1. Given the lines and direction as follows: AB BC- 91.5m N 68° E, CD - is to connect these three lines thus forming the center line of a new road. Compute the length of the common radius of the reverse 57.6m due East, 102.6m azimuth of 312°. A reverse curve
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- A certain horizontal curve has a central angle of 30.5°. If STA PI is 18+303 and STA PT is at 18+747, determine the station of PC.A certain horizontal curve has a central angle of 39.3°. If STA PC is 18+334 and STA PT is at 18+793, determine the station of PI.The tangents of a sag curve connects at the point of intersection having Station 18+410.48 and elevation 155.94 meters. Point PT at the end of the curve lies at Station 18+533.71 and elevation 172.06 meters. The grade of the back tangent is -5.19% . Determine the elevation of the lowest point in meters. Plot and show solution
- NEED ASAP The tangents of a sag curve connects at the point of intersection having Station 18+448 and elevation 153.7267 meters. Point PT at the end of the curve lies at Station 18+534 and elevation 157.5285 meters. The grade of the back tangent is -5.5207%. 1. Determine the value of the maximum offset in meters. 2. Determine g1. 3. Determine the horizontal length of cirve in meters.A 6.58° horizontal curve has a central angle of 48.4°. If the station of PT is at 25+198, determine the station of PC.The centerline of two parallel tracks are connected by a reversed curve. The angle of intersection of the first curve is 15 degrees and the distance between parallel tracks is 30m. If the radius of the second curve is 300m. a. Compute the length of the long chord from P.C. to P.T. ANS. 229.839m b. Compute the radius of the first curve. ANS. 580.432m please also include the drawing of the curve
- The back tangent of a vertical curve has a grade of 2% and a forward tangent of -3.3%. Compute the length (m) of the curve it the sight distance is 176m. Use h1 = 1.1m and h2 = 0.5m.The PC of a vertical parabolic curve is at station 154 + 104 of elevation 80.45 m and the grade of the PC is +2.5 %. The elevation of station 154 + 184 is 81.95 m. What is the elevation of 154 + 204? (INCLUDE ILLUSTRATION/FBD)A spiral of 250 ft length is used for transition into a 4° 00ʹ circular curve. The angle ∆ at PI station of 72+04.39 is 28° 00ʹ and the tangent distance Ts=482.56 ft. Compute length of the curve.
- A simple curve connects two tangents AB and BC with bearings N85°30’E and S68°30’E, respectively. Point D along line AB has a coordinate of 20100 N and 20100 E while point E along line BC has coordinates of 20086.55 N and 20184.27 E. Find the distance of line BD, the degree of the curve that shall be tangent to the three lines AB, DE, and BC, determine the stationing of PT if the stationing of D is at 1+052.87.Two tangents having an azimuth of 235° and 277° meet at station 30+28.212 are connected by a 330ft spiral curvewith a 5° circular curve used for light rail transit.a) Determine the difference between the external distance of the spiral and the central curve.b) Determine the minimum super-elevation of the curve at the two-thirds of the spiral with the railway width of 15ft. If g= 32.2 ft/s^2c) Determine the stationing of the S.T.The tangents of a sag curve connects at the point of intersection having Station 18+427.14 and elevation 157.50 meters. Point PT at the end of the curve lies at Station 18+571.93 and elevation 172.42 meters. The grade of the back tangent is -7.60%. Determine the value of the maximum offset in meters. Draw the curve and show the solution.