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The kinematics analysis, power flow analysis and meshing efficiency calculation are carried out for each gear. Kinematics analysis - solving the relative angular velocity of each component is convenient for analysis. The middle differential gear train is simply referred to as unit 1 from left to right. Unit 2, unit 3 and unit 41 are the same in the analysis of the different gears, so the most complicated third gear is analyzed. 1B3 is the third gear when braking, and the participating mechanism is as In order to clearly reflect the structure of the composite train and the connection relationship and power flow between the components, the schematic diagram of the discrete third gear mechanism of the mechanism is shown in the discrete diagram of the mechanism, and the triangle represents each differential gear train. Unit (because there are 3 basic components), the basic components are marked by 3 vertices, the components fixed together between the units are connected by thin solid lines, and the thin solid lines become the third flow of the flow channel in the invisible power flow. The arrow in the discrete diagram of the discrete diagram of the mechanism indicates the power flow direction. The arrows at the two joints of the input shaft I and the output shaft II can be drawn first to indicate the total power flow of the train. The remaining arrows need to be determined in each of the differential trains. Basic structure After the master-slave position, you can draw 1 according to the discrete graph, according to equation (2), which shows that the degree of freedom of the mechanism is 11 and the angular velocity ωH3=1 of the output axis II (ie, the arm H3) can be solved. Relative angular velocity of each member The angular velocity, moment of each wheel train unit member and the angular velocity of the master-slave member ω moment Mη0 = 1 when η0 < 1 power positive and negative (master-slave position in the train unit) unit 10.
When the wheel train rotates at a constant speed, that is, in the state of stress balance, the algebraic sum of the moment acting on several members connected by the thin solid line should be zero, so the following torque contact equation can be obtained: M2 MH2 M6=0, MH1 M4=0(16) In the equations consisting of 8 linear equations of the moment variable M from equations (10) to (16), there are 9 M variables, and 3 β variables 1 can be temporarily pressed. Η01=η02=η03=1, and let the output torque MH3=-1, solve the equations and find the relative moments of the components under ideal conditions (ie, regardless of the meshing friction loss). According to the angular velocity of each component. According to the torque, according to the power N=Mω, analyze whether the (relative) power transmitted by each component in the ideal state is positive or negative 1 and then the positive power is the input power of the train unit (arrow tip pointing unit), negative power For the output power of the train wheel unit (arrow tip away from the unit), the direction of the arrow in the arrow 1 is determined, and the power flow direction is clearly shown. 1 After the angular velocity and the master-slave position of each component are determined, each round can be analyzed. The meshing power transmitted by the "wheel a" of the unit To determine that the β value is equal to 1, or -11, according to equations (10) to (15) and equation (3), in unit 1, wheel 1 is "wheel a", and in unit 2, wheel 3 is " The wheel a" then solves the linear equation of the moment and solves the actual moment acting on each component (relative to MH1=-1). The angular velocity and actual moment of the component listed in the middle 1 can be used to calculate the actual power delivered, and then The efficiency is equal to the ratio of the output power to the input power. The efficiency of each wheel train unit is calculated. 1 The meshing efficiency of each wheel train unit is calculated to be less than 1, indicating that each β value is correct.
The direction of the arrow in the power flow direction analysis clearly reflects that the power flow in the train is divided into two paths from the input of the shaft I, one way through the wheel 3 into the wheel train unit 2, and the other through the wheel 5 into the wheel train unit 3; And a part of the power flowing out from the arm H2 of the train wheel unit 2 enters the train wheel unit 3 via the wheel 6, and is combined with the power entering from the wheel 5 and transmitted to the shaft II via the arm H3; the other part of the wheel 2 Entering the train unit 1, and then flowing out from the arm H1, returning to the train unit 2 via the wheel 4, this part of the power can not be output, only circulating in the closed loop formed by the train units 1 and 2, so called closed loop The closed flow power flow of power flow 1 will increase the load of the component and cause additional power loss, which will reduce the strength of the component and reduce the efficiency of the train, so it should be avoided or reduced as much as possible.