Lesson 1 - Leveling
Leveling - the process of finding the difference in elevation between two points by
measuring the vertical distance between the level surface through points.
Definition of Leveling Terms:
Horizontal line
A
Level line thru point A
(level surface)
horizontal line
Difference in elevation between A and B
Level line thru point B
B
(level surface)
Ground Surface
Elevation of Point B
Elevation of point A
Vertical line
Thru point A
Vertical line
thru Pt. B
Datum
(Mean Sea Level)
Where:
Level Surface - is a curved surface every element of which is normal to plumb line.
Level line - is a curved line in a level surface by which all parts are equidistant from the
earth center.
Horizontal line - is a straight line tangent to a level line.
Elevation - is a vertical distance above or below some arbitrarily assumed level surface or
datum.
Mean Sea Level - is the surface of the sea exactly midway between high and low tides.
Datum - is a surface of reference coincident or parallel with mean sea level to which all
elevations of a given region are referred.
Difference in elevation - is the vertical distance between an imaginary level surface
containing the high point and a similar surface containing the low point.
1
WORKSHEET NO. 1
Definition of Leveling Terms
Name:
Course/Year:
Date Submitted:
Define the following terms. Draw and indicate those terms in an illustration.
a. Level Surface
b. Level Line
c. Vertical Line
d. Horizontal Line
e. Elevation
f. Mean Sea Level
g. Datum
h. Difference in Elevation
i. Leveling
2
Lesson 2 - Differential Leveling
The procedure of running a line of differential levels follows:
a. Select a reference bench mark and a suitable route to be taken to reach the other point whose
elevation is to be established.
b. After the instrument is set up and leveled take a reading on the rod held on the bench mark.
c. Compute the elevation of the line of sight (HI) by adding the backsight reading to the bench mark
elevation.
d. Transfer the rod on a solid point which is as far from the instrument as the bench mark is and in
the general direction in which to advance. Take a foresight on this rod. Designate this point as a
turning point and compute its elevation by subtracting the foresight from the height of instrument.
e. Move the instrument to a new location from which a sight can be taken to the turning point just
established, and which is advantageously located for sighting another turning point.
f. From this new sight set up take a backsight to the turning point just established. The new height of
instrument can now be computed in the same way as when a backsight was taken to the reference
benchmark. With the height of instrument known, the elevation of the second turning point is now
determined by taking a foresight to it.
g. Repeat the above process until a foresight is taken on the final point whose elevation is to be
determined.
Leveling rod
Horizontal line
BSD
Level Station
FSD
Level Station
BS
FS
BS
FS
TP
L-1
HI1
BM-B
Elevation of TP
BM-A
Elev.
Elevation
of BM-A
of
BMB
(Unknown)
Datum (Mean Sea Level)
Example 1:
Complete the differential level notes shown below and perform the customary
arithmetic check.
Station
BS
HI
FS
Elevation
BM-A
3.23
1533.67m
TP1
2.97
0.42
TP2
4.87
0.23
BM - B
0.96
Tabulated Solution:
Station
BM-A
TP1
TP2
BM-B
BS
3.23
2.97
4.87
HI
1536.9
1539.45
1544.09
FS
0.42
0.23
0.96
∑BS = 11.07
∑FS = 1.61
DE1 = ∑BS - ∑FS = 9.46m
DE2 = Elev. of BM-B - Elev. of BM-A = 9.46
Since DE1 = DE2 the computation is correct.
Elevation
1533.67
1536.48
1539.22
1543.13
3
WORKSHEET NO. 2
Differential Leveling Terms
Name:
Course/Year:
Date Submitted:
Define the following terms used in differential leveling. Draw and indicate those
terms in an illustration.
a. Bench Mark
b. Backsight
c. Foresight
d. Backsight Distance
e. Foresight Distance
f. Turning Point
g. Height of Instrument
4
WORKSHEET NO. 3
Differential Leveling
Name:
Course/Year:
Date Submitted:
Complete the differential level notes shown and perform the customary arithmetic
check. Show your computation.
Station
BS
HI
FS
Elevation
BM 230
2.37
3,466.27m.
TP1
2.95
3.42
TP2
3.86
1.87
TP3
3.99
3.40
TP4
2.30
2.10
TP5
1.34
3.10
TP6
1.67
1.89
TP7
2.35
2.60
BM 240
2.56
5
WORKSHEET NO. 4
Differential Leveling (Double - Rodded Lines)
Name:
Course/Year:
Date Submitted:
Complete the differential level notes shown below and perform the customary
arithmetic checks.
Station
BS
HI
FS
Elevation
BM-45
3.456
2546.76m.
3.466
TP -1L
4.244
1.697
TP-1H
4.253
1.769
TP -2L
4.452
2.423
TP-2H
4.533
2.589
TP-3L
4.921
1.223
TP-3H
4.954
1.457
TP - 4L
3.454
1.996
TP-4H
3.589
1.999
TP-5L
4.232
2.243
TP-5H
4.337
2.542
BM-46
2.861
2.978
6
Example 2
Complete the differential level notes shown below and perform the customary
arithmetic check.
Station
BM-10
TP-1
TP-2
TP-3
TP-4
Backsight
Hair Rdgs.
Mean
3.779
3.068
2.355
9.134
7.941
6.748
5.619
4.724
3.825
8.499
6.871
5.245
2.997
2.395
1.795
HI
S
Elevation
S
567.68
3.664
2.951
2.238
6.235
5.042
3.845
4.757
3.860
2.965
7.250
5.623
3.995
5.225
4.626
4.025
BM-11
Tabulated Solution:
Backsight
Station
Hair Rdgs. Mean
3.067
BM-10
3.779
3.068
2.355
7.941
TP-1
9.134
7.941
6.748
4.723
TP-2
5.619
4.724
3.825
6.872
TP-3
8.499
6.871
5.245
2.396
TP-4
2.997
2.395
1.795
BM-11
Foresight
Hair Rdgs.
Mean
HI
S
1.424
570.747
2.386
575.737
1.794
575.419
3.254
578.43
1.202
575.203
Foresight
Hair Rdgs. Mean
Elevation
S
567.68
3.664
2.951
2.238
6.235
5.042
3.845
4.757
3.860
2.965
7.250
5.623
3.995
5.225
4.626
4.025
2.951
1.426
567.796
5.041
2.390
570.696
3.861
1.792
571.558
5.623
3.255
572.807
4.625
1.200
570.578
7
WORKSHEET NO. 5
Differential Leveling (Three-Wire Method)
Name:
Course/Year:
Date Submitted:
Complete the differential level notes shown below and perform the customary
arithmetic checks.
Backsight
Foresight
Station
HI
Elevation
Hair Rdgs.
Mean
S
Hair Rdgs.
Mean
S
BM-10
4.777
784.35m
4.064
3.353
TP-1
8.134
2.664
7.944
1.950
6.749
1.232
TP-2
6.629
5.335
5.734
4.142
4.835
2.945
TP-3
9.489
3.755
7.881
2.861
6.255
1.964
TP-4
3.996
8.252
3.375
6.625
2.695
4.997
BM-11
4.224
3.625
3.024
8
Lesson 3 - Profile Leveling
A column for intermediate foresights (IFS) is usually added to the standard format in the
preparation of profile level notes. This is for intermediate radiation or side shots and is handled
exactly like foresights except that they are just intermediate shots along the way and not part of
the main circuit.
Example:
Complete the following set of profile level notes and show the customary arithmetic check.
Station
BM-A
900+00
900+10
900+20
900+30
TP - 1
900+40
900+50
TP - 2
900+60
900+70
900+80
900+90
TP - 3
1000+00
1000 +10
1000 + 20
1000 + 30
1000 + 40
BS
2.34
HI
IFS
FS
Elevation
3524.76m
2.31
2.67
1.23
1.89
4.89
0.34
3.67
2.11
5.20
0.22
4.35
3.21
2.10
1.70
4.12
0.90
4.56
3.45
2.34
4.70
2.10
Tabulated Solution:
Station
BM-A
900+00
900+10
900+20
900+30
TP - 1
900+40
900+50
TP - 2
900+60
900+70
900+80
900+90
TP - 3
1000+00
1000 +10
1000 + 20
1000 + 30
1000 + 40
BS
2.34
HI
3527.1
IFS
FS
2.31
2.67
1.23
1.89
4.89
3531.65
0.34
3.67
2.11
5.20
3536.63
0.22
4.35
3.21
2.10
1.70
4.12
3539.85
0.90
4.56
3.45
2.34
4.70
2.10
Elevation
3524.76m
3524.79
3524.43
3525.87
3525.21
3526.76
3527.98
3529.54
3531.43
3532.28
3533.42
3534.53
3534.93
3535.73
3535.29
3536.40
3537.51
3535.15
3537.75
∑BS = 16.55
∑FS = 1.46
DE1 = ∑FS - ∑BS = 15.09m ; DE2 = Elev. Ref. BM - Last HI = 3524.76 - 3539.85 = 15.09m.
Since DE1 = DE2 the computation is correct.
9
WORKSHEET NO. 6
Profile Leveling
Name:
Date Submitted:
Course/Year:
Complete the following set of profile level notes and show the customary arithmetic check.
Station
BS
HI
IFS
FS
Elevation
BM-A
4.32
3452.78m
600+00
3.32
600+10
1.87
600+20
4.22
600+30
2.99
TP - 1
3.79
0.24
600+40
3.67
600+50
2.11
TP - 2
3.90
0.22
600+60
3.95
600+70
3.81
600+80
3.50
600+90
2.70
TP - 3
4.32
0.60
700+00
3.56
700 +10
4.47
700 + 20
3.37
700 + 30
5.10
700 + 40
3.10
10
Lesson 4 - Reciprocal Leveling
Reciprocal leveling is employed in topographic features such as rivers, lakes,
deep ravines and canyons using the method of reversion.
Example:
Reciprocal leveling were taken across a deep and wide river and between two
points BM 35 and BM 36 as follows: From the first instrument set up near BM 35: on
BM 35, 4.31, 4.34, 4.36 and 4.22 meters; on BM 36, 2.30, 2.35, 2.36 and 2.28 meters. For
the set up near BM 36 the readings are: on BM 36, 4.99, 4.90, 4.94 and 4.96 meters; on
BM 35, 6.89, 6.92, 6.96 and 6.93 meters. Determine the following:
a) The difference in elevation between the two bench marks.
b) The elevation of BM 36 if the elevation of BM 35 is 1345.64 meters.
Leveling rod
Leveling rod
bm bm'
am' am
BM 36
TDE = ?
BM 35
Elevation of BM 35
Elevation of BM 36
To datum
Solution:
Mean rod reading (am) on BM 35 with set up near BM 35:
am = (4.31 + 4.34 + 4.36 + 4.22) / 4 = 4.31m
Mean rod reading (bm) on BM 36 with set up near BM 35:
bm = (2.30 + 2.35 + 2.36 + 2.28) / 4 = 2.32m
Mean rod reading (bm') on BM 36 with set up near BM 36:
bm' = (4.99 + 4.90 + 4.96 + 4.93) / 4 = 4.95m
Mean rod reading (am') on BM 35 with set up near BM 36:
am' = (6.89 + 6.92 + 6.96 + 6.93) / 4 = 6.92
DE1 = am - bm = 4.31 - 2.32 = 1.99m
DE2 = am' - bm' = 6.92 - 4.95 = 1.97m.
True difference in elevation (TDE) between the two bench marks:
TDE = (DE1 + DE2) / 2 = (1.99 + 1.97) / 2 = 1.98m.
Elevation of BM 36 = 1345.64 + 1.98 = 1347.62m.
11
To datum
WORKSHEET NO. 7
Reciprocal Leveling
Name:
Date Submitted:
Course/Year:
Reciprocal leveling were taken across a deep and wide river and between two
points BM 43 and BM 44 as follows: From the first instrument set up near BM 43: on
BM 43, 5.11, 5.14, 5.19 and 5.22 meters; on BM 44, 1.10, 1.15, 1.06 and 1.18 meters. For
the set up near BM 44 the readings are: on BM 44, 5.99, 5.82, 5.94 and 5.96 meters; on
BM 43, 6.87, 6.90, 6.94 and 6.83 meters. Determine the following:
c) The difference in elevation between the two bench marks.
d) The elevation of BM 44 if the elevation of BM 43 is 11335.67meters.
12
Lesson 5 - Earths Curvature and Atmospheric Refraction
A
Point of Tangency
B
Horizontal Line
C
Line of Sight
D
O
Level Line
1. Curvature:
C = 0.667 M2 = 0.024(f/1000)2
Where:
C = is the curvature effect, in feet, of a level surface from horizontal line.
M = is the distance AB in miles
f = the same distance in feet.
2. Atmospheric Refraction:
R = 0.09 M2 = 0.003 (f/1000)2
Where:
R = is the refraction effect in feet
3. Combined Effect of Curvature and Refraction:
a) h'1 = 0.577M2
c) h'3 = 0.0671K2
b) h'2 = 0.021(f/1000)2
d) h'4= 0.0671(m/1000)2
Where:
M = distance of the object sighted from the point of tangency in miles.
f = same distance in feet
K = same distance in kilometers
m = same distance in meters
h'1 & h'2 = combined effect of curvature and refraction in feet
h'3 & h'4 = combined effect of curvature and refraction in meters
Example #1:
Determine the combined effect of the earth's curvature and mean atmospheric refraction in the following
distances: 500 ft, 3.0 miles, 250 meters, 10.0 kms.
Solution:
a) h' = 0.021(f/1000)2 = 0.021(500/1000)2 = 0.00525 ft.
b) h' = 0.577(M) 2 = 0.577(3.0) 2 = 5.193 ft.
c) h' = 0.0671 (m/1000) 2= 0.0671(250/1000) 2 = 0.00419 m.
d) h' 0.0671 (k) 2 = 0.0671(10) 2 = 6.71 m.
Example #2:
A woman standing on a beach can just see the top of a lighthouse 24.140 kms. Away. If her eye height above
sea level is 1.738 m., determine the height of the lighthouse above sea level.
Figure:
Woman on shore
horizontal line point of tangency
lighthouse
h'w
W
K1
K2
K
Solution:
hw' = 0.0671(K1)2 ; 1.738 = 0.0671(K1)2 ; K = 5.089 kms.
K2 = K - K1 = 24.140 - 5.089 = 19.051 kms.
h'L = 0.0671 (K2)2 = 0.0671(19.051)2 = 24.353 m.
13
L
h'L =?
WORKSHEET NO. 8
Effects of Curvature and Refraction 1
Name:
Date Submitted:
Course/Year:
Determine how far, in kms, out from shore an inter-island vessel will be when a
red light on its deck, 10.46 meters above the water, disappears from the sight of a child
standing on shore whose eye level is 1.32 meters above the water.
14
WORKSHEET NO. 9
Effects of Curvature and Refraction 2
Name:
Date Submitted:
Course/Year:
An observer standing on shore can just see the top of a lighthouse 15.0 miles
away. The eye height of the observer above sea level is 5.7 feet. What is the height of the
lighthouse above sea level?
15
Lesson 6 - Direct or Spirit Leveling
Leveling Rod
Leveling Rod
Horizontal
I
Instrument
h'b
h'a
Level Line Thru Point I
Hb
ha
Ha
Ground Surface
L
H
B
Level Line thru point A
Level Line thru point BS
A
hb
Elev. of A (known)
Elev.
of B
(Unknown)
Difference in Elevation between A and B:
H = Ha - Hb = (ha - ha') - (hb - hb')
Where:
A = point of known elevation
B = point of unknown elevation
ha = rod reading on A
hb = rod reading on B
ha' = effect of curvature and atmospheric refraction for the horizontal distance from I to B.
hb' = effect of curvature and atmospheric refraction for the horizontal distance from I to B.
Ha = difference in elevation between A and I.
Hb = difference in elevation between B and I.
H = difference in elevation between A and B.
Example:
Two points A and B, are 1000 ft. apart. The elevation of A is 615.03 ft. A level is set up on
the line between A and B and at a distance of 250 ft. from A. The rod reading on A 1s 9.15 and that on B is
2.07 ft. Making due allowances for curvature and refraction, what is the elevation of B?
Horizontal line
h'A
Ground Surface
hA=9.15' HA
h'B
HB
B
H
L
A
AL = 250'
AB = 1000'
BL = 750 '
Solution:
hA = 9.15 ft.
hB = 2.07 ft.
h' = 0.021(AL/1000) = 0.021(250/1000) = 0.0013 ft.
h' = 0.021(BL/1000) = 0.021(750/1000) = 0.0118 ft.
H = H - H = (h - h') - (h - h') = (9.15 - 0.0013) - (2.07 - 0.0118) = 7.09 ft.
Elev. B = Elev. A + H = 615.03 + 7.09 = 622.12 ft.
16
hB=2.07
WORKSHEET NO. 10
Effects of Curvature and Refraction 2
Name:
Date Submitted:
Course/Year:
Two points, M and N are 259.146 meters apart and the elevation of M is
1963.28 meters. A level is set up on line and distant 76.219 meters from M in the
direction of N. The rod reading on M is 2.416 meters and on N is 3.416 meters.
a. Neglecting the effects of curvature and refraction, determine the difference in
elevation between the two points, and the elevation of point N.
b. Making due allowance for curvature and refraction, determine the difference in
elevation between the two points, and the elevation of point N.
17
Lesson 7 - Trigonometric Leveling
Vertical (Plumb) Line
Vertical (Plumb Line)
Level Line thru B
B
hB
Line of Sight
C
h'B
H

Level Line Thru A
A
Elev. of Point A
(Known)
Elev. of
Point B
(Unknown)
D
(DATUM MSL)
Horizontal Line
Where:
H = difference in elevation between A and B, (H = hB  hB')
hB = vertical distance of point B above or below the horizontal line thru A, (hB = AC Tan )
h'B = correction for curvature and refraction
 = Observed vertical angle (positive or negative) from A to B.
A = point of known elevation
B = point whose elevation is to be determined.
Example:
Two points A and B, are each distant 2,000 ft. from a third point C, from
which the measured vertical angle to A is +2321' and that to B is +1032'. What is the
difference in elevation between A and B.
Level Line Thru A
Level Line Thru B
A
H
Level Line Thru C
B
hA
A = +23 21'
HA
B = +10 32'
hB
C
HB
h'B
h'A
HBC = 2,000 ft.
HAC = 2,000 ft.
Solution:
18
hA = (HAC)Tan A = 2000 Tan (2321') = 863.41 ft.
hA' = 0.021(HAC/1000)2 = 0.021(2000/1000)2 = 0.08 ft.
hB = (HBC)Tan B = 2000Tan (1032') = 371.88 ft.
hB' = 0.021(HBC/1000)2 = 0.021(2000/1000)2 = 0.08 ft.
H = HA - HB = (hA + hA') - (hB + hB') = (863.41 + 0.08) - (371.88 + 0.08)
H = 491.53 ft.
WORKSHEET NO. 11
Trigonometric or Indirect Leveling
Name:
Date Submitted:
Course/Year:
Two points A and B are 4,134.50 meters apart. From a third point, C, on the
line between A and B, and 2,992.25 meters from A. The measured vertical angle to A is
+3528' and that to B is -1514'. What is the difference in elevation between A and B
making due allowance for the effect of curvature and atmospheric refraction?
19
Lesson 8 - Borrow Pit
Volumes of Borrow-Pit Excavations:
V = A (h1 + h2 + h3 + h4)1/4
Where:
V = volume of square or rectangular truncated prism
A = area of the right or horizontal section, A = L(W)
h1, h2, h3, & h4 = the cuts or corner heights of the four corners of the prism.
V = A (h1 + h2 + h3)1/3
Where:
V = volume of a triangular truncated prism
A = area of the right or horizontal section, A =1/2 (b)h
h1, h2 & h3 = the cuts or corner heights of the three corners of the prism.
Example:
Determine the volume in cubic meters of the borrow-pit sketched below.
Corner heights are in meters.
3.2
5.0
A
4.7
5.4
5.0
B
15.0
C
5.2
3.9
6.2
5.0
D
E
3.8
4.7
4.0
3.3
4.0
4.0
12.0 M.
Solution:
VA = A(h1 + h2 + h3 )1/3 = 1/2 (4.0)(5.0)(3.2 + 5.4 + 4.7)1/3 = 44.34 cu.m.
VB = A(h1 + h2 + h3 + h4)1/4 = 4.0(5.0)(4.7 + 5.4 + 3.9 + 6.2)1/4 =101.0 cu.m.
VC = A(h1 + h2 + h3)/3 = 1/2 (4.0)(5.0)(4.7 + 6.2 + 5.2)1/3 =53.67 cu.m.
VD = A(h1 + h2 + h 3+ h4)1/4 = 4.0(5.0)(5.2 + 6.2 + 3.3 + 4.7)1/4 = 97.0 cu.m.
VE = A(h1 + h2 + h3)1/3 = 1/2 (4.0)(5.0)(5.2 + 4.7 + 3.8)1/3 = 45.67 cu.m.
V = VA + VB + VC + VD+ VE
V = 44.34 + 101.00 + 53.67 + 97.00 + 45.67 = 341.68 cu.m. say 342 cu.m.
20
WORKSHEET NO. 12
Volumes of Borrow-Pit Excavation
Name:
Date Submitted:
Course/Year:
Determine the volume in cubic meters of the borrow-pity sketched below.
Corner heights are in meters.
4.2
5.0
A
5.7
4.4
5.0
B
15.0
C
6.2
4.9
6.2
D
5.0
E
4.8
5.7
5.0
4.3
5.0
15.0 M.
21
5.0
Assembly of Prisms Method:
V = A (h1 + 2h2 + 3h3 + 4h4)1/4
Where:
V = volume of the excavation
A = area of one right or horizontal section in the assembly
h1 = sum of all heights applying to one square
h2 = sum of all heights common to two squares
h3= sum of all heights common to three squares
h4= sum of all heights common to four squares
Example:
Find the quantity in cubic meters of the borrow-pit given in the figure below. Corner heights (cuts)
indicated are in meters.
1
2
3
4
5
A
3.3
4.1
4.5
3.7
10.0
B
4.4
5.2
6.7
4.8
10.0
C
40.0 M
5.5
7.2
7.6
8.9
10.0
D
6.3
9.5
6.7
5.9
4.7
E
3.9
4.2
12.0
Tabulated Solution:
Corner
A-1
A-2
A-3
A-4
B-1
B-2
B-3
B-4
C-1
C-2
C-3
C-4
D-1
D-2
D-3
D-4
D-5
E-1
E-2
E-3
E-4
E-5
Sums
5.1
12.0
h1
3.3
H2
2.5
12.0
H3
h4
4.1
4.5
3.7
4.4
5.2
6.7
4.8
5.5
7.2
8.9
7.6
6.3
9.5
6.7
5.9
4.7
3.9
4.2
5.1
3.0
2.5
h1 = 18.1
A = (12.0)(10.00) = 120.0 sq.m.
V = A (h1 + 2h2 + 3h3 + 4h4)1/4
V = (120/4)[18.1 + (2)(49.5) + (3)(5.9) + (4)(44.2)]
V = 9,348 cubic meters
22
3.0
12.0
h2 = 49.5
h3 = 5.9
h4 = 44.2
Where: h1 = heights applying to one rectangle.
h2 = heights common to two rectangles
h3 = heights common to three rectangles
h 4 = heights common to four rectangles
WORKSHEET NO. 13
Trigonometric or Indirect Leveling
Name:
Date Submitted:
Course/Year:
Find the quantity in cubic meters of the borrow-pit given in the figure below.
Corner heights (cuts) indicated are in meters.
1
2
3
4
5
A
4.3
5.1
6.5
3.7
9.0
B
3.4
6.2
5.7
3.8
9.0
C
36 M
4.5
6.2
7.9
7.6
9.0
D
5.3
7.5
5.7
7.9
4.7
9.0
E
4.9
5.2
11.0
23
6.1
11.0
3.5
4.0
11.0
11.0
Lesson 9 - Principle in Stadia Measurement
Standard Symbols Used in Stadia Measurement:
f1
f2
stadia rod
A
Telescope of Instrument
b
b'
m
a'
I
A
c
Line of Sight
f
s
B
C
d
D
Plumb Line thru instrument center
Where:
f = focal length of the lens
f1 = image distance
f2 = object distance
i = distance or spacing between stadia hairs
c = distance from the center of the instrument to the center of the objective
lens.
C = Stadia constant or the distance from the center of the instrument to
Principal focus. C = c + f
d = distance from the focal point in front of the telescope to the face of the
Rod. d = (f/I)s
D = distance from the instrument center to the face of the rod.
D = d + (f + c) = (f/i)s + C = Ks + C
K = Stadia interval factor. K = f/I
S = Stadia or rod intercept.
24
Example 1
Determine the stadia interval factor of the instrument. Assume that the stadia
constant (C) is zero.
Stadia Hair Readings
Point
Distance from
transit to rod (m)
Upper (m)
Lower (m)
A
30
0.96
0.66
B
45
1.10
0.64
C
60
1.21
0.60
D
75
1.35
0.58
E
90
1.47
0.56
F
105
1.57
0.53
G
120
1.72
0.50
Figure:
1.72
1.47
0.96
1.10
1.21
1.57
1.35
0.66
0.64
0.60
0.58
0.56
0.53
a
30 m.
b
15 m.
c
15 m.
d
15 m.
e
15 m.
0.50
f
15 m.
g
15 m.
Solution:
Ka = da/sa= 30/(0.96-0.66) = 100.00
Kb = db/sb = 45/(1.10-0.64) = 97.83
Kc = dc/sc = 60/(1.21-0.60) = 98.36
Kd = dd/sd = 75/(1.35-0.58) = 97.40
Ke = de/se = 90/(1.47-0.56) = 98.90
Kf = df/sf = 105/(1.57-0.53) = 100.96
Kg = dg/sg = 120/(1.72-0.50) = 98.36
Ave. K = (Ka + Kb + Kc + Kd + Ke + Kf + Kg)/7
= (100.00 + 97.83 + 98.36 + 97.40 + 98.90 + 100.96 + 98.36)/7
Ave. K = 98.83
25
WORKSHEET NO. 14
Stadia Interval Factor
Name:
Date Submitted:
Course/Year:
A theodolite is set up at one end of a level base line 100.0m long. Stakes
marks the line at every 25.0 m. and a stadia rod is held at each stake. The stadia intercept
at each location of the rod is observed as follows: 0.298, 0.597, 0.894, 1.195 and 1.475
meters, respectively. Compute the stadia interval factor (K) for each distance and also
determine the average value of K.
26
Example 2
A dumpy level with an internal focusing telescope was set up on the left bank
of a river and the rod readings tabulated below were taken on a stadia rod held
successively at the left and right water edges. If the stadia interval factor of the
instrument is 100, determine the width of the river.
Hair Readings
Middle (c)
2.172m
2.173
Rod Position
Upper (a)
2.189m
2.277
Rod held at LWE
Rod held at RWE
Lower (b)
2.155m
2.069
Figure:
Dumpy level (K=100, C=0)
Stadia Rod held @ LWE
a1
c1
Stadia Rod held @ RWE
a2
c2
b2
b1
Water Surface
Left Water Edge (LWE)
Right Water Edge (RWE)
Ground Surface Underwater
d1
Width of river (W)
d2
Solution:
S1 = (a1 - b1) = (2.189 - 2.155) = 0.034m
S2 = (a2 - b2) = (2.277 - 2.069) = 0.208m
d1 = ks1 + C = 100(0.034) + 0 = 3.4m
d2 = ks2 + C = 100(0.208) + 0 = 20.8m
W = d2 - d1 = 20.8 - 3.4 = 17.4m.
27
WORKSHEET NO. 15
Horizontal Stadia Sights
Name:
Date Submitted:
Course/Year:
An automatic level with an internal focusing telescope was set up somewhere
at mid-length of a long-span steel bridge. The rod readings tabulated below were
observed on a stadia rod held successively at the vicinity of the concrete abutments in the
southern and northern approaches of the bridge. If the stadia interval factor of the
instrument is 98.5, determine the length of the bridge.
Rod Position
Rod at Southern Approach
Rod at Northern Approach
28
Upper (a)
2.98m
3.54
Hair Readings
Middle (c)
1.68m
2.02
Lower (b)
0.38m
0.49
Lesson 10 - Inclined Stadia Sights
Stadia rod
ID
d
a'
a
P
c
b

b' RR
f
c
o
N
m
VD

F
DE
i
HD
D
HI
M
Instrument Station
Where:
K = Stadia Interval factor, K = f/I
a = upper stadia hair reading
b = lower stadia hair reading
P = horizontal cross hair reading or rod reading, P = RR
S = stadia interval, S = a-b
 = Observed Vertical Angle of elevation or depression.
C = Stadia constant of the instrument, C = c + f, Assume C = 0 for internal
focusing telescopes.
ID = The inclined or slope distance from the center of the instrument O to the
Horizontal cross hair reading at P. ID = Ks cos  + C
HD = The horizontal distance between the instrument and the rod.
HD = Ks Cos2 + C cos 
VD = The vertical distance between the telescope axis at O and the
Horizontal cross hair reading at P. VD = Ks cosSin + C sin
DE = the difference in elevation between the instrument station at M and the
Distant stadia point at N. DE = HI + VD - RR.
29
Example #1
The following data were obtained by stadia measurement: vertical angle = +1823', and
observed stadia intercept = 2.20 m. The stadia interval factor of the instrument used is 95.5 and C = 0.30m.
If the height of instrument is 1.62m, and the rod reading is taken at 1.95m, determine the following:
a) horizontal distance (HD) from the instrument set up at A to the rod held at point B.
b) Vertical Distance (VD) from the center of the instrument to the point of the rod bisected
by the horizontal cross hair.
c) Inclined or slope distance (ID) from the instrument center to the point on the rod bisected
by the horizontal cross hair.
d) Difference in elevation (DE) between the point over which the instrument is set up and
the point on which the rod was held.
Stadia rod
ID
d
a'
a
P

b b' RR= 1.95m
c
f
K=95.5
C=0.30m
c
o
B
m
VD
 = 1823'
F
DE
i
HI = 1.62m.
HD
A
Instrument Station
Solution:
HD = Ks Cos2 + C cos 
= 95.5(2.20)Cos2 (1823') + 0.30Cos(1823')
HD = 189.48m.
VD = Ks cosSin + C sin
= 95.5(2.20)Sin(1823')Cos(1823') + Sin(1823')
VD = 62.97m
ID = Ks Cos + C
= 95.5(2.20) Cos(1823') + 0.30
ID = 199.68m.
Check:
ID = (HD) 2 + (VD) 2 = (189.48) 2 + (62.97) 2 = 199.67m.
DE = HI + VD - RR
= 1.62 + 62.97 -1.95
DE = 62.64m.
30
D
WORKSHEET NO. 16
Inclined Stadia Sights
Name:
Course/Year:
Date Submitted:
The following data were obtained by stadia observations: vertical angle =
+925', upper stadia hair and lower stadia hair readings are 2.352m and 0.995m,
respectively. The stadia interval factor is known to be 99.0 and C is 0.381m. The height
of instrument above the instrument station (point A) is 1.496m and rod reading is taken at
1.598m. Determine the following:
a) horizontal, vertical and inclined distances by exact stadia formulas.
b) Elevation of the point sighted (point B) if the elevation of point A is
776.545m.
c) Difference in elevation between the two points.
31
Example #2
The upper and lower stadia hair readings on a stadia rod held at station B
were observed as 3.50 and 1.00m, respectively, with the use of a transit with an internal
focusing telescope and having a stadia interval factor of 99.5. The height of the
instrument above station A is 1.45m. and the rod reading is taken at 2.25m. If the vertical
angle observed is -2334', determine the following:
a) horizontal, vertical, and inclined stadia distances.
b) Difference in elevation between the two stations.
c) The elevation of station B, if the elevation of station A is 155.54m above
mean sea level.
Instrument
K=99.5, C=0
Horizontal Line of Sight
 = -2334'
ID
Line of Sight
HI=1.45
a=3.5
VD
b=1.0
RR=2.25m
A
DE
Elev. of A
155.54m.
B
HD
Reqd. Elev.of B
Reference Datum (m.s.l)
Solution:
a)
S = (a-b) = 3.50 - 1.00 = 2.50m.
HD = Ks Cos2 + C cos  = 99.5(2.50)cos2 (2334') = 208.99m.
VD = Ks cosSin + C sin = 99.5(2.50)sin(2334')cos(2334') = 91.16m
ID = (HD) 2 + (VD) 2 = (208.99) 2 + (91.16) 2 = 228.00m
32
b)
DE = RR + VD - HI
DE = 2.25 + 91.16 -1.45 = 91.96m.
c)
Elev. of B = Elev. of A - DE
= 155.54 - 91.96
Elev. of B = 63.58m.
WORKSHEET NO. 17
Inclined Stadia Sights
Name:
Course/Year:
Date Submitted:
A stadia interval of 1.325m is observed on a stadia rod held over a turning
point using an engineer's transit with a stadia interval factor of 98.5 and a stadia constant
of 0.300m. The elevation of the instrument station is 379.246m. and the height of
instrument above the station is 1.245m. If the rod reading is 1.649m. and the vertical
angle observed is -1517', determine the horizontal, vertical, and inclined distances by
exact stadia formulas. Also determine the difference in elevation between the turning
point and the instrument station.
33
Example #3
Given the following set of stadia level notes, the instrument used has a stadia interval factor of 100 and equipped
with an internal focusing telescope. Completye the tabulation and perform the customary arithmetic check.
Backsight
Foresight
Sta
Change in
Elev.
Elevation
Intercept
Vertical
RR
VD
Intercept
Vertical
RR
VD
(s)
Angle()
(s)
Angle()
BMa
1.55
-525'
1.50
550.50m.
TP1
1.74
+815'
1.68
1.76
+1030'
1.48
TP2
0.95
-448'
1.77
1.98
+1208'
1.66
BMb
2.49
-1250'
2.53
1.06
722'
2.05
TP3
2.14
+1405'
1.79
2.67
-1532'
1.92
TP4
1.92
-941'
1.33
2.16
-759'
1.25
BMc
2.65
+732'
1.88
Stadia Rod
Level Line
Line of Sight
TP2
BMb
DE4
DE3
BMc
TP3
DE2
DE1
DE6
DE5
TP1
Elev. BMb
Elev. TP2
BMa
Elev. TP1
Elev. BMa
TP4
Elev. TP3
Elev. BMc
Elev. TP4
Datum (m.s.l.)
Solution:
A) Computation of Vertical Distances on backsights and foresights using VD = Ks CosSin
Station
Backsight
Foresight
BMa
VDbs= 100(1.55)cos525'sin525' = +14.57m.
TP1
VDbs = 100(1.74)(cos815')sin815' =-24.71m.
VDfs=100(1.76)cos 1030' sin 1030' =+31.54m.
TP2
VDbs = 100(0.95)cos 448'sin 448' = +7.92m.
VDfs = 100(1.98)cos128' sin128' =+53.93m.
BMb
VDbs = 100(2.49)cos 1250'sin1250' = +53.93m.
VDfs = 100(1.06)cos722'sin722' = +13.48m.
TP3
VDbs = 100(2.14)cos145'sin145' = -50.51m.
VDfs = 100(2.67)cos1532'sin1532' = -68.89m.
TP4
VDbs = 100(1.92)cos941'sin941' = +31.83m.
VDfs = 100(2.16)cos759'sin759' = -29.71m.
BMc
VDfs = 100(2.65)cos732'sin732' = +34.44m.
b. Computing change in elevation between backsights
c. Determining elevations of Turning Points and Bench Marks:
and foresights using CE = RRbs  VDbs  VDfs - RRfs
Elev TP1 = Elev. BMa  CE1 = 550.50 + 46.13 = 596.63m.
CE1 = 1.50 + 14.57 + 31.54 -1.48 = +46.13 m.
Elev TP2 = Elev. TP1  CE2 = 596.63 + 16.00 = 612.63m.
CE2 = 1.68 - 24.57 + 40.69 - 1.66 = +16.00 m.
Elev BMb = Elev. TP2  CE3 = 612.63 + 21.12 = 633.75m
CE3 = 1.77 + 7.92 + 13.48 - 2.05 = +21.12m.
Elev TP3 = Elev. BMb  CE4 = 633.75 - 14.35 = 619.40m
CE4 = 2.53 + 53.93 -68.89 - 1.92 = -14.35m.
Elev TP4 = Elev. TP3  CE5 = 619.40 - 79.68 = 539.72m
CE5 = 1.79 - 50.51 - 29.71 - 1.25 = - 79.68m.
Elev BMc = Elev. TP4  CE6 = 539.72 + 65.72 = 605.44m.
CE6 = 1.33 + 31.83 + 34.44 - 1.88 = +65.72m
Backsight
Foresight
Sta
Elev.
Change in
Elevation
Vertical
RR
VD
Vertical
RR
VD
Intercept
Intercept
Angle()
Angle()
(s)
(s)
BMa
1.55
-525'
1.50
+14.57
550.50m.
TP1
1.74
+815'
1.68
-24.71
1.76
+1030'
1.48
+31.54
+46.13
596.63
TP2
0.95
-448'
1.77
+7.92
1.98
+1208'
1.66
+40.69
+16.00
612.63
BMb
2.49
-1250'
2.53
+53.93
1.06
722'
2.05
+13.48
+21.12
633.75
TP3
2.14
+1405'
1.79
-50.51
2.67
-1532'
1.92
-68.89
-14.35
619.40
TP4
1.92
-941'
1.33
+31.83
2.16
-759'
1.25
-29.71
-79.68
539.72
BMc
2.65
+732'
1.88
+34.44
+65.72
605.44
DEb = 54.94
d) Arithmetic Check:
DEa = Elev. BMc - Elev. BMa = 605.44 - 550.50 = 54.94m. Note: Since DEa = DEb, the solution was assumed correct.
34
WORKSHEET NO. 18
Stadia Leveling
Name:
Date Submitted:
Course/Year:
Complete the stadia level notes shown below and perform an arithmetic
check. Assume K =100 and C = 0.0.
Station
BM-10
TP-11
TP-12
TP-13
BM-14
35
S
(m)
1.245
2.044
1.720
2.426
Backsight
RR

(m)
2.42
-425'
320'
1.08
1.46
-640'
2.08
-915'
VD
(m)
S
(m)
Foresight
RR

(m)
VD
(m)
DE
(m)
Elev.
(m)
1,552.35
1.515
1.438
1.025
1.326
818'
-325'
1005'
422'
1.55
3.06
0.45
1.45
WORKSHEET NO. 19
Stadia Leveling
Name:
Date Submitted:
Course/Year:
Following are the notes for a line of stadia levels. The elevation of BM-15 is
184.29m. The stadia interval factor is 100.00 and C = 0.381m. Rod readings are taken at
height of instrument. Determine the elevations of remaining points and tabulate values
accordingly.
STA
S
(m)
1.30
0.87
1.01
0.69
BM-15
TP-16
TP-17
TP-18
BM-20
Backsight

VD
(m)
-140'
-243'
055'
210'
S
(m)
0.97
0.83
1.36
0.94
Foresight

VD
(m)
DE
(m)
Elevation
(m)
228'
-406'
-036'
825'
Lesson 11 - Subtense Bar Method
Subtense bar is a convenient and practical device used for quick and accurate
measurement of distances. The bar, which is precisely 2 meters long, consists of a round
steel tube through which runs a thin invar bar. At each end of the frame are housed the
target marks which are made of special transparent material for easy illumination when
used at night. The figure shows the principle of the subtense bar:
X
Left Target Mark
Theodolite
/2

A
36
Subtense Bar
1.00
B
2.00
1.00
Right Target Mark
Horizontal Distance (HD)
Y
Tan (/2) = (XY/2)/(AB)
AB = (XY/2)/Tan(/2)
Since XY = 2.00m and HD = AB
HD = 1/Tan(/2)
Example 1:
A 2-meter long subtense bar was first set up at A and subsequently at B, and
the subtended angles to the bar, as read from a theodolite positioned somewhere along the
middle of line AB, were recorded as 44'32" and 52'12", respectively. Determine the
length of AB.
Figure:
Subtense bar
theodolite
subtense bar
a = 44'32"
b = 52'12"
2.00 A
B 2.00
da
db
Dab
Solution:
da = 1/tan(a/2) = 1/tan(44'32"/2) = 154.39 m.
db = 1/tan(b/2) = 1/tan(52'12"/2) = 131.71 m.
Dab = da + db = 154.39 + 131.71 = 286.10 m.
WORKSHEET NO. 20
Subtense Bar Method
Name:
Date Submitted:
Course/Year:
A subtense bar 2 meters long is set up near the middle of a traverse line MN.
At M, the angle subtended is 38'05", and at N the angle is 63'16". Find the length of MN.
37
Example 2
A traverse line AB is being measured by theodolite and subtense bar. At
station A, the horizontal axis of the theodolite is 1.48m. above the ground mark and the
horizontal angle subtended by a 2-meter subtense bar set up at station B is 42'18". The
subtense bar is 1.22m. above the ground mark and the vertical angle measured to it on the
theodolite is +1356'. Determine the length of the line AB and the difference in elevation
between the two ground station marks.
Figure:
Subtense bar
 = 42'18"
38
2.00m.
B'
HB=1.22
VD
A'
B
 = +1356'
DE
C
HI=1.48
A
HD
Solution:
a) HD = 1/tan(/2) = 1/tan (42'18"/2) = 162.54 m.
b) From triangle A'B'C
tan  = VD/HD
VD = (HD) tan = (162.54)tan(1356') = 40.32m.
Solution check:
ID1 = HD/cos = 162.54/cos(1356') = 167.47m.
ID2 = (HD)2 + (VD)2 = (162.54)2 + (40.32)2 = 167.47m
Since ID = ID it is assumed that the solutions are correct.
c) DE + HB = VD + HI
DE = VD + HI - HB = 40.32 + 1.48 - 1.22 = 40.58m.
WORKSHEET NO. 21
Subtense Bar Method
Name:
Date Submitted:
Course/Year:
A traverse is being measured by theodolite and subtense bar. At one station,
the theodolite axis is 1.36m above the ground mark and the horizontal angle subtended by
a 2-meter subtense bar placed at the second station is 42'48". The subtense bar is 1.23m.
above the grouind mark, and the vertical angle measured to it on the theodolite is
+2214'. Determine the horizontal and vertical distances between the two station marks.
39
Lesson 12 - Plane Table Surveys
Plane table is an oldest type of surveying instrument consisting of board used as a drafting table for
plotting map details and ground contours while conducting field works. The board is attached to a tripod in such a way
that it can be leveled or rotated to any desired direction. The advantages of the plane table method are the following:
1. In the actual survey the map is plotted that requires only significant points or details to be located.
2. Stream banks, contours and other irregular lines are represented accurately since the terrain is
viewed as the outlines are plotted.
3. No need to use field notes since plotting is done in the field. This makes errors in recording and
plotting in the office avoided.
4. Checks on the plotted points are readily made in the field.
5. Since office work is reduced mapping is done in a shorter period.
Some disadvantages of the plane table method are the following:
1. More time is spent on the field.
2. Unsuited for heavily forested areas since plane table equipment and accessories are cumbersome.
3. To gain proficiency, considerable time is required.
4. For precise plane table work, horizontal and vertical control must be plotted in advance.
5. The clearance of lines of sight above possible obstruction were lower since the height of the plane
table is usually set lower than a transit.
6. Limited to relatively open country and only to favorable weather conditions
The stepping or interval method is used to determine differences in elevation when the slope of the
ground is so small that assumed the horizontal distance practically equal to inclined distance.
Example:
The plane table and alidade is set up at A and the rod is held at point B. First, the alidade is sighted at
the rod held at B, and the stadia interval is observed as 1.39 m. The line of sight of the alidade is then rotated in a
vertical plane until the telescope is level, and the position of the horizontal line of sight on some clearly defined object
on the ground is noted at 1. The telescope is raised until the lower stadia hair cuts 1, and the position of the upper stadia
hair on the ground is noted at 2. The telescope is again rotated about the horizontal axis until the lower stadia hair cuts
2, and the position of the upper stadia hair on the ground is now noted at 3. Finally, the telescope is rotated until the
lower stadia hair cuts 3, when the upper stadia hair is seen to intersect the rod at 4, which is at rod reading of 0.40 m. If
the height of the instrument is 1.35 m., what is the difference in elevation between points A and B.
Figure:
level or stadia rod
S = 1.39 m.
4
RR = 0.40 m.
B
3
Alidade
2
D.E.
1
Horizontal line of sight
Plane Table
HI = 1.35m.
40
A
Solution:
D.E. = 1.35 + 3(1.39) - 0.40 = 5.12 m.
Note:
Similar procedure is followed for negative vertical angles.
Stepping method is usually limited to inclination less than 2 degrees and where the number of steps does not exceed
three.
WORKSHEET NO. 22
Stepping Method in Plane Table Survey
Name:
Date Submitted:
Course/Year:
When an alidade and a plane table were set up over a point the elevation of
The line of sight of the alidade was 406.29m. In order to determine the elevation of point
M, the stepping method was used. The interval intercepted between the stadia hairs on a
rod held on M was 1.95m. After the line of sight was elevated for five steps from the
horizontal line of sight, the rod reading at the horizontal cross hair was 2.62m. Determine
the elevation of point M.
41
WORKSHEET NO. 23
Stepping Method in Plane Table Survey
Name:
Date Submitted:
Course/Year:
In Worksheet No. 15, if the same values were observed but this time the line
of sight was instead depressed for five steps from the horizontal line of sight, determine
the elevation of point M.
42
Lesson 13 - Radiation by Plane Table
Radiation is one of the most common methods in plane table survey where a
series of points are located in relation to the plotted position of the table. In the figure, the
plane table is shown in position over station N. The plotted location of the plane table is
indicated at n on the plane table sheet. The alidade is pivoted at this point and while
sights are taken at points A, B and C, rays are drawn. With the distances plotted to scale
along the corresponding rays, points a, b and c are located on the map. HI above the
datum is determined by adding the measured distance from the ground to the telescope
axis of the alidade to the elevation of the ground station. From these quantities, the
corresponding elevations of all other points sighted can be determined and indicated on
the map.
Figure:
A
a
N
n
b
c
B
C
43
WORKSHEET NO. 24
Radiation With Plane Table
Name:
Date Submitted:
Course/Year:
A series of points are to be located by the principle of radiation using a plane table and an alidade set up
over a control station whose elevation is 147.75m. above mean sea level. The height of the instrument's line of sight
above the occupied station is 1.22m. The value of K is 100 and C = 0. The observed data are shown in the
accompanying tabulation. Plot the six points sighted and determine their respective elevations.
point
10
11
12
13
14
15
44
Hair Readings
upper
2.02
1.88
3.25
3.43
2.54
2.38
Middle
1.62
1.28
2.45
2.73
1.94
1.88
lower
1.22
0.68
1.65
2.03
1.34
1.38
Vertical
angle
+215'
+1618'
-1206'
+554'
-733'
+1604'
Azimuth
from south
10044'
17536'
20430'
22718'
32007'
1525'
Horizontal
distance
Elevation
Lesson 14 - Map
Map - a graphic representation of all portion of the earth's surface or other celestial body, by means of signs and
symbols or photographic imagery at some given scale or projection, to which lettering is added for identification.
Classification of Maps
1. Topographic Map - is a representation of the earth's surface in three dimensions.
2. Planimetric Map - is a representation of the earth's surface in the two horizontal dimensions only.
3. Computer-Generated Maps - Computer graphics system is used in the production of maps.
4. Photomaps - It is a reproduction of an aerial photgraph on which grid lines, contours, boundaries,
placenames and marginal information have been added or overprinted.
Map Scales
Scales refers to the relationship which the distance between any two points on the map bears to the
corresponding distance on the ground. Map scales are classified as a) Large- Scale Maps having scales of 1: 2000 or
larger and with contour intervals ranging fro 0.10 to 2.0 meters, b) Medium Scale Maps having scales ranging from
1:2,000 to 1: 10,000 and with contour intervals ranging from 1.0 to 5.0 meters, c) Small-Scale Maps having scales of 1:
10,000 or smaller and with contour intervals ranging from 5 to 2,000 meters. Map scales are portrayed by:
a) Words and figures or an equivalence
Examples:
1 inch = 1 mile
1 centimeter = 1 kilometer
3 inches = 200 feet
b) By scale ratio or representative fraction - the scale of a map can be expressed in ratio such as 1: 5,000
or as a fraction such as 1/5000. A fraction indicating a scale is termed the representative fraction (RF) or scale ratio
(SR).
SR = MD/GD
Where: MD = the map distance or the scaled distance between any
two selected points on the map.
GD = the corresponding distance on the ground.
c) Graphically.
1
Example:
1/2
0
1
2
3 Mile
Common Map Scales and their equivalents:
Map Scale
One inch represents
1: 2,000
1: 5,000
1: 10,000
1: 20,000
1: 25,000
1: 50,000
1: 75,000
1: 100,000
1: 125,000
1: 250,000
1: 500,000
56 yards
139 yards
0.158 mile
0.316 mile
0.395 mile
0.789 mile
1.18 miles
1.58 miles
1.97 miles
3.95 miles
7.89 mile
45
One
Centimeters
represents
20 meters
50 meters
0.1 km
0.2 km
0.25 km.
0.50 km.
0.75 km.
1.0 km.
1.25 km.
2.50 km.
5.0 km
One
Mile
represented by
31.68 inches
12.67 inches
6.34 inches
3.17 inches
2.53 inches
1.27 inches
0.845 inch
0.634 inch
0.507 inch
0.253 inch
0.127 inch
is
One Kilometer
represented by
50.0 cm.
20.0 cm
10.0 cm.
5.0 cm.
4.0 cm.
2.0 cm.
1.33 cm.
1.0 cm.
8.0 mm
4.0 mm
2.0 mm
is
1: 1,000,000
15.78 miles
10.0 km.
0.063 inch
1.0 mm
Example #1
Determine the scale of a sketch wherein one centimeter represents one hundred
meters on the ground.
SR = MD/GD = 1cm/ (100cm)(100cm/1m) = 1/10,000
Example # 2
If the equivalence scale of a map is 5 cm = 10 km, what is the scale ratio?
SR = MD/GD = 5cm/(10km)(1,000m/1km)(100m/1m) = 5/1,000,000 = 1/200,000
Example #3
The ground distance between two points on a map is 4 kilometers. If the distance
between the same two points on a map is 8 centimeters, determine the scale of the map.
SR = MD/GD = 8cm/(4km)(1,000m/1km)(100cm/1m) = 8/400,000 = 1/50,000
Example #4
On a map with a scale of 1 cm = 3,000m, the measured length of a power
transmission line is 5.25 cm. What is the equivalent ground length of the line?
GD = MD/SR = 1cm/(3,000m)(100cm/1m) = 5.25(3,000)(100)cm = 1,575,000 cm =
15,750 m.
Example #5
On a map with a scale of 1cm = 250 m, the measured length of an irrigation canal is
0.20m. Determine the equivalent ground length of the canal in kilometers.
GD = MD/SR = (0.20m)(100cm/1m)/(1cm/250m) = (250)(0.20)(100)m = 5,000m or
5 km.
Example #6
On a map of scale 1:75,000 the map distance measured between points A and B is
5.0 cm. Determine the scale of a sketch of the same area in which drawn distance between the
same points measured 12.5 cm.
For the map:
GDm = MDm/SRm = 5cm/(1/75,000) = 375,000cm
For the sketch:
GDs = MDs/SRs = 12.5cm/SRs
46
Since ground distances are equal for both map and sketch:
GDm = GDs
375,000cm = (12.5cm/SR)
SR = 12.5/375,000 = 1/30,000
WORKSHEET NO. 25
Map Scales
Name:
Date Submitted:
Course/Year:
The ground distance between two points is 25km. If the distance between the
same two points on topographic map is 50mm, determine the scale of the map.
47
WORKSHEET NO. 26
Scale Ratio
Name:
Date Submitted:
Course/Year:
The measured ground distance between two points is 3,525.65 m. If the
corresponding distance between the two points on a map is 2.35 cm, determine the scale
of the map.
48
WORKSHEET NO. 27
Equivalence Scales
Name:
Date Submitted:
Course/Year:
On a map with a scale of 1cm = 150m, the measured length of a concrete
fence is 0.20m. Determine the equivalent ground length in kilometers.
49
WORKSHEET NO. 28
Determining Unknown Scale
Name:
Date Submitted:
Course/Year:
On a map of scale 1:25,000, the map distance measured between points X and
Y is 32.750cm. Determine the scale of a sketch of the same area in which the drawn
distance between the same points measures 10.916 cm.
50
WORKSHEET NO. 29
Dimensions From Maps
Name:
Date Submitted:
Course/Year:
Compute the approximate are, in square meters, of a rectangular shaped tract
of land whose sides measure 29.67mm by 64.19mm on a map drawn to a scale of
1:25,000.
Lesson 15 - Topographic Surveys
Contour Line Representation of Terrain:
40
30
20
10
0
Elevation or Section Th ru Line AB
0
10
51
20
A
30
20
30
B
Index contour
Intermediate
contour
Plan or Outline of the Hill Represented by Contour Lines
Types of Contours
1) Index contour - this is presented by a heavier line at regular intervals in order to provide a
convenience in scaling elevations and to ease and speed in reading contours.
2) Intermediate contour - This is presented by lighter lines found between the index contours, not
usually labeled except where the terrain is relatively flat and their elevations are not readily obvious.
3) Depression Contour - This is drawn to show low spots such as excavations around which contours
close.
4) Supplemental Contours - They are drawn as dashed lines to show properly important breaks in the
terrain.
5) Approximate Contours - If contour accuracy cannot be definitely determined, the map maker has to
make an educated guess rather than have a blank area in the map.
Contour Interval for Different Map Scales:
Scale
Interval
Scale
Interval
1/500
0.50m
1/25,000
10m
1/2,000
1
1/50,000
20m
1/5,000
2
1/100,000
25
1/10,000
5 or 10
1/250,000
50
Characteristics of Contours:
1. All points on any one contour have the same elevation.
2. Every contour closes on itself either within or beyond the limits of the map.
3. A contour that closes within the limits of the map indicates either a summit or depression.
4. Contours on the ground cannot cross one another except where an overhanging cliff, a vertical ledge or wall
is represented on a map.
5. Contours are spaced evenly on a uniform slope.
6. Contours are straight and parallel to each other on a plane surface.
7. Irregular contours signify rough, rugged terrain.
8. The horizontal distance between adjacent contours indicates the steepness of the slope of the ground.
9. Contours cross curbs and a crowned sloping street in typical U-shaped curves.
10. As a contour approaches a stream, the contour turns upstream until it intersects the shoreline.
Locating Contour Lines by Interpolation:
The methods included in locating contour lines by interpolation are the following:
52
a)
b)
c)
d)
e)
by estimation - this method is suitable on small-scale maps where the ground form is not
too irregular.
Rubber band method - the rubber band is stretched between two plotted points so that
these points fall at scale divisions corresponding to their elevations.
Triangle and scale method - This method by which the engineer' scale and triangle are
used provides an accurate and rapid procedure for interpolating contour lines in which
mathematical computation is eliminated.
Template method - Contour template, consists of a series of equally spaced parallel lines
that are drawn on transparent film or paper, allow many interpolations to be made
quickly and accurately.
Analytical method - Arithmetical computations is well suited if high accuracy is desired
in drawing large scale maps.
By Similar Triangles:
hm/de = H/DE
hm = (de/DE )(H)
B
Ground Surface (Uniform Slope)
M
DE
Elevb
De
A
Elevm
Hm
Eleva
H
Datum (m.s.l.)
Where:
A & B = established points on the ground of known elevation.
M = a point on a contour crossing the lie between A and B.
hm = horizontal map distance from A to M. (cm)
de = difference in elevation between M and A (m).
DE = difference in elevation between A and B (m).
H = horizontal map distance between A and B (cm).
Example #1
The map distance between two benchmarks, A and B, is 15 cm. and their
elevations above sea level are 850.50m and 939.60m, respectively. Assuming uniformity
of slope between A and B, calculate the map distance from A where each of the
following contours will cross the line between the two benchmarks: 860, 880, 900, and
920m:
Figure:
h4
Elev = 939.60m
h3
B
4
3
h2
53
de4
2
h1
DE
de3
de2
1
de1
A
Elev. 850.50m
H = 15 cm
Solution:
a) Difference in elevation:
DE = elevb - eleva = 939.60 - 850.50 = 89.10m.
de1 = elev1 - eleva = 860.00 - 850.50 = 9.50m
de2 = elev2 - eleva = 880.00 - 850.50 = 29.50m
de3 = elev3 - eleva = 900.00 -850.50 = 49.50m
de4 = elev4 - eleva = 920.00 - 850.50 = 69.50m
b) Horizontal distance of contours from A:
h1 = (de1/DE)H = (9.50/89.10)(15) =1.599 say 1.6cm.
h2 = (de2/DE)H = (29.50/89.10)(15) = 4.966 say 5.0cm.
h3 = (de3/DE)H = (49.50/89.10)(15) = 8.333 say 8.3cm
h4 = (de4/DE)H = (69.50/89.10)(15) = 11.70 say 11.7cm.
WORKSHEET NO. 30
Interpolating Contours Analytically
Name:
Date Submitted:
Course/Year:
On a map the scaled distance between points A and B is 15.45 cm, and the
given elevation of the two points are 815.26m and 836.45m, respectively. Assuming
uniformity of slope between the two points, calculate the distance from point A on the
map where each of the following contour lines will cross the line joining the two points:
820, 825, 830, and 835m.
54
WORKSHEET NO. 31
Interpolating Contours Graphically
Name:
Date Submitted:
Course/Year:
The accompanying tabulation gives elevations of points over the ares of a 30m by 40m city lot. The
elevations were obtained by the Checkerboard Method using 10-m squares. Point A-1 is located at the northwest corner
of the lot and point E-4 at the southeast corner. All elevations are in meters.
Point
1
2
3
4
A
878.8
882.5
883.5
883.6
B
883.2
885.9
893.7
887.2
C
882.8
888.6
891.8
889.3
D
882.5
890.4
892.6
889.5
E
883.2
888.5
892.8
893.6
Requirements:
a)
b)
55
Plot the contours using a horizontal scale of 6cm = 10m and a contour interval of 1 meter.
Arrange the plotting paper so that the longer side is vertical (Use 81/2" by 11" bond paper)
c)
d)
e)
f)
All corner designations and corresponding elevations must be indicated, including all vertical and
horizontal lines.
Show the direction of north and indicate the scale used.
Indicate the elevations of the following index contours: 880, 885, 890, and 895m. The thickness of
the index contours must be drawn about twice the thickness of intermediate contours.
Use a pencil for the initial plot and ink the final plot using a technical pen.
WORKSHEET NO. 32
Slopes From Contours
Name:
Date Submitted:
Course/Year:
On a map of scale 1cm = 50m with a contour interval of 1.5m, two adjacent
contour lines are 1.37cm apart. What is the slope of the ground in percent?
56
Lesson 16 - Control Surveys
Adjustment of A Chain of Triangles.
The adjustment of triangulation data may be undertaken by the exact method called the method of least
squares which is based upon the theory of probability and the Approximate method which is a simple and convenient
method used for adjusting the angles and sides of triangulation systems which are of lower order of precision.
Example #1
For the given chain of triangles shown in the accompanying figure and tabulation, perform station and
figure adjustment by the approximate method.
j
a
q
A
D
m
b
F
l
k
r
c
g
p
57
e
d
f
B
h
o
C
I
E
h
Tabulation Data:
Angle
A
B
C
D
E
F
G
H
I
Observed value
24021'00"
6029'10"
5910'05"
30134'49"
5825'15"
6225'10"
5925'10"
6310'08"
17459'24"
Angle
J
K
L
M
N
O
P
Q
R
Observed Value
18909'51"
3905'01"
7140'02"
6005'10"
24935'30"
4510'20"
6514'10"
28419'47"
7540'19"
Tabulated Solution for station adjustment:
Station
A
B
C
D
E
F
Angle
A
B
C
Sum
D
E
Sum
F
G
H
I
Sum
J
K
L
M
Sum
N
O
P
Sum
Q
R
Sum
Observed Value
24021'00"
6029'10"
5910'05"
36000'05"
30134'49"
5825'15"
36000'04"
6225'10"
5925'10"
6310'08"
17459'24"
35959'52"
18909'51"
3905'01"
7140'02"
6005'10"
36000'04"
24935'30"
4510'20"
6514'10"
36000'00"
28419'47"
7540'19"
36000'06"
Correction
-05"
-05"
-05"
-15"
-02"
-02"
-04"
+02"
+02"
+02"
+02"
+08"
-01"
-01"
-01"
-01"
-04"
00"
00"
00"
00"
-03"
-03"
-06"
Adjusted Value
24020'55"
6029'05"
5910'00"
36000'00"
30134'47"
5825'13"
36000'00"
6225'12"
5925'12"
6310'10"
17459'26"
36000'00"
18909'50"
3905'00"
7140'01"
6005'09"
36000'00"
24935'30"
4510'20"
6514'10"
36000'00"
28419'44"
7540'16"
36000'00"
Correction
Value After
Figure
Adjustment
5909'52"
5825'05"
6225'03"
18000'00"
6029'17"
5925'23"
6005'20"
18000'00"
6310'00"
Tabulated Solution for Figure Adjustment:
Triangle
Angle
ABC
C
D
E
Sum
B
G
M
Sum
H
ACD
CDE
58
Angles From
Station
Adjustment
5910'00"
5825'13"
6225'12"
18000'25"
6029'05"
5925'12"
6005'09"
17959'26"
6310'10"
-08"
-08"
-09"
-25"
+12"
+11"
+11"
+34"
-10"
L
O
Sum
K
P
R
Sum
DEF
7140'01"
4510'20"
18000'31"
3905'00"
6514'10"
7540'16"
17959'26"
-11"
-10"
-31"
+11"
+11
+12"
+34"
7139'50"
4510'10"
18000'00"
3905'11"
6514'21"
7540'28"
18000'00"
WORKSHEET NO. 33
Adjustment of A Chain of Triangles
Name:
Date Submitted:
Course/Year:
Given in the tabulation are the observed angles of a chain of triangles as
shown in the accompanying sketch. Perform station and figure adjustments using the
approximate method.
a A
b
D
k
c
B
59
j
f
d
e
h
g
C
i
n
m
E
Angle
a
b
c
d
e
f
g
Measured Value
3120'24"
4759'40"
13710'29"
5702'42"
10040'25"
6506'20"
24105'13"
l
Angle
H
I
J
K
L
M
N
Measured Value
6653'30"
5210'20"
28900'30"
7059'30"
28043'58"
2718'12"
5157'53"
Lesson 17 - Adjustment of Quadrilateral
The following sequence of steps are performed in the adjustment of a quadrilateral:
1)
2)
3)
4)
The angles about each station are adjusted to total 360 degrees before the next adjustment is made.
The sum of the interior angles of the quadrilateral is adjusted to equal (n-2)180 or 360 degrees.
The opposite angles at the intersection of the diagonals should be equal.
The trigonometric condition is satisfied by means of computations involving the sines of the angles. The
angles are adjusted so that the computed length of an unknown side opposite a known side will be the same
regardless of which of the four possible routes is used.
C
d
B
c
b
a
A
60
h
e
f
g
D
Angle Condition Equations:
From Overlapping Triangles ABC, BCD, ABD, and ACD:
a + b + c + d = 180
c + d + e + f = 180
a + b + g + h = 180
e + f + g + h = 180
From the quadrilateral ABCD:
a + b + c + d + e + f + g + h = (n-2)180 = 360 , since n = 4
Considering the opposite angles at the intersection of the diagonals:
a+b=e+f
c+d=g+h
Corrections to the angles:
v1 = v2 = (1/4)I1 - (1/4)[I2 - (1/2)I3 - (1/2)I1]
v3 = v4 = (1/4)I1 + (1/4)[I2 - (1/2)I3 - (1/2)I1]
v5 = v6 = (1/4)I3 + (1/4)[I2 - (1/2)I3 - (1/2)I1]
v7 = v8 = (1/4)I3 - (1/4)[I2 - (1/2)I3 - (1/2)I1]
Where v1, v2, etc., are corrections to be determined for angles a, b, etc., and I1, I2, and I3 are the errors of
closure of the triangles.
Side Condition Equations:
CD = (AB)(sin b)( sin h)/(sin e)(sin g)
CD = (AB)(sin a)(sin c)/(sin d)(sin f)
(Sin a)(Sin c)(Sin e)(Sin g)/(Sin b)(Sin d)(Sin f)(Sin h) = 1
Considering the logarithms of both members of the equations:
(Log Sin a + Log Sin c + Log Sin e + Log Sin g) - Log Sin b + Log Sin d + Log Sin f + Log Sin h = 0
Example #1
The observed angles of a quadrilateral, after station adjustment, are given in the accompanying tabulation and sketch.
Adjust these angles by use of both the angle and the side equations.
Angle
Observed Value
Angle
Observed Value
A
4238'36"
E
6200'46"
B
6452'28"
F
4529'58"
C
4032'57"
G
3331'32"
D
3156'07"
H
3857'40"
C
B
d
e
c
Where:
A, B, C, and D are triangulation stations
of the quadrilateral ABCD and the
observed angles are a, b, c, d, and etc.
b
a
h
A
f
g
D
Solution:
a) Sum of angles:
For triangle ABC = a + b + c + d = 4238'36" + 6452'28" + 4032'57" + 3156'07" = 18000'08"
61
d1 = 180 - (a + b + c + d) = 180 - 18000'08" = -08"
For triangle BCD = c + d + e + f = 4032'57" + 3156'07" + 6200'46" + 4529'58" = 17959'48"
d2 = 180 - (c + d + e + f) = 180 - 17959'48" = +12"
For triangle CDA = e + f + g + h = 6200'46" + 4529'58" + 3331'32" + 3857'40" = 17959'56"
d4 = 180 - ( e + f + g + h) = 180 - 17959'56" = +04"
b) Determining the values of corrections:
k = (1/4)[d2 - d3/2 - d1/2] = d2/4 -d3/8 -d1/8 = (+12"/4) - (+04"/8) - (-08"/8) = 3.5"
v1 = v2 = d1/4 - k = (-08"/4) - 3.5" = -5.5"
v3 = v4 = d1/4 + k = (-08"/4) + 3.5" = +1.5"
v5 = v6 = d3/4 + k = (+04"/4) + 3.5" = +4.5"
v7 = v8 = d3/4 - k = (+04"/4) - 3.5" = -2.5"
c) Applying corrections to the angles:
a' = a  v1 = 4238'36" - 5.5" = 4238'30.5"
e' = e  v5 = 6200'46" + 4.5" = 6200'50.5"
b' = b  v2 = 6452'28" - 5.5" = 6452'22.5"
f'' = f  v6 = 4529'58" + 4.5" = 4530'02.5"
c' = c  v3 = 4032'57" + 1.5" = 4032'58.5"
g' = g  v7 = 3331'32" - 2.5" = 3331'29.5"
d' = d  v4 = 3156'07" + 1.5" = 3156'08.5"
h' = h  v8 = 3857'40" - 2.5" = 3857'37.5"
d) Side Adjustment
Angle
Log sine
Tabular diff. For I"
Angle
Log sine
Tabular difference
for I"
9.87
33.78
20.70
26.04
90.39
a'
9.830853449
22.86
b'
9.956825228
c'
9.812984077
24.61
d'
9.723428685
e'
9.945991446
11.19
f'
9.853247226
g'
9.742174071
31.78
h'
9.798501028
sums
9.332003043
90.44
9.332002167
Diff. = 304.3 - 216.7 = 87.6
Corr. = diff./(sum1+ sum2 = 87.6/(90.44 + 90.39) = 0.48"
Notes:
1) The tabular difference for I" is taken from logarithmic tables for trigonometric functions or by use of electronic calculator.
2) The correction is then added to each of the angles whose log sine is smaller, and added from each of the other angles.
e) The final adjusted Angles:
Angle
First Adjusted Value
Corr.
Finally adjusted value
a
4238'30.5"
+0.48"
4238'30.98"
b
6452'22.5"
-0.48"
6452'22.02"
c
4032'58.5"
+0.48
4032'58.98"
d
3156'08.5"
-0.48"
3156'08.02"
e
6200'50.5"
+0.48"
6200'50.98"
f
4530'02.5"
-0.48"
4530'02.02"
g
3331'29.5"
+0.48"
3331'29.98"
h
3857'37.5"
-0.48"
3857'37.02"
sums
36000'00"
36000'00"
WORKSHEET NO. 34
Adjustment of A Quadrilateral
Name:
Date Submitted:
Course/Year:
The observed angles of a quadrilateral are given in the accompanying tabulation and sketch. Adjust the
angles about each station by the approximate method.
l
i
D
k
C
62
g
j
a
e
b
c
d
B
f
A
base line
Angle
a
b
c
d
e
f
Observed value
5721'10"
3137'05"
27101'30"
2222'00"
5731'25"
28006'50"
Angle
G
H
I
J
K
L
Observed value
3001'55"
7115'05"
25842'45"
6800'00"
2150'30"
27009'30"
Lesson 18 - Strength of Figure
Expressions For Determining Strength of Figure:
R = [(D - C)/D][2 A + AB + 2 B] = (F)(2 A + AB + 2 B)
Where:
R = Relative strength of figure.
D = Number of directions observed (forward and back), not including the fixed or known side of a
given figure.
F = A factor for computing strength of figure = (D - C)/D
A, B = Tabular difference for I second, expressed in units of the sixth decimal place, corresponding to
angles A and B of a triangle.
( +  + ) = The summation of values for particular chain of triangle through which the
computation is carried from the known line to the line required.
C = Number of geometric conditions to be satisfied in a given figure.
C = (n' - s' +1) + (n - 2s + 3)
Where:
n' = Number of lines observed in both directions, including the fixed or known side of a given figure.
s' = Number of occupied stations.
n = Total number of lines in figure, including fixed or known line.
s = Total number of stations.
Example #1
Determine the strength of figure factors of the following:
63
a)
Single Triangle:
Known line
D=4
n' = 3
s' = 3
n=3
s=3
B
C = (n' - s' + 1) + (n - 2s + 3)
C = (3 - 3 + 1) + [3 - 2(3) + 3)
C=1
F = (D -C)/D = (4 -1)/4 = 0.75
C
A
b)
Completed Quadrilateral
D =10
n' = 6
s' = 4
n=6
s =4
D
C
C = (n' -s' + 1) + [n -2s +3]
C = (6 - 4 + 1) + [6 - 2(4) + 3]
C=4
F = (D - C)/D = (10 - 4)/10 = 0.60
A
Known line
c)
B
Triangle with interior station
D = 10
n' = 6
s' = 4
n=6
s=4
B
Known line
D
C = (n' - s' + 1) + (n - 2s + 3)
C = (6 - 4 + 1) + [6 - 2(4) +3]
C=4
F = (D - C)/D = (10 - 4)/10 = 0.60
C
A
central point
WORKSHEET NO. 35
Strength of Figure Factors
Name:
Date Submitted:
Course/Year:
Given the accompanying six-sided central-point figure with one diagonal JM. The triangulation stations are I, J, K
, L, M, N, and O. The known side is IJ. If all the stations were occupied and all lines, including the diagonal were
observed in both directions, determine the following strength of figure factors: C, D, and F.
J
K
I
O
N
64
L
M
Example 2
Given the quadrilateral ABCD in the accompanying figure and assume that the observed interior angles
have already been subjected to station and figure adjustment. If all triangulation stations were occupied and all
lines observed in both directions, determine the following:
a) Strength of figure factors
b) Distance angles and the equations for determining the length of CD by different routes.
c) Strength of figure for each chain of triangles.
d) Length of the check base CD using the strongest route if the base line AB is 1,586.85m.
Figure:
check base
D
C
40
36 53





47
43
40 37

A
Base line
Solution:
a) Determining strength of figure factors:
n' = 6
C = (n' - s' + 1) + (n -2s + 3)
s' = 4
C = (6 - 4 + 1) + [6 - (2)(4) + 3]
n=6
C=4
s=4
65
B
D = 10
F = (D - C)/D = (10 - 4)/10 = 0.60
b) Determining distance angles and length of CD by different routes:
Considering triangles ABC and ACD with AC as the common side: CD = (AB)(sin43)(Sin 40)/(sin 60)(sin36)
Considering triangles ABD and ACD with AD as the common side: CD = (AB)(sin90)(sin40)/(sin53)(sin104)
Considering triangles ABC and BCD with BC as the common side: CD = (AB)(sin77)(sin47)/(sin60)(sin89)
Considering triangles ABC and BCD with BD as the common side: CD = (AB)(sin37)(sin47)/(sin53)(sin44)
c) Determining strength of figure for each chain of triangles:
1) Consider triangles ABC and ACD with AC as the common side:
For distance angles 43 and 60 of triangle ABC the value of X is interpolated as follows: 2
5
40 -------11
a/3 = 2/5
[ +   + 2 ] = X +Y
3
a
2
A
A B
B
43 ------- X
a = 1.2
= 9.8 + 22.2
X = 11 - a =11 - 1.2
= 32
45 --------9
X = 9.8
R = [(D - C)/D]( +AB +2B)
For distance angles 36 and 40 of triangle ACD the value of Y is
= F(X +Y)
interpolated as follows:
b/1 =4/5
= 0.60(32)
5
23
b 4
b =0.8
R = 19.20
1 35
36
Y
Y = 23 - b = 23 -0.8
40
19
Y = 22.2
2) Consider triangles ABD and ACD with AD as the common side and using the same procedure of computation in 1): X = 2.4,
Y = 5.2, and R = 4.56
3) Consider triangles ABC and BCD with BC as the common side and using the same procedure of computation in 1): X = 2.0,
Y = 3.7, and R = 3.42
4) Consider triangles ABD and BCD with BD as the common side and using the same procedure of computation in 1): X = 15.2,
Y = 12.8, and R = 16.80
Tabulation of Data
2A + AB + 2 B
Dist
Relative Strength
Quadrilateral
Chain of triangles
Common sides
angles
of Figure ( R )
Each
Summation
ABC
AC
9.8
32.0
19.20
43; 60
ABCD
22.2
ACD
36 ; 40
ABD
AD
2.4
7.6
4.56
53 ; 90
ACD
40 ; 104 5.2
ABC
BC
2.0
5.7
3.42
60 ; 77
3.7
BCD
47 ; 89
ABD
BD
15.2
28.0
16.80
37 ; 53
12.8
BCD
44 ; 47
d) Determining length of CD using the strongest route.
Triangle ABC and BCD with BC as the common side provides the strongest chain. To compute CD, use therefore the equation:
CD = (AB)(sin77)(sin47)/(sin60)(sin89) = 1,305.94m.
WORKSHEET NO. 36
Strength of Figure
Name:
Date Submitted:
Course/Year:
For the quadrilateral CDEF shown below, calculate the value of the relative strength of figure ( R ).
Assume that the angles of the quadrilateral have already been adjusted, all stations were occupied, and all lines were
observed in both directions. If the base line CD is 3,254.26m long, determine the length of EF using the strongest
chain.
E
F
1040'
1620'
6130'
7710'
9130'
7550'
1410'
C
66
1250'
D
Lesson 19 - Reduction to Sea Level
Derivation of sea level reduction factor:
Measured length of line AB
d
B
A
s = (d - c)
h
B'
A'
Sea level
67
Length of Line AB
R
R+h
Where:
A&B = Two points on the earth's surface defining a line
Which is part of a triangulation system.
d = horizontal distance measured between the two points.
c = correction to be applied to the measured length.
s or A'B' = equivalent length of line AB at sea level or (d - c)
O
h = average elevation above sea level of the two points.
R = average radius of curvature of the earth at the vicinity of line AB.
Since the lengths of arcs are proportional to their radii:
(d - c)/R = d/(R + h)
the equation may be expressed into expanded form:
d/R - c/R = d/R - d(h/R2) + d(h2/R3) - d(h3/R4) + ….
Where:
c = d(h/R) - d(h2/R2) + d(h3/R3)
From the figure:
s = (d - c) = d - d(h/R) = d(1 - h/R)
Sea level reduction factor is given by:
FSL = (1 - h/R)
Sea-Level Reduction Factors for R = 6,372,226 m.:
Elevation
(h)
meters
100
150
200
250
300
Sea-level
Reduction
Factor fSL
0.9999 8430
0.9999 7646
0.9999 6861
0.9999 6076
0.9999 5292
Elevation
(h)
meters
600
650
700
750
800
Sea-level
Reduction
Factor fSL
0.9999 0584
0.9998 9799
0.9998 9014
0.9998 8230
0.9998 7445
Elevation
(h)
meters
1100
1150
1200
1250
1300
Sea-level
Reduction
Factor fSL
0.9998 2737
0.9998 1952
0.9998 1168
0.9998 0383
0.9997 9598
Elevation
(h)
meters
1600
1650
1700
1750
1800
Sea-level
Reduction
Factor fSL
0.9997 4891
0.9997 4106
0.9997 3321
0.9997 2537
0.9997 1752
350
400
450
500
550
0.9999 4507
0.9999 3722
0.9999 2938
0.9999 2153
0.9999 1368
850
900
950
1000
1050
0.9998 6660
0.9998 5876
0.9998 5091
0.9998 4306
0.9998 3522
1350
1400
1450
1500
1550
0.9997 8814
0.9997 8029
0.9997 7245
0.9997 6460
0.9997 5675
1850
1900
1950
2000
2050
0.9997 0967
0.9997 0183
0.9996 9398
0.9996 8613
0.9996 7829
Example 1
A line measures 6,844.89m. at an average elevation of 500m. If the average
radius of curvature in the area is 6,354,243m, determine the equivalent sea level length
by a) using the table for sea level reduction, b) the derived formulas.
Solution:
a) Using the table for sea level reduction and by interpolation:
150
68
50
450
500
600
-
2938
X
584
a
2354
a/2354 = 50/150
a = (50/150)(2354) = 785
X = 2938 - a = 2938 - 785 = 2153
FSL = 0.99992938 - 0.99992153
FSL = 0.99990785
S = d(fSL) = 6,844.89(0.99990785)
S = 6,844.26 m.
b) Using the formula:
fSL = (1 - h/R)
= (1 - 500/6,354,243)
fSL = 0.99992131
S = d(fSL)
= 6,844.89(0.99992131)
S = 6,844.35 m.
WORKSHEET NO. 37
Reduction To Sea Level
Name:
Date Submitted:
Course/Year:
The elevation of the two end points of a base line measures 3,123.45m and
3,133.22m, respectively. If the measured length of the base line is 4,566.79m and the
mean radius of the earth at the vicinity of the base line is 6,368,597m, determine the
equivalent sea level length of the line.
69
Lesson 20 - Intervisibility of Stations
In triangulation work where there are problems regarding the intervisibility of triangulation stations and
in the construction of observation towers can be solved using the following formula:
In metric system:
h = h1 + (h2 - h1)d1/d2 - 0.065d1d2
Where:
h = elevation of line of sight at obstruction in meters
h1 = elevation of the first station in meters
h2 = elevation of the second station in meters
d1 = distance between first station and obstruction in kms.
d2 = distance between obstruction and second station in kms.
d3 = distance from first station to second station or d1 + d2 in kms.
In English system:
h = h1 + (h2 - h2)d1/d3 - 0.574 d1d2
Where:
h, h1 and h2 are in feet; d1, d2 and d3 are in miles
2nd Station
B
Figure:
Obstruction
C
70
X
Line of Sight
h2-h1
1st Station
0.065d21
A
0.065(d1 + d2)2
h
h1
h2
h1
d2
d1
Derivation of the formula:
From the figure:
h = h1 + 0.065d 12 + X
Eq. 1
Also:
X/d = [(h2 - h1)d1 - 0.065(d1 + d2)2]/(d1 + d2)
Or:
X = (h2 - h1)d1/(d1 + d2) - 0.065d 12 - 0.065d1d
Eq. 2
Substituting the value of X to eq. 1:
h = h1 + (h2 - h1)d1/d2 - 0.065d1d2
Example:
Cathedral Hill, with an elevation of 1,138.253m is on a line between Aurora Hill and Mirador Hill (see
accompanying figure). It is 16.608km. from Aurora Hill and 11.878km. from Mirador Hill. The elevations of Aurora
Hill and Mirador Hill are 1,136.264m. and 1,152.428m, respectively. If a line of sight originating from Aurora Hill is
directed toward Mirador Hill, determine the following from Aurora Hill is directed toward Mirador Hill, determine the
following:
a) Whether Mirador Hill is visible from Aurora Hill.
b) Height of identical towers which could be constructed at Aurora Hill and Mirador Hill so that the
line of sight would clear cathedral Hill by 3 meters.
c) Height of a tower which could be constructed on Aurora Hill so that the line of sight would clear
the ground at Cathedral Hill by 3 meters, if no tower is to be constructed on Mirador Hill.
d) Height of a tower which could be constructed on Mirador Hill so that the line of sight would clear
the ground at Cathedral Hill by 3 meters, if no tower is to be constructed on Aurora Hill.
e) Of the three options stated in requirements b, c, and d, which would be the most logical choice.
Solutions:
a) Determining if Mirador Hill is visible from Aurora Hill.
Cathedral Hill
(Obstruction)
Line of Sight
Mirador Hill
(2nd Station)
de
71
Aurora Hill
(1st Station)
h3 = 1,138.253kms.
h
d = 11.878kms
h2 = 1,152.428m
d1 = 6.608km.
d3
h1 = 1,136.264km.
Datum (mean sea level)
Solution:
d3 = d1 + d2 = 16.608 + 11.878 = 28.486km.
h = h1 + (h2 - h1)d1/d3 - 0.065 d1d2
= 1,136.264 + (1,152.428 - 1,136.264)(16.608/28.486) - 0.065(16.608)(11.878)
h = 1,132.865m
de = h - h3 = 1,132.865 - 1,138.253 = -5.388m, therefore Mirador Hill is not visible from Aurora Hill.
b)
Determining height of identical towers at both stations:
Tower
Modified Line of Sight
Tower
3.0m (reqd. clearance)
H2
H
Original line of sight
de = 5.388m.
H1
Cathedral Hill
Mirador HIll
Aurora Hill
From the figure:
H1 = H2 = H3 = c + de
= 3.0 + 5.38
H1 = 8.388m
c) Determining height of a single tower at Aurora Hill:
Required Tower
Modified Line of Sight
c = 3.0m (required clearance)
de = 5.388
H1
Original line of sight
Cathedral Hill
Aurora Hill
d2 = 11.878km
d1 = 16.608km
d3 = 28.486 km
Solution:
72
Mirador Hill
H1/d3 = (c + de)/d2
H1 = d3/d2 (c + de)
= (28.486/11.878)(30 + 5.388)
H1 = 20.116m
d) Determining height of a single tower at Mirador Hill:
Required Tower
Modified Line of Sight
Original Line of Sight
c = 3.0m (reqd. clearance)
H2
de = 5.388m
Cathedral Hill
Aurora Hill
Mirador Hill
d2 = 11.878km
d2 = 16.608km.
d3 = 28.486km
Solution:
H2/d3= (c + de)/d1
H2 = d3/d1 (c + de)
= (28.486/16.608)(3.0 + 5.388)
H2 = 14.387m
e) The most logical option is to construct a single tower at Mirador Hill. The required tower is the shortest, it will be
the least expensive to construct than the other two options.
WORKSHEET NO. 38
Intervisibility of Stations
Name:
Date Submitted:
Course/Year:
It is desired to sight from Dimasalang Hill (elev. = 196.85m) to Humabon Hill
(elev. = 198.15m) 40 km. Away. Sikatuna Hill (elev. = 197.65m), 2.2km away,
apparently obstructs the line of sight. If towers of equal height on Dimasalang Hill and
Humabon Hill are to be constructed so that the line just clears Sikatuna Hill by 3.50m,
what will be the height of these towers? How high a tower would be required on
Humabon Hill alone? On Dimasalang Hill alone?
73
WORKSHEET NO. 39
Intervisibility of Stations
Name:
Date Submitted:
Course/Year:
Cathedral Hill is on a line between Constable Hill and Aurora Hill. Cathedral
Hill is 6.253 km away from the Constable Hill and Aurora Hill is 11.081 km from the
Constable Hill. The elevations of the hills are: Constable Hill, 1260.86m; Aurora Hill,
1266.44m; and Cathedral Hill, 1261.04m. Determine the following:
a) Elevation of the line of sight from Constable Hill at the vicinity of
Cathedral Hill.
b) Height of identical towers to be constructed at Constable Hill and Aurora
Hill so that the line of sight will clear cathedral Hill by 5.0m.
c) Height of tower to be constructed on constable Hill so that the line of
sight will clear the ground at Cathedral Hill by 5.0m, if no tower is to be
constructed on Aurora Hill.
74
d) Height of tower to be constructed on Aurora Hill so that the line of sight
will clear the ground at Cathedral Hill by 5.0m, if no tower is to be
constructed on Constable Hill.
Lesson 21 - Spherical Excess
The value of the spherical excess is used to test the closure of the triangle. It
is also used in making the least squares adjustment. A more exact value for determining
the spherical excess in seconds is given by the formula:
e" = Area/R2 Sin 01"
Where:
e" = Spherical excess in seconds of angle.
Area = Area of the triangle in square meters
R = mean radius of curvature of the earth in meters.
Example:
The interior angles in triangles ABC are A = 5830' 30", B = 6417'25", and C = 5712'17". The
distance from A to B has been found from preliminary calculations to be 36,285.55m. Assuming the average radius of
curvature to be 6,372,160m, determine the following:
a) Spherical excess in the triangle
b) Values of the angles corrected for spherical excess.
75
B
c
A
Given:
A = 5830'30"
B = 6417'25"
C = 5712'17"
c = 36,285.55m
R = 6,372,160m
a
b
C
Solution:
a)
Determining spherical excess:
By sine law:
a/Sin A = c/Sin C
a = c Sin A/Sin C
a = 36,285.55(sin5830'30")/Sin5712'17"
a = 36,808.07m.
Area = (1/2)acSin B = (1/2)(36,808.07)(36,285.55)(sin6417'25")
= 601,690,565.40 sq.m.
e" = Area/R2 sin01" = 601,690,565.40/(6,372,160)2sin 01"
e" = 3.06"
Correcting the angles for spherical excess:
Corr. = e"/3 = 3.06"/3 = 1.02"
A = 5830'30" - 1.02" = 5830'29.98"
B = 6417'25" - 1.02" = 6417'23.98"
C = 5712'17" - 1.02" = 5712'15.98"
Note:
18000'09.94"
The spherical excess is always subtracted. The remaining 09.94" can be considered random which
is to be compensated in the subsequent adjustment of the triangulation.
WORKSHEET NO. 40
Spherical Excess
Name:
Date Submitted:
Course/Year:
Calculate the spherical excess in a triangle two sides of which are 3,950.55m
and 6,245.25m with an included angle of 4330'. Assume the average radius of curvature
to be 6,372,160m.
76
WORKSHEET NO. 41
Spherical Excess
Name:
Date Submitted:
Course/Year:
For a given triangle ABC, the observed interior angles are A = 5730'25", B =
6547'35", and C = 5642'16". The distance from A to B as determined by EDM
equipment is 25,485.65m. Compute the spherical excess in the triangle and correct the
observed angles for spherical excess. Assume the earth's mean radius of curvature to be
6,372,160m.
77
WORKSHEET NO. 42
Spherical Excess
Name:
Date Submitted:
Course/Year:
Compute the number of square kilometers in a triangle on the surface of the
earth which has a spherical excess of 1.75 seconds. Use 6,372,157m as the mean radius
of curvature of the earth.
78
Lesson 22 - Arc and Time Measure
Arc and Time Conversion Table:

'
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
79
h.m.
m.s.
0 04
0 08
0 12
0 16
0 20
0 24
0 28
0 32
0 36
0 40
0 44
0 48
052
0 56
1 00
1 04
1 08

'
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
h.m.
m.s.
3 04
3 08
3 12
3 16
3 20
3 24
3 28
3 32
3 36
3 40
3 44
3 48
3 52
3 56
4 00
4 04
4 08

'
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
h.m.
m.s.
6 04
6 08
6 12
6 16
6 20
6 24
6 28
6 32
6 36
6 40
6 44
6 48
6 52
6 56
7 00
7 04
7 08

'
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
h.m.
m.s.
9 04
9 08
9 12
9 16
9 20
9 24
9 28
9 32
9 36
9 40
9 44
9 48
9 52
9 56
10 00
10 04
10 08

'
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
h.m
m.s.
12 04
12 08
12 12
12 16
12 20
12 24
12 28
12 32
12 36
12 40
12 44
12 48
12 52
12 56
13 00
13 04
13 08

'
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
h.m.
m.s.
15 04
15 08
15 12
15 16
15 20
15 24
15 28
15 32
15 36
15 40
15 44
15 48
15 52
15 56
16 00
16 04
16 08

'
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
h.m.
m.s.
18 04
18 08
18 12
18 16
18 20
18 24
18 28
18 32
18 36
18 40
18 44
18 48
18 52
18 56
19 00
19 04
19 08

'
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
h.m.
m.s.
21 04
21 08
21 12
21 16
21 20
21 24
21 28
21 32
21 36
21 40
21 44
21 48
21 52
21 56
22 00
22 04
22 08
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
1 12
1 16
1 20
1 24
1 28
1 32
1 36
1 40
1 44
1 48
1 52
1 56
2 00
2 04
2 08
2 12
2 16
2 20
2 24
2 28
2 32
2 36
2 40
2 44
2 48
2 52
2 56
3 00
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
4 12
4 16
4 20
4 24
4 28
4 32
4 36
4 40
4 44
4 48
4 52
4 56
5 00
5 04
5 08
5 12
5 16
5 20
5 24
5 28
5 32
5 36
5 40
5 44
5 48
5 52
5 56
6 00
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
7 12
7 16
7 20
7 24
7 28
7 32
7 36
7 40
7 44
7 48
7 52
7 56
8 00
8 04
8 08
8 12
8 16
8 20
8 24
8 28
8 32
8 36
8 40
8 44
8 48
8 52
8 56
9 00
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
10 12
10 16
10 20
10 24
10 28
10 32
10 36
10 40
10 44
10 48
10 52
10 56
11 00
11 04
11 08
11 12
11 16
11 20
11 24
11 28
11 32
11 36
11 40
11 44
11 48
11 52
11 56
12 00
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
13 12
13 16
13 20
13 24
13 28
13 32
13 36
13 40
13 44
13 48
13 52
13 56
14 00
14 04
14 08
14 12
14 16
14 20
14 24
14 28
14 32
14 36
14 40
14 44
14 48
14 52
14 56
15 00
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
16 12
16 16
16 20
16 24
16 28
16 32
16 36
16 40
16 44
16 48
16 52
16 56
17 00
17 04
17 08
17 12
17 16
17 20
17 24
17 28
17 32
17 36
17 40
17 44
17 48
17 52
17 56
18 00
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
19 12
19 16
19 20
19 24
19 28
19 32
19 36
19 40
19 44
19 48
19 52
19 56
20 00
20 04
20 08
20 12
20 16
20 20
20 24
20 28
20 32
20 36
20 40
20 44
20 48
20 52
20 56
21 00
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
22 12
22 16
22 20
22 24
22 28
22 32
22 36
22 40
22 44
22 48
22 52
22 56
23 00
23 04
23 08
23 12
23 16
23 20
23 24
23 28
23 32
23 36
23 40
23 44
23 48
23 52
23 56
24 00
Time and Arc Relations:
24h = 360
1m = 15'
1 = 4m
h
s
1 = 15
1 = 15"
1' = 4s
Example 1
Change an arc measure of 8544'36" to time measure by the use of the Arc and Time
Conversion Table.
Solution:
85 = 5h 40m 00.0s
44' = 02m 56.0s
36" =
02.4s
h
m
Sum = 5 42 58.4s
WORKSHEET NO. 43
Arc and Time Measure
Name:
Date Submitted:
Course/Year:
Using the arc and time conversion table, perform the following conversions:
a) Express in arc measure a difference in longitude of 2h 39 m 44 s, 8 h 42 m 29 s, 14 h
52 m 12 s, 20 h 40 m 35 s, and 23 h 09 m 10 s.
80
b) Express in time measure a difference in longitude of 515'18", 1338'35",
12008'20", 14552'10", and 17958'16".
Check your answers by a calculator.
Example 2
The longitude of Washington is 7704'00" West and that of San Francisco is
12225'45" West. Determine the following:
a) Difference in solar time between Washington and San Francisco.
b) Time in Washington when it is 9:03:00AM (9h 03m 00s) at San Francisco.
c) Time at San Francisco when it is 7:54:30 PM (19h 54m 30s) at Washington.
Figure:
81
TSF (Time in San Fransisco)
TW (Time in Washington)
Difference In Longitude
122 25' 45" W
(Longitude of San Francisco)
Longitude of Greenwich
(Reference Meridian)
77 04' 00"W
(Longitude of Washington)
Solution:
a) Diff. = Long. of San Francisco - Longitude of Washington
= + 12225'45" - (+7704'00")
= 4521'45" arc measure or 3h 01m 27s
Note: The difference in solar time between any two points on the earth's surface is
equal to the difference in their longitudes.
b) TW = T  Diff. = 9h 03m 00s + 3h 01m 27s
= 12h 04m 27s or 04m 27s
= 0:04:27 PM
Note: The solar time difference is added since Washington is East of San Francisco.
c) TSF = TW  Diff. = 19h 54m 30s - 3h 01m 27s
= 16h 53m 03s
= 4: 53: 03 PM
Note: The difference in solar time is subtracted since San Francisco is West of
Washington.
WORKSHEET NO. 44
Longitude and Time
Name:
Date Submitted:
Course/Year:
The longitude of Bogota, Colombia is 4h 58 m 29 s West and that of San Jose,
Nueva Ecija is 8 h 20 m 18 s East. Determine the time at San Jose when the time in Bogota
is 23 h 08 m 45 s on March 18, 1986.
82
Example 3
What is the Greenwich civil time (GCT) when at a given instant the standard
or mean time at:
a) 120 East longitude is 4:45 PM?
b) 120 East longitude is 9:30 AM?
c) 90 West longitude is 3:15 AM?
d) 75 West longitude is 4:45 PM?
e) 150West longitude is 10:05 PM?
Solution:
83
a)
tm = 4:45 PM or 1645 H
= 16h 45m or 16.75h
(Mean or standard time at 120 E longitude)
TZC = 120(1h/15) = 8h
(Time zone correction, 15 = 1h)
GCT = tm  TZC = 16.75h - 8h = 8.75h or 8:45 AM(Equivalent Civil Time on the same day)
Note: The TZC is subtracted since the Greenwich central meridian is 8 hours behind that of 120 East longitude.
b)
tm = 9:30 AM or 0930H
= 9h 30m or 9.50h
TZC = 120(1h/15) = 8h
GCT = tm  TZC = 9.50h - 8 h = 1.50 h or 1:30 AM
Note: The TZC is subtracted since the Greenwich central meridian is 8 hours behind that of 120 East longitude.
c)
tm = 3:15 AM or 0315 H
= 3h 15m or 3.25h
TZC = 90(1h/15) = 6h
GCT = tm  TZC = 3.25 h + 6 h = 9.25 h or 9:15 AM
Note: The TZC is added since the Greenwich central meridian is 6 hours ahead of 90 West longitude.
d)
tm = 4:45 PM or 1645 H
= 3.25 h + 6 h
= 16 h 45 m or 16.75 h
TZC = 75(1h /15) = 5 h
GCT = tm  TZC = 16.75 h + 5 h = 21.75 h or 9:45 PM
WORKSHEET NO. 45
Greenwich Civil Time
Name:
Date Submitted:
Determine Greenwich Civil Time when it is:
a) 8:35 AM at 90 West longitude.
b) 10:08 AM at 120 East longitude.
c) 9:39 PM at 150 East longitude.
84
Course/Year:
d) 11:53 PM at 150 West longitude.
e) 7:00 PM at 145 West longitude.
WORKSHEET NO. 46
Standard Time and Local Time
Name:
Course/Year:
Date Submitted:
Calculate the following conversions:
a) The longitude of Naguilian is 12025'15". What is the standard time (at 120 East)
when the local time is 16h 19m 47?
85
b) If the longitude of a place is 12513'15", what is the local time at the place when
the standard time is 13h 42m 17s on 13 June 1986?
WORKSHEET NO. 47
Greenwich Mean Time
Name:
Course/Year:
Date Submitted:
The longitude of Baguio is 8h 02 m 21s East of Greenwich, England.
Determine Greenwich Mean Time (GMT) when:
a) the standard time in Baguio is 15h 18m 30s.
86
b) The Baguio local mean time is 15h 39m 25s.
Lesson 23 - Astronomical (PZS) Triangle
A spherical triangle is the figure formed by joining any three points on the
surface of a sphere by arcs of great circle. The spherical triangle PZS in the figure is
called the astronomical triangle.
87
Parallactic Angle
Polar Distance (p)
Co-latitude
Bearing of Body
P
S
t
Zenith Distance (Z)
Declination (d)
Z
O
A
Latitude (Z)
M
Celestial Equator
Where:
P - the north pole of the celestial sphere.
Z - the observer's zenith which is a point on the celestial sphere found by
projecting the center of the instrument at the time of observation upward along the
direction opposite to that of gravity.
S - the heavenly body observed which may be the sun or any other known star
such as Polaris.
PS or p - the polar distance or codeclination. It is equal to 90 minus the
declination (d) of the observed heavenly body (S).
PZ or y - the colatitude of Z. It is equal to 90 minus the latitude (I) of the
observer.
ZS or z - the zenith distance or coaltitude of the observed heavenly body (S).
It is equal to 90 minus the true altitude (h).
Z - true azimuth of the heavenly body. Its value may exceed 90 but is always
less than 180.
S - the parallactic angle.
t - the meridian angle.
Example #1
During an instant of observation at latitude N1810.1', the sun's apparent
declination and true altitude (already corrected for refraction and parallax) were recorded
as 1410.5' and 3250.2', respectively. Determine the length (in angular units) of the three
sides of the astronomical or PZS triangle.
88
Figure:
P (Celestial Pole)
Polar distance (PS or p)
co-latitude of Z (PZ or y)
S
Zenith Distance (ZS or z)
Z (observer's Zenith)
Given:
I = N 1810.1' (latitude of place of observation)
d = 1410.5' (sun's apparent declination)
h = 3250.2' (true altitude of sun)
Solution:
PS = (90-d)
= (90 -1410.5')
PS = 7549.5'
PZ = (90 - I)
= (90 - 1810.1')
PZ = 71 49.9'
ZS = (90 - h)
= (90 - 3250.2')
ZS = 57 9.8'
WORKSHEET NO. 48
Astronomical Triangle
Name:
Course/Year:
Date Submitted:
During an instant of observation at a place whose latitude is N3829'42", the
sun's apparent declination (d) already corrected for refraction and parallax, and true
89
altitude (h) were determined to be N1724'29" and 3255'30", respectively. Determine the
length (in angular units) of the three sides of the astronomical triangle formed by the sun,
the observer's zenith, and celestial pole.
WORKSHEET NO. 49
True Altitude of Sun
Name:
Date Submitted:
90
Course/Year:
The mean altitude of the sun (h') after three sets of pointing measured 5222'.
If the observation was made at an elevation of 487m when the average temperature was
28C, determine the true altitude (h) and zenith distance (z) of the sun.
WORKSHEET NO. 50
Apparent Declination of Sun
Name:
Date Submitted:
91
Course/Year:
For zero hours (0h) GCT, June 19 and 20, 1977, the sun's apparent
declinations are N2325.1' and N2325.9', respectively. If the difference in declination
for one hour between the two given periods is 0.03', what is the sun's apparent declination
and polar distance if the observation was made at 9:35 AM, 19 June 1977 along 120
East longitude.
Lesson 24
Sun's Azimuth
Using the fundamental equations in spherical triangle may solve the
astronomical or PZS triangle.
92
P
y = 90- l
Z
p = 90-d
z =90h
S
From Cosine Law of Spherical Trigonometry:
Cos a = Cos b Cos c + Sin b Sin c Cos A
Cos (90-d) = Cos (90 - l) Cos (90-h) + Sin (90-l) Sin (90-h) CosZ
Sin d = Sin l Sin h + Cos l Cos h Cos Z
or:
Cos Z = (Sin d - Sin l Sin h)/(Cos l)(Cos h)
Cos Z = Sin d/(Cos l)(Cos h) - (Sin l)(Sin h)/(Cos l)(Cos h)
Cos Z = Sin d/(Cos l)(Cos h) - Tan l Tan h
AlternativeEquations For Determining Z:
Cos (Z/2) =  (Cos s)[Cos (s-p)]/(Cos l)(cos h)
Sin (Z/2) =  Sin (s-h) Sin (s-l)/(Cos l)(Cos h)
Tan (Z/2) = Sin (s-l) Sin (s-h)/(Cos s)[Cos (s-p)]
Sin (Z/2) =  Cos 1/2 (z + l + d) Sin 1/2 (z + l - d)/(Sin z) (Cos l)
Determining True Bearing of Sun:
Reference Formula: Cos Z = (Sin d - Sin h Sin l)/(Cos h Cos l)
A) MORNING OBSERVATION:
N
N
True Meridian
93
Sun
Z=
Z
W
E
W
(Instr. Sta) O
E
(Instr. Sta.) O


S
Sun
CASE I
S
CASE II
a) Value of Cos Z is negative.
b) Angle Z represents azimuth of sun
measured clockwise from north with a
value ranging from 90 to 180.
c) The sun is east of the meridian and falls
somewhere in the southeast quadrant.
d) Where  = 180-Z, and true bearing of
sun is S  E.
a) Value of Cos Z is positive.
b) Angle Z represents azimuth of sun measured clockwise
from north with a value ranging from 0 to 90.
c) The sun is east of the meridian and falls somewhere in
the northeast quadrant.
d) Where  = Z, and true bearing of sun is N  E.
B) AFTERNOON OBSERVATION:
Sun
N
N
Z=
True Meridian
Z
W
E
W
E
O (Instr. Sta)
O (Instr. Sta.)
S
N
Sun
CASE III
CASE IV
a) Value of Cos Z is negative.
b) Angle Z represents azimuth of sun
measured counter clockwise from north
with a value ranging from 90 to 180.
c) The sun is west of the meridian and falls
somewhere in the southwest quadrant.
d) Where  = 180 - Z, and true bearing of
sun is S  W.
a) Value of Cos Z is positive.
b) Angle Z represents azimuth of sun measured counter
clockwise from north with a value ranging from 0
To 90.
c) The sun is west of the meridian and falls somewhere in
the northwest quadrant.
d) Where  = Z, and true bearing of sun is N  W.
WORKSHEET NO. 51
True Directions
Name:
Date Submitted:
94
Course/Year:
Using the given values in a problem in worksheet no. 40, and assuming that
the clockwise horizontal angle from the line to the sun is 27515'35", determine the true
bearing of the sun and of the line if the observation was done in the morning. Employ
each of the following formulas for determining azimuth of the sun: a) cosine formula, b)
sine formula, and c) tangent formula.
WORKSHEET NO. 52
True Directions
Name:
Date Submitted:
95
Course/Year:
A solar observation is made in the afternoon and the following quantities have been
determined:
N1405' = latitude of place of observation (l)
4032' = true latitude of sun (h)
N0723' = sun's apparent declination (d)
The instrument is set up at point A and mark is sighted at point B. If the measured horizontal clockwise
angle from the line to the sun were recorded as follows: 22208', 22212', 22209', 22213', 22216', and
22220', determine the following:
a) the sun's azimuth from north
b) true bearing of the sun
c) true bearing of line AB.
WORKSHEET NO. 53
Solar Observation
Name:
Date Submitted:
96
Course/Year:
Given the following data for sightings made on the sun:
N 1722' = latitude of place of observation
23.89C = temperature during observation
1682.317m = elevation of place of observation
120east = meridian of place of observation
22 Jan 77 = date of observation
TELESCOPE POSITION
HORIZONTAL ANGLE
VERTICAL ANGLE
TIME
(Clockwise from mark to the sun)
Direct (Normal)
5754.0'
4348.0'
5812.5'
4325.8'
5822.0'
4315.9'
Reversed (Plunged)
5906.8'
4248.9'
5916.3'
4235.2'
5938.2'
4223.6'
The transit is set up at point A and a mark is placed on B. Determine the following:
a) Mean horizontal angle from mark to sun.
b) Mean vertical angle to sun.
c) Mean time at place of observation.
d) GCT when observation was made.
e) Sun's apparent declination on date and time of observation.
f) True altitude of sun corrected for refraction and parallax.
g) The value of angle Z and the sun's true bearing.
h) True bearing of the line AB.
Lesson 25 - Hydrographic Surveying
Locating Soundings:
The following are the various methods used for locating soundings:
1. Time interval along a range line.
2. Range line and an angle from shore.
97
9:20 AM
9:30 AM
3. Intersecting range lines.
4. One angle and stadia distance from shore.
6. Two angles from shore.
Scow Measurement:
The quantity of material dredged from any body of water can be determined either by
soundings or scow mwasurements.
Waterline of Loaded Scow
Waterline of unloaded or light Scow
Ld
Wd
Dredge Mtls. Loaded on Scow
Ls
Deck of Scow
Du
Dl
Lb
Lu
Ll
Formulas:
Vw = (1/2)(Lu + Lb)(Du)(Wd)
Disp = (1/2)(Ll + Lb)DlWd
Ww = (Disp - Vw)(WtDw)
Vm = Ww/WtDm
equation 1
equation 2
equation 3
equation 4
Where:
Lb = Length of scow bottom
Ld = Length of scow deck
Wd = width of scow deck
Ls = length of vertical side of scow
Ll = length of waterline when scow is loaded
Lu = length of waterline when scow is unloaded or ligth
Du = draft of unloaded or light scow
Dl = draft of loaded scow
WtDw = weight density of the water.
WtDm = weight density of the material loaded on the scow.
Notes:
1. Use equation 1 to determine the volume of water (cu.m.) displaced when the scow is
unloaded or light.
2. Use equation 2 to determine the displacement loaded (cu.m.).
3. Use equation 3 to determine the weight of the water displaced (kg) by the load.
4. Use equation 4 to determine the volume of material (cu.m.) loaded in the scow deck.
Example #1
A rectangular deck scow 30.50 m. long, 6.10 m. wide, and 3.66 m. high has a draft of 3.05 m
when loaded. The bottom length of the scow is 23.15 m. The waterline is 29.26m long when the scow is
loaded with rocks and 25.60 m. long when light. If seawater weighs 1026 kg/cum. and the loaded dredged
material weighs 3208 kg/cum., determine the following:
a. volume of water displaced when unloaded or light
b. displacement loaded
98
c. weight of the water displaced by the load
d. volume of the loaded rock.
Figure:
Waterline of Loaded Scow
Waterline of unloaded or light Scow
30.50
6.10m
Dredge Mtls. Loaded on Scow
3.66
Deck of Scow
1.22m
23.15m
25.60m
29.26m
Solution:
a) Vw = (1/2)(Lu + Lb)(Du)(Wd
= (1/2)(25.60 +23.15)(1.22)(6.10)
Vw = 181.40 cu.m.
b) Disp = (1/2)(Ll + Lb)DlWd
= (1/2)(29.26 + 23.15)(3.05)(6.10)
Disp = 487.54 cu.m.
d)
Ww = (Disp - Vw)(WtDw)
= (487.54 - 181.40)(1026)
Ww = 314,099.64 kg.
e) Vm = Ww/WtDm
= 314,099.64/3208
Vm = 97.91 cu.m.
WORKSHEET NO. 54
Three-Point Problem
Name:
Date Submitted:
99
Course/Year:
3.05
In the accompanying figure, triangulation stations A, B, and C are observed from P, a
hydrographic station whose location is to be established by the principle of the three-point problem. The
known data are: the angle BAC (the exterior angle) = 10245'20", alpha (angle BPA) = 2634'50", beta
(angle APC) = 4415'15", b or side AC = 6883.4m, and c or side AB = 6605.3m. By the analytical method
determine the values of X, Y, d, e, and m which are needed to locate P.
B
C
X
c
b
Y
A
d
e
m
P
WORKSHEET NO. 55
Discharge by Velocity-Area Method
Name:
Date Submitted:
100
Course/Year:
At a certain section of a river the left and right water edges are 3.0 and 45.0 meters
respectively from an initial reference point. Verticals are located at distances 7.0, 12.0, 16.5, 19.5, 23.0,
27.0, 32.0, 35.0 and 39.5 meters from the initial point. Depths of verticals are 1.8, 3.5, 4.6, 3.4, 5.8, 6.6, 5.7,
3.6, and 1.2 meters. Mean velocities in the verticals are 0.15, 0.30, 0.45, 0.50, 0.48, 0.55, 0.54, 0.48, and
0.20 meters per second, respectively. Determine the following:
a) Cross sectional area of the river (sq.m.)
b) Discharge of the river (cum/sec)
c) Average velocity of flow of the river (m/sec)
Sketch the cross section of the river. Assume that the discharge in the end zones to be
negligible or zero.
WORKSHEET NO. 56
Capacity of Reservoir by Contours
Name:
Date Submitted:
101
Course/Year:
A hydrographic survey of a lake produced the following approximate data:
13,340 sq.m. = area enclosed by the water line
8,550 sq.m. = area enclosed by contour 1
5,149 sq.m. = area enclosed by contour 2
2,088 sq.m. = area enclosed by contour 3
1,975 sq.m. = area enclosed by contour 4
If the vertical distance between contour levels is 3.0 meters determine the approximate total
volume of the lake above the level of contour 4.
Bibliography
102
Juny Pilapil La Putt. Elementary Surveying. Third Edition. Baguio City: Baguio
Research and Publishing Center. 1987
Juny Pilapil La Putt, Surveying Laboratory Manual. Baguio City: Baguio
Research and Publishing Center. 1985
Juny Pilapil La Putt. Higher Surveying. Baguio City: Baguio Research and
Publishing Center.
Venancio I. Besavilla Jr. Theory and Practice in Surveying. Cebu City: VIB
Publisher. 1981
103