Advertisement
®
©
SF-7200
(Original)
Void area
a
(Enlarge copy of x 2.0)
Void area
(Reduced copy of x 0.7)
Obtain the difference between the lead edge void area
A of the x2.0 copy and the lead edge void area B of the
x0.7 copy, and divide the result by 0.425.
If Ais same
as B, there
is no need
of performing
the following
computation.
In case the lead edge void area A of the x2.0 copy is
larger than the lead edge void area B of the x0.7 copy
(A>B):
(Ex-1)
In the case of A=22.5mm and B=16.0mm
(A—B)/0.425=S1
(A—B)/0.425=(22.5—16.0)/0.425=15.3
$1=15.3 (the value under the decimal point must be
rounded by 4/5).
In case the lead edge void area A of the x2.0 copy is
smaller than the lead edge void area B of the x0.7
copy (A<B):
(Ex-2)
In the case of A=5.0mm and B=7.5mm
(B—A)/0.425=S2
(B—A)/0.425=(7.5—5.0)/0.425=5.9
$2=5.9 (the value under the decima! point must be
rounded by 4/5).
Make
the
MS6
actuator
adjusting
screw
rotated as
much
as S1 or S2. The following
procedure
is not
required when A is same as B.
In case A is greater than
B, turn the MS6 actuator
adjusting screw counterclockwise as much as $1.
(Ex-3)
In the case of $1=15.3, the screw must be rotated
15 steps counterclockwise.
In case A is smaller than B, turn the MS6 actuator
adjusting screw clockwise as much as S2.
(Ex-4)
In the case of $1=5.9, the screw must be rotated 6
steps clockwise.
*Clockwise
turn
of the
MS6
actuator
adjusting
screw makes A increased and B decreased. Counter-
clockwise turn of the MS6 actuator adjusting screw
makes A decreased and B decreased.
Rotate one step.
MG6 actuator
Original table
MS6 adjusting screw
Perform Items (8) and (). This is not required if A is
same as B.
In case there is a difference between the lead edge void
area A of the x2.0 copy and the lead edge void area B
of the x0.7 copy, repeat the above Items (6), ©), @),
and @), until A becomes equal to B. This procedure
is not required if A is same as B.
_, Image loss (140.5mm)
eran
ene aed
1
(Normal copy of 1:1)
Void area (2.04 1.5mm)
a. Make a copy of the normal mode
(x1.0) and mea-
sure the lead edge void area X. If the result is within
the given rating, there is no need of performing b
thru d described next.
b. Obtain
the difference between
the lead edge void
area X measured
value and the lead edge void area
rating (2.0mm), and divide the result by 0.5.
@
In case the lead edge void area measured value X
is greater than 2.0 (X>2.0):
(Ex-5)
In the case of X=5.0:
X-2=Y1
Y1/0.5=2Z1
Y—2=5-—2=3
Y1=3
Y1/0.5=3/0.5=6
Z1=6 (the value under the decimal point must be
rounded by 4/5).
Advertisement
