Q1:A 600-ft equal-tangent sag vertical curve has the PVC at station 170 + 00 and elevation 1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationing and elevation of the PVI, the PVT, and the lowest point on the curve. (Use 1 decimal point) Ans. ft Ex: 999.4 (Do not add units) *
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A: PVCSTATION=PVISTA-L2PVCsta=3800-4002=36+00PVCELV=PVIELV-g1L2PVCelv=560+0.027×4002PVCelv=565.4Equatio…
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A: G1 = 7% = 0.07 G2 = -4.9% = -0.049 G1' = -4.9% = -0.049 G2' = 2.2% = 0.022 l1 = 113.61m
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- A 500-ft curve, grades of g₁ = +2.00% and g2 = -2.50%, VPI at station 96+80, and elevation 845.260 ft, stakeout at full stations. Part A Determine the elevation at the high point of the curve. Express your answer in feet to six significant figures. Elev = Submit Part B AΣo↓ vec Request Answer Determine the station at the high point of the curve. O 97+12.77 O 96+13.33 O 96+52.22 O 97+96.66 ? ftA-2.50% grade meets a +1.75% grade at station 44+25 and elevation 3385.74 ft, 400-ft curve, stakeout at half stations. Part A Determine the elevation at the low point of the curve Express your answer in feet to six significant figures. Elev= VAΣ1 vec + A ftB- Unequal-Tangent Parabolic (Vertical) Curve: Example: A grade gl of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and 743.24, respectively. A 400 vertical curve is to be extended back from the vertex, and a 600' vertical curve forward to closely fit ground conditions. Compute and tabulate the curve for stakeout at full stations. Stations 00 00 00 00 EVC (752.84) BVC (751.24) A CVC (748.04) (747.24) (747.56) -2.00% 400 +1.60% V(743.24) >と じ。 Solution: The CVC is defined as a point of Compound Vertical Curvature. We can determine the station and elevation of points A and B by reducing this unequal tangent problem to two equal tangent problems. Point A is located 200' from the BVC and Point B is located 300' from the EVC. Knowing this we can compute the elevation of points A and B. Once A and B are known we can compute the grade from A to B thus allowing us to solve this problem as two equal tangent curves. Point A STA 85+00: Elev. = 743.24+2 (2) = 747.24'…
- Question t Pow t Edge Calculate the length and radius for the following curve. The PC is at Sta 8+10. The PT is at Sta 10+50. The delta is 17d 32m 00s. t Excel t Word t Word t Pow t Pow t Wore t Pow3. An engineering mistake has resulted in the need to connect an already constructed tunnel and bridge with a vertical curve. The profile of the tunnel and the bridge is given in the figure below. Devise a vertical alignment to connect the tunnel bridge if the highest common design speed is 20mph for the sag and 10 for crest (equal tangents). STATION 4+00 PVT BRIDGE DECK E LEVATION 126.67 FT TUNNEL FLOOR E LEVATION 100FT. STATION 0+00 PVCPI 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 ? Note: I need right solution.. Don't copy from other expert solution.
- GIVEN THE HORIZONTAL CURVE BELOW. DETERMINE THE INTERSECTION ANGLE AT PI, THE RADIUS OF THE HORIZONTAL CURVE. THE LENGTH OF THE CURVE AND LOCATE THE PC STATION. PRAW THE FIGURE BELOW AND SHOW YOUR SOLUTIONS NO SHORTCUTS. PI IS LOCATED ON AN INACCESSIBLE AREA. FI FC STR STA 79+14 A STA 55+27 12' 28 147 B LENGTH OF CHORD A TO BIS 574-48FT PTa) compute the grade of the forward tangent. [Format: AX:XX%) A - sign [+ or -], X - numerical value b) compute the stationing of the highest point of the curve from the P.C. [Format: 0+000.00]c) compute the elevation of the highest point of the curve **USE 2 DECIMAL PLACES**An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0% (final). The PVI is at station 11 + 000.000 and at elevation 420 m. Due to a street crossing the roadway, the elevation of the roadway at station 11 + 071.000 must be at 421.5 m. Design the curve. Use the editor to format your answer
- A vertical summit curve is having tangents which intersects at point Pl at Station 12+617.78 and 134.19 meters elevation. The grade for the forward tangent is -6.28%. Station PC is at station 12+506.53 and having an elevation of 123.37 meters. Determine the elevation of the highest point in meters. CS Scanned with CamScannerThe simple curve has a deflection angle of 105 degrees and 15 minutes and the length of its tangent lines is 21.0 m, what is the stationing at PC if PT is 0 + 75.00. Express answer in the form of x +000.00. Your answerA +5.00% grade meets a +1.50% grade at station 4+200 and elevation 605.568 m, 180-m curve, stakeout at 30-m imcrements. Part A Tabulate station elevations for an equal-tangent parabolic curve for the data given above. Express your answers, separated by commas, to six significant figures from lowest to highest elevation. vec ? ft, ft, ft, ft, ft, ft, ft