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drylin® T - system design, vertical, 1 rail and 1 trolley
Step 2:
Check whether the distances between the applied forces are within the permissible values (see maximum permissible distances).
| Variant: 1 rail, 1 trolley | ||
|---|---|---|
| sy + sz | < | 2 Lx - Y0 |
| ay + az | < | 2 Lx - Y0 |
| sy | < | 5 Zm |
| sz | < | 5 Zm |
Step 3:
A = Coordinate origin
Calculating the required drive force
Firstly, four calculations must be carried out:
Variables in the calculations
| Fa: | drive force | [N] |
| Fs: | Mass force | [N] |
| Fy, Fz: | Bearing load in y or z direction | [N] |
| sx, sy, sz: | Distance of the inertial force in x, y or z direction | [mm] |
| ay, az: | Distance of the drive force in the y or z direction | [mm] |
| wx: | Distance between the wagons on a rail | [mm] |
| LX: | Size-dependent constant | [mm] |
| Zm: | Size-dependent constant | [mm] |
| Y0: | Size-dependent constant | [mm] |
| b: | Distance between the guide rails | [mm] |
| µ: | coefficient of friction, µ = 0 for static loads, µ = 0.2 for dynamic loads | |
| ZW: | Number of slides per rail |
Step 4:
Calculate the maximum bearing load
The constant values
| Order number | LX [mm] | ZM [mm] | Y0 [mm] |
|---|---|---|---|
| TW-01-15 | 29 | 16 | 11.5 |
| TW-01-20 | 35 | 23 | 15.0 |
| TW-01-25 | 41 | 25 | 19.0 |
| TW-01-30 | 49 | 29 | 21.5 |
Coefficients:
| 1 rail, 1 trolley | |
|---|---|
| K1 | |(ay+Y0)/Lx| |
| K2 | (sy+Y0)/Lx |
| K3 | |az/Lx| |
| K4 | |sx/Lx| |
| K5 | sz/Lx |
| K6 | |(sy+Y0)/Zm| |
| K7 | |sz/Zm| |
Step 5:
Check the maximum bearing load of the most heavily loaded bearing with the load calculated in step no. 4
Maximum permissible load
| Order number | Fymax, Fzmax [N] |
|---|---|
| TW-01-15 | 2,000 |
| TW-01-20 | 3700 |
| TW-01-25 | 5,000 |
| TW-01-30 | 7,000 |
Step 6:
graph to determine the maximum permissible speed for the determined bearing load.
X = centric bearing load [N]
Y = permissible average speed [m/s]
Determination of the maximum permissible speed for the load from step no. 4
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