Magnetic and gravity combined power generation device and design method thereof

Abstract

The invention relates to a magnetic and gravity combined power generation device and a design method thereof. The device comprises a gear box; the output end of the start-stop motor is connected with the first input end of the gear box; the transmission device comprises a transmission shaft and a mechanical arm, the first end of the mechanical arm is connected to the transmission shaft, and the transmission shaft is connected with the second input end of the gearbox; the gravity pendulum comprises a swing arm and a pendulum bob, the first end of the swing arm rotates around a rotation axis parallel to the axis of the transmission shaft, the pendulum bob is installed at the second end of the swing arm, and the second end of the mechanical arm is connected to the swing arm in a sliding mode and can do reciprocating rectilinear motion between the first end and the second end of the swing arm; the magnetic device comprises a magnetic rotor assembly and a magnetic stator assembly; and the generator set comprises a generator and a flywheel which is connected with the generator and the output end of the gear box. Through the combination of the gravity pendulum and the magnetic device, the energy conversion efficiency of the system is optimized, and the flywheel is utilized to realize energy storage and release, so that the power generation process is more stable and efficient.

Description

Technical Field

[0001] The present invention relates to the technical field of generators, in particular to a magnetic and gravity combined power generation device and a design method thereof. Background Art

[0002] At present, electricity is the core power source for the operation of modern society. The production method of electricity directly affects the global energy utilization efficiency and environmental sustainability. The existing electricity production methods mainly include thermal power generation, hydropower generation, solar power generation, wind power generation, nuclear power generation, hydrogen power generation and energy storage power generation. Each power generation method has its advantages and disadvantages in terms of technical realization, economy and environmental friendliness, but also faces its own limitations and challenges:

[0003] Thermal power generation: Relies on fossil fuels such as coal, oil and natural gas. Although thermal power generation has the advantages of mature technology, low cost and stable power generation, burning fossil fuels will produce a large amount of harmful gases such as carbon dioxide, sulfur dioxide and nitrogen oxides, leading to air pollution and greenhouse effect, which has a great impact on the environment.

[0004] Hydropower: Using the kinetic energy of river water to generate electricity has the advantages of being highly sustainable, pollution-free and low in operating costs. However, hydropower accounts for a relatively small share of global electricity production, and its development is constrained by the geographical environment and the availability of water resources, making it difficult to be widely used in arid areas.

[0005] Solar and wind power generation: As clean energy, solar and wind power generation play an important role in the global energy transformation, but their energy sources are limited by natural conditions and power generation is unstable. Solar energy needs to operate under sufficient sunshine conditions, while wind energy depends on changes in wind speed. These characteristics make it difficult for solar and wind energy to be used as basic load power sources to achieve all-weather power supply. In addition, the construction of solar and wind power stations usually requires a lot of land resources, especially large-scale photovoltaic power stations and wind farms.

[0006] Hydrogen power generation: Hydrogen is an efficient and clean energy carrier. The combustion of hydrogen only produces water and no carbon dioxide. However, hydrogen power generation requires the electrolysis of water to produce hydrogen, and this process itself consumes a lot of electricity, increasing the cost of power generation. In addition, the storage and transportation of hydrogen also face technical challenges and safety risks.

[0007] Nuclear power generation: With its high energy density and stable power generation, it has become the mainstay of basic power supply in some countries. However, nuclear power generation has problems such as complex technology, high construction cost, radioactive waste disposal and nuclear leakage risks, which make nuclear power generation less accepted by the society.

[0008] Energy storage power generation: Energy storage technology is usually used to balance the imbalance between electricity supply and demand. It stores electricity when electricity prices are low and releases it when electricity prices are high, thus achieving the function of peak load shifting and valley filling. Although energy storage technology helps to improve the stability and dispatchability of the power grid, it is not an independent power generation method. It still relies on other power generation methods in essence and cannot fundamentally solve the problems of energy consumption and pollution.

[0009] In summary, existing electricity production methods have more or less problems such as energy consumption, environmental pollution and power generation stability. In order to cope with the global energy crisis and environmental pollution problems, a greener, low-carbon and sustainable electricity production method is urgently needed, which can reduce dependence on fossil fuels while improving the stability and efficiency of power generation. Summary of the invention

[0010] To this end, the present invention provides a magnetic and gravity combined power generation device and a design method thereof. The invention converts gravitational potential energy into mechanical energy and then converts mechanical energy into electrical energy by utilizing the conversion of gravitational potential energy and the assistance of a magnetic device. The device can not only continuously generate electrical energy in a natural environment, but also does not consume any fossil energy during the power generation process, thus avoiding the emission of harmful gases and having a truly green energy production capability. By combining a gravity pendulum and a magnetic device, the energy conversion efficiency of the system is optimized, and a flywheel is used to store and release energy, making the power generation process more stable and efficient, thereby solving the energy consumption and environmental pollution problems faced by current power generation methods to a certain extent.

[0011] In order to solve the above technical problems, the present invention provides a magnetic and gravity combined power generation device, comprising:

[0012] Gearbox;

[0013] A start-stop motor, an output end of which is connected to the first input end of the gear box;

[0014] A transmission device, comprising a transmission shaft and a mechanical arm, wherein a first end of the mechanical arm is connected to the transmission shaft, and the transmission shaft is connected to the second input end of the gear box to drive the first end of the mechanical arm to rotate around the axis of the transmission shaft;

[0015] A gravity pendulum, comprising a swing arm and a pendulum, wherein the first end of the swing arm rotates around a rotation axis parallel to the axis of the transmission shaft, the pendulum is mounted on the second end of the swing arm, and the second end of the mechanical arm is slidably connected to the swing arm and can perform reciprocating linear motion between the first end and the second end of the swing arm;

[0016] A magnetic device, comprising a magnetic mover assembly and a magnetic stator assembly, wherein the magnetic mover assembly is mounted on the pendulum, the magnetic stator assembly is arranged in the circumferential direction of the swing of the magnetic mover assembly, and magnetic pole repulsion can be generated between the magnetic mover assembly and the magnetic stator assembly when they are close to each other; the magnetic mover assembly and the magnetic stator assembly include a plurality of groups of permanent magnets arranged correspondingly;

[0017] A generator set, comprising a generator, a flywheel connected to the generator and to an output end of the gearbox;

[0018] The gravity pendulum converts gravitational potential energy into mechanical energy by swinging, and the swing of the gravity pendulum can drive the mechanical arm to make a circular motion, so as to drive the transmission shaft to rotate, and further drive the generator to generate electricity.

[0019] The present invention also provides a design method for a magnetic and gravity combined power generation device, the design method comprising:

[0020] S1. Determine the rated power P of the generator n and the rated speed of the generator n N ;

[0021] S2, according to the rated power P of the generator n , and the torque M of the generator is obtained n ;

[0022] S3, according to the rated speed n of the generator N , get the speed n of the gearbox Z ;

[0023] S4, determine the length L of the mechanical arm, and obtain the length l of the gravity pendulum according to the length L of the mechanical arm;

[0024] S5. According to the length L of the mechanical arm and the length l of the gravity pendulum, the maximum swing angle θ of the gravity pendulum during operation is obtained;

[0025] S6. According to the rated power P of the generator n , the mechanical loss of the gearbox P1, the mechanical loss of the flywheel P2, the loss of the generator P3, the loss of the start-stop motor P4, and the required mechanical input power P of the gearbox are obtained. s ;

[0026] S7, according to the mechanical input power P of the gearbox s , we get the gearbox mechanical input power ∑W, which is the work done by the gravity pendulum in the unit;

[0027] S8. Determine the parameters of the swing arm and the pendulum in the gravity pendulum according to the mechanical input power ∑W of the gear box;

[0028] S9. Determine the output power P of the gravity pendulum according to the parameters of the swing arm and the pendulum in the gravity pendulum;

[0029] S10. Determine the total power loss P of the gravity pendulum 总损失 According to the output power P of the gravity pendulum and the total power loss, determine the unconverted excess power P of the gravity pendulum in the second half of the cycle d ;

[0030] S11, according to the gravity pendulum unconverted excess power P in the second half cycle d , determine the force F1 produced by the unconverted excess power of the gravity pendulum in the second half cycle;

[0031] S12. Determine the magnetic pole repulsion F of the magnetic device m , determine the number of permanent magnet groups N according to the force F1 generated by the unconverted excess power of the gravity pendulum in the second half of the cycle and the force F1 generated by the unconverted excess power of the gravity pendulum in the second half of the cycle g ;

[0032] S13, determining the maximum surplus work and the maximum deficit work of the gravity pendulum according to the parameters of the swing arm and the pendulum in the gravity pendulum;

[0033] S14, according to the maximum surplus work W of the gravity pendulum max and the maximum work loss W min , determine the parameters of the flywheel device.

[0034] In one embodiment of the present invention, the speed n of the gearbox Z The calculation is as follows:

[0035]

[0036] Where: n Z is the gearbox speed; n N is the rated speed of the generator; b1 is the first-stage speed ratio; b2 is the second-stage speed ratio.

[0037] In one embodiment of the present invention, the maximum swing angle θ of the gravity pendulum during operation is calculated as follows:

[0038] l=2L+0.3

[0039]

[0040]

[0041] Where: θ is the maximum swing angle of the gravity pendulum; L is the length of the robot arm; l is the length of the gravity pendulum.

[0042] In one embodiment of the present invention, the mechanical input power P of the gearbox s The calculation is as follows:

[0043] P s =P n +P1+P2+P3+P4

[0044] Where: P s is the mechanical power input to the gearbox; P n is the rated power of the generator; P1 is the mechanical loss of the gearbox; P2 is the mechanical loss of the flywheel, which is initially calculated as 2% of the total power, and then corrected when the flywheel parameters are determined; P3 is the loss of the generator; P4 is the loss of the start-stop motor.

[0045] In one embodiment of the present invention, step S8 includes determining the mass m1 of the pendulum, the diameter d of the pendulum, the height h of the pendulum, the length l1 of the swing arm, and the mass m2 of the swing arm; wherein the mass m1 of the pendulum is calculated as follows:

[0046] ∑W=m1gl(1-cosΦ)

[0047]

[0048] Where: ∑W is the work done by the potential energy of the gravity pendulum; m1 is the mass of the gravity pendulum; g is the acceleration of gravity; l is the length of the gravity pendulum; Φ is the representative swing angle;

[0049] The diameter d of the pendulum is calculated as follows:

[0050]

[0051] Where: d is the diameter of the pendulum; m1 is the mass of the gravity pendulum; ρ is the density of the gravity pendulum material; when the mass of the gravity pendulum is large, the coefficient takes a small value, otherwise it is the opposite;

[0052] The height h of the pendulum is calculated as follows:

[0053] h=(0.2~0.5)d

[0054] Where: h is the height of the gravity pendulum; d is the diameter of the gravity pendulum;

[0055] The length l1 of the swing arm is calculated as follows:

[0056]

[0057] Where: l1 is the length of the gravity pendulum arm; l is the length of the gravity pendulum; d is the diameter of the gravity pendulum;

[0058] The mass m2 of the swing arm is calculated as follows:

[0059] m2=l1b1c1ρ-(2L+0.1)b2c2ρ+0.02b1l1ρ

[0060] In the formula: m2 is the mass of the swing arm; l1 is the length of the swing arm; b1 is the width of the swing arm; c1 is the thickness of the swing arm; b2 is the width of the slide; c2 is the depth of the slide; ρ is the steel density; L is the length of the robotic arm.

[0061] In one embodiment of the present invention, in step S9, the output power P of the gravity pendulum is calculated as follows:

[0062] P=Fv

[0063] Where: P is the power of the work done by the pendulum; F is the force at the lowest point of the pendulum; v is the speed of the pendulum at the lowest point;

[0064] F=m1α

[0065] Where: F is the force at the lowest point of the pendulum; m1 is the mass of the pendulum; α is the angular acceleration of the pendulum;

[0066]

[0067] Where: α is the angular acceleration of the pendulum; v A is the speed of the pendulum at its lowest point; v0 is the starting speed of the pendulum; t is the time it takes for the pendulum to swing to its lowest point;

[0068]

[0069] Where: v A is the speed of the gravity pendulum at the lowest point; T is the kinetic energy; j o is the moment of inertia of the pendulum about its axis; m1 is the mass of the pendulum; m2 is the mass of the pendulum arm; r is the radius of the pendulum; ω is the angular velocity of the pendulum; g is the acceleration due to gravity; l is the length of the pendulum; and Φ is the pendulum angle.

[0070] In one embodiment of the present invention, in step S10, the total power loss P of the gravity pendulum is 总损失 The calculation is as follows

[0071] P 总损失 =P f1 +P m +P f2

[0072] Where: P 总损失 is the total power loss of the gravity pendulum; P f1 is the friction loss of rolling bearing; m is the friction loss power of the pendulum; P f2 Power is lost due to air resistance;

[0073] The unconverted excess power P of the gravity pendulum in the second half of the cycle d The calculation is as follows:

[0074]

[0075] Where: P d is the unconverted excess power of the gravity pendulum in the second half cycle; P is the power of the work done by the gravity pendulum; α1 is the angle of the gravity pendulum in the first half cycle; α2 is the angle of the gravity pendulum in the second half cycle.

[0076] In one embodiment of the present invention, the number of permanent magnet groups N g The calculation is as follows:

[0077]

[0078] Where: N g is the number of permanent magnet groups; F1 is the force generated by the unconverted excess power of the gravity pendulum in the second half cycle; F m is the repulsion force of the permanent magnet poles;

[0079]

[0080] Where: F m is the repulsion force of the permanent magnet poles; B δ is the air gap magnetic induction intensity of the permanent magnet; μ0 is the absolute magnetic permeability of vacuum, μ0=4π×10 -7 H / m; S m is the magnetic pole area of ​​the permanent magnet;

[0081]

[0082] Where: F1 is the force generated by the unconverted excess power of the gravity pendulum in the second half of the cycle; v a is the absolute velocity of the gravity pendulum;

[0083] P z =P d ≥P 总损失

[0084] Where: P z Generate power for the magnetic device; P d is the unconverted excess power of the gravity pendulum in the second half cycle; P s总 is the total power loss of the gravity pendulum.

[0085] In one embodiment of the present invention, in step S14, the moment of inertia J of the flywheel F The calculation is as follows:

[0086]

[0087] Where: J F is the flywheel moment of inertia; W n is the maximum profit or loss work; m is the difference between the maximum angular velocity and the minimum angular velocity; δ is the velocity non-uniformity coefficient;

[0088] W n =W max +W min

[0089]

[0090] Where: W max is the maximum surplus work; W min is the maximum work loss; m1 is the mass of the pendulum; m2 is the mass of the pendulum arm; g is the acceleration of gravity; l is the pendulum length; is the maximum swing angle of the gravity pendulum; is the angle at which the pendulum stops during operation; t1 and t2 are the time periods representing the maximum surplus work and the maximum deficit work respectively, which are obtained based on the period of the pendulum;

[0091] The average diameter D of the flywheel is calculated as follows:

[0092]

[0093] Where: D is the average diameter of the flywheel; v p is the permissible peripheral speed; n m is the flywheel speed;

[0094] The mass m0 of the flywheel is calculated as follows:

[0095]

[0096] Where: m0 is the flywheel edge mass;

[0097] The width b and thickness H of the flywheel rim are calculated as follows:

[0098]

[0099] Where: b is the width of the flywheel rim; J F is the moment of inertia of the flywheel; ρ is the flywheel density; k is the coefficient (1-2); D is the average diameter of the flywheel;

[0100] H = kb

[0101] Where: H is the flywheel rim thickness.

[0102] The above technical solution of the present invention has the following advantages compared with the prior art:

[0103] The magnetic and gravity combined power generation device and its design method described in the present invention realizes the conversion of gravitational potential energy into mechanical energy through a gravity pendulum, and then the mechanical energy drives the generator to generate electricity. The magnetic repulsion is used to compensate for the mechanical friction and air resistance losses, so as to realize the long-term stable swing of the gravity pendulum, further improve the energy conversion efficiency, and reduce the energy loss caused by friction and resistance.

[0104] The design method of this invention starts from the energy balance of the system, takes into account the rated power of the generator, mechanical loss, gearbox transmission ratio and other factors, and systematically determines the parameters of the gravity pendulum and its various components, so that the device can achieve the best operating state under different working conditions. By optimizing the design parameters such as the swing angle of the gravity pendulum, the number and position of the magnetic poles of the magnetic device, and the inertia moment of the flywheel, it can be flexibly adjusted according to different power generation needs, improving the adaptability and scalability of the system. BRIEF DESCRIPTION OF THE DRAWINGS

[0105] In order to make the contents of the present invention more clearly understood, the present invention is further described in detail below based on specific embodiments of the present invention in conjunction with the accompanying drawings.

[0106] Figure 1 It is a schematic diagram of the overall structure of one side of the magnetic and gravity combined power generation equipment of the present invention.

[0107] Figure 2 It is a schematic diagram of the overall structure of the other side of the magnetic and gravity combined power generation equipment of the present invention.

[0108] Figure 3 It is a structural schematic diagram of the gravity pendulum of the present invention.

[0109] Figure 4 It is a structural schematic diagram of the magnetic mover assembly of the present invention.

[0110] Figure 5 It is a schematic structural diagram of the first magnet base of the present invention.

[0111] Figure 6 It is a structural schematic diagram of the magnetic stator assembly of the present invention.

[0112] Figure 7 It is a schematic structural diagram of the second magnet base of the present invention.

[0113] Figure 8 It is a schematic diagram of the motion state of the gravity pendulum of the present invention at different positions.

[0114] Description of the Figures in the Specification:

[0115] 1. Start-stop motor; 11. Drive shaft; 12. First gear;

[0116] 2. Transmission device; 21. Transmission shaft; 22. Mechanical arm; 23. Slider; 24. Gear shaft; 25. Second gear; 26. First intermediate gear; 27. Second intermediate gear; 28. Gear box;

[0117] 3. Gravity pendulum; 31. Swing arm; 311. Slide; 32. Pendulum; 33. Swing shaft; 34. Swing shaft bearing seat; 35. Strengthening arm;

[0118] 4. Magnetic mover assembly; 41. First magnet base; 42. First permanent magnet; 421. First magnet surface; 422. First permanent magnet unit; 43. First permanent magnet intermediate partition; 44. First permanent magnet side baffle;

[0119] 5. Magnetic stator assembly; 51. Second magnet base; 52. Second permanent magnet; 521. Second magnet surface; 522. Second permanent magnet unit; 53. Stator fixing seat; 54. Stator reset holding mechanism; 541. Mounting seat; 542. First hinged arm; 543. Second hinged arm; 55. Stopper; 56. Stator bearing seat; 57. Second permanent magnet middle partition; 58. Second permanent magnet side baffle;

[0120] 6. Generator set; 61. Generator; 62. Flywheel; 63. Flywheel shaft; 64. Flywheel shaft gear;

[0121] 8. Electrical system; 81. Start / stop motor control cabinet; 82. Rectifier and battery cabinet; 83. Inverter AC output cabinet. DETAILED DESCRIPTION

[0122] The present invention is further described below in conjunction with the accompanying drawings and specific embodiments so that those skilled in the art can better understand the present invention and implement it, but the embodiments are not intended to limit the present invention.

[0123] In the present invention, if directions (up, down, left, right, front and back) are described, it is only for the convenience of describing the technical solution of the present invention, and does not indicate or imply that the technical features referred to must have a specific orientation, be constructed and operate in a specific orientation. Therefore, it cannot be understood as a limitation of the present invention.

[0124] In the present invention, "several" means one or more, "multiple" means more than two, "greater than", "less than", "exceed" and the like are understood to exclude the number itself; "above", "below", "within" and the like are understood to include the number itself. In the description of the present invention, if there is a description of "first" or "second", it is only used for the purpose of distinguishing technical features, and cannot be understood as indicating or implying relative importance or implicitly indicating the number of the indicated technical features or implicitly indicating the order of the indicated technical features.

[0125] In the present invention, unless otherwise clearly defined, the words "set", "install", "connect" and the like should be understood in a broad sense, for example, they can be directly connected or indirectly connected through an intermediate medium; they can be fixedly connected or detachably connected or integrally formed; they can be mechanically connected or electrically connected or able to communicate with each other; they can be the internal connection of two elements or the interaction relationship between two elements. Those skilled in the art can reasonably determine the specific meanings of the above words in the present invention in combination with the specific content of the technical solution.

[0126] Example 1

[0127] Reference Figure 1 As shown, a magnetic and gravity combined power generation device comprises:

[0128] Gear box 28;

[0129] The start-stop motor 1, whose output end is connected to the first input end of the gear box 28, and the start-stop motor 1 adopts a reluctance motor;

[0130] The transmission device 2 comprises a transmission shaft 21 and a mechanical arm 22, wherein a first end of the mechanical arm 22 is connected to the transmission shaft 21, and the transmission shaft 21 is connected to a second input end of the gear box 28 to drive the first end of the mechanical arm 22 to rotate around the axis of the transmission shaft 21;

[0131] The gravity pendulum 3 comprises a swing arm 31 and a pendulum 32. The first end of the swing arm 31 rotates around a rotation axis parallel to the axis of the transmission shaft 21. The pendulum 32 is mounted on the second end of the swing arm 31. The second end of the mechanical arm 22 is slidably connected to the swing arm 31 and can perform reciprocating linear motion between the first end and the second end of the swing arm 31.

[0132] A magnetic device, comprising a magnetic mover assembly 4 and a magnetic stator assembly 5, wherein the magnetic mover assembly 4 is mounted on the pendulum 32, and the magnetic stator assembly 5 is arranged in the circumferential direction of the swing of the magnetic mover assembly 4, and magnetic pole repulsion can be generated between the magnetic mover assembly 4 and the magnetic stator assembly 5 when they are close to each other; the magnetic mover assembly 4 and the magnetic stator assembly 5 include a plurality of groups of permanent magnets arranged correspondingly;

[0133] A generator set 6, comprising a generator 61, and a flywheel 62 connected to the generator 61 and to an output end of the gearbox 28;

[0134] The gravity pendulum 3 converts gravitational potential energy into mechanical energy by swinging. The swing of the gravity pendulum 3 can drive the mechanical arm 22 to make a circular motion, so as to drive the transmission shaft 21 to rotate, and further drive the generator 61 to generate electricity.

[0135] Due to the friction, resistance and other factors in the mechanical system (such as bearing friction, gear friction, air resistance, etc.), these will cause energy loss. The magnetic repulsion is provided by the magnetic device, and the additional magnetic energy is used to supplement these mechanical losses, ensuring that the gravity pendulum 3 can continue to swing, thereby maintaining the continuity of the power generation process. While using the magnetic device, the flywheel 62 is used for energy storage to ensure that the gravity pendulum 3 can operate stably and continuously during operation.

[0136] In one embodiment, referring to Figure 3 As shown, the gravity pendulum 3 also includes a pendulum shaft 33, and a pendulum shaft bearing seat 34 is provided on the pendulum mounting crossbeam to support the rotation of the pendulum shaft 33. The gravity pendulum 3 includes a swing arm 31 and a pendulum 32, and the first end of the swing arm 31 is connected to the pendulum shaft 33. It can be understood that the axis of the pendulum shaft 33 is the rotation axis. In order to make the gravity pendulum 3 swing stably, a reinforcing arm 35 is provided which is arranged parallel to the swing arm 31 along the axial direction of the pendulum shaft 33, the first end of the reinforcing arm 35 is connected to the pendulum shaft 33, and the second end of the reinforcing arm 35 is connected to the pendulum 32.

[0137] As shown in the figure, a slide groove 311 is provided along the direction from the first end to the second end of the swing arm 31 , and a slider 23 slidably connected to the slide groove 311 is installed at the second end of the mechanical arm 22 .

[0138] In one embodiment, a magnetic mover assembly 4 is disposed on both radial sides of the magnetic mover assembly 4 .

[0139] Reference Figure 4 , Figure 5 As shown, the magnetic mover assembly 4 includes a first magnet base 41 installed at the bottom of the pendulum 32, the first magnet base 41 is installed with a first permanent magnet 42, and the first permanent magnet 42 includes a first magnet surface 421 in a (circular) arc shape; wherein the first magnet surface 421 is convex outwardly in a direction away from the pendulum 32;

[0140] The radial side of the pendulum 32 is a stator mounting surface, and the magnetic mover assembly 4 further includes a second magnet base 51 movably connected to the stator mounting surface, and the second magnet base 51 is installed with a second permanent magnet 52, and the second permanent magnet 52 includes a (circular) arc-shaped second magnet surface 521; the second magnet surface 521 and the first magnet surface 421 have opposite magnetic properties and the same concave-convex direction;

[0141] Under the interaction between the first magnet surface 421 and the second magnet surface 521 , the second magnet base 51 can move relative to the stator mounting surface.

[0142] Specifically, the first magnet base 41 and the second magnet base 51 are both in an arc-shaped structure.

[0143] Specifically, the curvature of the first magnet surface 421 is the same as the curvature of the second magnet surface 521, and a predetermined gap, specifically -mm, is left between the first magnet surface 421 and the second magnet surface 521. When the first magnet surface 421 and the second magnet surface 521 are in concentric positions, the magnetic repulsion force generated is zero.

[0144] In one embodiment, referring to Figure 6 As shown, it also includes a stator fixing seat 53, a stator reset holding mechanism 54 and a stopper 55 respectively mounted on the stator mounting surface;

[0145] The stator fixing seat 53 is provided with a stator bearing seat 56;

[0146] A first rotating shaft is installed in the middle of the second magnet, a second rotating shaft parallel to the first rotating shaft is installed at one end of the second magnet, and the other end of the second magnet is limited by the stopper 55;

[0147] The first rotating shaft is rotatably connected to the stator bearing seat 56;

[0148] The two stator reset and holding mechanisms 54 are symmetrically arranged on both axial sides of the second magnet base 51, and each includes a mounting seat 541 installed on the stator mounting surface, a first hinged arm 542 with one end hinged to the mounting seat 541, and a second hinged arm 543 with one end hinged to the other end of the first hinged arm 542, and the other end of the second hinged arm 543 is hinged to the second rotating shaft.

[0149] In one embodiment, referring to Figure 5 As shown, the magnetic mover assembly 4 also includes a first permanent magnet middle partition plate 43 and a first permanent magnet side baffle plate 44 installed on the first magnet base 41, the first magnet base 41 is provided with a first arc-shaped mounting surface, the first permanent magnet 42 includes a plurality of first permanent magnet units 422 embedded and mounted on the first arc-shaped mounting surface, the plurality of first permanent magnet units 422 are arranged in multiple rows along the arc direction of the first arc-shaped mounting surface and in multiple columns along the axial direction of the first arc-shaped mounting surface, the first permanent magnet middle partition plate 43 is used to separate two adjacent columns of the first permanent magnet units 422, the two first permanent magnet side baffle plates 44 are respectively abutted against the two outermost columns of the first permanent magnet units 422 to limit the first permanent magnet 42, the two side ends of the bottom of the pendulum 32 are provided with mounting notches, and the first permanent magnet side baffle plates 44 are connected to the walls of the mounting notches.

[0150] In one embodiment, referring to Figure 7As shown, the magnetic stator assembly 5 also includes a second permanent magnet middle partition plate 57 and a second permanent magnet side baffle plate 58 installed on the second magnet base 51, the second magnet base 51 is provided with a second arc-shaped mounting surface, the second permanent magnet 52 includes a plurality of second permanent magnet units 522 embedded and mounted on the second arc-shaped mounting surface, the plurality of second permanent magnet units 522 are arranged in a plurality of rows along the arc direction of the second arc-shaped mounting surface and in a plurality of columns along the axial direction of the second arc-shaped mounting surface, the second permanent magnet middle partition plate 57 is used to separate two adjacent columns of the second permanent magnet units 522, and the two second permanent magnet side baffle plates 58 are respectively abutted against the two outermost columns of the second permanent magnet units 522 to limit the second permanent magnet 52.

[0151] It can be understood that the first permanent magnet unit 422 and the second permanent magnet unit 522 (ie, the distribution positions and sizes of the permanent magnets on the corresponding magnet bases) correspond one to one.

[0152] Specifically, in order to take advantage of the gravity potential energy, the swing angle of the gravity pendulum 3 is in the range of 45°-60°, and the swing angle is determined to be 47° through experiments. The larger the swing angle of the same weight pendulum, the greater the potential energy, that is, the greater the power generation, which improves the power generation efficiency.

[0153] Reference Figure 8 As shown, it is a schematic diagram of the motion state of the gravity pendulum 3 at different positions. Figure 8 . Exemplarily, the swing angle is 47° for explanation. During the entire power generation operation process, the gravity pendulum 3 rotates around the center (O1). Assuming that the clockwise rotation direction is the upper half cycle and the counterclockwise rotation direction is the lower half cycle, point A and point B are two inflection points of the circular operation. The angle of the corresponding mechanical arm 22 running on the circle in the upper half cycle of the gravity pendulum 3 is 274°, and the angle of the corresponding mechanical arm 22 running on the circle in the lower half cycle of the gravity pendulum 3 is 86°. This causes the work done by the gravity pendulum 3 in the upper half cycle and the lower half cycle to be different. Assuming that the gravity pendulum 3 does full work in the upper half cycle, the work done by the gravity pendulum 3 in the lower half cycle is part of the full work, and there is excess energy that is not utilized.

[0154] At point B, according to the velocity composition theorem in kinematics, the entrained velocity is normally zero, and the absolute velocity (Vr) of the gravity pendulum 3 is equal to its relative velocity (Va) along the direction of the slideway 311. However, at this time, due to the excess energy not being exerted, an additional entrained velocity is generated, and the direction is the same as Figure 8 The direction of the involved velocity (Ve) indicated in the figure is opposite. Due to the limitation of the structure of the mechanical arm 22, the mechanical arm 22, the slider 23 and the slide slot 311 collide with each other, which may cause a collision sound and a loud noise, or even cause mechanical damage. Therefore, it is necessary to absorb, transform and release the excess energy.

[0155] Since the magnetic stator assembly 5 can rotate around the first rotation axis, when the gravity pendulum 3 runs to Figure 8 Before the position of the dotted part on the right, the chord of the stator circumferential surface (second magnet surface 521) is parallel to the Z axis and in the vertical direction. When the first row of first permanent magnet units 422 in the front part of the magnetic mover assembly 4 enters the position of the magnetic stator assembly 5, there is a large angle between the magnetic mover assembly 4 and the magnetic stator assembly 5, that is, there is a large gap distance between the magnetic poles of the two. At this time, the repulsive force generated between the magnetic poles is small. As time goes by, the magnetic mover assembly 4 gradually approaches the magnetic stator assembly 5, and the magnetic pole gap between the mover and the stator also decreases, and the repulsive force gradually increases. This repulsive force and the force of the excess energy conversion that is not exerted offset each other, thereby playing a buffering role and avoiding the collision of the mechanical parts. When the gravity pendulum 3 is in the position to Figure 8 At the right dotted line position, the mover and the stator are completely corresponding (i.e., the second magnet surface 521 and the first magnet surface 421 are in a concentric position), the magnetic repulsion force generated is zero, and the magnetic device is in a balanced state. This state is transferred to the beginning of the first half of the next cycle. Due to the existence of the potential energy of the gravity pendulum 3, the balance of the magnetic device is instantly broken, and the gravity pendulum 3 swings in the clockwise direction. The energy at this time is the superposition of the potential energy of the gravity pendulum 3 and the magnetic potential energy of the magnetic device. This magnetic potential energy is used to supplement the mechanical loss (including the bearing of the pendulum shaft 33, the bearing in the gear box 28, the gear, the flywheel 62, the bearing of the flywheel shaft 63, the rotor of the start / stop motor, the rotor of the generator 61) and the energy loss caused by air resistance during the operation of the gravity pendulum 3. In this way, the gravity pendulum 3 is kept running uninterruptedly throughout the cycle, ensuring that the generator 61 continues to generate electricity. In addition, the second permanent magnet unit 522 of the front part of the magnetic device does not need to be installed in principle, but it can be determined according to the actual situation. If it is installed, the number of magnetic poles should be appropriately adjusted.

[0156] In one embodiment, referring to Figure 1 As shown, the generator 61 is installed on the generator 61 mounting seat 541, the flywheel shaft 63 is rotatably connected to the flywheel 62 mounting seat 541, the output end of the generator 61 is connected to the flywheel shaft 63, the flywheel 62 is installed on the flywheel shaft 63, and the generator 61 adopts a permanent magnet generator 61.

[0157] In one embodiment, a gear shaft 24 is provided in the gear box 28, the output end of the start-stop motor 1 is connected to the drive shaft 11 (through a coupling), a first gear 12 is installed on the drive shaft 11, a second gear 25 meshing with the first gear 12 is installed on the transmission shaft 21, a flywheel shaft gear 64 is installed on the flywheel shaft 63, and a first intermediate gear 26 meshing with the second gear 25 and a second intermediate gear 27 meshing with the flywheel shaft gear 64 are installed on the gear shaft 24.

[0158] It should be noted that the first gear 12 and the first intermediate gear 26 are gears with smaller diameters, and the second gear 25 is a gear with larger diameter. The drive shaft 11 and the gear shaft 24 rotate at a relatively high speed, while the transmission shaft 21 rotates at a relatively low speed.

[0159] Specifically, the start / stop device uses a start / stop motor installed on a motor mounting base 541 to play a start / stop role.

[0160] Specifically, the gear box 28 is installed on the gear box 28 mounting seat 541 , and the gear shaft 24 , the drive shaft 11 , and the transmission shaft 21 are all arranged in the gear box 28 .

[0161] It can be understood that the mechanical energy loss caused in the energy conversion process (including the energy consumption of the gravity pendulum shaft 33, the energy consumption of the gear box 28 bearing, the flywheel 62, the start / stop motor, and the rotor energy consumption of the generator 61) and the energy loss caused by the resistance encountered by the pendulum in the air, these energy losses need additional energy to be supplemented. By adopting the flywheel 62 and the energy release of magnetic potential energy for supplementation, the gravity pendulum 3 can continue to operate uninterruptedly, thereby ensuring the continuous operation and power generation of the generator 61.

[0162] The setting of the flywheel 62 releases energy at the two inflection points A and B in the figure, so that the slider 23 slides smoothly through the two inflection points in the slide groove 311, so that the whole system can operate stably, so that the speed of the gravity pendulum 3 is stable during operation, and excessive speed fluctuations during operation are avoided.

[0163] When the gravity pendulum 3 runs in a full cycle, the excess energy in the second half of the cycle is stored simultaneously by the magnetic device and the flywheel 62, and the energy is released in the first half of the cycle, which is equivalent to an energy storage device in effect.

[0164] Reference Figure 1 As shown, it also includes an electrical system 8, including a start / stop motor control cabinet 81, a rectifier and battery cabinet 82, and an inverter AC output cabinet 83.

[0165] Working principle of the present invention:

[0166] The start-stop motor 1 starts according to the preset program, driving the drive shaft 11 in the gear box 28 to rotate. The drive shaft 11 drives the transmission shaft 21 to rotate through gear transmission. As the mechanical arm 22 rotates, the slider 23 moves linearly along the slide groove 311, pushing the gravity pendulum 3 to start rising clockwise (or counterclockwise) from the original position (lowest point). When the gravity pendulum 3 rises to the preset angle, i.e. the turning point (point A in the figure), the start-up is completed at this time. The gravity pendulum 3 moves in the opposite direction under the action of gravitational potential energy, and the slider 23 will continue to move linearly upward along the slide groove 311. At this time, the gravity pendulum 3 enters the normal swing of the second half cycle.

[0167] After the startup is completed, the gravity pendulum 3 enters normal swing, and the slider 23 will make a linear motion in the slide slot 311 of the gravity pendulum 3. The linear motion of the slider 23 drives the mechanical arm 22 to make a circular motion. Since the mechanical arm 22 is connected to the transmission shaft 21, it drives the second gear 25 on the transmission shaft 21 to rotate, and transmits power to the gear shaft 24 and the flywheel shaft 63. The flywheel shaft 63 drives the flywheel 62 to rotate, and the flywheel shaft 63 drives the generator 61 to rotate. The generator 61 then generates three-phase alternating current, and the generated three-phase alternating current is rectified by the rectifier and the rectifier device of the battery cabinet. The rectified direct current is converted into standard three-phase alternating current for use by the load through the three-phase inverter in the inverter and AC output cabinet.

[0168] During the operation of the gravity pendulum 3, the slider 23 drives the mechanical arm 22 to move along a circular trajectory. In the first half cycle, the movement angle of the mechanical arm 22 is different from that of the second half cycle. Due to the asymmetric energy consumption of the upper and lower half cycles, the energy consumed in the second half cycle is relatively small, resulting in excess energy. The excess energy is absorbed, converted and released by the magnetic device and the flywheel 62 to compensate for the mechanical energy loss in the system and the energy loss caused by air resistance, thereby effectively alleviating the energy attenuation problem during the operation of the gravity pendulum 3 and ensuring the continuous operation of the system and the continuous production of green electricity.

[0169] Press the stop button in the start / stop motor cabinet to generate a braking torque in the start / stop motor, that is, the entire equipment stops working.

[0170] Example 2

[0171] A design method for a magnetic and gravity combined power generation device, characterized in that the design method comprises:

[0172] S1. Determine the rated power P of the generator 61 n and the rated speed n of the generator 61 N .

[0173] S2, according to the rated power P of the generator 61 n , the torque M of the generator 61 is obtained n ;

[0174]

[0175] Where: M n is the rated torque of the generator 61; n is the rated power of the generator 61; n N is the rated speed of the generator 61.

[0176] S3, according to the rated speed n of the generator 61 N , get the speed n of the gearbox 28 Z .

[0177]

[0178] Where: n Z is the speed of the gearbox 28; n N is the rated speed of the generator 61; b1 is the primary speed ratio; b2 is the secondary speed ratio.

[0179] S4. Determine the length L of the robot arm 22 (i.e., the radius of the circle when the robot arm 22 rotates), and obtain the length l of the gravity pendulum 3 based on the length L of the robot arm 22.

[0180] l=2L+0.3

[0181] S5. According to the length L of the mechanical arm 22 and the length l of the gravity pendulum 3, the maximum swing angle θ of the gravity pendulum 3 during operation is obtained.

[0182]

[0183] Wherein: θ is the maximum swing angle of the gravity pendulum 3; L is the length of the robot arm 22; l is the length of the gravity pendulum 3.

[0184] S6. According to the rated power P of the generator 61 n , the mechanical loss P1 of the gearbox 28, the mechanical loss P2 of the flywheel 62, the loss P3 of the generator 61, the loss P4 of the start-stop motor 1, and the required mechanical input power P of the gearbox 28 are obtained. s .

[0185] P s =P n +P1+P2+P3+P4

[0186] Where: P s P is the mechanical power input to the gearbox 28; nis the rated power of the generator 61; P1 is the mechanical loss of the gearbox 28; P2 is the mechanical loss of the flywheel 62, which is calculated at 2% of the total power in the initial calculation, and then corrected after the parameters of the flywheel 62 are determined; P3 is the loss of the generator 61; P4 is the loss of the start-stop motor 1. Among them:

[0187] P1=P f1 +P f2 +P w

[0188] Where: P f1 is the friction loss of the bearing of the gear box 28; f2 is the windage loss of the gear;

[0189] P w It is the oil churning loss of gear.

[0190]

[0191] P f is the friction loss of the rolling bearing; F′ is the rolling bearing load; d is the diameter from the center of the ball or roller of the rolling bearing to the center of rotation; v is the circumferential speed of the ball or roller center.

[0192]

[0193] Where: P f is the power loss due to wind resistance; C4=2.04×10 -8 ; t is the gear width; R is the gear pitch radius; n is the rotational speed; ρ and μ are the density and viscosity of the oil-gas mixture in the oil-gas space of the gear box 28.

[0194]

[0195] Where: P w The power lost by the oil stirring of the gear box 28; f g is the immersion factor of the gear; γ is the dynamic viscosity; n is the working speed of the gear; D is the outer diameter of the gear; L is the length of the gear; B is the width of the gear; A g is the configuration constant; R f is the roughness factor; β is the helix angle of the helical gear.

[0196] P2=P m +P f1 +P f2

[0197] Where: P m is the friction power loss of the flywheel 62; P f2 is the air resistance loss.

[0198]

[0199] Wherein: μ is the friction coefficient; ω is the angular velocity of the flywheel 62; m is the mass of the flywheel 62; and r is the radius of the flywheel 62.

[0200]

[0201] Where: P f2 is the air resistance loss of the flywheel 62; ρ is the air density; A is the surface area of ​​the flywheel 62; C is the air resistance coefficient; n is the rotation speed of the flywheel 62; and r is the radius of the flywheel 62.

[0202] P3=P cu +P Fe +P f1

[0203] Where: P cu P is the stator copper loss of the generator 61; Fe is the stator iron loss of generator 61

[0204]

[0205] Where: P cu is the stator copper loss of generator 61; m is the number of phases; I N is the current of generator 61; R is the resistance of the winding at 75°C.

[0206]

[0207] Where: P Fe is the stator iron loss; P Fej is the stator yoke iron loss; P Fet is the stator tooth iron loss; K d is the empirical coefficient of stator yoke iron loss, which can be obtained by looking up the table; p Fe / 50 is the iron loss value corresponding to the stator yoke silicon steel sheet in the iron loss curve table at 50Hz; K′ d G is the empirical coefficient of stator tooth iron loss; j is the stator yoke weight; G t is the stator tooth weight.

[0208] P3=P cu +P Fe +P f1

[0209] Where: P cu is the stator copper loss of the start-stop motor 1 (reluctance motor); P Fe is the stator iron loss of the start-stop motor 1; P z is the rotor loss of start-stop motor 1.

[0210]

[0211] Where: P cu is the stator copper loss of generator 61; m is the number of phases; I N is the current of generator 61; R is the resistance of the winding at 75°C.

[0212]

[0213] Where: P Fe is the stator iron loss; P Fej is the stator yoke iron loss; P Fet is the stator tooth iron loss; K d is the empirical coefficient of stator yoke iron loss, which can be obtained by looking up the table; p Fe / 50 is the iron loss value corresponding to Bj in the iron loss curve table of the stator yoke silicon steel sheet at 50Hz; K′ d G is the empirical coefficient of stator tooth iron loss; j is the stator yoke weight; G t is the stator tooth weight.

[0214] S7, according to the mechanical input power P of the gear box 28 s , the mechanical input power ∑W of the gear box 28, i.e., the work done by the gravity pendulum 3 per unit area, is obtained.

[0215]

[0216] Where: ∑W is the work done by the gravity pendulum in 3 unit time; P s is the mechanical input power of the gearbox 28; t is the unit time.

[0217] S8. According to the mechanical input power ∑W of the gear box 28, the parameters of the swing arm 31 and the pendulum 32 in the gravity pendulum 3 are determined, including the mass m1 of the pendulum 32, the diameter d of the pendulum 32, the height h (center of gravity height) of the pendulum 32, the length l1 of the swing arm 31, and the mass m2 of the swing arm 31; wherein the mass m1 of the pendulum 32 is calculated as follows:

[0218] ∑W=m1gl(1-cosΦ)

[0219]

[0220] Wherein: m1 is the mass of the gravity pendulum 32; g is the acceleration of gravity; l is the length of the gravity pendulum 3 (the distance from the pendulum shaft 33 to the center of the pendulum 32); Φ is the representative swing angle;

[0221] The diameter d of the pendulum 32 is calculated as follows:

[0222]

[0223] Wherein: d is the diameter of the pendulum 32; m1 is the mass of the gravity pendulum 3; ρ is the material density of the gravity pendulum 3; when the mass of the gravity pendulum 32 is large, the coefficient takes a small value, otherwise it is the opposite;

[0224] The height h of the pendulum 32 is calculated as follows:

[0225] h=(0.2~0.5)d

[0226] Wherein: h is the height of the gravity pendulum 32; d is the diameter of the gravity pendulum 32;

[0227] The length l1 of the swing arm 31 is calculated as follows:

[0228]

[0229] Wherein: l1 is the length of the gravity pendulum arm 31; l is the length of the gravity pendulum 3; d is the diameter of the gravity pendulum 32;

[0230] The mass m2 of the swing arm 31 is calculated as follows:

[0231] m2=l1b1c1ρ-(2L+0.1)b2c2ρ+0.02b1l1ρ

[0232] Wherein: m2 is the mass of the swing arm 31; l1 is the length of the swing arm 31; b1 is the width of the swing arm 31; c1 is the thickness of the swing arm 31; b2 is the width of the slide groove 311; c2 is the depth of the slide groove 311; ρ is the steel density; L is the length of the robot arm 22.

[0233] S9. Determine the output power P of the gravity pendulum 3 according to the parameters of the swing arm 31 and the pendulum 32 in the gravity pendulum 3.

[0234] The details are as follows:

[0235] Calculate the angular velocity and speed of gravity pendulum 3 at the lowest point.

[0236] Gravity pendulum 3 potential energy:

[0237] V=(m1+m2)gl(1-cosΦ)

[0238] Wherein: V is the potential energy of the gravity pendulum 3; m1 is the mass of the pendulum 32; m2 is the mass of the swing arm 31; g is the acceleration of gravity; 1 is the pendulum length (the distance from the pendulum shaft 33 to the center of the pendulum 32); Φ is the swing angle.

[0239] Work done by gravitational potential energy:

[0240] ∑W=(m1+m2)gl(1-cosΦ)

[0241] Kinetic energy of gravity pendulum 3:

[0242]

[0243] According to the kinetic energy theorem, we can get

[0244]

[0245]

[0246] Where: v A is the speed of the gravity pendulum 3 at the lowest point; T is the kinetic energy; j o is the moment of inertia of the pendulum about its axis; m1 is the mass of the pendulum 32; m2 is the mass of the swing arm 31; r is the radius of the pendulum 32; ω is the angular velocity of the pendulum; g is the acceleration due to gravity; l is the pendulum length (the distance from the pendulum axis 33 to the center of the pendulum 32); and Φ is the swing angle.

[0247] Calculate the acceleration, force and power of the gravity pendulum 3 at the lowest point. The details are as follows:

[0248] Angular acceleration of gravity pendulum 3:

[0249]

[0250] Where: α is the angular acceleration of the pendulum; v A is the speed of the pendulum at its lowest point; v0 is the starting speed of the pendulum; t is the time it takes for the pendulum to swing to its lowest point.

[0251] Force at the lowest point of the pendulum:

[0252] F=m1α

[0253] Where: F is the force at the lowest point of the pendulum; m1 is the mass of the pendulum 32; α is the angular acceleration of the pendulum.

[0254] The power of the work done by the pendulum:

[0255] P=Fv

[0256] Where: P is the power of the work done by the pendulum; F is the force at the lowest point of the pendulum; v is the velocity of the pendulum at the lowest point.

[0257] S10. Determine the total power loss P of the gravity pendulum 3 总损失 According to the output power P of the gravity pendulum 3 and the total power loss, determine the unconverted excess power P of the gravity pendulum 3 in the second half of the cycle. d .

[0258] The total power loss of the gravity pendulum 3 is calculated as follows:

[0259] P 总损失 =P f1 +P m +P f2

[0260] Where: P总损失 is the total power loss of the gravity pendulum 3; P f1 is the friction loss of rolling bearing; m is the friction loss power of the pendulum; P f2 Power lost due to air resistance,

[0261] Calculate the bearing friction loss energy of the gravity pendulum 3 during operation:

[0262]

[0263] Where: P f1 is the friction loss of the rolling bearing; F′ is the rolling bearing load; d is the diameter from the center of the ball or roller of the rolling bearing to the center of rotation; v is the circumferential speed of the ball or roller center.

[0264] The friction loss power of the gravity pendulum 3 is calculated with reference to the friction loss power of the flywheel 62 as follows:

[0265]

[0266] Where: P m is the friction power loss of the gravity pendulum 3; μ is the friction coefficient; ω is the angular velocity of the gravity pendulum 3; m is the mass of the gravity pendulum 3; l is the pendulum length.

[0267] Gravity pendulum 3 air resistance power loss

[0268]

[0269] Where: P f2 is the air resistance loss of the gravity pendulum 3; ρ is the air density; A is the surface area of ​​the gravity pendulum 3; C is the air resistance coefficient; n is the rotation speed of the pendulum; r is the radius of the pendulum.

[0270] S11, according to the unconverted excess power P of the gravity pendulum 3 in the second half cycle d , determine the force F1 generated by the unconverted excess power of the gravity pendulum 3 in the second half cycle.

[0271] Determine the unconverted energy of the gravity pendulum 3 in the second half cycle:

[0272]

[0273] Where: P d is the unconverted excess power of the gravity pendulum 3 in the second half cycle; P is the power of the work done by the gravity pendulum 3; α1 is the angle at which the gravity pendulum 3 travels in the first half cycle; α2 is the angle at which the gravity pendulum 3 travels in the second half cycle.

[0274] S12. Determine the magnetic pole repulsion F of the magnetic device m, according to the force F1 generated by the unconverted excess power of the gravity pendulum 3 in the second half cycle and the force F1 generated by the unconverted excess power of the gravity pendulum 3 in the second half cycle, determine the number of permanent magnet groups N g .

[0275] Calculate and determine the magnetic device:

[0276] P z =P d ≥P 总损失

[0277] Where: P z Generate power for the magnetic device; P d is the unconverted excess power of the gravity pendulum 3 in the second half cycle; P s总 is the total power loss of gravity pendulum 3.

[0278] P d =F1v a

[0279]

[0280] Where: F1 is the force generated by the unconverted excess power of the gravity pendulum 3 in the second half cycle; v a is the absolute velocity of the gravity pendulum 3.

[0281] Calculation of the number of permanent magnet groups:

[0282] Air gap magnetic induction intensity of permanent magnet

[0283]

[0284] Where: B δ k is the air gap magnetic induction intensity of the permanent magnet; m is the end face coefficient of permanent magnet; B r is the remanence of the permanent magnet; σ is the magnetic leakage coefficient; a m b is the short side length of the rectangular end face of the permanent magnet; m is the long side length of the rectangular end face of the permanent magnet; δ is the air gap length.

[0285] Permanent magnet pole repulsion:

[0286]

[0287] Where: F m is the repulsion force of the permanent magnet poles; B δ is the air gap magnetic induction intensity of the permanent magnet; μ0 is the absolute magnetic permeability of vacuum, μ0=4π×10 -7 H / m; S m is the magnetic pole area of ​​the permanent magnet.

[0288] Calculation of the number of permanent magnet groups:

[0289]

[0290] Where: N g is the number of permanent magnet groups; F1 is the force generated by the unconverted excess power of the gravity pendulum 3 in the second half cycle; F m is the repulsion force of the permanent magnet poles.

[0291] S13, determining the maximum surplus work and the maximum deficit work of the gravity pendulum 3 according to the parameters of the swing arm 31 and the pendulum 32 in the gravity pendulum 3;

[0292] S14, according to the maximum surplus work W of the gravity pendulum 3 max and the maximum work loss W min , determine the parameters of the flywheel 62 device.

[0293] In actual operation of the gravity pendulum 3, the maximum surplus work is the work done by the potential energy of the gravity pendulum 3, and the maximum deficit work is the total mechanical loss of the gravity pendulum 3. This work is also the work required between the stopping position of the gravity pendulum 3 when it is lifted from the lowest point and the maximum swing angle. The calculation of surplus and deficit work is as follows:

[0294]

[0295] Where: W max is the maximum surplus work; W min is the maximum work loss; m1 is the mass of the pendulum 32; m2 is the mass of the swing arm 31; g is the acceleration due to gravity; l is the pendulum length; is the maximum swing angle of gravity pendulum 3; is the angle at which the pendulum stops during operation; t1 and t2 are the times representing the maximum surplus work and the maximum deficit work respectively, which can be obtained based on the period of the pendulum.

[0296] The period of gravity pendulum 3:

[0297]

[0298] Maximum profit and loss:

[0299] W n =W max +W min

[0300] Calculate the moment of inertia of flywheel 62

[0301] Moment of inertia calculation:

[0302]

[0303] Where: J F W is the moment of inertia of the flywheel 62; n is the maximum profit or loss work; mis the difference between the maximum angular velocity and the minimum angular velocity; δ is the velocity non-uniformity coefficient.

[0304] Flywheel 62 average diameter:

[0305]

[0306] Where: D is the average diameter of the flywheel 62; v p is the permissible peripheral speed; n m It is the speed of flywheel 62.

[0307] Flywheel 62 mass:

[0308]

[0309] Where: m0 is the edge mass of the flywheel 62; J F is the moment of inertia of the flywheel 62; and D is the average diameter of the flywheel 62.

[0310] Flywheel 62 rim width and thickness:

[0311]

[0312] Where: b is the rim width of the flywheel 62; J F is the moment of inertia of the flywheel 62; ρ is the density of the flywheel 62; k is the coefficient (1-2); and D is the average diameter of the flywheel 62.

[0313] H = kb

[0314] Wherein: H is the thickness of the rim of the flywheel 62.

[0315] S13. In addition, the stiffness and strength of the rotating shaft (including the drive shaft 11, the transmission shaft 21, and the flywheel shaft 63) are calculated as follows:

[0316] Permissible torsion angle of the shaft

[0317]

[0318] Where: [θ] 0 / m is the permissible torsion angle; K is the overload factor; M n is the torque; G is the shear elastic modulus of the shaft; J p is the polar moment of inertia of the axis of rotation.

[0319] Polar moment of inertia of the shaft, when it is a solid shaft

[0320]

[0321] When it is a hollow shaft,

[0322] Where: α is the ratio of the inner and outer diameters of the hollow shaft; D is the outer diameter of the shaft.

[0323] Torsional stress of the shaft

[0324]

[0325] Where: W n is the torsional section modulus; K is the overload factor; M n is the torque; [τ n ] is the allowable stress

[0326] When the shaft is solid,

[0327] When it is a hollow shaft,

[0328] Where: α is the ratio of the inner and outer diameters of the hollow shaft; D is the outer diameter of the shaft.

[0329] The most dangerous shaft diameter when it is a solid shaft

[0330]

[0331] When it is a hollow shaft,

[0332]

[0333] S14, calculate the efficiency of the generator 61

[0334]

[0335] Where: η is the efficiency of the generator 61; P n is the rated power of the generator 61; P 整 is the power loss of the DC rectifier; P 逆 is the power loss of the inverter; P 控 is the power loss of the controller; these three power losses can be obtained from the product manual.

[0336] It should be noted that the above calculation formula can be compared with actual data obtained through experiments and then corrected through repeated calculations.

[0337] Finally, it should be noted that the above specific implementation methods are only used to illustrate the technical solution of the present invention rather than to limit it. Although the present invention has been described in detail with reference to examples, those skilled in the art should understand that the technical solution of the present invention can be modified or replaced by equivalents without departing from the spirit and scope of the technical solution of the present invention, which should be included in the scope of the claims of the present invention.

Claims

1. A magnetic and gravity combined power generation device, characterized in that: include: Gearbox; A start-stop motor, an output end of which is connected to the first input end of the gear box; A transmission device, comprising a transmission shaft and a mechanical arm, wherein a first end of the mechanical arm is connected to the transmission shaft, and the transmission shaft is connected to the second input end of the gear box to drive the first end of the mechanical arm to rotate around the axis of the transmission shaft; A gravity pendulum, comprising a swing arm and a pendulum, wherein the first end of the swing arm rotates around a rotation axis parallel to the axis of the transmission shaft, the pendulum is mounted on the second end of the swing arm, and the second end of the mechanical arm is slidably connected to the swing arm and can perform reciprocating linear motion between the first end and the second end of the swing arm; A magnetic device, comprising a magnetic mover assembly and a magnetic stator assembly, wherein the magnetic mover assembly is mounted on the pendulum, the magnetic stator assembly is arranged in the circumferential direction of the swing of the magnetic mover assembly, and magnetic pole repulsion can be generated between the magnetic mover assembly and the magnetic stator assembly when they are close to each other; the magnetic mover assembly and the magnetic stator assembly include a plurality of groups of permanent magnets arranged correspondingly; A generator set, comprising a generator, a flywheel connected to the generator and to an output end of the gearbox; The gravity pendulum converts gravitational potential energy into mechanical energy by swinging, and the swing of the gravity pendulum can drive the mechanical arm to make a circular motion, so as to drive the transmission shaft to rotate, and further drive the generator to generate electricity.

2. A design method for a magnetic and gravity combined power generation device according to claim 1, characterized in that: The design method comprises: S1. Determine the rated power P of the generator n and the rated speed of the generator n N ; S2, according to the rated power P of the generator n , and the torque M of the generator is obtained n ; S3, according to the rated speed n of the generator N , get the speed n of the gearbox Z ; S4, determine the length L of the mechanical arm, and obtain the length l of the gravity pendulum according to the length L of the mechanical arm; S5. According to the length L of the mechanical arm and the length l of the gravity pendulum, the maximum swing angle θ of the gravity pendulum during operation is obtained; S6. According to the rated power P of the generator n , the mechanical loss of the gearbox P1, the mechanical loss of the flywheel P2, the loss of the generator P3, the loss of the start-stop motor P4, and the required mechanical input power P of the gearbox are obtained. s ; S7, according to the mechanical input power P of the gearbox s , we get the gearbox mechanical input power ∑W, which is the work done by the gravity pendulum in the unit; S8. Determine the parameters of the swing arm and the pendulum in the gravity pendulum according to the mechanical input power ∑W of the gear box; S9. Determine the output power P of the gravity pendulum according to the parameters of the swing arm and the pendulum in the gravity pendulum; S10. Determine the total power loss P of the gravity pendulum 总损失 According to the output power P of the gravity pendulum and the total power loss, determine the unconverted excess power P of the gravity pendulum in the second half of the cycle d ; S11, according to the gravity pendulum in the second half of the cycle of the unconverted excess power P d , determine the force F1 produced by the unconverted excess power of the gravity pendulum in the second half cycle; S12. Determine the magnetic pole repulsion F of the magnetic device m , determine the number of permanent magnet groups N according to the force F1 generated by the unconverted excess power of the gravity pendulum in the second half of the cycle and the force F1 generated by the unconverted excess power of the gravity pendulum in the second half of the cycle g ; S13, determining the maximum surplus work and the maximum deficit work of the gravity pendulum according to the parameters of the swing arm and the pendulum in the gravity pendulum; S14, according to the maximum surplus work W of the gravity pendulum max and the maximum work loss W min , determine the parameters of the flywheel device.

3. The design method according to claim 2, characterized in that: Gearbox speed n Z The calculation is as follows: Where: n Z is the gearbox speed; n N is the rated speed of the generator; b1 is the first-stage speed ratio; b2 is the second-stage speed ratio.

4. The design method according to claim 2, characterized in that: The maximum swing angle θ of the gravity pendulum during operation is calculated as follows: l＝2L+0.3 Where: θ is the maximum swing angle of the gravity pendulum; L is the length of the robot arm; l is the length of the gravity pendulum.

5. The design method according to claim 2, characterized in that: Mechanical input power P of the gearbox s The calculation is as follows: P s =P n +P1+P2+P3+P4 Where: P s is the mechanical power input to the gearbox; P n is the rated power of the generator; P1 is the mechanical loss of the gearbox; P2 is the mechanical loss of the flywheel, which is initially calculated as 2% of the total power, and then corrected when the flywheel parameters are determined; P3 is the loss of the generator; P4 is the loss of the start-stop motor.

6. The design method according to claim 2, characterized in that: Step S8 includes determining the mass m1 of the pendulum, the diameter d of the pendulum, the height h of the pendulum, the length l1 of the swing arm, and the mass m2 of the swing arm; wherein the mass m1 of the pendulum is calculated as follows: ΣW=m1gl(1-cosΦ) Where: ∑W is the work done by the potential energy of the gravity pendulum; m1 is the mass of the gravity pendulum; g is the acceleration of gravity; l is the length of the gravity pendulum; Φ is the representative swing angle; The diameter d of the pendulum is calculated as follows: Where: d is the diameter of the pendulum; m1 is the mass of the gravity pendulum; ρ is the density of the gravity pendulum material; when the mass of the gravity pendulum is large, the coefficient takes a small value, otherwise it is the opposite; The height h of the pendulum is calculated as follows: h=(0.2~0.5)d Where: h is the height of the gravity pendulum; d is the diameter of the gravity pendulum; The length l1 of the swing arm is calculated as follows: Where: l1 is the length of the gravity pendulum arm; l is the length of the gravity pendulum; d is the diameter of the gravity pendulum; The mass m2 of the swing arm is calculated as follows: m2=l1b1c1ρ-(2L+0.1)b2c2ρ+0.02b1l1ρ In the formula: m2 is the mass of the swing arm; l1 is the length of the swing arm; b1 is the width of the swing arm; c1 is the thickness of the swing arm; b2 is the width of the slide; c2 is the depth of the slide; ρ is the steel density; L is the length of the robotic arm.

7. The design method according to claim 2, characterized in that: In step S9, the output power P of the gravity pendulum is calculated as follows: P=Fv Where: P is the power of the work done by the pendulum; F is the force at the lowest point of the pendulum; v is the speed of the pendulum at the lowest point; F=m1α Where: F is the force at the lowest point of the pendulum; m1 is the mass of the pendulum; α is the angular acceleration of the pendulum; Where: α is the angular acceleration of the pendulum; v A is the speed of the pendulum at its lowest point; v0 is the starting speed of the pendulum; t is the time it takes for the pendulum to swing to its lowest point; Where: v A is the speed of the gravity pendulum at the lowest point; T is the kinetic energy; j o is the moment of inertia of the pendulum about its axis; m1 is the mass of the pendulum; m2 is the mass of the pendulum arm; r is the radius of the pendulum; ω is the angular velocity of the pendulum; g is the acceleration due to gravity; l is the length of the pendulum; and Φ is the pendulum angle.

8. The design method according to claim 2, characterized in that: In step S10, the total power loss of the gravity pendulum P 总损失 The calculation is as follows: P 总损失 =P f1 +P m +P f2 Where: P 总损失 is the total power loss of the gravity pendulum; P f1 is the friction loss of the rolling bearing; m is the friction loss power of the pendulum; P f2 Power is lost due to air resistance; The unconverted excess power P of the gravity pendulum in the second half of the cycle d The calculation is as follows: Where: P d is the unconverted excess power of the gravity pendulum in the second half cycle; P is the power of the work done by the gravity pendulum; α1 is the angle of the gravity pendulum in the first half cycle; α2 is the angle of the gravity pendulum in the second half cycle.

9. The design method according to claim 2, characterized in that: Number of permanent magnet groups N g The calculation is as follows: Where: N g is the number of permanent magnet groups; F1 is the force generated by the unconverted excess power of the gravity pendulum in the second half cycle; F m is the repulsion force of the permanent magnet poles; Where: F m is the repulsion force of the permanent magnet poles; B δ is the air gap magnetic induction intensity of the permanent magnet; μ is the absolute magnetic permeability of vacuum, μ0=4π×10 -7 H / m; S m is the magnetic pole area of ​​the permanent magnet; Where: F1 is the force generated by the unconverted excess power of the gravity pendulum in the second half of the cycle; v a is the absolute velocity of the gravity pendulum; P z =P d ≥P 总损失 Where: P z Generate power for the magnetic device; P d is the unconverted excess power of the gravity pendulum in the second half cycle; P s总 is the total power loss of the gravity pendulum.

10. The design method according to claim 2, characterized in that: In step S14, the moment of inertia J of the flywheel F The calculation is as follows: Where: J F is the flywheel moment of inertia; W n is the maximum profit or loss work; m is the difference between the maximum angular velocity and the minimum angular velocity; ω is the velocity non-uniformity coefficient; IN n =In max +W min Where: W max is the maximum surplus work; W min is the maximum work loss; m1 is the mass of the pendulum; m2 is the mass of the pendulum arm; g is the acceleration of gravity; l is the pendulum length; is the maximum swing angle of the gravity pendulum; is the angle at which the pendulum stops during operation; t1 and t2 are the time elapsed representing the maximum surplus work and the maximum deficit work respectively, which are obtained according to the period of the pendulum; The average diameter D of the flywheel is calculated as follows: Where: D is the average diameter of the flywheel; v p is the permissible peripheral speed; n m is the flywheel speed; The mass m0 of the flywheel is calculated as follows: Where: m0 is the flywheel edge mass; The width b and thickness H of the flywheel rim are calculated as follows: Where: b is the width of the flywheel rim; J F is the moment of inertia of the flywheel; ρ is the flywheel density; k is the coefficient (1-2); D is the average diameter of the flywheel; H = kb Where: H is the flywheel rim thickness.

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