Errors in measurement are different from the mistakes committed in measurement. Errors occur when we measure the sizes using instruments like a ruler or a measuring tape. The errors in measurements depend on the instrument we use. The greatest possible error (GPE) that may occur in using a instrument is half the smallest unit the instrument is capable of measuring. Suppose a ruler is marked to the precision of one tenths of an inch. Then the greatest possible error that can occur using the ruler is ± 0.05.

The relative error is found by comparing the actual error occurred to the measurement taken and conveys the accuracy of the measurement taken.

Absolute error is the actual error occurred during measurement. It is the difference between the true value and measured value.
Absolute error = |True value - Measured value|Assuming that a person reads measurement using an instrument with at most care, the greatest possible error of the instrument is taken as the absolute error occurred. The absolute error is expressed in the same unit of the actual measurement.

For example, if the value is expressed in meters, then the absolute error is also expressed in meters. Absolute error is always positive.

Relative error is calculated comparing the absolute error to the actual measurement. Relative error is expressed either as a decimal or a percent without units. When the relative error is expressed as a percent it is called the percentage error.
The formula used for calculating relative error is as follows:
Relative error = $\frac{Absolute\ error}{ Actual\ measurement}$

Percent error = Relative error x 100%
Relative error is expressed without units. Note that while calculating the relative error the absolute error is written in the same units as that of the actual measurement.

Suppose the true value of the quantity is x and the absolute error occurred is Δx. Then,
The absolute error Δx = |x - x0| , where x0 is the measured value.

Relative error = $\frac{\bigtriangleup x}{x}$

Percentage error = $\frac{\bigtriangleup x}{x}$ . 100%
Step 1: The first step in calculating the relative error is finding the absolute error. When the true and measured values are known, the absolute error is the difference between the two measurements.
Example: If the True length of a cable is 35 meters and if the measured length is noted as 34.5 meters, then the absolute error = 35 - 34.5 = 0.5 meter.

Step 2: When the precision of the measuring instrument is given,
Absolute Error = Greatest Possible error for the instrument = Half the precision of the instrument.
If an instrument is capable of measuring the weight to the precision of 1 gm, then
the absolute error that can occur using the machine = $\frac{1}{2}$ x 1 gm = 0.5 gm

Step 3: After finding the absolute error, the relative error is calculated using the formula
Relative error = $\frac{Absolute\ error}{Actual\ measurement}$
In our example of cable, Relative error in measurement of the cable = $\frac{0.5}{35}$ = $\frac{1}{70}$ $\approx$ 0.0143

Step 4: If the precision of the instrument is used to calculate the absolute error, the actual measurement refers to the measurement taken using the instrument.
If an instrument with GPE 0.5gm (or precision 1 gm) shows the weight of the object as 48.5 gms, then
the relative error occurred in measuring = $\frac{0.5}{48.5}$ = $\frac{1}{97}$ $\approx$ 0.0103

Step 5: The percentage error is got by multiplying the relative error by 100.
Percentage error in the measurement of the length of the cable = 0.0143 x 100 = 1.43%
Percentage error in the measurement of the weight of the object = 0.0103 x 100 = 1.03%

Solved Examples

Question 1: Mrs.Simpson weighed 165 lbs in her Physicians scale. But the defective scale used by her at Home showed her weight as 160 lb. Calculate the relative and percentage error in the measurement of her scale.
If her daughter weighs 105 lb on the defective scale, find her actual weight nearest to the tenth of a lb.
Solution:
 
The actual error in the measurement of Mr. Simpson's weight = 165 - 160 = 5 lbs
Relative error in the measurement shown by the defective scale = $\frac{5}{165}$ = $\frac{1}{33}$
Percentage error = $\frac{1}{33}$ x 100 $\approx$ 3.03%

Let the actual weight of the daughter = x lbs
As Mrs.Simpson's weighing scale shows weight less than the actual weight x > 105.
Relative error in the measurement of defective scale = $\frac{x-105}{x}$ = $\frac{1}{33}$
33(x -105) = x (Cross multiplication of the proportion)
33x - 3465 = x
32x = 3465 $\Rightarrow$ x = 108.3 lb
The actual weight of her daughter = 108.3 lb
 

Question 2: A meter scale can measure to a precision of 1 mm. A rope and a small length of wire are measured using the scale as 57.5 cm and 4 cm. Calculate the relative errors occurred in both the measurements read using the scale. If it is required to measure the length of the wire to the same accuracy level as that found in the measurement of the rope, find the precision level required for the measuring instrument.
Solution:
 
The absolute error for the measurements taken using the scale = $\frac{1}{2}$ x precision level = $\frac{1}{2}$ x 1 mm = 0.5 mm = 0.05 cm

Relative error in the measurement of the rope = $\frac{0.05 }{57.5}$ = $\frac{1}{1150}$ $\approx$ 0.0009

Relative error in the measurement of the wire = $\frac{0.05}{4}$ = $\frac{1}{80}$ $\approx$ 0.0125

If the length of the wire is to be measured to the same accuracy level found in the measurement of the rope, the relative error in the
measurement of the wire should also be equal to $\frac{1}{1150}$.
Let the required precision = x.
The absolute error that can occur in the measurement = $\frac{x}{2}$

Thus the relative error that can occur in the measurement of length 4 cm = $\frac{\frac{x}{2}}{4}$ = $\frac{x}{8}$

$\frac{x}{8}$ = $\frac{1}{1150}$ $\Rightarrow$ x = $\frac{8}{1150}$ = 0.007 cm

Hence the required precision level = 0.07 mm.