BIZOMAT MAKEOS · Pneumatics
Vacuum Cup Holding Force
Will these suction cups hold this part during handling?
Holding force of vacuum suction cups from cup diameter, vacuum level and number of cups, compared with the force needed for horizontal lifting or vertical (shear) handling with acceleration and a safety factor.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Cup effective diameter | mm | Effective suction diameter of each cup. |
| Number of cups | cups | Cups sharing the load. |
| Vacuum level (below atmospheric) | kPa | Vacuum level achieved at the cups (e.g. 60 kPa below atmospheric for porous-free parts). |
| Part mass | kg | Mass of the part. |
| Acceleration during handling | m/s² | Peak acceleration of the gripper. |
| Load direction | choice | Whether the cups pull straight off the part or hold it in shear. |
| Friction coefficient cup to part (vertical) | Friction between cup and part for shear loading (dry metal about 0.5; oily parts much lower). | |
| Safety factor | Safety factor (2 for horizontal, 4 or more for vertical handling are common). |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Cup effective diameter
- What it is
- The effective suction diameter of each cup, the diameter of the area the vacuum acts on when the cup is sealed against the part.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 1000 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 40 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Holding force rises with the square of the effective diameter.
- Get it right
- Use the maker’s effective diameter, which is usually smaller than the cup’s outside diameter.
What this means: Number of cups
- What it is
- The number of suction cups sharing the load of the part.
- Units and format
- Enter a number in cups: cups. The form accepts 1 cups to 100 cups. Use a whole number. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 4 cups as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Total holding force rises in direct proportion.
- Get it right
- Count only cups that will be sealed on the part at the same time.
What this means: Vacuum level (below atmospheric)
- What it is
- How far the pressure inside the cups is below atmospheric pressure when they are sealed on the part, as read on a vacuum gauge.
- Units and format
- Enter a number in kPa: kilopascals. The form accepts 0.000001 kPa to 100 kPa. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 60 kPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Holding force rises in direct proportion to the vacuum level.
- Get it right
- Enter the difference below atmosphere, not an absolute pressure. Porous or rough parts reach lower vacuum.
What this means: Part mass
- What it is
- The mass of the part being handled.
- Units and format
- Enter a number in kg: kilograms. The form accepts 0.000001 kg to 100000 kg. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 5 kg as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The force required rises in direct proportion to mass.
- Get it right
- Enter kilograms; the tool adds gravity and acceleration to get the force.
What this means: Acceleration during handling
- What it is
- The peak acceleration of the gripper during lifting or moving, which adds to gravity in the force the cups must resist.
- Units and format
- Enter a number in m/s²: metres per second squared. The form accepts 0 m/s² to 200 m/s². Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 5 m/s² as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The force required rises with gravity plus this acceleration.
- Get it right
- Include the highest acceleration in the motion profile, including emergency stops if relevant.
What this means: Load direction
- What it is
- How the load acts on the cups: pulled straight off a horizontal part, or held in shear on a vertical part where friction must carry it.
- Units and format
- Choose one of 2 options from the list: Horizontal part, lifted vertically (tension); Vertical part (shear on the cups).
- Example
- The form opens with “Horizontal part, lifted vertically (tension)” selected as a worked example. Choose the option that matches your case.
- Get it right
- Shear loading relies on friction and needs much more holding force than a straight pull.
What this means: Friction coefficient cup to part (vertical)
- What it is
- The friction coefficient between the cup rubber and the part surface, used only when the part is held vertically in shear.
- Units and format
- Enter a plain number with no unit. The form accepts 0.05 to 2. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0.5 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- For vertical handling, the force required falls in inverse proportion to friction.
- Get it right
- Oily or wet surfaces have much lower friction; take the value from the cup maker for your surface. Ignored for horizontal parts.
What this means: Safety factor
- What it is
- The multiple of the calculated force that the cups must be able to hold, to allow for leaks, wear and uncertainty.
- Units and format
- Enter a plain number with no unit. The form accepts 1 to 10. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 2 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The force required rises in direct proportion.
- Get it right
- Take it from the cup maker’s guidance or your safety standard for the orientation used.
Method and formulas
- Holding force per cup = vacuum × π D² ÷ 4 (kPa × mm² ÷ 1000 = N).
- Horizontal: required = m (g + a) × safety factor.
- Vertical (shear): required = m (g + a) ÷ μ × safety factor.
- Example: 4 cups of 40 mm at 60 kPa hold 301.59 N against 148.07 N required.
Limitations
- Flat, non-porous, rigid surfaces; porous, curved or oily parts reduce holding force.
- Cup deformation and peel loads are not considered.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Size vacuum cups for a robot or lifter
- Check a gripper for faster motion
- Compare horizontal and vertical handling
Who it is for
- Automation engineers
- Robot integrators
What you get
- Holding force per cup and in total
- Force required
- Margin
Interpreting the result
- Use the cup maker’s theoretical force tables and add vacuum monitoring for safety.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 3032–3041. Topic reference; no source tables or recommended settings are reproduced.
Learn the Concepts Behind This Tool
These BIZOMAT Library readings explain the method this tool uses, with worked examples.
Method explained here · BIZOMAT original book
Pressure, vacuum and how much gas is left
What it covers: Absolute and gauge pressure, pascals, millibar and torr, the ideal gas law at the molecular scale, how much gas VC-200 holds at 200 Pa, the force on its door, and a vacuum specified in gauge units.
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