Newton's 1st Law and Torque

Newton's 1st law of motion can be stated as;
An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.

Newton's work usually focused on celestial bodies. With perpetual motion, a pendulum will remain at rest.


Why this matters is that what a PPM (perpetual motion machine) requires to work is torque. This allows for gravity (9.81m/s) to
use leverage to rotate a body. This is what allows for the linear force of gravity to be converted into linear or angular momentum
in a PPM device.
When leverage is used to rotate a body (weight) it is then using an external force to act on the bob at rest. This is the most basic
consideration when converting the acceleration of gravity, force, f = ma) into mechanical energy.

This gets into basic conservation of momentum and how the acceleration of gravity influences rotation.
While this is basic, at 40 inches or 1 meter, analytical trigonometry can explain how torque is converted into momentum.
The weight to the right is cosine 0º. When the weight to the right drops, the weight at the bottom becomes another force
acting on the body in motion.


This is because it becomes an opposing force. One aspect of a falling weight is that when it rotates from top center to
bottom center, its velocity will be the same as if it dropped straight down minus the resistance of the bearings. The difference
is that it's path is radius times π instead of the diameter. If the diameter is 80 inches or 2 meters then the path a rotating weight
follows is 40π or 125.663 inches or 6.28 meters.
This means the falling weight will take 1.57 times longer to accelerate than a weight that drops straight down.

Newton's 2nd law of motion
The acceleration of an object depends on the mass of the object and the amount of force applied.
The equation f = ma sums it up. force = mass x acceleration. An example of this is a 1kg weight moving at 1m/s. It will have 1 n-m or newton meter of force. A 1kg weight has
0 n-m if it's not moving. If it is moving at 9.81m/s then it has 9.81n-m of force. This can be a bit confusing because a 1kg weight at 1 meter has 9.81n-m of torque. This applies
the value of 9.81 as representing the force a 1kg weight has when it's not moving. Torque and force should view the relationship to a 1kg weight in the same way.  
Newton’s second law can be summed up in a simple mathematical equation where Force (F) is equivalent to the mass of an object (m) multiplied by its acceleration (a), or F=ma.
To put things simply, an object’s acceleration is dependent upon the amount of force that is applied to it, and its mass.
What this means is that for a PPM device, the greater the fall and the greater the overbalance, the greater the acceleration. With this example the under balanced weight is an opposing force. The acceleration the overbalanced weight can have is dependent on the net torque. If we use 9.81n-m of torque for the overbalanced weight and 8.92n-m of torque for the under
balanced weight (90%) then 0.98n-m of torque will be accelerating both weights.



This gets into moment of inertia which is I = lw. Simply stated, the average mass and the distance from the axle. For the time being, we'll go with a 1kg weight accelerating 2 weights.
The 0.91n-m of torque will be accelerating both weights. This means that 0.98n-m/2 = 0.49n-m. Basically it is like the overbalanced weight weighs 1/2. This is for an approximation.  It
gives the general idea of how much torque there will be. For people who don't mind doing more math, then 8.92/9.81 = 0.91.  The 91% would apply to the 0.49n-m of torque. This then
should allow for a more accurate rate of acceleration to be calculated. This because it would be allowing for a loss of acceleration due to resistance. A device will use bushings or bearings, right?
 With gravity, since it will be acting on both weights the same, then 1/2 of the net torque provides power and then what % of it relative to both weights? It's about 1/20 or 5%. It's possible then that acceleration will be 9.81m/s x 0.05 = 0.49m/s. And that is what the n-m of torque is as well.

Newton's 3rd law of motion
Whenever one object exerts a force on another object, the second object exerts an equal and opposite on the first.
A basic method to create an overbalance is this;




As the lever on the right drops with that weight, it can lift both the top and bottom weights. If a 4 to 1 ratio is used then about
1/2 of the work on the right does will be to lift weights. The other 1/2 of the work done will be to rotate the wheel. An example
is if the amount of overbalance is 5 inches or 12.5cm then it can lift both the top and bottom weights 5 inches or 12.5cm using
only 1/2 of its net torque.
As the drawing shows, the weight rotates (swings) downward and then there is the vertical drop. The vertical drop will always be
less than the long arced path the weight will follow.
This is a basic way to show how torque can perform work in lifting weights using 2 different methods at the same time.
When the top and bottom weights are lifted, that influences angular momentum of the wheel, the weight rotating upward is
angular momentum. This aspect of momentum can be viewed differently. When the Earth orbits the Sun, that is linear momentum.
As the Earth spins on its axis, that is angular momentum.
With a wheel, because the weight does not spin, it can be linear motion. Why I called it angular momentum is to separate it from the
linear momentum caused by gravity and when the weight moved closer to or away from the axle, that changes how angular
momentum is conserved. And then that conserved momentum can continue the cycle of allowing for acceleration due to gravity
which allows for an outside force to power the wheel.
With this design 1 can represent 1 lb. or 1 kg, it is to be used as a reference. If the lever being rotated downward is 10 long and the line
lifting the weight is 2 1/2 from the axle of the wheel, then if that is lift 1, then 25% of the torque of the wheel will lift the weight at top.
And if the weight at the bottom is connected to the top weight then both weights will be