Errors and Measurement Theory
All measurements made in every science are inexact. It is impossible to measure anything
physical exactly. All readings are approximations. Measurements are or are not correct according to how
these errors compare with the errors expected if the best possible measurement was made. The
precision of a measurement depends upon (a) the type and construction of the measuring device, (b) the
calibration of its scale, (c) reading errors of the experimenter, and (d) the number of measurements
made.
The inaccuracy in a measurement is normally specified by stating an error in the measurement.
Errors made are not usually mistakes but are rather the uncertainty in the measurement due to factors
such as (a), (b), and (d) above. Every measurement has an error. The better the measurement the
smaller the error.
Errors are expressed in two forms, absolute error and relative (sometimes referred to as
percentage) error. Absolute error is always expressed with the units of the quantity being measured
48.5 ± 0.5 cm
absolute error
Relative error can be expressed as either a fraction of the measured quantity or as a percentage, with the
latter most common. It does not have units.
48.5 cm ± 1%
relative error
Comparing quantities
If an experiment is performed and yields a value of g = 950 cm/sec2 (g is the acceleration due to
gravity), is this a "good" value? If the standard value is 979.5 cm/sec2, does this mean the student
performed a poor experiment? In other words, FOR THIS EXPERIMENT, how does the experimental
value compare with the accepted value?
This question is answered when the experimenter considers all possible errors in his experimental
apparatus and calculates the expected maximum error (by procedures discussed in next section) for his
particular experiment. If this yields
g = 950 ± 70 cm/sec2
then his value agrees with the standard value of 979.5 cm/sec2 ,WITHIN EXPERIMENTAL ERROR. His
value is "right." However, if his value were
950 ± 10 cm/sec2
then they do not agree and the experimenter must try to find the reason for the discrepancy and explain
why it occurred or else adjust the experiment and perform it again.
Calculation of errors
In most experiments calculations must be performed to achieve the final result of the experiment,
consequently the error in the final result will be some combination of the errors in individual
measurements.
(a) Addition and subtraction:
When numbers are added or subtracted, their absolute errors ALWAYS add. Consider:
98.7 ± 0.5 cm
98.7 ± 0.5 cm
- 13.3 ± 0.3 cm
+ 13.3 ± 0.3 cm
112.0 ± 0.8 cm
85.4 ± 0.8 cm
Notice that 97.8 ± 0.5 varies between 98.2 and 99.2 while 13.3 ± 0.3 varies between 13.0 and 13.6,
therefore the limits of the sum will be 98.2 ± 13.0 = 111.2 and 99.2 ± 13.6 = 112.8, i.e., 112.0 ± 0.8.
(b) Multiplication and division:
When numbers are multiplied or divided, their percentage or relative errors ALWAYS add.
(25.0 m ± 2%) x (4.00 ± 1%) = 100 ± 3%
(200.0 cm ± 5%) ) (40 cm ± 3%) = 5.0 ± 8%
(c) Exponentiation:
A number X raised to the nth power has a percentage error n times the percentage error in the
number X.
Linear Propagation of Errors:
In the discussion given above, a linear error propagation has been assumed. In more careful
analysis, this method gives errors that are too large, since one normally does not expect to get the
maximum error in every measurement. To calculate the probable error (to a close approximation)
instead of taking the sum of the errors, one takes the square root of the sum of the squares of the errors.
This is always less than the maximum error. In our error analysis, we will always use the maximum
error for simplicity.
Significant figures:
The error in a measurement does not always directly determine the number of figures in a
measurement. This is usually determined by the type of instrument used in making the measurement.
The number of digits one can read from an instrument are called the significant figures in that
measurement. The position of the decimal point does not affect the number of significant figures. If the
length of an object is found to be 10.23 mm or .01023 meters the number of significant figures is still four
in each case; only the position of the decimal point has changed. The zero at the end of a number is
significant. If we are dealing with a whole number it will be necessary to write it in scientific notation to
tell how many zeros are significant. If we say the radius of the earth is 4,000 miles and we mean this is
correct to the nearest 100 miles then we should write the number as 4.0 x 103 thus indicating that only
one zero is significant. A reading of 200 mm means that the measure is between 199.5 and 200.5 mm
while if the measure has been written 20. mm instead of 20.0 cm it would indicate that the measurement
was between 19.5 and 20.5 cm or 195 and 205 mm.
In computing with numbers that are not exact (and no measurement is exact or may be
considered an exact number) figures which are not significant should be noted at each stage of
computation and dropped before proceeding with the next step in the computation. This has two
advantages. One, it decreases the labor involved in the computation, and second, it gives a proper
concept of the accuracy of the result. Computing with figures beyond those that are significant may
cause one to imply an accuracy greater than actually exists.
In addition and subtraction significant figures in the answer are obtained only from columns in
which all digits are significant. For example consider the lengths, using ? marks to represent the first
non significant number:
21.41?
2.4?
0.013?
+ 241.3?
265.1???
From the sum it is seen that only 265 and the first number in the decimal is significant since one of the
addends is significant only to this place column. Thus, the sum would be 265.1 for these approximate
numbers.
It is advisable to round off to the same decimal position numbers which are to be added or
subtracted; for example, 33.6 minus 2.67 should be changed to 33.6 minus 2.7, giving the answer 30.9
The technique of using question marks in place of unrecorded digits applies in multiplying and
dividing. The question mark is handled just like a number:
(a) 4.1234?
2.3?
??????
123702?
82468?
9.4??????
(b)
2.7??
16?) 44.33?
32?
12??
112?
1??
In general, in multiplying or dividing, the answer can have no more significant figures than did the
number with the least significant figures used in computing the answer. If you’re measured quantities
were 4.12? and 9.51?, it would not be correct to write that 4.12 x 9.51 = 9.5172 since this would imply
that your numbers are accurate to five figures, the correct answer is 9.51, accurate to three figures.
IT IS COMMON TO MAKE READINGS TO THREE FIGURES BECAUSE MANY COMMON LAB MEASURING
DEVICES, BALANCES, MICROMETERS, ETC. GIVE THREE FIGURE RESULTS.
Instrument errors:
Meter stick: The length of a wooden meter stick may vary considerably with time and abuse.
Unless the meter stick has recently been calibrated against a laboratory standard, an error in overall
length of ± 2 mm is possible, corresponding to a calibration error of ± 0.2% of each reading. The
minimum total error in a quantity measured with a wooden rule is the calibration error of 0.2% plus the
reading error which is often ± 0.5 mm. For example, if you read 89.30 cm, the error in this number
(assuming it was read correctly) would be ± (0.05 cm + 0.2% of 89.30) = ± (0.05 cm + 0.1786 cm) =
± (0.05+0.18) cm or finally ± 0.23 cm. The reading would be recorded as:
89.30 ± 0.23 cm. or equivalently 89.30 cm ± 0.3%
This would be the minimum expected error due to the meter stick. If you could not read your meter stick
this closely, then you would have to assign a larger reading error leading to a larger total error.
Vernier caliper: The vernier caliper type used in the laboratory can be read to the nearest
tenth of a millimeter. Considering the uncertainty in reading the value and correcting for any zero errors,
the error should not exceed +0.01 cm. If the vernier is made by a reputable firm the calibration error will
be significantly less than this and can be neglected.
Micrometer caliper: If the micrometer is properly used by always using the ratchet to close
down the jaws and correcting for zero errors, an object can be measured to +0.001 cm.
Balances: The error in determining the mass of an object using a balance will originate in two
ways; first, in the calibration of the comparison or balancing weights and secondly, in the sensitivity of
the balance point, i.e., how much variation can be made in the masses without any detectable change in
balance. Normally in lab work we will assume a 0.3% calibration error in the standard masses since these
are digital devices with very small errors. There will also be the ubiquitous decimal error ± 1 digit
Electronic Calipers: These devices serve as replacements for both the vernier caliper and the
micrometer caliper. They should be used by pushing the on button and then carefully closing the jaws of
the instrument and pressing the zero button to set the device to 00.00. Usually you will have it set to
measure metric units (mm); however, consecutive presses of the Inch/Metric button will switch back and
forth between inches and mm measurement units. These devices have an accuracy of 0.01 % and a
repeatability of 0.01 mm. In addition, since these are digital readouts, they have an additional error of ±
1 digit. E.g., if you properly zero the instrument and then read 10.25 mm for a measurement, the total
error would be ± .01 (1 digit decimal error) and ± .0001 x 10.25 (calibration error) for a total of ± .01 ±
.0103 for a total error of ± 0.02 mm.
Accuracy and Absolute Error: At no point in this discussion was any distinction made between
accuracy and absolute error because this difference will not be emphasized in our error analyses. The
two quantities are in reality quite different. Accuracy refers to the repeatability of measurements. For
example, micrometer readings of the diameter of a ball are 1.515 cm, 1.515 cm, 1.516 cm, 1.515cm, and
1.514 cm. Since the values are repeated within ±.001 or 0.06%, this is a very accurate measurement.
However, if there were a ± 0.05 cm zero error in the micrometer and no allowance was made, then all
of these numbers would be high by ± 0.05 cm. If there were no other errors, the absolute error would
be ± 3%. Again, no distinction will be made in our error analyses between these quantities.