--- title: "Derive RL on parabolic vertical curve" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:parabolic-vertical-curve-rl-expression" source: https://www.bctengineer.com/q/semester_4/surveying_ii/parabolic-vertical-curve-rl-expression site: BCT Engineering --- # Derive RL on parabolic vertical curve Surveying II · Chapter 6: Curves - Asked in: 2072 Asoj, 6 marks - Video: [Solved Example of Vertical Curve (Crest Curve): Complete calculations of vertical parabolic curve](https://www.youtube.com/watch?v=wX8KNsVjoT0) — CIVIL Solved - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/parabolic-vertical-curve-rl-expression - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Derive the expression that in a parabolic shaped vertical curve, RL of any curve point 'P' is equal to $y_P = \frac{(g_2 - g_1)x^2}{200L} + \frac{g_1 x}{100} + \text{RL of BVC}$ where, $g_1$ and $g_2$ are percentage of grade of two tangents, $L$ is the total length of curve and $x$ is the chord distance taken from BVC. Also find the formula to determine lowest and highest point of the curve. ## Short answer For a parabolic vertical curve, rate of grade change constant yields RL at x from BVC as y = (g2−g1)x²/(200L) + g1 x/100 + RL_BVC, with grades in %. Highest/lowest point where grade zero: x = g1 L/(g1−g2); it lies on the curve only if g1 and g2 have opposite signs. ## Answer A parabolic vertical curve is used because the rate of change of grade is constant along the curve. This gives smooth riding and simple computation. #### Derivation Take the origin at BVC, $x$ horizontal along the chord distance, $y$ the level. The rate of change of grade is constant, so with grades in % (rise per 100 m): $$ \frac{d^2y}{dx^2} = \frac{g_2-g_1}{100\,L} $$ Integrate once. At $x=0$ the slope is the first grade, $dy/dx = g_1/100$: $$ \frac{dy}{dx} = \frac{g_1}{100} + \frac{(g_2-g_1)\,x}{100L} $$ Check: at $x = L$, $dy/dx = g_2/100$, as required. Integrate again, with $y = \text{RL of BVC}$ at $x=0$: $$ y_P = \frac{(g_2-g_1)\,x^2}{200L} + \frac{g_1 x}{100} + \text{RL of BVC} $$ The first term is the offset of the curve from the tangent, and $g_1x/100 + \text{RL}_{BVC}$ is the level on the first tangent. #### Highest or lowest point At the highest (summit) or lowest (sag) point the grade is zero, $dy/dx = 0$: $$ \frac{g_1}{100} + \frac{(g_2-g_1)\,x}{100L} = 0 \quad\Rightarrow\quad x = \frac{g_1 L}{g_1-g_2} $$ Substituting this $x$ in the level equation gives the RL of that point: $$ \text{RL}_{max/min} = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L} = \text{RL}_{BVC} + \frac{g_1 x}{200} $$ If $g_1 > g_2$ (summit curve) this is the highest point; if $g_1 < g_2$ (sag curve) it is the lowest point. A turning point exists on the curve only if $0 < x < L$, i.e. $g_1$ and $g_2$ have opposite signs. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is a parabola used for a vertical curve? A. Rate of change of grade is constant B. Grade is constant C. Radius is constant D. Offsets are constant Answer: A. d2y/dx2 is constant. (Key point: Why parabola) 2. Starting equation is A. dy/dx = (g2 - g1)/(100L) B. d2y/dx2 = (g2 - g1)/(100L) C. d2y/dx2 = g1/100 D. d2y/dx2 = (g2 - g1)x/L Answer: B. Constant change of grade over L. (Key point: Second derivative) 3. Integration constants come from A. y = 0 at x = L B. dy/dx = 0 at x = 0 C. dy/dx = g1/100 and y = RL BVC at x = 0 D. dy/dx = g2/100 at x = 0 Answer: C. Origin at BVC. (Key point: Boundary conditions) 4. Highest or lowest point is at A. x = L/2 always B. x = g2 L/(g1 + g2) C. x = (g1 - g2)/L D. x = g1 L/(g1 - g2), only if 0 < x < L Answer: D. Set dy/dx = 0; it needs g1 and g2 of opposite signs. 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