BESAVILLA

REVIEW CENTER

LECTURE 5

Instructor:
ENGR. RUPERTO T. FAELNAR, JR.

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REVIEW CENTER

Review Reference

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Topics
1. Simple Curve
2. Symmetrical Parabolic Curve

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 REVIEW PROBLEMS

PROBLEM 1

The longitude of San Francisco (USA) is 73° 57’

30’’ W and that of Cebu (Philippines) is
122°36’15’’ E. Determine the time in Cebu
when it is 2:30 AM in San Francisco.

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 REVIEW PROBLEMS

PROBLEM 1
The longitude of San Francisco (USA) is 73° 57’ 30’’ W and that of
Cebu (Philippines) is 122°36’15’’ East. Determine the time in Cebu
when it is 2:30 AM in San Francisco.
0° 122°36’15’’ E
73° 57’ 30’’ W

Greenwich
San Francisco

Cebu

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 REVIEW PROBLEMS

PROBLEM 1
The longitude of San Francisco (USA) is 73° 57’ 30’’ W and that of
Cebu (Philippines) is 122°36’15’’ East. Determine the time in Cebu
when it is 2:30 AM in San Francisco.
0°
Solution:
Diff. in longitude = 122°36'15'' +
73°57'30''
Diff in longitude = 196°33'45''
San Francisco
Note: 15° = 1 hr
Time measure = 13h 06m 15s Cebu
Time in Cebu = 2h 30m + 13h 06m 15s 73° 57’ 30’’ 122°36’15’’ E
= 15h 36m 15s
Time in Cebu = 3 : 36 :15 PM

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 REVIEW PROBLEMS

PROBLEM 2

Find the correction in latitude of line AB and

the correction of departure of line BC.

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PROBLEM 2

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PROBLEM 2

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PROBLEM 2

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PROBLEM 2

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 REVIEW PROBLEMS

PROBLEM 2
Find the
correction in
latitude of line
AB and the
correction of
departure of
line BC.

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 SIMPLE CURVE

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 SIMPLE CURVE

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 SIMPLE CURVE???

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 SIMPLE CURVE

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 SIMPLE CURVE

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 SIMPLE CURVE

PC – point of curvature P.T.

Lc

PT – point of tangency

Lc – length of curve
P.C. R

I – central angle/angle
R I
of intersection

R – radius of the curve

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 SIMPLE CURVE

SIMPLE CURVE
I T
Tangent Distance, T P.T.
Lc
T
tan(I/2) = T/R

T = R tan (I/2)
P.C. R

R I/2
I

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 SIMPLE CURVE

SIMPLE CURVE
T
External Distance, E P.T.
E
T
E = R sec(I/2) - R

E = R [sec (I/2) – 1]
P.C. R

R I/2
I

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 SIMPLE CURVE

SIMPLE CURVE
T
Middle Ordinate, M P.T.
T
M
M = R [1 – cos(I/2)]

P.C. R

R I/2
I

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 SIMPLE CURVE

SIMPLE CURVE
T
Length of Chord, C P.T.
T

P.C. R

R I

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 SIMPLE CURVE

SIMPLE CURVE
T
Length of Chord, C P.T.
T
sin(I/2) = (C/2) / R C/2

C = 2R sin(I/2) C/2
P.C. R

R I/2
I

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 SIMPLE CURVE

SIMPLE CURVE

Length of Curve, Lc P.T.

Lc
Lc 20

I D
20m
20 I
Lc  (M etric) P.C. R
D
100 I
Lc  (English) R D I
D

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 SIMPLE CURVE

SIMPLE CURVE

Degree of Curve, D
20m
D 360

20 2R
1145.916
D (M etric) R
R
5729.58
D (English) R
D
R
Note:
D → 20 m (Metric)
D → 100 ft (English)

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 SIMPLE CURVE

SIMPLE CURVE
T
Deflection angle, Ɵ P.T.
2 360 d1
 Ɵ
d1 2R x/2
90d1
 x/2
R R
P.C.

90
 (deflectio n angle per R
R Ɵ
Ɵ
unit length of curve)

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 SIMPLE CURVE

PROBLEM 3
A simple curve having a radius of 400 ft. has an angle
of intersection of 44°. Stationing of P.C. is 11 + 10.57,
P.I. = 12 + 72.18 and P.T. = 14 + 17.75

1. Determine the length of long chord from the P.C. to

station 12 + 00
2. Determine the perpendicular offset distance of STA
12 + 00 from the tangent passing thru the P.C.
3. Determine the perpendicular offset distance from
the tangent passing thru the P.C. from sta. 14 +
17.75.

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 SIMPLE CURVE

PROBLEM 3
A simple curve having a radius of 400 ft. has an angle of
intersection of 44°. Stationing of P.C. is 11 + 10.57, P.I. = 12 +
72.18 and P.T. = 14 + 17.75
P.T.
Lc= (14 + 17.75) – (11 + 10.57) 14 + 17.75
Lc= 307.18 ft
Lc= 307.18 ft

P.C.
11 + 10.57
I = 44°
R = 400 ft

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 SIMPLE CURVE

PROBLEM 3
A simple curve having a radius of 400 ft. has an angle of
intersection of 44°. Stationing of P.C. is 11 + 10.57, P.I. = 12 +
72.18 and P.T. = 14 + 17.75
Solution:
1. Determine the length of
P.T.
long chord from the P.C.
to station 12 + 00 12 + 00
14 + 17.75
d1 = (12 + 00) - (11 + 10.57) d1
d1 = 1200 - 1110.57 C
d1 = 89.43 ft. P.C.
11 + 10.57
I = 44°
R = 400 ft

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 SIMPLE CURVE

PROBLEM 3
A simple curve having a radius of 400 ft. has an angle of
intersection of 44°. Stationing of P.C. is 11 + 10.57, P.I. = 12 +
72.18 and P.T. = 14 + 17.75
Solution:
1. Determine the length of
long chord from the P.C. d1
6°24'18''
to station 12 + 00
d1 = (12 + 00) - (11 + 10.57)
d1 = 89.43 ft.
11 + 10.57

6°24'18''

6°24'18''

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 SIMPLE CURVE

PROBLEM 3
A simple curve having a radius of 400 ft. has an angle of
intersection of 44°. Stationing of P.C. is 11 + 10.57, P.I. = 12 +
72.18 and P.T. = 14 + 17.75
Solution:
1. Determine the length of
long chord from the P.C. d1
6°24'18''
to station 12 + 00

11 + 10.57
c = 2(400) Sin 6°24'18''

6°24'18''

6°24'18''

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 SIMPLE CURVE

PROBLEM 3
A simple curve having a radius of 400 ft. has an angle of
intersection of 44°. Stationing of P.C. is 11 + 10.57, P.I. = 12 +
72.18 and P.T. = 14 + 17.75
Solution:
2. Perpendicular offset
d1
distance, x of sta. 12 + 00 6°24'18''
from the tangent passing
thru the P.C.
11 + 10.57
x = C sin 6°24'18''
x = 89.244 sin 6°24'18''
x = 9.956 ft.
6°24'18''

6°24'18''

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 SIMPLE CURVE

PROBLEM 3
A simple curve having a radius of 400 ft. has an angle of
intersection of 44°. Stationing of P.C. is 11 + 10.57, P.I. = 12 +
72.18 and P.T. = 14 + 17.75
Solution: y
3. Perpendicular offset
P.T.
distance from the tangent
passing thru the P.C. from sta.
14 + 17.75
14 + 17.75.
x/2
x/2 = 400 sin 22°
x = 299.685 22°
x/2
P.C.
y = x Sin 22° 11 + 10.57
y = 299.685 Sin 20°
22°
y = 112.264 ft. 22°
R = 400 ft

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 SIMPLE CURVE

PROBLEM 4
Given data for a simple P.T.
curve: 11 + 50
14 + 17.75
I = 44°
R = 400 ft. x
P.C. = 11 + 10.57 P.C.
P.I. = 12 + 72.18
11 + 10.57
P.T. = 14 + 17.75
I = 44°
R = 400 ft
Compute the short
chord, x from P. C. to
station 11 + 50

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 SIMPLE CURVE

PROBLEM 4
Compute the short chord, x from P. C. to
station 11 + 50
Solution:
d1 = (11 + 50) - (11 + 10.57)
d1 = 39.43 ft. d1

11 + 10.57

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 SIMPLE CURVE

PROBLEM 4
Compute the short chord, x from P. C. to
station 11 + 50
Solution:

d1

11 + 10.57

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 SIMPLE CURVE

PROBLEM 5
Given data for a simple curve: P.T.
R = 400 ft. Ɵ
I = 44° 14 + 17.75
PC = 11 + 10.57
P.I. = 12 + 72.18 50 ft
P.T. = 14 + 17.75
P.C.
Compute the deflection angle
11 + 10.57
for 50 ft. intervals of the curve.
I = 44°
R = 400 ft

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 VERTICAL CURVES

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 VERTICAL CURVES

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 VERTICAL CURVES

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 SYMMETRICAL PARABOLIC CURVE

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 SYMMETRICAL PARABOLIC CURVE

H
y
H

L/2 L/2

y H

x 2
L  
2
2

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 SYMMETRICAL PARABOLIC CURVE

PROBLEM 6
Given:
PVI = 26 + 00 g1 = + 6.38%
Elev. PVI = 655.06 g2 = - 3.54%
L = 4 stations

Determine the vertical distance from PVI to the

curve.

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 SYMMETRICAL PARABOLIC CURVE

PROBLEM 6
Given:
PVI = 26 + 00 g1 = + 6.38%
Elev. PVI = 655.06 g2 = - 3.54%
L = 4 stations (400 ft)

Determine the vertical distance from

PVI to the curve.

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 SYMMETRICAL PARABOLIC CURVE

PROBLEM 7
Given:
Station of PVI = 26 + 00 g1 = + 6.38%
Elev. of PVI = 655.06 m. g2 = - 3.54%
L = 4 stations (400 ft)

Determine the curve elevation of station 25 + 50.

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 SYMMETRICAL PARABOLIC CURVE

PROBLEM 7
Given:
g1 = + 6.38%
g2 = - 3.54%
Station of PVI = 26 + 00
Elev. of PVI = 655.06 m. H
L = 4 stations (400 ft)

Determine the curve

elevation of STA 25 + 50

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 SYMMETRICAL PARABOLIC CURVE

Solution:

y H

x 2
L  
2
2

y 4.96
2

150 200 2
y  2.79 ft
Elev. of B = 655.06 - 50(0.0638)
Elev. of B = 651.87 ft. Elev. A = 651.87 - 2.79
Elev. A = 649.08 ft.

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