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How does contact area change compression?

Use the same force, mass and damper on the same ideal spring foundation. Change contact area to compare compression and oscillation.

Diagram and inputs ↓
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Same load, different supporting feet

A 1 kg equivalent load, 9.80665 N, uniform on a circular contact face.

All three diameters use one scale. Results are mean pressure, not indentation or failure predictions.

More numeric references and examples

Compare values with everyday references

Baseline: 1 kg on a uniform circular contact 1 cm across gives about 125 kPa

Compare values and familiar conditions, then apply a representative example.

Support 1 kg on a round footAbout 124.86 kPa

A circular contact 1 cm across, uniformly supporting 1 kg, or 1 L water equivalent.

Support 1 kg on a round footAbout 13.87 kPa

A circular contact 3 cm across, uniformly supporting 1 kg, or 1 L water equivalent.

Support 10 kg on a round footAbout 12.49 kPa

A circular contact 10 cm across, uniformly supporting 10 kg, or 10 L water equivalent.

More references (3)
Support 1 kg on a round footAbout 31.22 kPa

A circular contact 2 cm across, uniformly supporting 1 kg, or 1 L water equivalent.

Support 5 kg on a round footAbout 24.97 kPa

A circular contact 5 cm across, uniformly supporting 5 kg, or 5 L water equivalent.

Support 0.1 kg on a round footAbout 12.49 kPa

A circular contact 1 cm across, uniformly supporting 0.1 kg, or 0.1 L water equivalent.

Reference conditions

Pressure = force / area, assuming uniform contact pressure. Does not assess real feet, flooring or effects on the body.

Visual preview

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Spatial schematic · circular contact face

0.00 sReveal animation

p=F/A, K=kA×A, m x″+c x′+Kx=F. x=0 is the equilibrium under gravity/preload; additional force F is applied at t=0. Initial displacement and velocity are zero. This is an ideal foundation, not a prediction for a named material.

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Conditions

p=F/A, K=kA×A, m x″+c x′+Kx=F. x=0 is the equilibrium under gravity/preload; additional force F is applied at t=0. Initial displacement and velocity are zero. This is an ideal foundation, not a prediction for a named material.

Motion conditions and equations

Conditions represented by this motion

p=F/A, K=kA×A, m x″+c x′+Kx=F. x=0 is the equilibrium under gravity/preload; additional force F is applied at t=0. Initial displacement and velocity are zero. This is an ideal foundation, not a prediction for a named material.

Foundation stiffness per area kA, plunger mass m and damping c define displacement through the equation of motion. Displacement magnification is shared by A/B. Plasticity and lateral coupling within the foundation are excluded.

From pressure to pushing force
  1. 1. Match the unitsPressure unit converter →
  2. 2. Calculate with the required inputsPneumatic cylinder thrust →
  3. 3. See the motion or differencePressure and contact areaCurrent page

Also enter bore, rod diameter, back pressure and losses. Pressure alone does not determine thrust.

Each tool opens with its own conditions. Values are not transferred automatically.