SME
Transportation Engineering
Geometric design
Use this for “A road is designed for 80 km/h on level ground. With t = 2.5 s and a = 3.4 m/s², what stopping sight distance does a driver need?”
Diagram
Reacting: 0.278Vt = 0.278 × 80 × 2.5 = 55.6 m.
Braking:
V
2
254(a ÷ g + G)
=
80
2
254 × (3.4 ÷ 9.81 + 0)
=
72.7
m
on
the
level.
SSD = 55.6 + 72.7 = 128 m, all to one scale.
This is an example. Type your own values to start a new problem.
dᵣ (m)
=
0.278
×
V (km/h)
×
t (s)
db (m)
=
(
V (km/h)
)
2
÷
(254
×
(
a (m/s²)
÷
9.81
+
G
))
SSD (m)
=
dᵣ (m)
+
db (m)
Formulas
dᵣ
=
0.278
×
V
×
t
55.6 = 0.278 × 80 × 2.5
db
=
V
2
254 × (
a
÷ 9.81 +
G
)
72.7003
=
80
2
254 × (3.4 ÷ 9.81 + 0)
SSD = dᵣ + db
128.3003 = 55.6 + 72.7003
Assumptions
•
AASHTO Green Book 2018: reaction time 2.5 s and deceleration 3.4 m/s² for design.
•
The grade G is a decimal, + uphill and − downhill; g = 9.81 m/s².
•
V in km/h gives distances in metres: 0.278 turns km/h into m/s.
Step-by-step
WE KNOW
V = 80 km/h, t = 2.5 s, a = 3.4 m/s², G = 0
FIND
reaction distance (dᵣ), braking distance (db), stopping sight distance (SSD)
Step 1 · Find reaction distance (dᵣ)
Use dᵣ = 0.278 × V × t
While the driver sees the object and reaches for the brake, the car runs on at full speed.
dᵣ = 0.278 × 80 × 2.5
dᵣ = 22.24 × 2.5
dᵣ = 55.6 m
Step 2 · Find braking distance (db)
Use db = V² ÷ (254 × (a ÷ 9.81 + G))
The
brakes
take
the
speed
to
0
at
a
;
an
upgrade
helps
them
stop
and
a
downgrade
works
against
them.
db
=
80
2
254 × (3.4 ÷ 9.81 + 0)
db
=
6,400
254 × 0.3466
db
=
6,400
88.0326
db = 72.7003 m
Step 3 · Find stopping sight distance (SSD)
Use SSD = dᵣ + db
The driver must see the object at least as far ahead as the car travels before it stops.
SSD = 55.6 + 72.7003
SSD = 128.3003 m
Check
Put the numbers back in. Both sides should match.
55.6 = 0.278 × 80 × 2.5 ✓
72.7003
=
80
2
254 × (3.4 ÷ 9.81 + 0)
✓
128.3003 = 55.6 + 72.7003 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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