Visualize Pascal's principle force amplification in real time. Adjust piston diameters, input force, and stroke to compute output force, pressure, flow rate, and pump power instantly.
The core principle is Pascal's Law: pressure applied to a confined fluid is transmitted undiminished. This creates the force multiplication between two connected pistons.
$$P = \frac{F_1}{A_1}= \frac{F_2}{A_2}$$Where $P$ is the fluid pressure (Pa), $F_1$ and $F_2$ are the forces on the input and output pistons (N), and $A_1$ and $A_2$ are their respective areas (m²). Since area depends on diameter $d$, the force ratio is $F_2 = F_1 \cdot (d_2/d_1)^2$.
Because the fluid is incompressible, the volume displaced by the input piston must equal the volume received by the output piston. This governs the trade-off between force and distance.
$$A_1 \cdot s_1 = A_2 \cdot s_2 \quad \text{or}\quad s_2 = s_1 \cdot \frac{A_1}{A_2}$$Here, $s_1$ and $s_2$ are the stroke lengths (m) of the input and output pistons. The flow rate $Q$ (m³/s) is the volume displaced per unit time: $Q = A_1 \cdot v_1$, where $v_1$ is the input piston speed.
Automotive Lifts: Car jacks are the most common example. A mechanic applies a small force over a long stroke on the small piston, and the large piston lifts the vehicle with massive force over a short distance. The simulator's force ratio directly models this.
Hydraulic Presses: Used in manufacturing to mold metal or crush materials. They use the same principle on an industrial scale, with precisely controlled pressure to apply tons of force for shaping or assembly.
Construction Equipment: Excavators, bulldozers, and crane arms use hydraulic cylinders to generate powerful, controlled linear motion. The operator's gentle lever movement controls high-pressure fluid to move massive loads.
Aircraft Landing Gear & Brakes: Hydraulic systems are crucial in aviation for retracting landing gear and applying brake pressure. They provide reliable, powerful actuation in a compact system, where safety depends on precise force transmission.
There are a few key points you should be especially mindful of when starting to use this simulator. First is the point that "force amplification is not infinite". While increasing the output piston size raises the output force, real hydraulic cylinders always have an upper limit called the "rated pressure". For example, if the pressure rating of the pump or hoses is 21 MPa (approx. 210 atmospheres), the maximum output force you can generate there is determined by $F_{out} = P_{max} \times A_{out}$, no matter how large you make the output piston area. Indiscriminately increasing the piston size can sometimes just be a waste of cost and space.
Next is the simulator's assumption of "incompressibility". Actual hydraulic oil is slightly compressible, and hoses also expand. In machine tools requiring ultra-precise positioning, if you don't account for this "spring constant of the oil", you can run into trouble like the machine not stopping at the intended position under load. Also, while the calculation might make it seem like force is transmitted instantly, in reality, delays occur due to oil viscosity and pipe resistance. When designing a control system for emergency stops, it's dangerous not to consider this "transmission delay".
Finally, a misconception about efficiency. The "required pump power" calculated by the simulator is a theoretical value. In reality, losses occur from various factors: mechanical losses in the pump itself, motor efficiency, pressure losses in the piping, leaks at the cylinder seals, etc. When selecting actual components, the practical wisdom is to apply a safety factor of at least 1.2 to 1.5 times this theoretical value and choose motors and pumps with some margin.
A bottle jack with input piston d1=15 mm (area=176.7 mm²), output piston d1Num=80 mm (area=5,027 mm²), applied force f1=300 N, and stroke s1=50 mm. Input pressure = 300 N ÷ 0.0001767 m² = 1.70 MPa. Output force = 1.70 MPa × 0.005027 m² = 8,546 N (871 kgf). Input volume displaced = 176.7 mm² × 50 mm = 8,835 mm³. Lift height = 8,835 mm³ ÷ 5,027 mm² = 1.76 mm per pump cycle.