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- -3% grade is connected to a +1.2% grade by means of 200 m vertical curve. The PJ station is 100+ 00 and the Pl is at elevation 100 m above sea level. What is the design sight distance for the curve?A simple curve has a central angle of 35°. The offset distance from the PT to thetangent line passing thru the PC is 75m long. Compute the degree of curve. Useb arc basis.A vertical summit curve has tangent of +6% and -4%. If the stationing and elevation of the PT is 10+ 020 and 142.63 m respectively. Determine the maximum length of the curve if the design speed of the curve is 90kph sight distance of 100 m, driver's eye heigt is 1.08m and the object heght is 0.8 m
- PI sta = 18+00, PI elev = 300.00, L = 20 stations, incoming grade = +2%, outgoing grade = -3% The high point of the curve is at station ?a horizontal curve has design speed of 50mph. compute the degree of the curve if e =0.10 and f=0.18a.2.25b.3.10c.3.25d.2.98a 1600 ft lomg equal tangent sag vertical curve has the PVI at station 120+00 and elevation 1500 ft the intial grade is -3.5% and the final grade is +6.5% determine the elevation of lowest point A.1525 B.1520 c.1518.20 D, 1514
- A vertical summit curve has a back tangent of +2% and a forward tangent of -3% intersecting at station 10+220.60 and elevation of 200 m. The design speed of the curve is 80 kph. Assuming coefficient is 0.30 and a perception reaction time of 2.5 sec. Compute the length of the curve. Please it's important for me... UrgentThe angle between a 5-m full station of a certain curve is 10°. The curve has 7 intermediate stations between P.C. (10 + 300m) and P.T. After laying out the curve, it was found out that the P.T. falls in a property lot and it should be moved in a way that the P.T. will be free from crossing the lot. Compute all the necessary elements of the old curve and new relocated curve.A compound curve has the following properties: 1,= 25-degree, D,= 6.2-degree, 2= 36-degree, D2= |4.5-degree. Compute the stationing of PCC if PC is at station 10+430.
- An easement spiral curve has a design speed of 110 Kph. The radius of the central curve is 365 m with a permissible super elevation of 0.06. Determine the Following: a. Centrifugal acceleration so as not to cause discomfort to the driver in m/sec3. b. Length of the transition curve to limit the centrifugal acceleration. c. Length of the short tangent of the transition curve.A vertical summit curve has tangent grades of +2.5% and -1.5% intersecting at station 12 + 460.12 at an elevation of 150 m above sea level. If the length of curve is 182 m. Compute the length of the passing sight distance. a. 182 m b. 205 m c. 150 m d. 250 mA symmetrical vertical summit curve has tangents of +4% and -2%. The allowable change of grade is 0.3% per meter station. Stationing and elevation of P.T. is at 10+020 and 142.63 respectively. 1. Compute the length of curve. 2. Compute the distance of the highest point of the curve from the P.C. 3. Compute the elevation of the highest point of the curve.