Vacuum Gripper Calculator
Calculate the theoretical holding force of a suction cup and the required number of cups. Enter cup diameter, vacuum, workpiece mass, acceleration, friction coefficient, load case and safety factor – the calculator returns F_th per cup, the required holding force for the chosen load case, the rounded-up cup count and a traffic-light rating against the cups actually fitted, live with every input.
Suction Gripper Calculator
Results
Calculating …
Formulas and fundamentals
The theoretical holding force of a suction cup is F_th = A·Δp with the effective suction area A = π/4·d² and the vacuum Δp (difference between ambient and system pressure). For d = 40 mm and Δp = 0.6 bar this gives A = 12.566 cm² and F_th = 75.4 N per cup. The force actually achievable is lower because leakage, surface roughness and compliant bellows reduce the vacuum and the sealing area; this loss is covered by the safety factor.
The required holding force depends on the orientation of the cup relative to the force (load cases per Schmalz): Load case I (cup horizontal, force vertical) F_erf = m·(g + a)·S; load case II (cup horizontal, force horizontal, held by friction) F_erf = m·(g + a/μ)·S; load case III (cup vertical, force vertical or swivelling) F_erf = (m/μ)·(g + a)·S. Here m is the workpiece mass, g = 9.81 m/s², a the acceleration, μ the friction coefficient cup/workpiece and S the safety factor.
The required number of cups is n ≥ F_erf/F_th, rounded up. The safety factor per accident-prevention rules is at least 1.5 for smooth, tight surfaces, at least 2.0 for critical or porous materials and at least 2.5 for swivelling (tilting moments). The traffic light compares the available holding force n·F_th with the required holding force: green when the required safety is met, amber when the load is held but the safety factor is not reached, red when the load is not held.
Worked example
A 10 kg sheet is lifted flat with 40 mm cups at 0.6 bar vacuum and accelerated vertically at a = 5 m/s² (load case I). The theoretical holding force per cup is F_th = 75.4 N.
The required holding force is F_erf = m·(g + a)·S = 10·(9.81 + 5)·1.5 = 222.15 N. This gives a required number of cups n = 222.15/75.4 = 2.95, i.e. 3 cups rounded up.
With 4 cups fitted the available holding force is 4·75.4 = 301.6 N and the available safety is S_vorh = 301.6/148.1 = 2.04 – the required safety of 1.5 is met and the traffic light is green.
Frequently asked questions
Why is the real holding force smaller than the theoretical one?
F_th = A·Δp assumes a tight, flat surface and full vacuum. In practice leakage on rough or curved surfaces, the compliance of bellows cups and an incompletely built-up vacuum reduce the force. This loss is not calculated individually but covered by the safety factor S.
Which load case should I choose?
Load case I applies when the cup sits horizontally and the force acts perpendicular to the suction face (lifting a flat part). Load case II applies with a horizontal cup and a horizontal force held only by friction. Load case III applies with a vertical cup when the load hangs by friction on the suction face or is swivelled – this is the most unfavourable case.
Which safety factor should I use?
Per accident-prevention rules at least 1.5 for smooth, tight and dry surfaces, at least 2.0 for critical, porous, oily or moving materials and at least 2.5 for swivelling, because additional tilting moments arise. When in doubt choose the higher value.
How does the friction coefficient enter the calculation?
In load case I the force acts perpendicular to the suction face and friction is irrelevant. In load cases II and III the load is held by friction; a small friction coefficient greatly increases the required holding force (factor 1/μ). Typical values are μ = 0.3 to 0.6 depending on the workpiece and cup material.
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