* For the sake of simplicity, in all of the following examples the value of (g) will be assumed to equal 32 ft./sec.², all measurements will be in English units and all examples will take place at or near the Earth�s surface.

First, imagine a 1 lb., free falling weight. It�s gravitational and inertial masses are exactly the same. It will accelerate at the normal gravitational rate (g). Thus, if it falls for 16 ft., it will reach a velocity of 32 ft/sec. in a time of 1.0 sec. and it will have a kinetic energy of 16 lb.-ft.

Using the version of the kinetic energy formula best suited to the English measuring system:

K.E. = mV² / 2g

Now imagine two 1 lb. weights connected by a rope, one above the other with the rope taught. Their combined gravitational mass and combined inertial mass are equal (2 lbs. each). After a 16 ft. fall they will both be moving at 32 ft./sec. and will have a combined kinetic energy of 32 lb.-ft. (2 lbs. multiplied by 16 feet of fall).

* In all of the following examples, the rope and falling weight will seperate from the rest of the system after 16 feet of fall. All velocities will be measured at the instant of seperation.

Next imagine the same two 1 lb. weights connected by a rope. One of the weights is on a very large, level table, while the other is suspended by the rope, which is hanging over the edge of the table. The weight on the table is mounted on wheels which are aligned in the direction of the rope. Gravity will have no direct effect on the movement of the weight on the table; only the suspended weight is actually being accelerated by gravity, and it will be pulling the weight on the table. In this case, the gravitational mass of the system is only 1 lb. while it�s inertial mass is 2 lbs. What this means is that 1 lb. of force (the force being supplied by gravity through the gravitational mass) will be used to accelerate 2 lbs. of mass.

1 lb. of constant force applied to 2 lbs. of mass will produce a constant acceleration of 0.5g.

g (1 lb. of gravitational mass / 2 lbs. of inertial mass) = 0.5g

This works out perfectly with the law of kinetic energy. After the suspended weight falls 16 ft. at 0.5g, it will reach a velocity of 22.627 ft./sec. in 1.414 sec. Actually, both of the 1 lb. masses will have moved 16 ft. and both will have that same velocity, which in fact is the only velocity that two 1 lb. masses can have if they are moving at the same speed, while possessing 16 lb.-ft. of kinetic energy.

Now imagine that the same two 1 lb. weights are connected this way: The suspended weight is still connected directly to the rope, but the other end is connected to a pulley. Another piece of rope is anchored to a wall abutting the table on the side of the table opposite the falling weight. This rope is then looped through the pulley and back towards the wall. The free end is then connected to the wheel mounted weight on the table. Now when the suspended weight falls a distance of (1), the wheel mounted weight on the table will travel a distance of (2). What this means is that the wheeled weight on the table, having the same actual mass as the falling weight, but moving twice as fast, must possess four times the amount of kinetic energy, or 4/5 of the total. The system is now behaving as though it contained a total of 5 lbs. of mass, even though it actually contains only 2 lbs. Again, the solution can be worked out easily once the difference between gravitational and inertial mass is understood.

g (1 lb. of gravitational mass / 5 lbs. of inertial mass) = 0.2g for the acceleration of the falling weight.

After the weight falls a distance of 16 feet at 0.2g (6.4 ft./sec./sec.), it will have reached a velocity of 14.318 ft./sec. in 2.236 sec.,while the weight on the table will have reached a velocity of 28.636 ft./sec.

When the kinetic energy stored in the two masses is combined, it is equal to the amount of potential energy which has been released into the system.(16 lb.-ft.)

K.E. = ((m₁V₁²) + (m₂V₂²)) / 2g

The method for calculating the acceleration of a falling weight in a system comprised of a flywheel starting at rest, with a moment mass of 1 lb., and caused to move by a falling 1 lb. weight connected to the flywheel by means of a cord and a pulley sharing the same axis of rotation, and neglecting friction, air resistance, the weight of connecting pieces etc., is as follows:

The actual combined mass (m) of the system is 2 lbs. (1 lb. mass of the falling weight plus 1 lb. of flywheel moment mass)

The gravitational mass (M) for the system is 1 lb. (the amount of mass directly accelerated by gravity; the falling weight)

The inertial mass (I) for the falling weight in this type of system (its amount of resistance to a change in velocity or acceleration) is determined as follows:

The distance of moment travel is equal to: distance of weight fall X (moment radius / pulley radius)

I = (distance of moment travel / distance of weight fall)² + 1

Assuming an equal radius for the flywheel and pulley, the acceleration value (a) for the falling weight is therefore:

a = g (M / I) = 16 ft./sec.² = 0.5g

Thus, it can be seen that if the 1 lb. weight falls a distance of 16 ft., with an acceleration of (a), it will achieve a velocity of 22.627 ft./sec., and if the pulley has a radius of 10.186 ft., then the flywheel will have made 90 degrees of rotation, and it�s moment mass (in this example) will have the same linear velocity as the falling weight.

Ö(16 ft. / 0.5a) = (t) time of fall in seconds = 1.4142 sec.

ta = 22.627 ft./sec.

Further, it may be shown by the law of kinetic energy that the values of the two actual masses (1 lb. each), multiplied by their respective velocities squared, then combined and then divided by 2g, will yield a result of 16 lb.-ft. of kinetic energy, or 1 lb. multiplied by 16 ft. of fall.

K.E. = mV² / 2g

K.E. = ((m₁V₁²) + (m₂V₂²)) / 2g

K.E. = ((1 lb. X 22.627²) + (1 lb. X 22.627²)) / 2g = 16 lb.-ft.

This is an example of energy being conserved as it is transformed from one state (potential energy) to another state (kinetic energy).

Now consider a similar system, with the flywheel�s moment of inertia located a distance of 20.372 ft. from its rotational center, while the pulley radius remains 10.186 ft.

From the formulae above it can be seen that the gravitational mass of this new system remains 1 lb., while its inertial mass increases to 5 lbs., even though its actual mass remains 2 lbs. (In the first example, a = 0.5g. In this example a = 0.2g)

Again, following the same rules, it will be seen that the 1 lb. weight falls 16 feet in 2.236 sec. and reaches a velocity of 14.318 ft./sec. Since the moment mass of the flywheel is twice as far from the rotational center as the pulley radius, it is moving twice as fast as the other weight is falling; 28.636 ft./sec.

Again, the combined kinetic energy stored in the two masses is found to be equal to the potential energy released into the system; the mass of the falling weight (1 lb.) multiplied by the height of its fall (16 ft.).

So far, it has been demonstrated that there is a difference between gravitational and inertial mass (If there weren�t, then the accelerations, velocities, times of fall and kinetic energies of the previous examples would not work out). But, so far the only difference that has occured in the operation of the various systems is in the time taken to release the potential energy into the system. As the inertial mass increases, the acceleration of the components decreases; they take longer to cover the same distance but still contain the predicted amount of kinetic energy.

The following example will demonstrate that a somewhat similar system, but one with a varying inertial mass, creates an interesting paradox concerning the Laws of Motion and the 1st Law of Thermodynamics. The Laws of Motion will produce a particular result, but the 1st Law of Thermodynamics will predict a different one.

At the start, the planetary gear/pulley is directly centered over the sun gear. The rope around the pulley hangs straight down, on a tangent, and suspends a 1 lb. weight. The weight�s line of force (in the case of a weight suspended from a rope, the rope will always represent this line) now occurs at a distance from the fulcrum (the meshed gear teeth) equal to the radius of the pulley. When the weight falls a distance of (1) the center of the planet gear also moves a distance of (1). The moment radius is twice as far from the flywheel center as the center of the planet gear is. The moment will thus travel twice as far, accelerate twice as fast, achieve twice the speed, and possess four times the amount of kinetic energy compared to the falling weight. The system at this point has an inertial mass of 5 lbs., and just as in the prior examples of similar movement, the falling weight will accelerate at 0.2g.

After 90 degrees of flywheel rotation, the planetary gear/pulley center point will be level across from the flywheel center, and the weight will have fallen 16 feet. (Since the diameter of the pulley is 6.223ft., after 90 degrees of flywheel rotation the pulley will have made 1/4 of a rotation relative to the sun gear and 1/2 of a rotation relative to the Earth�s surface, paying out 9.777 ft. of rope. Additionally, the center of the pulley (and the tangent) will have fallen 6.223 ft. towards the Earth for a total of 16 feet of fall). At this point, the line of force (the rope) will be at a distance from the fulcrum equal to the pulley diameter. Therefore, the center of the pulley will move only half as far as the weight falls. Since the moment radius is twice as far from the flywheel center as is the pulley center, the moment mass will move twice as far and twice as fast as the pulley center. In other words, at this point the moment mass will be moving with a velocity equal to the falling weight. Since both the falling weight and the moment mass are being moved equally, the inertial mass of the system is now 2 lbs., and the falling weight will be accelerating at 0.5g.

Here�s where it gets interesting.

Since this system is designed to allow for exactly 16 feet of fall of a 1 lb. weight (just as in all of the previous examples), exactly 16 lb.-ft. of potential energy is released into the system. However, all of it cannot be accounted for. Let�s review:

The first flywheel system that was described had a constant acceleration for the falling weight of 0.5g. Both the falling weight and the moment mass achieved 22.627 ft./sec., and equally shared the total kinetic energy.

In the second flywheel example, the falling weight had a constant acceleration of 0.2g and achieved only 14.318 ft./sec., while the flywheel moment, which was twice as far from the flywheel center as the weight�s line of force, achieved twice that, (28.636 ft./sec.) and held 80% of the total kinetic energy.

In the most recent example, the falling weight began with an acceleration of 0.2g and ended with an acceleration of 0.5g, which means that it�s end velocity (measured after 16 feet of fall), must lie somewhere between 14.318 ft./sec. and 22.627 ft./sec. Since the falling weight was not subjected to a full 0.5g acceleration for the entire 16 feet of fall, nor anywhere along it�s fall was it ever subjected to more than that (the inertial mass was never less than 2 lbs., while the gravitational mass was never greater than 1 lb.), it must be moving slower than 22.627 ft./sec. (and so too the moment mass since both are now moving at the same speed). That is what the Laws of Motion indicate. A lower average acceleration over an equal distance will result in less velocity.

The 1st Law of Thermodynamics on the other hand, requires that the total energy of a closed system be conserved as it is transformed from one form to another. In this case from potential to kinetic energy. But, if 16 lb.-ft. of kinetic energy is stored in two 1 lb. masses that are moving at exactly the same speed, their velocities must be exactly 22.627 ft./sec. However, according to the Laws of Motion, the system as described above is not capable of producing that result.

As you�ve probably noticed, the system just described loses rather than gains total energy (in the end, the total kinetic energy in the system is less than the potential energy released into it; it is an under-unity device). But this is not bad. The missing energy has not been lost to friction or to the outside since it is a closed system. Since the Laws of Motion produce opposite results in reverse, reversing the operation of the system will cause potential energy to slip into existence rather than out of it, thereby providing a means of producing usable kinetic energy with no additional input.

Undoubtedly, some attempts will be made to locate this missing kinetic energy by calculating the velocity of the flywheel moment mass using its own predicted acceleration (as opposed to that of the falling weight). What that will show is that the moment mass is always at the exact linear velocity predicted by its relationship to the falling weight. While the falling weight has an increase in velocity and an increase in its rate of acceleration (from 0.2g to 0.5g), the flywheel moment has an increase in velocity but a decrease in its rate of acceleration (from 0.4g to 0.0g), and it can't attain 22.627 ft./sec. either. The end velocity for both masses is approximately 20.3 ft./sec., giving a total combined kinetic energy of approximately 12.75 lb.-ft. (Because the accelerations and velocities each change at different rates, the math used to determine the end velocity is quite complicated, and I don't have the computer skills to post the formulae here in the HTML format, however I will gladly fax them to any interested party.)

To understand how such a thing could be possible, imagine a tower 3,960 miles high. A weight dropped from this height would begin its fall with an acceleration of 0.25g. As it approached the surface of the Earth, both its acceleration and its velocity would increase until at the Earth�s surface its acceleration became 1g. Now imagine another weight of equal mass falling from the same tower, but with a constant 1g of acceleration. When the weight reached the surface of the Earth, its velocity and thus the kinetic energy it possessed, would be much greater than that possessed by the weight which fell with a changing acceleration.

If a weight fell a certain distance subject to a constant acceleration, and after its fall was completed, it was lifted back to its original height, but within a field of diminishing negative acceleration, then the kinetic energy produced as a result of the fall would be more than needed to return the weight to its original height.

The varying inertia flywheel system described above is actually a sub-system of a larger system that does much the same thing. By varying the inertial mass within the system (rather than the gravitational field surrounding it), during the lifting of a weight but not during its fall, the same net result is achieved. With this type of system, a weight may be raised to any given height, using less kinetic energy than it will produce in a more conventional system as a result of its falling from an equal height.

It is not at all neccessary to maintain the exact ratios that were used in the varying inertial mass flywheel example. They were chosen only to simplify the explanations; if other comparison ratios for weight and distance (the size of the flywheel, the ratio of the sun to planet gear size, etc.) are used, they simply produce different inequality ratios between the potential and kinetic energies. In theory, a device of this type could be constructed in such a way as to produce a 1:1 conversion ratio between potential and kinetic energy. But in practice, this would be like trying to balance a pencil on its point.

It is important to note that while gravity was used to more easily explain how this system works, gravity is not a requirement. The design will work equally well when a spring, a pneumatic cylinder with a piston or any other method capable of storing potential energy and accelerating a mass is used as a substitute for gravity. The non-linear acceleration produced by such methods will not affect the ability of the system to operate as designed, since it is equally non-linear in both directions.

It is also worth noting that the predicted increase in mass associated with extremely high velocity is almost certainly inertial in its nature. This would mean that with the proper system in place, the drag on a body�s acceleration might be reduced by altering its inertial mass, while that reduction in inertial mass is simultaneously being used to produce the energy needed for ever greater acceleration. If the mass increase in question is merely an increase of inertial mass, then there would be strong indication that the speed of light could possibly be exceeded by a material body.

Questions or comments?

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© 2005 H. Joseph Schiess III