IRI in South Africa: what the number tells you—and what to check next

When a number becomes an engineering question

A road profile is a signal before it is a score. What survives when we collapse it to one number?

A roughness result is useful because it compresses a longitudinal profile into a comparable response measure. That compression is also its limitation: a single reported value cannot show where every irregularity occurs, explain the cause of that irregularity, or select a rehabilitation treatment. A good report therefore allows the reader to move in both directions: from the profile to a summary, and from the summary back to the section and evidence it represents.

The International Roughness Index, IRI, is a standardised reference response to a longitudinal road profile, commonly reported in m/km. Its reference simulation uses a speed of 80 km/h. This describes the reference calculation; it is not a universal instruction to drive every survey instrument at that speed. Measurement-system capabilities and the applicable survey procedure still govern collection. (Sayers, Gillespie and Paterson, 1986)

For an asset manager, the practical question may be where to investigate next. For a pavement engineer, it may be whether a change is real, whether it is localised, and which other evidence would explain it. The same roughness record can contribute to both questions, provided its context survives the journey into the dashboard.

Put a suspension on the profile

Governing relationship: z̈s = −63.3(zs − zu) − 6(vs − vu).

zs, zu: sprung and unsprung displacement (m); r: road input (m).

Stiffness / sprung mass: 63.3 and 653 s⁻²; damping / mass: 6 s⁻¹.

Unsprung / sprung mass = 0.15; L = evaluation distance (m).

The film first plots road elevation (mm), then follows the calculation and plots relative velocity (mm/s). Integrate the absolute relative velocity, then divide by distance: R = 3.256 m/km. Signed cancellation is not allowed.

Dynamic response

Quarter-car response

The run-in response is an educational calculation, not a certified IRI result.

Download example data (CSV)

Model, calculations and transcriptEquations, worked steps, source notes and a timed reading guide

Understand the model

A road profile excites two connected masses. Their relative velocity tells us how much suspension travel accumulates over the evaluation distance.

Work through the inputs and equations below, then compare the calculated output with the film. All numerical inputs in this study are synthetic.

Inputs and symbol key

zₛ, zᵤ

Sprung and unsprung vertical displacements (m).

r

Road elevation input (m), evaluated along the travelling wheel.

t; a dot above z

Time (s); one dot is velocity, two dots are acceleration.

L; R

Evaluation distance (m); accumulated suspension travel per distance (m/km).

Values used in this film

• zs, zu: sprung and unsprung displacement (m); r: road input (m).
• Stiffness / sprung mass: 63.3 and 653 s⁻²; damping / mass: 6 s⁻¹.
• Unsprung / sprung mass = 0.15; L = evaluation distance (m).

The mathematical relationships

1. Equation 1

   Source and scope
   $$\ddot z_s=-63.3(z_s-z_u)-6(\dot z_s-\dot z_u)$$

   The first equation balances the sprung mass acceleration against spring and damper response. A dot means a time derivative.

2. Equation 2

   Source and scope
   $$\ddot z_u=\frac{63.3(z_s-z_u)+6(\dot z_s-\dot z_u)-653(z_u-r)}{0.15}$$

   The second equation includes the tyre response to the road and the unsprung-to-sprung mass ratio. Displacements are in metres and time in seconds.

3. Equation 3

   Source and scope
   $$R=\frac{1000}{L}\int |\dot z_s-\dot z_u|\,dt$$

   Take the absolute relative velocity before integrating, so upward and downward travel do not cancel. Divide travel by the actual evaluation distance; 1000 converts m/m to m/km.

Work through the example

Use the relationships above with consistent units. These substitutions are our worked example, not values quoted from the source.

1. Profile

   r(x) = 0.003 sin(2πx/10) + 0.001 sin(2πx/2)

   metres

2. Time coordinate

   t = x / (80/3.6); 200 m takes 9 s

   seconds

3. Suspension travel

   ∫ |vs − vu| dt ≈ 0.651269

   metres over approximately 200 m

4. Normalisation

   1000 × travel / evaluated distance = 3.256345

   m/km

Read the output

Integrate the absolute relative velocity, then divide by distance: R = 3.256 m/km. Signed cancellation is not allowed.

Check your understanding

Recompute with a smaller output step and compare the integrated response. Inspect the run-in separately, preserve the actual evaluated distance, and confirm that halving both amplitudes halves the response within solver tolerance.

Assumptions and limits

The run-in response is an educational calculation, not a certified IRI result.

Return to the full references and method notes to check the cited edition and its applicability.

Timestamped film transcript 25 short entries · select a time to revisit the film

1. 0:00

   Quarter-car response. Synthetic profile; 80 km/h; zero-state start; 100 m run-in.

2. 0:01

   MATHEMATICS · define the relationship and the meaning of each symbol before calculating.

3. 0:07

   zs, zu: sprung and unsprung displacement (m); r: road input (m). Stiffness / sprung mass: 63.3 and 653 s⁻²; damping / mass: 6 s⁻¹. Unsprung / sprung mass = 0.15; L = evaluation distance (m).

4. 0:14

   INPUT · a specified road profile · read the axis units and the numerical values.

5. 0:23

   Synthetic profile; 80 km/h; zero-state start; 100 m run-in.

6. 0:37

   Computed suspension travel / distance = 3.256 m/km over the 100–300 m evaluation interval.

7. 0:43

   OUTPUT · suspension relative velocity · read the axis units and the numerical values.

8. 0:52

   Integrate the absolute relative velocity, then divide by distance: R = 3.256 m/km. Signed cancellation is not allowed.

9. 0:59

   CALCULATION · substitute the defined inputs; keep units through each intermediate step.

10. 1:05

    Profile: r(x) = 0.003 sin(2πx/10) + 0.001 sin(2πx/2) metres

11. 1:08

    Time coordinate: t = x / (80/3.6); 200 m takes 9 s seconds

12. 1:11

    Suspension travel: ∫ |vs − vu| dt ≈ 0.651269 metres over approximately 200 m

13. 1:14

    Normalisation: 1000 × travel / evaluated distance = 3.256345 m/km

14. 1:17

    SENSITIVITY · change one declared input · read the axis units and the numerical values.

15. 1:26

    Half amplitude: 1.62817 m/km. Multiply both profile amplitudes by 0.5.

16. 1:29

    Reference: 3.25635 m/km. Amplitudes 3 mm and 1 mm; wavelengths unchanged.

17. 1:33

    Double amplitude: 6.51269 m/km. Multiply both amplitudes by 2; keep speed and coefficients fixed.

18. 1:36

    Check: Recompute with a smaller output step and compare the integrated response. Inspect the run-in separately, preserve the actual evaluated distance, and confirm that halving both amplitudes halves the response within solver tolerance.

19. 1:42

    ONE ROW · A road profile is a signal before it is a score. What survives when we collapse it to one number?

20. 2:00

    PARAMETER MAP · 24461 evaluated cells. The colour limits stay fixed across every row.

21. 2:10

    Read one row as a longitudinal profile. Moving between rows scales the same wavelengths; it does not change their spacing. For the linear, zero-state quarter-car calculation above,

22. 2:20

    scaling the entire profile by a multiplies suspension travel and the response index by |a|. The 3D height is road elevation, not vehicle acceleration.

23. 2:29

    VALUE LANDSCAPE · A 0–20 m close-up retains every half-metre sample. Height is profile elevation; the second axis is amplitude multiplier.

24. 2:41

    This is a prescribed profile family, not measured roughness. A different wavelength mix requires solving the vehicle response again; amplitude alone cannot describe an arbitrary road.

25. 2:49

    The run-in response is an educational calculation, not a certified IRI result.

Download the timed captions (VTT)

The transcript records the on-screen wording. Formula provenance and interpretation are explained in the source note above.

Work through a 600 m section

Consider six equal, contiguous 100 m reporting intervals. The figures below are invented to demonstrate aggregation; they are not RoadSense project measurements.

Chainage (m)	Reported IRI (m/km)	Length (m)
0–100	2.2	100
100–200	2.4	100
200–300	2.3	100
300–400	5.8	100
400–500	6.2	100
500–600	2.5	100

The arithmetic mean of these six reported values is (2.2 + 2.4 + 2.3 + 5.8 + 6.2 + 2.5) / 6 = 3.57 m/km, rounded to two decimals. It is a descriptive mean of interval results. It is not a claim that a fresh IRI calculation over the combined profile, with its own initialisation and processing, must produce exactly the same result.

Case review

The average is calm. Two intervals are not.

Where should the engineer look first on this 600 m section?

0–100 · 2.2 m/km

The discriminating check

Confirm wheelpath, interval support and survey acceptance before comparing the rise with earlier surveys.

Evidence supports the comparison

If the comparison is valid, locate the two intervals and inspect the supporting condition evidence.

Evidence is missing or incompatible

If the wheelpath or reporting support differs, reconcile those differences before calling the change deterioration.

What you can conclude. A roughness feature locates an investigation question. It does not identify its mechanism or select a treatment.

Original worked case. Read the source editions and method notes.

The two elevated intervals occupy 200 m, or one-third of the section. Their mean is 6.0 m/km; the remaining four intervals average 2.35 m/km. Reporting only 3.57 hides that contrast. Reporting only the maximum, 6.2, would hide the amount of the section to which it relates. The engineering value lies in keeping the summary, location and distribution together.

Spatial view · synthetic data

Follow the measurement along chainage

0–100 m2.2m/km100–200 m2.4m/km200–300 m2.3m/km300–400 m5.8m/km400–500 m6.2m/km500–600 m2.5m/km

0–100 m · 2.2 m/km

Notice what this example does not establish. It does not identify a failed layer, distinguish settlement from construction irregularity, or demonstrate the appropriate treatment. It identifies a localised part of the record that deserves inspection. The next step is to verify that the section identity, measurement quality and supporting observations are sound.

Unequal intervals need an explicit aggregation rule

Suppose two valid intervals have lengths 100 m and 300 m and reported values 2.0 and 4.0 m/km. Their simple unweighted mean is 3.0 m/km. A length-weighted descriptive mean is (100 × 2.0 + 300 × 4.0) / 400 = 3.5 m/km. Neither operation should be hidden behind a column labelled merely “average”.

A reporting specification may require a particular processing or aggregation method. Follow that method and name it. Keep excluded coverage visible: deleting a missing interval from a table does not make the surveyed length equal to the network length. The denominator is part of the reported result.

Compare surveys before declaring deterioration

A second survey might report 3.2 m/km where an earlier delivery reported 2.8. The difference is 0.4 m/km, but subtraction does not establish its cause. Begin with road identity, reference geometry, section limits, survey direction and wheelpath. Then compare measurement methods, relevant quality-control evidence and the rules used to produce the reported intervals.

If one delivery summarises a different length or excludes a local feature, the two numbers may answer different questions. If the acceptance record is unresolved, the comparison needs investigation before it supports a deterioration claim. Record the unresolved item and the action needed to settle it; do not silently convert uncertainty into either a pass or a rejection.

Calibration ranges are not treatment bands

TMH13 Part C discusses measurement, calibration and quality control. Table C.5 concerns the selection of calibration sections. Reusing its ranges as generic maintenance thresholds changes the purpose of the source. (COTO, 2016a)

A project may legitimately use interpretation classes, but the report must identify their source, applicability and boundary conventions. A colour legend should explain whether it represents a measured magnitude, a project screening rule, or an engineering decision. Those are different statements. In the heat map above, colour represents numerical magnitude only.

What a defensible conclusion sounds like

A useful conclusion from the worked example is: “Two adjacent reporting intervals account for the elevated roughness within this 600 m section. Review their source profile, location and quality evidence, then compare with the relevant observations before deciding what investigation is required.” It names a spatial finding and an action without inventing a diagnosis.

The accompanying profile illustration explains geometry only. It does not implement the IRI reference calculation. A numerical implementation would need its model, sampling, filtering, initialisation and reference comparisons checked. The engineering story remains the same at every level of detail: preserve what was measured, explain how it was summarised, and keep the decision traceable to evidence.

Solve the reference suspension response

The film now integrates a two-degree-of-freedom quarter-car response, using the reference normalised coefficients reported by Sayers, Gillespie and Paterson. The assumed road profile is r(x) = 0.003 sin(2πx/10) + 0.001 sin(2πx/2) metres, and speed is 80/3.6 m/s. (Sayers, Gillespie and Paterson, 1986)

z̈s = −63.3(zs − zu) − 6(żs − żu)
z̈u = [63.3(zs − zu) + 6(żs − żu) − 653(zu − r)] / 0.15
Response per distance = integral |żs − żu| dt / L

The subscripts s and u denote sprung and unsprung motion. The coefficients are normalised by sprung mass: the stiffness terms have units s⁻², the damping term s⁻¹, and the mass ratio is dimensionless. The differential equations are integrated from a zero state over 300 m. The first 100 m is a run-in, with the following 200 m used for the illustrative travel-per-distance result: 3.256 m/km. Halving the output sampling interval changes that result by less than 0.1% in the numerical check.

This deliberately specified run-in is not the standard initialisation and processing of a certified IRI implementation. The graph adds an 8 mm display offset to keep the three traces on one positive axis; the model itself uses signed displacements. Blue is sprung motion, mustard is unsprung motion and grey is the input road profile. The animation displays the first 30 m after run-in while the reported result uses the full stated evaluation interval. The earlier six-interval aggregation table remains a separate example.

Calculation laboratory

What does doubling the profile amplitude actually change?

Calculation step	Substitution or result	Units and interpretation
Profile	r(x) = 0.003 sin(2πx/10) + 0.001 sin(2πx/2)	metres
Time coordinate	t = x / (80/3.6); 200 m takes 9 s	seconds
Suspension travel	∫	vs − vu
Normalisation	1000 × travel / evaluated distance = 3.256345	m/km

The absolute value in the integral is essential. Positive and negative suspension velocities represent movement in opposite directions; both contribute to accumulated travel. Integrating the signed velocity would largely cancel that travel and answer a different question. The numerical model solves four first-order state equations for sprung displacement, sprung velocity, unsprung displacement and unsprung velocity. The plotted road is an input, while the two mass motions are calculated states.

For this linear system with zero initial state, scaling the entire input profile by a positive factor scales every state and the response by the same factor. Changing just one wavelength does not have that simple interpretation: it changes excitation frequency at the selected speed. A constant vertical datum shift is also different from an amplitude change. Consequently, a sensitivity result must identify precisely which transformation was applied.

The reference quarter-car context comes from Sayers, Gillespie and Paterson (1986). The zero-state initialisation and 100 m run-in used here are deliberately documented departures from a certified processing procedure. The demonstration explains response mechanics; use validated processing and measurement quality evidence for reported IRI. (Sayers, Gillespie and Paterson, 1986)

Controlled sensitivity experiment

Controlled case	Calculated output	What changes
Half amplitude	1.628173 m/km	Multiply both profile amplitudes by 0.5.
Reference	3.256345 m/km	Amplitudes 3 mm and 1 mm; wavelengths unchanged.
Double amplitude	6.512691 m/km	Multiply both amplitudes by 2; keep speed and coefficients fixed.

CALCULATION LABORATORY

What does doubling the profile amplitude actually change?

Compare the cases on a fixed scale. Select a case to inspect its input change and calculated output.

Sensitivity case1. Half amplitude2. Reference3. Double amplitude

Half amplitude: 1.628173 m/km. Multiply both profile amplitudes by 0.5.

Reproduce and challenge the result

Recompute with a smaller output step and compare the integrated response. Inspect the run-in separately, preserve the actual evaluated distance, and confirm that halving both amplitudes halves the response within solver tolerance.

These are original synthetic calculations. The sensitivity cases vary the stated input while retaining the other assumptions; they are not field observations or treatment thresholds. The film’s calculation and sensitivity chapters use the same numerical cases as this table.

Zoom out without losing the wavelengths

Read one row as a longitudinal profile. Moving between rows scales the same wavelengths; it does not change their spacing. For the linear, zero-state quarter-car calculation above, scaling the entire profile by a multiplies suspension travel and the response index by |a|. The 3D height is road elevation, not vehicle acceleration.

Reproduce the experiment. r(x, a) = a[3 sin(2πx/10) + sin(2πx/2)] mm. The downloadable table contains 24,461 evaluated cells, with the two input coordinates and output value retained on every row. These are parameter combinations, not that many independent field observations. The equation extends the worked example above; it is original teaching arithmetic, not a new standard or calibrated prediction.

Computed parameter atlas · 24,461 cells

A small ripple, repeated thousands of times

A road profile is a signal before it is a score. What survives when we collapse it to one number?

Download every calculated cell (CSV) · Open full-size explorer · Equation source and tests (ZIP)

This is a prescribed profile family, not measured roughness. A different wavelength mix requires solving the vehicle response again; amplitude alone cannot describe an arbitrary road.

References and method notes

1. Sayers, Gillespie and Paterson (1986). Guidelines for Conducting and Calibrating Road Roughness Measurements. World Bank Technical Paper 46. Washington, DC: World Bank. Original source (Accessed: 5 September 2026).Reference quarter-car and measurement calibration
2. COTO (2016a). TMH13: Automated Road Condition Assessments. Part C: Roughness. Committee Draft Final, May 2016. Original source (Accessed: 5 September 2026).Sections C.2–C.5; Table C.5 is for calibration-section selection

Use the source edition and project requirements that apply to your work. Educational examples are not site-specific design advice.