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ATTENTION!
For safety reasons, in accordance with DIN EN 1550, the centrifugal force may be a
maximum of 67% of the initial clamping force.
The formula for the calculation of the total centrifugal force F
F
= ∑ (m
r
)
·
·
c
b
s
n is the given speed in r.p.m.. The product m
.
M
c
M
= m
· r
[ kgm ]
c
B
s
In case of chucks with split chuck jaws, i.e. with base jaws and top jaws, for which the base
jaws change their radial position only by the stroke amount, the centrifugal torque of base jaws
M
and the centrifugal torque of top jaws M
cGB
M
= M
+ M
[ kgm ]
c
cGB
cAB
The centrifugal torque of the base jaws M
The centrifugal torque of the top jaws M
M
= m
· r
[ kgm ]
cAB
AB
sAB
The lathe chuck K11-80, 100, 125 has no base jaws and no top jaws.
Example for K11-125:
The centre of gravity radius r
the chuck)
Weight of a jaw = 0.058 kg
Centrifugal moment for one jaw
M
= 0.189 kg · 0.04925 m = 0.0093 kgm
c
The lathe chuck has 3 jaws.
= 0.0093 kgm · 3 = 0.0279 kgm
Calculation of the total centrifugal force at a rotational speed of 3800 rpm
F
= ∑ (m
r
)
·
·
c
b
s
3.14
(
=
0.0279 kgm ·
The total possible clamping force of the chuck at a standstill is ∑
of 100 Nm with the lathe chuck key.
An effective clamping forceF
- F
=
F
= ∑
17 kN - 4.4 kN = 12.6 kN
sp
s
c
see
Clamping force-speed diagram - Lathe chuck K11-80 on page 113
see
Basic safety instructions on page 112
GB
3440287│3442710│3442712
116
π
2
π
n
(
)
(
·
= ∑ M
·
30
c
30
B
cAB
can be found in the data of the lathe chuck.
cGB
is calculated.
cAB
of the jaw = 0.04925 m (jaw flush with the outer diameter of
s
2
π
π
(
n
)
(
·
= ∑ M
·
30
c
30
3800
)
·
2
= 4413.55 N = 4.4 kN
30
of 12.6 kN remains at the chuck.
sp
Translation of original instruction
is:
c
2
n
)
·
[N]
is described as the centrifugal force torque
· r
s
need to be added:
2
n
)
·
[N]
17 kN at a tightening torque
s
Version 1.0 - 2020-12-22
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