Share this

How to solve the acceleration problem of a moving object under PLC control?

2026-04-06 05:58:43 · · #1

In our practical work, when we encounter a variable that is controlled as an output result , such as a moving object, rotational speed, flow rate, pressure, etc., due to the limitations of physical parameters such as mechanical structure and materials, it is often necessary to reduce the impact during startup or shutdown, that is, to establish startup acceleration or shutdown deceleration to reduce the impact on the physical structure.

When a moving object starts, if the object is heavily loaded, such as a crane or a heavy vehicle, its physical inertia is too great. If the starting acceleration is too great, it will cause an impact on the power source and even damage the mechanical structure.

Flow control systems, such as large electric valves, can cause damage to pipelines and pumps if the valve is opened too quickly, leading to pipeline vibration, pump stoppage (caused by motor overcurrent protection), and so on.

Pressure control systems, such as hydraulic pipelines, tension, and strain gauges, can cause pipeline vibration and physical deformation of the stretched object if the starting acceleration is too high.

Through practice, the author has developed a simple yet accurate solution: First, determine the allowable values ​​for mechanical acceleration based on the physical parameters of the controlled object, such as acceleration, rate of change of flow rate per unit time, and rate of change of pressure (tension, tension) per unit time . Then, calculate the acceleration time (deceleration time) or change time, assuming this time is T, the maximum speed of the moving object is V, the maximum system pressure is P, and the maximum flow rate is Q. The corresponding analog output is 0-10V voltage, and the PLC is 15-bit, with an internal value of 0-32767. That is, when the output is 0V, the PLC internal value is 0 , and when the output is 10V, the PLC internal value is 32767. By controlling the upper and lower limits of the output value change from 0 to 32767 to equal T, uniform acceleration control can be achieved. The idea is as follows:

1. Select two timing control function blocks TON inside the PLC and make them work alternately and periodically, and select an adder;

2. Using an approximate piecewise linear value method, when the two timers work alternately and periodically, the adder will accumulate a constant C sequentially and send it to the PLC output.

III. Assume that timer 1 has a timing period of t1 and timer 2 has a timing period of t2, such that t1 = t2. When timer 1 is running, the adder adds a constant C. After timer 1 expires, timer 2 starts, timer 1 stops, and the adder stops. After timer 2 expires, timer 1 starts, the adder adds a constant C = 2C, and timer 2 stops. This process continues with 3C, 4C, and so on, until the input instruction is greater than or equal to the input instruction, at which point this section of the program ends.

When changing from large to small, simply replace the adder with a subtractor.

Here, by simply adding up the total number of timers t1 corresponding to the output upper limit value, we can obtain the acceleration time T.

Application examples;

• Gantry crane travel, load capacity 450 tons, trolley travel speed 0-5 m/min, overhead crane travel speed 0-4 m/min, frequency conversion speed regulation, Siemens PLC control, RS485 communication, travel height 9 meters, load height approximately 5-6 meters.

If the acceleration during travel is not controlled, once the load is hoisted to a height of 5 to 6 meters and starts moving, it will sway back and forth in mid-air once it stops moving. On a slope, this could even cause it to slide down, with very serious consequences.

Large-displacement water pumps require heavy-load startup. Under normal use, the motor power is far greater than the actual operating power, resulting in a significant impact on the power grid during startup. However, if an electric valve is added to the pump outlet, with a displacement control signal of 0-10V, and the motor is selected based on 1.2 times the actual power consumption, the outlet controlled by the electric valve is closed during startup. After the motor starts, the electric valve is slowly opened. This prevents the motor from overloading and reduces the impact on the power grid. Furthermore, the pump displacement can be adjusted according to actual needs.

In addition, frequent starts and stops of chain drives can easily cause deformation of the mechanical structure, making the chain longer and the keys wear.

In conclusion, by controlling high-speed or high-pressure transmission systems that frequently start and stop, many mechanical failures can be reduced, and significant energy-saving benefits can also be achieved.

Read next

CATDOLL 123CM Sasha TPE

Height: 123cm Weight: 23kg Shoulder Width: 32cm Bust/Waist/Hip: 61/54/70cm Oral Depth: 3-5cm Vaginal Depth: 3-15cm Anal...

Articles 2026-02-22