Stihl SR 5600 Notice D'emploi page 18

English
Multiply the required quantity of active
ingredient for 1 hectare by the area
determined in hectares. The result is the
quantity of active ingredient required for
the area to be treated.
Example:
According to the maker's instructions,
0.4 liters of active ingredient are
required per hectare to obtain a
concentration of 0.1%.
Quantity of active ingredient:
0.4 (l/ha) x 0.36 (ha) = 0.144 l
Determining quantity of solution
The quantity of solution required is
calculated as follows:
T
W
x 100 = T
B
K
T = Quantity of active ingredient in l
W
K = Concentration in %
T = Required quantity of solution in l
B
Example:
The calculated quantity of active
ingredient is 0.144 liters. According to
the maker's instructions, the
concentration is 0.1%.
Quantity of solution:
0.144 l
x 100 = 144 l
0.1 %
Determining walking speed
Carry out a trial run with the machine
fueled and the container filled with
water. Operate the spray tube (swing it
16
back and forth) as for the real run
described below. Determine the
distance walked in one minute.
Also use the trial run to check the
selected working width. The best
working width for low-growing crops is
4–5 m. Mark the working width with
stakes.
Dividing the distance walked in meters
by the time in minutes gives you the
walking speed in meters per minute
(m/min).
Example:
The distance covered in one minute is
10 meters.
Walking speed:
10 m
= 10 m/min
1 min
Determining discharge rate
The setting of the metering unit is
calculated as follows:
V (l) x v (m/min) x b(m)
a b
2
A (m )
V = Quantity of solution
a
v = Walking speed
b
V = Discharge rate
c
b = Working width
A = Area
Example:
The values determined above and a
working width of 4 meters require the
following setting on the metering unit:
144 l x 10 (m/min) x 4 m
3,600 m
Hectares (ha) have to be converted into
2
m (ha x 10,000 = m
To adjust the required discharge rate
see "Metering Unit".
= V (l/min)
c
= 1.6 l/min
2
2
).
SR 5600