Issue link: https://onenorgren.uberflip.com/i/1463811
136 VAC U U M C U P S & F I T T I N G S 6 For more information or for additional configurations, please contact us at NASInsideSales@imi-precision.com VACUUM CUPS Atmospheric Pressure Atmospheric Pressure Non Porous II. Determine Type of Material to be handled: Non-Porous, Porous, Flexible/Non-Porous Materials being handled in pick & place applications can be grouped into three categories – non-porous, porous and flexible. It is important to determine what type of material you are working with in order to determine the cup type, and the fitting choices. Norgren offers a variety of cup styles – including bellows, multi-bellows, round, oval, flat (with and without cleats), cups with removable fittings and cups with permanent fittings. Non-Porous Materials: steel, glass, laminated chipboard, rigid plastic, semiconductors, etc. Handling non-porous materials is the easiest application for choosing a vacuum cup and vacuum generator because there is no vacuum flow (leakage). The cup seals to the surface of the object enabling the generator to reach its maximum vacuum level. Typically, flat cleated cups are used for non-porous applications because the rigid, low profile design resists peeling away. In horizontal applications, where there is a large array of cups, bellows cups may be an option as they offer the pliability needed to ensure that all cups make contact with the object(s) being handled. Vacuum Cup Selection Guide Example: Holding Force Calculation for Non-Porous Materials Application: lift a 100 lb [45.36 kg] steel plate, 1/8" [3mm] thick, measuring 4' x 4' [121.9cm X 121.9cm] from a horizontal stack and place into a press. Norgren recommends an "H" series generator when handling non-porous materials. All "H" series generators generate 14 PSI [28"Hg, 0.965 bar]. F = P * A Force = 200 lbs [90.72 kg](weight x safety factor/horizontal lift or 100 lbs [45.36 kg] x 2) Pressure = 14 PSI [.965 bar] (convert 28"Hg to PSI by dividing by 2) If F (200 lbs [90.72 kg]) = P (14 PSI [.965 bar]) * A (Solve for A) A = 200/14 [90.72/.965] which is 14.3 in 2 [94.01 cm 2 ] – "A" represents the total area of the cup or all the cups combined to lift this load horizontally Determine the Number of Cups Needed to Determine the Diameter of each Cup Whereas the metal is only 1/8" [3mm] thick, it will tend to droop. Norgren recommends using 2 rows of 3 cups each for a total of 6 cups. Therefore, 14.3 in 2 [94.01 cm 2 ] divided by 6 cups = 2.38 in 2 [15.67 cm 2 ] is the area per cup Solve for the diameter (d) using the equation: A = p * [ d 2 / 4] or pr 2 [A = p ( * d 2 / 4) or pr 2 ] d 2 = 4 x 2.38 / p or d 2 = 3.03 in 2 [d 2 = 4 x 15.67/ p or d 2 = 19.96 cm 2 ] d = sq. root of 3.03 or 1.74 in [d = sq. root of 19.96 or 44.7mm] Solution: Choose a flat cup with cleats with a diameter of 1.75" [44.45mm] or greater. With plenty of space on the steel plate to position cups, choosing a larger cup will add to the holding force and take into account any acceleration or deceleration loads during transfer.