File: Introduction to Lesson 3: Conversions within Measurement Systems
/ Introduction
When working with measurement problems, it is often necessary to convert from one unit to another within a measurement system or between measurement systems. In this lesson, these types of conversions will be examined.
The following table of conversions will be used in this lesson. You may wish to print this page for reference as you work through this lesson.
File: Outcomes
/ Outcomes
When you have completed this lesson, you will be able to:
- convert between units of linear measure in the metric system
- convert between units of linear measure in the imperial system
- solve problems involving conversion within a measurement system
File: Unit Conversion Ratio
/ Unit Conversion Ratio
To convert a measurement given in one unit of measure to another unit of measure, a unit conversion ratio can be used. A unit conversion ratio is a fraction equal to 1. Examples of unit conversion ratios taken from the table on the previous page are:
The conversion factor can be written with either value in the numerator or the denominator. For example:
When converting between units of measure, it is best to write the conversion factor as follows:
- The numerator of the ratio consists of the required unit of measure (the unit to which you want to convert).
- The denominator of the ratio consists of the given unit of measure (the original units in which the measurement was taken).
File: Conversions – Example 1
/ Conversions – Example 1
Consider the following example.
Example 1
A piece of paper measures 765 mm on one side. What is the length of the paper in centimetres?
Solution
Step1
The given units of measure are mm and the required units of measure are cm. Therefore write the conversion factor as:
Step 2
Multiply the 765 mm by the unit conversion ratio as follows:
This procedure works for all unit conversions, including metric conversions and imperial conversions.
It also works for conversions from one measuring system to another, although students are not responsible for these conversions in this course.
Shortcut
Recognizing that the unit conversion ratio indicates that there are 10 mm to 1 cm, the conversion can be completed as: 765 ÷ 10 = 76.5 cm
File: Conversions – Example 2
/ Conversions – Example 2
Consider the following example involving a conversion in the imperial system.
Example 2
A plank measures 6 ft, 4 in. How many inches does the plank measure?
Solution
Step1
The given units of measure are feet and inches and the required units of measure are inches. Therefore write the conversion factor as:
Step 2
Multiply six feet by the unit conversion ratio.
Step 3
Add the two inch measures as follows:
6 ft, 4 in
= 72 in + 4 in
= 76 in
Shortcut
Recognizing that the unit conversion ratio indicates that there are 12 in to 1 ft, the conversion can be completed as: 6 x 12 + 4 = 76 in.
File: Conversions – Example 3
/ Conversions – Example 3
Consider the following example.
Example 3
A living room has a length of 5 yards, 2 feet. What is the length of the room in inches?
Solution
Step 1
In this example, two unit conversion ratios are needed.
One to convert yards into inches:
One to convert feet to inches:
Step 2
Multiply the 5 yards and 2 feet by the appropriate unit conversion ratios.
and
Therefore, 5 yards, 2 feet:
= 180 + 24
= 204 inches
Shortcut
Recognizing that the unit conversion ratios indicate that there are 36 in to 1yd and 12 in to 1 ft, the conversion can be completed as:
= (5 x 36) + (2 x 12)
= 180 + 24
= 204 in
File: Conversions – Example 4
/ Conversions – Example 4
Consider the following example.
Example 4
Perform the following calculations in the metric system: 2 km – 820 m
Solution
Step1
Convert 2 km to m as follows:
Step 2
Subtract the two measures as follows:
= 2 km – 820 m
= 2000 m – 820 m
= 1180 m
Shortcut
Recognizing that the unit conversion ratio indicate that there are 1000 m to 1 km, the conversion can be completed as:
= 2 x 1000 – 820
= 2000 – 820
= 1180
File: Conversions – Example 5
/ Conversions – Example 5
Consider the following calculation in the imperial system.
Example 5
Perform the following calculation: 3 ft, 7 in + 4 ft, 11 in
Solution
Step 1
Add the measurements in feet together and then the measurements in inches together.
Step 2
Convert 18 inches to feet and inches: 18 inches = 1 ft, 6 in.
Step 3
Add the two measures.
= 7 ft + 1 ft, 6 in
= 8 ft, 6 in
File: Conversions – Example 6
/ Conversions – Example 6
Consider the following calculation in the imperial system.
Example 6
Perform the following calculation:
Solution
Step 1
Because 1000 ft cannot be subtracted from 250 ft, it is necessary to rewrite the first measure as follows:
Step 2
Complete the subtraction as follows:
File: Conversions – Example 7
/ Conversions – Example 7
Consider the following example.
Example 7
A roll contains 13 m of wire. You are required to cut off individual lengths of 8 cm wire.
- How many lengths can you cut from the roll?
- How much wire is left on the roll?
Solution
- Convert 13 metres to centimetres as follows:
Determine the number of lengths that can be cut from the roll by dividing 1300 cm by 8 as follows: 1300 ÷ 8 = 162.5
Therefore 162 complete lengths of wire can be cut from the roll.
- The number of centimetres in 162 lengths of wire is:
= 162 x 8
= 1296 cm
The amount of the wire left on the roll is:
= 1300 – 1296
= 4 cm
File: Introduction to Lesson 4: Accuracy versus Precision
/ Introduction
In Lesson 2, various figures were measured using both the metric and imperial systems of measure. In this lesson, the accuracy and precision of those measurements will be considered.
File: Outcomes
/ Outcomes
When you have completed this lesson, you will be able to:
- distinguish between the terms “accuracy” and “precision”
- perform calculations involving precision
File: Accuracy, Precision and Uncertainty of Measurements
/ Accuracy, Precision and Uncertainty of Measurements
A measurement can be examined in terms of its accuracy and its precision. In everyday language, the terms “accuracy” and “precision” are often used interchangeably. However, in mathematics and science, a distinction is made between the terms.
The term “accuracy” is defined to be how close a measurement comes to its true value.
- The difference between a measurement and its true value involves a measurement error.
- The closer a measurement is to its true value, the greater its accuracy.
- The more closely a measuring device is to the actual measure, the more accurate its measurements can be.
- The precision of a measuring instrument is an indication of its ability to distinguish and is usually stated as the smallest scale division of measurement.
- The smallest scale division on a measuring device is taken to be the uncertainty of the device.
- The greater the number of divisions of a measuring device, the greater its precision.
File: Precision of Various Measurement Devices
/ Precision of Various Measurement Devices
A common metric ruler divided into millimetres is said to have a precision of 1 millimetre. A common imperial ruler divided into 1/16 of an inch is said to have a precision of 1/16 in.
In Lesson 5, the Vernier caliper will be examined. The metric Vernier caliper has a precision of 1/10 of a millimetre while an imperial Vernier caliper has a precision of 1/128 of an inch.
In Lesson 6, even more precise measurements will be made using a micrometer. The metric micrometer has a precision of 1/100 of a millimetre, while an imperial micrometer has a precision of 1/100 of an inch.
File: Comparing Accuracy and Precision “A”
/ Comparing Accuracy and Precision “A”
It is possible to have a measuring device which is precise but not accurate. Consider the example of a ruler bowed out of shape.
Since this ruler gives the same measurement every time, according to the definition of precision it is a precise measuring device.
However, since it does not give a measure close to its true value, it is not an accurate measuring device.
File: Comparing Accuracy and Precision “B”
/ Comparing Accuracy and Precision “B”
The difference between the terms “accuracy” and “precision” can be further illustrated by the following target practices where five shots are fired at each target.
In A, all the shots are to the right of the centre, and the average accuracy is poor. As all five shots are quite close together, the precision is good.
In B, the average of the shots is relatively close to the centre, and the average accuracy is good. As all five shots are quite far apart, the precision is poor.
In C, all shots are very near the centre, and the average accuracy is good. As all the shots are close together, the precision is good.
In D, all shots are to the right of the centre, and the average accuracy is poor. As all five shots are quite far apart, the precision is poor.
File: Precision and Uncertainty of Measurement – Example 1
/ Precision and Uncertainty of Measurement – Example 1
Consider the following example.
Example 1
A radius of a circle is measured to be 28 millimetres. State the precision of the measuring device.
Solution
The measuring device is precise to the nearest millimetre. The actual measure of the circle could be from 27.5 mm – 28.5 mm.
The uncertainty of the measuring device, unless otherwise stated, is considered to be one-half the precision of the device. In this instance, the precision is 1 millimetre and the uncertainty is therefore ±0.5 of a millimetre.
File: Precision – Example 2
/ Precision – Example 2
Consider the following example involving the measurement of strings.
Example 2
Using the given ruler, measure the following three strings. How precise are each of the measurements?
Solution
String A measures 5 centimetres. As the smallest scale division of the ruler is one centimetre, the measurement is precise to the nearest centimetre.
Using the given ruler, strings B and C also measure 5 centimetres. The smallest scale division of the ruler is one centimetre. Therefore, the measurement can be precise only to the nearest centimetre.
In the above example only the concept of precision is involved. Accuracy is not involved. The accuracy would depend on how closely the given ruler is to the actual measure.
File: Precision – Example 3
/ Precision – Example 3
Consider the following example involving the measurement of strings.
Example 3
- Measure the above string, using the given ruler.
- How precise is your measurement?
Solution
- The string measures 36 millimetres.
- The measurement is precise to the nearest millimetre.
File: Precision – Example 4
/ Precision – Example 4
Sometimes a level of precision is needed that is less than that indicated by a particular measuring device. Consider the following example where a string is measured to various indicated levels of precision using an imperial ruler calibrated to 1/16 of an inch.
Example 4
Measure the length of the following string precise to the nearest indicated unit, using the given imperial ruler.
- 1/2”
- 1/4”
- 1/8”
Solution
The length of the string is between 2 1/2” and 3”.
Since the string is closer in length to 2 1/2”, the length of the string precise to the nearest 1/2” is 2 1/2”.
The length of the string is between 2 1/2” and 2 3/4”.
Since the length of the string is closer in length to 2 3/4”, the length of the string, precise to the nearest 1/4”, is 2 3/4”.
The length of the string is between 2 5/8” and 2 6/8 (2 3/4).
Since the length of the string is closer in length to 2 5/8”, the length of the string, precise to the nearest 1/8”, is 2 5/8”.
File: Precision – Example 5
/ Precision – Example 5
Consider the following example.
Example 5
A homeowner wishes to build a fence around her property. The dimensions of her property, precise to the nearest metre, are 65 m by 32 m.
- How precise are the dimensions?
- What is the perimeter of her property?
Solution
- The dimensions are precise to the nearest metre.
- The perimeter, or the distance around the property is:
= 65 + 32 + 65 + 32
= 194 m
Note that in the above example, the length of 65 m is precise to the nearest metre. It could be as small as 64.5 m or as large as 65.5 m.
The width of 32 m is also precise to the nearest metre, and could be as small as 31.5 m or as large as 32.5 m.
Taking the lower values gives a perimeter:
= 64.5 + 31.5 + 64.5 + 31.5
= 192 m
Taking the higher values gives a perimeter:
= 65.5 + 32.5 + 65.5 + 32.5
= 196 m
Therefore, the answer of 194 m in the example is more uncertain than the original dimensions. The answer is less precise than the original measurements.
File: Precision – Example 6
/ Precision – Example 6
Because calculations involving measurements can introduce more uncertainty, an answer cannot be more precise than the least precise factor. This is further illustrated by the following example.
Example 6
A painting has dimensions 7.8 cm by 9.4 cm.
- Calculate the area using the given dimensions.
- State the precision of each dimension.
- Calculate the least possible area.
- Calculate the greatest possible area.
- Determine a reasonable answer for the area of the painting.
- Would the area found in part (a) be precise to the nearest hundredth of a centimetre?
Solution
- Using the given dimensions, the area is:
= 7.8 x 9.4
= 73.32 cm2
- Each dimension is precise to the nearest tenth of a centimetre.
- The shortest length could be 7.75 cm and the shortest width could be 9.35 cm.
The least possible area is:
= 7.75 x 9.35
= 72.4625 cm2
- The longest length could be 7.85 cm and the longest width could be 9.45 cm.
The greatest possible area is:
= 7.85 x 9.45
= 74.1825 cm2
- An area of 73 cm2 that is rounded to the nearest unit is about halfway between the smallest and largest possible area.
- No, the area in part (a) would not be precise to the nearest hundredth of a centimetre.
The above example illustrates how calculations involving measurements introduce less precision. Although the original dimensions are precise to the nearest tenth of a centimetre, the area could be as small as 72.4625 cm2 or as large as 74.1825 cm2.
Therefore, the area is less precise than the nearest tenth of a centimetre. It is certainly less precise than the nearest hundredth of a centimetre.