16
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Details
Angular position of profile rail β
= 90° (vertical axis)
Mass
m
= 200 kg
Initial speed V0
= 0.5 m/s
Valve switching time tV
= 0.016 s
Control unit switching time tSV
= 0.020 s
Available operating pressure
= 6 bar
1. Preselection of braking force
Freq. = m × g
[N]
0.5
Freq. =
200 × 9.81
= 3924
[N]
0.5
Selected: SHB 2 Size 5.1,
Type 3850.0_0_ _ at 6 bar,
Nominal holding force FNom = 6000 N
(from Section 5.2 Table "Technical data")
2. Calculation of stopping distance/stopping time
Checking the selected brake size
Acceleration of the load
aB = g x sin(β) = 9.81x sin(90°) = 9.81 [m/s2]
Acceleration of the load
SSys = V0 × tSys + aB × tSys2 × 0.5 [m/s2]
SSys = 0.5 × 0.079 + 9.81 × 0.0792 × 0.5 [m/s2]
SSys = 0.071 [m/s2]
tSys = t50 + tV + tSV = 0.043 + 0.016 + 0.02
tSys = 0.079 [m/s2]
Braking distance
SBr = Vmax2
[m]
2 × ( FNtot - aB )
m
Switching times
Size
Brake switching time Type 3850/1 t50 [s]
0.035
0.035
0.040
5.0
5.1
5.3
Stopping distance
SBr
= 1.272
=0.04 [m]
2 ×20.19
Vmax
= V0 + aB × tSys[m/s]
Vmax
= 0.5 + 9.81 × 0.079 =1.27 [m/s]
SKo =SBr + SSys [m]
SKo
=0.04 + 0.071
=0.11 [m]
Stopping time
tKo =tBr + tSys [s]
tKo
=0.063 + 0.079
=0.142 [s]
tBr
= Vmax = 1.27 =0.063 [s]
FNtot
- aB
20.19
m
Deceleration (for system dimensioning
aV = FNtot×2.5 --g = 6000×2.5 -9.8 =65.1 [m/s2]
m 200
Load
= aV = 65.19 = 6.64
[g]
g
9.81
6 Correct use
6.5.1 Calculation example (dynamic braking)
