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Author Topic: Milkovic's Pendulum  (Read 6178 times)

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #30 on: January 22, 2022, 06:57:55 PM »
Why is oil still not pumped using help of Milkovich's double pendulum?


  It might be interesting if they tried it. With your Maxwell pendulum, is the line securely attached?
You might need to use a screw or drill a hole through the rod that you're using. Also your weight
has to be secured to the rod.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #31 on: January 22, 2022, 06:59:52 PM »
Hello John,
I'm finishing this week a proto that I believe in a lot, which uses a principle not far from yours, but simpler, it's about mounting a weight.
It is in the organization of the weights that we are very different.
I can't show you anything at the moment, but at the end of the week we will know. I am associated in the idea with a member that you probably know is Robinhood46.
If my proto falls through you can count on me.


A++




  Bon chance.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #32 on: January 26, 2022, 08:23:07 AM »
 thx4 and Robinhood46, some numbers for you guys to consider. This is with the counterweight
and the fulcrum for the pendulum both 50 cm from the fulcrum for the crossbar using a .5kg weight.
 If the counterweight drops 5 cm then the fulcrum for the crossbar rotates 11.5°. How work would
be transferred. If the swinging weight is inward at a 45° and it is 50 cm from its fulcrum, the
counterweight will need to drop 3 cm to lift the pendulum. This is by calculating work. The pendulum
would be 15 cm from the fulcrum of the crossbar. It's mass would be factored as 30° of the counterweight.
 This leaves 7 cm of drop for the counterweight. For the swinging pendulum to be rotated upwards
at a 1:1 ratio, the counterweight couldn't lift it. And this is where if the counterweight rotates it 2
to 3 cm then the ratio becomes 7:2 or 7:3. That is where free energy could be realized.
 This is an example of the math. It could be converted into n-m if you guys are familiar with doing that.
I already have a unique design for the gears that control the pendulum because it needs to be caught
then released.
 This is where when you guys would like to discuss this further just let me know. Then it can be diagrammed
in a way that you readily understand it. Also, the counterweight would be heavier because as we all know, a
swinging pendulum has inertia which is in addition to gravity.

p.s., I think you guys know the real work starts when you have to work out the mechanics and the details so it will work.

Floor

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Re: Milkovic's Pendulum
« Reply #33 on: January 26, 2022, 09:38:45 AM »

Couple of basic questions.


1. Do pendulums remain in motion longer than do balanced spinning wheels ?

2. Do balanced spinning wheels remain in motion longer than do
unbalanced spinning wheels ?


Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #34 on: January 26, 2022, 04:17:17 PM »
Couple of basic questions.


1. Do pendulums remain in motion longer than do balanced spinning wheels ?

2. Do balanced spinning wheels remain in motion longer than do
unbalanced spinning wheels ?


  And yet you don't understand that inertia increases the work that a swinging pendulum can do?
Myself I think it would be pretty cool to see a perpetual pendulum working. Then it would be a
counterbalanced pendulum that has a single oscillation.
 Hopefully you're aware that when a swinging pendulum is moving downward it is moving in the
direction of gravity and not perpendicular to it? This is why mv^2/r increases it's force.
 If a pendulum were spinning around a tether ball pole then it would have inertia but no influence
from gravity. This is where the basic math is g * sinx + i = f. See how easy that is? And with a
tether ball orbiting its pole it is i = mv^2/r = f.
 This is why a swinging pendulum can lift a heavier counterweight. And that creates a source of free energy.
And I would like to apologize to you but I am not a math tutor. Maybe you could Google one?

p.s., i in the equations I stated is mv^2/r or mass * velocity squared divided by the radius. And g = gravity
gives a 1 kg weight 9.81 newton's of mass. If it's not moving then it has no force because force = mass * acceleration
which is mass * d/t (distance divided by time). Kind of why m/s is used.
 And with sinx it is sine times the variable x which is the angle in degrees relative to its axis of rotation.

Floor

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Re: Milkovic's Pendulum
« Reply #35 on: January 26, 2022, 05:36:30 PM »
1. Do pendulums remain in motion longer than do balanced spinning wheels
in which the plane of their rotation is aligned with the force of gravity ?

2. Do balanced spinning wheels in which the plane of their rotation is aligned
with the force of gravity remain in motion longer than do unbalanced spinning wheels
in which the plane of their rotation is aligned with the force of gravity ?

Simple questions.

The answers are no, unless there are friction losses or losses due to energy
being translated into unproductive direction by vibration in one case more so
than the other.

Are you aware of this ?

And yet You don't understand that a swinging pendulum can do no work except that which
is equal to the work that was put into it.  Neither do you understand that a pendulum
in its self, does no work that is not then undone by its own motion.

It is true that a pendulum without these kinds of losses and others, would probably
stay in motion for practically forever.  This does not change the fact that it contains
no energy that was not put in to it by the initial lifting of the pendulum.  Nor the fact
that any energy which is extracted from its motions will decrease the energy content
of the pendulum.

You need to be more creative here than any thing you have presented so far, because
energy does not come from that aspect of inertia which is momentum.

Rest assured, I am already aware that you are not capable of answering simple
direct questions.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #36 on: January 26, 2022, 06:20:28 PM »
1. Do pendulums remain in motion longer than do balanced spinning wheels
in which the plane of their rotation is aligned with the force of gravity ?

2. Do balanced spinning wheels in which the plane of their rotation is aligned
with the force of gravity remain in motion longer than do unbalanced spinning wheels
in which the plane of their rotation is aligned with the force of gravity ?

Simple questions.

The answers are no, unless there are friction losses or losses due to energy
being translated into unproductive direction by vibration in one case more so
than the other.

Are you aware of this ?


  And when I read your question, I decide to think about something meaningful so I had mashed potatoes,
mixed vegetables and chicken for lunch. And yes, I did enjoy my lunch.
 As for you, do you know how to convert Pi into radians, or degrees or vice versa? If you can't consider
leverage and work then what are you talking about? Your opinion? And yet in both threads specific designs
have been shown which the work that allows for them is based on math.
 If you consider how you refuted your own answer which it dawned on you they will be built differently and
thus will work differently. And my mixed vegetables had diced potatoes while I also had mashed potatoes.
 And you need to remember, I am working towards realizing perpetual motion and proving that Bessler was successful.
And other people in here share the same goal. There are a couple of tests that thx4 and Robinhood46 can try which
would help them to understand what I am talking about.

Floor

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Re: Milkovic's Pendulum
« Reply #37 on: January 26, 2022, 07:00:44 PM »
' ' ' '

Offline Merg

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Re: Milkovic's Pendulum
« Reply #38 on: January 27, 2022, 03:22:48 PM »
FROM VELJKO MILKOVIC'S NEWSLETTER:

Veljko Milkovic would like to share with you his newest video where he presented his latest achievements in the field of two-stage oscillator development – a new improved model with an elastic (flexible) pendulum and elastic (flexible) lever which is far more efficient than the previous oscillator versions.

This fast and superior model has been significantly improved with elastic oscillations that give a better result with both the pendulum and the lever. The work is still underway on further technical improvements.

New Fast and Improved Two-Stage Oscillator (VIDEO)
https://www.youtube.com/watch?v=7wn15yJ9JYY

* English subtitles available in the video player settings *


This advanced technology is being investigated on all continents. There are now over 500 companies from Southeast Asia that are producing, developing and practically applying the machines based on the principle of Veljko Milkovic’s two-stage oscillator.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #39 on: January 28, 2022, 06:26:39 PM »
FROM VELJKO MILKOVIC'S NEWSLETTER:

Veljko Milkovic would like to share with you his newest video where he presented his latest achievements in the field of two-stage oscillator development – a new improved model with an elastic (flexible) pendulum and elastic (flexible) lever which is far more efficient than the previous oscillator versions.

This fast and superior model has been significantly improved with elastic oscillations that give a better result with both the pendulum and the lever. The work is still underway on further technical improvements.

New Fast and Improved Two-Stage Oscillator (VIDEO)
https://www.youtube.com/watch?v=7wn15yJ9JYY

* English subtitles available in the video player settings *


This advanced technology is being investigated on all continents. There are now over 500 companies from Southeast Asia that are producing, developing and practically applying the machines based on the principle of Veljko Milkovic’s two-stage oscillator.


 It's actually a single stage oscillator. It has only one swinging pendulum. That is where I have let his research team know about this thread.
A 2 - stage oscillation requires 2 pendulums. Veljko deserves credit for the interest he has raised in this because it does help people as well.
 By making a perpetual machine inspired by his work would be a new invention. In Ancient Egypt they irrigated their fields using something
similar. The weight on the other end of the lever wasn't swung. It allowed water to be pulled from the Nile River more easily. With a perpetual
motion design, more work could be done. This would be torque vs power. And for crushing either grain or ore, power is necessary.
 Old stone mills used actual stones for crushing grain. This is why windmills and water wheels powered them. And this is what I have suggested,
harnessing the inertial force of a pendulum. If Velko and his team wish to work with thx4, Robinhood46 and myself on this, it would be out of
respect for what he has done. And who knows, maybe no one will be interested in what I have proposed. And that is okay if that is what people
decide.


Offline kolbacict

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Re: Milkovic's Pendulum
« Reply #40 on: January 28, 2022, 08:14:13 PM »
Quote
It has only one swinging pendulum.
Isn't a long horizontal stick a second pendulum?
It has its own oscillation frequency, determined by its mass and length.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #41 on: January 28, 2022, 08:40:10 PM »
Isn't a long horizontal stick a second pendulum?
It has its own oscillation frequency, determined by its mass and length.


  It's a counterweight or counterbalance. It would need to be swinging to have an oscillation. When
a pendulum swings, it can be graphed. https://www.youtube.com/watch?v=uUfr8WtJuyg
 With a graph you can see its oscillations. move or swing back and forth at a regular speed. With the
graph, at the top of each change in up or down direction is a cycle which is an oscillation. The graph
allows for both frequency and amplitude (vertical change in movement) to be known. And with the
distance between each peak on the graph, that represents time.

Offline norman6538

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Re: Milkovic's Pendulum
« Reply #42 on: January 28, 2022, 08:42:34 PM »
This machine is a variable lever assisted by a pendulum to change the leverage. When the big weight is close to the fulcrum there is less mechanical advantage and when it is further away the large weight is further from the fulcrum it has enough mechanical advantage to lift the long arm. And it is assisted by the swinging pendulum...BUT WHERE ARE THE WORK IN AND OUT MEASUREMENTS? They are easy to get. 1. the weight of the long arm x distance moved x lifts = work out. 2. work in - weight lifted x distance traveled....

The flexible fulcrum adds a little to the variable lever mechanical advantage.

My magnet assisted pendulum drops  from 2 o'clock and travels past 9 and 10 to noon and then slowly drops back down to 5 then stops at 6 - where does that 2+ hrs of travel come from?
No fuel or elec or water or wind power.....

Norman

Offline kolbacict

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Re: Milkovic's Pendulum
« Reply #43 on: January 28, 2022, 09:49:45 PM »
And why then does Milkovich himself call his rocking chair a double pendulum?
the first pendulum is a classic pendulum. The second "pendulum" is a lever.
And if you hang a second pendulum on the other end of the lever, it will be a triple pendulum. :)
I think so...

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #44 on: January 29, 2022, 12:42:01 AM »
And why then does Milkovich himself call his rocking chair a double pendulum?
the first pendulum is a classic pendulum. The second "pendulum" is a lever.
And if you hang a second pendulum on the other end of the lever, it will be a triple pendulum. :)
I think so...


  He can call it whatever he likes. The work it performs is a stroke like in an engine and not an oscillation.
It can be graphed like a swing or pendulum can and that might be why. With a pendulum, because it has
a bidirectional swing it should probably be graphed with a positive and a negative z axis to show which
direction it is swinging.
 With this video, I made it at the end of 2019 because I was bored and had access to both SketchUp and
Final Cut Pro 10. Since y is the amplitude or the difference in how high and low a pendulum swings when
considering its radius is 1 as a reference. If a z axis is used then x will be less than 2r (2 x radius). Then we
can consider something like the Pythagorean theorem where a^2 + b^2 = c^2 or 1 because 1^2 is 1.
 Then z^2 + x^2 = π ( 3.142) would allow for the radius of 1 to be used as an example. And this video goes
along those lines but I might've said z^2 + x^2 = y^2.  The 2nd statement would be wrong because the relationship
between π and radius would be lost. I haven't actually tried proving it, am just using the Pythagorean theorem as a basis.
 An example is if x = 2 then z would = 1.142. Then sqrt both numbers and x = 1.414 and z = 1.068 and y =1. With x, in
a 2 dimensional graph, x = 2 if the radius is 1.

https://www.youtube.com/watch?v=crcKfIcVSzk

And if you consider Einstein's light from a distant star bending around the Sun, is light 2 dimensional or 3 dimensional?
This gets into space being distorted or warped like Einstein wrote in his paper on the subject. It was proven later by astronomers.
« Last Edit: January 29, 2022, 03:22:33 AM by Johnsmith »