Library / Hydraulics, open channels and fluid power / HYD-41
Hydraulics, open channels and fluid power
Net extension force of a double-acting cylinder
Force on the rod of a double-acting cylinder during extension, from the cap-side pressure on the full piston area and the rod-side pressure on the annular area.
- Ref
- HYD-41
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- F [N]
During extension, the cap-side pressure acts on the full piston area, and the rod-side pressure acts on the smaller annular area (the piston area less the rod area). The net force on the rod is F = pc Ap − pr(Ap − Ar). A positive value pushes the rod out. Friction and the inertia of the load are not included, so the force at the load is smaller by the seal friction.
Force on the rod of a double-acting cylinder during extension, from the cap-side pressure on the full piston area and the rod-side pressure on the annular area. Positive force extends the rod. Friction and inertia are excluded.
Calculator
- Annular area, Ap − Ar0.004 m²
- Force with no back pressure, pc Ap50,000 N
- Force lost to back pressure, pr(Ap − Ar)8,000 N
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The cylinder pushes with 42,000 N (42 kN), after losing 8,000 N to back pressure on the rod side.
- Cutting the back pressure to zero would raise the force to 50,000 N, so the return line should have little resistance when the force matters.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Ap Piston area | 0.0025 to 0.5 m² | 22,000 | 4,002,000 | 3,980,000 | +10% | +9.52 % |
| pc Cap-side gauge pressure | 10,000 to 4e7 Pa | −7,950 | 192,000 | 200,000 | +10% | +11.9 % |
| pr Rod-side gauge pressure | 0 to 4e7 Pa | −110,000 | 50,000 | 160,000 | +10% | −1.9 % |
| Ar Rod area | 1e-5 to 0.0025 m² | 40,020 | 45,000 | 4,980 | +10% | +0.476 % |
Sweep of Rod-side gauge pressure
| Rod-side gauge pressure (Pa) | Net extension force (N) |
|---|---|
| 0 | 50,000 |
| 5,000,000 | 30,000 |
| 1e7 | 10,000 |
| 1.5e7 | −10,000 |
| 2e7 | −30,000 |
| 2.5e7 | −50,000 |
| 3e7 | −70,000 |
| 3.5e7 | −90,000 |
| 4e7 | −110,000 |
Sweep one variable
Variables and how each one contributes
Net extension force
What the equation gives you. Everything else feeds it.
Cap-side gauge pressure
The cap-side gauge pressure, in Pa. Raises the force in proportion to the piston area. Raising it by 10 % (to 1.1×107) moves the result by +11.9 %, from 42,000 to 47,000 N.
Rod-side gauge pressure
The rod-side gauge pressure (back pressure), in Pa. Lowers the force in proportion to the annular area. Raising it by 10 % (to 2,200,000) moves the result by −1.9 %, from 42,000 to 41,200 N.
Piston area
The full piston area, in m². Raises the force, mainly through the cap-side pressure. Raising it by 10 % (to 0.0055) moves the result by +9.52 %, from 42,000 to 46,000 N.
Rod area
The rod cross-section area, in m². A larger rod raises the force a little, as it shrinks the annular area carrying the back pressure. Raising it by 10 % (to 0.0011) moves the result by +0.476 %, from 42,000 to 42,200 N.
Worked example with the values above
- Cap-side force: pc Ap = 1×107 × 0.005 = 50,000 N.
- Back-pressure force: pr(Ap − Ar) = 2,000,000 × 0.004 = 8,000 N.
- F = 42,000 N.
Assumptions and limits
- An incompressible fluid with no leakage, except where an efficiency is given.
- Gauge pressures and steady operation, with friction and inertia not included unless stated.
Common mistakes
- Using the piston area where the annular area is needed, or the reverse.
- Mixing rev/min and rad/s.
- Forgetting back pressure on the other side of the piston.
Statistics and uncertainty
Real actuators lose force to seal friction and flow to leakage, and oil bulk modulus drops sharply with entrained air, so the stiffness and speed here are best-case figures.
Where it is used
- Sizing hydraulic cylinders, pumps and motors.
- Estimating drive speed, stiffness and natural frequency of a hydraulic actuator.
Maths you need
- Pascal’s law: pressure acts equally in all directions in a closed fluid
- Displacement: volume moved per revolution or stroke
- Bulk modulus: B, the pressure per unit volume change of the fluid