TMH9 visual assessment: keep degree, extent and evidence together

Why two assessment fields deserve separate attention

Would you send a crew to one continuous defect or sixty tiny interruptions? The total length alone cannot answer.

A visual assessment describes what is present and how it is distributed. TMH9 distinguishes degree from extent and provides guidance for representing observations within an assessment segment. The applicable pavement and distress guidance matters: the words are part of a method, not interchangeable labels for a photograph that looks severe. (COTO, 2016b)

In plain terms, degree concerns the assessed seriousness of a distress, while extent concerns how much of the assessed segment is affected under the relevant convention. A short concentrated defect and a widely distributed pattern can call for different investigation questions even when a summary hides the distinction.

A professional record therefore preserves the original observation, assessment fields, segment definition and supporting evidence. It allows a reader to understand the judgement and an assessor to revisit it.

Count the affected length without losing its pattern

Governing relationship: Extent = affected length / assessed length × 100.

lj: non-overlapping affected patch length (m); L: assessed length (m).

Extent is a proportion. Degree and spatial arrangement are separate inputs.

The film first plots length (m), then follows the calculation and plots extent (%). Both sections have 30% extent. The sum preserves length but discards where the patches occur.

Spatial accounting

Distribution of distress

Equal extent does not imply equal degree, pattern or cause.

Download example data (CSV)

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

Understand the model

Coverage tells us how much of a declared section is affected. It discards the arrangement of those affected patches.

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

lⱼ

Unique affected length of patch j (m).

L

Total assessed section length (m).

e

Longitudinal coverage (%); not a TMH9 ordinal extent rating.

Values used in this film

• lj: non-overlapping affected patch length (m); L: assessed length (m).
• Extent is a proportion. Degree and spatial arrangement are separate inputs.

The mathematical relationships

1. Equation 1

   Source and scope
   $$e=100\frac{\sum_j l_j}{L}$$

   Sum unique, non-overlapping affected lengths and divide by the assessed length. Both lengths must use the same units.

2. Equation 2

   Source and scope
   $$100\frac{30}{100}=100\frac{6(5)}{100}=30\%$$

   A continuous 30 m patch and six separate 5 m patches occupy the same proportion of a 100 m section. Equal coverage does not imply equal severity.

Work through the example

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

1. Continuous pattern

   30 / 100 × 100 = 30

   percent

2. Distributed pattern

   (5 + 5 + 5 + 5 + 5 + 5) / 100 × 100 = 30

   percent

3. Overlapping records

   [0,20] and [10,30]: union length = 30, not 40

   metres

4. Longer denominator

   30 / 120 × 100 = 25

   percent

Read the output

Both sections have 30% extent. The sum preserves length but discards where the patches occur.

Check your understanding

Test disjoint, overlapping and fully nested intervals. The merged affected length must lie between zero and the assessed length. Retain degree, distress identity and the original interval coordinates alongside the percentage.

Assumptions and limits

Equal extent does not imply equal degree, pattern or cause.

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

Timestamped film transcript 24 short entries · select a time to revisit the film

1. 0:00

   Distribution of distress. Two 100 m sections; 30 m total affected length in each.

2. 0:01

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

3. 0:07

   lj: non-overlapping affected patch length (m); L: assessed length (m). Extent is a proportion. Degree and spatial arrangement are separate inputs.

4. 0:14

   INPUT · patch lengths in the distributed section · read the axis units and the numerical values.

5. 0:23

   Two 100 m sections; 30 m total affected length in each.

6. 0:33

   30/100 = 30% in each section. Six patches × 5 m = the same length, a different pattern.

7. 0:43

   OUTPUT · extent, same denominator · read the axis units and the numerical values.

8. 0:52

   Both sections have 30% extent. The sum preserves length but discards where the patches occur.

9. 0:59

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

10. 1:05

    Continuous pattern: 30 / 100 × 100 = 30 percent

11. 1:08

    Distributed pattern: (5 + 5 + 5 + 5 + 5 + 5) / 100 × 100 = 30 percent

12. 1:11

    Overlapping records: [0,20] and [10,30]: union length = 30, not 40 metres

13. 1:14

    Longer denominator: 30 / 120 × 100 = 25 percent

14. 1:17

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

15. 1:26

    100 m section: 30 % coverage. Thirty unique affected metres divided by 100 m.

16. 1:29

    120 m section: 25 % coverage. Same affected length; denominator increases to 120 m.

17. 1:33

    200 m section: 15 % coverage. Same affected length; denominator doubles.

18. 1:36

    Check: Test disjoint, overlapping and fully nested intervals. The merged affected length must lie between zero and the assessed length. Retain degree, distress identity and the original interval coordinates alongside the percentage.

19. 1:42

    ONE ROW · Would you send a crew to one continuous defect or sixty tiny interruptions? The total length alone cannot answer.

20. 1:55

    PARAMETER MAP · 6000 evaluated cells. The colour limits stay fixed across every row.

21. 2:05

    Each row distributes an ideal 30 m affected length into n equal patches over 100 m. The map samples their starts at one-metre intervals. Thin patches can fall between samples:

22. 2:16

    count the coloured cells and you will see aliasing rather than a change in the underlying continuous coverage. That is a useful warning about reporting support.

23. 2:25

    INTERPRETATION · These are discrete states. Read the map cell by cell; a smooth surface would invent intermediate categories.

24. 2:34

    Equal extent does not imply equal degree, pattern or cause.

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 two 100 m sections

Consider two invented sections of equal length. In Section A, the affected longitudinal length occupies 0–30 m. In Section B, six separate 5 m patches occur at different positions. Both have 30 m of affected length, giving 30 / 100 × 100 = 30% under this deliberately simple longitudinal-coverage definition.

This definition is not an assertion of TMH9's extent-rating procedure. It lets us discuss the spatial information that a total alone can lose.

Comparison	Section A	Section B
Section length	100 m	100 m
Affected length in this example	One 30 m length	Six 5 m lengths
Total affected length	30 m	30 m
Longitudinal coverage	30%	30%
Spatial question	Why concentrated here?	Why distributed across the section?

Case review

One repair location or six?

Can equal affected length conceal a different maintenance problem?

Continuous patch · 30 Unique affected length (m)

The discriminating check

Return to the located observations: retain degree, extent assessment, defect type and patch boundaries separately.

Evidence supports the comparison

If the mapped boundaries are reliable, investigate the spatial pattern and the relevant defect mechanism.

Evidence is missing or incompatible

If overlapping records double-count the same patch, reconcile the interval union before summarising coverage.

What you can conclude. The 30% geometric coverage example is not a conversion into a TMH9 ordinal extent rating.

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

The chart breaks Section A into five 20 m reporting windows: 20, 10, 0, 0 and 0 m affected. The first two windows contain the entire observation. In Section B, the same total is spread out. A total can support a summary; the located pattern supports a different conversation about investigation.

The denominator changes the statement

Suppose a reviewer mistakenly reports Section A against a 200 m segment while retaining the same 30 m affected length. The illustrative coverage becomes 30 / 200 × 100 = 15%. The road has not changed. The reported denominator has.

This is why chainage limits and segment identity belong in an assessment record. A boundary correction can alter what is represented without constituting a new observation of distress. The revision should identify the changed boundary, the affected fields and the reason for reassessment. Keeping only the latest number makes that distinction difficult to recover.

What the image establishes, and what the assessor adds

An image can establish visible evidence at a particular location and time. Its framing, resolution, viewing conditions and coverage determine what can reasonably be seen. An assessor applies the relevant guidance to that evidence; an index calculation, where required, then follows its own defined method.

A suggested record links distress type, degree, extent, chainage, direction, assessor, date and image identifiers. Where the evidence is inadequate, the record should say so. A blank assessment is not automatically a finding of no distress, and a missing image should not be replaced by a confident description invented from neighbouring frames.

Separate disagreement from data loss

Two assessors may need to discuss a rating. That is different from a record that has lost its location or supporting image. The first needs a review of the assessment against the method; the second needs recovery or clarification of the evidence. Combining both problems into a general “quality issue” makes resolution harder to assign.

For a review, retain the initial assessment, the revised assessment and the rationale. A useful note explains whether the difference came from corrected segment limits, better evidence, a method clarification or a changed assessment judgement. These distinctions make the resulting dataset more defensible without turning every adjustment into a lengthy narrative.

Resist the shortcut from ratings to a condition index

An extent percentage, a degree rating and a condition index are not interchangeable quantities. Multiplying convenient numbers does not create a validated index. If the project requires an index, name its source edition, input rules, missing-data treatment and aggregation. Keep the original assessment fields available alongside the result. (COTO, 2016b)

The coverage figures retain the affected lengths and section denominator. The Manim film contrasts concentrated and distributed patterns. Neither assigns a TMH9 rating or computes VCI. The worked example's conclusion is narrower and more useful: keep degree, extent, spatial distribution and evidence together so that the next engineering judgement starts from a faithful record.

Calculation laboratory

Why is overlap a mathematical issue before it is a rating issue?

Calculation step	Substitution or result	Units and interpretation
Continuous pattern	30 / 100 × 100 = 30	percent
Distributed pattern	(5 + 5 + 5 + 5 + 5 + 5) / 100 × 100 = 30	percent
Overlapping records	[0,20] and [10,30]: union length = 30, not 40	metres
Longer denominator	30 / 120 × 100 = 25	percent

Coverage is the measure of the union of affected intervals divided by the assessed length. Adding raw patch lengths is valid only when patches do not overlap. Sort the intervals by start chainage, extend the current interval when the next start lies within it, and sum the merged lengths. This retains a defensible denominator and prevents duplicate photographs of one defect from increasing coverage.

A boundary change can alter coverage even when the pavement has not changed. The sensitivity cases deliberately retain the same thirty affected metres while extending the assessment section. They therefore describe an aggregation effect, not improvement. Comparing surveys requires consistent boundaries or a documented reaggregation onto a common reference.

TMH9 distinguishes degree and extent. The continuous percentage calculated here is a geometric example, not an automatic conversion into the manual’s ordinal extent categories. Select the appropriate distress-specific measurement and rating rules before coding an assessment. Longitudinal coverage also cannot stand in for affected area when the applicable measure requires area. Two patterns with equal coverage can demand different inspection questions because their location and concentration differ. (COTO, 2016b)

Controlled sensitivity experiment

Controlled case	Calculated output	What changes
100 m section	30.000000 % coverage	Thirty unique affected metres divided by 100 m.
120 m section	25.000000 % coverage	Same affected length; denominator increases to 120 m.
200 m section	15.000000 % coverage	Same affected length; denominator doubles.

CALCULATION LABORATORY

Why is overlap a mathematical issue before it is a rating issue?

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

Sensitivity case1. 100 m section2. 120 m section3. 200 m section

100 m section: 30 % coverage. Thirty unique affected metres divided by 100 m.

Reproduce and challenge the result

Test disjoint, overlapping and fully nested intervals. The merged affected length must lie between zero and the assessed length. Retain degree, distress identity and the original interval coordinates alongside the percentage.

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.

When the sampling grid misses the patch

Each row distributes an ideal 30 m affected length into n equal patches over 100 m. The map samples their starts at one-metre intervals. Thin patches can fall between samples: count the coloured cells and you will see aliasing rather than a change in the underlying continuous coverage. That is a useful warning about reporting support.

Reproduce the experiment. Affected if x mod (100/n) < 30/n; sample at 1 m spacing. The downloadable table contains 6,000 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 · 6,000 cells

Thirty metres can live in many places

Would you send a crew to one continuous defect or sixty tiny interruptions? The total length alone cannot answer.

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

The grid is a sampling demonstration, not a TMH9 extent or degree rating. Exact union length stays 30 m; sampled affected-cell percentages need not equal 30%.

References and method notes

1. COTO (2016b). TMH9: Manual for Visual Assessment of Road Pavements. Part A: General. Committee Draft Final, May 2016. Original source (Accessed: 5 September 2026).Section A.2: degree and extent

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