Renewable energy system (RES)

Abstract

This invention consists of two separate systems (System 1, System 2). The purpose of this invention is to generate electrical energy using the force of gravity. This objective is achieved through the operation of System 1, which involves the downward flow of a quantity of water from a high level to a low level (as further explained in the operation of System 1). To maximize the benefits of this invention, System 2 has been designed to raise the water from the low level back to the high level, and this process occurs in multiple stages. As a result, this invention offers significant benefits since both System 1 and System 2 operate mechanically without the use of any type of fuel and with minimal human intervention, except for monitoring and supervising the operation of both systems (System 1, System 2).

Description

[0001] Renewable Energy System (RES)

[0002] Technical Field

[0003] [1] Engineering

[0004] Background Art

[0005] [2] None

[0006] Summary of Invention

[0007] This invention consists of two separate systems (System 1, System 2). The purpose of this invention is to generate electrical energy using the force of gravity. This objective is achieved through the operation of System 1, which involves the downward flow of a quantity of water from a high level to a low level (as further explained in the operation of System 1). To maximize the benefits of this invention, System 2 has been designed to raise the water from the low level back to the high level, and this process occurs in multiple stages. As a result, this invention offers significant benefits since both System 1 and System 2 operate mechanically without the use of any type of fuel and with minimal human intervention, except for monitoring and supervising the operation of both systems (System 1, System 2).

[0008] Operation summary of system (No 1)

[0009] - The system generates electrical energy using gravitational force.

[0010] - Two different masses, W1 (No1) and W2 (No2), are connected together using flexible horizontal and vertical metal cables (No4). These cables pass over smooth iron rollers (No8).

[0011] - A solid metal body, (No5), is attached to the horizontal cables, and it moves horizontally along with the movement of the horizontal cables (No4) in a path with a length of ml2 during the system's operation.

[0012] - Two parallel metal rods, (No6), are attached to the solid metal body (No5) by hinges on one end, and these two metal rods (No6) are connected to a large metal disc (No7) on the other end. This setup converts the straight horizontal motion of the metal body (No5) into circular motion for the large metal disc (No7), as shown in the drawings. - The large metal disc (No7) is connected to smaller metal gears to engage with the dynamo's drive shaft responsible for generating electricity.

[0013] - After transmitting the circular motion of the large metal disc (No7) to the dynamo's drive shaft through the required torque and speed, electrical energy is generated, which is the primary goal of system number 1.

[0014] - During the operation of this system, a quantity of water is transferred from a higher level to a lower level, as explained in detail in the system's operation description.

[0015] This summary provides an overview of how the system works and its main purpose, which is to generate electrical energy using gravitational force and the movement of water between different levels.

[0016] Operation Summaryof system

[0017] This system operates mechanically to lift water from a low level (No 11) to a high level (No 12) without using any fuel. It employs an iron lever (No 1) that pivots on a horizontal axis, consisting of a specified diameter steel rod. This lever operates vertically to raise the water and is connected on one end to a metal container (No 2) responsible for lifting water from the low water tank (No 11). On the other end, this lever is connected to flexible iron cables (No 3). These cables are attached to an iron cylinder (No 4) fixed with gears at both ends.

[0018] Motion and torque are transferred between the gears (No 4) to the assembly of ring frames and gears (No 5) and vice versa, according to the motion path.

[0019] Assembly (No 5) is connected to weights Wl(No 7) and W2(No 8)through iron cables and rollers.

[0020] These weights and iron cables work in a vertical plane perpendicular to the vertical level of the lever (No 1).

[0021] At the end of the path (No 1) for the operation of the system, transferring water from the primary water tank (No ll)to the final water tank (No 12) with the highest level will be done . Lever (No 1) then returns in path 2 to its initial position, and the mechanical process repeats.

[0022] Technical Problem

[0023] [3] electrical power plants rely on non-renewable resources and extremely expensive motors.

[0024] Solution to Problem

[0025] [4] produce a reasonable amount of clean energy that is maybe similar to the electrical energy produced from traditional electrical power plants.

[0026] Advantageous Effects of Invention

[0027] [5] The components of this invention can all be locally procured in any country and are all readily available. Moreover, they are cheap as well as affordable in terms of cost.

[0028] [6] Maintenance costs for this invention are basically negligible in comparison to the maintenance costs of traditional power plants.

[0029] Brief Description of Drawings.

[0030] * Total invention drawings = 36

[0031] * Drawings of system (No-1) = 14

[0032] * Drawings of system (No-2) = 22

[0033] Brief description of the drawing

[0034] Important note

[0035] Omitting some elements of the system in one of the drawings does not mean that they do not exist, but since these elements do not have a function at this stage of the system's operation, they have been hidden to simplify the drawings and make them easier to understand.

[0036] System (No-1)

[0037] 1 . Drawing (No-1 ) (first step of the first path)

[0038] 2. Drawing (No-2) (second step of the first path)

[0039] 3. Drawing (No-3) (third step of the first path)

[0040] 4. Drawing (No-4) (fourth step of the first path)

[0041] 5. Drawing (No-5) (fifth step of the first path)

[0042] 6. Drawing (No-6) (final step of the first path)

[0043] 7. Drawing (No-7) (first and second steps of the second path)

[0044] 8. Drawing (No-8) (third step of the second path)

[0045] 9. Drawing (No-9) (fourth step of the second path)

[0046] 10. Drawing (No-10) (fifth step of the second path)

[0047] 11. Drawing (No-11 ) (final step of the second path)

[0048] 12. Drawing (No-12) (details of elements No 2, 9, 19)

[0049] 13. Drawing (No-13) (details of elements No 3, 12, 13, 14, 15, 16, 17, 18, 20)

[0050] 14. Drawing (No-14) (details of elements No 4, 5, 6, 7, 21 , 22, 23) System (No-2)

[0051] 1 . Drawing No (1 -A) (plan of the system)

[0052] 2. Drawing No (1 -B) (sec-(2-2), (3-3))

[0053] 3. Drawing No (1 -C) (details of elements 1 ,2)

[0054] 4. Drawing No (1 -D) (details of elements 14, 16, 17, 18, 19, 20, 21 , 22, 24)

[0055] 5. Drawing No (1 -E) (details of elements 8, 15, 23)

[0056] 6. Drawing No-2 (starting point for the first path and final step for the second path)

[0057] 7. Drawing No-3 (first step of the first path)

[0058] 8. Drawing No-4 (second step of the first path)

[0059] 9. Drawing No-5 (third step of the first path)

[0060] 10. Drawing No-6 (fourth step of the first path)

[0061] 11 . Drawing No-7 (fifth step of the first path)

[0062] 12. Drawing No-8 (sixth step of the first path)

[0063] 13. Drawing No-9 (seventh step of the first path)

[0064] 14. Drawing No-10 (eighth step of the first path)

[0065] 15. Drawing No-11 (final step of the first path and starting point for the second path)

[0066] 16. Drawing No-12 (first step of the second path)

[0067] 17. Drawing No-13 (second step of the second path)

[0068] 18. Drawing No-14 (third step of the second path)

[0069] 19. Drawing No-15 (fourth step of the second path)

[0070] 20. Drawing No-16 (fifth step of the second path)

[0071] 21. Drawing No-17 (sixth step of the second path)

[0072] 22. Drawing No-18 (seventh step of the second path)

[0073] Detailed description of the invention

[0074] This invention consists of two systems as we mention previously

[0075] System (No 1) :- which is responsible for electricity generation using gravitational force of earth . System (No 2) :-which is responsible for re-raise the water from the low level back to the high level .

[0076] So we will discuss each system of them by detailed description , operation method and calculations.

[0077] Firstly : system (No 1)

[0078] ❖ If we have 2 different weights W1 & W2 connected together by a rope passing by 2 rollers on the same vertical level as shown in Figure (1), in this case, one of the two weights will be heavier than the other one. The entire system will move towards the heavier weight and the horizontal rope between the 2 rollers will move horizontally towards the heavier weight.

[0079] ❖ If the heavier weight loses a part of its weight such that it becomes less than the other weight, then the system will still move with a negative acceleration till the velocity of the system reaches zero. Then, the system will change its direction towards the other weight.

[0080] ❖ If we could make the system stay in continuous motion in two opposite directions with the same distance so the horizontal rope between the 2 rollers will stay also moves towards the left and the right in continuous motion , then this horizontal motion may be transferred to circular motion in one direction by a diameter that is equal to each single path in one direction.

[0081] ❖ On a larger scale, by changing the fine rope to an iron cable (flexible cable) that has an iron body fixed to it and this iron body, in turn, is connected through a hinge with another metal rod which is also connected through a hinge to the iron body from one side and connected with another hinge with a metal circular disc on the other side; such that this iron body moves horizontally in a straight line (with the cable) and during this motion, the metal rod can rotate the circular metal disc of a diameter equaling the length of path of the iron body.

[0082] ❖ Through this method, we can transfer the linear horizontal motion into a circular motion (as shown in figure 2). We can use this circular motion, which rotates the metal disc with the force, and velocity that would be sufficient to produce electrical energy.

[0083] System (No 1) consists of two weights w1 (No 1) & w2 (No 2) (as shown in the drawing)

[0084] > The right weight W1 (No 1):

[0085] It is a solid iron mass with a certain shape and weight that moves vertically (up and down) with the total system motion.

[0086] > The left weight I / V2 (No 2):

[0087] To determine the weight and the shape of W2 (No 2), we must take 2 terms into account and they are as follows:

[0088] I. The volume of the mass W2 (No 2) must be exactly equal to the weight of the mass. Moreover, the shape of the moving mass W2 must be suitable for vertical movement inside water to assist it to move with minimal water resistance.

[0089] II. The mass movement of W2 (No 2) must be vertically straight without any rotation around its axis. Therefore, we need to make the mass W2 (No 2) move inside a vertical iron cylinder (No 3) with a diameter bigger than the diameter of the W2 and the external cylinder (No 3) is equipped with II shapes protrusions inside it to ensure that the W2 (No 2) doesn’t rotate while moving inside the cylinder.

[0090] ❖ The external iron cylinder (No 3), which surrounds the W2 (No 2), is connected from the lower part of its body by a water tank (No 11) with a constant water level (the water level must be kept constant during the operation of the system). The external iron cylinder (No 3) is connected with the lower water tank (No 11) through an iron gate (No 12), which can open and close to control the flow of water.

[0091] ❖ this cylindrical iron gate( drainage gate ) (No 12) which is installed inside the external iron cylinder (No 3) moves up and down within a range of ( 0.6 m ). As when this gate moves up allowing the excess water inside the external cylinder (No 3) to drain into the lower water tank (No 11) until the water level inside the external iron cylinder (No 3) will be the same level as of that of the lower tank (No 11).

[0092] ❖ This gate (drainage gate No 12) is connected with a metal circular disc (No 14) installed on the wall of the external iron cylinder (No 3), located before the end of the path by 60 cm. This metal disc (No 14) can move up and down within a 60 cm range inside the external iron cylinder (No 3) and the connection between the drainage gate (No 12) and the upper metal disc (No 14) is linked through 8 cables of 6mm each in diameter. ❖ When the circular metal disc (No 14) moves up with a distance of 0.6 m, the drainage gate (No 12) travels the same distance which in turn makes the water level inside the external iron cylinder similar to the water level in the lower tank (which called drainage stage) (as shown in the drawing).

[0093] ❖ Also, the external iron cylinder (No 3) is connected with the other water tank (No 10), which has a higher water level than the lower tank (No 11) as per the distance determined in the drawings. Also, the connection between them is through a filling gate (No 15) but this gate is installed around the external iron cylinder (No 3) from the outside. This gate is connected by the lower metal disc (No 18) (which contains the gate keys) inside the external iron cylinder through 8 cables (No 16) of 6mm each in diameter. So, during the descent of the body w2 (No 2), the four iron protrusions (No 19) at the base of the lower cone of w2 (No 2) press on the lower metal disc (No 18) which is located very precisely at this level, this causes the iron cables (No 16) connected to the filling gate (No 15) to move downward, opening the filling gate (No 15) to allow water to flow from the upper water tank (No 10) into the external cylinder (No 3) until the water levels on both sides are equal (which is called filling stage).

[0094] ❖ The 2 weights W1 & W2 are connected together with 4 flexible iron cables (No 4) with a diameter of 16mm each, passing through 8 iron rollers (No 8) (ensuring that the rollers and the beams are as smooth as possible). Each side has 4 iron rollers and the distance between the two groups of rollers on both sides must be longer than the total path by about 1.5 m

[0095] ❖ The 4 iron cables (No 4) between the 2 groups of rollers are connected together by a solid metal body (No 5) of specific dimensions, which is fixed in the 4 cables and with a clear distance between each 2 cables. Consequently, this solid metal body (No 5) will move horizontally with cables over a steel Beams (No 21 & 22) with the help of small circular wheels (ball bearings)(No 23) installed at the bottom of the solid metal body (No 5)

[0096] ❖ Two parallel bars (No 6) are fixed through a hinge from one side in the solid metal body (No 5) and attached to a large metal disc (No 7) (which has a diameter equal to the total path length) with a hinge from the other side (as shown in the detailed drawings).

[0097] ❖ These 2 parallel bars (No 6) are used to convert the straight motion of the solid metal body (No 5) to circular motion of the large metal disc (No 7) and this motion in turn will be used to rotate the gears of the dynamo with a specific speed and sufficient power to generate electric power (as shown in the detailed drawings).

[0098] ❖ We chose the weight W2 (No 2) to be a cylindrical empty iron body of total length 5.80 m and the outside diameter of 0.84m .this cylindrical empty body is connected from its top and bottom by two empty cones with a height of 2.10 m and have the same diameter of the cylindrical empty body. So that the total height of the body w2 (No 2) equal (10.0 m), and the cylindrical part of the body w2 (No 2) has 8 protrusions of 2cm, which move inside the U-shaped grooves (No 20) within the external iron cylinder (No 3) that surrounds the moving cylinder (No 2) (as shown in the detailed drawings).

[0099] ❖ There are 4 metal arms (No 9) fixed at the top of the upper cone of the body w2 (No 2) and these arms work to apply upward pressure on the upper metal disc (No 14) consequently this disc (No 14) moves upward and raises the drainage gate (No 12) to drain excess water from the external cylinder (No 3) to the lower water tank (No 11) until the water level is equal at both ends.

[0100] ❖ Note that the volume of the body w2 (No 2) equals (1.270087r)m3. So the weight of the body w2 (No 2) including protrusions and metal arms must equal (1.27008TT) ton and must be distributed uniformly.

[0101] ❖ We chose a path length of 12m distributed as follows: a) D4= 0.60m (required distance to open and close the gates) b) D2= 7.072m c) D3= 1.49m (The required distance which the body w2 (No 2) moves so that the lower cone would be completely immersed inside the water) d) D4= 0.748m (the required distance which the body w2 (No 2) moves so that the cylindrical part of the body with a length of 5.80m would be completely immersed inside the water) e) Ds= 1.49m (the required distance which the body w2 (No 2) moves so that the upper cone can be completely immersed inside the water) f) D6= 0.60m (the required distance to open and close the gates and also used for filling and draining water) ❖ We chose the external iron cylinder (No 3) that will surround the moving iron body (No 2) to have an internal diameter of 0.90m so that the distance between the moving body (No 2) and the wall of the external iron cylinder (No 3) will be 3cm.

[0102] ❖ This external iron cylinder (No 3) is connected with the lower tank (No 11) by the drainage gate (No 12) and connected with the upper tank (No 10) by the filling gate (No 15).

[0103] ❖ Note that: we chose iron rollers, wires and Beams as well as paints of the highest quality. These choices are to take all the precautions so that the friction dynamic force during the motion would not exceed more than 100kg on average. Also, the choice of shape for the moving cylinder (as shown in the detailed drawing) is so that the water resistance during the motion would not exceed 115kg.

[0104] ❖ Operation of System No. 1:**

[0105] ❖ - System ( No. 1) begins its operation from the starting point, where Body W1 (No 1) is at its lowest point in its path, and Body W2 (No 2) is at its highest point in its path, compressing the upper metal disc (No 14) upwards using the four metal arms (No 9) fixed at the top of the upper cone of Body W2 (No 2). As a result, the metal cables (No 13) connected to the upper metal disc (No 14) are lifted, causing the drainage gate (No 12), also connected to these cables from below, to fully open. The water level inside the external cylinder (No 3) becomes equal to that of the lower water tank (No 11) after the excess water drainage process from the external cylinder to the lower water tank is completed.

[0106] ❖ - At the starting point, the solid metal body (No 5) is at the beginning of its path, which starts on the right side, next to Body W1 (No 1). The two parallel metal rods (No 6), connected by hinges to both the metal body (No 5) and the large metal disc (No 7), are in a completely horizontal position.

[0107] ❖ - The forces acting on Body W 1 (No 1) are:

[0108] ❖ 1. The weight of Body W1 (No 1), acting downward.

[0109] ❖ 2. The tensile force in metal cables (No 4), acting upward.

[0110] ❖ - The forces acting on Body W2 (No 2) are:

[0111] ❖ 1. The weight of Body W2 (No 2), acting downward.

[0112] ❖ 2. The tensile force in metal cables (No 4), acting upward.

[0113] - The forces affecting the entire System No. 1 are: ❖ 1. The weight of Body W 1 acting downward.

[0114] ❖ 2. The weight of Body W2 acting downward.

[0115] ❖ 3. The frictional resistance force of the assembly, always opposing the direction of movement, initially static friction resistance before motion starts, then transitioning to dynamic friction resistance during movement, with measures taken to minimize it as much as possible, not exceeding an average of 100 kg during system operation.

[0116] ❖ 4. The force (F) required to rotate the large metal disc (No 7), always opposing the direction of movement.

[0117] ❖ 5. The weight of the drainage gate (No 12), which affects the same direction as the weight of W2 (No 2) and is equal to 50 kg.

[0118] ❖ - By studying the forces affecting the movement of the system (W1 & W2) using Newton's laws of motion, we can calculate the acceleration of the system and its velocity at the end of each step, as well as the time duration of each step in the path of System No. 1.

[0119] ❖ - In the initial phase, Body W1 (No 1) moves upward, Body W2 (No 2) moves downward, and the upper metal disc (No 14) descends under the influence of the weight of the drainage gate (No 12), covering a distance of 0.60 m. In this case, the drainage gate (No 12) is fully closed to prevent water from passing from the external cylinder (No 3) to the lower water tank (No 11) and vice versa.

[0120] ❖ - At this point, the water level inside the external cylinder (No 3) is equal to the water level in the lower water tank (No 11).

[0121] ❖ - The solid metal body (No 5) moves horizontally to the left along with the horizontal cables (No 4), pulling the two parallel metal rods (No 6) to move in the same direction. This results in the rotation of the large metal disc (No 7), which is connected to the rods by hinges, as mentioned.

[0122] ❖ - It is important to note that the rotation direction of the large metal disc (No 7) must be carefully controlled and should not allow it to rotate in the opposite direction. The rotation direction is determined according to the rotation direction of the dynamo shaft connected to it through several gears, as mentioned.

[0123] ❖ - The movement of System No. 1 continues with Body W1 (No 1) ascending and Body W2 (No 2) descending, following the steps of the first path, until Body W2 (No 2) begins to descend into the water. As a result of Body W2 (No 2) descending into the water, the water level inside the external cylinder (No 3) rises, depending on the size of the submerged body. When W2 (No 2) descends into the water by a distance of 1.49 m, the water level inside the external cylinder (No 3) rises to completely cover the lower cone of Body W2 (No 2) to a height of 2.10 m.

[0124] ❖ - Then, in the next step, Body W2 (No 2) continues its descent by a distance of 0.748 m, causing the water level to rise further. During this step, the middle cylindrical part of Body W2 (No 2) with a height of 5.80 m is completely submerged in the water.

[0125] ❖ - The system proceeds with Body W2 (No 2) descending further by a distance of 1.49 m, and the water level continues to rise inside the external cylinder (No 3) to completely submerge the upper cone of body W2 (No 2) by an elevation of 2.10 m at the end of this step .As When W2 (No 2) submerged by a total height of 10.0 m, the water level inside the external cylinder (No 3) has risen by 6.272 m above the level of the lower water tank (No 11).

[0126] ❖ - During the movement of Body W2 (No 2) in the water, it is subjected to additional forces, including buoyancy (Fb) and water resistance force (W.R).

[0127] ❖ - By studying the forces affecting the entire System (W 1 & W2) during the movement of Body W2 (No 2) in the water, we observe that the forces can be summarized as follows:

[0128] ❖ a- The weight of Body W 1 acting downward.

[0129] ❖ b- The weight of Body W2 acting downward.

[0130] ❖ c- The buoyancy force (Fb) affecting Body W2 (No 2), always acting upward.

[0131] ❖ d- The water resistance force acting against the movement of Body W2 (No 2), always opposing its direction.

[0132] ❖ e- The dynamic friction resistance force during the movement of the system, always opposing the direction of movement.

[0133] ❖ f- The force (F) required to rotate the large metal disc (No 7), always opposing the direction of movement.

[0134] ❖ g- Additionally, the weight of the drainage gate (No 12) during the drainage step or the weight of the filling gate (No 15) during the filling step should be considered.

[0135] ❖ - After completing the previous step, in which Body W2 (No 2) is fully submerged in the water with a height of 10.0 m, the final step in the first path begins, which is the filling step, with a length of 0.60 m.

[0136] ❖ At the beginning of the last step:

[0137] ❖ - Body W2 (No 2) begins by pressing on the lower metal disc (No 18) using the four metal protrusions (No 19) created at the base of the lower cone of Body W2 (No 2). Consequently, the lower metal disc (No 18) moves downward, pulling the metal cables (No 16) with it. As a result, the filling gate (No 15) rises to allow water to pass from the upper water tank (No 10) into the external cylinder (No 3) until the water levels on both sides are equal, according to the levels specified in the drawings.

[0138] ❖ - At the end of this step, after a distance of 0.60 m, the entire system's speed reaches zero. Body W1 (No 1) reaches its highest point in its path, Body W2 (No 2) reaches its lowest point in its path, and the filling gate (No 15) is fully open. The solid metal body (No 5) also moves to the left to its final point in its horizontal path, which is located at a distance of 12.0 m, pulling with it the two parallel metal rods (No 6) responsible for rotating the large metal disc (No 7).

[0139] ❖ - At this moment, the two metal rods (No 6) are in a completely horizontal position, and the large metal disc (No 7) has rotated half a circle around its axis. After this moment, the system begins its second path.

[0140] ❖ Start of the second path of the system (1):

[0141] ❖ - After the system's velocity reaches zero at the end of the first path, and since Body W2 (No 2) is fully submerged in water, it is subjected to buoyancy force (Fb) based on Archimedes' principle, which is equal to the weight of Body W2 (No 2). The two forces, (W2) and (Fb), act in opposite directions.

[0142] ❖ - When studying the motion of the assembly (W1 & W2), we find that the resultant forces on the assembly make the system move in the opposite direction to its movement in the first path. Body W1 (No 1) moves downward, Body W2 (No 2) moves upward, and, as a result, the solid metal body (No 5) starts moving horizontally to the right along with the horizontal cables, causing the large metal disc (No 7) to continue rotating in the same direction as in the first path.

[0143] ❖ - As Body W2 (No 2) moves upward, the lower metal disc (No 18) also moves upward under the influence of the weight of the filling gate (No 15). Consequently, the filling gate (No 15) gradually closes to prevent water from passing from the upper water tank (No 10) to the external cylinder (No 3) and vice versa.

[0144] ❖ - The movement of the system continues with Body W1 (No 1) descending and Body W2 (No 2) ascending inside the water in the external cylinder (No 3). The solid metal body (No 5) also continues its horizontal movement to the right, compressing the two metal rods (No 6), which are connected to it by hinges, responsible for rotating the large metal disc (No 7). ❖ - The system's movement remains continuous with increasing velocity for a distance of 7.072 m. At this point, the upper cone of Body W2 (No 2) reaches the water level in the external cylinder (No 3), and it begins to emerge from the water.

[0145] ❖ - As a result of Body W2 (No 2) emerging from the water, the water level inside the external cylinder (No 3) decreases according to the size of the body that is emerging. When Body W2 (No 2) moves upward by a distance of 1.49 m, the water level inside the external cylinder (No 3) drops so that the upper cone of Body W2 (No 2), with a height of 2.10 m, has completely emerged from the water.

[0146] ❖ - When Body W2 (No 2) continues to move upward by another distance of 0.748 m, the water level inside the external cylinder (No 3) drops further, and the middle cylindrical part of Body W2 (No 2), with a height of 5.80 m, is completely out of the water.

[0147] ❖ - As Body W2 (No 2) continues its upward movement by another distance of 1.49 m, the water level inside the external cylinder (No 3) drops further, and the lower cone of Body W2 (No 2), with a height of 2.10 m, has completely emerged from the water. By the end of this step, the entire Body W2 (No 2) with a height of 10.0 m is completely out of the water.

[0148] ❖ - With the completion of this step and the continuous descent of Body W1 (No 1) and ascent of Body W2 (No 2), along with the horizontal movement of the solid metal body (No 5) to the right, compressing the two metal rods (No 6) responsible for rotating the large metal disc (No 7), the final step in the second path begins, which is the drainage stage.

[0149] ❖ - At the beginning of this final step, Body W2 (No 2) rises, compressing the upper metal disc (No 14) using the four metal arms (No 9) fixed at the top of the upper cone of Body W2 (No 2). Consequently, the metal cables (No 13) rise along with it, pulling up the drainage gate (No 12) until this gate is fully opened at the end of this step, which has a length of 0.60 m.

[0150] ❖ - As a result of opening the drainage gate (No 12), excess water is drained from the external cylinder (No 3) into the lower water tank (No 11) until the water levels on both sides are equal. It should be noted that the amount of water drained equals the amount of water filled during the filling stage.

[0151] ❖ - At the end of this step, Body W1 (No 1) reaches its lowest point in its path, and Body W2 (No 2) reaches its highest point in its path. The solid metal body (No 5) also reaches the end of its second path to the right horizontally. ❖ - At the end of this step, the two metal rods (No 6) are in a completely horizontal position, and the large metal disc (No 7) has completed its full rotation. Consequently, the entire system returns to the starting point, where the first path begins again, and the system's operation repeats.

[0152] ❖ According to the forces affecting the system, all the forces are known except the weight W1 and the force F that are required to rotate the large metal disc. If we consider ( W1&F ) are unknown forces and we want to determine theses forces

[0153] So, we study the path of the motion in both directions based on the initial velocity in the beginning of the path, which is zero, and the final velocity at the end of the path (12m), which also equals zero.

[0154] ❖ By solving the equations of motion in the 2 paths using Newton’s laws of motion and Archimedes principle, we get the following:

[0155] We found that the required weight W1 = 2.0289237 Ton

[0156] And the Force F= 1.0075181 Ton

[0157] Archimedes Principle

[0158] Fb= mg = (pV)g

[0159] Whereas:

[0160] ■ p = liquid density

[0161] ■ V= volume of immersed body

[0162] ■ g =9.81m / sec2

[0163] Fb= buoyant force

[0164]

[0165] ❖ The body W2( No. 2 ) moves down by a distance 0.60m from the initial point of the path. During this distance the drainage gate( No. 12 ) is closing.

[0166] ❖ We study the forces affecting the system during this step (as average forces during each step) to calculate the acceleration of the system and the final velocity of the system at the end of each step and also the time required for each step.

[0167] ❖ So, we found the following a = acceleration m / sec2v = Velocity m / sec t = time (sec)

[0168] ❖ The bodyW2 ( No. 2 ) continues its motion down by a distance of 7.072m till it reaches the water level inside the external iron cylinder( No. 3 ), which is the same water level of the lower tank ( No. 11 ).

[0169] ❖ We study the forces affecting the system to calculate the following (a2&v2&t2)

[0170] ❖ The body W2 ( No. 2 ) moves inside the water covering a distance of 1.49m so that the water level inside the external iron cylinder ( No. 3 ) will rise till it reaches 0.61m over the lower tank level (No. 11 ) (note that the gate which connects the external iron cylinder and the lower tank are both closed) ❖ This means that the bottom cone of the moving body W2 ( No. 2 ) is completely immersed in water at the end of this step.

[0171] ❖ Also we study the forces affecting the system during this step to calculate (a3&v3&t3)s* step

[0172] ❖ The body W2 (No. 2 ) moves inside the water covering a distance of 0.748m, so the water level inside the external iron cylinder (No. 3 ) will rise such that the cylindrical part of the body W2 ( No. 2 ) with a length of 5.8m will be completely immersed in water at the end of this step.

[0173] ❖ By the same method, we calculate (a4&v4&t4) step

[0174] ❖ The body W2( No. 2 ) moves inside the water covering a distance of 1.49m. So, the water level inside the external iron cylinder( No. 3 ) will rise such that the total body W2 will be completely immersed at the end of this step. (Note that: when the total moving body W2 will be completely immersed, it will be exposed to a buoyant force which will equal pVg according to Archimedes principle. This will finally equal (1.27008TT) Ton, which is the same number of the weight W2.

[0175] ❖ This means that there are 2 equal and opposite forces affecting the moving body resulting in zero force. Using the previous method, we can calculate the following (a5&v5&t5)

[0176] ❖ By the beginning of this step and during the descent of the body w2 (No 2), the four iron protrusions (No 19) at the base of the lower cone of w2 (No 2) press on the lower metal disc (No 18) which is located very precisely at this level, this causes the iron cables (No 16) connected to the filling gate (No 15) to move downward, opening the filling gate (No 15) to allow water to flow from the upper water tank (No 10) into the external cylinder (No 3) until the water levels on both sides are equal (which is called filling stage).

[0177] ❖ By the end of this step, the body’s velocity will be close to zero.

[0178] ❖ And using the same method, we can calculate (a6&v6&t6)

[0179] pathfall the steps are illustrated in detailed drawings)

[0180] ❖ The body W2( No. 2 ) begins to rise with a zero velocity and with the beginning of motion, the metal disc ( No. 18 ) goes up and the filling gate ( No. 15 ) moves down gradually until the filling gate is completely closed by the end of this step.

[0181] ❖ We study the forces affecting the system taking the gate force into consideration so

[0182] ❖ The body W2 ( No. 2 ) continues its motion up inside the water covering a distance of 7.072m until it reaches the water level inside the external iron cylinder ( No. 3 ) , which is the same level of the upper tank ( No. 10 ).

[0183] We study the forces affecting the system to determine (a2&v2&t2)

[0184] ❖ The body W2 ( No. 2 ) begins to come out of the water so that by the end of this step, the body W2 (No. 2 ) would have moved up 1 ,49m and the water level inside the external iron cylinder ( No. 3 ) is going down to the extent that the upper cone of the body W2 (No. 2 ) with a height of (2.10m) will be completely outside the water.

[0185] ❖ We study the forces affecting the system to determine the following: (a3&v3&t3) Step f D.t '™ 0.748m I

[0186] ❖ The body W2 ( No. 2 ) moves up covering a distance of 0.748m and the water level inside the external iron cylinder ( / No. 3 ) will reduce to the extent that the cylindrical part of the body of 5.80m in height will be completely outside the water.

[0187] ❖ Using the same method as earlier, we determine the following: 5th(StepD5=1.49 )

[0188] ❖ The body W2 (No. 2 ) moves up covering a distance of 1.49m and the water level inside the external iron cylinder ( No. 3 ) will go down so that the total moving body will be completely outside the water.

[0189] ❖ Using the same method, we determine the following: (a5&v5&t5)

[0190] ❖ By the beginning of this step, the 4 metal arms( No.9 ) which are fixed on the top of the upper cone of the moving body W2 ( No. 2 )rises, raising the metal disc( No. 14 ) which existing inside the external iron cylinder ( No.3 ) at a certain level.

[0191] ❖ This metal disc( No. 14 ) is connected with the drainage gate ( No. 12 ) by 8 cables ( No. 13 ) so that when the metal disc ( No. 14 ) is going up, the drainage gate ( No. 12 ) goes up covering the same distance of 0.6m to allow the water to pass from the external iron cylinder ( No. 3 ) to the lower tank ( No. 11 ) through the drainage gate ( No. 12 ) so that the water level in the external iron cylinder ( No. 3 ) and the lower tank ( No. 11 ) are equal.

[0192] ❖ We study the forces affecting the system during this step to calculate (a6&v6&t6) and note that (v6) will bezero by the end of the path. Important notes

[0193] 1) The 4 main variable elements in our system are:-

[0194] > (W1, W2, path length, F (force required for rotating the metal disc) ) as we mentioned before.

[0195] > We can assume any 2 variables of them and estimate the other 2 variables.

[0196] > So we can control all the variables according to our availability. For instance, if we want a specific force for rotating the metal disc and a specific path length, we can assume these 2 variables and estimate W1 & W2.

[0197] > This means we can control the distance between the upper and lower tanks and also control the volume of water, which will pass from the upper tank to the lower tank, by controlling the path length and the diameter of the mass W2.

[0198] > In general, we can control any variable according to our availability and to our need.

[0199] 2) The direction of rotation of the metal disc may be clockwise or counterclockwise according to our need.

[0200] Calculations of system ( No. 1 )

[0201] As we mentioned before that there are 4 main variable elements in our system, which are:

[0202] 1. The weight W1

[0203] 2. The weight W2

[0204] 3. The total path length

[0205] 4. The force F (the required force for rotating the large disc which is responsible for the generation of electricity)

[0206] ❖ In our method we assume specific values for 2 of the 4 variables and by using Newton’s Laws of motion and Archimedes principle. We can specify the 2 other variables depending on the main clause that ‘the initial velocity of the system in each 2 paths is equal zero and also the final velocity is equal to zero’ (using trial and error method).

[0207] ❖ In our system, we take the shape of the mass W2 by a specific form, which has a volume of (1.270087r)m3and a weight of (1.27008TT) Ton. We also assume that the total path length = 1 ,20m divided as we mentioned before.

[0208] ❖ So, we know 2 variables of the 4 variables and it becomes easier to specify the other 2 variables (W1 & F).

[0209] ❖ We found that o W1 = 2.028923665 Ton o F = 1.007518055 Ton

[0210] ❖ But the Force F which is responsible for rotating the large disc is not constant along the path as we know that the component of the force (F) in the direction tangent to the circle (Ft) is the responsible force to rotate the disc which is variable as it depends on the angle between the direction of the Force (F) and the radius at the effecting point on the circle.

[0211] ❖ But we can get the average (Ft) during each step of the total path using integral calculus for each step (as shown in the following calculations).

[0212] ❖ By using integral calculus, we found that the average (Ft) during the 1ststep (as

[0213] = 0.60m) = 0.164171 F and also the final step (as D6= 0.60m ) =0.266471 F.

[0214] ❖ It means that we need a big force of (F) in the 1stand the final steps. If we assume that the minimum moment required to rotate the disc is 1.5 t.m. so we found that minimum average Ft0.25 ton

[0215] ❖ So, minimum F in the 1ststepr1.522805 Ton o And minimum F in the final stepr0.938189 Ton

[0216] ❖ As the total work done by the force (F) during 1 path = o 12 * 1.007518055 = 12.09021665 Ton.m.

[0217] ❖ So the remaining work = o 12.09021665 - 0.6 (1.522805 + 0.938189) = 10.61362026 Ton.m

[0218] ❖ So the average Force (F) during the remaining path = o (10.61362026 -F 10.80) = 0.982742614 Ton

[0219]

[0220] ■ F = 1 .522805 Ton (for the 1ststep as D1= 0.60m)

[0221] ■ F= 0.938189 Ton (For the final step as D6= 0.60m)

[0222] ■ F= 0.982742614 Ton (For the remaining steps)

[0223] ■ F = 0.938189 Ton (for the 1ststep as D1= 0.60m)

[0224] ■ F = 1 .522805 Ton (For the final step as D6= 0.60m)

[0225] ■ F = 0.982742614 Ton (For the remaining steps)

[0226] CALCULATIONS OF THE WORK DONE BY THE TANGENTIAL FORCE

[0227] ❖ To calculate the total work done by the tangential force (Ft) during one path we must do the following:

[0228] ❖ Firstly, we must calculate the tangential force (Ft) during the path, which is variable as it depends on the angle between the direction of the force (F) and the radius at the effecting point on the circle.

[0229] ❖ So we calculate the average tangential force (Ft) for each step of the path using integral calculus (as shown in the following calculations)

[0230] o (1.25)2= (6)2+ (7.1)2- 2 * 6.0 * 7.1 cos 9 1 o θ1= 145.O566O54°&02= 34.94339458° o θ3= 18.98566515° = 0.3313623676 (rad)

[0231] ❖ Assume that (F) is constant during the path step = 0.60m but Ftis variable during this step.

[0232] ❖ To find the average Ftduring this step D1= 0.60m o Ft=F sin 03

[0233] By calculating area under the curve We find 0.0544 F o Average Ft during step b 1 b

[0234] T1 Total work of one path = (0.164171 * 1.522805) * 3.659263723

[0235] + (0.773386541 * 0.982742614) * 8.727891157

[0236] + (0.974388999 * 0.982742614) * 1.405385735

[0237] + (0.8990786774 * 0.982742614) * 0.7830801183

[0238] + (0.7026564057 * 0.982742614) * 2.04592216

[0239] + (0.2664708222 * 0.938189) * 2.228013028

[0240] = Ft* 6TT = 11.55579859 Ton. m

[0241] Average Tangential force during path = 0.61305416 Ton

[0242] Total work during 2 paths = total work done during one circle = 23.1116 Ton. m

[0243] Total time for 2 paths = 9.94841805 sec = 22790.04 Watts

[0244] = 22.79 Kilowatts

[0245] = 30.975 horsepower

[0246] It means that we can produce electrical energy equals 22.79 kilowatts from system ( No. 1 )

[0247] 3. Dynamic Friction force < 0.10 Ton = 0.1 * 103* 9.81 Newton

[0248] 4. Gate force = 0.05 Ton =0.05 * 103*9.81 Newton

[0249] 5. F = required force for rotating disc

[0250] = 1.522805 * 103* 9.81 Newton (for 1ststep)

[0251] = 0.982742614 * 103* 9.81 Newton (for 2nd, 3rd, 4thand 5thstep)

[0252] = 0.938189 * 103* 9.81 Newton ( for final step)

[0253] 6. Water resistance (W.R.)

[0254] Any body moves inside a fluid faces resistance during its motion and this resistance depends on several items: a) Velocity of the fluid and its direction (it doesn't exist in our case) b) Velocity of the moving body (V) as W.R. oc (V2) c) Fluid density (p) (in our case p= 1000 kg / m3) as W.R. oc p d) (Ap) the projected area of the moving body on the plan which is perpendicular to the movement direction of the body as W.R. oc Ap

[0255] e) The shape factor (f)

[0256] The shape of the moving body is very effective in (W.R.) as if the shape is tapered enough in the direction of the motion, it will face less water resistance. o Note: the 3 bodies have the same Ap. However, the water resistance is less in the body, which has the most tapered shape as the total projected area( Ao ) in Fig.l is more than the total projected area in Fig.2, which is more than the total projected area in Fig.3.

[0257] Ap

[0258] > So, we can consider that the shape factor f=

[0259] Ao o As (Ao) is the total area of the body, which is projected to find Ap on condition that the vector line 1 to the body must have a component in the direction of motion.

[0260] > In our case, o Ap = n(o. 422)m2o Ao= total area of the body that is projected to find Ap

[0261] = total surface area of the cone

[0262] ❖ The average velocity of the moving body can be taken about 3.0 m / sec

[0263] ❖ Water resistance in our case will be less than 100kg taken 115 kg o W.R. = 0.115 Ton = 0.115 * 103* 9.81 Newton ❖ Total path length = 12.0 m

[0264] ❖ Total time during path 2 = 4.610199073 sec

[0265] ❖ Total time during path 1 = 5.338218977 sec

[0266] ❖ Total time for path 1 and path 2 = 9.94841805 sec

[0267] ❖ The average straight velocity of the elements of the total system [the right side load w1 & the left side load (moving cylinder) W2 & the cables (vertical cables& horizontal cables) & the base of iron rod- (iron rod rule)]

[0268] Average straight velocity a a }

[0269] ON THE OTHER SIDE

[0270] The tangential velocity of the rotating disc at the point, which the iron rod is connected with the disc (impact point force)

[0271] ❖ Average tangential velocity 3.789457947 m / sec 3.79 m / sec a a}

[0272] ❖ Note that: the relationship between the straight velocity and the tangential velocity (at the impact point force) always

[0273] ❖ While the tangential velocity at any point on the rotating disc depends on the position of the pt (the distance between the pt and the origin of the rotating disc (7?\)) and always equal.

[0274] So the average tangential velocity of the end point of the disc which has R\=6.25m 3.94735203 m / sec Second: - system ( No. 2 )

[0275] Description of System 2:

[0276] System 2 consists of 24 elements, as clarified in the system's claims, which can be divided into two groups:

[0277] First Group: Elements from (1) to (7) operate in a vertical plane.

[0278] These elements comprise an iron lever (No 1) with a length of ( 15.0 m ) in the shape of the letter (U) and a width of ( 1.0 m ) to allow water to pass through it.

[0279] The lever pivots on a horizontal axis that enables it to rotate in a vertical plane. The right arm of Lever (No 1) measures ( 6.0 m ) in length, while the left arm measures ( 9.0 m ). The right arm of Lever (No 1) is connected to a metallic container (No 2) in the shape of a half-circle with a radius of ( 1.8 m ), having the same width as Lever (No 1). These two elements (1) and (2) function as a single unit. When Container (No 2) is filled with water, the water is transferred through Lever (No 1) while both the lever and the container rotate together. This action results in pouring the water at the end of the path (No 1) into the upper water tank (No 13).

[0280] The left arm of Lever (No 1) is connected to three flexible iron cables (No 3) with a diameter of (16 mm) at a distance of (8.4 m) from the lever's pivot axis. When the left arm of the lever is pulled from the point where the cables are anchored using Cables (No 3). Lever (No 1) begins to rotate around the horizontal pivot axis. Cables (No 3) are also connected to an iron cylinder (No

[0281] 4) with a diameter of (0.64m), equipped with two iron gears at its ends with a diameter of (1.075 m), and the net distance between these two gears is (135 cm).

[0282] These gears (No 4) engage with two other gears with a diameter of (5.05m) (No

[0283] 5) that rotate around a horizontal axis made of steel with a diameter of (10 cm). The net distance between Gears (No 5) is also (135 cm). The steel axis around which Gears (No 5) rotate is equipped with three iron ring frames with a diameter of (2.466m) and a width of (8 cm) for each frame. Each of these iron ring frames is connected to a flexible iron cable with a diameter of (16 mm) (No 6).

[0284] These three cables (No 6) are connected to the three iron ring frames (No 5) on one end and to the weight (Wl) (No 7) on the other end. When Weight (Wl) (No 7) moves downward, the three frames and Gears (No 5) rotate clockwise. This motion is transferred through Gears (No 4) in a counterclockwise direction, causing the cables (No 3) to wind around the circumference of Iron Cylinder (No 4).

[0285] This action makes Lever (No 1) and Container (No 2) rotate in a vertical plane until Lever (No 1) contacts the circumference of Iron Cylinder (No 4) at the end of Path (No 1). At that point, the angle between the Lever axis and the horizontal axis is

[0286] (1.85 degrees).

[0287] In this way, water is transferred from the primary water tank (No 11) through Container (No 2), then through Lever (No 1), and ultimately poured into the upper water tank (No 13) at the end of the first path.

[0288] Second Group:

[0289] Consisting of elements [6,7,8,14,15,16,17,18,19,20,21,22,23,24], these elements operate in a vertical plane perpendicular to the plane of operation of the first group, intersecting vertically with it along the axis of flexible iron cables (No 6). It's worth noting that the common elements between the two groups are (6) and (7). Elements (11), (12), and (13) are water tanks at different levels, with the initial tank (No 11) having a lower water level, and the goal is to lift the water from it to the final water tank (No 12), which is positioned (1.10 m) higher than the water level in tank (No 11). This is the purpose of this system's operation.

[0290] As for Upper tank (No 13), it is an auxiliary water tank with a water level higher than that of the final tank (No 12) by a magnitude of 5 meters. Its total area does not exceed 50 square meters, and its storage capacity is no more than 30 cubic meters. This tank receives water from Lever (No 1) and then distributes it to two iron cylinders (No 14) according to the system's operation.

[0291] * Body W2 (No 8) consists of two cylindrical bodies, each with a hollow structure, having a length of 1.6 meters and an external diameter of 1.10 meters. Both the upper and lower ends of these cylindrical bodies are connected to a hollow cone with the same external diameter as the cylindrical body.

[0292] The cone has a height of 1.65 meters, resulting in a total height for Body W2

[0293] (No 8) of 4.90 meters.

[0294] Each component of Body W2 (No 8) is connected to Body W1 (No 7) through two flexible iron cables (No 9) with a diameter of 16mm, and these cables pass over smooth metal rollers (No 10).

[0295] * Four metal arms (No 15) are attached to the upper cone of each component of Body W2 (No 8). These metal arms press upward on the drainage gate keys during the system's operation to raise and lower the drainage gate as needed, allowing excess water to drain. * Each component of Body W2 (No 8) moves vertically along a track with a length of 9.90 meters inside two external iron cylinders (No 14). These external cylinders have an internal diameter of 1.14 meters, and the clear distance between the cylindrical body W2 (No 8) and the inner wall of the external iron cylinder (No 14) is 2 cm.

[0296] * Each external cylinder component (No 14) is connected from the bottom to the final water tank (No 12) through the drainage gate (No 16) and approximately from its middle to the upper water tank (No 13) through the filling gate (No 19).

[0297] * Drainage Gate (No 16) consists of two cylindrical metal gates, with one gate installed inside each external iron cylinder (No 14) from the inside. These gates move vertically upward to open, allowing excess water to drain from the external cylinder (No 14) to the final water tank (No 12), and they move downward to close the gate, preventing water from flowing in both directions. It should be noted that the gate must be capable of sliding up and down without causing water leakage between the gate and the wall of external iron cylinder (No 14).

[0298] * Each gate of the drainage gates (No 16) is connected to eight flexible iron cables (No 17) with a diameter of 6mm. These cables are installed on the inner wall of external cylinder (No 14) inside metal casings that are vertically attached to the wall of external cylinder (No 14), allowing the iron cable (No

[0299] 17) to move easily within these casings.

[0300] * Iron cables (No 17) are connected from below to the drainage gate (No 16) and from the top to the gate keys, which are fixed to the upper metal disc (No

[0301] 18). This metal disc is installed in the upper part of external cylinder (No 14) from the inside and can move vertically up and down within a distance of 0.50 meters.

[0302] When the upper metal disc (No 18) is pressed upwards by the metal arms (No 15), this disc rises and lifts Drainage Gate (No 16) upwards to allow excess water to exit from the external cylinder (No 14) into the final water tank (No

[0303] 12). When the upper metal disc (No 18) moves downwards, it closes the drainage gate, thus preventing water from flowing in both directions.

[0304] As mentioned earlier, the external cylinder (No 14) is connected to the upper water tank (No 13) through the filling gate (No 19), with one gate installed in each cylinder. Filling Gate (No 19) is a cylindrical metal gate installed approximately at the midpoint of external cylinder (No 14) from the outside, as depicted in the drawings. It moves vertically upward to open, allowing the external cylinder (No 14) to be filled with water from the upper water tank (No

[0305] 13) to the desired level. It moves downward to close, preventing water from flowing in both directions.

[0306] Filling Gate (No 19) is connected to eight flexible iron cables (No 20) with a diameter of 8mm. These cables are connected to the gate on one end and to the gate keys on the other end. They pass through smooth metal rollers (No 21). These gates must be capable of sliding up and down without causing water leakage between the gate and the wall of external cylinder (No 14).

[0307] The iron cables (No 20) that connect to Filling Gate (No 19) from one end and to the gate keys on the other end are installed in each external cylinder (No 14). These gate keys are fixed to a metal disc (No 22) that is installed in the lower part of external cylinder (No 14) at the specified level indicated in the drawings. It should be noted that each cable of the iron cables (No 20) that connects to Filling Gate (No 19) installed outside external cylinder (No 14) is placed inside a metal casing attached to the exterior wall of the cylinder (No 14). Then, they pass through two smooth metal rollers (No 21) mounted at the top of the cylinder (No 14) before descending vertically inside the cylinder (No 14) within a casing fixed to the interior wall of the cylinder (No 14). They are then connected to the gate keys, which are fixed to the metal disc (No 22).

[0308] When pressure is applied to the lower metal disc (No 22) and it is pressed downwards by the moving body W2 (No 8), the cables (No 20) move along with it, lifting Filling Gate (No 19) upwards to allow water to flow from the upper water tank (No 13) into external cylinder (No 14) until the desired level is reached. When the pressure on this disc is released, Filling Gate (No 19) moves downward under its weight, closing the gate and preventing water from flowing in both directions.

[0309] It should be noted that the metal disc (No 22) can move vertically up and down within a distance of 0.50 meters.

[0310] Since the lower metal disc (No 22) is pressed downwards by the moving body W2 (No 8) while moving downwards, four metal protrusions (No 23) were created at the base of the lower cone of the moving body W2 (No 8) to apply the required pressure, as shown in the drawings.

[0311] To ensure that Body W2 (No 8) moves vertically inside the external cylinder

[0312] (No 14) without rotating around its axis during the movement, iron protrusions in the shape of a (U) (No 24) were created, and their location inside the external cylinder (No 14) is indicated in the drawings.

[0313] All the elements of System 2, in their specified shapes and sizes as shown in the drawings, work together to lift a quantity of water equal to 5.09 cubic meters from the initial tank (No 11) to the final tank (No 12) in a duration of 9.12 seconds, which is the duration of one cycle. This is achieved without any human intervention or the use of any type of fuel.

[0314] Operation of System Number (2)

[0315] Firstly: Starting Point

[0316] The position of System Number (2) at the beginning of its operation is referred to as the starting point. At this point, Lever (No 1) is positioned entirely vertically, and Container (No 2) is fully submerged and filled with water, with a volume of 5.09 m3. In this state, the gates of Container (No 2) are closed by means of the metal springs attached to the gates from one side and to the inner side of the container from the other side. In this case, the water pressure on the gate is equal to zero, as the water level inside Container (No 2) is exactly the same as the water level in Primary Tank (No 11). Body W1 (No 7) is at the highest point of its path, while Body W2 (No 8), which consists of two identical bodies in shape, size, and weight, is at the lowest point of its path and fully submerged in water. The two elements, Numbers (7) and (8), are connected to each other by flexible iron cables (No 9), which pass through smooth metal rollers (No 10).

[0317] Since the components of Body W2 (No 8), which consists of two movable bodies, are fully submerged in water, they are affected by an upward force known as buoyant force, according to Archimedes' principle of buoyancy. Therefore, Element (No 8) is affected by three forces:

[0318] 1. Weight with a value of (3.6x 103x9.81) Newtons, acting downward.

[0319] 2. Buoyant force, equal to (0.81675x n x 103x 9.81) Newtons acting upward. 3. Tension force in cables (No 9).

[0320] As for Body W1 (No 7), it is affected by three forces:

[0321] 1. Weight with a value of (5.573914xl03x9.81) Newtons, acting downward.

[0322] 2. Tension force in cables (No 6), which equals the tension force (T) in cables (No 3) multiplied by the reduction factor (FR).

[0323] 3. Tension force in cables (No 9).

[0324] Upon studying the motion of the group W1 (No 7) & W2 (No 8) under the influence of the forces mentioned above, it is noted that, in addition to the a forementioned forces, there is dynamic friction resistance to motion and water resistance force on Body W2 (No 8) during its movement in water. Both of these forces act in the opposite direction of motion. According to the calculations of the force study provided, the sum of the forces acting on the group (wl&w2) causes the group to move in an increasing acceleration motion, with Body W1 (No 7) moving downward and Body W2 (No 8) moving upward.

[0325] Secondly: Path Number (1) [All path steps are illustrated in the drawings]

[0326] The motion begins with Body W1 (No 7) moving downward, which causes the three cables (No 6) to unwind, and thus the three iron ring frames in Element (No 5) start rotating clockwise. Due to the interlocking of the gears in Element (No 5) with the gears in Element (No 4), the motion is transferred between them. The gears in Element (No 4) begin to rotate counterclockwise, and as a result, the three cables (No 3) are wound around the iron cylinder (No 4). Consequently, the lever (No 1), [which is connected to Container (No 2)] rotates around its horizontal axis, causing Container (No 2) to rise out of Primary water Tank (No 11). The water inside Container (No 2) starts moving towards lever (No 1) based on the angle of inclination of the lever (No 1) with the vertical axis until lever (No 1) reaches a completely horizontal position. At this stage, most of the water inside the lever (No 1) has been poured into the upper water tank (No 13). At the end of Path Number (1), when the axis of the lever (No 1) is inclined at an angle of (1.85 degree), all the water has been poured into the upper water tank (No 13).

[0327] Path Number (1) will be divided into (9) steps based on the forces acting on the group fwl&w2l and the position of the lever (No 1) during the path, as follows:

[0328] Step 1 (DI = 0.50 m):

[0329] During this step, Body W1 (No 7) moves downward, while Body W2 (No 8) moves upward, as explained in the starting point. As a result of the upward movement of Body W2 (No 8), the lower metal disc (No 22) also moves upward under the influence of the weight of the filling gate (No 19), which weighs (50 Kg). The forces acting on the group [w2&wl] are studied, including the force of compression of the lower metal disc on Body W2 (No 8) with a value of (50 Kg) over this step of length (0.5 m). Through the calculation of the forces acting on the group [w2&wl], the acceleration (al) is calculated, as well as the velocity (VI) at the end of step 1 and the time duration (tl).

[0330] Step 2 (D2 = 2.26 m):

[0331] Body W1 (No 7) continues to descend while Body W2 (No 8) continues to rise, and the lever (No 1) continues to rotate counterclockwise. At the end of this step, Container (No 2) has completely exited the Primary water Tank (No 11). The forces acting on the group [wl&w2] are studied to calculate the acceleration (a2), velocity at the end of this step (V2), and the time duration (t2).

[0332] Step 3 (D3 = 2.24 m):

[0333] During this step, lever (No 1) continues to rotate counterclockwise, and Body W1 (No 7) continues to descend, while Body W2 (No 8) continues to rise within the water until the upper point of Body (w2) reaches the water level inside the external cylinder (No 14), which is equal to the water level in the upper water tank (No 13). The forces acting on the group [wl&w2] are studied to calculate the acceleration (a3), velocity at the end of this step (V3), and the time duration (t3).

[0334] Step 4 (D4 = 1.14 m):

[0335] Body W1 (No 7) continues to descend while Body W2 (No 8) continues to rise, and the lever (No 1) continues to rotate counterclockwise. During this step, the upper cone of Body W2 (No 8) starts emerging from the water, with a length of (1.65 m), completely exiting the water by the end of this step. The forces acting on the group [wl&w2] are studied to calculate [(a4), (V4), (t4)] .

[0336] Step 5 (D5 = 0.11 m):

[0337] The system's movement continues, and during this step, the cylindrical middle part of the body W2 (No 8) starts to emerge from the water, completely exiting the water with a length of (1.60 m) at the end of this step. By studying the forces acting on the group [wl&w2], we can calculate [(a5), (V5), (t5)]. It's important to note that (a5) is negative during this step, causing the system's velocity to start decreasing.

[0338] Step 6 (D6 = 1.14 m):

[0339] The system's movement continues, and during this step, the lower cone of the body W2 (No 8) begins to emerge from the water, completely exiting with a length of (1.65 m) by the end of this step. Consequently, W2 (No 8) has completely exited the water with a total length of (4.9 m). As a result of (w2) exiting the water, the water level inside the external cylinder (No 14) decreases by (2.51 m) below the water level in the upper tank (No 13). By studying the forces acting on the group [wl&w2], we can calculate [(a6), (V6), (t6)], and the acceleration of the group (a6) is negative, so the velocity of the group (V6) continues to decrease.

[0340] Step 7 (D7 = 2.01 m):

[0341] The system's movement continues, and during this step, the lever (No 1) continues to rotate counterclockwise. The flow of water inside the lever continues until it reaches the end of the left arm of the lever by the end of this step, after which the water is poured into the upper water tank (No 13). The body W1 (No 7) continues to descend, and the body W2 (No 8) continues to ascend. By studying the forces acting on the group [wl&w2], we can calculate [(a7), (V7), (t7)], and the acceleration of the group is negative during this step, so the velocity of the group (V7) continues to decrease.

[0342] Step 8 (D8 = 0.26 m):

[0343] The system's movement continues with the descent of element W1 (No 7), the ascent of element W2 (No 8), and the counterclockwise rotation of lever (No 1). During this step, water is poured from lever (No 1) into the upper water tank (No 13), and at the end of this step, lever (No 1) becomes completely horizontal. While element W2 (No 8) ascends, it exerts pressure on the upper metal disc (No 18) using the metal arms (No 15) attached to the head of element (No 8). Consequently, metal disc (No 18) moves upwards and raises the drainage gate (No 16) to drain excess water from the external cylinder (No 14) into the final water tank (No 12). It's important to note that the amount of water discharged is equal to:

[0344] 2 x n x 2.49 x (0.57)2= 5.08 m3

[0345] And this quantity is equal to the amount of water lifted from the initial water □ tank (No 11), which is (5.09 m ), considering the slight difference between them to account for water loss.

[0346] By studying the forces acting on the group [wl&w2], we can calculate [(a8), (V8), (t8)], and the velocity of the group continues to decrease.

[0347] The Last Step (D9 = 0.24 m):

[0348] The system's movement continues with the descent of element W1 (No 7), the ascent of element W2 (No 8), and the continued counterclockwise rotation of lever (No 1). During this step, water continues to drain from the external cylinder (No 14) into the final water tank (No 12) until the water level is equal at both ends. The system's velocity continues to decrease until it reaches zero at the end of this step. At the end of the path, the total distance moved by each of elements (wl,w2) is equal to (9.90 m), and the angle between the axis of lever (No 1) and the horizontal axis is (1.85 degree). Third: Path Number (2) (All steps of the path are illustrated in the drawings)

[0349] After the system's velocity reaches zero at the end of path number (1), where lever (No 1) makes an angle of (1.85 degree) with the horizontal axis, and cables (No 3) are fully wound around the circumference of iron cylinder (No 4).

[0350] Element W1 (No 7) is at its lowest point in its path, with cables (No 6) fully extended, while element W2 (No 8) is at the highest point in its path, compressing the upper metal disc (No 18) to be at the specified height of (0.50 m) in its path. Drainage gate (No 16) is fully open at this moment. The system starts its second path, with element W2 (No 8) descending and element W1 (No 7) ascending. Lever (No 1), which is now completely free of water and connected to container (No 2), rotates clockwise.

[0351] At the beginning of second path, the vertical distance between the lower head of element W2 (No 8) and the water level in the external cylinder (No 14) is equal to (5.0 m).

[0352] Due to the rotation of lever (No 1), cables (No 3) unfold, causing the iron cylinder and gears (No 4) to rotate clockwise. Motion is transferred between gears (No 4) and (No 5), causing gears and the three iron ring frames (No 5) to rotate counterclockwise. As a result of the ascent of element W1 (No 7) and the rotation of the three ring frames (No 5), cables (No 6) are wound around the three ring frames (No 5), and the system continues to operate in the second path.

[0353] The second path is divided into (8) steps based on the forces acting on the group [wl&w2] and the position of lever number (1) during the path as follows:

[0354] Step 1 (DI = 0.50 m): During this step, element W1 (No 7) starts ascending, and element W2 (No 8) starts descending. The upper metal disc (No 18) also moves downward under the influence of the weight of drainage gate (No 16), which is (50 kg), working in the same direction as the weight of element W2 (No 8). Gradually, drainage gate

[0355] (No 16) is locked completely by the end of this step, preventing water from flowing from the external cylinder (No 14) to the final water tank (No 12), and vice versa. Lever (No 1) also starts rotating clockwise, causing the cables (No 3) to unfold and the iron cylinder and gears (No 4) to rotate clockwise. Motion is transferred to gears and the three ring frames (No 5) in the opposite direction, winding cables (No 6) as element W1 (No 7) ascends. By calculating the forces acting on the group (wl, w2), the acceleration of the group (al) is determined. Using Newton's laws of motion, the velocity of the group (wl, w2) (VI) at the end of this step and the time duration of this step (tl) are calculated.

[0356] Step 2 (D2 = 1.91 m):

[0357] The system continues its operation:

[0358] - Element Wl (No 7) ascends.

[0359] - Element W2 (No 8) descends.

[0360] - Lever (No 1) rotates clockwise.

[0361] - By analyzing the forces acting on the group (wl, w2), the group's acceleration (a2), the group's velocity (V2) at the end of this step, and the time duration of this step (t2) are calculated.

[0362] Step 3 (D3 = 2.59 m):

[0363] The system continues its operation.

[0364] - By analyzing the forces acting on the group (wl, w2), the group's acceleration (a3), the group's velocity (V3), and the time duration of this step (t3) are calculated.

[0365] Step 4 (D4 = 1.14 m): The system continues its operation.

[0366] - At the beginning of this step, the lower head of element W2 (No 8) reaches the water level inside the external cylinder (No 14).

[0367] - The lower cone of element W2 (No 8) starts descending into the water.

[0368] So the water level inside the external cylinder rises as when the element W2 moves down a distance of (1.14 m) the lower cone of the element W2 (No 8) of length (1.65 m) will be completely submerged.

[0369] According to Archimede's principle the buoyant force starts to work against the submerged body, which equal the weight of the fluid displaced by the body.

[0370] - By analyzing the forces acting on the group (wl, w2), the group's acceleration (a4), the group's velocity (V4), and the time duration of this step (t4) are calculated.

[0371] Step 5 (D5 = 0.11 m):

[0372] The system continues its operation.

[0373] - Element W2 (No 8) continues to descend in the water.

[0374] - As known, depending on the size of the submerged body, the water level inside the external cylinder (No 14) rises, as when the element (w2) moves down a distance of (0.11 m) the cylindrical part of length (1.60 m) of the body W2 (No 8) will be completely submerged.

[0375] - By analyzing the forces acting on the group (wl, w2), [ (a5), (V5), (t5)] are calculated. Note that the group's wheel is negative during this step, causing the group's velocity to start decreasing.

[0376] Step 6 (D6 = 1.14 m):

[0377] - The system continues its operation.

[0378] -Element W2 (No 8) continues to descend in water.

[0379] -The water level inside the external cylinder (No 14) rises, as when the element (w2) moves down a distance of (1.14 m) the upper cone of the element W2 (No 8) of length (1.65 m) will be completely submerged. Consequently W2 (No 8) with a total length (4.9 m) has completely submerged in the water.

[0380] -As a result of (w2) submerged in the water, the water level inside the external cylinder (No 14) increases by (2.51 m) over the water level in the final tank (No 12) which is the same level of water inside the external cylinder (No 14) in the beginning of the path number (2).

[0381] - By analyzing the forces acting on the group (wl, w2), [ (a6), (V6), (t6)] are calculated. Note that the group's acceleration is negative during this step, causing the group's velocity to continue decreasing.

[0382] Step 7 (D7 = 2.01 m):

[0383] The system continues to operate in this step, with element Wl (No 7) ascending and element W2 (No 8) descending entirely within the water. During this step, element W2 (No 8) is subjected to the resistance force of water during its motion in addition to buoyancy force. Both of these forces act upward, opposing the downward motion of element (w2) and its own weight.

[0384] The lever (No 1) and container (No 2) continue to rotate clockwise. At the beginning of this step, container (No 2) reaches the water level in the primary water tank

[0385] (No 11) and starts descending into the water. This exposes it to water resistance as it descends in the primary water tank (No 11).

[0386] The value of this resistance (250 kg) during this step is estimated based on the size and shape of container (No 2), the area of the part of the container that encounters water during this step, and the speed of container (No 2) when it reaches the water surface. Additionally, the movable gates at the base of container (No 2), which open under the pressure of water as it moves, allowing water from the primary water tank (No 11) to enter, reduce the water resistance to the movement of container (No 2).

[0387] The optimal solution to accurately estimate the water resistance to the movement of container (No 2) inside the water of tank (No 11) is to create a prototype model of container (No 2) with the same shape, movement method, and speed to calculate the water resistance to its movement accurately. By analyzing the forces acting on the group (wl, w2), including the additional forces such as water resistance to the movement of container (No 2), which affects the tension force (T) in cables (No 3), [ (a7), (V7), (t7)] are calculated. It should be noted that the group's acceleration is negative during this step, causing the group's velocity to continue decreasing as it approaches reaching zero velocity by the end of path number (2).

[0388] The Eighth [Final] Step (D8 = 0.50 m):

[0389] Element Wl (No 7) continues to ascend, and element W2 (No 8) continues to descend within the water. Lever (No 1) and container (No 2) also continue to rotate clockwise. Container (No 2) fills with water through the gates at its base until it is completely filled with water by the end of path number (2).

[0390] During the descent of element W2 (No 8), the four iron protrusions (No 23) at the base of the lower cone of element W2 (No 8) press on the lower metal disc (No 22), which is located very precisely at this level. This causes the iron cables (No 20) connected to the filling gate (No 19) to move downward, opening the filling gate to allow water to flow from the upper water tank (No 13) into the external cylinder (No 14) until the water levels on both sides are equal.

[0391] □ The total amount of water filled into the external cylinder (No 14) [5.08 m ] is equal to the amount of water transferred from the primary water tank (No 11)

[0392] □ through the lever and containers numbers (1) and (2) [5.09 m ], taking into account the water loss.

[0393] The velocity of the system continues to decrease until it reaches zero at the end of this step. At the end of path number (2), the total distance moved by each of elements (7) and (8) is equal to (9.90 m), which is the same as the total distance of path number (1).

[0394] By analyzing the forces acting on the group (wl, w2), including the additional forces such as the weight of the filling gate ((50 kg)) and the water resistance to the movement of container (No 2) (250 kg) during this step, [ (a8), (V8), (t8)] are calculated.

[0395] Atthe end of the entire cycle [path number (1), path number (2)], a total of (5.09 m3) of water is transferred from the primary water tank

[0396] (No 11) to the final water tank (No 12) in a time duration of (9.12) seconds.

[0397] Calculation of system ( No. 2 )

[0398] Calculation of tension force of cables (No3)

[0399] * We calculate the tension force of iron cables (No3) in system (2) as a static system and use these values of the tension force (T) in the calculation of system (2) after multiplied these values by the reduction factor ( / r) which is equal the ratio between the diameter of the iron cylinder (No4) and the diameter of iron ring frame (No5).

[0400] * We calculate (T) for all steps of 1stpath and 2ndpath of system (2) as follows.

[0401] Weights and C.G of lever parts

[0402] Weight and C.G of the left side of the lever.

[0403] Wt of S.I.B (500 x 158 mm) = 93.654 kg / m Taken 95 kg / m for bracing.

[0404] *W1= 8.55 X 0.095 x 3.0 = 2.43675 Ton

[0405] * W2 = 9.0 x 1.02x 0.01 x 7.85 = 0.72063 Ton

[0406] *W3 = 2 x 9.0 x 0.20 xO.Olx 7.85= 0.2826 Ton i

[0407] *W4 = 2 x- x 0.485 x 9.0 x 0.01 x 7.85 = 0.3426525 Ton

[0408] 2 o tt

[0409] ❖ 2.43675 x — + (0.72063 + 0.2826) x 4.5 + 0.3426525 x 3.0

[0410] = 3.7826325 x x 4.2192 m

[0411] 2.43675 x 0.25 + 0.72063 x 0.505 + 0.2826 x 0.61 + 0.3426525 x 0.8717 = 3.7826325 x Y Y = 0.382 M

[0412] Weight and C.G of the right side of the lever

[0413] W1 = 3.0 x 6.0 x 0.095 = 1.71 Ton

[0414] W2 = 6.0 x 1.02 x 0.01 x 7.85 = 0.48042 Ton

[0415] W3 = 2x 4.0 x 0.685 x 0.01 x 7.85 = 0.43018 Ton 1

[0416] W4 = -x 2 x 4.0 x 0.215 x 0.01 x 7.85 = 0.06751 Ton 2 W5 = 2x 2.0 x 0.9 x 0.01 x 7.85 = 0.2826 Ton

[0417] 1

[0418] W6 = 2 x— x 2.0 x 0.90 x 0.01 x 7.85 = 0.1413 Ton 2

[0419] Resultant Force = 3.11201 Ton

[0420] ❖ (1.71+ 0.48042) x 3.0 + 0.43018 x 2.0 + 0.06751 4.0

[0421] + 0.2826 x 5.0 + 0.1413

[0422] = 3.11201 x X => X=3.1421 m

[0423] (1.71+ 0.48042) x 1.71 x 0.25 + 0.48042 x 0.505+ 0.43018 x 0.8525

[0424] 19

[0425] + 0.06751 x — + 0.2826 x 0.96 + 0.1413 x 1.71 15

[0426] =3.11201 x Y ¥=££255 m

[0427] Weight and C.G of container tank

[0428] *Weight of container plates = wl

[0429] W1 = [n (1.805) x 1.02] xO.Ol x7.85 = 0.454043 Ton

[0430] Distance from C.G of plates to origin = 0.63662 R

[0431] = 1.1491 m

[0432] *W2 = weight of S.I.B = 3 x (JI x 2.06) x 0.095 = 1.84443 Ton

[0433] *Distance from origin to C.G OF S.I.B weight

[0434] = 0.63662 x 2.06 =1.31144 m

[0435] *Weight of plates in the 2 sides of container tank W3 = [2(nx(1.8)2 / 2) x O.Olx 7.85 = 0.799033 Ton

[0436] *Distance from origin 0.763944 m

[0437] *Total weight of container tank = 3.097506 Ton

[0438] ❖ 0.454043 x 1.1491 + 1.84443 x 1.31144 + 0.799033 x 0.763944

[0439] = 3.097506 x X

[0440] X = 1.146412 m

[0441] Values of tension (T)

[0442] 1stpath

[0443] DO = 0.0

[0444] *Total cable length = 12.07 m

[0445] * This case occurs in the beginning of the 1stpath and in the final point of the 2ndpath.

[0446] 3.112 x 0.5255 + 3.0975 x 2.31 + 3.783 x 0.382 = T cos(45.925)x 8.40 = 1.752 Ton

[0447] 1ststep

[0448] DI = 0.50 m

[0449] * 0.50 x^^ = 0.61 m

[0450] 9.9

[0451] * cable length (No3) = 12.07 - 0.61 = 11.46 m

[0452] * Angle between the lever arm and the vertical line = 5.83°

[0453] Distances

[0454] * 3.1421 sin(5.83) + 0.5255 cos(5.83) = 0.842 m

[0455] * 7.5105 sin(5.83 + 17.91 ) = 3.024 m

[0456] * 4.2192 sin(5.83)- 0.382 cos(5.83)= 0.049m

[0457] ❖ 3.112 X 0.842 + 3.0975 x 3.024 + 0.9542 x 2.938

[0458] = 3.783 x 0.049 + T cos(43.01) x8. 40 T = 2.378 Ton

[0459] • Case of the 2ndpath when the container tank is empty.

[0460] • 3.112 x 0.842 + 3.0975 x 3.024 = 3.783 x 0.049 + T cos(43.01) x 8.40

[0461] T = 1.921 Ton

[0462] 2ndstep

[0463] D2 = 2.26 m

[0464] Note that when the angle between the lever arm and the vertical line = 29.4155°

[0465] The container tank (No2) is just raised outside the water tank (Noll).

[0466] In this case the length of the cable (No3) = 8.71 m

[0467] So: 12.07 - 0.61 - 8.71 = 2.75 m

[0468] . 9.90

[0469] ❖ 2.75 x - = 2.26 D2 = 2.26 m

[0470] 12.07

[0471] Distances

[0472] • 3.1421 sin(29.4155)+0.5255cos(29.4155)= 2.001 m

[0473] • 7.5105 sin(29.4155 + 17.91) = 5.522 m

[0474] • 4.2192 sin(29.4155) - 0.382 cos(29.4155)= 1.739 m ❖ 3.112X 2.001 + 3.0975 x 5.522 + 0.873 x 3.742 + 4.258 x 5.192 - 0.0365 x 5.862 = 3.783 xl.739 + T cos(31.21725)x 8.40 T = 5.834 Ton

[0475] 3rdstep

[0476] D3 =2.24 m

[0477] Total D = 0.50 + 2.26 + 2.24 = 5.0 m

[0478] „ 12.07

[0479] 5.0 x - = 6.096 m

[0480] 9.90

[0481] So: - cable length (No3) =12.07- 6.096 = 5.974 m

[0482] For this length: - the angle of lever = 50.19°

[0483] Distances 0.5255 cos(50.19)= 2.75 m 17.91)= 6.969 m

[0484] * 4.2192 sin(50.19) - 0.382 cos(50.19)= 2.9965 m

[0485] ❖ 3.112 X 2.75 +3.0975 x 6.969 + 1.63126 x 4.786 +

[0486] 3.6703 x 6.476 - 0.21344 x 6.941 = 3.783 x 2.9965 + T cos(20.83 )x 8.40

[0487] T = 6.229 Ton

[0488] 4thstep

[0489] D4 =1.14 m

[0490] Total D = 6.14 m

[0491] „ „ 12.07

[0492] So: 6.14 x - =7.486 m

[0493] 9.90

[0494] ❖ cable length = 12.07 - 7.486 = 4.584 m For this length the angle of lever arms = 60.18°

[0495] * Distances 0.5255 cos(60.18)= 2.987 m

[0496] * 7.5105 sin( 60.18 + 17.91) = 7.349 m

[0497] * 4.2192sin(60.18) — 0.382 cos(60.18) = 3.471 m

[0498] ❖ 3.112 X 2.987 + 3.0975 x 7.349 + 2.114 x 4.919 + 3.388 x 6.844 - 0.405 x 7.239 = 3.783 x 3.471

[0499] +T cos(15. 835) x 8.40

[0500] T = 6.136 Ton

[0501] 5thstep

[0502] D5 = 0.11 m

[0503] Total D = 6.25 m

[0504] .c n r12.07

[0505] ♦♦♦ 6.25 x - = 7.62 m

[0506] 9.90

[0507] ❖ cable length (No3) = 12.07 - 7.62 = 4.45 m

[0508] ❖ For this length we find the angle of lever arms = 61.13°

[0509] Distances

[0510] *3.1421sin(61.13) + 0.5255 cos(61.13) = 3.005 m *7.5105 sin(61.13 + 17.91) = 7.3735 m

[0511] * 4.2192sin(61.13 — 0.382) cos(61.13) = 3.5104 m

[0512] *3.112 x 3.005 + 3.0975 x 7.3735 + 2.142 x 4.928 + 3.361 x 6.861 - 0.41 x 7.241 = 3.783 x 3.5104 +

[0513] Tcos(15.36) x 8.40

[0514] T = 6.118 Ton 6thstep

[0515] D6 = 1.14 m

[0516] Total D = 7.39 m

[0517] „ ™ 12.07 „

[0518] So: 7.39 x - = 9.01 m

[0519] 9.90

[0520] ❖ cable length (No3) = 12.07 - 9.01 = 3.06 m

[0521] For this length we find that the angle between the lever arm and the vertical line = 70.86°

[0522] Distances

[0523] *3.1421 sin(70.86) + 0.5255 cos(70.86) = 3.141 m

[0524] *7.5105sin(70.86 + 17.91) = 7.509 m

[0525] *4.2192 sin(70.86) — 0.382 cos(70.86) = 3.861 m

[0526] *3.112 x 3.141 + 3.0975 x 7.509 + 2.7385 x 4.732 + 2.7247 x

[0527] 6.952 - 0.3768 x 6.972 = 3.783 x 3.861 + T cos(10.495) x 8.40

[0528] T = 5.775 Ton

[0529] 7thstep

[0530] D7 = 2.01 m

[0531] Total D = 9.40 m

[0532] „ „ 12.07 „ „

[0533] So: 9.40 x - = 11.46 m

[0534] 9.9

[0535] ❖ cable length (No3)= 12.07 - 11.46 = 0.61 m

[0536] For this length the angle of the lever arm with vertical line = 87.7096° Note that: this case occurs when the water reaches the end of the left arm at angle 87.7096°

[0537] Distances

[0538] *3.1421sin( 87.7096) + 0.5255cos(87.7096) =

[0539] 3.161 m

[0540] * 7.5105 sin(87.7096 + 17.91) = 7.233 m

[0541] * 4.2192 sin(87.7096) - 0.382 cos( 87.7096) = 4.201 m

[0542] *3.112 x 3.161 + 3.0975 x 7.233 + 4.5 x 1.01 + 1.4284 x 6.53 - 0.8384 x 6.50 = 3.783 x 4.201 + T cos(2.07 )x 8.40 T = 2.951 Ton

[0543] Case of the 2ndpath when the container tank (No2) is empty.

[0544] 3.112 x 3.161 + 3.0975 x 7.233 = 3.783 x4.201 + T cos(2.07)x 8.40

[0545] T = 1.948 Ton

[0546] 8thstep

[0547] D8 = 0.26 m

[0548] In this case the lever arm will be horizontal and the container tank will be completely empty.

[0549] *cable length = 0.29 m

[0550] *The angle of the lever with vertical line = 90°

[0551] *The angle between the cable (No3) and the vertical line 0.925°

[0552] *3.112 x 3.1421 + 3.0975 x 7.1464 = 3.783 x 4.2192 +

[0553] TCOS(0.925)X8.4

[0554] T = 1.899 Ton

[0555] Final step

[0556] D9 = 0.24 m

[0557] *total D = 9.9 m

[0558] * cable length (No3) = 0.0

[0559] 12 07

[0560] As: 0.24 x — : — = 0.29 m (the same distance in the step 8) 9.9

[0561] • The angle between the lever arm and the vertical line=91.85° which is the same angle between the cable direction and the lever arm at the end of the 1stpath.

[0562] Distances

[0563] *3.1421 sin(91.85) + 0.5255 cos(91.85) = 3.123 m *7.5105 sin(91.85 + 17.91) = 7.068 m

[0564] *4.2192 sin(91.85) - 0.382 cos(91.85) = 4.229 m

[0565] *3.112 x 3.123 + 3.0975 x 7.068 = 3.783 x 4.229 +

[0566] Tcos(1.85)x8.40

[0567] T =1.86 Ton

[0568] 2ndoath

[0569] D=0.0 nd

[0570] This case is beginning of the 2 path which is the final step of the 1stpath. As T = 1.86 Ton

[0571] Dl=0.50 m

[0572] In this case the position of lever arms are similar to the position of the 7thstep in path (1) but when the container tank (No2) is empty

[0573] We find T = 1.948 Ton

[0574] 2ndstep

[0575] D2= 1.91 m

[0576] Total D =0.5 + 1.91 =2.41 m

[0577] 12 07

[0578] So cable length (No3) = 2.41x“^“ — 2.936 m

[0579] In this case the angle between the lever arm and the vertical line = 71.72 °

[0580] Distances

[0581] *3.1421sin(71.72) + 0.5255cos(71.72) = 3.148 m

[0582] *7.5105sin(71.72 + 17.91) = 7.51 m *4.2192sin(71.72) — 0.382 cos(71.72) = 3.886 m

[0583] *3.112 x 3.148 +3.0975 x 7.51 = 3.783 x 3.886 +

[0584] T cos(10.065)x8.40

[0585] T = 2.22 Ton

[0586] 3rdstep

[0587] D3 = 2.59 m

[0588] Total D = 0.5 + 1.91 + 2.59 = 5.0 m

[0589] So cable length 6.096 m In this case the angle between the lever arm and the vertical line = 49.30°

[0590] Distances

[0591] *3.1421sin(49.30) + 0.5255cos(49.30) = 2.725 m

[0592] * 7.5105 sin(49.30 + 17.91) = 6.924 m *4.2192sin(49.30) — 0.382cos(49.30) = 2.95 m

[0593] *3.112 x 2.725 + 3.0975 x 6.924 =3.783 x 2.95 + T cos(21.275)x 8.40

[0594] T = 2.398 Ton

[0595] 4Thstep

[0596] D4 = 1.14 m

[0597] Total D = 6.14 m

[0598] 12 07

[0599] So: cable length (No3) = 6.14 X —:— = 7.486 m

[0600] 9.9

[0601] In this case the angle between the lever arm and the vertical line= 38.93°

[0602] Distances

[0603] *3.1421sin(38.93) + 0.5255 cos(38.93) = 2.383 m

[0604] *7.5105sin(38.93 + 17.91) = 6.287 m *4.2192sin(38.93) - 0.382cos(38.93) = 2.354 m

[0605] *3.112 x 2.383 + 3.0975 x 6.287 = 3.783 x 2.354 + T cos(26.46)x 8.40

[0606] T = 2.392 Ton

[0607] 5thstep

[0608] D5 =0.11 m

[0609] Total D= 6.25 m 12 07

[0610] So: cable length (No3) = 6.25x^~=7.62 m

[0611] In this case the angle between the lever arm and the vertical line = 37.904°

[0612] Distances

[0613] *3.1421sin(37.904) + 0.5255 cos(37.904) = 2.345 m

[0614] *7.5105 sin(37.904 + 17.91) = 6.213 m

[0615] *4.2192 sin(37.904) - 0.382cos(37.904)) = 2.291 m

[0616] *3.112 x 2.345 + 3.0975 x 6.213 = 3.783 x 2.291 +

[0617] T cos(26.973)x 8.40

[0618] T = 2.388 Ton

[0619] 6thstep

[0620] D6 = 1.14 m

[0621] Total D = 7.39 m

[0622] 12 07

[0623] So: cable length (No3) = 7.39 x — : — = 9.01 m

[0624] 9.9

[0625] In this case the angle between the lever arm and the vertical line = 26.98°

[0626] Distances

[0627] *3.1421sin(26.98) + 0.5255 cos(26.98) = 1.894 m

[0628] *7.5105sin(26.98 + 17.91) = 5.301 m

[0629] *4.2192 sin(26.98) - 0.382cos(26.98) = 1.574 m

[0630] *3.112 x 1.894 + 3.0975 x 5.301 = 3.783 x 1.574 +

[0631] Tcos(32.435)x 8.40

[0632] T= 2.308 Ton

[0633] 7thstep

[0634] D7= 2.01 m

[0635] Total D = 9.40 m In this case the position of the lever arms are similar to the position of the 1ststep in path (1). But when the container tank (No2) still filling water from the primary water tank (Noll) as its gates are open.

[0636] We find T= 1.921 Ton

[0637] Final step

[0638] D8= 0.50 m

[0639] Total D = 9.90 m

[0640] This is the final step of path (2) which is similar to the initial point of the 1stpath

[0641] We find T= 1.752 m

[0642] Total path length =9.90 m & Total wire length = 12.07 m 1stpath

[0643] 2ndpath Calculations of the system (No2)

[0644] • We study system (No2) as follow

[0645] A- Determine the main forces affecting the system during its motion in the 1stpath and 2ndpath.

[0646] B- Using Newton's laws of motion and Archimedes principle to determine the acceleration, velocity and the time of each step in the two paths of the system. As the system begins its motion in the 1stpath by a velocity equal zero and in the final step of path (1) also equal zero.

[0647] The same thing in the 2 path.

[0648] So after we determine the time of each step in both path (1) and path (2) we can know the total time of the complete turn of the system.

[0649] A- Forces affecting the system (No2)

[0650] *we chose the weight (w2) (No8) to be cylindrical empty iron body of total length 4.90 m and the outside diameter of 1.10 m. The upper and the lower parts of the body are 2 cones with a height of 1.65 m. so the volume of body equals(0. 816757r)m3and we assume the weight of this body = 3.60 Ton

[0651] *we chose a path length of the 9.90 m which is equal the total length of the weight (No8)4.90m plus the distance between the water levels of the (2)tanks (Nol2),(Nol3) * we chose the external iron cylinder (Nol4) that will surround the moving iron cylinder (No8)to have an internal diameter of 1.14m so that; the distance between the moving cylinder (No8) and the wall of the external iron cylinder (Nol4) will be 2.0cm. * The friction dynamic force during the motion would not exceed more than 150kg on average.

[0652] * The water resistance force on the moving body (No8) during its motion inside water will be about 125 kg on average that according to its shape and its average velocity inside water as we explained in calculations of system (Nol)

[0653] * The tension force (T) in cables (No3) which we had calculated before will be multiplied by a factor of reduction (fr) which reduce the tension force (T) by a ratio equal the diameter of the iron cylinder (No4) to the diameter of iron ring frame (No5)

[0654] * So; we know: - Volume of w2 = 0.81675n m3

[0655] - Weight w2 =3.60 Ton

[0656] - Friction dynamic force = 0.15 Ton -water resistance = 0.125 Ton

[0657] - Tension force in wires (No3)

[0658] - Total path length = 9.90 m

[0659] - The buoyant force affecting the body W2 (No8) when this body is immersed in water. That according to Archimedes principle Whereas *p= liquid density

[0660] *V = volume of immersed body

[0661] * g = 9.81 ml sec1

[0662] *according to the forces affecting the system. All forces are known except the weight (wl) and the reduction factor (fr) [Note that: the elements (No4) and (No5) are the responsible for reducing tension (T) in cables (No3) to a tension force equals (Txfr) in cables (No6)].

[0663] So, we study the path of motion in both directions based on the main principle that the initial and final velocity of each path must equal zero.

[0664] *By solving the equations of motion in the 2 paths using Newton's laws of motion and Archimedes principle.

[0665] We found that the required weight wl (No7) = 5.573914 Ton and fr= 0.259531

[0666] *After we found (fr) we can determine the diameter of the iron cylinder with horizontal axis (No4) [DI] and the diameter of the iron ring frame (No5) [D2]

[0667] So take: - Dl= 0.64 m

[0668] * D2 = 2.466 m

[0669] And also we can determine of the (2) gears (No4) D3 and the (2) gears (No5) D4

[0670] 9 9

[0671] As D3 = D4 x f

[0672] Jrr x— 12.07

[0673] * D3 = 1.075 m— > D4 = 5.05 m

[0674] Now after we had got all values of the main forces affecting the system as follow:-

[0675] *W1 (No7) = 5.573914 xlO3X 9.81 Newton

[0676] *\N2 (No8) = 3.60 xlO3X 9.81 Newton

[0677] * Average dynamic friction force = 0.15 xl03x 9.81 Newton

[0678] * Average water resistance force = 0.125 xlO3x 9.81 Newton

[0679] * Average tension force in wires (No3) = T [as shown in the table]

[0680] * Average tension force in cables (No6) = T*fr * Gate force = 0.05 x 103x 9.81 Newton

[0681] [Gate force equals the weight of drainage gate and which equals 50 kg and also the filling gate equals 50 kg] *volume of the body (w2) No8 = 0.81675nm3

[0682] *Total path Length = 9.90 m

[0683] *Length of wire (No3) =12.07 m

[0684] So we can calculate the acceleration, velocity and the time of each step in the two paths to calculate the total time for one turn of the system (No2)

[0685] Volume & weight of body “\N2"

[0686] *n(0.572) X 3.0 = 0.9747 n m3

[0687] 0.9747 =(X + 1.65) 0.572

[0688] Drop (1) = 3.0 - 1.86 = 1.14 m n n

[0689] *0.9747 = (x + 1.65 +1.60)0.572- 0.552(

[0690] Total drop = 3.0 - 1.75 =1.25 m =>Drop (2) = 1.25 - 1.14 = 0.11 m

[0691] *0.9747 = (X + 4.90)0.572- 0.5 =0.61 m

[0692] Total drop = 3.0 - 0.61 = 2.39 m „ Drop (3) = 2.39 - 1.25 = 1.14 m

[0693] *W2 =3.60 Ton &volume = 0.81675nm3

[0694]

[0695] *(Cone +cylinder) volume = n x (0.552)(^|^ +1.6) = 0.650375 n

[0696] [0.81675 n + 0.650375 n] / 2 = 0.7335625 n

[0697] [0.650375 n+ 0.166375 n] / 2 = 0.408375 n

[0698] [0.166375 n + 0] / 2 = 0.0831875 n Acceleration, velocity and time of each step in the (2) paths of system (No2)

[0699] 1stpath

[0700] 1ststep

[0701] DI = 0.50 m

[0702] _(5.573914- 2(3.6 - 0.816757T + 0.125)-2.065 X 0.259531- 0.15 + 0.05)xl03x9.81 CL't —

[0703] 1(2 X 3.60 + 5.573914 )xl03

[0704] =2.011911445 m / sec2

[0705] V2= 702+ 2a1D1= 0 + 2arX 0.5

[0706] = 1.418418642 m / sec 7050105 m / se&>

[0707] 2ndstep

[0708] D2 = 2.26 m

[0709] (5.573914 - 2(3.60 - 0.81675TT + 0.125) - 0.15 - 4.106 X 0.259531) X 9.81

[0710] ’ (2 x 3.60 + 5.573914)

[0711] = 1.566716324 m / sec2 015537967 m / sec

[0712] V2= Vi + a2t2t2===>1.0194056 sec

[0713] 3rdstep D3 = 2.24 m

[0714] _(5.573914-2(3.6 - 0.816 + 0.125)-0.15 - 6.032X0.259531)X9.81

[0715] Go — " "

[0716] 5(2X3.60+5.573914)

[0717] =1.182840694m / sec2

[0718] 732= V22+ 2a3£)373=^3.793757443 m / sec

[0719] V3= V2+ a3t3t3^&6579241654 sec

[0720] 4thstep

[0721] D4 = 1.14 m

[0722] _ (5.573914 - 2(3.6 - 0.7335625TT + 0.125) - 0.15 - 6.182 X 0.259531) X 9.81

[0723] 14“ (2 x 3.60 + 5.573914)

[0724] = 0.7515386692 m / sec2

[0725] V2= [Z32+ 2 X a4D4V4=^..013241047 m / sec

[0726] V4= V3+ a4t4t4= 0^2^^0457 sec

[0727] 5thstep

[0728] D5 = 0.11 m

[0729] _ (5.573914 - 2( 3.6 - 0.408375TT + 0.125 ) - 0.15 - 6.127 x 0.259531) X 9.81

[0730] “5“ (2 x 3.60 + 5.573914)

[0731] = -0.8066284469 m / sec2(-ve) y52= y42+ 2a5D575=^=03^70714 m / sec

[0732] V5= V4+ a5t5t5= 0.0271052 sec

[0733] 6thstep

[0734] D6 = 1.14 m 81

[0735] V6= V5+ a6t6t6= 0 >.3132^604 sec

[0736] >

[0737] 7thstep

[0738] D7 = 2.01 m

[0739] > (5.573914 - 2 X 3.60 - 0.15 - 4.363 X 0.259531) X 9.81 a?~ ( 2 X 3.6 + 5.573914 )

[0740] = —2.233583049 m / sec1(— ve)

[0741] V72= y62+2a7D7V7 > = 1.35^98803 m / sec

[0742] K7= V6+ a7t7t7= 0. >865Q4-1 sec

[0743] 8thstep

[0744] D8 = 0.26 m

[0745] > (5.573914 - 2 x 3.60 - 0.15 - 2.425 x 0.259531 - 0.05) x 9.81 aQ~ (2 X 3.6 + 5.573914)

[0746] = —1.885714238 m / sec2(— ve)

[0747] V82= y2+2a8£>8V8> ^44235865767 m / sec

[0748] V8= V7+ a8t8t8= 0-22.8302 sec

[0749] > Final step

[0750] D9 = 0.24 m

[0751] > (5.573914 - 2 x 3.60 - 0.15 - 1.88 x 0.259531 - 0.05 ) x 9.81 a9~ (2 X 3.60 + 5.573914)

[0752] = -1.777088995 m / sec2

[0753] >

[0754] V92= V82+ 2a9D9V9= 0.003074 = 0.0 m / sec

[0755] V9= V8+ a9t9t9- n 51 Q71 Q sec

[0756] Total time of the 1stpath

[0757] = 4.62905 sec

[0758] 2ndoath

[0759] 1ststep

[0760] DI = 0.50 m

[0761] > (2 X 3.60 - 5.573914 - 0.15 + 1.904 X 0.259531 + 0.05) X 9.81 ai~ (2 X 3.60 + 5.573914)

[0762] = 1.551481086 m / sec2

[0763] V2= y02+ 2a1£>1V1> =4^45-584636 m / sec 2ndstep

[0764] D2 1.91 m

[0765] > (2 x 3.60 - 0.15 - 5.573914 + 2.084 x 0.259531) x 9.81 a2~ (2 x 3.60 + 5.573914)

[0766] = 1.548958746 m / sec2

[0767] V22= V2+ 2a2D2I72- > 2.732856288 m / sec

[0768] V2= Vi + a2t2t2= n QApi 75 sec

[0769] 3rdstep

[0770] D3 = 2.59 m

[0771] > (2 x 3.60 - 0.15 - 5.573914 + 2.309 X 0.259531) X 9.81 a3~ (2 x 3.60 + 5.573914)

[0772] = 1.593804029 m / sec2

[0773] V2= V2+ 2a3D3V3> =3.965^101413 m / sec

[0774] 4thstep

[0775] D4= 1.14 m a4

[0776] (2 (3.60 - 0.083187571 - > — ") - 0.15 - 5.573914 + 2.395 X 0.259531") X 9.81

[0777] _ V V _ 2 J _ _

[0778] (2 X 3.6 + 5.573914)

[0779] = 1.113543306 m / sec1

[0780] Z4= V3+ a4t4t4= 0.276734 sec 5thstep

[0781] D5= 0.11 m

[0782] _ (2(3.60 - 0.408375TT - 0.125) - 0.15 - 5.573914 + 2.39 x 0.259531) X 9.81

[0783] “5“ (2 x 3.60 + 5.573914)

[0784] = -0.5525789717 m / sec2(-ve)

[0785] V52= [Z2+ 2a5D575- 4.25ft34)9771 m / sec

[0786] V5= V4+ a5t5t5= O.C2573«-sec

[0787] 6thpath

[0788] D6 = 1.14 m

[0789] _ (2(3.60 - 0.7335625TT - 0.125 ) - 0.15 - 5.573914 + 2.348 x 0.259531) X 9.81

[0790] “6“ (2 x 3.60 + 5.573914)

[0791] = —2.130079388 m / sec2(—ve) v62= 14? + 2a6D6V6T 3.644878423 m / sec 0.25) X 0.259531) X 9.81 (2 x 3.60 + 5.573914)

[0792] = -2.627752433 m / sec2(-ve)

[0793] Notice that: - we assume the water resistance against the container tank (No2) equals 250kg as average during its motion inside water.

[0794] V72= y2+ 2a7£)7V7=1.649719351 m / sec Final step

[0795] D8= 0.50 m as

[0796] (2(3.60 - 0.816757T - 0.125) - 0.15 - 5.573914 + (1.837 - 0.25)0.259531 - 0.05) X 9.81

[0797] “ (2 X 3.60 + 5.573914)

[0798] = —2.72155984 mlsec1( — ve)

[0799] = 4.492749 sec

[0800] *Total time of one turn = 4.62905 + 4.492749 = 9.1218 sec

[0801] *notice that the total time of one turn of system (No2) which equals (9.1218 sec) is less than the total time of one turn of system (Nol) which equals (9.94842 sec ).

[0802] As we must ensure that the required water for system (Nol) is always sufficient.

Claims

Claims

1. System Number (1) contains 24 component work together to generate electrical energy.

2. Weight with a specific weight (Wl): It moves vertically up and down within the system and is represented in the drawings with the number (1).

3. Weight with a specific weight (W2): This is a hollow cylindrical body connected from top and bottom with an empty conical shape with the same diameter as the cylinder, as shown in the drawings. This body must have the same volume as its weight, and it moves vertically up and down within the system. During the ascent and descent, it moves a specified distance inside the water, causing a reduction in its weight during movement, and is represented in the drawings with the number (2).

4. External iron cylinder: This is an iron cylinder in which the body W2 (No 2) moves during the system's operation. It is a vertical cylinder with a larger diameter than the movable body's diameter (No 2), connected from the bottom to a lower water tank, and almost from the middle to an upper water tank through drainage and filling gates. It is represented in the drawings with the number (3).

5. Vertical and horizontal iron cables: These are flexible iron cables with a diameter of (16mm) connected to the bodies (Wl) & (W2), and they move with their movement according to the system's motion. These cables pass over smooth iron rollers, with these rollers being at the same horizontal level, and the distance between them is greater than the length of the path of motion by (1.50 m). They are represented in the drawings with the number (4).

6. The solid metal body attached to the horizontal cables: It moves horizontally on a small circular wheels (Ball Bearings) (which are installed in the bottom of the solid body) to the right and left with the movement of the horizontal cables (No 4) above smooth iron beams. This body is connected by two parallel metal rods (as a hinged connection). The distance between the two parallel rods being greater than the thickness of the large metal disc (No 7) to be rotated. These rods are connected to the solid metal body that moves horizontally on one side and connected to the large metal disc (No 7) on the other side. It is represented in the drawings with the number (5).

7. The two parallel metal rods connected to the solid metal body: They are represented in the drawings with the number (6).

8. The large metal disc: It is rotated by the two parallel metal rods (No 6) connected to it. This large metal disc is used to transfer the circular motion to electrical energy through gears. It is represented in the drawings with the number (7).

9. Smooth iron rollers: Through which the flexible iron cables (No 4) pass, represented in the drawings with the number (8).

10. Four metal arms: Fixed to the upper head of the body W2 (No 2) which shaped like a cross (+) and apply upward pressure to the upper metal disc (No 14) to open and close the drainage gate. They are represented in the drawings with the number (9).

11. Upper water tank: Supplies water to the external iron cylinder through the filling gate, represented in the drawings with the number (10).

12. Lower water tank: Drains excess water from inside the external iron cylinder (No 3) through the drainage gate. It is represented in the drawings with the number (11).

13. Drainage gate: It is a cylindrical metal gate installed inside the external iron cylinder (No 3). It moves vertically (within a distance of 0.6 m) upward to open and drain excess water from the external cylinder to the lower water tank (No 11) and downward to close and prevent water from flowing in both directions. It is connected to flexible metal cables with a diameter of (6mm) that lift it up and down. It is represented in the drawings with the number (12).

14. Cables connected to the drainage gate: These are flexible iron cables with a diameter of (6mm), eight in number, connected from the bottom to the cylindrical drainage gate and connected from the top to the keys of the drainage gate. They are represented in the drawings with the number (13).

15. Upper Metal Disc: It is a metal disc with a diameter smaller than the inner cylinder's diameter, (No 3). It is installed inside the external cylinder (No 3) and has (8) gate keys installed in it. It can move up and down within a distance of (0.60 m) to open and close the drainage gate. This metal disc is installed insidethe external iron cylinder (No 3) at the top of the cylinder at the specified level shown in the drawings, and is represented by the number (14) in the drawings.

16. Filling Gate: It is a cylindrical metal gate that is installed directly outside the external iron cylinder (No 3) as shown in the drawings. It moves vertically (within a distance of 0.6 m) upward to open and allow the cylinder to fill with water from the upper water tank to the required level. It moves downward to close and prevent the flow of water in both directions. This gate is connected to (8) flexible metal cables with a diameter of (6 mm) that raise and lower the gate. It is represented by the number (15) in the drawings.

17. Iron Cables Connected to the Filling Gate: These are flexible iron cables with a diameter of (6 mm), (8) in total, connected to the filling gate on one end and to the filling gate keys on the other end. They pass through smooth metal rollers and are represented by the number (16) in the drawings.

18. Smooth metal rollers: Through which the flexible iron cables (No 16) connected to the filling gate pass. They are represented in the drawings with the number (17).

19. Lower metal disc: This is a metal disc installed inside the external iron cylinder (No 3). This disc contains (8) keys for the filling gate (No 15) and it can move up and down within a range of (0.60 m). This movement opens and closes the filling gate (No 15). This metal disc is installed at the bottom of the external cylinder (No 3) from inside and at the specified level in the drawings. This is represented as number (18) in the drawings.

20. Protrusions for pressure of the movable cylinder: These are four protrusions at the base of the lower cone of the movable body W2 (No 2) as shown in the drawings. Their function is to apply pressure to the lower metal disc (No 18) to open and close the filling gate. They are represented as number (19) in the drawings.

21. Iron Protrusions: Iron protrusions in the shape of the letter (U) are created on the external iron cylinder (No 3) from the inside to control the movement of the body (w2) vertically without rotating around its axis during motion. The location of these protrusions inside the iron cylinder (No 3) is illustrated in the drawings and represented as number (20).

22. (2) Iron Beams: These beams are in the shape of the letter (I) and are horizontally fixed. The solid metal body (No 5) moves on them, and another beam in the shape of the letter (U) is fixed above each beam to maintain the path of the solid metal body (No 5) during motion. They are represented as number(21) in the drawings.

23. (2) Iron Beams: Shaped like the letter (U) with an internal width of (22 cm), they are welded above the iron beams (B.F.I) (No 21). The solid metal body (No 5) moves inside these beams as shown in the drawings and is represented as number(22).

24. Small Circular Wheels (Ball Bearings): These are attached to the bottom of the solid metal body (No 5) to facilitate the movement of the solid metal body with minimal friction. They are represented as number (23) in the drawings.

25. All elements of system number (1 ) work together to convert the straight horizontal motion of the solid metal body (No 5) into circular motion of the large metal disk (No 7) to generate electrical power.

26. System Number (2) contains 24 component work together to re-raise water from the lower tank to the upper tank mechanically without any kind of fuel.

27. Metal lever: It consists of iron plates with a thickness of (10mm) arranged in the shape of the letter (U). Its clear internal width is (Im), and its height is variable, starting from (20cm) and reaching a maximum height of (180cm) as shown in the drawings. These iron plates are supported by three iron beams (S.I.B 500*158 mm) through welding. These iron beams are supported by a sturdy steel rod with a suitable diameter perpendicular to the direction of the beams, allowing the lever to rotate around this rod vertically. The iron plates and beams are fastened together using a bracing method to make the lever function as a single unit. It is represented as number (1) in the drawings.

28. Container: The metal lever is connected to a semi-circular metal tank (container) with a width of (1.0m) and a radius of (1.80m) as shown in the drawings. This metal container is made of iron plates with a thickness of (10mm) and a clear width of (1.0m), which is the same as the lever's width. Hinged gates are created on the container's body, opening inwards. When the container descends into the water, these gates open under the pressure of the water to allow water to enter thecontainer. These gates are hinged to the container's body and connected to a metal spring on one side and to the inner body of the container on the other side, ensuring that the gates remain closed in the absence of external forces, as shown in the drawings. This container is supported by three iron beams (S.I.B 500*158 mm) through welding. A bracing system is used to connect the container to the beams, allowing the container and beams to function as a single unit during the system's operation. It is represented as number (2) in the drawings.

29. Lever Tension Cables: These are (3) flexible iron cables with a diameter of (16mm) connected to the iron beams of the lever (No 1) on one end and to an iron cylinder with a specified diameter, (No 4) which rotates around its axis, on the other end. These cables wind around the iron cylinder during the lever's rotational movement in the first path and unwind when the iron cylinder rotates in the opposite direction in the second path. These cables are always under tension throughout the system's operation, as shown in the drawings. They are represented as number (3) in the drawings.

30. Iron Cylinder with Horizontal Axis: This is a cylinder with a specified diameter, and it has three cables number (3) attached to it. The cables wind around its circumference in the first path and unwind in the second path as this cylinder can rotate around its axis in both directions. It is supported by (2) iron gears with a larger diameter than the cylinder's diameter, as shown in the drawings. Each gear is fixed at one end of the cylinder, and the distance between these two gears is (135cm). These gears rotate around the same axis and take the same direction as the iron cylinder, and they are represented as number (4) in the drawings.

31. Gear Reduction for Tensile Force and Motion Path Adjustment: This consists of two iron gears with a diameter larger than the gears referred to as (No 4) in the drawings. These two gears rotate together around a horizontal axis, which is a steel rod with a diameter of 10 cm. Each gear is fixed to one end of the rod in such a way that the clear distance between the two gears is 135 cm, allowing lever (No 1), which has an inner width of 1.0m, to move between these two gears. The lever (No 1) moves vertically along its path, starting in a completely vertical position. It then follows a circular motion vertically perpendicular to the axis of iron cylinder (No 4) until it reaches the end of the first path, where it tilts horizontally at an angle of 1.85 degrees, as illustrated in the drawings. During thispath, it passes between the two gears, making contact with the outer surface of iron cylinder (No 4) at the end of path number (1). It then returns in path number (2), rotating in the opposite direction to its rotation in path number (1) until it reaches the vertical position again. Since lever (No 1) passes between the two gears, the clear distance between the gears has been set to 135 cm, ensuring a minimum distance of 10 cm between the lever's side and the adjacent gear. Each of these two gears is connected to a gear specified as (No 4), and the motion and torque are transmitted through both gears together by engaging their teeth. When one of the interlocking gears rotates clockwise, the other gear rotates counterclockwise. Three iron ring frames are vertically installed along the steel axis at equal distances between the two gears, as specified in the drawings, with each iron ring frame having a width of 8 cm. The three iron ring frames and the two gears rotate together around the same steel axis. Each of the three iron ring frames is fitted with a flexible iron cable with a diameter of 16 mm, and all three cables are connected to weight W1 (No 7). When this weight moves downward, the iron ring frames rotate, and the cables are unwound. When the weight moves upward, the iron ring frames rotate in the opposite direction, and the cables are wound around the periphery of these ring frames. The combination of the three ring frames and the two gears in system number (2) reduces the tension in the cables (No 3) by a factor of the diameter of iron cylinder (No 4) compared to the diameter of the iron ring frame. This combination also ensures the alignment of the length of cable path (No 3), which is 12.07 m, with the length of the body's motion path W 1 (No 7), which is 9.90m. This assembly of frames and gears is represented as number (5) in the drawings.

32. Vertical Iron Cables Connected to Body (Wl): There are three iron cables with a diameter of 16 mm, connected to body (Wl) from one end and to the three ring frames (No 5) from the other end. They are represented as number (6) in the drawings.

33. Body (Wl): It is a body with a specific weight, defined by calculations, and also with a specific shape in the drawings. It is connected to the three cables (No 6) and also to (4) other cables, (2) on each side, as shown in the drawings. This body moves vertically up and down along a path with a length of 9.9 meters within system number (2), represented as number (7) in the drawings.

34. Body (W2): It consists of two bodies with specific shapes and weights. Each of them consists of a cylindrical body connected from the bottom and top to a hollow conical shape with the same diameter as the hollow cylinder, as shown in the drawings. These bodies move up and down within system number (2) along a path of 9.90 meters. We must note that part of this path is submerged in water, which causes a decrease in weight during movement, essential for the system's operation. They are represented as number (8) in the drawings.

35. Vertical and Horizontal Iron Cables: There are four flexible iron cables with a diameter of 16 mm. Each pair of cables is connected to one side of body W1 (No 7) on one end and to components of body W2 (No 8) on the other end, passing through smooth iron rollers. These rollers are at the same horizontal level, as shown in the drawings and are represented as number (9).

36. Smooth Iron rollers: These are four smooth iron rollers through which cable number (9) passes. They are represented as number (10) in the drawings.

37. Primary Water Tank: It is a low-level water tank from which water is lifted to the final water tank with a higher level. Container (No 2) is filled from this tank, which is ultimately placed in the final water tank (No 12). It is represented as number (11) in the drawings.

38. Final Water Tank: It is the water tank with the higher level, into which water is lifted from the primary water tank (No 11). It is represented as number (12) in the drawings.

39. Upper Water Tank: It is a water tank with a capacity of no more than 30m3. This tank receives water from tank (No 11) through the lever (No 1), and then it distributes the water to two external iron cylinders (No 14). These cylinders are the ones in which body W2 (No 8) moves. Finally, this water reaches the final water tank (No 12). It is represented as number (13) in the drawings.

40. Two External Iron Cylinders: These are two iron cylinders in which body W2 (No 8) moves during the operation of system number (2). Each cylinder is connected from the bottom to the final water tank (No 12) and roughly from the middle to the upper water tank (No 13) through drainage and filling gates. They are represented as number (14) in the drawings.

41. Metal Arms: There are four metal arms attached to the upper part of each movable body (No 8), shaped like a cross (+), as shown in the drawings. These metal arms apply upward pressure to the drainage gate keys to lift the gate and allow water to flow. They are represented as number (15) in the drawings.

42. Drainage Gates: These are two cylindrical metal gates installed, with one gate in each external iron cylinder (No 14). They move vertically upwards to open and allow excess water to flow from the external cylinder (No 14) into the final water tank (No 12). Then, they move downwards to close the gate, preventing water from flowing in both directions. Each gate is connected to eight flexible iron cables with a diameter of 6mm, which raise and lower the gate. They are represented as number (16) in the drawings.

43. Iron Cables Connected to the Drainage Gate: There are (8) flexible iron cables with a diameter of (6 mm) for each external cylinder number (14). They are connected from below to the cylindrical drainage gate (No 16) and from above to the drainage gate keys. They are represented by the number (17) in the drawings.

44. Upper Metal Disc: It consists of (2) metal discs, with one disc installed in each external cylinder (No 14) from the inside. Each of them has (8) gate keys. They can move up and down within a distance of (0.50 m) to open and close drainage gate (No 16). This metal disc is installed at the top of each external cylinder (No 14) from the inside, at the specified level shown in the drawings, and is represented by the number (18) in the drawings.

45. Filling Gate: There are (2) cylindrical metal gates installed, with one gate in each external iron cylinder (No 14), but from the outside, as clearly shown in the drawings. They move vertically upward to open and allow filling of cylinder (No 14) with water from the upper water tank (No 13) to the required level. They move downward to close and prevent the flow of water in both directions. Each gate is connected to (8) flexible iron cables with a diameter of (6 mm) that raise and lower the gate. They are represented by the number (19) in the drawings.

46. Flexible Iron Cables Connected to Filling Gates: These are eight flexible iron cables with a diameter of 6mm in each external cylinder (No 14). They are connected to the cylindrical filling gate (No 19) from one side and connected tothe keys of the filling gate from the other side. They pass through smooth metal rollers and are represented as number (20) in the drawings.

47. Smooth Iron rollers: These are the rollers through which the flexible iron cables (No 20) connected to the filling gate (No 19) pass. They are represented as number (21) in the drawings.

48. Lower Metal Disc: There are two metal discs installed, with one disc in each external iron cylinder (No 14) from the inside. Each disc contains eight keys for the filling gate (No 19), and it can move up and down within a range of 0.50 meters. This movement opens and closes the filling gate (No 19). These metal discs are installed at the bottom of the external cylinder (No 14) from the inside and at the specified level in the drawings. They are represented as number (22) in the drawings.

49. Protrusions for Pressure of the Movable Cylinder: These are four metal protrusions at the base of the lower cone of the movable body (No 8), as shown in the drawings. Their function is to apply pressure to the lower metal disc (No 22) to open and close the filling gate. They are represented as number (23) in the drawings.

50. Metal Protrusions: Iron protrusions in the shape of a "U" are created on the inside of the external iron cylinder (No 14) to control the vertical movement of body W2 (No 8) without rotation around its axis during vertical movement. The location of these protrusions inside the iron cylinder (No 14) is indicated in the drawings. They are represented as number (24) in the drawings.

51. All elements of system number (2) work together to lift water from the low- level tank (No 11) to the high-level tank (No 12) without human intervention or the use of any type of fuel.

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