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                  Technical
                  Discussion Home>>Technical >>Discussion
                  Selection Calculation Of Reducer
                  The last section is just a preliminary selection of a form of reducer, such as planetary gear, this section to discuss is according to the requirements of the application, the selected reducer for calculation and review, to ensure that the performance, parameters can meet the requirements of the application.
                  Here, only need to consider to choose the appropriate reduction ratio and the size of the gear box, we know that if a torque is not enough, we can choose a larger motor and gear box, and the speed is not big enough, or the moment of inertia is not match but will affect the basic properties of the machinery and equipment, so normally, should make sure to slow down than I. According to the rated speed of the selected motor and the load speed required by the application, the reduction ratio can be calculated as follows:

                  i=/=/

                  Because of servo motor rated power is reflected in the working point of rated speed, in order to make the motor has the highest efficiency, often to the motor rated speed as input, sometimes customers need is a very low output speed, such as the application of single crystal furnace, even chose the biggest reduction ratio, also cannot meet the requirements, you need to input speed, in turn, accounting, said earlier, Too low input speed will lead to low efficiency of the gearbox. In this case, increasing the motor power appropriately is a solution.
                  Some works mention the selection of the best deceleration ratio under certain requirements, such as selecting the deceleration ratio according to the maximum angular acceleration:
                  Assuming that the frictional load is ignored, the angular acceleration of the load axis is:

                  iη=ηi 2+

                  = iηηi 2+

                  I take the derivative of this with respect to I, which is d/d I =0

                  i=

                  So if I change this, I get

                  =/ηi 2

                  Assuming η≈1, at this deceleration ratio, half of the torque of the motor is used to accelerate the load and the other half is used to accelerate the rotor of the motor.

                  Next, we need to discuss the matching problem of the moment of inertia. The moment of inertia of the load of the servo system refers to the combined moment of inertia of the transmission system and the rotating part of the load, which is expressed in symbol. Let's discuss the relationship between it and the system's open-loop cutoff frequency Ω c, electromechanical time constant Tm, and low-speed stationary tracking performance. (The three formulae are taken from "Precision Tracking Measurement Radar Technology" and their derivation has not been verified.)

                  1         The relationship between the moment of inertia and the system cutoff frequency ωc:

                  ωc=

                  A high system cutoff frequency ωc indicates that the system has good transient response but poor stability. As shown in the above equation, the system cutoff frequency ωc will become smaller when the moment of inertia is large.

                  2   Relationship between moment of inertia and electromechanical time constant Tm:

                  As shown in the above formula, when the moment of inertia is large, the electromechanical time constant Tm will become large, leading to the increase of overshoots in the transition process.

                  2    Relationship between rotational inertia and low-speed stationary tracking performance:

                  When some servo systems are applied at low speeds, there will be uneven jumping and crawling, which leads to the acceleration of this phenomenon:

                  εL=

                  This is due to the difference between the static friction moment and the Coulomb friction moment, as shown in the above equation, the moment of inertia is large,εL

                  Thus, the low speed characteristics of the system are improved.

                  To sum up, in general, it is desirable to be smaller, especially in the application of robotic arms, but smaller is not always better, especially in situations where low speed and smooth operation are required.

                  In general, the match between the moment of inertia of the motor Jm and the moment of inertia of the load is expressed by the matching coefficient λ :

                  λ=

                  Matching coefficientλ=15

                  At this point, we have been able to determine the deceleration ratio I, while ensuring the speed required by the application, and check the matching problem of moment of inertia.
                  Before choosing the speed reducer size, we need to evaluate the operation condition, that is, we need to know whether it is working in S1 or S5.
                  According to the customer's movement requirements to make the work drawing is shown on the right.
                  Calculate the running cycle time of a single cycle:
                  And duty cycle:

                  P-B-23

                  IfED60%,At the same timeEZ20min(There is also a manufacturer's label15min),The working mode is S5, otherwise, even if one of the conditions is not met, it is S1 working mode.

                  To evaluate the acceleration impact, count the number of cycles:

                  It is possible that the working diagram is more complex than Figure B-23, such as having multiple different shaped runtime areas that can be computed and summarized in the same way. In general, the application of interpolation operation should basically belong to S1 working mode.
                  Most applications can use the following quick method:

                  S1 Working mode:

                  A simple method only needs to work out the output torque amplified by the reducer according to the rated torque of the motor:

                  T2m=Tm×i

                  Then select a gear box that satisfies T2m≤T2N (rated output torque of the gear box)
                  Finally, check whether the motor shaft diameter is less than or equal to the maximum input diameter of the selected gearbox. (This step is important because some manufacturers give a large output torque rating but specify a disproportionately small maximum allowable input diameter.)
                  S5 working mode:
                  The process is the same as in S1 mode, just change the rated torque T2M of the motor to the peak acceleration torque T1B of the motor, and the rated output torque T2N of the gearbox to the maximum acceleration torque T2B of the gearbox

                  T2b=T1b×i

                  Meet T2bT2B

                  Of course, we still need to check the axle diameter

                  The gearbox selected by the above quick method may be too large and the number of cycles must be less than 1000 / h

                  It needs to be more objective, it needs to be more sophisticated:

                  S1 Working mode:

                  This is represented by the average output moment T2m

                  T2m=

                  Then select a gear box that meets T2M ≤T2N (rated output torque of the gear box). If necessary, verify the service life of the bearing L10. Of course, we still need to check the axle diameter.

                  S5 Working mode:

                  If the number of cycles is ≥1000 times/hour, find out the coefficient fs according to the figure on the right, and calculate as follows:

                  T2b=T1b×i×fs×η

                  Meet T2bT2B

                  If there is external impact load T2NOT, check that T2NOT ≤ T2NOT = Emergency Stop Torq.
                  If necessary, verify the service life of the bearing L10. Of course, we still need to check the axle diameter.

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