Jumil JM 2640 EX PD, Руководство оператора

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Manual de Instruções JM 2640 EX PD
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In this case:
A - is the area to be fertilized, expresses in m2(square meters);
B - is the spacing among the rows of tillage in millimeters;
C - is the amount of fertilizer that wants to distribute in the area in
subject;
D - is the space to travel for the test of fertilizer debit;
X = it is the amount, in grams, that should fall by row, after traveling
the certain space.
Exemplifying, if you want to distribute 350kg/Ha, in the tillage
with 0,80m of spacing among rows; it should come in the following way:
Like this, in 50 traveled meters 1.400 g/row will fall.
If you want to do the prove, come in the following way:
In a hectare, in other words, in 10.000m2 planted the 0,80m among
rows, there are 12.500 lineal meters (10.000m2/0,80m = 12.500m lineal). If
in 50 traveled meters fell 1.400g of fertilizer, in 12.500m will fall350kg, that
is the intended amount.
To make this test, you must dedicate special attention to the fact
that all the endless fertilizer threads should be provisioned and, only after it
should begin the test and picked up the fertilizer in plastic sacks that should
be identified and weigh.
This test should be accomplished at the place where the planting will
be made, with the same speed.
It will also be able to be done at the hangar, , giving n turns in the
wheel, corresponding to the space that will be traveled.
Example: if the perimeter of the wheel goes 2 meters, 25 turns will be
given to be equal to 50 lineal meters, being picked up the fertilizer that fell
during those turns.
This test is not usually rigorous, for the difficulty of maintain a
continuous pulse to the wheel, as well as to maintain the speed of planting.
Formula
B x C
A
x D X =
B x C
A
x D X =
800 x 350
10000
x 50 X = 1.400g X =