Module 1 : Traffic Stream Characteristics

Lecture 02 : Fundamental Relations of Traffic Flow

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Derivation of the relation

The relation between time mean speed and space mean speed can be derived as below. Consider a stream of vehicles with a set of sub-stream flow $q_1$, $q_2$, ... $q_i$, ...$q_n$ having speed $v_1$,$v_2$, ...$v_i$, ...$v_n$. The fundamental relation between flow($q$), density($k$) and mean speed $v_s$ is,

$$q = k\times v_s$$ (1)

Therefore for any sub-stream $q_i$, the following relationship will be valid.

$$q_i = k_i\times v_i$$ (2)

The summation of all sub-stream flows will give the total flow $q$:

$$\Sigma q_i = q.$$ (3)

Similarly the summation of all sub-stream density will give the total density $k$.

$$\Sigma k_i = k.$$ (4)

Let $f_i$ denote the proportion of sub-stream density $k_i$ to the total density $k$,

$$f_i = \frac{k_i}{k}.$$ (5)

Space mean speed averages the speed over space. Therefore, if $k_i$ vehicles has $v_i$ speed, then space mean speed is given by,

$$v_s = \frac{\Sigma k_i v_i}{k}.$$ (6)

Time mean speed averages the speed over time. Therefore,

$$v_t = \frac{\Sigma q_i v_i}{q}.$$ (7)

Substituting 2, $v_t$ can be written as,

$$v_t = \frac{\Sigma {k_i}{{v_i}^2}}{q}$$ (8)

Rewriting the above equation and substituting 5, and then substituting 1, we get,

$$v_t = k\Sigma\frac{k_i}{k}v_i^2$$

$$= \frac{k\Sigma {f_i}{{v_i}^2}}{q}$$

$$= \frac{\Sigma {f_i}{{v_i}^2}}{v_s}$$

By adding and subtracting $v_s$ and doing algebraic manipulations, $v_t$ can be written as,

$$v_t = \frac{\Sigma {f_i}{(v_s+(v_i-v_s))^2}}{v_s}$$ (9)

$$= \frac{\Sigma{f_i}{{(v_s)^2}+{(v_i-v_s)}^2+2.v_s.(v_i-v_s)}}{v_s}$$ (10)

$$= \frac{\Sigma {{f_i}{v_s}^2}}{v_s} + \frac{\Sigma{f_i}{{(v_i-v_s)}^2}}{v_s} + \frac{2.v_s.{\Sigma{f_i}{(v_i-v_s)}}}{v_s}$$ (11)

The third term of the equation will be zero because $\Sigma f_i(v_i-v_s)$ will be zero, since $v_s$ is the mean speed of $v_i$. The numerator of the second term gives the standard deviation of $v_i$. $\Sigma f_i$ by definition is 1.Therefore,

$$v_t = v_s\Sigma{f_i} +\frac{\sigma^2}{v_s}+0$$ (12)

$$= v_s + \frac{\sigma^2}{v_s}$$ (13)

Hence, time mean speed is space mean speed plus standard deviation of the spot speed divided by the space mean speed. Time mean speed will be always greater than space mean speed since standard deviation cannot be negative. If all the speed of the vehicles are the same, then spot speed, time mean speed and space mean speed will also be same.

Numerical Example

For the data given below,compute the time mean speed and space mean speed. Also verify the relationship between them. Finally compute the density of the stream.

speed range	frequency
0-10	5
10-20	15
20-30	20
30-40	25
40-50	30

Solution

	speed	mid interval	flow
No.	range	$v_i=\frac{v_l+v_u}{2}$	$q_i$	$q_iv_i$	$v_i^2$	$q_iv_i^2$	$q_i/v_i$
	$v^l<v<v^u$
1	0-10	5	6	30	25	150	6/5
2	10-20	15	16	240	225	3600	16/15
3	20-30	20	24	600	625	15000	24/25
4	30-40	25	25	875	1225	30625	25/35
5	40-50	30	17	765	2025	34425	17/45
	total		88	2510		83800	4.3187

The solution of this problem consist of computing the time mean speed $v_t = \frac{\Sigma{q_iv_i}}{\Sigma{q_i}}$,space mean speed $v_s = \frac{\Sigma{q_i}}{\frac{\Sigma{q_i}}{v_i}}$,verifying their relation by the equation $v_t = v_s+\frac{\sigma^2}{v_s}$,and using this to compute the density. To verify their relation, the standard deviation also need to be computed $\sigma^2 = \frac{\Sigma{qv^2}}{\Sigma{q}}-v_t^2$. For convenience,the calculation can be done in a tabular form as shown in table 0.1.1.
The time mean speed($v_t$) is computed as:

$$v_t = \frac{\Sigma{q_iv_i}}{\Sigma{q_i}}$$

$$= \frac{2510}{88} = 28.52$$

The space mean speed can be computed as:

$$v_s = \frac{\Sigma{q_i}}{\frac{\Sigma{q_i}}{v_i}}$$

$$= \frac{88}{4.3187} = 20.38$$

The standard deviation can be computed as:

$$\sigma^2 = \frac{\Sigma{qv^2}}{\Sigma{q}}-v_t^2$$

$$= \frac{83800}{88}-28.52^2 = 138.727$$

The time mean speed can also $v_t$ can also be computed as:

$$v_t = v_s+\frac{\sigma^2}{v_s}=20.38+\frac{138.727}{20.38} = 27.184$$

The density can be found as:

$$k = \frac{q}{v} = \frac{88}{20.38} = 4.3~{\mathrm{vehicle/km}}$$