elect one: OA. Summit curve B. Spiral curve OC. Compound curw Reverse curve
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- Quiz 2: 5. A horizontal curve is laid out with the point of curve, PC, station and the length of long chord, LC, as shown. The curve radius, R, is 620 ft. With stationing around the curve, the stationing of the point of tangent, PT, is most near to (A) sta 6+20 (B) sta 10+42 PT (C) sta 14+02 LC = 980 ft (D) sta 14+64 sta 4+22 PC I 96°18' R 620 ftGiven: Stationing B= 18 + 685 Stationing C= 19 + 125 Tangent Lines: AB: Azim= 224° 24' BC: Azim= 264° 29' CD: Azim= 3150 28' A proposed highway curve is to connect these three tangents. Find: Radius of simple curve that connects these tangents. b. Sta PC. c. Length of the curve from PC to PT. a.The common point in a reversed curve O PC PCC PRC SC
- 2 existing highways are to be connected by a simple curve of 2° (100marc basis). The first highway has a North azimuth of -18°26’34” whilethe second highway has a South azimuth of -47°21’08”. Determine therequired length of curve to be constructed and the station of the endof the curve given that the theoretical intersection of both highwayswas found to be at STA. 3+016.SHOW THE COMPLETE SOLUTION AND ILLUSTRATION. In a compound curve, the line connecting the P.I. at point V and the P.C.C. is an angle bisector. AV is 270m and BV is 90m. The stationing of A is 6+421 and that of B is 6+721. Point A is a along the tangent passing thru P.C. while point B is passing thru P.T. The P.C.C. is along line AB.a. Compute the radius of the first curve passing thru P.C.b. Compute the radius of the second curve passing thru P.T.c. Determine the length of chord from P.C. to P.t.In a railroad layout, the centraline of two parallel tracks are connected with a reversed curve of unequal radil. The central angle of the first curve is 16° and the distance between paralel tracks is 27,60m Stationing of PC is (15-420) and the radius of the second curveis 290m, Compute the length of the long chord from PC to PT and the total length of the reversed curve
- please include illustrations. ASAP Traverse lines MN, NO, OP are center lines of a portion of a proposed highway. Respectively the bearings and distances are; MN due North, 277.60 m; NO N26°40°E, 107 m; OP N 61°40° E, 200 m. A previously designed compound curve connected these three tangent lines with the PCC at a station 2+012.00. It is desired to revise the system into a single circular curve that will still be tangentto the three lines. a. Determine the radius of the simple curveb. Determine the stationing of the new PC if station N is at 1+975c. Determine the stationing of the new PTA sag curve has a length of 500 ft. If the tangent lines intersect at station (PVI) 475+00 at an elevation of 300.00 ft. Determine the station of BVC. 478+00 470+00 472+50 477+50Situation 1] A circular road is proposed to be laid out with tangent AB whose bearing is N85º54’E intersecting another tangent line BC whose bearing is N48º30’E. If the degree of the curve is 4.5º, using arc basis, determine the following: Tangent distance of the curve, T Long chord of the curve, C Middle ordinate, M External distance, E Length of the road, LC If the station of point B 10+584, determine the station of PT.
- 7. Vertical curves are used to connect stretches of road that go up or down at a constant slope. These lines of constant slope are called Grade Tangents Slope A. Grade В. C. Radius D.Traverse lines MN, NO, OP, are centerlines of a portion of a proposed highway. Respectively, the bearings anddistances are: MN, due North, 277.60 m; NO, N 26°40’ E, 107.00 m; OP, N 61°30’ E, 200.00 m. A previously designedcompound curve connected these three tangent lines with PCC at Station 2+012.00. It is desired to revise the systeminto a single circular curve that will still be tangent to the three lines. (HINT: Convert the Compound Curve into a singularSimple Curve, taking R1 and R2 as equal, taking N as new PC, and O as new PT).Vertical Curve