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Passing Sight Distance

PSD = d_{1} + d_{2} + d_{3} + d_{4} ≫ SSD

On two-lane two-way highways, passing sight distance is much longer than SSD because the maneuver occupies the opposing lane. The classical AASHTO four-term model sums: d1 initial maneuver, d2 occupation of the left lane, d3 clearance to the oncoming vehicle, and d4 oncoming-vehicle travel during the pass. Current Green Book values (e.g. 710 ft at 50 mph vs SSD ≈ 425 ft) are shorter than the old 4-term tables but still well above SSD. Crest vertical curves for passing use a larger object height (oncoming car) and a different constant than the SSD crest formula.

PSD

(ft)

Passing sight distance

d_{1}

(ft)

Initial maneuver distance

d_{2}

(ft)

Distance occupying opposing lane

d_{3}

(ft)

Clearance length

d_{4}

(ft)

Opposing-vehicle distance

Step-by-step solved example

Design speed 50 mph. Compare AASHTO Green Book PSD (≈ 710 ft) with SSD (≈ 425 ft). By what ratio is PSD longer?

Passing sight is much longer than SSD on two-lane highways.

1. Tabulated values
PSD ≈ 710 ft, SSD ≈ 425 ft at 50 mph (AASHTO).
$PSD \approx 710 ft$
2. Ratio
710/425 = 1.67, so PSD is about 67% longer than SSD — and older 4-term PSD at 50 mph was ~1600–1800 ft (about 4× SSD).
$PSD / SSD \approx 1.67$

Answer: PSD ≈ 710 ft ≈ 1.67 × SSD (50 mph)

10 practice questions

0/10 correct

1.Passing sight distance is used primarily on

2.Relative to SSD at the same speed, PSD is

3.Term d2 in the four-component model is the distance

4.AASHTO Green Book PSD at 60 mph is nearest

5.No-passing (solid yellow) zones are warranted where

6.Crest vertical curves designed for passing (vs stopping) are

7.On a one-way or divided facility, passing sight

8.The oncoming-vehicle term d4 grows with

9.If SSD = 425 ft and PSD = 710 ft at 50 mph, a crest that barely meets SSD

10.Decision sight distance, compared with SSD, is