End-of-Arm Tooling and Grippers

Worked example: a 2.1 kg part

• Two fingers (n = 2), steel parts with rubber-faced fingers, μ = 0.3 (use a low value; oil lowers it), safety factor 2, and a move that accelerates the part at 10 m/s² (about 1 g) upward: N = 2.1 × (9.81 + 10) × 2 ÷ (0.3 × 2) = 139 N per finger.
• A pure static hold, with no acceleration, would need only 69 N: the move, not the weight, sets the force.
• With a high-friction urethane finger face (μ = 0.6) the force halves to 69 N: friction is the cheapest way to cut the force, the gripper size and the weight.
• A pneumatic gripper with a 20 mm piston at 6 bar gives F = P A = 188 N before losses; with a typical efficiency of 80 % split over two fingers that is about 75 N per finger at the cylinder, less at the fingertip if the fingers are long (the force falls with the distance from the pivot or the guide). Use the gripper maker’s force chart at your finger length.
• Two rules: grip force must be ample but not crush the part; and the force the gripper makers quote is at the base of the finger, at the stated pressure, so check the finger length and the supply pressure (a gripper at 4 bar, not 6, delivers two thirds of the force).

Worked example: one cup, then a box

• One 40 mm cup at a vacuum of −0.6 bar (60 kPa below atmosphere): area = π ÷ 4 × 0.04² = 1 257 mm², force = 60 000 × 0.001257 = 75.4 N.
• Vertical lift with a safety factor of 2: mass = 75.4 ÷ (9.81 × 2) = 3.84 kg per cup. Fast horizontal moves or friction-only holding with a factor of 4 or more: 1.92 kg.
• A 5 kg box carried on 4 cups of 30 mm: each gives 42.4 N, together 170 N. With 30 % lost to leakage and tilt, the usable force is 119 N. The load, lifted at 5 m/s² with a factor of 2, is 148 N: not enough.
• Sideways forces are carried by friction, and are small: the cups hold 59 N sideways at μ = 0.5, against a 30 N inertia force at 6 m/s² for the box. Slow the horizontal acceleration, or use cups with a lip, stops or a gripper with side supports.
• Real cup force is lower than the theoretical because of seal leakage, surface roughness and cup flex: use the cup maker’s rated force and keep a factor of 4 for rough, oily or tilted loads.

Worked example: how long to build vacuum?

An ejector with a free-air flow of 25 L/min evacuates the cups and hose. For a volume V = 0.05 L it needs about t = (V ÷ Q) ln(p_atm ÷ p_target) = (0.05 ÷ 25 × 60 s) × ln(101.3 ÷ 41.3) = 0.11 s to reach −60 kPa. Doubling the line volume to 0.12 L (a long hose) takes 0.26 s. That is gripper time added to every cycle (see Payload, Inertia, Speed and Cycle Time): keep the ejector close to the cups and the hose short.

Worked example: how far must the fingers open?

• A part 60 mm wide ±0.5 mm is picked from a nest. The fingers must open to at least the largest part plus a clearance on each side: 60.5 + 2 × 3 = 66.5 mm.
• They must also close on the smallest part: 59.5 mm. The stroke between those two settings, 7.0 mm, is the real working stroke; choose a gripper whose adjustment range and tolerance of the pick location cover it, with the clearance larger than the robot’s placing error.
• Fingers of aluminum instead of steel: a finger of 120 × 20 × 10 mm weighs 65 g in aluminum and 188 g in steel; for two fingers 0.1 m from the wrist axis the difference in inertia is 0.0025 kg·m². Small, but it is mass at the end of the lever; drill out the fingers and use plastic or aluminum where the loads allow.

Worked example: inserting a 10 mm peg

• The peg has 0.1 mm total clearance in its hole, but the robot and the part position are uncertain to ±0.4 mm (see Vision, Calibration and Accuracy). Without help the peg would land on the hole’s edge.
• A lead-in chamfer on the hole of at least 0.4 − 0.1 = 0.35 mm per side lets the peg find the hole; a compliant device lets it follow the chamfer without jamming.
• Jamming: a rigidly held peg that enters the hole with a sideways offset can wedge when the depth of insertion is more than about 1 ÷ (2 μ) times the diameter. With μ = 0.2 that ratio is about 2.5: beyond it, pushing harder only locks the peg. Compliance and a chamfer keep the push along the hole’s axis.
• Force sensing or motor-current monitoring allows a search (a small spiral) and detection of seating (see Vision, Calibration and Accuracy).

Worked example: what does a changeover cost?

If a tool change takes 3 s to put one tool back and take another, and the cell changes tool 12 times an hour (twice per change: drop and pick), it spends 72 s an hour, or 2.0 % of the time, changing tools. Cheap if the cycle is 20 seconds, expensive if the tool changes every cycle: then use a multi-function tool with two grippers on one plate.

Worked example: holding after the air fails

A vacuum reservoir of 0.5 L at 6 bar feeds a valve that must keep above 4 bar. With a steady leak of 0.1 L/min of free air (the volume at atmospheric pressure), the usable air is V (p₁ − p₂) = 0.5 × 2 = 1.0 L of free air, which lasts about 10 minutes. Plan the response (a controlled stop and lowering the load) within that time, and check the actual leak rate on the real tool.