1.3
Calculations
….
Throughout this course, unless otherwise stated, it is
assumed that the given parameters are all known to within
three significant digits. Under this assumption, you can
perform accurate calculations by adopting the following
practice: When writing down calculated numbers, always
retain at least five significant digits. The answers will nearly
always be accurate to within three significant digits.
Figure 1.3 – 1
Plate IV and some of Plate V11, from “Watchmaking,”
George Daniels, ISBN 0 85667 497 4
Ideally, it’s best at intermediate steps to store numbers
in the calculator to avoid the five digit truncation error.
However, this isn’t always very convenient. When writing
down the calculated numbers, the question also arises how to
retain the five digits. The sixth and higher digits could simply
be truncated, reducing the number to five digits. A more
accurate way, however, is to round-off the numbers to five
significant digits. To perform the round-off, simply truncate the
sixth and higher digits when the sixth digit is 4 or less. When
the sixth digit is 5 or greater, increase the fifth digit by 1 and
truncate the sixth and higher digits.
Caption:
A high level of precision is needed to make delicate
instruments, like the mechanical watch shown above.
Physical parameters are always measured with some
error and therefore known only approximately. The level of
accuracy often needs to be specified. There are several ways to
express a parameter and its level of accuracy. The most precise
way is to use statistical notation. For example, consider a 700
foot road that has been measured to an accuracy of 10 feet.
Using statistical notation, the length would be written as 700 
5 feet. The percent error is about 1.43. The length of the road
could also be written in scientific notation as 7.0102 in which
the number of digits is taken as the number of significant
digits. One would say the length of the road is known to within
two significant digits. By writing the number in scientific
notation, with the rule that the number of digits shown is equal
to the number of significant digits, the 10 foot accuracy does
not need to be explicitly stated and the ambiguity of writing a
number unscientifically is avoided. However, scientific notation
is less refined than statistical notation because the precise error
in scientific notation is not specified. Also, writing out numbers
using scientific notation can be tedious, particularly when there
are a lot of calculations. Engineers generally adopt a different
approach.
This engineering practice is easy to follow and very
efficient. As a cautionary note, though, situations can arise in
which the final answer is not accurate to three significant digits.
For instance, when a lot of calculations are being written down,
you may suspect that the accuracy of the answer could fall
below the required three significant digits. To check the
accuracy of the answer, recalculate the answer from beginning
to end while retaining six or more significant digits instead of
five. The difference between the two answers exposes the
calculation error.
The Nature of Calculation Error
Many of the calculations that you perform are arithmetic
operations, namely, addition, subtraction, multiplication,
division, and powers. These are not the only operations that you
deal with. For example, non-arithmetic operations include
differentiation, integration, and trigonometric operations. But,
the arithmetic operations are the most common. The following
describes the nature of calculation error produced by each of
the arithmetic operations.
Engineering Calculation
Addition and Subtraction
The engineering approach is described as follows. Assume that
a problem is being solved and that all of the given parameters
are known to at least s significant digits. Then, at least s + 2
significant digits are retained with each new calculation that is
written down until the answer is found. The answer will nearly
always be accurate to s significant digits. This approach is
similar to the way calculations are performed on the computer.
In computer calculations as many as 16 or 32 digits are retained
(even more when floating point arithmetic is used), which is
many more significant digits than the given parameters have.
The answer is nearly always as accurate as the given
parameters. When using a calculator, retaining only s + 2 digits
with each calculation that is written down reduces the number
of numerals that are written down, which saves you time and
reduces the number of careless errors that you make.
Let's add the numbers a  ea and b  eb , where ea is
the error in a and eb is the error in b. When a and b are added,
we get a + b = c. The actual answer, though, is
(1 – 1)
(a  ea ) + (b  eb) = (a + b) + ( ea  eb) = c  ec
so the error in c is
(1 – 2)
ec =  e a  eb
Since ea and eb are positive values that really indicate the
extreme limits of the errors in a and b, the extreme limit of the
error in c is
1
(1 – 3)
ec = ea + eb
Powers
Let's now raise the number a  ea to the n-th power.
The number a raised to the n-th power is an = c. The actual
answer, though, is
Equation (1 – 3) states that the error produced by adding the
numbers a and b is equal to the sum of the errors in a and b.
When two numbers are subtracted the same result is found.
Thus, the error produced by adding or subtracting two
numbers a and b is equal to the sum of the errors in a and b.
(1 – 8)
in which the terms (...) are products of ea2 . Assuming that the
error ea is much smaller than a, the terms (....) can be neglected,
so
The nature of the errors that you get when multiplying
and dividing is quite different.
Multiplication and Division
(1 – 9)
Multiply the numbers a  ea and b  eb. The
multiplication of a and b yields a  b  c. The actual answer,
though, is
(1 – 4)
(a  ea )n  a n  n  a  ea  (....)  c  ec
ec  n  a  ea
Multiplying Eq. (1 – 9) by 100/|c| yields
(a  ea )  (b  eb )
 a  b  (a  eb )  (b  ea )  (ea  eb )  c  ec
(1 – 10)



100 ec   n100 ea


c 
a



.


so the error in c is
(1 – 5)
ec  (a  eb )  (b  ea )  (ea  eb )
Equation (1 – 10) states that the percent error produced by
raising the number a to the n-th power is equal to n times
the percent error in a.
It is generally the case that the percent errors are small, in
which case the term ea  eb in Eq. (1 – 5) is much smaller than
the other terms in Eq. (1 – 5) and can therefore be neglected.
Furthermore, since ea and eb are positive values that indicate
extreme limits of errors, the extreme limit of the error in c is
actually
(1 – 6)
Notice that the error produced by multiplying two
identical numbers (letting a = b in Eq. (1 – 7)) degenerates to
the error produced by squaring a number (letting n = 2 in Eq. (1
– 10)).
Key Terms
ec  a  eb  b  ea
Algebraic Operations, Calculation Error, Percent Error,
Round-off, Scientific Notation, Significant Digits, Statistical
Notation.
Multiplying Eq. (1 – 6) by 100/|c| yields
(1 – 7)

 
100 ec   100 ea

c  
a

 

  100 eb 
 
b 
 
Review Questions
1. What type of notation places together a number with its error.
2. How is calculation error reduced in a minimally cumbersome
way?
3. Describe how calculation error is checked?
4. In this course, the given parameters are assumed to be
accurate to how many significant digits?
5. In this course, how many significant digits are retained with
each calculation that is written down?
6. What is round-off?
7. Describe the error that results from adding or subtracting two
numbers.
8. Describe the error that results from multiplying or dividing
two numbers.
9. Describe the error that results from raising a number to a
power.


e  
e 
e 
where 100 a  , 100 b  , and 100 c  represent the percent





a  
b
c 


errors in a, b, and c, respectively. Equation (1 – 7) states that
the percent error produced by multiplying the numbers a and b
is equal to the sum of the percent errors in a and b. When two
numbers are divided the same result is found. Thus, the
percent error produced by multiplying or dividing the
numbers a and b is equal to the sum of the percent errors in
a and b.
It is said that additive error is additive and that
multiplicative percent error is additive.
2
Article: Tuning a Violin
Figure 1
Figure 2
The violin and the bow
Close-ups of the strings on a violin
Violinists need to periodically tune their violins. This
means tuning the instrument’s strings using the pegs and fine
tuners. The pegs and fine tuners are turned and then, either the
bow is drawn across the open string or the string is plucked by a
finger to determine if the string is in tune. This process is
called “tuning.”
The violin has four strings, labeled G, D, A, and E
(See Figure). These letters refer to the notes the strings emit
when there is nothing depressing them. The strings are kept in
tension on one end by pegs that have a diameter of 7 mm, and
by the violin’s tail on the other. A completely un-tensioned
string is tensioned in about 2 turns. The tensions of the strings,
in order for them to produce the right notes, need to be about 17
lb, 13 lb, 11 lb, and 10 lb, respectively. The tuning of the
strings needs to be much more precise than the tuning of the
hairs, because the strings are being tuned to notes with specific
frequencies and the hair is not. A good violinist would only be
satisfied with strings that have been tuned to an accuracy of less
than 1/100 of a turn of the peg. Therefore the tensions in the
strings need to be much more precise than the tension in the
hair. For example, the tension in the G string changes by (17
lb)/(2 turns)·0.01 turns = 0.085 lb. The required accuracy of the
G string is 0.085 lb.
The bow consists of a stick that holds the hair. The
hair is actually strands of horse hair (the horse’s tail hair) or
nylon threads. The hair is kept in about 5 lb of tension. A
tensioning screw keeps the hair in the right amount of tension.
The thread on a typical tensioning screw is 3 mm in diameter.
Forty turns of the screw move it one inch. A completely untensioned hair is tensioned in about 2 turns. When turning the
screw, the violinist can be satisfied with a setting to within an
accuracy of about 1/8 of a turn.
It follows from these facts that the tension increases
2.5 lb per turn which is equal to 2.5/8 = 0.3125 lb per 1/8 turn.
The required accuracy of the tension in the hair is 0.3125 lb.
The violin and its bow are delicate instruments that
have been designed with the right balance among all of the
parameters to give it the best performance. The tensioning
screws and pegs were designed to give the violinist enough
sensitivity to enable the tensioning to be done with a sufficient
level of accuracy.
Figure 3
Close-ups of the bow tensioning screw and
the string tensioning peg
3