survey review & problems part i

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~ X A M PR~PARATION

CIVIL ~NG IN[;~RINGSURV l;Y R !;V I(;W

Third ~dition

Jack Liu, PE Civil Eng.

~ AEC EDUCATION



This publication is designed to provide accurate and authoritative information inregard to the subject

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08 09 10 L O 9 8 7 6 5 4 3 2

j



Preface_ ..__ _ "",,,,,,-_._---_."-_

The California Special Civil Engineer Exam is a four-hour examination consisti

of two sections, Seismic Principles and Engineering Surveying. As of Decemb

2005, the Test Plan for the surveying portion cites four primary content areas:

• Engineering surveying equipment and field activities (15%)

• Engineering surveying field measurements (11%)

• Engineering surveying calculations (33%)

• Engineering surveying office procedures (41%)

The Test Plan further specifies knowledge areas within these four conte

areas. For optimal preparation, exam candidates can obtain these details fro

the California Board for Professional Engineers and Land Surveyors

www.dca.ca.gov /pels/e jplan-civsurvey.pdf.

Although the survey exam is open book, there is no time to read referen

material in detail during the examination hours. The examination is very fa

paced. Solving the problems in this book is a good method for preparing the exa

Read the review text and follow through the solved examples carefully. Then so

the sample problems at the end of each chapter without first looking at

solutions.

http://www.dca.ca.gov/

http://www.dca.ca.gov/



TABLE of CONTENTS

PREFACE i

TA BLE of C ON 1EN TS iii

1.HORIZONTAL CURVE 1·1

S am ple Problem s 1 1·18

2. V ER TICAL CURVE 2·1

S am ple P roblem s 2 2·13

3.1RAVERSE 3·1

S am ple P ro blem s 3 3·14

4 .AREA 4-1

S am ple P roblem s 4 4-15

5. TOPOGRAPlDC SURVEY 5·1

Sam ple Problem s 5 5·8

6 . PHOTOORAMMETRY 6 - 1

7. CONSTRUCT ION SURVEY 7~1

Sam ple Problem s 7 7-7

8 . LEVELING 8·1

Sam ple Problem s 8 8 · 6

9. E NG IN EE RIN G PRACT IC E 9-1

Sam ple Problem s 9 9-8

ANSWERS to Sample Problems A-I

lNDEX



50 100

There are two types of horizontal curves: Circular Arcsand Spirals. A circular arc connecting two tangents is a simplecurve. Easement curves are used to lessen the sudden change incurvature at the junction of a tangent and a circular curve.A spiral makes a good easement curve.

There are two definitions of 'Degree of Curve':The CHORD definition is used in railroad practice.The ARC definition is used for highway work.

l~a f t

R R

Figure

CHORD definition ARC definition

R ;: =

sin (0/2) 27tR

R = (Eq.1-2)5729.58

The DEFLECTION ANGLE formed by a tangent and chord is equalto one-half the angle of the intercepted arc. This isillustrated in the following figure:

o 1-2



1. HORIZONTAL CURVE

Symbols

PI Point of Intersection.I Intersection angle at PI (= Central angle).T Tangent distance from BC to PI (or from EC to PI).E External distance, distance from PI to midpoint of curve.M Middle ordinate, distance from midpoint of curve to

midpoint of long chord.L Length of curve, distance from BC to EC along the arc.BC Beginning of Curve. (also termed PC)EC End of Curve. (also termed PT)o Radius Point (center of radius).LC Length of the Long Chord.R Radius of the curve

Figure I - A

0

T = R tan (1/2)

L = R I (1 in radians)

LC = 2 R sin (1/2)

R I I

= cos . E = R (sec - 1),R + E 2 2

R - M I I

= cos i M = R (1 - cos -) (Eqs.1-1)R 2 2

1-1



Example 1-1

Given a curve with a radius R = 350 ft, calculate thedeflections to each full station and the E.C.(End of Curve).

The B.C. (Beginning of Curve) is station 12+13.22

Station Deflection

B.C. 12+13.2213+0014+0015+00E.C. 15+19.96

solutions:L L

R = =

a. Arc length = L =9 :::

27tR

DO 1 3600L 900L 90l1L

b. Deflection = = ( ):::

=2 2 21tR }(R 3507(

The deflections to each station:

Station L (ft) Deflection

B.C. 12+13.22 0.00 O ( )O O' OO . .

13+00 86.78 71 106' 11"14+00 186.78 15"17'17"15+00 286.78 23028'24"E.C. 15+19.96 306.74 25°06'25"

1-3



Example 1-2

An existing Sa curve connects B.C. and E.C. as shown in Fig.l-C.In order to join a proposed highway, the tangent of the curveis relocated. The new tangent line is 150 feet away, and isparallel to the existing tangent as shown in Figure I-C. Find

(A) the existing B.C., E.C. station.

(B) the radius of curvature for the new curve.

(C) the E.C., middle ordinate, and external distanceof the new curve.

(0) the station at the intersection of the existing curveand the new tangent.

Curve

B . C ..---_ . . . . . . . . .

Ne" Curve --~

New Tangent

Figure l-CSolutions:

( A) E xi st in g curve, from Eq.(1-2),

5729.58 5729.58R :: ::1145.916'::

T :: R tan (I/2) = (1145.916) tan(6if/2) = 661.6'

o 0

L = 6. (R) :: 2lC (60/360) (1145.916) = 1200'

Station:

B.C. = P.1. T :: (66+00) - (6+61.6) = I 59+38.4 IE.C. = B. C. + L :: (59+38.4) + (12+00) = I 71+38.4 I

1-4



(B) New curve

From Fig.l-D,

sin 60' = l50'/(A to P.I.)

A to P.I. = 173.205'

New Tangent Tn = T - (A to P. I. )

Til = = 651.6 173.205 = = 488.4'

Tn 488.4New Radius Rn = = =

tan (1/2) tan 30°845.93 ft

(C) New curveJ\ 0 0

Ln = L . : : : : . (Rn) = 27((60/360)(845.93) = 885.856'

New E.C. = B.C. + Ln = (59+38.4) + (8+85.86)

= 68+24.261

I

M idd le or din at e, M = Rn(l - cos ---)2

(Eqs.l-1)

= 845.93 (1 cos 30°) = 113.33 it

I

E xt er na l d is ta nc e, E = Rn(sec - 1)2

1

= = 845.93 ( - 1) = =

cos 300130.87 ft

1-5



B e

Existing Curve

New Cur-v.

R = 1 1 4 S . 9 1 6 f t

(D) Station K (at the intersection of the existing curveand the new tangent)

R - 150 1145.916 - 150cos Of = " " " * « x » 29.646(1=

R 1145.916

Curve length from K to E.C.

LK = 0( (R) = 2]((29.646°/360°)(1145.916) :< 592.91'

Station K = E.C. - L K = (71+38.4) - (5+92.91)

=65+45.5

Example 1-3

A highway center line begins at point 'A' (B.C.) which is duesouth of the center of a lOOO-foot radius curve, thence alongthe curve concave to the northwest to point 'B' (P.R.C.),

thence on a curve concave to the south (radius 600 ft) to point'c' (E.C.), station 104+00, thence on a tangent to point '0',station 109+00. The bearing of C.D. is S 75~E and point '0'

is directly east of point 'A'.

REQUIREMENT

It is proposed to relocate the highway center line to a newl oc at io n w hi ch is a straight line between paints 'A' and '0'.what is the length AD of this new alignment?

1-6

I



C IA::

OB::

1000' (radius)

CF :: BF :: 600' (radius)

CD = (109+00) - (104+00)

= 500'

OF = OB + SF = 1600'

DOA

x =

y = Of

z = fA

~-C=- __-*__~__~ ~_D

~I

solutions:

AD=x+y+z

a. Find x

E

z

x = CD/(cos 15Q) = 500 /(COS 15°) = 517.638'

b. Find y

CO = CD tan 15° = 500(tan 15°)= 133.975'

Y = (CF - CO) sin ISO = (600 - 133.975) sin 15° = 120.616'

c. Find z

_2 2.-'2

QE + z = OF

OE = QA + AE i OF = OB + BF

AE = Ff = y/(tan 15°): 120.616 /(tan 15°)= 450.145'

- -2 1 -2-( OA

+ AE) +Z = (

OB + BF )(1000 + 450.1451 + z~ = 1600

2

z = 676.076'

d. AD = x + y + z = 1314.33 ft

1-7



Example 1-4

Line ABCD in the Figure below represents a center line of asurvey for a certain part of a construction project. Curve B e

has a radius of 500', with center at o . A no th er c on st ru ct io nline EG, intersects line ABCD at F.

The elevation of B is 100 ft and the grade toward C is + 1.0 %.The elevation of E is 93 ft and the grade toward F is + 0.5 %.

All other data is shown on the figure.

REQUIRED

What is the difference in elevation between the twoconstruction lines at point F?

D

C : X = 4 4 1 . 5 2

Y = 5 2 8 . 3 1

N

A

N)eo

\ Bg,/,'

B", \X=15~. "" \

Y = 3 2 1 1 J . <, \

",\

'~O : X = 5 6 9 . 3 4

y= 47.68

w E

s

E : X = 7 9 9 . 1 5

Y = 2 4 0 . S 2

1-8



D

o 001t N~l'

V - t E

sli78°

..I.v \ 88 '11

o~<,

B <,

, g > X = l S 8 . <,

~<, E:X=799.15

Y = 3 2 B .Y = 2 4 B . S 2

A

East

O : X = 5 6 9 . 3 4

y= 4 7 . 6 8

Figure l-ESolutions:

(A) Calc ulate t he elevatio n of F on line BF:

The grade from B toward F is 1%, length BF = 500 ( L BOF)

a. Find L BOF

L BOF = 1800 - 33D - L FOE - LEON

b. Find LEON-I

LEON = tan240.52 - 47.68 -I 192.84

= tan = 40°799.15 - 569.34 229.81

c. Find L FOE

L FO E = 1800 - L EFD - L FE D

L FEO = LEON + (900 - 70°) = 40° + 200 = 60'

1-9



FO = 500 ;

sin (L FOE)

FE

L FOE : 88.6936~;

::

sin ( L FEO)

FO

L FEO = 60 t1

d. Find L EFO ( Se e Fig .1-E )

sin ( L EFO) sin ( L FEO)In ~ FOE, = =

OE FO

DE = = )(229.81/ + (192.84l = 300 . FO = 500,

sin (L EFO) sin (60(1)Therefore, =

300 500

L EFO = 31.3064°

e. Calculate the elevation of F on line BF:

From the above results,

L FOE = 1800 - L EFO - L FED = 88.6936(1

L BOF = 180~ 33° - L FOE - LEON = 18.3064°

SF = 500 (L BOF) = 500 (18.306i')(21()/360(l= 159.75 ft

The grade from B toward F is 1%

The elevation of F on line BF:

100 + 0.01(159.75) = 101. 598 ft

(B) Calculate the elevation of F on line EF:

In LFOE,

Therefore, FE = 577.2 ft

The grade from E toward F is 0.5%

The elevation of F on line EF:

95.886 ft _ ,93 + 0.005(577.2) =

(C) the difference in elevation between line BF and line EFat point F is

101.598 - 95.886 = 5.712 it

1-10



Example 1-5

The centerline of an aqueduct was originally laid out as areversed curve as indicated by the existing ~ in the planbelow. A construction project in the vicinity requires theaqueduct to be realigned. The new alignment will connect the

existing tangents with a 1200 ft radius curve. This realign-ment will move the curve part of the ~ back away from theproposed project and it will also replace the reversed curvethat now exists.

REQUIREMENT

(A) What is the central angle of the new curve?

(B) Locate the B.C. of the new curve using the existing B.C.as the point of reference.

(C) Locate the E.C. of the new curve using the existing E.C.as the point of reference.

(0) What is the saving in distance between the new and the oldalignments? Base your calculations on the distance from theB.C. of the new curve to the E.C. of the existing alignment.

/'.,/

/

rNa" AIIgnment

R = 1200-

LAqueduct ExIst Ing I t .

Curve 1

/ t:Curve 2

--..,L_ -...---- /

, _ . . . . - /;"""

RI= 400 ' //

~/ I

Tangent

E . e .

1-11



Existing B.C.: point G; Existing E.C.: point F

R,= GO = 400'; R,= OF = 300'; 00 = 400' + 300' = 700'

0: Center of radius 300'

0: Center of radius 400'

r = LQOG = 98Q30'

L2 = 200' = R1(q)

q = L FQO

y

o

solutions:Figure 1-F

- -(A) Draw DE /I QF

d = The central angle of the new curve = r-q

LZ 360 200 360

q = - ( ) = ( ) = 38° 11. 8'

R2, 2Jt 300 27C.

60°18.2'

1-12

B

x



(B) New curve, T = R tan(o/2) = 1200(tan 30°9.1')= 697.062'

From the new B.C. to the existing B.C. = T - AG

and AG = OP - QK - FD (See Fig.1-F)

a. Find QP (See Fig.1-G)

OP = aO sin ( L OOP) i LOOP =

x;,

OP = 700 sin(81.So)= 692.311'

r = L OOG = 98°30'

b. Find OK (See Fig.1-G)

OK = OF cos ( L FOK) i

o p

F19IN l-tr

L FQK = q LOOK

L OQK = r - 90° = 8°30', L FQK = q 8°30' ~ 29Q41.8'

OK = 300(cos 29°41.8')= 260.598'

c. Find FD (See Fig. I-F)

FD = AD tan ( L BAD) ;

AD = R,+ OP + KF

O F = 00 cos(L OOP) = 700 cos(81.50)= 103.467'

KF = OF sin(L KOF) = 300 sin(29°41.8')= 148.622'

Therefore, AD = 400 + 103.467 + 148.622 = 652.089'

FD = 652.089 (tan 29°41.8') = 371.895'

d. AG = OP - OK - FD = 692.311 260.598 - 371.895 = 59.818'

From the new B.C. to the existing B.C.

637.24 It )= T - AG : 697.062 - 59.818=

1-13

b



AF= = = 750.684'

(C) From the new E.C. to the existing E.C. = AF - T (See Fig.1-F)

AD 652.089

cos ( L BAD) cos (29~41.8')

AF - T = 750.684 - 697.062 = 53.622 ft

(D) Distance saving:

(Base on the distance from the new B.C. to t he e xist in g E.C.)

a. New curve

L3= 1200(a) = 1200(21C)(60.:f) /360Q= 1262.92'

From the new B.C. to the existing E.C.

= (New curve L3) + ( Ne w E .C . to existing E.C. )

= 1262.92 + 53.622

eL,= 400(r) = 400(21f..)(98.5) /360'J= 687.66'

= 1316.54 ft

b. Existing c ur ve

From the new B.C. to the existing E.C.

= ( Ne w B .C . to G ) + (Existing curve L,) + L2

= 637.24 + 687.66 + 200

= 1524.9 ft

c. Saving in distance

1524.9 - 1316.54 = I 208.36 ft

1-14



Example 1-6

A loop ramp at an interchange was designed as a broken-back curve.See Fig.l-H. The short tangent section is to be eliminated bycompounding the two given curves with a curve of 200' radius.

REQUIRED

Calculate the stationing of the points of compoud curvature(P.C.C.), and show the equation in stationing at the secondP . c . c .

, . _ _---- . . . . . . . . . - . . . . . . . . . . . . . . -......_----------oE-

------.----------

Figure I-H

1-15



Solutions:

(A) Stationing along broken-back curve.

a. Curve of 100' radius

Intersection angle I ~ 180Q - 8920' - 52°30' = 119Q10'

L = I (R) = 21(119.1670/360°)(100) = 207.99'

E.C.1 = (1+00) + (2+7.99) = 3+7.99

b. Curve of 150' radius

B.C.2 = (3+7.99) + (0+50) ~ 3+57.99

(B) Stationing the points of the compound curve:

B. Draw the following figure

In order to calculate the stationing of the P.C.C.,angles p and q are needed.

1-16



Given: a = 50'b = 100'c = 50(j2) = 70.7106'

b. Find angles A and B

Law of Cosines ~ 2 2100 + 70.7106 - 50

cos A =

2 b c 2 (100) (70.7106)

Law of Sines

a c 50 70.7106

= =sin A sin C sin 27.886~ sin C

~ C = 41.409° :: 41()24'34"

B = 1800 - A C = 1101)42'18"

c. Find curve length L( , L2

p - 1801)

q = 1800

L I ;: P (1 00) ;: 2 Tt ( 17. 114 q /360) (100) = 29. 87'

L2 = q (150) = 27C.(24.295~/36o")(l50) = 63.604'

d. Fin d sta tio n F, and G along broken-back curve

Station F = E.C.l - L, = (3+7.99) - (0+29.87)

= 2+78;12

Station G = B.C.2 + L1 = (3+57.99) + (0+63.6)

= 4+21.59

e. Find station G along the compound curve

L3 = C (200) = 2K(41.4090/360j(200) = 144.55'

Station G = Station F + L3 = (2+78.12) + (1+44.55)

= 4+22.67

1-17



Sample Problems 1

Problem A

A horizontal curve is shown in the Figure below._FI islocated in the lake and inaccessible. Length of AB hasbeen found to be 614.7 5 ft.

(0) 4 Q 23'

REQUIRED

1. Determine the intersection angle at PI.

2. Find the radius of the horizontal curve (ft).

(F) 2031.30 (G) 2030.30 (H) 2030.03 (J) 2033.31

3. Determine the length of the horizontal curve (tt).

(A) 1445.74 (B) 1474.54 (e) 1447.54 (D) 1454.74

4. Find the PT station.

(F) 75+89.75

(H) 79+60.54

(G) 79+58.74

(J) 79+67.74

P o i n t A : S t a t i o n 6 9 + 7 5

1-18



Problem B

A highway right of way cuts across the west end of a propertyparcel. Half the width of the right of way is 50' as shown inthe sketch below. The radius to the highway curve ~ is 400'and the total central angle is 30°. The lot lines along thenorth and the south boundaries are parallel, are 100' apart.and lie on a line due East - West. The east property linelies on a bearing of N 300E as shown. The intersection ofthe east right of way line and the south property line be-comes the PC of a circular curve which will become the newwestern lot boundary.

1-19

E



REQUIRED

5. What is the central angle PC - 0 - 8 1

(A) 15.0° (8) 15.6° (D) 16.61 1(C) 16.0°

6. What is the arc length PC - B which becomes the newwestern property line?

(F) 101.0' (G) 101.4' (J) 105.1'(H) 103.5'

7 . What is the length of the north property line of thes ho rt en ed l ot ? (Be =?)

(A) 336.8' (B) 335.2' (D) 337.0'(C) 338.2'

8 . What is the length of the east property line? (CO=?)

(F) 115.5' (G) 114.5' (J) 116.5'(H) 111.5'

9. What is distance measured along the north property

line from the highway~ to the east R/W line? (EB =?)(A) 50.2' (B) 51.0' (e) 51.9' (0) 53.0'

Problem C

In the figure below, the indicated thus J1 are0

angles 90 .

10. Calculate the length of the line marked b (ft).

(A) 137.02 (B) 137.12 (C) 137.36 (D) 136.84

11. Calculate the length of the line marked a (ft).

(F) 146.80 (G) 146.98 (H) 146.33 (J) 146.19

150'

N o t t o Seale

1-20



Problem D

Line BCD in the Figure below represents a center line ofa survey for a certain part of a construction project.Curve B e has a radius of 100.5 ft, with center at O.Another construction line AS, intersects line BCD at B.

REQUIRED

12. Find the coordinates of point O.

(A) X = 159.92, y = 20.82 (B) X = 0.0, y = 0.0

(C) X := 34.10, y - 0.40 (D) X := 68.2, y := 0.2

13. Find the length OA (ft) .

(F) 255.47 (G) 265.53

14. Find the angle L BAO.

0(A) 14 12'35"

(C) 19°22'35"

15. Find the angle L ABO.

(F) 112°30'40"

0

(H) 122 30'11"

16. Find the Arc length Be ( ft ) .

(A) 126.58 (B) 129.12

(H) 317.83 (J) 279.95

(B) 14°35'12"

(D) 19045'12"

o(G) 52 19'45"

(J) 57°29'49"

(C) 121.38 (D) 122.74

AX : a - 6 2 . 7 1 Ny= 2 4 7 . 6 5

v 4 -Lu E. . .&

S. . . .0

n

z CD = 170.

ft ( S 70

0

10' E )

C

B

1-21



Problem E

A reversed curve is composed of two simple curves turningin opposite direction, as shown below. Both simple curveshave the same degree of curvature.

17. Find the radius (ft) of the simple curve.

REQUIRED

(A) 866.025 (B) 1082.53 (e) 2366.025 (D) 2957.532

18. Find the station of PC.

(F) 18+66.03 (G) 16+33.98 (H) 20+00.00 (J) 16+66.03

19. Find the station of PRe.

(A) 41+11.68 (B) 38+66.03 (C) 44+77.70 (D) 43+66.03

20. Find the station of PT.

(F) 57+16.55 (G) 51+04.88 (H) 50+00.00 (J) 53+50.53

2l. Find the station of PI2.

(A) 36+33.98 (B) 47+16.55 (e) 47+45.65 (D) 50+00.00

(Stat Ion

5+00

1-22



Problem F

The tangent of a horizontal curve is relocated 20' away, and isparallel to the existing tangent as shown in the Figure below.

REQUIRED

22. Find the radius of curvature R' for the new curve.

(A) 966.35' (B) 965.36' (C) 956.36' (D) 963.65'

23. What is the deflection angle from B.C. to the center ofthe new curve (Point A) ?

(F) 300

(G ) 60° (H) 15° (J) 90"

o to Scale

ANS l.A 2.G 3 . e 4.H 5.D6.G 7.C B . F 9 . e 10,B

I1.F 12.B 13.F 14.C IS.H16.A 17.C 18,G 19.A 20.J21.D 22.B 23.H

1-23



2. VERTICAL CURVE

Vertical curves are used to provide a smooth transitionbetween grade lines of a highway or railroad. An equal-tangentparabolic curve is illustrated in Fig. 2-A. It has the property

that the vertex (V) is midway between the beginning of thevertical curve (BVC) and its end (EVC) measured horizontally.Parabolic curves can be calculated by two different procedures:Tangent-offset method or Chord-gradient method.

T an ge nt-o ff se t m et ho d

Fig. 2-A shows: the center of a parabola is midwaybetween the vertex and the long chord. Parabolic curve hasthe property that offsets from a tangent to a parabola areproportional to the squares of the horizontal distances from

the point of tangency (See Eq.2-1)."

o : : : : : t CD 00 C \J+i C~ . . . . . IS l OJ c o C 0CD c .o (J) r" r" r-, r-, to to

"C \J (\J ru C\ J C\ J ru OJ ru ru (7\~. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C 0

V0 >

~X'

. . _~

lJJ

Chord _--Long _--

_ - B

_ __ -1 <

L

> 1 <L

> f2 c2 0

S lS I lS I lSI lSI ISl lSI lS I lSIIS l 11'\ ISl If) ISl If) lS I If) lSI -t--l+ + + + + + + + + 0.-t rl C\ J C\ J Cl") Cl") c : : t c : : t 1 1 ' \ +i. . . . . . . . . . .-t . . . . . . . . . . .-t . . . . . . ....-t . . . . . (.I)

Figure 2-ATerminology

2-1

L

g. ,g2-B v eE v eV

r

e

Xa

Horizontal Length of curve; in stations.Grade rates; in percentages.Beginning of the Vertical Curve.End of the Vertical Curve.Vertex; point of intersection of two tangents.Rate of change of grade; per station.Mi dd le or di na teHorizontal distance from BVC (or EVe) to point a.



Table 2-1 Equal-Tangent Vertical Curve

Station Point Tangent Tangent Elevation

Elevation Offset of Curve11+00 BVC 1264 0.00 1264.0011+50 1266 0.19 1265.8 112+00 1268 0.7 5 1267 .2512+50 127 0 1.69 1268.3113+00 V 1272 3.00 1269.0013+50 127 1 1.69 1269.3114+00 127 0 0.7 5 1269.2514+50 1269 0.19 1268 .8 115+00 EVC 1268 0.00 1268 .00

1. Tangent offset

Offset a Xa '1

= (--) (Eq.2-1)Offset V L/2

Calculate the tangent offset of Table 2-1.

(1/4)1- x 3 = 0.19 (here, offset V = 3)

(1/2)" x 3 = 0.75

(3/4)'1. x 3 = 1.69

9, - g22. Middle ordinate e = L (Eq.2-2)

8

4 - (-2)From Fig.2-A, e = 4 :: 3

8

1 Ysvc + Yeve

3. External distance AS = ( - Yv)2 2

1 1264 + 1268From Fig.2-A, AS :: ( - 1272 ) = 3

2 2

4. Offset V = 3 = e = AB (ckecked)

2-2



Chord-gradient method

The following Equations define Equal-Tangent Vertical curve.

rXl

y = Yeve. + gIX +

g2. - g,

(Eq.2-3)2

r = = (Eq.2-5)L

where Y is the curve elevation of any point P on the parabola,r is the rate of change of grade per station of the curve,g, the tangent grade through the BVC, and X is the horizontaldistance from the BVC to point P. The HIGH or LOW point on avertical curve will occur at the turning point where the tangentto the curve is horizontal and its slope is equal to zero. Theturning point is not located directly above or below vertex (V).It is obtained by equating to zero the first derivative of Eq.2-3:

dY= = r X + gl = = 0

dX-s, gj (L)

Turning paint, X = = (Eq.2-4)r 9, - 9z

For example, calculate X and y of Figure 2-A:

9" - 9, -2 - 4

r = = = = = -1.5L 4

At station 13+00

X = (13+00) - (11+00) = = 2+00

rX~ -1.5 (2 )2Y ::: Y e v e + g, X + = 1264 + 4(2) + = 1269

2 2

At the turning point,

-9, 9, (L) -4X = = = = 2.667

r9,-

92,

-1.5

r X' -1.5 (2.667 )2

Y = Ysv c + g, X + = 1264 + 4(2.667) +2 2

= 1269.33

The turning point is at station: (11+00) + (2+66) = 13+66

2-3



Example 2-1

A 600 ft vertical curve with grades g1 =(+)0.6% and g2 =(-)1.0%that intersect at station 42+7 0 with elevation 8 20.00 ft. Find

(A) the stations and elevations of the BVC and Eve.

(B) the station and elevation at the high point of the curve.

s"+Nq

L = 6~~ f t

Solutions:

( A ) B v e station:(42+70) - (L/2) = (42+70) - (3+00) = 39+70

B v e elevation:820 - g , (L/2) = 820 - 0.6(3) = I 818.2 ft IE v e station:(42+70) + (L/2) = (42+70) + (3+00) = I 45+70

E v e elevation:820 + 9

2(L/2) = 820 - 1.0(3) = 817.0 ft

(B) The high point station:g2. - g, -1 -0.6

From (Eq.2-5), r = = = -0.267L 6

x =-gI

e, (L) -0.6

= = = 2.25r -0.267

y = Y ave + g,x +

= I 818.87 ft

The high pOint is

= 818.2 + 0.6(2.25) +

1-0.267 (2.25)

22

(elevation)

at station:

41+95 I(39+70) + (2+25) =

2-4



Example 2-2

A vertical curve with grades g1 =(-)0.8 % and g2 =(+)1.0% that

start at station 42+7 0 with elevation 8 20.ft. At 48 +7 0, thereis an overpass with an underside level of 8 35.2 ft and thecurve is designed for a 15-foot clearance under the overpassat this pOint. Find the required curve length.

seve: t 82B itlid'

I:o-~D+-'(.;1

v c8 2 3 . 2 ft_

J, . . . . .+aJc : : : r

L = ?

Solutions:

(A) Method I ( Ch or d-g ra di en t m et ho d) :

g2- - g, 1 - (-0.8) 1.8From (Eq.2-5), r= = =

L L L

At station 48+70:

X = (48+70) - (42+70) = 6+00

rX2 1 1.82

Y = YaVG

+ g, X + = 820 + (-0.8) (6) + -. -( ) (6)

2 2 L

= 835.2 - 15. = 820.2 it

32.4~ 815.2 + ( ) = 820.2

L~ L = 6.48

The required curve length = 648 tt (longest)

2-5



(B) Method II (Tangent-offset method):

a. At station 48+70:

x = (48+70) - (42+70) = 6+00

Y = 835.2 - 15. = 820.2 ft

T an ge nt e le va ti on = Yavt + g,X

= 820 - 0.8(6) = 815.2 ft

Tangent offset = 820.2 - 815.2 = 5.ft

b. Ta nge nt o ff se t

Offset a2.

= ( - - )Xa

(Eq.2-1)Offset v L/2

Here, Offset V = e ( Mi dd le o rd in at e)

e =1-(-0.8) 1.8

c. M idd le or din ate L = L = L

8 8

d. Re quir ed c urv e length

Offset a Xa2

= ( ) (Eq.2-1)Offset V L/2

Here, Offset V = e = 1.8 L/8

5 6 l= (

1.8 L/8 L/2

40 144=

1.8 L L2-

~ L =6.48

The required curve length = 648 it (longest)

2-6



Determine (A) the longest length of the vertical curve.

(B) the station a nd e lev at io n of the low point onthe curve.

1 4 6 8 . S f t

fL/21 + H . 3 (L/21 - 3.3

Example 2-3

A vertical curve with grades gl =(-)3.0% and g2 =(+)2.0%that intersect at station 8 +50 with elevation 1450.06 ft.It is necessary to pass under an overhead structure atstation 8 +8 0 with a 16-foot clearance. The lowest point of

overhead struc,ture is 1468 .50 ft at station 8 +8 0.

a. Tangent offset

E v e

Solution:

(A) lSI

It1+CD

IS:!

OJ+CD

Offset a2.

= ( - - )Xa

. . .

(Eq.2-1)Offset V L/2

At station 8+80, d = (8+80) - (8+50) = 0.3, then

Offset AB L/2 - 0.3'2

Offset AB Offset V= ( ) => =

Offset V L/2 (L/2 - 0.3/ (L/2)2

Offset AC L/2 + 0.3 2 - Offset AC Offset V= ( ) ~ =

Offset V L/2 (L/2 + O.3/- (L/2/-

Therefore, =Offset V

(L/2/-

ACAB

= 2[(L/2) + 0 . 3 J[(L/2) - 0.3]"'

2-7



(A)

AB ACb. =

[(L/2) - 0.3)'2. «L/2)+

0.3)1

AB = 1452.5 - (1450.06 + 0.3Xg1) = 1.84'

AC = 1452.5 - (1450.06 + 0.3x~) = 3.34'

Substitue AS = 1.84', AC = 3.34' into the above equation,

=> 0.3 ( J]:84 + }3.34) = L ( )3.34 - .[1:84)/2

> L = 4.055

The longest length of the vertical curve:: 405.5 ft

(8)B. BVC station:

(8+50) - (L/2) = (8+50) - (2+2.75) = 6+47.25

BVC elevation:1450.06 - gJ (L/2) ::1450.06 + 3(2.0275) ::1456.143 ft

b. The low point station:g 2 .

From (Eq.2-5), r::

- gl 2 - (-3):: = 1.233

L 4.055

-(-3):: :: 2.433

1.233

-g, g, (L)X :: =

r g, - ga

rXl

y :: fgVC + s , X +2

= 1456.143 + (-3)(2.433) +

1.233 (2.433l

2

:: I 1452.49 it (elevation)

The low point is at station:

(6+47.25) + (2+43.3) = 8+90.55

2-8



Example 2-4

A 300 ft vertical parabolic curve with grades 91 =(+)5.0%,

92 =(-)2.5%, and the elevation of the high point on the curveis 123.45 ft. Visibility is to be improved over this roadby replacing this curve by another 600 ft parabolic curve.Find the depth of excavation required at the mid-point ofthe curve.

v

3BB ft

Solution:

Offset V = M id dl e o rd in at e e = L (Eq.2-2)

g, - gl 5 -(-2.5)Lf = 300.ft, e = L, = (300) = 2.8125'

I

8 8

g, - g 2 . . 5 -(-2.5)

L2. = 600.ft, e2,= L" = (600) = 5.625'8 8

The depth of excavation required at the mid-pointof the curve:

e1- el

= 5.625' - 2.8125' = 2.81 ft

2-9



Example 2-5

A 8 00 ft vertical curve with grades g1 =(+)2.0%, g2 =(-)3.0%and the data shown in the figure below.

(A) Find the station and elevation at the Bve, Eve, and theHIGH point on the curve.

(8) A new highway with an underside level 302 ft is runningperpendicular to the existing curve. The new highway mustmaintain a 25-foot clearance over the existing curve below.Find the minimum and maximum station where the new highwayshould not be constructed.

v X·?

B v e

I+CD. . . .

L iii 9 8 1 1 rt

Station 18+00

Station 36+00

Elev.=260 ft

Elev.=264 ft

2-10



271.2 it

solutions:

(A)

a. Distance between Station (36+00) and (18+00) :::l8DD.it

Distance between V (vertex) and Station 36+00 = X

Elevation from station (18+00) to (36+00):

Y/8+o0 + s, (1800 - X) + g2 (X ) = Y 36+o0

> 260 + 0.02(1800 - X) - 0.03(X) :::264

- - > X::: 640.ft

Vertex station:(36+00) - X = (36+00) - (6+40) = 29+60

Vertex elevation:264 - 9 X = 264 + 0.03(640) = = 283.2 ft

2

BVe station:(29+60 ) - (L/2) ::: (29+60) (4+00) = I 25+60 ]

Bve elevation:

283.2 - 91(L/2) = = 283.2 - 2(4) ::: I 275.2 it I

Eve station:(29+60) + (L/2) = (29+60) + (4+00) = I 33+60 IEve elevation:

283.2 + g2(L/2) = 283.2 - 3(4) =

b. The high pOint station:

g, - g, -3 - 2From (Eq.2-5), r = = ::: = -0.625

L 8

-g, g, (L) -2X ::: ::: = ::: 3.2

r g, - g2 -0.625

rXl -0.625 (3.2/Y = Ysvc + 9 X + ::: 275.2 + 2(3.2) +

I2

2(elevation)= [ 278.4 ft

The high point is at station:

(25+60) + (3+20) = 28+80

2-11



B v e~

271.2 ft

L = B S B ft

(B) New highway location:

Maintain a 25-io ot cl ea ran ce :

Y = 302 - 25 = 277 it

rX2.

Y = YSVC + s, X + = 275.2 + 2(X) +

'2-0.625 (X)

22

:;277.ft

~ X = 1.083' or 5.317'

The new highway should not be co nst ru ct ed bet we en:

Station: (25+60) + (1+8.3) = I 26+68.3 I (Min)

(25+60) + (5+31.7) ~ I 30+91.7 , (Max)

2-12



(A)200'

(B)

400'(e)

600'(D)

800'

Sample Problems 2

Problem A

A vertical curve with grades g1 =(+)0.8 % and g2 =(-)0.4%that intersect at station 18 +00 with elevation 8 20.00 ft.The maximum allowable change in grade per station is 0.2.

REQUIRED

1. What is the m~n~mum length of a vertical curveconnecting these two grades ?

2. Find the elevation (ft) of the B v e .

(F) 817.6 (G) 818.8 (H) 821.2 (J) 822.4

3. Find the elevation of the mid-point of the long chord.

(A) 817.8' (B) 818.2' (e) 821.2' (D) 819.1'

4. Find the elevation (ft) of the mid-point of the verticalcurve between the BVC and EVC.

(F) 817.8 (G) 818.2 (H) 8 21.2 (J) 819.1

5. Find the station of the summit of the curve.

(A) 17+50 (B) 18+50 (C) 19+00 (D) 19+50

6. Find the elevation (ft) of the summit of the curve.

(F) 8 19.0 (G) 818.2 (H) 821.2 (J) 819.2

7 . Find the middle ordinate (ft) of the curve.

(A) 0.3 (B) 0.6 (C) 0.9 (D) 1 . 2

8. Find the tangent offset (ft) at station 17 +00.

(F) 0.1 (G) 0.4 (H) 0.6 (J) 0.9

2-13



Problem B

A vertical curve with grades gl =(+)1.25% and g2 =(-)2.7 5%that intersect at station 8 +00 with elevation 8 20.00 ft.The length of the curve is to be 600 ft.

REQUIRED

9. Find the elevation (ft) of the BVe.

(A) 823.75 (B) 818.125 (C) 816.25 (D) 8 11. 7 5

10. Determine the rate of change of grade per station ( % ) •

(F) +0.667 (G) -0.25 (H) +0.25 (J) -0.667

11. Find the station at the high point of the curve.

(A) 6+87.5 (B) 6+78.5 (C) 8+00 (D) 9+12.5

12. Find the elevation (ft) at the high point of the curve.

(F) 817.25 (G) 817.42 (H) 818.25 (J) 818.42

ANS l.C6.J

l1.A

2.F7 .C

12.G

3.BB.G

4.J9.C

5.C10.J

2-14



3.TRAVERSE

A traverse is a series of consecutive lines or courses onwhich the lengths and directions have been determined. A traversewhich comes back to its starting point is called a closed traverse.

Closed traverses provide checks on the angles and distances.An open traverse is a series of lines which do not return to thestarting point. Open traverses are sometimes used on route survey.

Meridian: The direction of a line may be expressed as an anglefrom an established reference line. This reference lineis called reference meridian and may be expressed asone of the following:

1. Magnetic meridian (influenced by magnetic pole).

2. True meridian (passing the north and south pole).

3. Assumed meridian (arbitrarily chosen).

Bearing: The bearing of a line is the ACUTE horizontal anglebetween a reference meridian and the line.

Bearing AB: N 30°,E N B

Bearing BA: S 30' w

Back Bearing of AB: S 30' w

A

Azimuth: Azimuth is angle measured clockwise froma reference meridian.

Azimuth is from 0° to 360°. N

AAzimuth AS (from the south) : 30 C 1

Azimuth AS (from the north) : 2100

B

Back Azimuth AS (from the north): 30°

3-1



Latitude: The projection of a course on the north-south direction.

Departure: The projection of a course on the east-west direction.Latitude = L (cos c J . ) N B

Departure = L (sin ~ ) Latitude

L =length A S

Q( = bearing or azimuthA

Departure

Angle Adjustment

The first step in traverse calculation is to balance theangles. The sum of the interior angles of a polygon with n sidesis (n - 2) X (18 0°). Angles of a closed traverse can be adjustedto the correct geometric total by arbitrary correction or anaverage correction (even distribution). Unlike correction forlinear measurement, the angle adjustment is independent of thesize of an angle. An example is illustrated below.

Arbitrary Correction Average CorrectionMeasured

Point Angle Adjust- Adjusted Adjust- Adjustedment Angle ment Angle

A 105"20' 5' 1050' 15' l' 105°19'

B 86"56' a 860'56' l' 86° 55'

C 8 2· 48 ' 0 82"48' 1' 8 2°47'

D 1590'31' a 159°31' l' 159030'

E 1050'30' a 105030' l' 105029 '

Total 5400 a s ' 5' 540000 ' 5' 540°00'

Table 3-1

3-2



Traverse .'Adjustment

There are five methods for traverse adjustment:

(1) Compass rule.(2) Least squares method.(3) Arbitrary method.(4) Transit rule.(5) Crandall method.

The compass rule (or Bowditch rule) is the most commonly usedmethod in practice. It is suitable for survey where the anglesand the distances are measured with equal precision. The other

four methods are seldom used.

Using the compass rule, corrections are made as follows:

correction in latitude of leg length of leg= =

closure in latitude traverse perimeter

correction in departure of leg length of leg= =

closure in departure t ra ve rs e p er im et er

The linear error of closure of a traverse is as follows:

Linear error of closure

1 . d 2. 1 ·1· d2.(c osure ~n eparture) + (c osure 1n at~tu e)=

The relative error of closure (or precision) for a traversei s e xp res se d by

linear error of closurePrecision =

traverse perimeter length

3-3



An example for traverse adjustment is illustrated below:

SITUATION:

The data obtained in a partially completed survey of anindustrial building site are giving in Figure 3-A below.

REQUIREMENTS:

(A) Compute the bearing of lines BC, CO, DE, and EA.

(B) Calculate the corrections and precision fortraverse adjustment.

Solutions:

(A) Compute the bearing of lines BC, CD, DE, and EA

Line Bearing

Be S SlQ26'ECD S 41Q38'WDE N 41alO'WEA N 20041'W

N8

A

N O T T O S C A L E

c

N

w E

D s

Figure 3-A

3-4



(B) Traverse Adjustment (By Compass rule)

-

Line Bearing Distance Latitude Departure(cos) (sin)

AS N 53"49'E 567.84 335.24 458.32

B e s 51"26'E 810.76 -505.45 633.92

CD S 41'38'W 922.75 -689.67 -613.04

DE N 41"10'W 411.72 309.94 -271.02

EA N 20Q41'W 588.16 550.25 -207.74

Total 3301. 23 +0.31 +0.44

M :: -

0.31

3301. 23

0.44N - -

3301. 23

Line Corrections Adjusted

Latitude Departure Latitude Departure

AB M ,}(AB . . . -0.05 N X AB = -0.08 335.19 458.24

BC M X BC . . . -0.08 N X BC . . . -0.11 -505.53 633.81

CD M X CD . . . -0.09 N XCD . . . -0.12 -689.76 -613.16

DE M )(DE . . . -0.04 N X DE :: -0.05 309.90 -271.07

EA M x EA = -0.05 N X EA . . . -0.08 550.20 -207.82

Total -0.31 -0.44 0.0 0.0

Linear error of closure = )0.312- + 0.442- = 0.54 ft

0.54 1 1Precision

,..". . . = . . .3301. 23 6113 6200

3-5



Example 3-1

What is the bearing of each side of Figure 3-B?

110QlS'lS"80"S3'02"89"01'45"79"49'58"

Given: 1 =

2 =

3 =

4 =

N

Not to Scale

FIgure 3 - 8

3-6



Solution:

Check the sum of the interior angles:

110015'15H

+ 80Q53'02" + 89001'45" + 79°49'58" = 360°0'0"

1800 X (n - 2) = 1800 x (4 - 2) = 3600 (Checked)

Side Line Bearing

A 12 5 59044'55"E

B 23 5 39Q22'03"W

C 34 N 49039'42"W

Line 12

N

o110 15'15"50°30'20"

59°44'55"

8 Line 23

180°00'00"59°44'55"

80~53'02H

39"22'03"

Line 34

89°01'45"- 39°22'03"

49° 39'42"

3-7



Example 3-2

A contractor has driven a tunnel heading from point B to paint Aand another from point B to point O. He has requested you, asthe resident engineer of the construction contract, to providehim with the correct bearing and distance 60 that he can connectpoints A and 0 with a connecting tunnel.

Points A and 0 are connected by a surface traverse as follows:

Course Bearing Distance (level) Elevations

AS S 12Q14'E 1291.30 ft At A, 357.6 ftB e N 83012'E 1317.40 ft At B, 454.7 ft

co s 89"41'E 18 19.8 0 ft At 0, 892.4 ft

REQUIRED:

(A) What is the bearing of the connecting tunnel A-D?

(8) What is the length of the connecting tunnel A-D?

(C) What is the slope of the connecting tunnel A-D?

3-8



A El~. 357.6 f tN

V i E. . . . . . . . . . . . . .

- . . . . . . . . .. . . . . . . . . . . . . .

. . . . . . .. . . . . . . . . . . . . .

. . . . . . . -.. . . . . . . . . . . . . .

-.. . . . . . . . . . . . . . . . . . . .. . . . . .

. . . . . . . . . . . . . .

-.. ................._D EI.,,~892.4 ft

-. . . . . . . . . . . . . . .. . . . . . . . . . . . . .

. . . . . . .B

solutions:

Using point A as the origin (0,0),calculate the coordinates as follows:

L. Bearing Dista. Latitu. Depart. P. y X

(cos) (sin)

A 0.0 0.0AB S 12°14'E 1291. 3 -1261. 98 273.62

B -1261. 98 273.62Be N 83()12'E 1317.4 155.99 1308.13

C -1105.99 1581.75

CD S 89°41 'E 1819.8 -10.06 1819.77 D -1116.05 3401.52

(A) Bearing AD-I 3401.52

~ tan =-1116.05

(B) Distance AD = )1116.051

+ 3401.522

= 3579.93 ft (level)

Ve rt ic al d ist an ce = 892.4 - 357.6 = 534.8 ft

Length A D =)3579.93k

+ 534.8~ = 3619.66 ft

534.8(C) Slope = ::: 14.94%

3579.93

3-9



Exa mp le 3-3

You are provided, by an individual authorized to practice lands ur ve yi ng , t he f ol lo wi ng p ro pe rt y d es cr ip ti on :

Beginning at most westerly corner of Lot 1, Tract No.2 as shownbelow.

Thence N 55'37 'E 567 .34 feet;thence S 49 D38 'E 8 10.7 6 feet;thence S 43026'W 922.7 5 feet;thence N 39022'W 411.7 2 feet;thence N 18 '53'W 58 8 .16 feetto the point of beginning.

REQUIRED

(A) C al cu lat e al l in te ri or a ng le s.

(B) Check the sum of the interior angles.

(C) Calculate the defiection angle on each station.

NB

' I ( , . "

h' ,~N' \ r o

~ . ocO : J ' V ~

S

W E

i

S[

rt~

I

1

3-10l

Lot 1

T r a c t N o . 2 c



Solutions:

(A) The interior angles:

L A

180°00'- 55°37'- 18°53'

L B

55°37'+ 49°38'

N8

105°15'

Lot 1

T r a c t N o . 2

h'~~~~ ~

f t . . . 0

'V0 ) t ? J ~'?

S

W

L E L D

18°53' 43°26'

+ 180000' + 39° 22'

- 39° 22'82°48'

159031 '

L c

180000 '

- 49°38'- 430 26'

86°56'

c

N

E

s

(B) Check the sum of the interior angles:

105~15' + 105030' + 159°31' + 82048' + 86°56' = 540°0'

o 0 0180 ~ (n - 2) = 180 X (5 - 2) = 540 (Checked)

3-11



(C) Show the deflection angles.

L E

39°22'

- 18 ° 53'

20°29'

L A55°37'

+ 180

53'

74°30' /'/'

/'

L B

180°00'

- 55°37 '- 49°38'

74°45'\

\

\

N

wL C

/

/

49°38'+ 43°26'

s

/

3-12



Example 3-4

The following figures show the double verniers. Determine thereading for the two sets of numbers on each circle.

40320

Scale

IS

20 Scale 10340 350

Solution:

Figure (A) :° 38° 47'The reading for the inner set is 38 30' + 17' =

The reading for the outer set is 321°00' + 13' = 321°13'

Figure(B) :

17° 17°05'30"The reading for the inner set is + 05'30" =

The reading for the outer set is 3430

05'30" = 342°54'30"

3-13



!

trI

IIII,

Sample Problems 3

Problem A

Two separate traverses start from Station A, which is locatedat the portal of a tunnel. Traverse #1 was run in the tunnel,

and traverse #2 was run on the surface. The field measurementsare reproduced as follows:

Note: Horizontal distances and north azimuths are used throughout.

Tunnel Traverse * 1:

Station A to Station B, Azimuth = 300°00',Distance = 300.00 feet, Grade = +1 per cent

Station B to breast of tunnel, Azimuth = 30°00',Distance = 100.00 feet, Grade = +2 per cent

Tunnel Traverse # 2:

Station A to Station C, Azimuth = 30°00',Distance = 200.00 feet, Grade = +2 per cent

Station C to Station D, Azimuth = 330°00',Distance = 300.00 feet, Grade = +25 per cent

A vertical shaft is to be sunk at Station D, and the breastof the tunnel is to be connected with the shaft by a drifthaving a +3 per cent grade. (+3% from breast of the tunnel)

REQUIRED

1. Determine the azimuth of the drift

(A) 39° 3 8 ' (B) 39008 ' (C) 3 8 ° 0 8 ' (D) 39048 '

2. Determine the required depth of shaft (f t) .

(F) 66.40 (G) 64.60 (H) 68.60 (J) 60.40

3. Determine the slope length of the drift (ft) .

(A) 250.33 (B) 258.33 (e) 255.53 (D) 253.33

4. If the coordinates of Point A are 1025 N, 157 5 Wand thoseof point Bare 425 N, 97 5 w , the bearing of course AB is :

(F) N 450

Eo

(J) N 45 W(H) S 45° E

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c

Problem B

You are provided, by an individual authorized to practice landsurveying, the following property description:

Beginning at most westerly corner of Lot 1, Tract No.2 as shownbelow.

Thence N 37°04'E 514.1 feet;thence S 68 °32'E 1395.6 feet;thence S 55° 40'W 961.3 feet;thence N 32° 48 'w 243.3 feet;thence N 57 ° 1 7 ' W 816.5 feetto the point of beginning.

REQUIRED

5. Calculate the interior angle at point A.

6. Calculate the sum of the interior angles.(F) 539°00' (G) 539°30' (H) 540°00' (J) 540°30'

7 . Find the linear error of closure of the traverseby the Compass rule.

(A) 5.0' (B) 5.5' (e) 4.5' (D) 6.0'

B. Find the precision (relative error of closure) ofthe traverse by the Compass rule.

(F) 1/786 (G) 1/714 (H) 1/873 (J) 1/655

I

~ ''+ No J o . . . . · 0 t-~

-t, , ' >

c :,

W E

S

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9. A backsight with inverted telescope is sighted at a pointthat bears N 38 °E from the transit. The telescope isplunged (reversed) and a right deflection angle of 6So

is turned. The bearing of the new line is

10. When set up at the B.C. (Beginning of Curve) of a simplecurve, the angle to turn off the P.I.(Point of Intersection)to hit the radius point is :

(H) one radian

(G) delta

(J) 90°

(F) one half delta

11. The following figures show the transit verniers.

80 S~lle

280

30

For a clockwise angle, determine the reading of thed ou ble v er ni er .

(A) 86°07'

For a clockwise angle, determine the reading of thed ir ec t v er ni er .

(F) 24°13'20" (G) 21°13'20"

survey review & problems part i

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