CIVIL SURVEYING SAMPLE EXAMS

```CIVIL
SURVEYING
SAMPLE
EXAMS
for the
California Special
Civil Engineer
Examination
Second Edition
Peter R. Boniface, PhD, PLS
Professional Publications, Inc. • Belmont, CA
Sample Exam 1
1
4
(6 points)
What is the area of a sector with a central angle of 47◦
within a curve that has a radius of 1100.00 ft?
(A)
(B)
(C)
(D)
2
10.3
10.8
11.0
11.4
(A)
(B)
(C)
(D)
5
boundary surveys
construction staking
infrastructure design
all of the above
3
(4 points)
(A)
(B)
(C)
(D)
Which of the following professional services can a civil
engineer legally solicit in the state of California?
Problems
If the degree of curvature (arc deﬁnition) is 1.1459◦ ,
what is most nearly the radius?
ac
ac
ac
ac
(4 points)
•
3200
3800
4200
5000
ft
ft
ft
ft
(6 points)
Point P lies on a 4000 ft radius curve. Point Q lies on
the tangent to the curve, and line PQ is perpendicular
to the tangent. The distance AQ is 100 ft. What is
most nearly the distance PQ?
(A)
(B)
(C)
(D)
1.3
2.5
3.0
9.5
ft
ft
ft
ft
Refer to the following illustration for Probs. 3 through 5.
6
V
T
direction of
increasing
stationing
and bearing
I
C
A
B
P
(4 points)
An aerial photograph was taken using a camera with a
focal length of 6 in. The plane was ﬂying at an altitude
of 4000 ft above mean sea level, and the mean ground
elevation was 400 ft above mean sea level. What is the
scale of the photograph?
(A)
(B)
(C)
(D)
7
not to scale
3
O
(4 points)
If the tangent distance, T , is 2658.55 ft and the deﬂection angle, I, is 56◦ , what is most nearly the radius of
the curve?
(A)
(B)
(C)
(D)
1790
3010
5000
5660
ft
ft
ft
ft
1
1
1
1
in:10 ft
in:50 ft
in:600 ft
ft:10 ft
(4 points)
In a boundary survey involving the reestablishment of
corner monuments, if the survey measurements agree
with those on an existing recorded map, the surveyor
should prepare a
(A)
(B)
(C)
(D)
corner record
record of survey
tract map
parcel map
Professional Publications, Inc.
Sample Exam 1
1
5
A
I
=
2
πR
360◦
Rearranging to solve for A,
◦ I
47
2
πR
π(1100 ft)2
A=
=
360◦
360◦
= 496,284 ft2
2
x = R sin α
100 ft = (4000 ft) sin α
Rearranging,
3
1
= 1.433◦
40
y = R(1 − cos α) = (4000 ft)(1 − cos 1.433◦ )
= 1.25 ft (1.3 ft)
α = arcsin
6
focal length
altitude − ground elevation
6 in
=
4000 ft − 400 ft
1 in
=
(1 in:600 ft)
600 ft
scale =
I
2
56◦
2658.55 ft = R tan
2
Rearranging to solve for R,
100 ft
1
=
4000 ft
40
Solving for α,
A licensed civil engineer can legally solicit work that
his license does not cover, provided the work is ultimately performed under the supervision of a person
with the appropriate license. (The codes are B&P 6735
and 6738.)
15
Using tangent oﬀsets,
sin α =
496,284 ft2
= 11.39 ac
ft2
43,560
ac
Solutions
Convert this area to acres.
A=
•
T = R tan
2658.55 ft
56◦
tan
2
= 5000.01 ft (5000 ft)
R=
7
In order to simplify the preparation of maps where there
is no discrepancy in the surveyed data, the surveyor can
prepare a corner record, which can be hand-drawn on a
standard form.
8
4
The true bearing of the line today is
Using the conversion equation expressing degree of curvature, D, as a function of radius, R,
N 12◦ 32 E + 05◦ 00 E = N 17◦ 32 E
The magnetic bearing of the line 100 years ago was
◦
D=
(180 )(100 ft)
πR
N 17◦ 32 E − 02◦ 00 E = N 15◦ 32 E
Rearranging to solve for R,
(180◦ )(100 ft)
(180◦ )(100 ft)
R=
=
πD
π(1.1459◦ )
= 5000 ft
9
The U.S. Public Lands System would deﬁne the area as
township 3 North, range 1 West.
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