16. Appendix 1 Calculation in general form of net torque of the device analogous to "Indian wheel"

Let us consider again Figure 2, shown in the main text.

Fig. 2

Denotations:

Rmin — radius of the circumference, along which are uniformly distributed centers of gravity of movable loads in the state of their minimum distance from the axis of rotation of the disk;

O — center of the disk (the axis of rotation of the disk);

Rmax — radius of the circumference, along which are uniformly distributed centers of gravity of the movable loads in the state of their maximum distance from the axis of rotation of the disk (Rmax = kRmin, where k – coefficient that indicates the degree of exceeding the length Rmax with respect to Rmin);

β°(Rmin) — angular distance (angle) between the radiuses Rmin, outgoing from the center of disk (point “O”) and ingoing in the centers of gravity of two any adjacent loads;

β°(Rmax) — angular distance (angle) between the radiuses Rmax, outgoing from the center of disk (point “O”) and ingoing in the centers of gravity of two any adjacent loads;

Fg — force of gravity, acting on any movable load;

Mg — torque, generated by a load under the action of only gravitational force;

Mg — total torque, produced by all the movable loads under the influence of gravity;

n — total amount of loads.

Initial data for calculation of the device, which corresponds to Figure 2:

1. Total amount of loads — n = 16.
2. The force of gravity acting on any movable load — Fg.
3. The minimum distance of the centers of gravity of loads from the axis of rotation of the disk — Rmin.
4. The maximum distance of the centers of gravity of loads from the axis of rotation of the disk — Rmax = 2Rmin.
5. β°(Rmax) = β°(Rmin) = 360°/16 = 22.5°.
6. The rotation of the disk counterclockwise accepted as the positive direction of rotation.
7. The angle of incline of the trajectory of the path along which spherical load can slide due to impact of gravity assumed to be 45°. This angle can, for example, be counted (in Fig. 2) as the angle between the axis of the guide tube, starting from the point “a3” and the ray emanating from the center of the disk (from the point “O”) and passing through the point “a3”.

We supplement the initial data by introducing Cartesian coordinates, the origin of which is the point “O” (Fig. 2).

Calculation procedure

Consider the location of the centers of gravity loads shown in Fig. 2. The point a1 is convenient to take as the starting point for calculation the lengths of the lever arms of torques generated by the loads, whose centers of gravities are from the rotational axis at the distance Rmin, and the point A4 - for the loads at maximum distance from the axis of rotation (at the distance Rmax).

The center of gravity of the load a1 is on the axis Y, passing through the axis of rotation of the disk (through the point “O” ), that corresponds equality to zero for the length of the lever arm (da1) of this load (da1 = Rmin·cos90° = 0).

Calculating the length of the lever arm dÀ4, i.e. the coordinate X for the center of gravity of the load A4, performed in the following order:

1. The coordinate Y is calculated for the point a3.
From Fig. 2 it follows that the radius Rmin, coming from the axis of rotation of the disk (point “O”) to the point a3, has been rotated about the axis Y by an angle equal to 2·β°(Rmin) = 2·22.5° = + 45°. This enables us to calculate the coordinate Y of the point a3 in the general form:

Y(a3) = Rmin·sin45° = 0.701Rmin.

2. The coordinate Y is calculated for the point A3.
As mentioned above, the radius Rmin, coming from the axis of rotation of the disk (point “O”) to the point a3, is rotated about the axis Y by the angle equal to + 45°. The angle of incline of the trajectory of the path along which spherical load can slide due to gravity assumed to be +45°. Hence, the line a3A3 is parallel to the axis X. Therefore, the coordinate Y of the point A3 is equal to the coordinate Y of the point a3, i.e.

Y(A3) = 0.701Rmin.

3. The coordinate X is calculated for the point A3.
The point A3 is located on the circumference, the radius of which is Rmax. The equation of the circumference in Cartesian coordinates, whose center coincides with the origin of coordinates is: x2 + `y`2 = `R`2. By solving this equation for the coordinate X(A3), we find:

X(A3) = √( R2maxY2(A3)) = √(4R2min − 0.7072R2min) = Rmin√(4 − 0.5) = 1.871Rmin.

4. Calculating of the angle between direction of the radius Rmax (corresponding to the point A3) and direction of the coordinate X(A3):

φ°(A3) = arcos(X(A3)/Rmax>) = arcos(1.871Rmin/Rmax ) = arccos(1.871/2) = 20.69°.

5. Calculating of the angle between direction of the radius Rmax (corresponding to the point A4) and direction of the coordinate X(A4):

φ°(A4) = ∠β°(Rmax) − ∠φ°(A3) = 22.5° − 20.69° = 1.81°.

6. The length of the lever arm dÀ4, i.e. coordinate X for the center of gravity of the load A4, is calculated as cathetus of a right triangle with the known length of the hypotenuse Rmax = 2Rmin and value of adjacent corner ∠φ° (A4) = 1.81°:

dÀ4 = 2Rmin· cos1.81° = + 1.999Rmin.

Results of the calculation

The calculation of the net torque produced of all 16 loads was made for the six fixed states of rotation of the disk. The initial state corresponds to Fig. 2. The remaining five states correspond to the results of a consistent rotation of the disk counterclockwise at angles that are listed below:

• β°(Rmin)×1/8 = 22.5° ×1/8 = 2.8125°;
• β°(Rmin)×1/4 = 22.5° ×1/4 = 5.625°;
• β°(Rmin)×1/2 = 22.5°×1/2 = 11.25°;
• β°(Rmin)×3/4 = 22.5°×3/4 = 16.875°;
• β°(Rmin)×7/8 = 22.5°×7/8 = 19.6875°.

The results of the calculation of the torques provided by each from the 16 loads at these turnings of the disk, and the net torques, corresponding to these states, are shown in Tables 1 and 2. In Table 3 is summarized the results of calculations to assess the behavior of the net torque in the process of turning the disk in the range 0° – 22.5°. Because of the fact that the inclined paths for displacement of the loads are distributed equally and symmetrically about the axis of rotation of disk, the behaviour of the net torque should be repeated in each sector of the turning disk on angle 22.5°.

Table 1
State of the disk
corresponds to Fig. 2
The disk has been rotated counter-clockwise
on the angle
β°(Rmin)× 1/8 = 22.5°×1/8 = 2.8125°
The disk has been rotated counter-clockwise
on the angle
β°(Rmin)× 1/4 = 22.5°× 1/4 = 5.625°
a1 FgRmincos270° = 0 a1 FgRmincos87.1875° = +0.049FgRmin a1 FgRmincos84.375° = +0.099FgRmin
a2 FgRmincos292.5° = +0.383FgRmin a2 FgRmincos64.6875° = +0.428FgRmin a2 FgRmincos61.875° = +0.472FgRmin
a3 FgRmincos315.5° = +0.707FgRmin À3 Fg2Rmincos17.8775° = +1.903FgRmin À3 Fg2Rmincos15.065° = +1.931FgRmin
À4 Fg2Rmincos1.81° = +1.999FgRmin À4 Fg2Rmincos4.6225° = +1.993FgRmin À4 Fg2Rmincos7.435° = +1.983FgRmin
À5 Fg2Rmincos24.31° = +1.823FgRmin À5 Fg2Rmincos27.1225° = +1.780FgRmin À5 Fg2Rmincos29.935° = +1.733FgRmin
À6 Fg2Rmincos46.81° = +1.369FgRmin À6 Fg2Rmincos49.6225° = +1.296FgRmin À6 Fg2Rmincos52.435° = +1.219FgRmin
À7 Fg2Rmincos69.31° = +0.707FgRmin À7 Fg2Rmincos72.1225° = +0.614FgRmin À7 Fg2Rmincos74.935° = +0.520FgRmin
À8 Fg2Rmincos91.81° = −0.063FgRmin À8 Fg2Rmincos85.3775° = −0.161FgRmin À8 Fg2Rmincos82.565° = −0.259FgRmin
À9 Fg2Rmincos114.31° = −0.823FgRmin À9 Fg2Rmincos62.8775° = −0.912FgRmin À9 Fg2Rmincos60.065° = −0.998FgRmin
À10 Fg2Rmincos136.81° = −1.458FgRmin À10 Fg2Rmincos40.3775° = −1.524FgRmin À10 Fg2Rmincos37.565° = −1.585FgRmin
À11 Fg2Rmincos159.31° = −1.871FgRmin a11 FgRmincos42.1875° = −0.741FgRmin a11 FgRmincos39.375° = −0.773FgRmin
a12 FgRmincos157.5° = −0.924FgRmin a12 FgRmincos19.6875° = −0.942FgRmin a12 FgRmincos16.875° = −0.957FgRmin
a13 FgRmincos180° = −1.000FgRmin a13 FgRmincos2.8125° = −0.999FgRmin a13 FgRmincos5.625° = −0.995FgRmin
a14 FgRmincos202.5° = −0.924FgRmin a14 FgRmincos25.3125° = −0.904FgRmin a14 FgRmincos28.125° = −0.882FgRmin
a15 FgRmincos225° = −0.707FgRmin a15 FgRmincos47.8125° = −0.672FgRmin a15 FgRmincos50.625° = −0.556FgRmin
a16 FgRmincos247.5° = −0.383FgRmin a16 FgRmincos70.3125° = −0.337FgRmin a16 FgRmincos73.125° = −0.290FgRmin
∑Mg = (+6.988 −8.153)FgRmin =
−1.165FgRmin
∑Mg = (+8.063 −7.192)FgRmin =
+0.871FgRmin
∑Mg = (+7.957 −7.295)FgRmin =
+0.665FgRmin

<
Table 2
The disk has been rotated counter-clockwise
on the angle
β°(Rmin)×½ = 22.5°×½ = 11.25°
The disk has been rotated counter-clockwise
on the angle
β°(Rmin)× 3/4 = 22.5°×3/4 = 16.875°
The disk has been rotated counter-clockwise
on the angle
β°(Rmin)× 7/8 = 22.5°× 7/8 = 19.6875°
a1 FgRmincos281.25° = +0.195FgRmin a1 FgRmincos73.125° = +0.290FgRmin a1 FgRmincos70.3125° = +0.337FgRmin
a2 FgRmincos303.75° = +0.556FgRmin a2 FgRmincos50.625° = +0.634FgRmin a2 FgRmincos47.8125° = +0.672FgRmin
À3 Fg2Rmincos350.56° = +1.971FgRmin À3 Fg2Rmincos3.815° = +1.996FgRmin À3 Fg2Rmincos1.0025° = +2.000FgRmin
À4 Fg2Rmincos13.06° = +1.948FgRmin À4 Fg2Rmincos18.685° = +1.895FgRmin À4 Fg2Rmincos21.4975° = +1.861FgRmin
À5 Fg2Rmincos35.56° = +1.627FgRmin À5 Fg2Rmincos41.185° = +1.505FgRmin À5 Fg2Rmincos43.9975° = +1.439FgRmin
À6 Fg2Rmincos58.06° = +1.058FgRminÀ6 Fg2Rmincos63.685° = +0.887FgRmin À6 Fg2Rmincos66.4975° = +0.798FgRmin
À7 Fg2Rmincos80.56° = +0.328FgRmin À7 Fg2Rmincos86.185° = +0.133FgRmin À7 Fg2Rmincos88.9975° = +0.035FgRmin
À8 Fg2Rmincos103.06° = −0.452FgRmin À8 Fg2Rmincos71.315° = −0.641FgRmin À8 Fg2Rmincos68.5025° = −0.733FgRmin
À9 Fg2Rmincos125.56° = −1.163FgRmin À9 Fg2Rmincos48.815° = −1.317FgRmin À9 Fg2Rmincos46.0025° = −1.389FgRmin
À10 Fg2Rmincos148.06° = −1.697FgRmin À10 Fg2Rmincos26.315° = −1.793FgRmin À10 Fg2Rmincos23.5025° = −1.834FgRmin
a11 FgRmincos146.25° = −0.831FgRmin a11 FgRmincos28.125° = −0.882FgRmin a11 FgRmincos25.3125° = −0.904FgRmin
a12 FgRmincos168.75° = −0.981FgRmin a12 FgRmincos5.625° = −0.995FgRmin a12 FgRmincos2.8125° = −0.999FgRmin
a13 FgRmincos191.25° = −0.981FgRmin a13 FgRmincos16.875° = −0.957FgRmin a13 FgRmincos19.6875° = −0.942FgRmin
a14 FgRmincos213.75° = −0.831FgRmin a14 FgRmincos39.375° = −0.773FgRmin a14 FgRmincos42.1875° = −0.741FgRmin
a15 FgRmincos236.25° = −0.556FgRmin a15 FgRmincos61.875° = −0.471FgRmin a15 FgRmincos64.6875° = −0.428FgRmin
a16 FgRmincos258.75° = −0.195FgRmin a16 FgRmincos84.375° = −0.098FgRmin a16 FgRmincos87.1875° = −0.049FgRmin
∑Mg = (+7.683 −7.687)FgRmin =
−0.004FgRmin
∑Mg = (+7.340 −7.927)FgRmin =
−0.587FgRmin
∑Mg = (+7.142 −8.019)FgRmin =
−0.877FgRmin

Table 3
¹
of the source
table
The angle of rotation disk
counter-clockwise
relative to state of Fig. 2
∑Mg
1 State of the disk
corresponds to Fig. 2
∑Mg = (+6.988 −8.153)FgRmin = −1.165FgRmin
1 β°(Rmin)× 1/8 = 22.5°×1/8 = 2.8125° ∑Mg = (+8.063 −7.192)FgRmin = +0.871FgRmin
1 β°(Rmin)× 1/4 = 22.5°× 1/4 = 5.625° ∑Mg = (+7.957 −7.295)FgRmin = +0.665FgRmin
2 β°(Rmin)× 1/2 = 22.5°×1/2 = 11.25° ∑Mg = (+7.683 −7.687)FgRmin = −0.004FgRmin
2 β°(Rmin)× 3/4 = 22.5°×3/4 = 16.875° ∑Mg = (+7.340 −7.927)FgRmin = −0.587FgRmin
2 β°(Rmin)× 7/8 = 22.5°× 7/8 = 19.6875° ∑Mg = (+7.142 −8.019)FgRmin = −0.877FgRmin

Findings

The results of calculations show the following.

The rotation of the disk is possible only when another force acts on it. We call it the "external" force (external with respect to gravity). After the cessation of such exposure the disk must to stop. And it should stop in position corresponding to the turning, regarding the state shown in Fig. 2, by the angle β°(Rmin)×1/2 = 22.5°×1/2 = 11.25°. And this is entirely consistent with the feature which determined by potential nature of the gravitational field, which consists in the fact that work on the movement of any physical object in a closed path is zero. In the system, corresponding to Fig. 2, the trajectory of the center of gravity of each load - closed. This device, like any other, like it, after the cessation of the external force action, inevitably comes to a state of stable equilibrium[32]. At this, in the device according to Fig. 2, there are 16 states in which it is possible to ensure of stable equilibrium.

Such a state can be compared with the state of the ball what is shown in the Figure 15.

Fig. 15

The results of calculation in a general form of the net torque allows us to select the magnitude and direction of the optimal annex of external force action to ensure the long lasting rotation of the disk with partial use of the kinetic energy of the gravitational field.

The Figures 2 and 3 of the main text clearly show and the results of calculation (that are listed in Tables 1-3) numerically confirm in the following:

1. The loads, marked in Fig. 2 by the symbols: A4a13, A5a14, A6a15 and A7a16, constitute couples of forces with lever arms of unequal length relative to the axis of disk (point “O” in Fig. 2). They create the total torque of disk directed counter-clockwise.
2. The load in position a1 does not create a torque drive, since its center of gravity lies on the vertical line passing through the center of the disk “O”.
3. Torques generated by a pair of loads in position a2 and a16 are mutually cancel each other out, as these loads are located on opposite sides and equidistant from the vertical line perpendicular to the axis of rotation of the disk and passing through the center of the disk axis “O”. The same thing happens with the torques which caused by the loads in position a3 and a15.
4. Loads A8, A9, A10, A11, whose centers of gravity are at the greatest possible distance (Rmax) from the axis of rotation of the disk, and the load a12, the center of gravity of which is removed from axis of rotation of the disk at the minimum possible distance (Rmin), create negative (clockwise) torque drive that completely compensates the total positive rotatory torque (directed counter-clockwise), created by the loads mentioned in paragraph 1.

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It is easy to see, that if by using an external force will be possible to counteract the gravitational impact on the loads, moving in the sector A8OA11 (Fig. 2), the effect of gravity on this section of the trajectory movement of loads will be reduced to zero. Equilibrium in such a state of the system will be broken. And during the time of action of this external force the resultant rotating effort will be determined by the total torque created by the loads specified in paragraph 1. The disk will be able rotate. At the same time a prerequisite for ensuring the possibility of rotation is that the magnitude of the resulting torque must ensure overcoming the total resistance to the rotation produced by useful load on the motor shaft, friction, heating, and other possible impacts impeding to rotation.

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Of course, the bulky calculation, resulted above, does not can add anything new to the fundamental concept of the physics - to the Law of Conservation of Energy, as well as to the feature of a potential character of gravity field of the Earth, concerning the equality to zero of the work of the force acting on a body, moving along the closed trajectory. However, this calculation allows understand, where the most appropriate to apply external force effort, in order to provide in such a system the rotation with partial use of the kinetic energy of the gravitational field of the Earth. The results of calculation "in general form" also allow to obtain numerical estimates of the contribution of each movable load in creating of a common torque by substituting into received symbolic expressions the real numeric values of weight of the movable loads and the values of the corresponding lever arms of rotation.

Successful utilizations of the kinetic energy of the Earth's gravity jointly with the influence of other forces of nature are used by people for a long time. Suffice it to recall, that hydroelectric power plants work by using the kinetic energy of water falling from great heights. But the rise of the water to great heights occurs due to the influence of other forces of nature. Or skier, sliding down to his great satisfaction from the mountains due to the influence of gravity, is forced to use the energy of his muscular effort to rise on mountain.