08Aerrors

= Errors In Measurement =

Error (or Rounding Error)
When we measure anything, there will always be a difference between the value we obtain and the real distance. This is because no measuring device can give a perfectly accurate result (it would need to show an infinite numer of decimal places to do so).

The difference between the value we obtain and the real distance is called the ** Error **. (or sometimes the ** Rounding Error **)


 * Note: ** An error is different from a mistake. A mistake is where the person reads the scale incorrectly or the measuring device is used incorrectly. A mistake is avoidable, the error in measurement is unavoidable.

Error is the absolute value (or the difference between) the measured value and the actual value. Therefore an error cant be negative. We are only interested in the magnitude (or size) of the error and not its sign(positive or negative).

math \text{Error} = \left| \text{measured value} - \text{actual value} \right| math

Absolute Relative Error (or Relative Error)
The importance of the error can be judged by considering the ** Absolute Relative Error ** (often simply called the Relative Error). This is sometimes converted to a percentage to get the Absolute Percentage Error.

math \text{Relative Error} = \dfrac{\text{Error}}{\text{Actual Value}} math

math \text{Percentage Error} = \text{Relative Error} \times 100\% math

Example 1
If a distance is measured as 8cm to the nearest cm and the actual distance is 8.22cm a) Find the Error b) Find the Relative Error c) Find the Percentage Error

a) Find the Error
 * Solution **

math \\ \text{Error} = \left| \text{measured value} - \text{actual value} \right| \\ \\ .\qquad = \left| 8 - 8.22 \right| \\ \\ . \qquad = \left| -0.22 \right| \\ \\ . \qquad = 0.22 \textit{ cm} math


 * b) Find the Relative Error **

math \\ \text{Relative Error} = \dfrac{\text{Error}}{\text{Actual Value}} \\ \\ . \qquad = \dfrac{0.22}{8.22} \\ \\ . \qquad = 0.0268 math


 * c) Find the Percentage Error **

math \\ \text{Percentage Error} = \text{Relative Error} \times 100\% \\ \\ . \qquad = 0.0268 \times 100\% \\ \\ . \qquad = 2.68\% math

Degree of Accuracy
The ** Degree of Accuracy ** is the smallest unit that your measuring device uses..

For example, a typical ruler is accurate to the nearest milimetre so the degree of accuracy is 1mm or 0.1cm My old blackboard ruler only has marks every 5cm so the degree of accuracy is 5cm.

Maximum Rounding Error
The ** Maximum Rounding Error ** will be __**half**__ the Degree of Accuracy.

For example, a typical ruler mentioned above (degree of accuracy 1mm or 0.1cm) would have a maximum rounding error of 0.5mm or 0.05cm.
 * This means that if we measure something with a standard ruler and get a result of 3.4cm,
 * the actual length will be somewhere in the range (3.4 ± 0.05) cm
 * ie somewhere between 3.35cm and 3.45cm

My old blackboard ruler (degree of accuracy 5cm) would have a maximum rounding error of 2.5cm.
 * This means that if we measure something with my blackboard ruler and get a result of 65cm,
 * the actual length will be somewhere in the range (65 ± 2.5) cm
 * ie somewhere between 62.5cm and 67.5cm

Errors in Calculations
When calculating using measurements, the errors are carried on through the calculations.
 * When adding or subtracting measurements, the errors are also added.
 * When multiplying measurements, multiply the smaller limits together and the larger limits together to get the new maximum error. (see Example 4)


 * Example 2 **

Two lengths of wood are joined end to end. The first piece is measured as (30 ± 2) cm and the second piece is measured as (40 ± 1) cm. a) Find the total length including the total error b) State the maximum and minimum possible total length

Solution (30 ± 2) + (40 ± 1) = (30 + 40) ± (2 + 1) cm = (70 ± 3) cm
 * a) **

b) Minimum length = 70 – 3 = 67cm Maximum length = 70 + 3 = 73 cm

Note: Even when __**subtracting**__ measurements, we __**add**__ the error!!


 * Example 3 **

A length of wood measure 2.35m has a piece cut from it measuring 15cm (all measured correct to the nearest cm). What is the length of the remaining piece of wood?

Solution:

(235 ± 0.5) – (15 ± 0.5) = (235 – 15 ) ± (0.5 + 0.5) = 220 ± 1 cm


 * Example 4 **

A rectangle is measured to have side lengths of 8cm and 10cm (correct to the nearest cm). a) Find the maximum rounding error for the measurements. b) State the minimum and maximum values for length and width c) Calculate the Area of the rectangle using the stated lengths d) Use the same calculation process to find the minimum possible area and the maximum possible area. e) Hence state the maximum error for the Area

Solution

a) Find the maximum rounding error for the measurements Degree of accuracy for both is 1cm, so maximum rounding error for both is 0.5cm

b) State the minimum and maximum values for length and width Length = 10cm, Min length = 9.5cm. Max length = 10.5cm Width = 8cm, Min width = 7.5cm, Max length = 8.5cm

c) Calculate the Area of the rectangle usig the measured lengths Area = 10 × 8 = 80 cm 2

{To find the minimum area, use the minimum length and the minimum width} Minimum Area = 9.5 × 7.5 = 71.25 Maximum Area = 10.5 × 8.5 = 89.25
 * d) Use the same calculation process to find the minimum possible area and the maximum possible area. **

e) Hence state the maximum error for the Area. {The maximum error for the Area will be the larger out of the difference between the stated area (80cm 2 ) and the minimum or maximum area} Diff for minimum area = 80 - 71.25 = 8.75cm Diff for maximum area = 89.25 - 80 = 9.25cm

Maximum Error for Area = 9.25cm

Tolerance
The idea of tolerance is used widely in manufacturing and production. Machines don't give perfect accuracy. The more accuracy you want, the more the cost of the machine. So an engineering/cost decision has to be made about what level of error is acceptable for the particular.

For example, a machine to fill cereal boxes with 650g of cornflakes might have a tolerance of ± 5g. If it puts too much cereal in each box, then that costs extra money, if it puts too little cereal in each box then the customers will complain. So the quantity of cereal in the box would be described as 650 ± 5. This means the minimum acceptable quantity is 645g and the maximum acceptable quantity is 655g.

A typical manufacturing process would include a checking process where random boxes are selected and the contents checked to make sure the machine is filling the boxes within the chosen limits.

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