1. Calculation of elevations of curve every 50 feet and plotting the curve to scale and determining the maximum elevation and horizontal distance from the PVC In this question, given the following information: Positive grade of 5% intersects a grade of -3%.The length of the curve is 450 feet. VPI elevation is 100.25 feet. To plot a vertical curve using a 10:1(H: V) magnification for the vertical scale, use a 1 inch =50 feet for horizontal scale. Here is the detailed solution: First, let's find out the PVC elevation, PVI elevation, and the length of the vertical curve as: PVC Elevation: 100.25 ft Grade at PVC: 5% PVI Elevation: 100.25 + [(5/100)*450] = 123.25 ft VPI Elevation: 100.25 + [(-3/100)*450] = 86.75 ft Length of Vertical Curve: VPI to PVI = (123.25 - 86.75) = 36.50 feet Now, let's find out the elevations of the curve every 50 feet:
The horizontal distance can be calculated using the formula:
`H = (L^2)/(2f)`
where L is the length of the curve and f is the algebraic difference between the initial and final grades.
We know that L = 450 feet and f = (5 - (-3)) = 8.
So, `H = (450^2)/(2×8) = 15,937.5 ft`
We can now plot the curve using the above information. The tangent grades are shown as well as horizontal distances.
Maximum Elevation:
The maximum point on the vertical curve is the midpoint of the curve, which is 18.25 ft away from the PVI. The elevation of this point can be calculated using the formula:
`E = PVI + [(fH^2)/(8L^2)]`
So, `E = 123.25 + [(8×18.25^2)/(8×450^2)] = 124.60 ft`
The horizontal distance from the PVC to this point can be calculated as:
`Hd = H/2 + [(fL^2)/(8H)]`
So, `Hd = 15,937.5/2 + [(8×450^2)/(8×15,937.5)] = 739.06 ft`
2. Determination of the depth below the surface of the curve to the top of the pipe, station of the highest point on the curve, and elevation of the highest point on the curveA vertical curve crosses a 4-ft diameter pipe at a right angle. The pipe is located at station 110+85, and its centerline is at elevation 1091.6 ft. The PVI of the vertical curve is at station 110+00, and elevation 1098.4 ft. The vertical curve is 600 feet long and connects an initial grade of +1.20% to a final grade of -1.08%. Here is the detailed solution:First, let's find out the PVC elevation, PVI elevation, and the length of the vertical curve as:
PVC Elevation: 1098.4 + [(1.20/100)*600] = 1106.4 ft PVI Elevation: 1098.4 + [(-1.08/100)*600] = 1082.16 ft Length of Vertical Curve: VPI to PVI = 600 feetNow, let's find out the depth below the surface of the curve to the top of the pipe:
We know that the pipe is located at station 110+85 and its centerline is at elevation 1091.6 ft.
So, the distance from the PVC to the pipe can be calculated as:
`D = 85 - 00 = 85 feet`
The elevation of the curve at station 110+85 can be calculated using the following formula:
`E = PVI + [(fD^2)/(8L^2)]`
So, `E = 1082.16 + [((1.20 - (-1.08))×85^2)/(8×600^2)] = 1092.03 ft`
Therefore, the depth below the surface of the curve to the top of the pipe is:
`1091.6 - 1092.03 + 2 = 1.57 ft`
Next, let's find out the station of the highest point on the curve:
The highest point on the vertical curve is the midpoint of the curve, which is 300 feet away from the PVI. So, the station of the highest point on the curve is:
`110 + 300 = 110+30`
Finally, let's find out the elevation of the highest point on the curve:
`E = PVI + [(fH^2)/(8L^2)]`
So, `E = 1098.4 + [((1.20 - (-1.08))×(600/2)^2)/(8×600^2)] = 1100.55 ft`
Therefore, the elevation of the highest point on the curve is 1100.55 feet.