Find the question you need
Define the average · Derive harmonic mean · Check both journeys · Traffic measurement · Practice and further reading
A trip covering equal distances at 60 km/h and 30 km/h averages 40 km/h, not 45. The difference is not a trick: it comes from the distance and time that each speed actually represents.
Chapter index
Chapters 1–3
1. Define the journey before calculating its average · 2. Use the quantity that defines trip-average speed · 3. Derive the harmonic mean for equal distances
Chapters 4–6
4. See why equal times produce a different answer · 5. Compare the two journeys side by side · 6. Generalise to unequal distances
Chapters 7–9
7. Separate travel speed from point-observed speed · 8. Add stopped time without pretending the car kept moving · 9. Avoid averaging corridor summaries carelessly
Chapters 10–12
10. Answer four frequently asked questions · 11. Practise with new speed and time conditions · 12. Link this calculation to wider transport systems
A speed average is not meaningful until we know what is being averaged. Are we looking at one journey through two road sections, several vehicles traversing the same corridor, or spot speeds observed at a roadside sensor?
This article starts with one fictional 120-kilometre journey in two equal-length sections. The first 60 km is travelled at a constant 60 km/h and the second at a constant 30 km/h. These speeds are idealised for an exact mathematical calculation, not traffic observations from a named road.
A common first reaction is to average 60 and 30 to obtain 45 km/h. That arithmetic mean describes the average of the two numbers with equal numerical weight. It does not describe the distance divided by the time spent on this particular journey.
Travel speed for a completed journey is total distance divided by total elapsed travel time, provided the chosen time boundary is clear. This definition comes before any rule about arithmetic or harmonic means.
For the first section, time=60 km ÷ 60 km/h=1 hour. For the second, time=60 km ÷ 30 km/h=2 hours. Total distance=120 km and total time=3 hours.
Therefore the trip-average speed is 120/3=40 km/h. The slower 60 km section uses twice as much time as the faster section. That extra exposure is precisely what the simple arithmetic average of 60 and 30 fails to represent.
Contents · Previous chapter · Next chapter
Let the two equal-distance legs each be d and have positive average speeds u and v. Their travel times are d/u and d/v. The combined trip-average speed is 2d/(d/u+d/v).
Cancel the common distance to get 2/(1/u+1/v)=2uv/(u+v). This is the harmonic mean of the two leg speeds. For u=60 and v=30, the result is 2×60×30/(60+30)=40 km/h.
The harmonic formula is not a new definition of speed. It follows from total distance divided by total travel time when the compared legs are equal in distance. Recognising this condition prevents the formula from being applied where its assumptions do not hold.
Contents · Previous chapter · Next chapter
Now consider a different trip: travel one hour at 60 km/h and one hour at 30 km/h. The respective distances are 60 km and 30 km. Total distance=90 km and total time=2 hours.
This journey’s average speed is 90/2=45 km/h, the arithmetic mean of 60 and 30. The speeds are the same as before, but the time allocation, travelled distances and reporting question have changed.
Do not call either answer an error without describing the journey. Forty km/h is correct for two equal-distance legs; forty-five km/h is correct for two equal-time legs under the stipulated constant-speed model. The weights belong to the travelled distance or elapsed time, not to personal preference.
Contents · Previous chapter · Next chapter
The equal-distance journey travels 60 km at each speed, with leg durations of one and two hours. The equal-time journey spends an hour at each speed but covers 60 km and 30 km. Both use the same two speed labels, yet their aggregate speeds differ.
The table below shows why the denominator must be constructed before selecting an averaging rule. Anyone can check each number using distance=speed×time, but only after deciding which quantities remain fixed.
Transport engineering often needs an answer that describes how long people or goods take to move through a corridor. A number computed from an unexamined average of reported speeds can disguise delay and distort comparisons of alternatives.
| Quantity | Equal distances | Equal times |
|---|---|---|
| At 60 km/h | 60 km in 1 hour | 60 km in 1 hour |
| At 30 km/h | 60 km in 2 hours | 30 km in 1 hour |
| Total distance | 120 km | 90 km |
| Total time | 3 hours | 2 hours |
| Trip-average speed | 40 km/h | 45 km/h |
Contents · Previous chapter · Next chapter
When leg distances differ, use V=(d1+d2)/(d1/u+d2/v). For example, let the first leg be 20 km at 60 km/h and the second leg 40 km at 30 km/h. Their times are 1/3 hour and 4/3 hours.
The total is 60 km in 5/3 hours, giving V=36 km/h. It is neither the unweighted arithmetic mean of 45 km/h nor the equal-distance harmonic mean of 40 km/h. Distances and times set the weights.
A distance-weighted harmonic mean gives the general expression: V=(d1+d2)/(d1/u+d2/v). If one recorded speed describes a whole segment including stops and intersections, its time basis must already incorporate those delays consistently.
Contents · Previous chapter · Next chapter
Transportation analysts distinguish a vehicle’s traversal speed over a segment from speeds measured at a single point. FHWA’s Traffic Monitoring Guide defines segment space-mean speed through segment length divided by traversal time.
Under suitable sampling assumptions, equal-distance vehicle traversals can be aggregated using their travel times, giving a harmonic-mean relation for the corresponding speeds. But a set of spot speeds sampled at a fixed detector is not automatically an unbiased record of all segment traversal times.
The FHWA Traffic Flow Theory discussion notes the representativeness issues created by the location and timing of observations. A formula does not replace an appropriate measurement design.
Contents · Previous chapter · Next chapter
Suppose our original 120 km journey also includes a fifteen-minute stop at a signal or rest point. If the metric is door-to-door elapsed-time speed and the stop lies within the reporting boundary, total time becomes 3.25 hours.
The average is then 120/3.25≈36.92 km/h. If the metric is moving speed excluding stops, the original 40 km/h calculation may describe that different quantity. Neither label should be assigned without stating whether stopped time was included.
For a bus or delivery vehicle, this distinction matters because dwell time, traffic queues and signal delay are part of the passenger or customer experience. An apparent increase in ‘moving speed’ can coexist with unchanged or worse total journey time.
Contents · Previous chapter · Next chapter
Suppose one corridor report covers ten trips and another covers one hundred. Combining their two reported speeds as if they described equal quantities can give the smaller dataset disproportionate influence. The correct aggregation depends on the travel-time and distance information behind both numbers.
If all traversals cover the same segment distance D, an overall speed based on aggregate distance divided by aggregate time is nD divided by the sum of all n trip times. This equals D divided by the mean trip time, not necessarily the arithmetic mean of individual trip speeds.
A credible report retains the numbers of trips, distances, time definitions, direction, time-of-day window and any exclusions. It should say whether the headline is a mean trip time, a travel speed, a median or a percentile. Those are not interchangeable measurements.
Contents · Previous chapter · Next chapter
Is 45 km/h wrong? Not universally. It is correct for equal-time legs at 60 and 30 km/h, and it is the arithmetic mean of those two speeds. It is not the average speed of the stated equal-distance trip.
Why does the slow leg matter more? Covering the same distance at half the speed takes twice the time. Trip-average speed uses total elapsed time in the denominator.
Is the harmonic mean always the right traffic speed? No. It applies to the specific equal-distance aggregation and to certain space-mean-speed calculations under stated sampling conditions. Study design matters.
What happens if one leg has zero speed? If a positive-distance leg truly cannot be completed, its travel time is unbounded and the finite two-leg formula no longer describes a completed journey.
Contents · Previous chapter · Next chapter
For two equal 50 km sections travelled at 100 km/h and 25 km/h, the leg times are 0.5 and 2 hours. Total distance=100 km and time=2.5 hours, giving an average speed of 40 km/h. The simple arithmetic mean would incorrectly give 62.5 km/h for this equal-distance route.
Now travel one hour at 100 km/h and one hour at 25 km/h. The distances become 100 and 25 km, so the average speed is 125/2=62.5 km/h. The arithmetic mean reappears because the time intervals are equal.
For a third check, retain the original equal-distance 60 and 30 km/h trip, but reduce the slow section to 40 km while increasing the fast section to 80 km. The time is 80/60+40/30=8/3 hours, and 120÷(8/3)=45 km/h. Explain why the original 40 km/h has changed.
Contents · Previous chapter · Next chapter
Start with The Gold Standard of Transportation Engineering and The Gold Standard of Transportation. For road design and practical movement, see The Gold Standard of Roads.
Queueing and congestion introduce additional travel-time variability. Explore Queueing Theory: Equal Means, Different Waits and the connected Singapore Roads, Expressways and Junctions guide.
For definitions and measurement limitations consult the FHWA Traffic Monitoring Guide and FHWA traffic measures of effectiveness. These references do not report our invented journeys. Return to The Gold Standard hub for other worked standards.