This rather unusual theoretical problem prompted the author to analyze it on the basis of a gravitational-hydraulic generator [1], but it can be extended to any other similar systems [2]. The author does not claim that the calculations made here are unambiguous, and if a worthy refutation is found, the author is ready to publish it in this work as an alternative.

The general meaning of the problem is that the path of the body can be decomposed into two (or more) sections, on one of which the body picks up a certain speed, and on the other it moves relative to this previously acquired speed. Moreover, one of the forces acting on the body must be free (for example, gravity). Then a theoretical paradox arises, in which the energy to move the body to the starting point may be less than the energy that the body will gain as a result of the described movement. By the way, a similar result is achieved with the Oberth effect [3], and in the two rockets paradox.

A diagram of the gravitational-hydraulic principle for obtaining additional energy, or COP more than one [4], is shown in Figure 1. Above is a tank with liquid (1), usually water, at the bottom of which there is a hole (2), through which this liquid flows out at a speed \(v\). The flow of falling fluid (3) accelerates to the speed \(\vartheta\), after which it reaches the tray (4), where we will calculate the total kinetic energy. This figure does not show the reverse process - the rise of liquid back into the tank, although in a real device it will have to be present. Let's analyze this circuit, make a mathematical model for it, and calculate its COP. We only note that losses, for example, friction, we neglect here.

To do this, we divide the tank into small sections in height, consider each of them separately, and then sum them up again. Let's start the calculation with the well-known rule about the total kinetic energy of a system of points, which is found as the sum of the kinetic energies of each point [5]: \[A = \sum A_i, \quad A_i = {m_i \vartheta_i^2 \over 2} \tag{1}\] As such a point, we consider a sufficiently thin layer of liquid with a height \(\Delta h\), the mass of each layer of which will be found as follows: \[ m_i = m_0 {\Delta h_i \over h_1} = m_0 {\Delta h \over h_1} \tag{2}\] where: \(m_0\) is the mass of a full tank. Since we consider the fluid to be incompressible, all layers will have the same mass. Each layer moves into the tank opening, while the outflow rate from it will be as follows [6]: \[ v_i = \sqrt{2\, g\, h_i}, \quad h_i = i\,\Delta h \tag{3}\] The height of the water level in the tank is denoted here by \(h_i\), moreover, the index \(i\) changes from a larger value to a smaller one, because the tank is first completely filled with water, and then its level decreases: \[ i \in N \ldots 0, \quad N\, \Delta h = h_1 \tag{4}\] Here: \(N\) is the number of points for dividing the tank height into layers.

Further, falling from height \(h_2\), the water jet with mass \(m_i\) picks up additional speed \(\sqrt{2\, g\, h_2}\), which is summarized with the already existing one [6]: \[ \vartheta_i = v_i + \sqrt{2\, g\, h_2} \tag{5}\] For the time being, we neglect air resistance. We now write the kinetic energy of the layer \(\Delta h\) based on (1.2): \[ A_i = {m_0 \over 2} {\Delta h \over h_1} \left( \sqrt{2\, g\, h_i} + \sqrt{2\, g \, h_2} \right)^2 \tag{6}\] or \[ A_i = {m_0 g \over h_1} \left( \sqrt{h_i} + \sqrt{h_2} \right)^2 \Delta h \tag{7}\] To accurately determine the sum of the kinetic energies of the points, it is necessary to reduce the layer thickness to an infinitely small height, and the number of layers to infinity. This means the transition from the sum (1) to the following integral: \[A = {m_0 g \over h_1} \int \limits_0^{h_1} \left( \sqrt{h} + \sqrt{h_2} \right)^2 \partial h \tag{8}\] Taking this integral, we are surprised to find that the kinetic energy of the movement of the liquid down (at the lowest point) differs from that spent on its rise: \[A = m_0 g \left( {h_1 \over 2} + h_2 + \frac43 \sqrt{h_1 h_2} \right) \tag{9}\] And to lift the liquid into the tank, energy is required, which is found according to the classical formula: \[A_C = m_0 g \left( h_1 + h_2 \right) \tag{10}\] In this scheme (Fig. 1), such a rise is not considered, but it is implied. In addition, this does not take into account the height of the pallet, which is unknown to us, but which can also make small adjustments to the calculation.

Dividing the energy received by the energy spent, we immediately find the COP \[C\!O\!P = {A \over A_C} = {1 + \frac12 x + \frac43 \sqrt{x} \over 1 + x}, \quad x = {h_1 \over h_2} \tag{11}\] and represent this dependence on the following graph:

As can be seen from the graph, the maximum COP, subject to the optimal ratio \(h_1/h_2\), is about 1.46. Now let's remember that we didn't take losses into account here. Basically, they will consist in the operation of the compressor to lift water and mechanics, if the received energy will be removed using, for example, buckets. Then the losses can be up to 60%, which means that the final COP will be less than one: \[C\!O\!P_{in} = C\!O\!P_{max} \cdot 0.6 = 0.88 \tag{12}\] It can be assumed that this is the reason for many of the failures of the inventors of free energy devices working in this area. Bringing the circuit to perfection, with obtaining a high efficiency of components and parts of such a device, allows, in theory, to overcome this barrier!