Hydraulic jack is a tool that is often used. Sometimes you need to know the pushing and pulling force of hydraulic jack and the speed of operation. Now, TorcStark will detail how to calculate the push-pull force and movement speed of a hydraulic jack when the cylinder bore, rod diameter, hydraulic supply pressure and flow are known.

The main parameters of hydraulic jacks are: cylinder diameter, rod diameter, stroke, hydraulic pressure, flow, thrust, pull, and movement speed.

The first three are its structural parameters, which have been determined at the factory and are fixed values.

The latter parameters are related to their working conditions, that is, the supply pressure and flow determine the thrust, pull, and movement speed.

Different pressure and flow have different corresponding thrusts, pulling forces and moving speeds. Of course, the pressure of the hydraulic system has a rated value and cannot be infinitely high. The relationship between them is determined by two simple but important formulas.

Push and pull calculation formula: F=PS
F: push force, pull force, unit N, kN
P: hydraulic pressure, unit Pa, MPa
S: the cross-sectional area of the cylinder body and the rod body, unit s2

This formula actually evolved from another formula: P=F/S .

This is the definition formula of pressure. When the applied pressure F and the force area S are known, the pressure P can be obtained. This formula is universal.

For example, why are our feet easy to sag when we step on the soil, and a wooden board is placed on the soil, and it will not sag when we step on it again. This is the reason for the different pressures.

If it is a hydraulic system, P is the same everywhere, that is, F1/S1=F2/S2, which is the famous Pascal’s law, which is an important theoretical basis for the work of the hydraulic system.

It should be pointed out that the corresponding relationship of the above units of quantity, F, P, and S corresponds to N, Pa, s2, you must pay attention to the calculation.

Rod and cylinder speed v=Q/S, V unit, m/s

Q: flow m3/s

S: Rod, cylinder cross-sectional area m2

Note: The rod body and the cylinder body are different. The cylinder body refers to the outer casing of the hydraulic cylinder and the inner part of the rod body hydraulic cylinder. The movement of the two is relative, it can be that the rod is stationary and the cylinder is moving, or the cylinder is stationary and the rod is moving. According to different needs, choose different ways.

Through this formula, we can know that under the condition of the same flow, the cross-sectional area increases, the movement speed decreases, and vice versa. When designing, we can choose to change the flow rate or cross-sectional area to change the movement speed to achieve our purpose.

The above two formulas are important formulas to solve the problem of hydraulic cylinder and rod motion. They are relatively simple and very useful. They are important formulas for designing and manufacturing hydraulic components, not only for hydraulic cylinders but also for the design and calculation of hydraulic control valves. Today, only the hydraulic cylinder will be explained.

Examples are as follows:
As shown in the figure below, the cylinder diameter of a hydraulic jack is D=160mm, the rod diameter is d=85mm, and the hydraulic pressure
P=40.47MPa, flow 0.2m3/min.

What are the pulling force and the pushing force? What are the extension speed and the retraction speed?

The answering process is as follows

Push and pull calculations







Extend and retract speed calculation






From the above calculations, for a hydraulic cylinder with a fixed structure, the thrust is greater than the pull, and the relationship between the two is a constant ratio relationship, which depends on the diameter ratio; but the retraction speed is greater than the extension speed, which is also a constant ratio relationship.

According to the above theory, different forms of jacks or different connection forms can be made to meet different application occasions.