E
NCODING THE
LH 1000 LH 1000 O
PERATORS
& F
ITTING MANUAL
8 LH A
GRO
Example:
In the following example we assume that we use a tractor with a wheel
circumference of 500 cm and 2 magnets fitted to the wheel:
Measurement shows that the effective wheel circumference = 500.6 cm.
There are 2 magnets fitted so the effective wheel circumference needs to be
divided with 2.
500.6 : 2 = 250.3 cm
Encode this as 250
WHEEL CIRCUMFERENCE WITH A MAGNET FITTED TO CARDAN SHAFT
A magnet can be fitted to the cardinal shaft if, for some reason, it is not possible to
fit any magnets onto a wheel. To determine the value to be entered as the "wheel
circumference" use the following procedure:
Step Key Procedure
1
Enter a working width of 9999 as described on page 7.
2
Press the area total key.
3 Drive slowly forward until 1 appears on the display STOP
IMMEDIATELY.
4 Make a mark on the field and on the tire.
5 Drive until the display changes from 10 to 11 STOP
IMMEDIATELY.
6 Make a mark on the field again.
7 Measure the distance between the 2 marks on the field. Divide this
distance with 10.
8
Press the wheel circumference key and an >0< appears on the left
of the display indicating that "wheel circumference" has been
selected.
9 Encode the value determined in steps 1 to 7 using the numerical
keys.
10 The encoded value is stored automatically until another wheel
circumference is encoded; press any other key to leave this
encodement.
Remember to encode the correct working width for the implement again!
To Next Page