1. **State the problem:** We need to analyze an unsymmetrical vertical curve between the Point of Curvature (PC) and Point of Tangency (PT) with given grades and elevations. 2. **Given data:** - Grade 1 ($G_1$) = -5% = -0.05 - Grade 2 ($G_2$) = +6% = 0.06 - Station at PC = 15+940 = 15940 m, Elevation at PC = 356.8 m - Station at PT = 16+540 = 16540 m, Elevation at PT = 354.8 m - Total length of curve $L = 16540 - 15940 = 600$ m 3. **Formulas and rules:** - The length of the curve $L = L_1 + L_2$ where $L_1$ and $L_2$ are lengths from PC to the lowest point and from lowest point to PT respectively. - The lowest point occurs where the slope is zero, i.e., where the tangent to the curve is horizontal. - The elevation at any station $x$ along the curve can be found using the vertical curve equation: $$y = y_{PC} + G_1 x + \frac{(G_2 - G_1)}{2L} x^2$$ where $x$ is the distance from PC. 4. **Calculate $L_1$ and $L_2$:** - The slope changes from $G_1$ to $G_2$ over length $L$. - The lowest point occurs where slope = 0: $$0 = G_1 + \frac{(G_2 - G_1)}{L} L_1$$ Solving for $L_1$: $$L_1 = -\frac{G_1 L}{G_2 - G_1} = -\frac{-0.05 \times 600}{0.06 - (-0.05)} = \frac{30}{0.11} = 272.73\text{ m}$$ - Then $L_2 = L - L_1 = 600 - 272.73 = 327.27$ m 5. **Verify $L_1 + L_2 = 600$ m:** $$272.73 + 327.27 = 600$$ Correct. 6. **Calculate elevation at lowest point:** - Distance from PC to lowest point is $L_1 = 272.73$ m. - Using vertical curve formula: $$y = 356.8 + (-0.05) \times 272.73 + \frac{(0.06 - (-0.05))}{2 \times 600} \times (272.73)^2$$ Calculate stepwise: - $G_1 x = -0.05 \times 272.73 = -13.6365$ - $\frac{(G_2 - G_1)}{2L} = \frac{0.11}{1200} = 0.00009167$ - $x^2 = (272.73)^2 = 74482.64$ - Quadratic term = $0.00009167 \times 74482.64 = 6.825$ - Elevation at lowest point: $$356.8 - 13.6365 + 6.825 = 350.0\text{ m (approx)}$$ 7. **Summary:** - $L_1 = 272.73$ m - $L_2 = 327.27$ m - Lowest point elevation = 350.0 m at station $15940 + 272.73 = 16212.73$ 8. **Mark stations every 40 m:** - Stations from PC (15940) to PT (16540) at intervals of 40 m: 15940, 15980, 16020, ..., 16540 - Elevations can be calculated using the vertical curve formula for each station. **Final answers:** - $L_1 = 272.73$ m - $L_2 = 327.27$ m - Lowest point elevation = 350.0 m at station 16212.73