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

My Gyroscope Experiments Part II

Torque output for rate input

This is a very speculative post, concerning the ability of a gyroscope to produce a torque output when given a rate input.

Fig 1.  Steady state conditions for a single axis gyroscope.
Ref: Analysis and Design of the Gyroscope for Inertial Guidance, Ira Cochin.

I've posted the above image before, when discussing the non-gyroscopic case (d). I now draw attention to the first case (a), which notes that a gyroscope produces a torque output when given a rate input (i.e. a change of angle over time). Obviously if the torque output is allowed to turn through some finite angle, then an output of energy must be delivered. The question is: could it be possible to separate this effect out from the "converse" effect noted in the second case, i.e. that a torque input gives a rate output?

Fig 2. A gyroscope with a connection between its gimbals

This question is shown in the 3D drawing above, which has a connection, shown in very schematic form, between the inner and outer gimbals of the gyroscope. Arms are joined to these gimbals, which are connected together via components in a "black box" (or a "black sphere" in this case). So, is it possible that this connection could be made using only passive components, such as links, springs, dampers, energy storing/delivering flywheels etc, in such a way that a net energy output could be delivered with only a rate input — i.e. without much torque input?

I haven't got much further with this idea beyond confirming that there is no net energy production for simple mechanical links between the gimbals.

Review

I'll now review, with some brief final comments, some of the discrepancies already discussed between actual versus predicted gyroscopic performance.

1. Torque applied to the output axis of a gyroscope
     (post of 17 January 2015).

Textbooks (e.g. case (d) in Fig 1 above) and finite-element computer analysis agree that the result should be non-gyroscopic, i.e. that it should make no difference whether the gyro wheel is spinning or not. This does not agree with experiment. (I suspect, but cannot confirm that finite-differences analysis, in which joints and bearings generally have some "flexibility", would give a more realistic result).

2. Prof. Laithwaite's "double-joint" experiment
     (posts of 14 and 21 February 2015).

The result obtained by computer analysis is different from either the result predicted by Prof. Laithwaite, or the result he obtained in his experiment.

3. Force-precessing, then lifting a heavy gyroscope
     (post of 31 January 2015).

It would be easy enough to show, from a strict energy-conservation point of view, that a strong enough experimenter could initially expend sufficient energy to force-precess a gyroscope significantly faster than its natural precessional speed, in which case the subsequent easy lift would be explained. But I'm far from convinced that either Prof. Laithwaite, or the experimenter in the Australian replication video was really doing that. A well-instrumented physical experiment would be needed to make progress in this area.

I have the building and testing of a heavy gyroscope on my "to-do" list, but unfortunately that will be well into the future.

4. Reduced centrifugal force for a precessing gyroscope
     (posts of 7 and 14 March 2015).

I have shown that Prof. Laithwaite's claims of reduced centrifugal force were quite correct, provided that his gyroscopes were nutating as well as precessing, as they surely would have been. However I don't agree that this effect could be developed into a net imbalanced-force propulsion system.

This concludes my series of posts on gyroscopes.

My Gyroscope Experiments Part I

Before leaving the topic of gyroscopes, I'll mention two more ideas of my own. One is very speculative. The other, the subject of this post, boils down to a warning about errors that can occur from step change transitions in finite-element analysis programs.

Negative vertical force for a looping-nutation gyroscope

The above graph, for a gyroscope undergoing looping nutation, includes the vertical reaction force Fz on the joint between the gyro shaft and the central axle (blue). Examination of this force shows that it goes negative, i.e. there is a net upwards force on the joint, over the upper portion of each loop.

An intermittently-rising looping-nutation gyroscope

I decided to check what would happen if the gyroscope as a whole was allowed to rise during these times of upwards force, while remaining held stationary for downwards force.

Graph of gyroscope tower vertical position vs time

I used a model similar to the one analysed in my post of 7 March 2015, with a generalized translational t-function joint between the tower and Base0, again using theoretically calculated curves (above) for the tower locus.

Abrupt, "step-change" transitions in tower position

Firstly, I tried the experiment with the simple "step-change" graph (black) for the tower vertical position vs time. This graph consists only of straight lines, either sloped or horizontal. Each time the vertical joint force Fz goes negative, the tower rises by 100 mm (for a full loop); otherwise it is stationary.

The results of this experiment were a surprise: although the tower rises by a total of 800 mm over one revolution of looping precession, the gyro wheel shows no overall drop in its average vertical position with respect to the tower! There is also no reduction at all in the value of the initially-assigned looping precession (of -2 rad/s). In more detail:—

r12.z (change in height between Wheel and Tower) at the top of each loop [m]:—

0, -4.484E-7, -4.027E-7, +1.710E-5, +2.305E-5, +2.900E-5, +3.494E-5, +4.088E-5, +7.764E-7 (last value recorded; not the full peak).

om.z (Hub angular velocity) at the top of each loop [rad/s]:—

-2, -2.000072, -2.000169, -2.000265, -2.000251, -2.000459, -2.00043, -2.000442, -2.0000019 (last value recorded; not the full peak).

So in this experiment there is no loss of kinetic or potential energy of the gyro wheel with respect to the rising tower during the cycle of looping nutation. Yet the 30kg gyro wheel alone would gain net potential energy of mgh = 30 × 9.80665 × 0.8 = 235.3596 joules, in 4.46 seconds. It could fall back to earth in another 0.4039 seconds, giving a power output of 235.3596 ÷ (4.46 + 0.4039) = 48.389 watts!

See the video of this experiment below.

Smooth transitions in tower position

The above result is obviously too good to be true. Sometimes results like that can be worse than clear-cut null results, because it's not always easy to find out the reason for them. But in this case I had at least a fairly obvious suspect — the "step-change" transitions in the tower position.

From my background in finite-differences dynamic analysis, I would not have expected step-changes like these to give an erroneous result. But to be fair, warnings about this issue are generally given to users of finite-element programs.

I modified the graph for the tower vertical position vs time so that each transition was now smoothly, sinusoidally curved (green) and repeated the experiment, with the tower still rising by 100 mm each time the vertical joint force Fz goes negative. This time, the results were very different, but much more credible.

I haven't analysed this experiment in detail, but the video of it (below) shows significant initial loss of energy associated with the looping behaviour as the tower rises. A gain in this energy does then occur, but that will certainly only be coming from the forced, on-going rise of the tower. I would not expect any net energy gain in this more realistic experiment.

Warnings — specific and general

A specific warning can be given here — when working with computer dynamic analysis programs, particularly finite elements ones, it is very important to avoid functions incorporating "step-change" transitions. (Mathematicians would call these functions "piecewise non-linear"). Although the curved transitions are not as simple to create and analyse as the step-changes, they are essential to achieve a true result. 

This warning can also be generalised:— whether computer analysis is involved, or just old-fashioned pencil-and-paper physical calculations, it is essential not to over-simplify the analysis; otherwise it may be impossible to achieve a true result.