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# Commentary 9

At an intersection where gap acceptance applies, where major road traffic arrives randomly from each direction and where a right-turning vehicle from a minor road seeks different sized critical gaps in the traffic flows from the two different directions on the major road, the theoretical absorption capacity can be developed as follows:

Assume that the total major traffic stream is made up of the traffic flow from the left, with volume qL, and the flow from the right, with volume qR. Each gap or headway in the total traffic stream starts with the arrival of a vehicle from one direction and ends with the next arrival, which may be from the same or the opposite direction. Therefore, during a significant period of time H, the number of headways in the total major traffic stream will be:

 Total number of headways = $$H (q_{L} + q_{R} )$$ C62

Now assume that a queue of minor traffic stream vehicles is waiting to turn right and must give way to this major traffic stream, which has a negative exponential headway distribution in each direction. Let the combination of a gap of at least tL,i in the major traffic from the left with a gap of at least tR,i in the major traffic from the right be the minimum condition required to allow i minor traffic stream vehicles to make the right turn within the one gap, where i = 1, 2, 3,....

The probability of the simultaneous occurrence of a headway greater than or equal to $$t_{L,i}$$ in the major road flow from the left and a headway greater than or equal to $$t_{R,i}$$ in the major road flow from the right is:

 Pr ($$h_{L} \ge t_{L,i}$$| $$h_{R} \ge t_{R,i}$$) $$= e^{-q_{L} t_{L,i} } . e^{-q_{R} t_{R,i} } = e^{-(q_{L} t_{L,i} + q_{R} t_{R,i} )}$$ C63

Hence, during the significant period of time H, the number of major traffic stream headways large enough to allow at least $$i$$ minor traffic stream vehicles to make the right turn, i = 1, 2, 3,... is:

 No. of headways ($$h_{L} \ge t_{L,i}$$| $$h_{R} \ge t_{R,i}$$) $$= H (q_{L} + q_{R} ) e^{-(q_{L} t_{L,i} + q_{R} t_{R,i} )}$$ C64

Therefore, the number of major traffic stream headways that allow exactly i minor traffic stream vehicles to turn right, i = 1, 2, 3,... is:

 $\mathrm{n}_{\mathrm{i}}\mathrm{ }\mathrm{=}\mathrm{ }\mathrm{\text{H.}}\left( \mathrm{q}_{\mathrm{L}}\mathrm{+}\mathrm{q}_{\mathrm{R}} \right)\mathrm{.}\left\lbrack \mathrm{e}^{\mathrm{- (}\mathrm{q}_{\mathrm{L}}\mathrm{t}_{\mathrm{L,i}}\mathrm{+}\mathrm{q}_{\mathrm{R}}\mathrm{t}_{\mathrm{R,i}}\mathrm{)}} - \mathrm{e}^{\mathrm{- (}\mathrm{q}_{\mathrm{L}}\mathrm{t}_{\mathrm{L,i + 1}}\mathrm{+}\mathrm{q}_{\mathrm{R}}\mathrm{t}_{\mathrm{R,i + 1}}\mathrm{)}} \right\rbrack$ C65

Hence, the total number of minor stream vehicles able to turn right during the period $$H$$ is:

 N = $$\sum _{i=1}^{\infty }i.n_{i}$$ C66 = $$H (q_{L} + q_{R} )\sum _{i=1}^{\infty }i . \left[e^{-(q_{L} t_{L,i} + q_{R} t_{R,i} )} - e^{-(q_{L} t_{L,i+1} + q_{R} t_{R,i+1} )} \right]$$ = $$H (q_{L} + q_{R} )\sum _{i=1}^{\infty }e^{-(q_{L} t_{L,i} + q_{R} t_{R,i} )}$$

Now assume that a critical gap TL in the major traffic flow from the left is the minimum that will allow one minor stream right-turner to cross that stream, and that a critical gap TR in the major traffic flow from the right is the minimum that will allow one minor stream right-turner to join that stream. Further, assume that an additional follow-up headway T0 is sufficient to allow one additional minor stream vehicle to follow in undertaking the manoeuvre. This implies that:

 $t_{L,i} = T_{L} + (i - 1) T_{0} \ and \ t_{R,i} = T_{R} + (i - 1) T_{0} , i = 1, 2, 3, ...$ C67

Substituting from Equations C67 into Equation C66:

 N= $$H (q_{L} + q_{R} )\sum _{i=1}^{\infty }e^{-(q_{L} T_{L} + q_{L} (i - 1)T_{0} + q_{R} T_{R} + q_{R} (i - 1)T_{0} )}$$ = $$H (q_{L} + q_{R} ) e^{-(q_{L} T_{L} + q_{R} T_{R} )} \sum _{i=1}^{\infty }e^{-(q_{L} + q_{R} ) (i - 1) T_{0} }$$ = $$H (q_{L} + q_{R} ) e^{-(q_{L} T_{L} + q_{R} T_{R} )} \sum _{k=0}^{\infty }\left(e^{-(q_{L} + q_{R} ) T_{0} } \right)^{k}$$ = $$\frac{H (q_{L} + q_{R} ) e^{-(q_{L} T_{L} + q_{R} T_{R} )} }{1 - e^{-(q_{L} + q_{R} ) T_{0} } }$$ C68

Thus the theoretical maximum rate at which minor stream vehicles can turn right, that is, the theoretical absorption capacity, is obtained as:

 $C = \frac{N}{H} = \frac{(q_{L} + q_{R} ) e^{-(q_{L} T_{L} + q_{R} T_{R} )} }{1 - e^{-(q_{L} + q_{R} ) T_{0} } }$ C69

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