Example of a complicated case
> Determine and record the lengths.
> Move the jack so that the load to be lifted is lifted
horizontally over the opposite rotating edge. Use the
jack to raise the load to be lifted to a maximum of
1 mm.
> Determine and record the pressure on all four sides
in the same way.
NOTE
To avoid overloads in the event the centre of gravity is
unknown, a single jack must be able to lift the entire
transport load by itself.
The total weight can exceed the specified total weight
of the original machine (weight on the nameplate) due
to later attachments.
1
3
1500 mm
Example:
Side (1) pressure = 230 bar
Side (2) pressure = 70 bar
- 50 -
1
3
2
4
4
2
230 bar + 70 bar = 300 bar
Length of sides (1) and (2) = 1000 mm
Side (3) pressure = 200 bar
Side (4) pressure = 100 bar
200 bar + 100 bar = 300 bar
Length of sides (3) and (4) = 1500 mm
> Set the result of the pressures (bar) of side (1) and
side (2) relative to the measured length (mm) of the
parallel sides.
> Also set the result of the pressures (bar) of side (3)
and side (4) in relation to the measured length
(mm) of the parallel sides.
> Using the rule of three, calculate where the centre
of gravity is located: for this purpose, use the higher
of the two pressure values of the parallel sides as
the reference value.
300 bar ≙ 1000 mm
230 bar ≙ 767 mm
Example sides (1) and (2):
300 bar ≙ 1500 mm
200 bar ≙ 1000 mm
Example: Sides (3) and (4):
> Enter the results.
The determined centre of gravity is at the
intersection of the two lines.
Before lifting the load, the lifting side, the axis of rota-
tion and the exact lifting position must be determined.
Top view, example 1 ideal case
© JUNG Hebe- und Transporttechnik GmbH
Jacks of the JH / JH EX / JHS series