Michael and Fletcher...
I had a LOT of trouble trying to understand what you were either trying to ask me or tell me. But here's my best answers to the parts of your posts that I think I understood:
Michael...
It is a well-known fact that, according to the Special Theory of Relativity, an object with some initial mass can not ever reach the velocity of light as it is accelerated toward that velocity. The reason is that, according to the equations of Einstein's revolutionary theory, the object would have to be supplied with an
infinite amount of energy to achieve light velocity because, as energy is supplied to the object to accelerate it, more and more of that energy will go into
increasing the mass of the object as its velocity increases toward that of light. As the mass of the object continues to increase, even more energy must be supplied per extra mile per second of increased velocity obtained.
Studying the equations that govern this effect shows that the increased mass of the object is given by the expression:
M (increased) = M (original) / [ (1 - {(v^2)/(c^2)}^)1/2 ] which shows that, as v begins to equal c in value, the denominator of the right hand quotient goes toward a value of zero. Well, when one divides M (original) by a number that is approaching zero, then M (increased) will go toward an infinite value!
The point is that the usual expression for the kinetic energy of an object must be, according to relativity theory, modified to take this effect of mass increase into account. When that is done, it is seen that the doubling of velocities so that the final velocity is close to that of light will result in having to supply the object with much more than quadruple the kinetic energy it had when it had its original velocity.
You then state:
Problem. Energy is a result of a masses velocity. Change mass or velocity, change energy. Still going with Ken's thoughts, the weight will loose mass. Kens view is it will gain energy from gravity. The truth though is it must loose energy as well because it has lost mass.
The falling object will loose mass by converting it into kinetic energy that is then used to accelerate the object. This effect will take place even if the object, say, due to air resistance, falls at a constant velocity. In this case, the energy being derived at the expense of the mass of the object is not increasing the velocity of the descending object, but, rather, is being used to push the air along its descent path out of the way.
I think that what is confusing you about this matter is that you do not realize that the amount of mass lost by a falling object is
VERY tiny. I present a problem at the end of this post in which I calculate the mass lost by weights which will show you how very small the mass loss really is.
Fletcher wrote:
Since weights travel in a closed system around a closed path always reaching the same gravitational potential each cycle, would they not loose & then gain the same mass each cycle & therefore there would be no net loss of mass ? The "what goes up, must come down" theory.
That is the whole point of the 4th Law of Motion. It shows how it is possible for weights moving around a closed path to continuously lose gravitational potential energy or, as I like to say, mass energy. This law states that in a running overbalanced gravity wheel the
average vertical velocity component of the weights will always be negative in value which is equivalent to saying that, on average, the weights are dropping.
The Earth's gravity field will then produce the
same reduction in rest mass in the weights as they move around their
closed path
inside of the overbalanced gravity wheel as it would if they were
outside and actually
dropped straight down toward the Earth along a path whose
length equalled that of the perimeter of their
closed orbit within the gravity wheel! This is really an amazing relevation and I wonder why it had not occurred to me years ago. Anyway, if this is still confusing to you, then I suggest you study my earlier post where I attached my "Bessler's 4th Law of Motion Demonstrator". You will see from the sinusoidal graph produced by a single moving weight within an overbalanced gravity wheel that, indeed, it's average
vertical velocity component during a single rotation is
downward.
You then wrote:
This would seem at odds with the motive force theory where a weights vertical velocity & horizontal displacement do not factor (just the gravitational potential argument).
I am not sure what you mean by the "motive force theory". My new law of motion only uses the properties of the
vertical motions of the weights in an overbalanced wheel to rationalize the ability of the wheel to output energy and then use this energy to accelerate itself and, if there is enough of an output to more than overcome air resistance and bearing friction, perform useful work in the wheel's environment.
I think like Michael, you should study the material that I am now going to add to this post.
Well, I've been playing around with the new law of motion like a kid playing with a new toy and want to show off some interesting new expressions that I derived from it.
But, first, let me briefly recap Bessler's 4th Law of Motion and some of the relationships immediately associated with it.
The primary equation of the 4th Law of Motion is:
Vaver = - 2πdω
where Vaver is the
average vertical velocity component of the weights as they move around their orbit inside the wheel, π is, of course, pi which is about 3.1415927, d is the
horizontal distance from an overbalanced gravity wheel's axle to the CG of its rotating weights, and ω is the rate of rotation of the wheel in rotations per unit of time. This equation assumes that the CG of the weights is directly horizontally to the left or right of the wheel's axle. If, however, the CG is below the height of the axle so that it makes an angle, φ, with the horizontal, then a modified form of the previous expression must be used to compensate for this:
Vaver = - 2πdωcosφ
From the fundamental equation that is applied when the CG is directly
horizontally to the left or right of the overbalanced wheel's axle, I was able to derive an expression for the
change in gravitational potential energy, ΔPEgrav, that the weights inside of the wheel's drum experience per wheel rotation. That expression was:
ΔPEgrav = - 2πMgd
where M is the total mass of the weights that drive the wheel, and g is the acceleration due to Earth's average surface gravity which is about 980.665 cm-sec¯².
And, from this, I showed that the
continuous power output of the wheel, P, when it is rotating, is given by:
P = - 2πMgdω
And, for the case where the CG is below the height of the axle, this is modified to give:
P = - 2πMgdωcosφ
I then wondered if I could derive any other useful expressions from these and came up with the following.
As everybody knows by now, I promote the idea that the energy that an overbalanced gravity wheel outputs actually comes from the
loss of the rest masses of its weights as they rotate around the axle during the wheel's operation. Equating Einstein's famous mass-energy equivalence equation with my expression for the change in the gravitational potential energy of a weight moving
vertically in a planet's gravity field yielded an expression for the amount of rest mass
lost by the driving weights per rotation of the overbalanced gravity wheel that they attached to. That expression is:
ΔM = -4πMgd/c²
where c is, of course, the velocity of light in a vacuum which is equal to 29 979 245 800 cm-sec¯¹
From this it's a simple matter to obtain an expression for the
total number of rotations, Σ, that the gravity wheel can complete before it has extracted all of the mass energy of its weights. That expression is:
Σ = c²/4πgd
And, finally, an expression for the total time, T, required for the wheel to complete the Σ rotations so that it's driving weights will be completely massless:
T = c²/4πgdω
Once I had these extra expressions, I decided to apply them to a "typical" overbalanced gravity wheel type problem. I decided to make up a
fictional scenario involving Bessler's 12 foot diameter two-directional Kassel wheel to see if these equations could give me some insights into it's potential. What I found was most interesting!
PROBLEM: A document is found that appears to have been written by a close friend of Count Karl. The writer claims that he was once confidentially told by the count that Bessler's two-directional wheel was actually made up of
two one-directional wheels contained within a
single drum and that the great wheel was driven in either direction by
only one of the two wheels it contained while the other wheel was disabled by having a mechanism lock its weights into positions that placed their CG at the axle of their component wheel.
Furthermore, each one-directional wheel contained 32 4 lb. weights and the CG of the
driving wheel, when the drum was turning at its
maximum terminal rotation rate of 26 rpm, dropped
below the level of the axle so that it was only displaced 1 inch horizonally from a vertical line passing through the axle (note: since we know how far the CG is displaced horizontally from the axle, we need not use the modified equations that contain the dip angle, φ).
From this information, one must determine the amount of mass lost, per wheel rotation, by the weights of the one-directional wheel driving
both component one-direction wheels and the drum that contains them, the total number of rotations that will be required to
completely drain the mass energy from this enabled one-directional wheel's weights, and the
total time, in
years, required to do this.
Assume that the wheel is indestructible and that, despite constant air resistance and bearing friction, the terminal rotation rate of the wheel remains constant until the weights of the single component one-directional wheel driving it in any one direction are completely drained of the energy associated with their rest masses.
SOLUTION: We start by converting the weight of a component one-directional wheel's weights from lbs. to gms (remember that the
total weight of the
two component one-directional wheels' weights inside of the Kassel wheel's hollow drum will be
double this amount).
32 x 4 lbs. = 128 lbs.
128 lbs. = 58 059.823 gm
Then, using the expression for the loss of mass of the weights per wheel rotation, we write, after using a value of 2.54 cm for a
horizontal displacement distance, d, of 1 inch:
ΔM = -4πMgd / c²
ΔM = -4(3.1415927)(58 059.823 gm)
(980.665 cm-sec¯²)(2.54 cm) /
(29 979 245 800 cm-sec¯¹)²
ΔM = - 2.02208 x 10¯¹² gm
Thus, after a single rotation, the weights that drove the Kassel wheel in one direction would have lost only about 2
picograms of mass!
The
total number of rotations of the wheel in
either direction that are required to completely exhaust the mass energy of the weights of the one component wheel driving the hollow drum is given by:
Σ = c² / 4πgd
Σ = (29 979 245 800 cm-sec¯¹)² / 4(3.1415927)(980.665 cm-sec¯²)(2.54 cm)
Σ = 2.87129 x 10^16 rotations
Thus, the wheel will have to complete about 28 712.9
million million rotations to completely use up the mass energy of the weights in
either one of its two one-directional component wheels.
Finally, we determine the time required to complete this enormous number of rotations. After determining that a wheel rotation rate, ω, of 26 rpm is equal to 0.4333333 rotations per second, we write:
T = c² / 4πgdω
T = (29 979 245 800 cm-sec¯¹)² / 4(3.1415927)(980.665 cm-sec¯²)(2.54 cm)(0.4333333 sec¯¹)
T = 6.62605 x 10^16 sec.
and, since
one year equals 31 556 926 seconds, we find that the
total time required is:
T = 2.0997 x 10^9 years.
That's slightly over 2
BILLION years before
either of the two one-directional wheels within the Kassel wheel's drum would be unable to maintain the constant 26 rpm terminal rotation rate of the great wheel!
Such a timespan is actually on the order of the time required for sentient life to evolve on the surfaces of terrestrial planets! And, after that time period had expired, one could just give the giant wheel a push in the other direction and it would then continue to rotate in that new direction for
another 2 billion years!
So, we see that Bessler's 4th Law of Motion leads us to the conclusion that an overbalanced gravity wheel is
not really perpetual in an
absolute sense, but, with a running time in the billions of years, it certainly is "perpetual" for all practical purposes.
Would it be possible to construct a single machine that could, in fact, run non-stop for 2 billion years
without the need for maintenance and without suffering a critical malfunction?
As an optimist, I think it might be possible, but it would have to be a
very special piece of machinery that was carefully located where it would not be disturbed for the billions of years that it continuously operated. I'm envisioning a machine made almost completely of precious metals that would use the best non-volatile silicone lubricants on boron nitride bearings and would then be enclosed in a virtually indestructible acrylic plastic chamber from which all air had been flushed out and replaced with nitrogen gas at ambient pressure.
This chamber and the great wheel it protected would then be carefully located in some sort of mined out cavern below the surface of land that would experience little or no seismic activity or flooding during the billions of years of theoretical running time. Over that vast expanse of time, even the slow drift of the Earth's tectonic plates would have to be taken into account as continents collided, mountain ranges formed, and oceans floors expanded.
Perhaps some automated mechanism could be attached to the giant wheel that would sense when it had stopped rotating in one direction and would then give it a short duration torque (provided by a dropping weight) in the other direction to again cause the great wheel to continue its continuous motion for another 2+ billion years. That would then mean that it would take a total of 4+ billion years to completely drain all of its weights in both of its component one-directional wheels of their last erg of mass energy.
Would the wheel last that long?
Probably not. Approximately 4 billion or so years from now our Sun will swell up into a Red Giant as it completes its life cycle. In the process, the inner terrestrial planets of our solar system will be vaporized. Of course, by then all of what humanity evolves into will have been long gone in search of fresh worlds to colonize in other star system of our galaxy or even in
other neighboring galaxies.
Perhaps the great wheel would still be quietly spinning away as the walls of the cavern surrounding it turned to vapor and were blown away into space by the expanding surface of our dying Sun...
Anyway, I hope that everybody found this interesting and gained some new insights into the subject of overbalanced gravity wheels from it. I know there are a lot of mobilists out there that are, unfortunately, "allergic" to things mathematical, so I tried to keep my presentation as simple, yet accurate, as possible. Although, the mathematics are rather elementary, they may, at first, be a little intimidating to a person who is not used to solving math problems.
The important thing to remember is what Bessler's 4th Law of Motion says about the
average vertical velocity of the weights inside a running chronically overbalanced wheel. Basically, this new law of motion tells us that, because the average
vertical velocity component of the rotating weights in an overbalanced wheel is
negative, the weights
interact with the Earth's gravity as though they were
constantly dropping.
This action causes them, by the physics of the 19th century, to continuously
lose gravitational potential energy and, by the physics of the 20th century, to continuously
lose rest
mass. This is where the energy that the wheel outputs comes from and, as I've shown above, although huge, it is
not an inexhaustible supply of energy.
ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ