SME
Transportation Engineering · Problem type
Length, tangent and external distance of a curve
Use this for “Two tangents meet at a PI with a deflection of 30°, joined by a 400 m radius curve. Find L, T and E.”
Diagram
T
=
R
tan(
Δ
2
)
=
400
×
tan(30°
÷
2)
=
107
m;
L
=
RΔπ
180
=
209
m;
E
=
R(
1
cos(Δ ÷ 2)
−
1)
=
14.1
m.
Drawn to scale.
This is an example. Type your own values to start a new problem.
L (m)
=
R (m)
×
Δ
°
×
π
÷
180
T (m)
=
R (m)
×
tan(
Δ
°
÷
2)
E (m)
=
R (m)
×
(1
÷
cos(
Δ
°
÷
2)
−
1)
Formulas
L
=
R
×
Δ
×
π
180
209.4395
=
400
×
30
×
π
180
T
=
R
×
tan(
Δ
2
)
107.1797 = 400 × tan(30° ÷ 2)
E
=
R
×
(
1
cos(
Δ
÷ 2)
−
1)
14.1105
=
400
×
(
1
cos(30° ÷ 2)
−
1)
Assumptions
•
A simple circular curve from the PC to the PT between two straight tangents.
•
Δ is the deflection angle at the PI in degrees, the same as the angle the arc turns through.
•
L runs along the arc; T and E are straight lines from the PI.
Step-by-step
WE KNOW
R = 400 m, Δ = 30°
FIND
length of curve (L), tangent length (T), external distance (E)
Step 1 · Find length of curve (L)
Use L = R × Δ × π ÷ 180
An
arc
is
the
radius
times
its
angle
in
radians;
π
180
turns
degrees
into
radians.
L
=
400
×
30
×
π
180
L
=
12,000
×
π
180
L
=
37,700
180
L
=
209.4395
m
Step 2 · Find tangent length (T)
Use T = R × tan(Δ ÷ 2)
The
radius
to
the
PC
and
the
tangent
to
the
PI
make
a
right
angle;
the
centre
angle
there
is
Δ
2
.
T
=
400
×
tan(30°
÷
2)
T
=
400
×
tan(15°)
T
=
400
×
0.2679
T
=
107.1797
m
Step 3 · Find external distance (E)
Use E = R × (1 ÷ cos(Δ ÷ 2) − 1)
From
the
centre
to
the
PI
is
R
cos(
Δ
÷ 2)
;
the
curve’s
middle
is
R
from
the
centre,
so
take
R
away.
E
=
400
×
(
1
cos(30° ÷ 2)
−
1)
E
=
400
×
(
1
cos(15°)
−
1)
E
=
400
×
(
1
0.965926
−
1)
E
=
400
×
(1.03528
−
1)
E
=
400
×
0.03528
E
=
14.1105
m
Check
Put the numbers back in. Both sides should match.
209.4395
=
400
×
30
×
π
180
✓
107.1797 = 400 × tan(30° ÷ 2) ✓
14.1105
=
400
×
(
1
cos(30° ÷ 2)
−
1)
✓
Topic check
See what you know: 5 questions on this topic
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