Check whether the clearances of the aggressive forces lie within the permitted values (see Maximum permitted clearances).
Option: 1 rail, 1 carriage | ||
---|---|---|
sy + sz | < | 2 Lx - Y0 |
ay + az | < | 2 Lx - Y0 |
sy | < | 5 Zm |
sz | < | 5 Zm |
Calculate the required driving power
Initially four calculations should be carried out:
The driving power Fa corresponds to the calculated maximum value:
Calculate the maximum bearing load
Maximum bearing load in y-direction
Maximum bearing load in z-direction
A = Point of origin
Fa: | Driving power | [N] |
Fs: | Mass force | [N] |
Fy, Fz: | Bearing load in y- or z-direction | [N] |
sx, sy, sz: | Clearance of the mass force in x-, y- or z-direction | [mm] |
ay, az: | Clearance of the driving power in y- or z-direction | [mm] |
wx: | Clearance of the carriage on a rail | [mm] |
LX: | Constant based on the installation size | [mm] |
Zm: | Constant based on the installation size | [mm] |
Y0: | Constant based on the installation size | [mm] |
b: | Clearance of the guide rails | [mm] |
µ: | Coefficient of friction, µ = 0 in static loads, µ = 0.2 in dynamic loads | |
ZW: | Number of carriages per rail |
Order no. | LX | ZM | Y0 |
---|---|---|---|
[mm] | [mm] | [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 |
1 rail, 1 carriages | |
---|---|
K1 | |(ay+Y0)/Lx| |
K2 | (sy+Y0)/Lx |
K3 | |az/Lx| |
K4 | |sx/Lx| |
K5 | sz/Lx |
K6 | |(sy+Y0)/Zm| |
K7 | |sz/Zm| |
Check the maximum bearing load of the most strongly stressed bearing with the calculated load in Step No. 4.
Order no. | Fymax, Fzmax [N] |
---|---|
TW-01-15 | 2000 |
TW-01-20 | 3700 |
TW-01-25 | 5000 |
TW-01-30 | 7000 |
Determination of the maximum permitted speed for the load in Step No. 4
Diagram for the determination of the maximum permitted speed for the determined bearing load.
X = centric bearing load [N]
Y = permitted mean velocity [m/s]
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