Interlude — and a Change in Direction

Mechanical devices — finished

Although there is more, I have now said all that I wish to say on mechanical devices, for the foreseeable future. I suppose it could be argued, correctly, that I haven't really contributed much to our existing knowledge of mechanical perpetual motion — although as I pointed out at the time, I think my ideas on the Casimir Effect Force Generator (posts of 2 to 17 September 2014) and the "Perpetual Force" Air Motor (posts of 22 November to 13 December 2014) could well be worth further investigation.

More support for the Casimir Effect Force Generator

As the years go by, I keep a look-out for support for my unorthodox technological ideas. On page 5 of its Issue 96, March/April 2011, Infinite Energy magazine published a letter from Wm. Scott Smith titled "Can We Make a Casimir-Cavity ZPE Thruster?" This letter, citing eleven peer-reviewed references from 1997 onwards, argued that an array of sufficiently small open-ended cavities would experience a net Casimir force. The only physical difference from the Casimir Effect Force Generator in my specification drawing of October 1994 was that the proposed open-ended cavities had straight rather than sloping sides. Although I found no-one who could make such a device in 1994, Smith claimed "A macroscopic array of nanoscopic cavities can be constructed quite easily with existing [2011] nanotechnology." If that's correct, I'd certainly agree that this would be an experiment well worth trying (for both straight and sloping-sided cavities).

This writer also said "Sometimes it is astonishingly difficult to reflect carefully enough on something we already know." I agree entirely with regard to Casimir force, and also, as will be seen, on some electrostatic matters, including Gauss's law.

Electrostatics

A modern electrostatic motor. Data: size 65 dia × 65 length; max rpm 10,000;
power 100W; weight 0.2kg; power-to-weight 500W/kg; efficiency 95%.
See http://www.shinsei-motor.com/English/techno

As mentioned at the outset of my very first post, I originally intended this blog to be mostly about my investigations into electromagnet/permanent magnet interactions. However, before getting into those, I'll first write up a few ideas on electrostatic devices.

In this series, I'll first discuss why electrostatics gives a wider scope than mechanics does for developing perpetual motion machines. There is at least one well-known technology which could in theory, and perhaps even in practice, be developed to give excess energy. There are also at least two other quite well-proven electrostatic perpetual motion machines that have already been built and demonstrated in the real world. These will all be discussed in due course.


A reminder about definitions

For now, I'll repeat the definition I first gave on 30 March 2014:—

       [quote begins]

Since this blog is titled "Perpetual Motion in the 21st Century", let's look at how "perpetual motion" is defined these days. In my opinion, the most authoritative English-language dictionary of all is the Oxford English Dictionary. Here is its definition:—

"Perpetual Motion: Motion that goes on for ever, spec. that of a hypothetical machine, which being once set in motion should go on for ever, or until stopped by some external force or the wearing out of the machine." 

(Reference: The Oxford English Dictionary, Second Edition, Clarendon Press, Oxford, Vol XI, p586).

This is the definition of Perpetual Motion to which I adhere.

There is an on-line version of the Oxford dictionary, at http://www.oxforddictionaries.com. It has this abbreviated definition:—

"the motion of a hypothetical machine which, once activated, would run forever unless subject to an external force or to wear: the age-old quest for the secret of perpetual motion"

The remarkable fact is that the Oxford dictionary is the only one that has not felt obliged to "modernise" the definition to incorporate some reference to energy, and thereby reduce it to an almost worthless banality (along the lines of "You can't get energy from nothing"). All other English-language dictionaries, and on-line sources such as Wikipedia have now done this, as far as I know.

       [quote ends]

I know I'm more or less a lone voice at present trying to preserve a valid and useful definition, but I've never been convinced that the appropriate response to the criticism that has been levelled at "perpetual motion" was to change its name (and to re-define the old name — badly). I'm reminded of the current fiasco concerning what was called "cold fusion" by Fleischmann and Pons. There are now well over a dozen different names for that topic, which generally only add confusion, rather than clarity.

However, it's true that in electrical technology there are cases where machines can deliver energy without needing any moving components. Unless the flow of electric charge carriers is being considered, "perpetual motion" is not a very appropriate term for such machines. So from now on, I'll probably use more modern terms like "excess energy" or "free energy" for these machines at least.

An electrostatic voltmeter (center) with two high-voltage power supplies:
a 2.5kV 12kHz supply for a corona-discharge ozone generator (right) and
a homemade 5kV 50Hz Cockroft-Walton voltage multiplier (left).

The "Laboratory" Frame Part IV

A part-inertial/part-laboratory experiment

Revisiting an earlier idea

When discussing the idea shown in the above figure, in my post of 25 April 2015, I assumed that the spring to be used to return the mass over portion B to C of the operating cycle would be a constant-force spring of force equal to twice the weight of the mass. Let's now look at this in more detail.

Does the mass-spring system act in full gravity?

First, recall that I have assumed that gravity always acts vertically, which it does, to an extremely close approximation, over the small time intervals of these experiments.

The value of the constant-force spring was calculated as:—

2W = 2mg = 2 × 10 × 9.80665 = 196.133 N.

Note that I used the same (standard) value for gravity, g = 9.80665 m/s² as I had used to calculate the behaviour of the mass between A and B. This is correct for a mass hanging freely from a suspension point and behaving fully "inertially", disconnected from Earth between B and C (as it was between A and B). However it would be possible to constrain the mass between B and C to radial movement only, e.g. in a vertical (radial) tube rigidly fastened to Earth. It could be argued that this should then modify the value to be used for gravity, to:—

g' = g - (Earth's centrifugal acceleration).

Hence, at the equator, g' = (9.80665 - 0.033916) m/s² , and the value of the constant-force spring becomes:—

2W = 2mg' = 2 × 10 × (9.80665 - 0.033916) = 195.45468 N.

So there would be a net gain of energy over the originally-calculated value, of:—

 (196.133 - 195.45468) × 0.271326 = 0.184046 joules, 

as the mass was returned from B to C, with the suspension point falling from B' to C' as before.

A net energy gain is calculated

This is a very low-power result, i.e. 0.184046J/10kg/8s  ≈  2.3 milliwatts of power per kilogram of active mass, over the operating cycle of 8 seconds. But nevertheless, it is a positive result as calculated.

I have set myself a simple "rule of thumb" — I will not even consider building any physical prototype machine that could not reach at least one watt of power per kilogram of active mass.

For now, I leave as an open question whether a gain like this could really be achieved in a system as discussed, where a mass is designed to behave truly "inertially" over part of its cycle, and is constrained into a laboratory frame for the rest of the cycle; or whether it just indicates some error, e.g. a breakdown of the assumption that gravity is always vertical. Two relevant points to bear in mind are:—

1. As soon as the mass is fired-off from Earth at point A, it no longer has any physical connection with Earth, and must act "inertially", albeit still under the influence of Earth's gravity.

2. The mass can certainly be constrained into a physical connection with Earth, forcing it to move radially in a laboratory frame over B to C, as discussed above. But gravity really acts radially, anyway! 

More experiments

There are obviously many more experiments that could be done, as developments of the ideas already discussed in this post and the three previous ones. I have done quite a few myself, including some that had a more clear-cut separation between "inertial" and "laboratory" behaviour than in the above introductory example. For example, a rotating-wheel gyroscope prefers to orient itself in an inertial frame as far as possible, rather than a laboratory frame. And the gyroscope can take other forms — such as the halter gyroscope, which many flying insects possess in some form. But to discuss these experiments further now would be to go too far out of order in what was always intended to be a roughly chronological-order blog.

Crane fly, with halteres visible behind the wings
Image from http://en.wikipedia.org/wiki/Halteres

The "Laboratory" Frame Part III

Once again we shall set up our inertial frame ABC, and this time we'll examine the behaviour of an unequal pair of masses in it (Fig 4).

Fig 4.  An unequal pair of masses

Centrifugal Force and Kinetic Energy data

The unequal mass-pair of Fig 4 is joined together by a rod (black) of negligible mass which pivots and rotates about the center of rotation as shown. For the values given we have:—

10kg mass:   Centrifugal Force = mv²/r  = 10 × 1² / 0.1 = 100 N
                          Kinetic Energy   = ½mv²  = ½ × 10 × 1² = 5 J

1kg mass:    Centrifugal Force = mv²/r   = 1 × 10² / 1    = 100 N
                         Kinetic Energy   = ½mv²  = ½ × 1 × 10² = 50 J

So we can have equal, balanced centrifugal forces, even though the masses have very different energies. By causing the mass-pair to travel along the inertial line ABC, we can achieve different start and finish velocities in the laboratory frame for both masses. Will that deliver any net energy?

Fig 5.  Mass-pair travelling along the (straight) inertial line ABC

Analysis

We shall assume that the mass-pair is "weightless", e.g. it has a constant-force spring to Earth (not shown) acting at the center of rotation, and neutralising its weight in the inertial frame. Since the spring is always active from A to C, and undergoes no net change in length, there can be no net energy change in it.

We start the mass-pair at point A, rotating as before, such that its center of gravity (at the center of rotation) is following the straight-line (purple) trajectory ABC in the inertial frame. For simplicity, we shall start it travelling downwards at Vy = -1m/s in the laboratory frame. Then, in that frame, we have:—

10kg mass:   Kinetic Energy = ½mv² = ½ × 10 × (1 + 1)² = 20 J

1kg mass:      Kinetic Energy = ½mv² = ½ × 1 × (10 - 1)²  = 40.5 J

When the mass-pair reaches point C, where its center of gravity is travelling upwards at Vy = 1m/s in the laboratory frame, we have:—

10kg mass:   Kinetic Energy = ½mv² = ½ × 10 × (1 - 1)²  = 0 J

1kg mass:      Kinetic Energy = ½mv² = ½ × 1 × (10 + 1)² = 60.5 J

So, although the masses each have quite different start and finish energies, there is no net energy loss or gain. More generally, in the inertial frame ABC, all we did was to take the rotating mass pair from a given energy environment at A, into another environment at C that had the same energy as A.

The "Laboratory" Frame Part II

In the next few posts, I'll look at two more ideas for extracting energy from the rotating Earth. In both cases, a straight-line inertial frame ABC is set up, which does not rotate, (but it does move along with Earth's center). We analyse the operating cycle in this truly inertial frame, and also check whether there is any overall difference in energy in the laboratory frame, which is the only frame where it would be accessible.

Fig 3.  A mass is fired off from Earth at point A, passes through B, and is collected against Earth at C

Drawing difficulties; gravity can be vertical

The first difficulty we face is that it is impractical to try to draw these ideas to scale, and so the drawing has to be greatly compacted-up in the horizontal direction (Fig 3). This in turn tends to exaggerate the curvature of the Earth's surface (shown above as the curved black line from A to C). From our previous calculations in Fig 2, we saw that in say 4 seconds, the Earth turns through an angle of only 0.00029168 radians = 0.016712 degrees. That is a negligibly small deviation, which would cause negligible error if we assume that gravity always acts vertically, over time intervals of this order.

Analysis

In Fig 3, assume that the mass is fired upwards from point A on the Earth's surface at a velocity of Vy = 19.6133 m/s in the inertial ABC frame. This will cause it to initially rise and then fall in Earth's gravity, crossing point B after 4 seconds, again at 19.6133 m/s. Then, using a spring attached to Earth, its fall at B is slowed and reversed, until it rises back to point C at the Earth's surface after another 4 seconds, again reaching 19.6133 m/s there in the ABC frame.

Let's look at this in more detail. Firstly, from symmetry, there will be no net gain or loss of velocity Vx in the horizontal direction, so that can be ignored. Next, in the laboratory frame, which is moving upwards at A at 0.135663 m/s, we would only have to give the mass a velocity of (19.6133 - 0.135663) = 19.477637 m/s. Then, by similar reasoning when it is collected at C, we would gain another 0.135663 m/s, to get (19.6133 + 0.135663) = 19.748963 m/s.

For a 10kg mass, we would gain energy of:— 

½ × 10 × (19.748963² - 19.477637²) = 53.21598 joules.

The flaw

The flaw in this is obvious enough, but let's do a thorough check:—

There is nothing wrong with the first part of the cycle, from A to B. The problem is in the second part, from B to C. The simplest spring to be used there (theoretically) would be a constant-force spring of force equal to twice the weight of the mass. Then the trajectory of the mass from B to C would be a "mirror-image" of its trajectory from A to B. However, such a spring would have to give up energy as its attachment point on Earth fell vertically between B' and C' i.e. through Δh = 0.271326m in the ABC frame. It would lose:—

2W × Δh = 2 × 10 × 9.80665 × 0.271326 = 53.21598 joules.

So there is no net energy gain in this case.

"Shifted time"

Could anything be done to avoid the energy given up by the spring? I used to think that the concept of "shifted time" associated with remote viewing of Bessler's wheel might refer to something like this (see my post of 14 June 2014 on this blog). For example, if the spring could be active between D and E, rather than between B' and C' there would be no net loss in the energy stored in it. I still think the general concept of "shifted time" has merit (in general terms: carrying out some action at a point in the operating cycle that only makes sense in terms of extracting energy from the rotating Earth). However I have come to modify my earlier views somewhat about this.

The "Laboratory Frame" Part I

The laboratory frame of reference is not truly inertial

A "laboratory" frame of reference is one that is attached to the surface of the Earth, and so it rotates along with the Earth. Normally we regard a laboratory frame as truly inertial, in which Newton's laws of motion can be applied in their simplest form. We generally do only that, as any errors that occur as a result are almost always small enough to be completely negligible.

Fig 1.  A silux macro that forces an object (o1) to behave as a point on the Earth's surface
at the equator would behave, as seen in a non-rotating frame attached to the center of the Earth.
(Note the centimetre-gram-microsecond system of units).

Back on 4 October 2014 I posted Figure 1 above, showing how an object in a laboratory frame at the Earth's equator is seen to move as measured in a truly inertial frame set at the center of the Earth. (The Earth's orbital motion is disregarded). From a macro like this, or just from the basic physics involved, results can be derived for the upwards displacement and velocity for a truly inertial "weightless" mass initially placed on the equator (Fig 2). I used values of 0.00007292115 radians/sec for the Earth's rotational velocity, and 6,378,137 meters for its equatorial radius.

An observer in a laboratory frame who stays attached to the Earth would see the weightless mass rise upwards, and drift to the West, slowly at first, then ever more rapidly.

Fig 2.  Spreadsheet of data for a weightless mass "floating" up from a point
on the Earth's equator.

Energy extraction?

From Figure 2, if a weightless mass rises upwards at the equator for say 5 seconds, it could then impact at a velocity of 0.16958 m/s against any structure fastened to Earth, after having risen through a distance of 0.423946 m. If the mass was say 10kg, that impact would deliver 
½ × 10 × 0.16958² = 0.14379 joules of energy. Could we really do that?

There are two problems:— achieving a weightless mass, and returning it back to the Earth's surface. The first could be easily solved e.g. with a constant-force spring from the mass to the earthed structure, but the second problem is much harder, if we hope to gain energy overall. We must distinguish carefully between a mass that is made weightless in a laboratory frame, which could be returned to the Earth's surface with no energy penalty, and one that is weightless in a truly inertial frame, which is required here. The latter would inevitably give an energy penalty equal to the centrifugal force caused by Earth's rotation (i.e. 0.033916 Newtons/kilogram at the equator) multiplied by the distance required to return it.

In the above case we would have a penalty of 
0.033916 N/kg × 10 kg × 0.423946 m = 0.14379 N-m = 0.14379 joules.

So the energy gained equals the energy lost, and this particular idea is ruled out.

I think that anyone who looks seriously into the idea of extracting energy from the rotating Earth will soon become as familiar with that figure of 0.033916 N/kg for equatorial centrifugal force as they probably already are with 9.80665 N/kg for (standard) gravitational force! (Assuming they site their thought experiments and models at the equator, as I always do).

[Postscript] — Calculation

The centrifugal force caused by Earth's rotation on a 1kg mass at the equator can be calculated thus:—

Centrifugal force = mω²r = 1 × 0.00007292115² × 6378137 = 0.033916 N