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 The answer is (D). 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 The answer is (C). focal length altitude − ground elevation 6 in = 4000 ft − 400 ft 1 in = (1 in:600 ft) 600 ft scale = The answer is (C). I 2 56◦ 2658.55 ft = R tan 2 Rearranging to solve for R, 100 ft 1 = 4000 ft 40 Solving for α, The answer is (D). 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 The answer is (A). 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 answer is (A). The answer is (D). The true bearing of the line today is The answer is (D). 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 answer is (C). The U.S. Public Lands System would deﬁne the area as township 3 North, range 1 West. Professional Publications, Inc.
© Copyright 2018