PRESENTATION 4
Precision, Accuracy,
and Tolerance
RANGE OF A MEASUREMENT
• Example: What is the degree of
precision and the range for 2
inches?
•
•
The degree of precision of 2 inches is to
the nearest inch
The range of values includes all
numbers equal to or greater than 1.5
inches or less than 2.5 inches
RANGE OF A MEASUREMENT
• Example: What is the degree of
precision and the range for 2.00
inches?
•
•
The degree of precision of 2.00 inches
is to the nearest 0.01 inch
The range of values includes all
numbers equal to or greater than 1.995
inches or less than 2.005 inches
ADDING AND SUBTRACTING
• The sum or difference cannot be
more precise than the least precise
measurement number used in
computations
ADDING AND SUBTRACTING
• Example: Add 15.63 in + 2.7 in +
0.348 in and round answer to degree
of precision of least precise number
•
Since 2.7 in is the least precise measurement,
round the answer to 1 decimal place
SIGNIFICANT DIGITS
• Rules for determining the number of
significant digits in a given measurement:
•
•
•
•
All nonzero digits are significant
Zeroes between nonzero digits are significant
Final zeroes in a decimal or mixed decimal are
significant
Zeroes used as place holders are not significant
unless they are identified as significant (usually with a
bar above it)
SIGNIFICANT DIGITS
• Examples:
812 has 3 significant digits
•
(all nonzero digits are significant)
14.3005 has 6 significant digits
•
(zeroes between nonzero digits are significant)
9.300 has 4 significant digits
•
(final zeroes of a decimal are significant)
0.008 has 1 significant digits
•
(zeroes used as place holders are not significant)
ACCURACY
• Determined by the number of significant
digits in a measurement
• The greater the number of significant
digits, the more accurate the number
• Product or quotient cannot be more
accurate than the least accurate
measurement used in the computations
ACCURACY
• Examples:
•
•
•
•
•
Number 2.09 is accurate to 3 significant digits
Number 0.1250 is accurate to 4 significant digits
Number 0.0087 is accurate to 2 significant digits
Number 50,000 is accurate to 1 significant digit
Number 68.9520 is accurate to 6 significant digits
•
Note: When measurement numbers have the same number of
significant digits, the number that begins with the largest digit is
the most accurate
ACCURACY
• Examples:
•
•
Product of 3.896 in × 63.6 in = 247.7856, but
since least accurate number is 63.6, answer
must be rounded to 3 significant digits, or
248 in
Quotient of 0.009 mm  0.4876 mm =
0.018457752 mm, but since least accurate
number is 0.009, answer must be rounded to
1 significant digits, or 0.02 mm
TOLERANCE (LINEAR)
• Tolerance (linear) is the amount of
variation permitted for a given
length
• Tolerance is equal to the difference
between maximum and minimum
limits of a given length
TOLERANCE (LINEAR)
• Example: The maximum permitted
length (limit) of a tapered shaft is 134.2
millimeters. The minimum permitted
length (limit) is 133.4 millimeters. Find
the tolerance.
•
Tolerance = maximum limit – minimum limit
= 134.2 mm – 133.4 mm = 0.8 mm
•
•
•
UNILATERAL AND BILATERAL
TOLERANCE
A basic dimension is the standard size from which
the maximum and minimum limits are made
Unilateral tolerance means that the total tolerance
is taken in only one direction from the basic
direction
Bilateral tolerance means that the tolerance is
divided partly above and partly below the basic
dimension
UNILATERAL AND BILATERAL
TOLERANCE
• When one part is to move within
another, there is a clearance
between the parts
• When one part is made to be
forced into the other, there is
interference between parts
UNILATERAL AND BILATERAL
TOLERANCE
•
Example: This is an illustration of
a clearance fit between a shaft
and a hole using unilateral
tolerancing. Refer to the diagram
and determine:
a. Maximum and minimum shaft
diameter
b. Maximum and minimum hole
diameter
c. Maximum and minimum
clearance
UNILATERAL AND BILATERAL
TOLERANCE
•
•
•
•
Maximum shaft diameter
1.385″ + 0.000″ = 1.385″
Minimum shaft diameter
1.385″ – 0.002″ = 1.383″
Maximum hole diameter
1.387″ + 0.002″ = 1.389″
Minimum hole diameter
1.387″ – 0.000″ = 1.387″
UNILATERAL AND BILATERAL
TOLERANCE
• Maximum clearance equals maximum hole
diameter minus minimum shaft diameter
1.389″ – 1.383″ = 0.006″
• Minimum clearance equals minimum hole
diameter minus maximum shaft diameter
1.387″ – 1.385″ = 0.002″
PRACTICAL PROBLEM
• Determine the maximum and
minimum permissible wall thickness
of the steel sleeve shown below.
PRACTICAL PROBLEM
• Maximum thickness:
[(26.08 mm + 0.05 mm) – (20.50 mm –
0.01 mm)] ÷ 2
= (26.13 mm – 20.49 mm)
= 5.64
PRACTICAL PROBLEM
• Minimum thickness:
[(26.08 mm – 0.05 mm) – (20.50 mm +
0.01 mm)]
= (26.03 mm – 20.51 mm)
= 5.52