Module Outline

This module will cover the following topics and subtopics:

I. Errors, Mistakes & Corrections

a. Stretch / Pull / Tension
b. Observation / Sag
c. Temperature
d. Slope
e. Sea Level Reduction
f. Standard Deviation, Probable Error, Standard Error, Relative Precision
g. Curvature and Refraction
II. Adjustment of Weighted Observation
III. Measurement of Distances
a. Pacing
b. Ratio of Error
IV. Surveying Instruments
a. Two-Peg Test
i. Conventional Method
ii. Alternate Method
V. Leveling
VI. Settlement of Rod and Instrument
VII. Tachymetry
VIII. Bearing and Azimuth
IX. Closed Traverse
a. Balancing Traverse
i. Compass Rule
ii. Transit Rule
b. Area of Closed Traverse
X. Earthworks
a. Cut and Fill
XI. Curves
a. Horizontal Curves
i. Simple
ii. Compound
iii. Reverse
b. Vertical Curves
I. Errors, Mistakes & Corrections

Common Errors in Measurement Mistakes in Measurement

– Tape not Standard in Length – Adding/Dropping a full Tape Length
– Alignment – Adding a cm at the fraction part at the
– Level end of the line
– Stretch / Pull – Recording numbers incorrectly
– Observation / Sag – Reading wrong meter mark
– Temperature

General Formula:
𝒆
(±)𝑪𝑫 = 𝑴𝑫 𝟏 ±
𝑻𝑳

TL = Tape Length
CD = Correct Distance Sign Convention
MD = Measured Distance Too Long / Add = “+”
e = Error of Tape Too Short / Subtract = “-“

Corrections or Errors:
 Correction due to Stretch / Pull / Tension
(𝑷 − 𝑷𝒐 )𝑳
(±)𝑪𝒑 =
𝑨𝑬

Cp = Correction due to pull Constants

P = Actual Pull ESteel = 200 GPa
Po = Standard Pull = 200,000 MPa
L = Length = 29,000 ksi
A = Area = 2.1 x 106 kg/cm2
E = Modulus of Elasticity

 Correction due to Observation / Sag

𝒘 𝟐 𝑳𝟑 𝑾
(−)𝑪𝒔 = 𝒘=
𝟐𝟒𝑷𝟐 𝑻𝑳
Cs = Correction due to sag W = Weight of Tape
w = Weight per Length of Tape TL = Tape Length
L = Length (Unsupported)
P = Actual Pull

 Unsupported Length
Held at Both Ends
𝑳 = 𝑻𝑳

Held at Both Ends and Midspan

𝑻𝑳
𝑳=
𝟐

Held at Both Ends and Middle Thirds

𝑻𝑳
𝑳=
𝟑
 Correction due to Temperature
(±)𝑪𝒕 = 𝜶𝑳∆𝑻 ∆𝑻 = 𝑻 − 𝑻𝒐
Ct = Correction due to Temperature Constant Thermal Expansion Coefficient
α = Coefficient of Thermal Expansion ΑSteel = 11.6 x 10-6 / oC
L = Length
To = Standard Temperature
T = Actual Temperature

 Correction due to Slope

𝒉𝟐
(−)𝑪𝒉 =
𝟐𝑺
Ch = Correction due to Slope
h = Vertical Height
S = Slope Distance

 Correction due to Sea Level

𝑺 𝑳
𝜽= =
𝑹 𝑹+𝒉

L = Measured Length at elevation, h

h = Distance from Sea Level
S = Measured Length at Sea Level
R = Radius of Earth
= 6,400 km

Sample Problems:
1. A line was determined to be 2395.25 m when measured with a 30 m steel tape supported
throughout its length under a pull of 4 Kg at a mean temperature of 35 deg C. Tape used is of
standard length of 20 deg C under a pull of 5 Kg. A= .03 cm2, K = 0.0000116/𝐶° and E = 2x106
kg/cm2.
a. Determine the error due to change in temperature (+ 0.4168 m)
b. Determine the error due to tension (- 0.0399 m)
c. Determine the Corrected Length of Line (2395.6269 m)
2. A 50m tape weighing 1.075 kg has a standard pull of 8 kg. The tape’s cross-sectional area and
modulus of elasticity are 0.05 cm2 and 200 GPa respectively. What pull is required in order that
the effect of sag will be eliminated when the tape is supported at the end points only. (197.40 N)

Errors and Mistakes

 Standard Deviation, SD
∑𝒙
∑ 𝑽𝟐 𝒙=
𝑺𝑫 = 𝑽=𝒙−𝒙 𝒏
𝒏−𝟏

V = Variance
x = Mean
n = Number of data
 Probable Error on a Single Observation, PE
𝑷𝑬 = 𝟎. 𝟔𝟕𝟒𝟓(𝑺𝑫)
 Probable Error of Mean, PEM
𝑷𝑬 𝟎. 𝟔𝟕𝟒𝟓(𝑺𝑫)
𝑷𝑬𝑴 = =
√𝒏 √𝒏
 Standard Error, SE
𝑺𝑫
𝑺𝑬 =
√𝒏
 Relative Precision, RP
𝑷𝑬𝑴
𝑹𝑷 =
𝒙
Note:
 When calculating for the Relative Precision, the final answer must be expressed as a UNIT
FRACTION. Unit Fraction is a fraction wherein the numerator is always 1.

Sample Problem:
1. From the measured values of AB, the following trials were recorded.
TRIALS DISTANCE MEASURED
1 120.68
2 120.84
3 120.76
4 120.64

(a) Find the Standard Deviation. (± 0.0887)

(b) Find the Probable Error of the Mean. (± 0.0299)
(c) Find the Standard Error. (± 0.0443)
(d) Compute the Relative Precision. ( )

Correction due to Curvature and Refraction

Principles:
1. Horizontal Distance is practically EQUAL to Level Distance
2. Horizontal at one point is PERPENDICULAR to the vertical on another point.

General Formula:
𝟔 𝒌𝟐
𝒉𝒄𝒓 =
𝟕 𝟐𝑹
hcr = Correction due to Curvature and Refraction
k = Distance
R = Radius of Earth

If R = 6,400 km,
𝒉𝒄𝒓 = 𝟎. 𝟎𝟔𝟕𝒌𝟐
Note:
 When using the formula above, see to it that the unit of the Distance, k is in KILOMETERS;
while the unit of hcr is in METERS.

Sample Problems:
1. A man’s eyes 1.75 m above sea level can barely see the top of a lighthouse which is at a certain
distance away from a man.
 a. What is the elevation of the top of the lighthouse above sea level if it is 20 Km away
from the man? (14.8533 m)
b. What is the height of the tower at a distance 20 Km away from the man that will just be
visible without the line of sight approaching rarer than 1.75 m to the water? (28.55 m)

2. A is the point having an elevation of 130 m above datum, and B and C are points of unknown
elevation, B is in between A and C. By means of an instrument set 1.22 m above B, the vertical
angles are observed, that to A being -14045’ and that to C being 8032’. The horizontal distance AB is
547.20 and the horizontal distance BC is 923.25 m. Making due to allowance for earth’s curvature
and atmospheric refraction.
a. Compute the difference in elevation A and B. (142.8256 m)
b. Determine the difference in elevation between B and C. (139.8069 m)
c. Determine the elevation of C. (412.6326 m)

II. Adjustment of Weighted Observation

Principles
 Weights are inversely proportional to the square of probable errors
𝒌
𝑾= 𝟐
𝒆
 Weights are proportional to the number of observations
𝑾=𝒏
 Errors are directly proportional to the square root of distances
𝒆 = 𝒌√𝑫
 Weights are inversely proportional to the distance
𝒌
𝑾=
𝑫
Computation of the Mean:
Same Value Related Values
𝒙=𝒙±𝑪
∑ 𝑾𝒙 𝑪 = 𝑫𝑭 × 𝒆
𝒙= 𝟏
∑𝑾
𝑫𝑭 = 𝑾
𝟏
∑
𝑾
W = Weights C = Correction
x = Number / Data DF = Differential Factor
x = Mean

Table Format
x n/e/D W Wx 1 DF C Corrected
𝑊
1
2
3
1
𝑊 𝑊𝑥
𝑊
Sample Problem:

1. The following data are observed in between BM1 and BM2 by running a line of levels over four
different routes
ROUTE DIFFERENCE PROBABLE
IN ERROR
ELEVATION
1 340.22 2
2 340.30 4
3 340.26 6
4 340.32 8

(a) What is the weight of route 2 assuming weight of route 1 is 1.

(b) Determine the probable value of the difference in elevation. (340.2416 m)
(c) If the elevation of BM1 is 650.42 what is the elevation of BM2 assuming it is higher than
BM1. (990.6616 m)

2. The following interior angles of a triangle traverse were measured with the same precision
ANGLE OBSERVED NUMBER OF
VALUE MEASUREMENTS
1 41 5
2 77 6
3 63 2

(a) Determine the most probable value of A. (40O 46’ 9.23”)

(b) Determine the most probable value of B. (76O 48’ 27.69”)
(c) Determine the most probable value of C. (62O 25’ 23.08”)

III. Measurement of Distances

Pacing
𝑳 𝑷𝟏 + 𝑷𝟐 + 𝑷𝟑 + ⋯ + 𝑷𝒏
𝑷𝑭 = 𝑷𝒂𝒗𝒆 =
𝑷𝒂𝒗𝒆 𝒏
PF = Pace Factor
L = Length or Distance
Pave = Average Pace

Ratio of Error
𝑹𝑫 − 𝑭𝑫 𝑹𝑫 + 𝑭𝑫
𝑹𝒂𝒕𝒊𝒐 𝒐𝒇 𝑬𝒓𝒓𝒐𝒓 = 𝑨𝑫 =
𝑨𝑫 𝟐
RD – Distance in Reversed Direction
FD – Distance in Forward Direction
AD – Average Distance

Sample Problems:
1. A 100 m long was paced by a surveyor for four times with the following data: 142,145, 145.5 and
146. Then another line was paced four times again with the following results: 893, 893.5, 891 and
895.5.
(a) Determine the Pace factor. (0.691 m/pace)
 (b) Determine the number of paces for the new line. (893.25 paces)
(c) Determine the distance of the new line. (617.236 m)

2. The length of a series of lines is 3427.62 m in forward direction and 3427 m in the reversed direction.
What is the ratio of the error? (-1.809 x 10-4)

IV. Survey Instrument

Two-Peg Test – Use to check the TILTATION of the instrument.

 Conventional Method – Instrument is set near the level rods.

a’, b’, c’, d’ = Correct Rod Readings

TDE = True Difference in Elevation

Note: Because the instrument are set near the level rod, reading a is EQUAL to a’ and c is EQUAL to c’.

Formulas:
𝑫𝑬𝒂 = |𝒂 − 𝒃|

𝑫𝑬𝒃 = |𝒄 − 𝒅|

If DEa = DEb, Line of Sight is Level. TDE = DEa = DEb.

DEa +DEb
If DEa ≠ DEb, Line of Sight is Tilted. TDE= .
2

For Positive Gradient, Line of Sight is tilted For Negative Gradient, Line of Sight is tilted
Upward. Downward.
TDE + b > a TDE + b < a
d > c + TDE d < c + TDE
Sample Problem
1. In the two peg test method of a dumpy level, the following observations were taken. Line of sight
tilted upwards.

(a) What is the difference in elevation between A and B? (0.53 m)

(b) If the line of sight is not in adjustment, determine the correct rod reading on A with the
instrument still set up at B? (0.919 m)
(c) Determine the error in the line of sight. (0.019 m)

INSTRUMENT NEAR A INSTRUMENT NEAR B

ROD READING ON A 1.505 m 0.938 m

ROD READING ON B 2.054 m 1.449 m

 Alternate Method – Instrument is set at the middle of the 2 rods and then, outside of the level
rods.

Formulas and Constants:

𝒆𝑨 = 𝒆 𝑩

𝑻𝑫𝑬 = |𝒂 − 𝒃|

𝒆𝑵 𝒆𝑭 𝒆𝑨 𝒆𝑩
= = =
𝑿 𝑿+𝑫 𝑫 𝑫
𝟐 𝟐

Sample Problem:
1. In the two peg test using model Wild NA2 dumpy level, the following observations were taken. Point
C is equidistant from A and B, and D is 12 m from A and 72 m from B. Line of sight tilted
downwards.
 INSTRUMENT NEAR A INSTRUMENT NEAR B

ROD READING ON A 1.103 m 0.568 m

ROD READING ON B 0.991 m 0.289 m

(a) What is the true difference in elevation between A and B? (0.112 m)

(b) With the level in the same position at D, to what rod reading on B should the line of sight be
adjusted? (0.4894 m)
(c) What is the corresponding rod reading on A for a horizontal line of sight with instrument still at
D? (0.6014 m)

V. Levelling

𝑯𝑰 = 𝑬𝑳𝑬𝑽𝟏 + 𝑩𝑺 𝑬𝑳𝑬𝑽𝟐 = 𝑯𝑰 − 𝑰𝑭𝑺

𝑬𝑳𝑬𝑽𝟐 = 𝑯𝑰 − 𝑭𝑺 𝑩𝑺 − 𝑭𝑺 = 𝑬𝑳𝑬𝑽𝟐 − 𝑬𝑳𝑬𝑽𝟏

HI = Height of Instrument IFS = Intermediate Foresight

BS = Backsight ELEV = Elevation
FS = Foresight

Sample Problem:

1. From the given data of a differential levelling as shown in the tabulation.

a. Find the difference in elevation of Station 7 and 5. (10.09 m)
b. Find the difference in elevation of Station 7 and 4. (11.77 m)
c. Find the elevation of station 3. (392.61 m)
Station BS FS Elevation
1 5.87 392.25
2 7.03 6.29
3 3.48 6.25
4 7.25 7.08
5 10.19 5.57
6 9.29 4.45
7 4.94
VI. Settlement of Rod and Instrument

Formula:
𝑪𝒐𝒓𝒓𝒆𝒄𝒕𝒆𝒅 𝑬𝒍𝒆𝒗𝒂𝒕𝒊𝒐𝒏 = 𝑬𝒍𝒆𝒗𝒂𝒕𝒊𝒐𝒏 − 𝒏𝒊 ∆𝒉𝒊 − 𝒏𝒕 ∆𝒉𝒕

𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒏𝒕 = 𝒏𝒊 − 𝟏 − (𝒏 − 𝟐)
𝒏𝒊 =
(𝑫𝑩𝑺 + 𝑫𝑭𝑺 )

ni = Number of Instrument Setup Δhi = Settlement of Instrument

nt = Number of Turning Points Δht = Settlement of Rod

Sample Problem
1. A line of levels 10 Km long was run over soft ground. Starting from BM 1 with elevation 22.5 M, the
elevation of BM2 was computed to be 17.25 M. It was found out however that the level settles 5 mm
between the instant of every backsight reading and the rod settles 2 mm. If the backsight and foresight
distance have an average 100 M, find the corrected elevation of BM 2. (16.902 m)

VII. Tacheometry
 A method of surveying to determine the horizontal and vertical distance between two points.

UHR

MHR

LHR

Formula:
𝑰𝑫 = 𝒌𝑺(𝒄𝒐𝒔𝜶) + 𝑪 𝑯𝑫 = 𝑰𝑫(𝒄𝒐𝒔𝜶)
𝑺 = 𝑼𝑯𝑹 − 𝑳𝑯𝑹 𝑽𝑫 = 𝑰𝑫(𝒔𝒊𝒏𝜶)
ID = Inclined Distance C = Stadia Constant
k = Stadia Interval Factor  C = 0 – Internal Focusing Telescope
 If not given, k = 100  C = 0.30 – External Focusing Telescope
S = Stadia Intercept / Interval HD = Horizontal Distance
α = Vertical Angle (Angle of Elevation or VD = Vertical Distance
Depression)

Sample Problem
A survey party proceeded to do their stadia work as follows:
 The transit was set up at a point A and with the line of sight horizontal, took rod readings with
the rod at points B and C, which were then measured to have taped distances from A to 200 m
and 60 m respectively.

ROD B ROD C
Stadia Interval 2.001m 0.600m
 Then they went on to survey two other points, with some of the data recorded as follows: with the
transit at point D, two points E and F were sighted.

ROD E ROD F
Stadia Interval 2.120m 3.560m
Vertical Angle 4022’ -3017’

a) Compute the stadia interval factor. (100)

b) Compute the horizontal distances DE and DF. (210.771 m, 354.832 m)
c) Compute the differences in elevation between points D and E and points D and F. (16.095 m,
20.356 m)

VIII. Bearing and Azimuth

Bearing
 Can be either Clockwise or Counter clockwise rotation
 Angles are measure from the Meridian Direction (North or South)
Notation:
o NβW o NβE
o SβW o SβE
Azimuth
 Clockwise rotation only
 Angles are measured from one Meridian Direction in a traverse
Notation: θ

 From North – Used for Land  From South – Used for Navigation
Surveying

IX. Closed Traverse

𝑳𝒂𝒕 = 𝑳 𝐜𝐨𝐬 𝜷 Sign Convention

𝑫𝒆𝒑 = 𝑳 𝐬𝐢𝐧 𝜷 Latitude Departure
𝑳 = 𝑳𝒂𝒕𝟐 + 𝑫𝒆𝒑𝟐 N+ E+
𝑫𝒆𝒑 S– W–
𝐭𝐚𝐧 𝜷 =
𝑳𝒂𝒕

L = Length or Distance Dep = Departure

β = Bearing Lat = Latitude
For a Closed Traverse,
𝑳𝒂𝒕 = 𝟎 𝑫𝒆𝒑 = 𝟎

Disclosures:
∑ 𝑳𝒂𝒕 ≠ 𝟎 (Latitude Disclosure) ∑ 𝑫𝒆𝒑 ≠ 𝟎(Departure Disclosure)

𝟐 𝟐
𝑳𝑬𝑪 = 𝑳𝒂𝒕 + 𝑫𝒆𝒑 𝑳𝑬𝑪
𝑹𝑷 =
𝑷

LEC = Linear Error of Closure

RP = Relative Precision of Closure
P = Perimeter

Adjustment:
𝑳𝒂𝒕 = 𝑳𝒂𝒕 − 𝑪𝑳𝒂𝒕 𝑫𝒆𝒑 = 𝑫𝒆𝒑 − 𝑪𝑫𝒆𝒑

 Balancing Closed Traverse

Compass Rule:
- Angular accuracy = Linear accuracy

𝑳𝑪𝒐𝒖𝒓𝒔𝒆 𝑳𝑪𝒐𝒖𝒓𝒔𝒆
𝑪𝑳𝒂𝒕 = 𝑳𝒂𝒕 × 𝑪𝑫𝒆𝒑 = 𝑫𝒆𝒑 ×
𝑷 𝑷

Transit Rule:
- Angular accuracy > Linear accuracy

|𝑳𝒂𝒕|𝑪𝒐𝒖𝒓𝒔𝒆 |𝑫𝒆𝒑|𝑪𝒐𝒖𝒓𝒔𝒆
𝑪𝑳𝒂𝒕 = 𝑳𝒂𝒕 × 𝑪𝑫𝒆𝒑 = 𝑫𝒆𝒑 ×
∑|𝑳𝒂𝒕| ∑|𝑫𝒆𝒑|

Area Calculations

a. Area by the rule of Thumb

𝟐𝑨 = [𝒚𝟏 (𝒙𝟏 − 𝒙𝟐 ) + 𝒚𝟐 (𝒙𝟏 − 𝒙𝟑 )+𝒚𝟑 (𝒙𝟐 − 𝒙𝟒 ) + 𝒚𝟒 (𝒙𝟑 − 𝒙𝟓 ) + 𝒚𝟓 (𝒙𝟒 − 𝒙𝟏 )]
b. Area by Coordinates
𝒙𝟏 𝒙𝟐
𝟐𝑨 = 𝒚 𝒚
𝟏 𝟐
c. Area by Double Meridian Distance Method (DMD) or Double Parallel Distance Method (DPD)

Steps – DMD Steps – DPD

1. DMD1st = Dep1st 1. DPD1st = Lat1st
2. DMDNext = ∑ (DMDPrev, DepPrev, 2. DPDNext = ∑ (DPDPrev, LatPrev, LatNext)
DepNext) 3. DPDLast = LatLast (For Checking)
3. DMDLast = DepLast (For Checking) 4. 2A = DPD x Dep
4. 2A = DMD x Lat 5. 𝐴 = |∑ 2𝐴|
5. 𝐴 = |∑ 2𝐴|
Sample Problem
1. From the given area of closed traverse,
Lines Lat Dep
AB -36.13 -25.77
BC 74.56 -115.93
CD 12.82 0.39
DE 19.90 61.74
EA -68.40 69.57

Using Transit Rule of balancing traverse,

a. Determine the corrected bearing of BC (N 56.6192O W)
b. Determine the corrected bearing of EA (S 46.1452O E)
c. Compute the Area of the Traverse (6753.992 m2)

X. Earthworks

Cut and Fill

Cut – Section of the existing ground level ABOVE the proposed new ground level.
Fill – Section of the existing ground level BELOW the proposed new ground level.

Volume Computation
o End-Area Method, VEA
𝑳
𝑽𝑬𝑨 = (𝑨 + 𝑨𝟐 )
𝟐 𝟏
o Prismoidal Formula, VP
𝑳
𝑽𝒑 = (𝑨 + 𝟒𝑨𝑴 + 𝑨𝟐 )
𝟔 𝟏
o Prismoidal Correction, Vcp
𝑽 = 𝑽𝑬𝑨 − 𝑽𝒄𝒑
𝑳
𝑽𝒄𝒑 = (𝑪 − 𝑪𝟐 )(𝑫𝟏 − 𝑫𝟐 )
𝟏𝟐 𝟏

A1, AM, A2 = Areas of Cut or Fill section

C1, C2 = Center Height of the section
D1, D2 = Horizontal Width of the Section

Sample Problem
1. Given the cross section notes of an earthwork between stations 10 + 100 to 10 + 200. Assume
both stations to have the same side slope and width of the base.

Sta 10 + 100
Left Center Right
6.45 0 4.5
+2.3 +1.5 +1.0
 Sta. 10 + 200
6.0 0 6.9
+2.0 𝑥 +2.6

a. Compute the side slope of both sections. (1.5H: 1V)

b. Compute the value of x at station 10 + 200 if it has a cross sectional area of 14.64 m 2. (1.2m)
c. Compute the volume between the stations using end area with prismoidal correction.
(1394.365 m3)

XI. Curves

Horizontal Curves

Simple Curves
- A circular arc, extending from one tangent to another.
- Can be easily solved by Trigonometry

PT - Point of Tangency
PI - Point of Intersection
PC - Point of Curvature
R - Radius of the Curve
D - Degree of Curvature
I - Angle of Intersection
T - Tangent
E - External Distance
M - Middle Ordinate
Lc - Length of Curve
C - Length of Chord

Formula:
Route Surveys:
𝒔𝒕𝒂. 𝑷𝑪 = 𝒔𝒕𝒂. 𝑷𝑰 − 𝑻 𝒔𝒕𝒂. 𝑷𝑻 = 𝒔𝒕𝒂. 𝑷𝑪 + 𝑳𝑪

Degree of Curvature
Arc Basis: Chord Basis
𝟐𝟎 𝑳𝑪 𝑫
= 𝟐𝟎 = 𝟐𝑹 𝐬𝐢𝐧
𝑫 𝑰 𝟐
Length of Curve External Distance
𝜋 𝐼
𝐿 = 𝑅𝐼 𝐸 = 𝑅 sec − 1
180 2
Tangent Distance Middle Ordinate
𝐼 𝐼
𝑇 = 𝑅 tan 𝑀 = 𝑅 1 − cos
2 2
Long Chord Distance
𝐼
𝐶 = 2𝑅 sin
2

Sample Problem
 A simple curve has a central angle of 36 deg and a degree of curve of 6 deg.
o Find the nearest distance from the midpoint of the curve to the point of intersection of
tangents. (9.8286 m)
o Compute the distance from the midpoint of the curve to the midpoint of the long chord
joining the point of curvature and point of tangency. (9.3475 m)
o If the stationing at the point of curvature is 10 + 020, compute the stationing of a point on
the curve which intersects with the line making a deflection angle of 8 deg with the
tangent through PC. (10+073.33)

Compound Curves
- Combination of 2 simple curves

PCC – Point of Compound Curvature

𝐼 =𝐼 +𝐼

𝑠𝑡𝑎. 𝑃𝐶 = 𝑠𝑡𝑎. 𝑃𝐼 − 𝑥 − 𝑇
𝑠𝑡𝑎. 𝑃𝐶𝐶 = 𝑠𝑡𝑎. 𝑃𝐶 + 𝐿
𝑠𝑡𝑎. 𝑃𝑇 = 𝑠𝑡𝑎. 𝑃𝐶𝐶 + 𝐿

Sine Law
𝑥 𝑇 +𝑇
=
sin 𝐼 sin 𝜃

Sample Problem
 The common tangent AB of a compound curve is 76.42 m with an azimuth of 268’30’. The
vertex B being inaccessible. The azimuth of the tangents AV and VB was measured to be
247’50’ and 282’50’ respectively. If the stationing of A is 43 + 010.46 and the degree of the
first curve was fixed at 4 deg based on the 20 m chord. Using Chord Basis.
 o Determine the stationing of the PC. (42+958.2152)
o Determine the stationing of the PCC. (43+061.5695)
o Determine the stationing of the PT. (43+109.6674)

Reversed Curves

PRC – Point of Reverse Curvature

𝑠𝑡𝑎. 𝑃𝐶 = 𝑠𝑡𝑎. 𝑃𝐼 − 𝑇

𝑠𝑡𝑎. 𝑃𝑅𝐶 = 𝑠𝑡𝑎. 𝑃𝐶 + 𝐿

𝑠𝑡𝑎. 𝑃𝐼 = 𝑠𝑡𝑎. 𝑃𝑅𝐶 + 𝑇

𝑠𝑡𝑎. 𝑃𝑇 = 𝑠𝑡𝑎. 𝑃𝑅𝐶 + 𝐿

Four types of Reversed Curves

1. Parallel Tangents & Equal Radii

2. Parallel Tangents & Unequal Radii
3. Converging Tangents & Equal Radii
4. Converging Tangents & Unequal Radii

Sample Problem
 In a railroad layout, the center line of two parallel tracks is connected with a reversed curve of
unequal radii. The central angle of the first curve is 16 deg and the distance between the parallel
tracks is 27.60 m. The PC is at station 15 + 420 and the radius of the second curve is 290 m.
o Compute the length of the long chord from the PC to PT. (198.3142 m)
o Compute the radius of the first curve. (422.4731 m)
o Compute the stationing of the PT. (15+618.96)
Vertical Curves

PC – Point of Curvature h2 – Vertical distance between PT and the

PT – Point of Tangency highest or lowest point of the summit or sag
PI – Point of Intersection of tangents curve
L – Length of parabolic curve g1 – Grade in percent of back tangent
S1 – Horizontal distance from PC to the highest g2 – Grade in percent of forward tangent
or lowest point of the summit or sag curve A – Change in grade from PC to PT
S2 – Horizontal distance from PT to the highest a – Vertical distance between PC and PI
or lowest point of the summit or sag curve b – Vertical distance between PT and PI
h1 – Vertical distance between PC and the H – Vertical distance between PI and the
highest or lowest point of the summit or sag parabolic curve
curve

Formulas:

𝟏 𝒈𝟏 𝑳 𝟏 𝑳
𝒂= 𝒈 𝑳 𝑺𝟏 = 𝒉𝟏 = 𝒈 𝑺 𝑯= (𝒈 + 𝒈𝟐 )
𝟐 𝟏 𝒈𝟏 + 𝒈𝟐 𝟐 𝟏 𝟏 𝟖 𝟏
𝟏 𝒈𝟐 𝑳 𝟏 𝑳 𝟐
𝒃 = 𝒈𝟐 𝑳 𝑺𝟐 = 𝒉𝟐 = 𝒈𝟐 𝑺𝟐 𝒙 𝟐
𝟐 𝒈𝟏 + 𝒈𝟐 𝟐 = 𝟐
𝒚 𝑯

Sample Problem
- A 3% grade at Sta. 10+100m is to be connected to a -2% grade at Sta. 10+420m using a symmetrical
parabolic curve. If the tangent grades intersect at elevation 152m, evaluate the following:
a) The location and elevation of the summit. (sta. 10+292, El. 150.08m)
b) The elevation and grade of the curve at Sta. 10+200. (El. 149.4189m, g = 1.4375%)
c) The elevation of the Vertical Point of Tangency or end of vertical curve. (El. 148.8m)