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Author Topic: Milkovic's Pendulum  (Read 14430 times)

Offline kolbacict

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Re: Milkovic's Pendulum
« Reply #15 on: January 17, 2022, 04:47:17 PM »
Well, two pendulums can swing sinhronization on the one side of the lever arm.
not necessarily at opposite ends.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #16 on: January 17, 2022, 07:25:13 PM »
Well, two pendulums can swing sinhronization on the one side of the lever arm.
not necessarily at opposite ends.

  They can. With what I am talking about is that one pendulum does not swing. It is a counterweight. If you consider how
a draw bridge is lifted, with this hinged table, it is an example.
 If the weight on the right were a pendulum that did not swing, it would be the same thing as having an idle pendulum.
I probably will need to continue with design so everyone can better understand it. After I have my camera, I might make
a video to show what I am talking about. It would be to show what I mean by timing and when the swinging pendulum
would need to be lifted.

Offline kolbacict

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Re: Milkovic's Pendulum
« Reply #17 on: January 17, 2022, 08:28:46 PM »
Okay, tomorrow make a Maxwell pendulum on the balance beam.
And pulling the lever at certain points in time, its oscillations will become undamped.
I hope.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #18 on: January 17, 2022, 11:21:03 PM »
Okay, tomorrow make a Maxwell pendulum on the balance beam.
And pulling the lever at certain points in time, its oscillations will become undamped.
I hope.

  I may need to stay with the drawings. I do have my build that I am doing. Next month I might
be able to do a demonstration. It will be more work for me to do. And this gets into engineering.
I have worked out a lot of the design. And to try for a working build like any build will be
challenging. It will need to be done in steps. And it might be better if someone else that is
interested tried this. I do need surgery and so I need to focus on that goal.

p.s., the purpose of this thread was to get other people interested in how something might work.
The math for the period of a pendulum is easy enough to do. And then a weight on the end of a
piece of string will let them actually see the math in motion. The top of the string could be dict
taped in a door way. It doesn't need to be anything fancy to understand the basic principles involved.
 That is the first thing that someone would need to understand. Then after that, how could
leverage and pulleys lift the pendulum higher? If taken one step at a time then they might actually
understand it.
« Last Edit: January 18, 2022, 01:26:11 AM by Johnsmith »

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #19 on: January 18, 2022, 04:28:35 AM »
 With that said, I can go over the math and explain how to do it on a calculator. With knowing the period
of a pendulum, you can know its velocity. And then you can calculate its inertia. Basically if you have 2
pendulums that have the same weight but only one will swing, the one that swings will develop more
downward force lifting the counterbalance pendulum.
 And when the pendulum that can swing is getting ready to change directions, the pendulum that is the
counterbalance will lift the other pendulum. I can make some basic diagrams and show how math influences
a double oscillating pendulum. Once someone becomes familiar with each step of the math, they'll find out
that it's not that difficult. It's just working a couple of problems will show you that you can do it.
 So over the next couple of days when I have time that is what I'll do. I think with diagrams and by showing
the math then who wants to can work the problems on their calculator then people will understand why such
a pendulum rocks from one side to the other.

 This is the first example. I thought why wait, okay?
The crossbar is 50 cm to each side from the middle fulcrum. Each pendulum is 50 cm as well. Or everything is
20 inches. Same thing. And the inward angle of the swinging pendulum is 22.5º. With this online calculator,
 side c is .50 (meters) and angle a is 22.5. And side b is .46 (meters). This means that the bob (weight) on the
pendulum will drop 4 cm. Side c minus side b.
 With a 45º angle, it's about a 15 cm drop and will swing faster. What needs to be known is when
a pendulum swings from the left side to the right side and back again, how far is it from its starting point?
 With calculating the acceleration of a pendulum, it's 9.81 m/s * sin x. And when put into a calculator for
22.5 it's 3.75 m/s and for 45 it.s 6.93 m/s.
« Last Edit: January 18, 2022, 08:17:13 AM by Johnsmith »

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #20 on: January 18, 2022, 06:30:47 PM »
 With the attached image, if you look at the top row of buttons to the right you'll see cos, sin and tan.
When using the sine function, it is from the center line of the fulcrum, this means straight up and
straight down.
 Then when 50sin22.5 = 19.13, it is 50 cm * sin 22.5º = 19.13 cm. And since gravity accelerates at
9.81 m/s, if we multiply 9.81 m/s sin 22.5º = 3.75 m/s. That is the initial acceleration because of gravity.
Because at bottom center there is no acceleration, by dividing 3.75 by 2 (3.75/2) = 1.875 m/s.
 Why this matters is that the average acceleration is what will allows us to calculate maximum inertia.
This is what will cause the swinging pendulum to drop and the idle pendulum to be lifted. A counterweight
attached to the crossbar serves the same function. What the pendulum is swinging from is also weight.
 This is probably why in some videos you'll see where the counterbalance weight can be moved on the
crossbar. And when inertia lifts the counterbalance, gravity is not doing that work. This is important. This
is because when the counterbalance drops, gravity is doing that work. And that is the work that can
be used to either accelerate or lift the swinging pendulum. And this is where timing matters.
 I will give people time to try some of the math. I think for this you'll find it's not that difficult.

p.s., if sqrt4 = is Googled then their calculator will be the top search result. It's functions are on the left side.
« Last Edit: January 18, 2022, 09:29:59 PM by Johnsmith »

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #21 on: January 19, 2022, 03:20:53 PM »
 With this math example, I'll be using 45º as an example. The rate of acceleration that I'll be using is 9.81 * sin45 = 6.94 m/s.
Then divide by 2 and we have 3.47 m/s. Next what needs to be known is how far the pendulum will travel. Pi or 3.142 is 1/2
of a circle and 45º is 1/4 or .25 of Pi.
 Then .5 (radius) * 3.142 = 1.57 meters. And 1.57 divided by 4 (1.57/4) is 0.39 meters or 39 cm. The pendulum's downward
swing is about 39 cm. Then we divide 3.47 by .39 =  8.9. This is a fraction of the acceleration rate, .39/3.47. By dividing both
the top and bottom number we reduce it to a smaller number which is 1/8.9.Then 1/8.9 = 0.11 seconds or how long it
takes the pendulum to reach bottom center.
 After this we multiply 3.47 m/s * 0.11 = 0.41 m/s. That's the velocity that pendulum will have at bottom center. We need to
know this so we can calculate centripetal force. If it were a ring or solid disc we'd be calculating inertia. With a single weight
it is centripetal force. And that formula is f = mv^2/2 or mass * velocity squared divided by 2.
 Then we have .5 kg * 0.41^2/2 = 0.5 kg * 0.17/2 = 0.5 kg * 0.205 = 0.1025 kg of force. This is in addition to the 0.5kg the
bob (weight) weighs. It represents a 20% increase in downward force. An example is if a 5kg weight is used, it will have 6kg
of downward force. This means that if the counterweight weighs 5.5kg, it will be lifted. This means that a swinging lighter
weight can lift an idle heavier weight which is one form of free energy.
 And the 2 designs that I'll be showing for a pendulum will be to use the potential energy a weight has after it has been lifted.
This is because the crossbar will lock into certain positions so potential energy (a weight that can drop) can be exploited/used
to perform meaningful work. And I know the math is a headache for some. This is why I showed each step.
 With centripetal force, as the radius of the pendulum increases so will the velocity that a pendulum swings at. This also decreases
centripetal force so that regardless of what height a pendulum swings from or what its radius is, if it swings from 45º it will always
allow for the same 20% increase in the downward force of a weight. This is where someone might say if I get a 20% increase with
a radius of .5meter then I'll increase the radius and go faster. And then they'll be surprised when they did more work and nothing
has changed as far as lifting a heavier weight goes. Where knowing the math can save a person from doing work that isn't needed.
 And over the next few days I'll go into how the 2 designs I've mentioned will be an attempt to exploit how centripetal force allows
a weight to do more work than what work = mass * distance allows for. And for me I like using * as times and x for as a variable
that can be any number like y sinx = z. y = radius, sinx or sin x = degrees and z = the answer. Any symbol representing a variable
is associated with what is being shown. Anytime math is explained, the variables will usually be made known what they are.
 I would apologize for such a lengthy post but IMO this math is what has been overlooked when considering a perpetually swinging
pendulum. How this can be tested is quite simple. Have a pendulum on one side of a pendulum and a counterweight on the other
side. Then by knowing the weight on each side, torque in n-m (newton meters) can be known.
 And a .5 kg weight .5 meter from the fulcrum is .5 * .5 = 0.25. Multiply this by 9.81 = 2,45 nm of torque. A swinging pendulum will
be .6kg * .5 = 0.3. And 0.3 * 9.81 = 2.94nm of torque. And by knowing how much torque is generated, how is it to be used? This is
where I'd suggest watching some videos of a double oscillating or Milkovic pendulum. Then when you see, not many examples on
YouTube. With this video, notice how the opposing weight is lifted higher when the pendulum is swinging outward
 I think that is what can be exploited. And if so then it will show where doing the math shows how that can happen. And with my work
on Bessler's wheel, centripetal force is generated by an overbalanced weight just as it is with a pendulum. The retraction line that
wraps around the disc cancels out this force when the weight is moved towards the axle of the wheel, why I think it will work.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #22 on: January 19, 2022, 09:34:22 PM »
 When the swinging pendulum goes past bottom center (blue to red) is when the crossbar will need to lock in
that position. This way the work the swinging pendulum has done can be stored/conserved. Then the counterweight
becomes potential kinetic energy. A tab can easily hold the crossbar tilted.
 I know with my own project that I like to have one of everything on each side. This is because this prevents
twisting of any moving part. And with tabs, one on each side would help to keep the crossbar properly aligned. Some
of the basic elements can be used for a design that uses pulleys or gears while other components will need to be
specific to the concept.
 With the toggle, it might be as simple as the 2nd drawing. When the swing pulley is lifted, it will release the toggle.
The toggle would have a line connecting it to the lock mechanism for the swinging pendulum. The swinging pendulum
will need to be held in position while the left side starts to drop. This is basically when the "magic" would start happening.
With the toggle, what's to the left of it would be weight so it would be in that position unless the bar to its the right is
pulled down. When I mention timing, this is when it becomes important.

p.s., with the pendulum on the left, it is a counterbalance and serves no other person. A weight can be attached to the
crossbar that can be moved. That would allow it to be adjusted. And in case people do not understand what n-m or a newton meter
is, it is 1 n-m = 8.87 inch lbs. or 0.737 ft. lbs. With inch and foot lbs., the amount of weight placed at that distance.
 With the metric system, 9.81 n-m is 1 kg at 1 meter. For what I'll be discussing, I'll be using n-m as torque/leverage values.

 What might need to happen is for the dropping counterbalanced weight to transfer its downward force. It'd act like a capacitor of sorts.
I'll probably do a walk through of the concept so everyone will know what needs to happen so it can work.
« Last Edit: January 20, 2022, 12:19:18 AM by Johnsmith »

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #23 on: January 20, 2022, 05:14:44 PM »
 The stop would be better placed at the end of the counterbalance. If the crossbar is 1 meter in total length then a lift and drop
of 5 to 10 cm might be all that is needed. As the drawing shows, the stop functions at a 2:1 ratio and when the top is rotated outward,
the crossbar will be able to drop without it interfering.
 If a counterweight weighs .5 kg then that is the force that will be on the lock. And for it to move 5mm then it will require
4.9nm/200 = 0.0245 nm of work. This then will allow how the swinging pendulum is held at the top of its inward swing to perform that work.
It is a machine and any work performed will need to come from the swinging pendulum itself except for when the counterbalanced weight
drops. Then the work it has conserved from the swinging pendulum lifting it will perform work.
 The lock for the stop can be attached to the stand of the pendulum as shown in the drawing. With the stop, it can be moved further outward
and have an extension that holds the crossbar in its up position.

 The line going down from the end of the crossbar shows the clearance the stop has. And with this, the weight of the swinging pendulum
will need to drop about 1 cm. If the counterweight drops 10 cm then it can easily lift or rotate the swinging pendulum upward 3 or 4 cm. It
will need to be determined how much motion is lost when the pendulum swings outward and then back in again.  It only needs to be lifted
to its original starting position.
« Last Edit: January 20, 2022, 09:10:19 PM by Johnsmith »

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #24 on: January 21, 2022, 03:43:51 PM »
 This is what hopefully some people have been waiting for. Not everything is in the design. This is where if someone has a computer
that can run the SketchUp program, I am using the free version. And this would allow for file sharing. This would allow for a better
examination of certain details.
 The tab to the right of the swinging pulley is best seen 3 dimensionally. This is because it will "catch" the swinging pendulum. And
when the pendulum drops in the catch, it will release the counterbalance stop. This is why the 2 pulleys above the cross bar are at
a 2:1 ratio from the fulcrum/axis of the crossbar. It uses basic leverage.
 And then when the crossbar is released, then it will lift the swinging pendulum. The tab on the catch will move away from a pin in
the swinging pendulum allowing it to swing again. And for any design I post, if someone is interested in trying it then we can go into
more detail.
 And tomorrow I will show how gears can be used. They'll transfer the same force/work in a different way.
This link is to a larger image.

p.s., This is where the math that I've been posting matters. It is how to verify that these mechanical relationships can work. And if
someone does want to try this, then they'll know that this design as well as the math can be scaled and adjusted as needed.

p.s.s., am kind of bragging in a way because I bought my computer with using SketchUp in mind. And it has a graphics card
with at least 1 GB of memory. It does require a lot of memory and a decent processor to run SketchUp. And if some wants to try such
a build but can't run SketchUp, I can probably make exploded views with a part separated by color. Then a person would see how the
different pieces fit together. Then if a person wants to, they could use some cheap wood to make a model with before building. That
would probably be best so they could get an idea of how everything works and what it would take to do an actual build.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #25 on: January 21, 2022, 04:43:10 PM »
 An email that I received from Milkovic's research team. With that said, if the designs I post work, they would
be considered as an invention. And if someone does build one of them then credit for the invention would
need to be shared. This is about not giving business a free product.
 And with Milkovic, his work is about something like this
As he mentions, some people don't have a way to pump water. This is where such work can benefit those
interested as well as those who don't have as it good as some of us.
 I also know that some people who might help to invent one of these designs might need the money as well.
It does cost to live and there are some non-profit groups that might raise money to help build these for what
Milkovic has in mind. This is basically mixing capitalism with humanitarianism.
 I will let Milkovic's research team know that a perpetually working double oscillation pendulum is a new
invention which is based on his invention and Bessler's work. And that as such, credit will need to be shared.
Otherwise why would someone want to improve an idea at their expense? And it is this thinking that encourages
people to improve upon existing ideas.
 What will need to be accepted is that between the builder, myself and Velko the patent rights will need to be shared equally.
Anyone building my design will need to agree with this to prevent greed from taking over. This is because as I said earlier, if
clocks are powered by this then the value of the patent would become worth money. After all, how much energy does a
pendulum clock need to run? Very little is the answer so this concept and the one I'll be posting tomorrow could be scaled down
to power a clock.

Offline Johnsmith

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Re: Milkovic's Pendulum
« Reply #26 on: January 21, 2022, 08:07:00 PM »
 I went ahead and modified the design so it uses gears. A gear will be placed at the fulcrum for the crossbar. And then
2 lines, one on top and below it will connect to another gear next to the fulcrum of the swinging pendulum. This is where
the math really comes into play. This is because the gears will not be your typical gears. The gear next to the outside
fulcrum will have an axle location that will allow for it to be lifted.
 And then after it rotates the swinging pendulum upwards, it will release the gear that controls the swinging pendulum.
And this is if anyone is interested on working on either design, let me know. After all, I am working on Bessler's Wheel
and that is a lot of work as any build is. And for a larger image of the gear type pendulum;

p.s., hopefully everyone understands that the images that I've shown are illustrations, are to scale and that with an
actual build dimensions will change. And if these designs work (the math says they will), there are other ways as well.
 And as for running a clock with either double oscillation pendulum, they can be scaled down. Think about the Atmos
clock. This would be comparable. This is because it would be difficult to manufacture an inexpensive one. And this
would be where the patent might be worth something. And until that happens it is speculation.

p.s.s., the gear at the fulcrum for the crossbar would not rotate. This is why it would rotate the gear next to the fulcrum
for the swinging weight. And this is where understanding gear ratios is necessary. And with how it is connected to the
other gear, a belt could work. This is where discussing things with the builder on either design will determine how they
are built if someone chooses to work on these designs.

 @thx4 and kolbacist, since I was thinking about what you both have done, it is up to both of you if you'd be willing to work with me.
And if not then it will be who is.

Offline kolbacict

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Re: Milkovic's Pendulum
« Reply #27 on: January 22, 2022, 10:18:47 AM »
Why is oil still not pumped using help of Milkovich's double pendulum?

Offline thx4

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Re: Milkovic's Pendulum
« Reply #28 on: January 22, 2022, 10:46:50 AM »

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.


Offline kolbacict

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Re: Milkovic's Pendulum
« Reply #29 on: January 22, 2022, 06:57:20 PM »
I can’t still make a successful Maxwell pendulum in any way.
I would never have thought it would be so difficult.