Milkovic's Pendulum

[Johnsmith]

 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.

[Johnsmith]

 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
https://www.youtube.com/watch?v=Y1cKWIAFT0I
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.

[Johnsmith]

 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.

[Johnsmith]

 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.

[Johnsmith]

 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. https://photos.app.goo.gl/t3EzyZ8w1nVPyS8B7

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.