1.4 Absolute value and distance

Signed measurements describe direction relative to a reference, while many engineering questions require only the magnitude of a difference.

Absolute value. The absolute value of a real number \(a\), written \(|a|\), is the distance between \(a\) and zero on the number line.

More formally,

\[|a|=\begin{cases}a,&a\ge0,\\-a,&a<0.\end{cases}\]

Absolute value measures distance from zero, so \(|a|\ge0\). The distance between \(a\) and \(b\) is

\[d(a,b)=|b-a|.\]

Absolute value is not an instruction to remove a negative sign mechanically. Evaluate the expression inside the bars first, then take its distance from zero. Thus \(|3-8|=|-5|=5\). The order inside the bars matters, but the distance between two values does not: \(|b-a|=|a-b|\).

For elevations \(-2.4\) m and 5.1 m, the distance is

\[|5.1-(-2.4)|=7.5\ \text{m}.\]

Worked Example: Distance between survey readings

Two readings relative to a benchmark are \(-2.4\) m and 5.1 m. Their distance is

\[d=|5.1-(-2.4)|=7.5\ \text{m}.\]

Distance is positive even though one position is below the benchmark.

Try It. Find the distance between \(-3.2\) m and 1.9 m.
Check Your Work \(|1.9-(-3.2)|=5.1\) m.

For real numbers \(a\) and \(b\), we have

\[|ab|=|a||b|, \qquad \qquad |a/b|=|a|/|b|, \text{ when } b\ne0\]

Worked Example: Measuring departure from a pH target

A process target is pH 7.20 and the measured value is 6.85. The signed deviation is \(6.85-7.20=-0.35\), while the magnitude of the deviation is

\[|6.85-7.20|=0.35.\]

Try It. Find the absolute deviation of a pH reading of 7.62 from a target of 7.25.
Check Your Work \(|7.62-7.25|=0.37\).

Worked Example: Comparing two pipe elevations

The invert elevations of two pipe ends are 412.84 m and 410.19 m. Their vertical separation is

\[|410.19-412.84|=2.65\ \text{m}.\]

Try It. Find the separation between elevations 506.30 m and 503.75 m.
Check Your Work \(|503.75-506.30|=2.55\) m.

Triangle inequality For real numbers \(a\) and \(b\), we have \(|a+b|\le|a|+|b|\).

Worked Example: Applying the triangle inequality

A tank level rises by \(0.42\) m and then falls by \(0.31\) m. The magnitude of the net change is

\[ |0.42+(-0.31)|=|0.11|=0.11\text{ m}. \]

The sum of the magnitudes of the individual changes is

\[ |0.42|+|-0.31|=0.42+0.31=0.73\text{ m}. \]

Therefore,

\[ |0.42+(-0.31)|\leq |0.42|+|-0.31|. \]

The net change cannot be greater than the total amount of movement.

Absolute value reports magnitude without direction. A signed deviation can show whether a reading is above or below target, while its absolute value shows how far it is from target.

Worked Example: Interpreting a deviation

A sensor reads 0.8 units below its reference, so its signed deviation is \(-0.8\). The magnitude of the deviation is \(|-0.8|=0.8\) units.

Try It. A reading differs from its target by \(-1.25\) mg/L. Find the magnitude of the difference.
Check Your Work \(|-1.25|=1.25\) mg/L.

Applied Problem: Checking a chemical-feed tolerance. A controller is set to deliver 24.0 L/h. During four inspections, the measured rates are 23.4, 24.3, 25.1, and 23.8 L/h. Calculate the absolute deviation from the set point for each reading and identify the largest deviation.

Check Your Work The absolute deviations are 0.6, 0.3, 1.1, and 0.2 L/h. The largest deviation is 1.1 L/h, produced by the 25.1 L/h reading.

1.4.1 Practice Problems

  1. Evaluate \(|-12|\).
    Check Your Work 12.
  2. Evaluate \(|4-11|\).
    Check Your Work 7.
  3. Find the distance between \(-6\) and 9.
    Check Your Work 15.
  4. Solve \(|x|=5\).
    Check Your Work \(x=-5\) or \(x=5\).
  5. Two elevations are 102.4 m and 98.7 m. Find their separation.
    Check Your Work 3.7 m.
  6. A chlorine residual is 0.18 mg/L below a 0.75 mg/L target. Find the magnitude of the deviation.
    Check Your Work 0.18 mg/L.
  7. Find the vertical separation between pipe elevations 284.65 m and 281.90 m.
    Check Your Work \(|281.90-284.65|=2.75\) m.
  8. A flow reading is 42.6 L/s and its reference value is 45.0 L/s. Find the signed deviation and absolute deviation.
    Check Your Work The signed deviation is \(42.6-45.0=-2.4\) L/s, and the absolute deviation is 2.4 L/s.