(CESURVE) M09 Simple Curve

FUNDAMENTALS
OF SURVEYING
INTRODUCTION TO HORIZONTAL
CURVE AND STATIONING
SIMPLE CURVE
[CESURVE] FUNDAMENTALS OF SURVEYING PRC 2020

HORIZONTAL CURVE
A curve parallel to earth’s surface…
[CESURVE] SURVEYING MEASUREMENTS PRC 2020 2

SIMPLE CURVE
Circular
arc 
The simple curve is
an arc of a circle. The
radius of the circle
determines the
sharpness or flatness
of the curve. 
R
Straight road
sections
COMPOUND CURVE
Circular arcs
R1
R2
Straight road
sections
This curve normally consists of two

simple curves joined together and
curving in the same direction.

BROKEN BACK CURVE
Circular arc
Straight road
sections
A broken back curve consists of two simple curves separated by a straight road.

REVERSE CURVE
Circular arcs
Straight road
sections
A reverse curve consists of two simple
curves joined together, but curving in
opposite direction.

SPIRAL CURVE
R = Rn
R=
The spiral is a curve that has a varying
radius. It is used on railroads and most
modern highways. Its purpose is to provide
a transition from the tangent to a simple
Straight road curve or between simple curves in a
compound curve
section
ELEMENTS OF SIMPLE CURVE
▣ BACK
TANGENT.
The first
tangent line
on the
simple
curve.

FORWARD
TANGENT.
The se cond
tange nt line on
the simple
curve .

PC: POINT OF
CURVATURE. The
point of curvature is
the point on the back
tangent where the
circular curve be gins.
It is sometimes
designated as BC
(beginning of curve)
or TC (tangent to
curve).

PI: POINT OF
INTERS ECTION.
The point of
interse ction is the
point where the back
and forward tangents
interse ct.
S ometimes, the point
of intersection is
designated as V
(vertex).

PT: POINT OF
TANGENCY. The
point of tangency is
the point on the
forward tangent
where the curve
ends. It is sometimes
designated as EC
(end of curve) or CT
(curve to tangent).

D: DEGREE OF
CURVE. The angle
subtended by 1 full
station. 1 full station
= 20m or 100 ft. This
defines the
sharpness or
flatness of the curve.

I: INTERS ECTING
ANGLE. The
interse cting angle is
the deflection angle
at the PI. Its value is
either computed from
the preliminary
traverse angles or
measured in the
field.

D: CENTRAL
ANGLE. The central
angle is the angle
formed by two radii
drawn from the
ce nter of the circle
(O) to the PC and
PT. The value of the
ce ntral angle is equal
to the I angle. S ome
authorities call both
the interse cting
angle and ce ntral
angle either I or ∆.

R: RADIUS . The
radius of the circle of
which the curve is an
arc, or se gment. The
radius is always
perpendicular to
back and forward
tangents.

POC: POINT OF
CURVE. The point of
curve is any point
along the curve.

Lc: LENGTH OF
CURVE. The length
of curve is the
distance from the PC
to the PT, measured
along the curve.

T: TANGENT
DISTANCE. The
tangent distance is
the distance along
the tangents from the
PI to the PC or the
PT. These distances
are equal on a
simple curve.

LC: LONG CHORD.

The long chord is the
straight-line distance
from the PC to the
PT.

C: Full Chord. The

full-chord distance
betwe en adjacent
stations (full, half,
quarter, or one-tenth
stations) along a
curve.

S C1: S ubchord 1.
The subchord
distance betwee n the
PC and the first
station on the curve .

S C2: S ubchord 2.
The subchord
distance betwee n the
last station on the
curve and the PT.

E: EXTERNAL
DISTANCE. The
external distance
(also called the
external se cant) is
the distance from the
PI to the midpoint of
the curve. The
external distance
bisects the interior
angle at the PI.

M: MIDDLE
ORDINATE. The
middle ordinate is
the distance from the
midpoint of the curve
to the midpoint of the
long chord. The
extension of the
middle ordinate
bisects the central
angle.

PROPERTIES OF A CIRCLE

PROPERTIES OF A CIRCLE

DEGREE OF CURVE
Arc Basis (Metric)
20m
Taking proportion:
20 2R
=
One Full Station D 360
DEGREE OF CURVE
Arc Basis (English)
5(20) 2R
=
100 ft D 360
5(20)(360)
R=
2D
5(1145.916)
R=
D
5(1145.916)
D=
R
One Full Station
DEGREE OF CURVE
Chord Basis (Metric)

D 10
sin =
20m
10m
2 R
20 m
10
R=
D
sin
2
One Full Station
DEGREE OF CURVE
Chord Basis (English)

D 50
sin =
2 R
20 m
50
R=
D
sin
2
One Full Station
TANGENT DISTANCE
 I T
tan = tan =
2 2 R
I
T = R tan
2
=I

LONG CHORD
I LC / 2
sin =
LC 2 R
LC/2
I
LC = 2 R sin
2
=I

SUB CHORD
 SC / 2
sin =
SC 1 2 R
SC/2

SC = 2 R sin
2

LENGTH OF CURVE
From arc definition,

Lc = R( 0
)
180

Lc = RI ( )
=I 1800

EXTERNAL DISTANCE
E = OPI − R
I
OPI = R sec
2
I
E = R sec − R
2
I
E = R(sec − 1)
2
MIDDLE ORDINATE
M = R − OF
I
OF = R cos
2
F
I
M = R − R cos
2
I
M = R(1 − cos )
2
STATIONING
▣ Metric English
STA : 0 + 020 STA : 2 + 00
{
{
Three digits after + symbol Two digits after + symbol
Meaning: Meaning,
The point is at station The point is at station

20m from the starting 200 ft from the
point. starting point.
STATIONING OF PI
If STA PC is known,
STA PI = STA PC + T
If STA PT is known,
STA PI = (STA PT – Lc) + T

STATIONING OF PI
Angle that the chord joining the
PC and POC makes with the
tangent from the PC.
d
From:
Angles whose sides
perpendicular each to each are
equal.
1
d= 
2

EXAMPLE 1
A simple curve has an external distance of 15m and a

central angle of 30o. Find the following:
a.) radius of the curve
b.) degree of the curve
c.) stationing of PC and PT if STA PI is at 8 + 036.17
d.) stationing of the first even station
e.) deflection angle of the first even station
f.) middle ordinate

EXAMPLE 1
A simple curve has an external distance of 15m and a central angle of 30o.
PT
d.) stationing of the first even station PI
15 m
f.) middle ordinate
PC R
30°
O

EXAMPLE 1
PT
b.) degree
Using right of the curve
∆PC-PI-O,
Rd.) stationing
= PI-O of the first even station
* cos (15°) PI
15 m
e.) deflection angle
where of the
length first= even
PI-O R + Estation
f.) middle ordinate
R = (R + E) * cos (15°) 90°
PC R
R
R = R cos (15°) + E cos (15°)
R – R cos (15°) = E cos(15°)

R
R = E cos (15°) / (1 – cos (15°)) 15°

30°
or using the formula E = R (sec (I/2) -1), R = E (sec (15°) – 1)
O
R = 15 cos (15°) / (1 – cos (15°)) = 425.216 m
EXAMPLE 1
PT

Using arc basis, where 1 arc is 20 m 15 m
f.) middle ordinate
20 2𝜋𝑅
=
𝐷 360
PC 425.216
1145.916
𝐷=
𝑅
1145.916
𝐷= = 2°41′ 42"
425.216 425.216
30°
O

EXAMPLE 1
PT

Using chord basis, where 1 chord is 20 m 15 m
f.) middle ordinate
𝐷 10
sin =
2 𝑅 PC 425.216
10
𝐷 = 2 ∗ sin−1
𝑅
10 425.216
𝐷 = 2 ∗ sin−1 = 2°41′ 43"
425.216
30°
O

EXAMPLE 1
PT
f.) middle
Sta. ordinate
PC = Sta. PI – T 8 + 036.17 PI
15 m
T
T = PC-PI
PC 90° 425.216
Using right ∆PC-PI-O,
T = R tan (I/2)
T = 425.216 tan (15°) = 113.936 m
Sta. PC = 8 + 036.17 – 113.936 = 7 + 922.234 425.216

15°
30°
O

EXAMPLE 1
PT
f.) middle ordinate 8 + 036.17 PI
Sta. PT = Sta. PC + Lc 15 m
Lc = R I (π / 180°) 113.936
7 + 922.234 PC 425.216
Lc = 425.216 * 30° * (π / 180°)
Lc = 222.643 m
425.216
Sta. PT = 7 + 922.234 + 222.643 = 8 + 144.87
30°
O

EXAMPLE 1
8 + 144.877 PT
From Sta. PC,
f.) middle located at 7 + 922.234,
ordinate
8 + 036.17 PI
15 m
the first even station (i.e. point on
curve (POC) located at stations that
are multiples of 20 m) is at 113.936
7 + 922.234 PC 425.216
7 + 940.000
Additionally, the second even station

is at 7 + 960.000
425.216
and the next, at 7 + 980.000 30°

O

EXAMPLE 1
8 + 144.877 PT
f.) middle ordinate
Deflection angles are always half of the
subtended angle.
8 + 036.17 PI
15 m
The subtended angle of the first even
station is that from PC to POC1. 113.936
POC1 7 + 940.000
7 + 922.234 PC 425.216
The subtended angle can be computed
from the arc from PC to POC1.
The arc length between PC and POC1

can be computed from the stationing 425.216
S = Sta. POC1 – Sta. PC 30°

S = 7 + 940.000 – 7 + 922.234 = 17.766 O

EXAMPLE 1
8 + 144.877 PT
f.) middle ordinate
Using the arc length, the subtended
angle of the first even station can be
computed as 8 + 036.17 PI
15 m
𝑆 180°
𝜃= ∗
𝑅 𝜋 113.936
POC1 7 + 940.000
7 + 922.234 PC S = 17.766 425.216
17.766 180°
𝜃= ∗ = 2°23′38"
425.216 𝜋
Going back,
Deflection angles are always half of the 425.216
subtended angle.
θ 30°
2°23′38"
𝑑= = 1°11′ 49" O
2

EXAMPLE 1
f.) middle ordinate
8 + 144.877 PT
If we set point P as the midpoint of the
longchord (i.e. chord from PC to PT),
8 + 036.17 PI Q
and point Q as the midpoint of the curve,
P
90°
then the middle ordinate, M, is the 113.936
distance from P to Q, and 7 + 922.234 PC 425.216
M + PO = R
Using right ∆PC-P-O,

425.216
15°
PO = R cos (I/2) 30°
PO = 425.216 (cos (15°)) = 410.727 m O

EXAMPLE 1
f.) middle ordinate
8 + 144.877 PT
8 + 036.17 PI Q
P
M + PO = R 113.936
90°
7 + 922.234 PC
M = R – PO 410.727 425.216
M = 425.216 – 410.727 = 14.489 m
425.216
15°
30°
O

EXAMPLE 1
425.216 m PT
2°41’42"
PI
c.) stationing of PC and PT if STA PI is at 8 + 036.17 15 m
Sta. PC 7 + 922.234
Sta. PT 8 + 144.877
d.) stationing of the first even station PC R
Sta. POC1 7 + 940.000
1°11′ 49"
f.) middle ordinate R
14.489 m
30°
O

EXAMPLE 2
A simple curve has a central angle of 60o. The length of the
curve is longer than the long chord by 13.5m. Compute the
following:
a.) length of the curve
b.) long chord
c.) radius of the curve
d.) degree of the curve
e.) deflection angle of the midpoint of the curve
f.) middle ordinate
g.) tangent distance

EXAMPLE 2
A simple curve has a central angle of 60o. The length of the curve is longer than the long chord by
13.5m. Compute the following:
a.) length of the curve PT
b.) long chord
c.) radius of the curve PI
f.) middle ordinate
R
Lc = LC + 13.5
Lc = RI (π/180°)
LC = 2 R sin (I/2)
PC
RI (π/180°) = 2 R sin (I/2) +13.5

R = 13.5 / [I (π/180°) - 2 sin (I/2)] R 60°
R = 13.5 / [60° (π/180°) – 2 sin (30 °)] = 286.032 m O

EXAMPLE 2
b.) long chord
c.) radius of the curve PI
Lc
e.)= deflection
RI (π/180°)angle of the midpoint of the curve
f.)=middle
Lc 286.032 (60° (π/180°) = 299.532 m
ordinate
286.032
LC = 2 R sin (I/2)
LC = 2 * 286.032 sin (30°) = 286.032 m
PC
286.032 60°
O

EXAMPLE 2
d.) degree of the curve PT
f.) middle ordinate PI
Using arc basis,
g.) tangent where 1 arc is 20 m
distance
20 2𝜋𝑅
=
𝐷 360
286.032
1145.916
𝐷=
𝑅
1145.916
𝐷= = 4°0′ 23"
286.032 PC
286.032 60°
O

EXAMPLE 2
d.) degree of the curve PT
f.) middle ordinate PI
Using chord basis,
g.) tangent where 1 chord is 20 m
distance
𝐷 10
sin =
2 𝑅
286.032
10
𝐷 = 2 ∗ sin−1
𝑅
10
𝐷 = 2 ∗ sin−1 = 4°0′26" PC
286.032
286.032 60°
O

EXAMPLE 2
e.) deflection angle of the midpoint of the curve PT
f.) middle ordinate
g.) tangent distance PI
Deflection angles are always half of the

subtended angle.
The subtended angle to the midpoint of

the curve is half the central angle. 286.032
d
θ = I / 2 = 60° / 2 = 30°
d = θ / 2 = 30° / 2 = 15° PC
θ
286.032 60°
O

EXAMPLE 2
f.) middle ordinate PT
Ifg.)
wetangent distance
set point P as the midpoint of the
longchord (i.e. chord from PC to PT), PI
and point Q as the midpoint of the curve, Q
P
then the middle ordinate, M, is the
distance from P to Q, and 286.032
90°
M + PO = R
Using right ∆PC-P-O, PC
30°
PO = R cos (I/2) 286.032 60°
PO = 286.032 (cos (30°)) = 247.711 m O

EXAMPLE 2
f.) middle ordinate PT
PI
M + PO = R P
286.032
90°
M = R – PO
247.711
M = 286.032 – 247.711 = 38.321 m
PC
30°
286.032 60°
O

EXAMPLE 2
g.) tangent distance PT
PI
Using right ∆PC-PI-O,

T 286.032
T = R tan (I/2)
T = 286.032 tan (30°) = 165.141 m

90°
PC
30°
286.032 60°
O

EXAMPLE 2
A simple curve has a central angle of 60o. The length of the curve is longer than the long chord by 13.5m.
Compute the following:
299.532 m
b.) long chord PI
286.032 m
c.) radius of the curve
286.032 m
R
4°0′’23“ (arc basis)
4°0′26" (chord basis)
15°
PC
f.) middle ordinate
38.321 m
g.) tangent distance R 60°
O
165.141 m

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FUNDAMENTALS

[CESURVE] FUNDAMENTALS OF SURVEYING PRC 2020

A curve parallel to earth’s surface…

[CESURVE] SURVEYING MEASUREMENTS PRC 2020 2

This curve normally consists of two

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LC: LONG CHORD.

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C: Full Chord. The

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Chord Basis (Metric)

Chord Basis (English)

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From arc definition,

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STA : 0 + 020 STA : 2 + 00

The point is at station The point is at station

STA PI = (STA PT – Lc) + T

[CESURVE] SURVEYING MEASUREMENTS PRC 2020 39

[CESURVE] SURVEYING MEASUREMENTS PRC 2020 40

A simple curve has an external distance of 15m and a

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R – R cos (15°) = E cos(15°)

R = E cos (15°) / (1 – cos (15°)) 15°

c.) stationing of PC and PT if STA PI is at 8 + 036.17

[CESURVE] SURVEYING MEASUREMENTS PRC 2020 44

c.) stationing of PC and PT if STA PI is at 8 + 036.17

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Sta. PC = 8 + 036.17 – 113.936 = 7 + 922.234 425.216

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Additionally, the second even station

and the next, at 7 + 980.000 30°

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The arc length between PC and POC1