will work to lock and unlock the device.
: The device will be in rest position as per Left side sketch.once ball falls down then it will turn like right side sketch.
: The wonder is that if the device's initial position is as per Left side sketch then there will be need of lifting uup the device only 1 cm.
: So now input will be only 2.6 Joule but output will be 200 Joule as ball is falling down from 2 meter height.
After falling down the device will be positioned like left sketch and there will be again need of just 2.6 Joule energy.
So input will be 2.6 Joule +1 Joule for lock mechanism.
Input=3.6 Joule
Now you will say that we will have to apply 3.6 Joule energy from outsource.
But not as you are forgetting that the ball has been stopped after transferring it's energy to a piston generator and this piston generator will work to generate energy to tilt the device again and again.
Though it will be a Oscillating device but main purpose is to get more output than input or overunity.
The impact energy of ball will do two work.
First it will provide momentum through pushing the piston generator and second the generator will work to generate energy due to kinetic energy of ball.
There are no issues like friction,heat air resistance in this mechanism as input is minimal and output is far greater in first effort.if first effort is providing overunity then it doesn't matter that it is Oscillating .
In this mechanism the ball is working to get overunity not the whole device.
Trying to simplify this so the math is more straightforward
Let us have a pendulum of the counterweight
So it hangs at bottom.
And the tube is to sit exactly at the pivot for the ball.
So that any motion left or right changes the ball
If there is extra energy, the ball will keep the pendulum moving.
and pendulum, in turn keeps the ball falling, tick-tock like a
self-powered clock.
Dear Sm0ky2 Sir,
You are correct it is like a self powered clock but what about input as input is very minimal(3.6 Joule) and kinetic energy of ball is 200 Joule.
There is a big difference between input and output.
As far as I think there is no need of calculating the energy at construction time .so we can get output as per our desire.
The device is getting back its initial position again so again input is 3.6 Joule and output 200 Joule.
We will have to extract impact energy of falling ball using generator .so the device will oscillate forever without extra energy.
Please reconsider it.
Dear Sir,
I did an experiment with a ball and a spring.when I released the ball from some height then the tube moved down side and spr8 get compressed.it is clearly showing that the ball will work to press the piston as well as lift up counterweight.
The lever analysis even proving overunity in it .
The initial position as per sketch ( left side) has solved all problems in it.
In a clock system if we connect a load then it will stopped immediately but in this mechanism load is not a problem.as ball is free to generate energy + providing momentum.
So it is not a pendulum based clock.
Dear Sir,
If a 10 kg ball is falling down from 2 meter height and again getting it's initial height using 3.6 Joule energy and again Falling down and getting back its initial height then why there is no overunity.
The whole game is depend on lock system and latch pin in this mechanism.
My earnest request to you that you please reconsider it.
I shall be very grateful to you.
The mechanism is in an unusual format
But the more I analyze different test models
I see it similar to this system
https://youtu.be/hesfXBcgV0s (https://youtu.be/hesfXBcgV0s)
the small difference in the arc path vs straight drop
should result in roughly the same velocity with similar
masses.
The energy analysis should also be similar.
Dear Sir,
You are correct.but in this mechanism the impact energy or velocity of ball is important not the velocity of whole device as ball is working to create energy.
We will have to ignore the energy at the time of construction as the input is almost 0 as per Left side sketch as the device is already in this position that even we can drop the ball with our breath.
Once we ignore the energy at the time of construction the energy analysis will also support overunity in this mechanism as this device will oscillate forever.
You please ignore the input energy.
I'm insisting to ignore the input energy because there will be need to set up the device only one time as after that the device will get back this position due to itself power.
So energy analysis will also support overunity.
Dear Sir,
I did an experiment regarding input.this time effort was so less that we can drop the ball using minimal input.i just had to tilt the tube only 1 cm.and ball is falling down.
Clearly showing Overunity and solving all problems.
Quote from: vikram_gupta11 on February 09, 2018, 01:10:11 AM
Dear Sir,
I did an experiment regarding input and got success.this time effort was so less that we can drop the ball using minimal input.i just had to tilt the tube only 1 cm.and ball is falling down.
The input can be reduce more.
Clearly showing Overunity and solving all problems.
What materials do you need to complete the construction of a 'self-running' system?
Springs or elastic materials can be obtained from recycled items.
Or you could use secondary lever mechanisms to store the kinetic energy of the ball,
which then tilt back and reset the main lever.
should be a simple next-step, to test if it can run itself.
Hello,
Overunity is being proven mathematically as well as technically.
See the video.
https://youtu.be/isfw6JMfUbw
See the sketch as per my video.
The ball weight is 400 gram
Distance from fulcrum of tube 0.35 meter
The ball is resting 1 meter
Now use pythagoras equation to calculate exact distance of ball.
It will be 1.060 m.
Now use torque formula
Torque= rf=rmg
So torque of ball is 0.400*1.060=0.424
So counterweight will be only 424 gram to Balanced the seesaw.
But I took 500 gram to take advantage of heavy counterweight. So seesaw will be tilted .
Now if I am releasing the ball from 1 meter height and it will travelled 2 meter.
Calculate bthe kinetic energy.
Mgh=0.400*10*2=8 Joule
And counterweight will be rotate more than 180 degree.
The reason is that the ball will have to lift up only 24 gram weight
As counterweight is 424 gram so the will do two work at first it will work to balance the seesaw then the energy will work to lift up only 24 gram weight.
This is the main reason of overunity in it.
.i will have to tilt the counterweight only 91 degree from rest position then there will be no need as due to heavy weight of counterweight the device will get back it's initial position without any external influence.
If the device tilt 91 degree then take 40 cm.
Now use mgh
0.5*10*0.4=2 Joule
Impact is 8 Joule
Now tell me where I am wrong.
See the sketches carefully and work of latch pin.the latch pin will work to hold the ball and prevent it from reverse falling down.
The latch pin will be spring based designed
Latch the ball inside the cup when you drop it in,
then the rotation will be less.
The ball falls out, therefore you are not lifting it back up.
The ball doesn't have to just reach the 12 o'clock position of the arm,
but has to go all the way back up to where you dropped it from.
I have tested it .the ball is getting more height than from where I dropped it.
Your next test should be to place the arm on
an axle, so you don't hold it
then you can build your latch mechanisms
Changes are essential as the experiment s are indicating that I will have to do some changes to get overunity.
This is the final design.
See the sketch.
As per sketch there will two arms and two fulcrum.both arms will be separated with each other by minimal distance.
The counterweight is 1.2 kg and ball weight is 1 kg.
The tube length is 50 cm.the total arm length is 70 cm.
Now mathematical proof.
Consider in it an inelastic collision.
Initial data:
m1=1 kg
M2=1.2kg
V1=3.33m/s
Kinetic energy=5.54joule
Final value:
V2=1.51m/s
Kinetic energy=2.52 joule
Amount of kinetic energy lost in collision is=3.02 joule
Ratio of kinetic energy before and after Collision=0.45
Fraction of kinetic energy lost in the collision=0.54
So input energy will be
Mgh=0.2*10*0.35
Input=0.70 joule
Why,because the arm length and mass distance is equal from fulcrum but there is only 200 gram weight difference.
There is an another very interesting point.
The counterweight will compress a spring when the ball is in air .once ball hit the bottom of tube then the spring will release its energy and device will turn with great momentum.
Dear Sir,
There are two fulcrum A and B.
An orange color ball is in resting position on fulcrum B and a green color counterweight is resting on fulcrum A as per sketch.
The mass of ball is 1 kg.
The mass of counterweight is 1.2 kg.
There are two spring based brake system and only counterweight will hit with these spring brakes.
How it will work?
The initial position of device will be 178 degree.
So when we tilt it at 180 degree then the ball will be dropped down and hit with the bottom of tube of fulcrum A.
Due to impact a " spring based latch mechanism" will work to hold the ball from reverse falling down at the time when the arm will rotate clockwise.
Counterweight will also move from arm A to arm b.
In this way the ball will get back it's initial or even more height than the dropping point.
Now see the second sketch.
The ball will fall down again and the same thing will be happened but the arm will rotate anticlockwise.
The friction heat air resistance are not issues as both mass are exchanging their position from one system to another.
In this way this mechanism will work.
The latch pin mechanism will work in this way that it will lock and unlock due to mechanical power.
The center of gravity will be always in line with fulcrum due to changing position of fulcrum so the design will work.
As per video the seesaw will be mounted on this kind of rotating wheel/ disc.
https://youtu.be/SikOJblbhV0
The disc will rotate back and forth.
Now see the sketch.
The counterweight will be some heavy than the ball.
The tube of counterweight will be like "U" shape.
Now see the design of tube of ball .
There are sliding point on the both side of tube .
The design of long tube will be work in this way that the ball will fall down after getting a certain angle.
The device will be mounted in middle of disc.
Measurements.
The length of long tube is =50 cm.
The length of tube of counterweight is ,
The curved arms =20 cm.
The straight arm=10cm.
The counterweight will be also a ball.
Mechanism:
When ball will fall down then it will hit the bottom of tube an due to it's kinetic energy the device will rotate and the ball will get back it's initial position or even more height than the dropping point.
The curved shape of tube of counterweight will work in this way that the movement of counterweig ht ball will work to provide the momentum.
The same mechanics will work again when ball will again fall down.
The disc will rotate clockwise and anticlockwise.
I would like to insist on the rotating mechanism of disc as it is very important in it as there is no pivot point in it.
I have tested it and it will work.
If you could design the disc using with ball bearing then it will work beyond our expectation.
At first you please test it without design the tubes with a long tube and drop a ball from some height.
The ball will get back it's initial height.
I want that you please use spring based lock or brake system to prevent the device from further movement.
Though we can use secondary lever system instead of spring but at first test it with spring.
All things are same except there is a disc which will oscillate due to impact energy of ball.
I tested it and ball is getting more height than the dropping point.
https://youtu.be/ajQVngShPkc
https://youtu.be/PFBS9ddiNRE
Please don't consider the lifting point as there is no brake to prevent the falling of counterweight.
The initial position of device will be 179 degree.
The initial position of device will be 179 degree so there will be need of tilting the device only 1 degree more.once the device get 180 degree from 179 degree then the ball fall down and overunity will be started in this mechanism.
Input is now very minimal or almost zero in it.
The device can be mounted outside around the disc also.
There will be need of resetting both masses and brakes and it will rotate at 360 angle also.
But I want to keep it simple that's why I'm insisting at 180 degree to 180 degree angle.
Hello Respected Members,
I think I have achieved overunity as the ball is getting back more height than the dropping point or static point and again getting ready to get more height.
The interesting point is that counterweight is some heavy than the ball.
I will present a working model very soon.
QuoteI will present a working model very soon.
Looking forward to it 8)
truesearch
Quote from: vikram_gupta11 on April 19, 2018, 01:02:40 AM
I will present a working model very soon.
Doubt it, seen claims like this 1000000 times and nothing ever comes through. I am 99.9 % sure you don't have a working model.
P.S.
I will go ahead and say this in advanced to those following this this thread, I TOLD YOU SO.
If I'm wrong I apologize in advanced. :-P
Free energy,
I will present something solid proof to prove this mechanism .most probably by the evening.the ball is getting back more height than the dropping point.
I did some changes.
As per sketch there will be two balls and these ball will work to rotate the device one by one.
The sketches are nice but we need to see it self running on video.
If you don't have a self runner just yet at least show on video that the output is greater than the input.
peace
I will show you very soon that it will work.though it is working in a simple model but let me mount it on a fixed pivot.
Quote from: vikram_gupta11 on April 22, 2018, 11:23:51 PM
I will show you very soon that it will work.though it is working in a simple model but let me mount it on a fixed pivot.
I have been disappointed way too many times in the past and I will NOT get my hopes up just yet.
I have been a member here since 2005, nothing surprises me anymore.
On a positive note, I am glad you're sharing this openly, it shows that you really care for humanity, maybe this this can finally free us.
peace
Free Energy,Dear Sir,
This time you will not disappoint as you will take very much interest in it.
My main problem is lack of resources to build it completely.
But whatever I made is clearly showing Overunity in this mechanism.
This time problem is going to be solved.
Hello Respected Members,
I did this simple work to prove that output is greater than input.
At first sketch details. There is a two tube mounted on a arm .each tube has ball of 1 kg. So total tube sidemass is 2 kg.i have not included tube weight. The counterweight is 1 kg. There is a latch mechanism in both tubes which will allow to fall the both balls verticallly after the arm get 180 degree. The device will oscillate clockwise and anticlockwise from 180 to 180 degree. A brake or lock system will work to prevent the device from further movement and only counterweight will hit this brake. Now as per video. If I drop the both balls from a certain height then the device move and trying to get 180 degree due to impact energy of falling ball s. And yes it will get it but there are some reasons due to the arm is not getting initial position of 180 degree. 1) the device is not using impact energy of the mass of lower tube. 2) the axle is wobbling 3)the dropping mass is dragging down it as there is no latch mechanism to hold the ball or mass. ...But ok. If the each ball mass is 1 kg.and each ball is falling down from 0.5 meter then the kinetic energy will be Using mgh formula Mgh= 1*9.8*0.5=4.9 joule Mgh= 1*9.8*0.5=4.9 joule Total energy will be 9.8 joule.
https://youtu.be/zHggwGWcK3w
But in the video there is need of only 5 cm. to get the initial position then if I use two piston generator and if each generator works with 40 % efficiency then I have 4 joule in my hand as I will use the impact energy of both falling balls. But if I lift up the device only 5 cm then the device will get back it's initial position so there will be need of only 0.5 Joule energy as I will have to lift up only 1 kg. The out put energy is 4 joule and input is only 0.5 joule. Why this is happening? The interesting point is that there will be need of only 5 cm more to lift up the device to reset it on initial position as mass is not an issue in it. Please tell me.
See these two videos.the handheld weight is not falling down reversely and arm is rotating at 180 to 180 degree.
just want to know can I use the impact energy of falling weights if yes then why it shouldn't work?
..and if no,then why????
In the video one stroke will generate 4 times energy as the handheld weight has been increased.
In second video I have hold the device to avoid wobbling but remember I am using impact energy of only one weight .if both weight fall down at same time then?
https://youtu.be/eqqOPZT9m2M
https://youtu.be/eJYi3uSOQyw
just want to know can I use the impact energy of falling weights if yes then why it shouldn't work?
..and if no,then why????
In the video one stroke will generate 4 times energy as the handheld weight has been increased.
In second video I have hold the device to avoid wobbling but remember I am using impact energy of only one weight .if both weight fall down at same time then?
I have increased the handheld weight.
Suppose the handheld weight is 1.5 kg
So the energy generate is 7.5 joule as it is falling down from 50 cm.height
The energy generate by lower tube weight 1 kg
Is 5 joule as it is also falling down from 50 cm.height.
Total energy from initial stroke is 12.5 joule.
Now the device is getting back its initial position of 180 degree so if the first stroke energy is being consumed then I can use second stroke energy,the same 12.5 joule .
So why it shouldn't work as there is clear cut gain in it.??????
interestingly when I increased the weight of upper tube then the arm is rotating more than 180 degree without releasing the weight but releasing the brakes but not getting back 180 degree.
When I release the weight then the arm is rotating more than releasing the brakes and device is getting back more than 180 degree.
So it is clearly showing Overunity in this mechanism.
I did it just now
I increased the weight of upper tube and at first I am releasing the device.
The device is getting more than 180 degree.
But it is not getting back 180 degree.
But when I release the weight then it is getting 180 degree.
https://youtu.be/WoubwuRco6I
MAIN PROBLEM:
It is ready to getting back more than 180 degree but only problem is lack of latch as weight is dragging down the device due to reverse falling down.
Once latch mechanism attached then ,no doubt,it will work.
The weight was luckily got stuck when I did first attempt and didn't fall down reversely.so it is clearly showing that there is need of latch in it.
I think (Rather know) the problem is how you're suppose to make that weight reset its position so it can fall down again. As you said, when the weight falls down, the pendulum moves beyond 180°, but not on its way back - due to lack of a working latch. I can't under any circumstances see how you're suppose to pull this off without spending your hands energy to initiate the first and next coming drops of that weight.
You have to look at this in a very simple way:
The weight must reach a higher level without putting energy in to do so. Without a higher level, and then more potential energy into the weight for each cycle, you have a dead device that isn't capable of sustaining its motion, and possibly be able to deliver useful work without stopping.
Vidar
See the video.the weight or arm is getting back its initial position.
https://youtu.be/erAEKdJErsw
I am releasing it like a pendulum but we all know that a pendulum cannot get back it's initial position after releasing from a point.
But in this mechanism the pendulum is getting back its initial position because of potential explosion of free falling of weight.
But when the weight is not falling down to then it is not getting it's initial position.
There is need of latch mechanism to hold the weight and no doubt it will work.
I do some changes in it.a bearing and mount it on a rigid foot and the pendulum will get back more height than the initial position.
All mathematics is supporting overunity in this mechanism.
I will present a second video very soon.
The weight is falling down twicely in this mechanism and again getting back its initial position.
I fix a bearing and mount it on a rigid foot and the device will get back more height than the initial position.
As per sketch I did some work and got success .
The arm is getting more than 180 degree after releasing the ball and again getting back more then previous height.
In each cycle the height of arm is getting more and more and it is clearly proving overunity.
Soon I will present a video.
I just will have to mount it on a rigid foot .
It is working even when ball is rubbing with side wall of tubes.
There are two options to build it.
I am considering both options to build it.
A possibly promising design in the objective eye... is a variation on Elon Musk's energy storage system using weights on upward-angled train tracks.