Intersection sight distance numerical

Given: Major road: $V_1 = 80$ km/h, grade 2%. Minor road: $V_2 = 60$ km/h, grade 3%. Uncontrolled right-angle intersection.

Principle: At an uncontrolled intersection each driver must be able to see the other vehicle in time to stop. The approach sight triangle is therefore formed with legs equal to the stopping sight distance of each approach (IRC/AASHTO). The grade is included in the braking term.

Assumptions: reaction time $t = 2.5$ s; $f = 0.35$ (NRS/IRC design value; no further reduction is made for the rain); the approach is taken on the downgrade, which is the worse case:

$$SSD = 0.278\,V\,t + \frac{V^2}{254\,(f - G)}$$

Major road (80 km/h, G = 2%)

$$SSD_1 = 0.278(80)(2.5) + \frac{80^2}{254(0.35-0.02)} = 55.6 + 76.35 = 132.0\ \text{m}$$

Minor road (60 km/h, G = 3%)

$$SSD_2 = 0.278(60)(2.5) + \frac{60^2}{254(0.35-0.03)} = 41.7 + 44.3 = 86.0\ \text{m}$$

Answer: the minimum intersection sight distance is 132 m along the major road and 86 m along the minor road (a sight triangle with legs 132 m and 86 m must be kept clear of obstructions). If the approaches are upgrades the values would be 123.7 m and 79.0 m; on level ground, 127.6 m and 82.2 m.