The math for PM

[Leafy]

A two segments pendulum is connected to a balanced wheel.

Pendulum start on the left from rest with some height.

When it hit bottom it breaks into two segments and continues.

The question is will it stop at position A or position B.

Position A has the same potential as when it starts, but the correct position should be B where energy is gained.

[Tarsier79]

This is easy enough to test for anyone with a basic test wheel.

[Soon]

B28B3161-A36E-4C7C-8421-1AF34F0FE76F.jpeg



A two segments pendulum is connected to a balanced wheel.

Pendulum start on the left from rest with some height.

When it hit bottom it breaks into two segments and continues.

The question is will it stop at position A or position B.

Position A has the same potential as when it starts, but the correct position should be B where energy is gained.

I made a diagram to help you with the math. Walt might help, I've heard he's really good at math. In the diagram I used
a 3:1 ratio and 30º angles. This is an easy trig calculator to use; http://www.carbidedepot.com/formulas-trigright.asp
It's a trig calculator search on Google and is carbide depot if the link is bad.
With what you drew, if the weight at top right is rotated forward from its current position, it might take about twice as
much energy as what was conserved by the weight at bottom right retarding. This means that a net force from 30º
after top center to 30º before bottom center might be possible.
If you let the weight at top right rotate until it advances on its own then the design is symmetrical and nothing would be
gained. Since I have my own work that I'm doing I'll leave you to what you're working on. It's where I started, something
like this.

p.s., on the right side I show the right triangles when the weights retard their motion. Then if you use Pythagorean's
Theorem or a trig calculator you can know where the end of your arm is positioned. Then you'll know how far to the right it is.
The left side is easier because it's only one right triangle.
With the bottom right triangle, a becomes c. Then with the trig calculator, angle a is 30º and use a from the other triangle as c
and that is subtracted from the arm if it's straight like on the left side. It'll take a little getting used but it's easy to grasp if you
limit yourself to 30 or 45 degree angles until you're used to working the math that way.

[Leafy]

Thx Soon

[Soon]

Thx Soon

Ur Welcome. One thing you'll realize is that the line on the right side will be different from the weights to the axle.
They'll still be considered as that arm. The pivot point is where the force will actually be so if it's on a different
line that's okay.

[Leafy]

This is easy enough to test for anyone with a basic test wheel.

Or you can logic the math. On the left side the work impart to the wheel is ——- torque x angle displaced.

On the right side, the wheel must give back the same work, but the torque is halved so the angle displaced must be doubled.

Therefore, the right side has doubled the angle of the left and end up with position B.

[Soon]

This is easy enough to test for anyone with a basic test wheel.

Or you can logic the math. On the left side the work impart to the wheel is ——- torque x angle displaced.

On the right side, the wheel must give back the same work, but the torque is halved so the angle displaced must be doubled.

Therefore, the right side has doubled the angle of the left and end up with position B.

With the 2 drawings, the arm on the right is a cos 0º. The bent arm is a cos 30º. In the 2nd image when the crossbar
is at a 30º angle then the logic you referred to would be a cos 30º + b cos 30º. If one part of the arm is 3x and the other is 1x
then 3 *cos30 + 1*cos(30 + 30) = 2.598 + 0.5 = 3.098.
And the straight arm opposite would be 4 * cos30 = 3.464.
With the bent arm, the arm it is attached to is at a 30º angle. And then when it is bent at a 30º angle, cos30 + cos30 = cos60.
This is where if you were to draw a line from the center of the axle to the center of the weight on the right, that'd be a different
vertex, Yet that force would be acting on where the arm is bent so it'd be opposite of the opposing arm.
With math for something like this it's best to be organized. That way it's easier to understand what you're doing.
You'll need to be mindful that with this application, cosine is moving up or down from the level of the axle. That'd be the x axis.
With the y axis or sine, just draw a straight line up and down from the axle. That is also the center line and all torque is
considered as an x value whether it's + or -. With this, x is + or - and is to the left or right of the center line while cosine is the x value.
For what I'm building, I think you saw some of the spiral notebooks I filled with going over math. I'm showing you how to simplify
it so you'll do a lot less work. It can be more formula oriented rather than trying to figure out what math works best. And I'm sure
some of these guys wouldn't mind working with you on your project. About every pc and smart device has a calculator and there
are many online. This should help you to have an easier go of things.

[Leafy]

Large scale power generation

[Tarsier79]

Or you can logic the math. On the left side the work impart to the wheel is ——- torque x angle displaced.

On the right side, the wheel must give back the same work, but the torque is halved so the angle displaced must be doubled.

Therefore, the right side has doubled the angle of the left and end up with position B.

My gut feeling is A. A is already a larger angle from the start. Twice the torque doesn't mean twice the distance/angle. If it did, I have a see-saw design that might come to life:)

Sorry I can't help further right now...Ill pop back later to see if you solved it.

[Leafy]

My gut feeling is A. A is already a larger angle from the start. Twice the torque doesn't mean twice the distance/angle. If it did, I have a see-saw design that might come to life:)

Sorry I can't help further right now...Ill pop back later to see if you solved it.

I know it’s true but I can’t think of a design. I love to see your see-saw idea.

[Tarsier79]

Hi Leafy

Here is my old See-Saw design. It was back when I was looking into gravity leverage designs. It doesn't work. I have done the math.

Drop a weight A, We can lift a weight B the same distance, but up a long ramp. The ramp is OB and falls. This turns the A weight into reset.etc.

[Leafy]

Below picture shows the car on the left able to lift the same car on the right straight up.

But the car on the left run horizontal and therefore required little input.

Left torque = right torque

Left car travel 1 distance horizontal = right car move 1 distance up

[Tarsier79]

That is a much more dynamic problem. The leverage changes based on the angle of the string compared to gravity. Gravity doesn't care about the mechanism, gravity only wants mass to fall in a system. Try it yourself with a piece of string and two weights.

[Leafy]

There is no string on the left, it just a reference line to prove the torque is the same.

Actually the left car moves 1 big diameter D = the right car move up 1 small diameter d

The force required to sustain left car in its position is ————Weight x sin(theta)

The force acting on the right car is its weight W

Energy required for 1 revolution of the left car is —- force x diameter D = weight W x sin(theta) x D

Energy required for 1 revolution of the right car is —- force x diameter d = W x d

Efficiency = Energy out/ energy in = energy right car/ energy left car = (W x d)/(Wsin(theta) x D) = 1/ sin(theta) x (d/D)

Pretty much when angle theta approach 0, efficiency becomes infinite. Theta gets smaller as wheel diameter gets bigger.

[Soon]

That's not far off from what I'm doing. As the wheel rotates the drum lifts a weight. And what you're saying is the drum rotates.
With the mechanics that you're discussing, they would help you to understand how to advance a weight as it nears top center.
Using with what you originally posted, if a weight rotates 30º while its arm rotates 30º, how far has it traveled? That is w = m*d.
And when an arm rotates 30º and its weight retards its movement 30º relative to its arm, how much has that weight rotated?
What you were thinking might work but how would you advance the movement of a weight when it's lever needs to be rotated?
What you're thinking about now would get into that. Myself I'd say your subconsciousness is having a fight with itself. Perpetual
motion is impossible and it can't work. And I don't want to be attacked like Soon so I can ask outlier questions and then I'll be
safe.
I mean seriously, your car rotating a drum and the other car moves. That sounds to me like as your wheel rotates, its arm rotates
as its arm does. You did use 2 cars, right? And your idea had an arm rotating that had its own arm that rotated, right? You are
wise to play it safe. After all, I am a sinner who needs a shepherd. I do need to accept salvation when it is offered to me otherwise
I am the fool.

If you like your idea, math will help as well as a moderator and forum members who do not support or encourage toxic behavior.
With the image, what I am building has a rotating hub. Actually the wheel rotates around the hub and the 2 pulleys shift the swing
of the pendulum which is a bob swinging from its fulcrum.

basically the image is 2 thoughts at one time and they make no sense. How can you be sure you're not trying to reconcile 2
different thoughts? And it is up to you what you decide.