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

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.