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# Commentary 1

## C1.1 Space Mean Speed and Time Mean Speed

For any traffic stream, space mean speed is always less than or equal to time mean speed because slower vehicles occupy any given segment of road for a longer period of time than faster vehicles, and therefore receive a greater weighting in the calculation of space mean speed than they do in the calculation of time mean speed. This is illustrated by the following simple numerical example:

Measured spot speeds at the start of a 1 km length of road record 50% of vehicles travelling at 30 km/h and 50% at 60 km/h. Hence, the time mean speed is vt = 45 km/h.

Assume that all vehicles maintain constant speed. Then, travel times over the 1 km length are two minutes for those travelling at 30 km/h and one minute for those at 60 km/h. Assume 3 s headways at the measurement point, with every second vehicle travelling at 30 km/h and every other vehicle 60 km/h. Then, in any two minute period, 40 vehicles enter the 1 km segment, 20 travelling at 30 km/h and 20 at 60 km/h. At the end of the two minute period, all 20 travelling at 30 km/h are still within the segment but the first 10 of those at 60 km/h have already left it – that is, at this instant, there are 30 vehicles within the 1 km segment, 20 travelling at 30 km/h and 10 at 60 km/h. Thus the space mean speed is:

 $$\text{ v }_{s} = \frac{20 (30) + 10 (60)}{30} = \frac{1200}{30} = 40$$ km/h

Note that this is the inverse of the average travel time over the 1 km segment, for all vehicles whose spot speeds are recorded at the measurement point. That is, 1.5 minutes/km is equivalent to 40 km/h.

Now, the variance of space speeds can be obtained by noting that, at any instant, 20 vehicles within the segment have a speed of 30 km/h and 10 have a speed of 60 km/h, that is, they have deviations from the 40 km/h space mean speed of -10 km/h and +20 km/h respectively. The variance of space speeds can thus be calculated as:

 $$\sigma _{s}^{2} = \frac{\sum _{i=1}^{N}(\text{ v }_{i} - \text{ v }_{s} )^{2} }{N - 1} = \frac{20 (-10)^{2} + 10 (+20)^{2} }{(30 - 1)} = 206.9$$ (km/h)2

Wardrop’s relationship (Wardrop 1952) would then estimate the time mean speed as:

 $$\text{ v }_{t} = \text{ v }_{s} + \frac{\sigma _{s}^{2} }{\text{ v }_{s} } = 40 + \frac{206.9}{40} = 45.2$$   km/h

which is close to the value of 45 km/h calculated as the arithmetic mean of the spot speeds.

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