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.