Perpetual Motion is Impossible

[Grimer]

A gradient gravitational field is one in which the acceleration towards a point on the earth changes with height.

All gravity fields change with height, ...

Can't argue with that. :-)

The point I was trying to make is that for objects the size of Bessler's Wheel the change with height is so negligible that for all practical purposes it doesn't change with height. That means we can't use that negligible change to get our wheel to rotate.

In contrast, for your horizontal wheel the change is large. EG goes to zero at the centre. So you have the possibility of using that gradient to get your wheel to rotate.

Interestingly enough, from the frame of reference of an observer at a particular point on the equator the apparent NG above his head goes to zero at the "stationary" satellite orbit.

[daxwc]

Jim:

A Newton's cradle is a horse of a different color.
A Newton's cradle isn't operating in a rotating environment.

It still should show a gain if your math works in reality. I look at it as the Newton’s Cradle is the rotating environment at zero.

The total momentum of a system is equal to the combination of angular momentum and linear momentum.
If the total kinetic energy of the system is increasing then setting one to zero shouldn’t affect your claim should it?

[jim_mich]

So please tell me EXACTLY what is is that you propose to set at ZERO? Are you proposing to set the wheel's rotational speed to ZERO? Doing so negates the PM effect.

[MrVibrating]

Blink... Blink... Blink...
My eyes hurt. You make things way too complex.

The solution is minimal input to get things rotating. Then no more input. Only output. Perpetually. Until motion is stalled by excessive load or the component parts fail.

Yes. Make wheel GO. Go FAST. Free. Yes.

[MrVibrating]

I agree with Jim's recent summary of MrV's dissertation: OU = input F*d < output F*d

Can anybody tell why a gradient gravitational field would be beneficial in achieving pepertual motion?

Thank you, and the answer is because gravity is a 100% efficient boomerang for reaction mass.

Invert the momentum of your reaction mass, and the net system momentum increases.

So in essence, by accelerating against a ballistically-flung mass, that reaction mass is returned to us with all of its momentum now vectored in the same direction as us; our net momentum has increased in an otherwise closed system.

The system can be regarded as closed because all of the input energy has been distributed soley between the two masses - although we still have the usual infinitesimal bobbing of the Earth in relation to the rising and falling reaction mass, this interaction remains entirely symmetrical and incidental... it's not a gravitational asymmetry we're targetting, but an inertial one.

Thus gravity's role is simply that of a 100% conservative spring to bounce our reaction mass off of. It mirrors it back at us, with its sign inverted. But whereas a physical spring would require a stator - it would have to use the Earth as reaction mass, like any normal rotor, precluding a net change in momentum - using gravity means the masses need only interact with one another, and so keeping the system's net momentum isolated.

TL;DR - you can get good kicks by chucking stuff up in the air, innit.

[MrVibrating]

Mrv, jim is right , your posts are unnecessarily wordy and over-analytical for Bessler's wheels.
When I said any interaction is asymmetric in the "loss" column, what I meant was any interaction will produce friction. This energy is dissipative and guarantees any interaction will have less output than input, i.e. < 100% efficiency, continuing the argument I was making. Sorry I confused things there.
I think, but I'm not sure because I 'm not going to wade through your entire post, my eyes are old too, that we are saying the same thing in different ways.

One day someone, somewhere will post a one-line comment so misconceived it'll spawn a whole new forum, or even diverging internets, and possibly civil war..

But what i tried to explain, if a little too exhaustively, is that input losses are included as part of our input energy, while output losses are included as part of our output energy.

The individual work conversion efficiency of either net input or net output is irrelevant to their combined balance of energy.

- "interaction" = full closed-loop exchange; input AND output

- all conventional interactions are symmetrical - net input = net output

Dissipative losses are wholly incidental to the net balance of an interaction's energy. Non-dissipative losses are critical.

If you need further clarification you'll probably find it in that text bomb...

[MrVibrating]

Blink... Blink... Blink...
My eyes hurt. You make things way too complex.

The solution is minimal input to get things rotating. Then no more input. Only output. Perpetually. Until motion is stalled by excessive load or the component parts fail.

Soz but there's a slight misconception there - no matter what the nature of the asymmetry, it still requires cycling... with no further input there can be no further output.

Otherwise it's a one-shot deal... like a SMOT (if SMOT's actually worked).

"Unity", and over or under, refers to the symmetry of input to output integrals, or lack thereof. So if we want to keep harnessing gains over successive cycles, then we need to keep inputting energy.

[MrVibrating]

When a mass attached to a lever rotates a rotating disc, it will act like a pendulum in a rotating frame.
It 'pendulates' towards the centrifugal acceleration.
This centrifugal acceleration acts like a gradient gravitational field, a=w^2*r.
Hence my question earlier.

That's a good point - chucking stuff inwards, against CF, so that it's flung back out again, would also seem a good way of inverting the sign of a reaction momentum..

Will have to spend some time considering this...


ETA: an obvious, if circumstantial, objection is that Bessler's wheels were vertical - he'd undoubtedly have realised if CF could be substituted for gravity and presumably produced horizontal wheels too...

Still could be a key dynamic tho - at least for the bi-directional wheels, which reportedly only engaged above a threshold RPM...

[jim_mich]

Soz but there's a slight misconception there - no matter what the nature of the asymmetry, it still requires cycling... with no further input there can be no further output.

Otherwise it's a one-shot deal... like a SMOT (if SMOT's actually worked).

The misconception is on your part.

Remember, the initial input is wheel rotation. And the output is accelerated forceful wheel rotation. Thus, no further input is needed. The existing output provides the wheel rotation needed for the process to keep going.

[daxwc]

Jim:

The formula for inertial momentum is IM = m × v.
The formula for kinetic energy is KE = 0.5 × m × v^2.

The result is that as the IM increases in a linear fashion, then the KE increased exponentially.

Your inertial momentum IM = m × v of your example is of non-rotating linear mass. It is not the inertial momentum of a rotating frame of reference. The total momentum of a system (the wheel) is equal to the combination of angular momentum and linear momentum. Which tends to negate itself and just doing one part of the math produces fictions forces.

[jim_mich]

Come on daxwc, use your brain. Or did it fall out of your head.

Angular momentum is the same as linear momentum. They are not different critters. If a cat is running in a straight line as 10 MPH and another cat is chasing his tail in a five foot diameter circle running at 0.9337 RPS (56.02 RPM), both cats are running at a same speed. Both cats have a same momentum. Momentum is momentum is momentum. It does not magically change from linear to rotational simply because a weight is moving in a circle versus moving in a straight line.

Any weight moving in a circle has an instantaneous velocity. If you cut the weight free from rotating, then the weights will be moving at that velocity. If you measure the velocity of the weights as it spins around the wheel center, then its speed will be the same as when cut free. Its momentum before and after will also be the same.

Thus the basic formula for inertial momentum is IM = m × v. The formula for rotational momentum is derived from this same basic formula.

So don't rag on me about my formula being for linear motion rather the for a rotating frame of reference. Again, Momentum is momentum is momentum.

[Grimer]

WAS THIS EXAMPLE TOO HARD TO FOLLOW ???
The formula for inertial momentum is IM = m × v.
The formula for kinetic energy is KE = 0.5 × m × v^2.

The result is that as the IM increases in a linear fashion, then the KE increased exponentially.

You don't really mean exponentially, do you?......... :-)

[ME]

When a mass attached to a lever rotates a rotating disc, it will act like a pendulum in a rotating frame.
It 'pendulates' towards the centrifugal acceleration.
This centrifugal acceleration acts like a gradient gravitational field, a=w^2*r.
Hence my question earlier.

That's a good point - chucking stuff inwards, against CF, so that it's flung back out again, would also seem a good way of inverting the sign of a reaction momentum..

Will have to spend some time considering this...

The rotary-variant of
"the importance of raising weights", or the "trading height for width"-dilemma
could be something like:
"the importance of compacting angular momentum", or "trading radius for angular velocity"?

Anyhoo, attached is a simulation of a rotating Newton's Cradle to decorate this forum. The left-one is the normal frame of view, the right-one is the same but in a rotating frame of view. (a bit small to fit max. filesize)
The green and orange weights are not participating and merely added as additional decoration, or otherwise showing a rotating pendulum.

[jim_mich]

You don't really mean exponentially, do you?.

Yes I do. This is where the extra energy comes from. You have a first weight and a second weight riding a rotating wheel. The second weight "transfers" most of it momentum to the first weight. The second weights slows down. The first weight speeds up. The exchange of momentum is very close to equal, with the wheel picking up any slight difference.

The velocity of the first weight has increased. The velocity of the second weight has decreased. Conservation of momentum is conserved.

Pick any value for the initial velocity of the two weights. Pick a value for the velocity change. Then do the calculations...
Back on Dec 14, in post #139011 I posted the following...

As an example. Assume you have two weights moving a same speed. Transfer momentum from one weight to the other weight. This can be done spontaneously in a rotating environment of a wheel with little or no effect on the rotation. Do the math. Assume both weights were moving at say 20 feet/second and both weights have a same mass of two units. The weight's KE is 1/2 × 2_mass × V^2. Since 1/2 × 2 equal 1, we can make the formula very simple as KE = V^2. Thus KE of each weight equals 20^2 = 400 KE units. Total KE combined of two weights is 800 KE units.

Now one weight spontaneously gives up half its velocity to the other weight.
Weight #1 V = 10 ft/s. Its KE = 10^2 = 100 KE units.
Weight #2 v = 30 ft/s. Its KE = 30^2 = 900 KE units.
Total KE combined of two weights is 1000 KE units.

Thus we gained 200 KE units simply by allowing motion to transfer from a weight that is slowing down to a weight that is speeding up. We added no energy. Momentum was conserved. The weights changed their velocities spontaneously without consuming any energy.

Note that I expect the decelerating weight to give up close to 90 percent of its velocity to the accelerating weight.

Weight #1 V = 2 ft/s. Its KE = 2^2 = 4 KE units.
Weight #2 v = 38 ft/s. Its KE = 38^2 = 1444 KE units.
Total KE combined of two weights is 1448 KE units.
An increase of 648 KE units.
A gain of about 81 percent..

[ME]

Jim, what's the result of your formula when you normalize to a rotating frame of reference?

I guess/hope you have some precautions in case your model spins out of control?