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English
Examples of setting linear compensation amounts.
As the mechanical error is measured, set the compensation amount with reference to the
following examples.
Addition or subtraction to or from the displayed value for the displacement
L:
Length of block gauge
a
s
:
Displayed value for the distance between the surfaces A and C
When L>
s
, add a compensation amount to the displayed value.
Set an appropriate positive compensation amount.
•
Example in millimeter operation
Where L = 250 mm,
s
= 249.9960 mm, the difference between L and
s
is 0.004 mm. The amount
χ
to be compensated per meter (1000 mm) is:
0.004mm
χ
250 mm
1000 mm
The compensation amount, therefore, is 0.016 mm.
Set “015” as the closest compensation amount.
•
Example in inch operation
Where L = 9.84252" and
s
=9.84236", the difference between L and
s
is 0.00016". The amount
χ
to be compensated per inch is:
0.00016"
χ
9.84252"
1"
The compensation amount, therefore, is 0.000016". Set “015” as the closest compensation amount.
When L <
s
, subtract a compensation amount from the displayed value.
Set an appropriate negative compensation amount.
•
Example in millimeter operation
Where L=250 mm,
s
=250.0040 mm, the difference between L and
s
is 0.004 mm. The amount
χ
to be compensated per meter (1000 mm) is:
0.004 mm
χ
250 mm
1000 mm
Therefore the compensation amount is –0.016 mm.
Set “–015” as the closest compensation amount.
•
Example in inch operation
Where L=9.84252" and
s
=9.84268", the difference between L and
s
is 0.00016". The amount
χ
to be compensated per inch is:
0.00016"
χ
9.84252"
1"
The compensation amount, therefore, is –0.000040".
Set “–015” as the closest compensation amount.
/
χ
= 0.000016"
/
χ
= 0.000016"
/
χ
= 0.016 mm
/
χ
= 0.016 mm
