Unit B: Measurement

Nominal Value and Tolerance

The mass of a product indicated on the label is not necessarily the mass of product in the bag. Frito-Lay® makes a bag of plain chips. The company wants to ensure that there are at least 400 g of chips in each bag produced. Frito-Lay® is targeting for each bag to have 410 g with a minimum mass of 405 g and a maximum mass of 415 g.

405 g is the minimum mass, which is 5 g lower than the ideal mass.

Ideal

415 g is the maximum mass, which is 5 g higher than the ideal mass.

The measurement goal of the manufacturer of a product is called the nominal value. Although a manufacturer may state that the product is a certain size, variation may occur. Most often, the manufacturer has an allowable range of measurements for the product. This acceptable range is called tolerance. This is the target of the manufacturing process (the bull's eye). The tolerance is the range around the nominal value that is acceptable.

Tolerance is the range from the minimum to the maximum measurement and can be calculated using the equation:

Tolerance = maximum measurement – minimum measurement

Tolerance can be written two different ways:

nominal value $\pm \frac{1}{2}$ (tolerance)
$\begin{matrix} maximum value \\ minimum value \end{matrix}$

Example 1

Write the mass of the bag of chips, in two different ways, using the information in the diagrams above.

Solution

For the bag of chips, the nominal value is 410 g.

The tolerance can be calculated as follows:

\begin{array}{rcl} Tolerance & = & maximum measurement - minimum measurement \\ = & 415 g - 405 g \\ = & 10 g \end{array}

The mass of the chips can be written in the following ways:

$\begin{array}{rcl} nominal value \pm \frac{1}{2} (tolerance) & = & 410 g \pm \frac{1}{2} (10 g) \\ = & 410 g \pm 5 g \end{array}$
$\begin{array}{rcl} maximum value & = & 410 g + 5 g \\ = & 415 g \\ minimum value & = & 410 g - 5 g \\ = & 405 g \end{array}$

Example 2

A nominal temperature of a child is 37.00 °C. An oral thermometer has a tolerance of 0.1 °C. Determine the range of temperatures that the child could have.

Solution

The nominal value is 37.00 °C, and the tolerance is 0.1 °C.

$\begin{array}{rcl} nominal value \pm \frac{1}{2} (tolerance) & = & 37.00 ° C \pm \frac{1}{2} (0.1 ° C) \\ = & 37.00 ° C \pm 0.05 ° C \end{array}$
$\begin{array}{rcl} maximum value & = & 37.00 ° C + 0.05 ° C \\ = & 37.05 ° C \\ minimum value & = & 37.00 ° C - 0.05 ° C \\ = & 36.95 ° C \end{array}$

The child's temperature could be between 36.95 °C and 37.05 °C.

Example 3

A measurement on an engine drawing has a nominal value of 4.5 cm and a tolerance of 0.2 cm. Write this measurement in two different ways.

Solution

The nominal value is 4.5 cm, and the tolerance is 0.2 cm.

$\begin{array}{rcl} nominal value \pm \frac{1}{2} (tolerance) & = & 4.5 cm \pm \frac{1}{2} (0.2 cm) \\ = & 4.5 cm \pm 0.1 cm \end{array}$
$\begin{array}{rcl} maximum value & = & 4.5 cm + 0.1 cm \\ = & 4.6 cm \\ minimum value & = & 4.5 cm - 0.1 cm \\ = & 4.4 cm \end{array}$

Example 4

The metal balls inside ball bearings must be uniform in size to operate effectively. To describe the precision of the metal balls, an international grading system has been devised that ranges from 2000 to 3 (the smaller the number, the more precise the measurements.)

A grade of 3 indicates a diameter tolerance of ± 0.00003 inches whereas a grade of 1 000 has a diameter tolerance of ± 0.005 inches.

Give the maximum and minimum range of a grade 3 ball bearing with a nominal value of 0.50000 inches.
State the tolerance. Give the maximum and minimum range of a grade 1000 ball bearing with a nominal value of 0.750 inches.

Solution

The nominal value is 0.50000 in, and the tolerance is 0.00003 in.

$\begin{array}{rcl} maximum value & = & 0.50000 in + 0.00003 in \\ = & 0.50003 in \\ minimum value & = & 0.50000 in - 0.00003 in \\ = & 0.49997 in \end{array}$

The range of the grade 3 ball bearings is 0.49997 in to 0.50003 in.
The nominal value is 0.750 in, and the tolerance is 0.005 in.

$\begin{array}{rcl} maximum value & = & 0.750 in + 0.005 in \\ = & 0.755 in \\ minimum value & = & 0.750 in - 0.005 in \\ = & 0.745 in \end{array}$

The range of the grade 1000 ball bearings is 0.745 in to 0.755 in.

Example 5

A 2 L bottle of cola has a nominal value of 2.07 L and a tolerance range of ± 0.03 L. Marika measures the volume of 5 bottles of cola and records them in the table below.

Bottle	Volume (L)
1	2.09
2	2.04
3	1.99
4	2.06
5	2.12

Determine the maximum and minimum acceptable volumes of cola.
Which bottle(s) fall within the acceptable range?
Which bottle(s) do not fall within the acceptable range? Explain.
Why is the nominal value for the volume of cola in the bottle not equal to 2.00 L?

Solutions

The nominal value is 2.07 L, and the tolerance is 0.03 L.

$\begin{array}{rcl} maximum value & = & 2.07 L + 0.03 L \\ = & 2.10 L \\ minimum value & = & 2.07 L - 0.03 L \\ = & 2.04 L \end{array}$
The acceptable volumes are between 2.04 L and 2.10 L.
Bottles 1, 2, and 4 fall in the acceptable range of 2.04 L and 2.10 L.
Bottle 3 has a volume of 1.99 L, which is below the minimum value of 2.04 L, and Bottle 5 has a volume 2.12 L, which is above the maximum value of 2.10 L.
Since the range of tolerance is ± 0.03 L, the volume in the cola bottle would be between 1.97 L and 2.03 L. Therefore, some bottles of cola would have a volume that will be less than the amount of 2.00 L indicated on the bottle.