1. **Problem statement:** We have two grades intersecting at station 14+262 (14262 m) with elevation 153.76 m. The first grade is +3% (0.03 slope) and the second is -2% (-0.02 slope). A vertical curve is formed with 130 m length backward and 80 m length forward from the intersection point, creating an unsymmetrical summit curve. We need to find the elevation of the summit. 2. **Formula and explanation:** The elevation $y$ along a vertical curve is given by: $$y = y_{PI} + g_1 x + \frac{(g_2 - g_1)}{2L} x^2$$ where: - $y_{PI}$ is the elevation at the point of intersection (PI), here 153.76 m - $g_1$ is the initial grade (backward), here 3% or 0.03 - $g_2$ is the final grade (forward), here -2% or -0.02 - $L$ is the total length of the vertical curve, $L = L_1 + L_2 = 130 + 80 = 210$ m - $x$ is the distance from the PI along the curve where elevation is calculated The summit occurs at the vertex of the parabola, where the derivative of elevation with respect to $x$ is zero: $$\frac{dy}{dx} = g_1 + \frac{(g_2 - g_1)}{L} x = 0$$ 3. **Find $x$ coordinate of the summit:** $$g_1 + \frac{(g_2 - g_1)}{L} x = 0$$ $$0.03 + \frac{-0.02 - 0.03}{210} x = 0$$ $$0.03 + \frac{-0.05}{210} x = 0$$ $$0.03 - \frac{0.05}{210} x = 0$$ $$\frac{0.05}{210} x = 0.03$$ $$x = \frac{0.03 \times 210}{0.05}$$ $$x = \frac{6.3}{0.05} = 126$$ 4. **Check if $x$ is within the curve length:** Since $x=126$ m is measured from PI forward (positive $x$), and forward length is 80 m, but $126 > 80$, the summit lies beyond the forward curve length. However, since the curve is unsymmetrical, $x$ can be negative (backward) or positive (forward). Here, $x=126$ m forward is outside the forward curve length, so the summit is beyond the curve. But the problem states the curve lengths are 130 m backward and 80 m forward, so the total curve length is 210 m, and $x$ is measured from PI with positive forward and negative backward. Since $x=126$ m is positive and less than total length 210 m, but greater than forward length 80 m, the summit is beyond the forward curve length, which is not possible physically. So we must consider $x$ measured from the start of the curve (backward end). Alternatively, measure $x$ from the start of the curve (backward end): - Backward length $L_1=130$ m - Forward length $L_2=80$ m - Total length $L=210$ m The vertex position relative to the start of the curve is: $$x_v = L_1 + x = 130 + 126 = 256$$ which is beyond total length 210 m, so this is inconsistent. Therefore, $x$ is measured from PI, positive forward, negative backward. Since $x=126$ m is beyond forward length 80 m, the summit is beyond the curve forward end. This suggests the summit is at the start of the curve backward end or at the PI. 5. **Reconsider the sign of $x$:** The formula assumes $x$ positive forward from PI. The vertex formula is: $$x = -\frac{g_1 L}{g_2 - g_1}$$ Plugging in values: $$x = -\frac{0.03 \times 210}{-0.02 - 0.03} = -\frac{6.3}{-0.05} = 126$$ So $x=126$ m forward from PI. Since forward length is only 80 m, the summit lies beyond the forward curve length, which is impossible. 6. **Conclusion:** The summit is at the start of the curve backward end (at $x = -130$ m) or at the PI. 7. **Calculate elevation at $x=126$ m (forward) anyway:** $$y = 153.76 + 0.03 \times 126 + \frac{-0.02 - 0.03}{2 \times 210} \times 126^2$$ Calculate each term: $$0.03 \times 126 = 3.78$$ $$\frac{-0.05}{420} = -0.00011905$$ $$126^2 = 15876$$ $$-0.00011905 \times 15876 = -1.89$$ Sum: $$y = 153.76 + 3.78 - 1.89 = 155.65$$ 8. **Calculate elevation at $x=-130$ m (backward end):** $$y = 153.76 + 0.03 \times (-130) + \frac{-0.05}{420} \times (-130)^2$$ Calculate each term: $$0.03 \times (-130) = -3.9$$ $$(-130)^2 = 16900$$ $$-0.00011905 \times 16900 = -2.01$$ Sum: $$y = 153.76 - 3.9 - 2.01 = 147.85$$ 9. **Calculate elevation at PI ($x=0$):** $$y = 153.76$$ 10. **Since the summit elevation is highest point, and $y=155.65$ m at $x=126$ m is highest, but $x=126$ m is beyond forward curve length, the summit is at the forward end of the curve ($x=80$ m):** Calculate elevation at $x=80$ m: $$y = 153.76 + 0.03 \times 80 + \frac{-0.05}{420} \times 80^2$$ $$0.03 \times 80 = 2.4$$ $$80^2 = 6400$$ $$-0.00011905 \times 6400 = -0.7619$$ Sum: $$y = 153.76 + 2.4 - 0.7619 = 155.398$$ 11. **Final answer:** The summit elevation is approximately **155.398 m** rounded to 3 decimal places.