Let's solve each of the problems one by one:
### Problem 1:
You have a symmetrical curve of 200 meters length, with the grades of the tangent intersecting at station 10+100 with an elevation of 103.55. The grades are -3% and -7%.
Symmetrical Curve Vertex Calculation:
1. The vertex of a symmetrical curve is equidistant from both tangent points. Since the curve length is 200 meters, the vertex will be at the midpoint:
\[ \text{Length to vertex from P.C. or P.T.} = \frac{200}{2} = 100 \text{ meters} \]
2. Compute the station of the vertex from the point of intersection (10+100):
\[ \text{Station of the vertex} = 10+100 + 100 = 11+200 \]
Check for the Highest Point:
- Given the grades (-3% and -7%), both are descending towards the vertex. Therefore, the highest point of the curve will be at the vertex:
\[ \text{Station of the highest point} = 11+200 \]
### Problem 2:
You have a 400m parabolic curve, with grades of 3% ascending and 6.5% descending at the point of intersection (4+600) and an elevation of 102.5.
Parabolic Curve Lowest Point Calculation:
1. For a parabolic curve, the lowest or highest point (vertex) is where the first derivative (slope) changes sign or is zero. The formula to find the stationing of the lowest or highest point is:
\[ x = \frac{L(W_2 - W_1)}{W_2 + W_1} \]
Where \( L \) is the total length of the curve, \( W_1 \) and \( W_2 \) are initial and final grades, respectively.
2. Plugging in the values:
\[ x = \frac{400(6.5 - 3)}{6.5 + 3} = \frac{400 \times 3.5}{9.5} \approx 147.37 \text{ meters from the P.C.} \]
3. Calculate the station of the lowest point:
\[ \text{Station of the lowest point} = 4+600 + 147.37 = 5+747.37 \]
### Problem 3:
You have a 300m parabolic curve located at 156.17m on PVC (Point of Vertical Curvature) with a slope of back tangent -4.2%.
Slope of Forward Tangent Calculation:
1. Using the symmetry of a parabolic curve:
\[ L = 300 \text{ meters} \]
The forward tangent's slope (given no external constraints) should create a symmetrical setup for elevation changes over the curve:
\[ x = 300 - 156.17 = 143.83 \text{ meters from PVC to PVT (Point of Vertical Tangency)} \]
2. Assuming a symmetrical vertical curve, the slope change is equally distributed:
\[ \text{Slope of forward tangent} = -\text{Slope of back tangent} = 4.2\% \]
The answers for your problems are:
1. 11+200
2. 5+747.37
3. 4.2% (assuming the curve is symmetric and there's no additional data suggesting otherwise).
PH