d1 | d2 Thread | l1 | d3 | d4 | l2 min. | max. | l3 | l4 | s Square | Static load | Stiffness R | Max. compression |
---|---|---|---|---|---|---|---|---|---|---|---|---|
3.15 | M 12 x 1.25 | 5.24 | 2.36 | 2.83 | 1.38 | 1.81 | 1.26 | 0.39 | 0.28 | 1124 lbf (5000 N) | 14275 lbf/in (2500 N/mm) | 0.08 |
4.72 | M 16 x 1.5 | 5.67 | 3.15 | 4.29 | 1.57 | 2.01 | 1.44 | 0.39 | 0.35 | 2248 lbf (10000 N) | 22841 lbf/in (4000 N/mm) | 0.10 |
6.30 | M 20 x 1.5 | 7.40 | 3.94 | 5.91 | 1.97 | 2.48 | 1.71 | 0.39 | 0.47 | 4496 lbf (20000 N) | 51391 lbf/in (9000 N/mm) | 0.09 |
7.87 | M 20 x 1.5 | 7.80 | 5.12 | 7.32 | 2.36 | 2.87 | 2.15 | 0.39 | 0.47 | 8992 lbf (40000 N) | 85652 lbf/in (15000 N/mm) | 0.11 |
Interference frequency [Hz]: Static load F [lbf or N]: Degree of insulation [%]: Compression s [in or mm]: Stiffness R [lbf/in or N/mm]: |
The first step is to determine the static load F per leveling foot. With appropriately arranged leveling feet and thus an evenly distributed load F, this value is calculated using the following equation: Force due to weight of the machine [lbf or N] / Number of leveling feet = Static load F [lbf or N] per leveling foot Use the calculated static load F to select a leveling 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 leveling foot is also taken from the table. Static load F [lbf or N] per leveling foot / = Actual compression s [in or 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 chart shown above. To optimize the achievable degree of insulation, the number of leveling feet may be changed such that the static load F for each leveling foot is as close as possible below a load capacity value given in the table. This will increase the compression s which, in turn, results in a higher degree of insulation. |
Static load F (machine weight) = 48,000 N, number of feet = 4, ergo: static load per foot = 12,000 N 12,000 N (static load/foot) / = 1.3 mm With an interference frequency of 20 Hz (1,200 rpm), the above chart shows a degree of insulation of only about 20 %. To optimize, the number of feet may be increased to 5, resulting in a static load per foot of 9,600 N. A leveling foot whose static load capacity is closer to the new result may now be selected. 9,600 (static load/foot) / = 2.4 mm With the same interference frequency of 20 Hz (1,200 rpm), the above chart now shows a degree of insulation of approximately 75 %. |
Min | Max | Price |
---|---|---|
1 | 6 | $67.51 |
7 | 99 | $65.06 |
100 | 249 | $61.38 |
≥ 250 | $55.24 |
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