Instructions on use
- The Vernier caliper is an extremely precise measuring instrument; the
reading error is 1/20 mm = 0.05 mm.
- Close the jaws lightly on the object to be measured.
- If you are measuring something with a round cross section, make sure that
the axis of the object is perpendicular to the caliper. This is necessary
to ensure that you are measuring the full diameter and not merely a chord.
- Ignore the top scale, which is calibrated in inches.
- Use the bottom scale, which is in metric units.
- Notice that there is a fixed scale and a sliding scale.
- The boldface numbers on the fixed scale are centimeters.
- The tick marks on the fixed scale between the boldface numbers are
- There are ten tick marks on the sliding scale. The left-most tick mark
on the sliding scale will let you read from the fixed scale the number of
whole millimeters that the jaws are opened.
- In the example above, the leftmost tick mark on the sliding scale is
between 21 mm and 22 mm, so the number of whole millimeters is 21.
- Next we find the tenths of millimeters. Notice that the ten tick marks
on the sliding scale are the same width as nine ticks marks on the fixed
scale. This means that at most one of the tick marks on the sliding scale
will align with a tick mark on the fixed scale; the others will miss.
- The number of the aligned tick mark on the sliding scale tells you the
number of tenths of millimeters. In the example above, the 3rd tick mark
on the sliding scale is in coincidence with the one above it, so the
caliper reading is (21.30 ± 0.05) mm.
- If two adjacent tick marks on the sliding scale look equally aligned
with their counterparts on the fixed scale, then the reading is half way
between the two marks. In the example above, if the 3rd and 4th tick
marks on the sliding scale looked to be equally aligned, then the reading
would be (21.35 ± 0.05) mm.
- On those rare occasions when the reading just happens to be a "nice" number
like 2 cm, don't forget to include the zero decimal places showing the
precision of the measurement and the reading error. So not 2 cm, but
rather (2.000 ± 0.005) cm or (20.00 ± 0.05) mm.
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