A | With two-hole flange (d1 = 60 / 90 / 113) |
B | With four-bolt flange (d1 = 113 / 126) |
1 | Without tear-off lock |
2 | With tear-off lock |
d1 | d2 | d3 | d4 Type A | d5 Type B | h | s | b Type A | l1 Type A | l2 Type B | m1 Type A | m2 Type B |
---|---|---|---|---|---|---|---|---|---|---|---|
60 | M 10 | 78 | 9 | - | 30 | 2 | 78 | 128 | - | 110 | - |
90 | M 12 | 106 | 13 | - | 39 | 3 | 110 | 170 | - | 140 | - |
113 | M 16 | 150 | 12,5 | 12,5 | 52 | 4 | 150 | 216 | 168 | 184 | 132 |
126 | M 20 | 177 | - | 13 | 63 | 4 | - | - | 184 | - | 150 |
F1 = static load in vertical direction (pressure) Stiffness R: Equation for calculating the stiffness: R = F/S |
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The table below gives details on the maximum static load F, the maximum rated compression and the resulting stiffness R. |
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Interference frequency [Hz]: Static load F [N]: Degree of insulation [%]: Compression s [mm]: Stiffness R [N/mm]: |
The first step is to determine the static load F for each levelling foot. With favourably arranged levelling feet and therefore an evenly distributed load F, this value is calculated using the following equation: Force due to weight of the machine [N] / Number of levelling feet = static load F [N] for each levelling foot Use the calculated static load F to select a levelling foot from the table, making sure that the static load F lies as close as possible to the static load capacity without exceeding it. The associated stiffness R of the selected levelling foot is also taken from the table. Static load F [N] per levelling foot / = actual compression s [mm] Starting from the calculated actual compression s, the achievable degree of insulation as factor of the interference frequency can now be taken from the graph shown above. |
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