Author Topic: The Paradox Engine (Read 122290 times)

Tusk · « **Reply #120 on:** January 08, 2014, 12:23:40 AM »

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I may have this wrong, which is why I am bringing it up.

Ok webby1 I'll try to retrace your steps and point to any differences between my interpretation and your own.

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If the disc is not allowed to rotate about its center of mass and the arm were to make one rotation then the arm would see the disc make one rotation, and the disc would see the arm make one rotation.

This first set of conditions is a little unclear; if you mean that the disk is prevented by inertia from rotating as the rotor arm rotates then yes, the observation is correct.

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To have the arm make one rotation and have the disc make one rotation about is center of mass while it is orbiting the pivot for the arm, the arm would see the disc make 2 rotations and the disc would see the arm make 2 rotations.

Again, based on my interpretation of your first premise this is correct for one direction of rotation but not the other; reverse motion to the opposite you had in mind and each would see the other as motionless (which point actually lends a little more clarity).

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From the PoV of the applied force of acceleration against the disc, the edge of the disc will move twice as far as the arm.

Correct.

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I interpret this then that the disc is spun up with full force over twice the distance of reaction from the arm,

And this is where things get tricky; you seem clear on the point that the disk will receive 50% of it's rotation from inertia (due rotor arm motion) and unclear how to proceed regarding determination of work done since this relates to force applied over a distance, but our disk is covering twice as much distance with a 50% assist from inertia.

So here's the interesting part then, since it's no simple matter:

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the arm having an equal force of motion but at 1\2 the distance and the disc is left with an increase in potential due to its rotation which when recovered will impart the full force into the arm over another 1\2 distance.

My thoughts on this lately have me inclined to slow the rotation of the rotor arm using a geared output to a generator. This sidesteps these issues quite tidily

Your meaning on this last point is again a little unclear, but if you are defining the motion by distance (which you seem to be doing) then it sounds like it might be correct for the current single disk PE apparatus. Your method of resolving it differs from my own. Also note that everything changes with two disks and a lightweight rotor arm.

Tusk · « **Reply #121 on:** January 08, 2014, 04:36:52 AM »

Ok this inertial rotation of the disk seems to be causing some concern; let's add another line of code then, and demonstrate the potential to sidestep and maybe even use our opponent's weight against himself

If we recover energy at the disk axis (i.e. a geared generator) no secondary reaction manifests and any retarding force on the rotor arm from the disk axis (due to resistance from the generator) is minimised. This gives us the option to recover energy from the disk without significantly reducing rotor arm motion.

Am I right in suggesting this? I'm off the reservation again here, so keep a close eye on things

Additional: I assume this is so due to the disk axis being located at a large enough radius from the rotor arm axis that the (essentially) two opposing forces at the generator will have nearly equal moment arms (depending on the relative size/diameter of the generator with regard to the radius) while noting that some bias will manifest; and that against our purpose.

First glance at this suggests the possibility of 'spinning up' as per usual, then recovery of total rotor arm FoR disk energy at the disk axis (as above) without significant reduction of rotor arm motion. By this (bold print) I refer to the full disk rotation rate as it appears in the frame of reference of the rotor arm. As we already noted this can be as much as double the rotation rate in the observer FoR.

So having brought the disk to a stop (FoR rotor arm) we can now recover energy from the rotor arm, which will induce yet another rotation of the disk (FoR rotor arm) due to inertia (much like our other little issue, but this time in 'reverse'). Recovery of this new disk rotation.... etc etc. I've lost count but this looks like a promising combination, at first glance. I seriously doubt it can go on ad infinitum without some additional input (potentially from all that output) but it seems like an interesting new FoR manipulation.

Tusk · « **Reply #122 on:** January 09, 2014, 02:34:26 AM »

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I can orientate the magnet any where, not just next to the axle, and have the same reaction to either acceleration or deceleration of the rotor.

I had a quick look for an example of this Tom, a veritable junkyard of spinning tops and swinging doors; no luck so far though

Anyway you almost certainly realise that what you are seeing is a simple case of lever arm bias. The rotor goes one way, the stator, case, mounts etc go the other way (let's refer to these as simply 'the case'). Since the 'case' is attempting to rotate regardless of being mounted on a rotor arm, that part furthest from the rotor arm axis will have a greater lever arm advantage than the nearside, resulting in rotation of the rotor arm.

With the PE apparatus (as telecom kindly pointed out recently) the secondary reaction at the disk axis has no force pair, except inertia; unless we provide some other opposition such as that incurred when turning a generator. If we choose not to, and allow the secondary motion to spend itself uselessly, it might easily be interpreted as a lever arm bias with the various forces and motions in balance and accounted for. Two distinct and separate yet equal reactions in the same direction and originating from a single applied force must create an imbalance, or bias (as I prefer it).

So while it seems unlikely we would see OU with a simple lever arm bias, the secondary force which manifests at the disk axis as described owes nothing in terms of the force which created it, that debt having been paid in full with the reaction at and rotation of the disk.

It's a 'freebie'

telecom · « **Reply #123 on:** January 09, 2014, 02:55:18 AM »

Since this "free" force at axis of the disc doesn't have a reaction pair,
it may be an ideal candidate for a propulsion force!

Tusk · « **Reply #124 on:** January 09, 2014, 04:02:48 AM »

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it may be an ideal candidate for a propulsion force!

I have to smile at that telecom, since this is full circle on how the investigation began. I was thinking about possible inertial drive systems myself

Here's a rough sketch of my original idea, which immediately brought forth the paradox (bearing in mind that no paradox can survive it's own solution).

A quick description then; this is a twin disk system in equilibrium, the disks motivated equally by twin EM drive units as shown. The applied forces (on the disks innermost edges) are indicated in red, with the reactions to same indicated in black at the drive units. Also I have indicated the rotation of the disks.

The question then becomes: does the device accelerate, if so which way and if not then why?

telecom · « **Reply #125 on:** January 09, 2014, 04:49:42 AM »

Hi Tusk,
it appears to me that according to a paradox effect you have discovered,
there should be 2 axial forces in the direction and a magnitude of the
EM forces, which should accelerate the whole assembly to the left.
In fact, if they were equal to the weight of the apparatus, they could lift the device
upward, if desired.

Tusk · « **Reply #126 on:** January 10, 2014, 01:50:28 AM »

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there should be 2 axial forces in the direction and a magnitude of the
EM forces, which should accelerate the whole assembly to the left

Afraid not telecom. If this were so we'd be in breach of CoM and the PE apparatus would not function as stated. It's a perplexing little puzzle is it not? Took me months to work it out, and a variety of inconclusive experiments before I finally came up with the peg pendulum experiment:

http://www.youtube.com/watch?v=zi8k3PMUM6k&feature=youtu.be

So minutia aside (and there have been some objections) the bottom line is this: the spring must impart the same force in each direction (Newton's Third) yet on the one side we observe only linear motion, while the other clearly has both an equal linear motion and rotation. Any dismissal of this outcome based on reference to angular momentum simply fails to address the fact that the force applied to the end of the peg results in more total mass in motion than the other, and therefore more energy.

Note that I do not claim more energy here than is provided by the spring. We require a frame of reference manipulation to take advantage of this phenomenon and achieve OU.

Thus the peg pendulum experiment finally provided good evidence for the secondary effect, which I had already postulated since my supposed inertial drive system failed to move even a gnat's whisker

Which made this seem all the more likely:

A force applied at any point on a body in equilibrium results in an equal and parallel reactive force at the centre of mass of the body acting in the direction of the applied force.
This reaction causes such linear motion of the body as would occur if the original force were applied at the centre of mass, independent of any rotational motion produced by the moment of the applied force.

Perhaps not quite a law, maybe a footnote to a rule?.... at any rate, in the absence of guidance it might help to keep things pointed in the right direction.

There should be enough here to demonstrate why the 'inertial propulsion system' didn't move; I've added the relevant forces in red to the sketch, in the interests of clarity:

telecom · « **Reply #127 on:** January 10, 2014, 02:12:58 AM »

So,
all the forces, red and black are balanced, and what we get extra is the
rotational momentum of the discs?

Tusk · « **Reply #128 on:** January 10, 2014, 03:25:40 AM »

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what we get extra is the rotational momentum of the discs

That's one way to view it telecom; not the best perspective for anyone unfamiliar with the phenomenon but yes, probably more accurate. Once you have a good grasp of the secondary reactive force your explanation becomes the more preferable I suspect.

I think you have to admit it's quite elegant. The conventional reasoning for it is in angular momentum, but the reality is that by simply shifting our point of applied force (which costs us nothing) we get significantly more mass in motion (thus KE) than for the same applied force at the centre of mass. Who knew?

It took a full year before someone found an MIT reference to support these findings. I have no idea where MIT referenced it from (probably some dusty old leather bound tomb) but I had some expectation that with these 'runs on the board' we could move forward and tackle the frame of reference phenomenon in a more receptive mood

For what it's worth I believe that frame of reference manipulation is really the key to 'mass in motion' OU. There seems little doubt (to me at least) that other methods will be found which have no dependence on the secondary reactive force. KE is simply a matter of the relationship between mass in motion and FoR, and the opportunity exists for various advantageous discoveries in this area.

The PE apparatus merely demonstrates mass in motion above and beyond expectations according to static 'rotor arm secure' flywheel tests. We know from the above that (as telecom put it) the rotational momentum of the disk is 'extra'; in this view we have 'paid' for the rotor arm motion and the disk energy is 'free'. But since we achieved disk rotation in the 'rotor arm secure' tests (albeit not as energetic) I actually prefer to say that the rotor arm motion is 'free'

telecom · « **Reply #129 on:** January 10, 2014, 04:37:27 AM »

Hi Tusk,
does this mean that all the linear forces are balanced by the linear reactions,
and only angular moments are unbalanced, no matter which one is "free"
or "paid for"?

telecom · « **Reply #130 on:** January 10, 2014, 04:50:42 AM »

Just want to add that in your apparatus with one disc, all the angular moments are balanced as well!

Tusk · « **Reply #131 on:** January 10, 2014, 07:35:27 AM »

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does this mean that all the linear forces are balanced by the linear reactions,
and only angular moments are unbalanced, no matter which one is "free" or "paid for"?

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all the angular moments are balanced as well!

Thanks telecom, I think you are now seeing 2 sets of 'force pairs', is that correct? A simple analysis might be:

1. applied force on disk has it's reaction on the drive unit (which is essentially bench mounted to this line of force)

2. the secondary reaction at the disk axis (being unique) must be considered a special case, being essentially a reaction to the inertia of the opposite side of the disk

A more thorough analysis should probably take inertia into account for both 'force pairs'. The bench mounted drive unit resists the reaction to the applied force with the inertia of the planet; likewise the disk resists the applied force with the inertia of it's own mass.

So the full story for instance 1. is:

1. inertia of planet on drive unit : reaction to applied force on drive unit : applied force on disk : inertia of disk (total mass)

and when we apply a similar comprehensive analysis on instance 2 we get:

2. inertia of planet on drive unit : reaction to applied force on drive unit : applied force on disk : inertia of opposite side of disk X2 due lever arm : secondary reaction at disk axis : inertia of rotor arm

I think that either clears things up a little, or not; depending on your initial level of comprehension. But I would definitely call instance 2. a special case.

Importantly (if we allow the introduction of energy in a simplistic yet conventional sense) we should note that the applied force on the disk provides the energy for both motions, or acts as a mechanism of transmission of it, if you prefer. As such it must surely be clear that here at least energy has spontaneously twinned itself to motivate two distinct motions since the applied force is singular, yet double, but not halved.

Thus the paradox, and once again knowing how uncomfortable this can get I stand ready to address the more familiar concerns while the elephant quietly dances around the room lol

telecom · « **Reply #132 on:** January 11, 2014, 12:49:04 AM »

applied force on disk : inertia of opposite side of disk X2 due lever arm : secondary reaction at disk axis : inertia of rotor arm

THis part is not absolutely clear to me.
Why it comes out as X2? opposite side?

Tusk · « **Reply #133 on:** January 11, 2014, 04:40:36 AM »

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Why it comes out as X2? opposite side?

My apologies telecom, I may not have fully explained the origins of the secondary reactive force (I mentioned it in passing but not in great depth). Bear in mind that in the absence of relevant material in the literature the following explanation derives from my own experimentation and analysis, but I believe it to be correct.

In equilibrium then:

In the sketch below (and for simplicity) a rod of zero mass has two objects A and B of equal mass mounted one at each end. A force is applied to one end (A) as shown.

The applied force must motivate both masses. Therefore the force required to motivate the opposite mass (B) is half the applied force. Thus the force of inertia opposing that force is also equal to half the applied force.

The applied force must act through the centre of mass in order to motivate the opposite side, therefore the centre of mass acts as a pivot point situated halfway between the applied force and the resistance of inertia.

Also the inertia of mass A opposes the applied force as shown, such that only half the applied force acts through the centre of mass (see blue annex in diagram).

Thus the resulting force on the 'pivot point' (centre of mass) is the sum of half the applied force and the inertial force on B, or (for simplicity) twice the force of inertia on B; and therefore equal to the applied force.

Here then we see the origins of the secondary motion which, while not in violation of CoE or CoM in it's own right yet offers up opportunities in the quest for OU. With this new 'line of code' might we not reprogram reality with a fresh outlook on the manipulation of frames of reference?

Consider that a constant force applied to a mass in equilibrium results in a constant acceleration, and since Ek = ½ mv² then by application of a constant force the kinetic energy of the mass increases exponentially; a tantilising hint of OU if we could only reduce or eliminate the cost of motivating the point of force to keep pace with the acceleration. This cost of point of force motion under normal circumstances erodes all advantage. Yet we see here in a simple peg pendulum experiment a reaction, indeed an unexpected additional force manifesting remotely from the point of applied force.

This allows us, with careful design and engineering, to apply a force from outside a frame of reference wherein the secondary reaction motivates a mass additionally to the primary motivation caused by the applied force, without the usual requirement to accelerate the point of applied force.

telecom · « **Reply #134 on:** January 11, 2014, 04:55:08 PM »

Thus the resulting force on the 'pivot point' (centre of mass) is the sum of half the applied force and the inertial force on B, or (for simplicity) twice the force of inertia on B; and therefore equal to the applied force.

So, the rotational movement is activated by the 1 unit of the applied force, and the linear motion by the 1/2 unit of the applied force + the inertia of the mass B?
Which also totals to 1 unit of the applied force?
In this case we are getting the inertia B working for us for free? And it is equal 1/2
of the applied force?

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