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Vertical curve -3.5% and +0.5%, IP 1300 m

Short answer

Equal-tangent vertical curve length 200 m. BVC at chainage 2500 m (RL 1303.5 m), EVC at 2700 m (RL 1300.5 m). Lowest point at chainage 2675 m, elevation 1300.438 m. Curve passes through 1301 m at the IP chainage.

A grade of (-)3.5% meets with another grade of 0.5%. The elevation and chainage of point of intersection are 1300 m and 2600 m respectively. Compute the suitable equal tangent vertical curve for full stations elevation assuming that vertical curve should pass through a point of elevation 2601 m at a chainage 1300 m. Take peg interval = 30 m.

In the question, the elevation and chainage of the given curve point appear to be interchanged (a chainage of 1300 m would be far away from the IP). The data are therefore taken as: IP at chainage 2600 m with RL 1300 m, and the curve passes through RL 1301 m at the IP chainage (2600 m).

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate ee of the curve.

Step 1: Length of the curve

The curve is a sag curve, so the curve lies above the IP:

e=1301−1300=1 m ⇒ ∣e∣=1 me = 1301 - 1300 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}

For an equal-tangent parabola, e=L ∣g2−g1∣800e = \dfrac{L\,|g_2-g_1|}{800} (grades in %), hence

L=800 e∣g2−g1∣=800×1∣0.5−(−3.5)∣=8004=200 mL = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|0.5-(-3.5)|} = \frac{800}{4} = 200\ \text{m}

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths L/2=100L/2 = 100 m.

Ch. of BVC=2600−100=2500 mCh. of EVC=2600+100=2700 mRL of BVC=1300−(−3.5)100×100=1303.5 mRL of EVC=1300+(0.5)100×100=1300.5 m\begin{aligned} \text{Ch. of BVC} &= 2600 - 100 = 2500\ \text{m}\\ \text{Ch. of EVC} &= 2600 + 100 = 2700\ \text{m}\\ \text{RL of BVC} &= 1300 - \frac{(-3.5)}{100}\times 100 = 1303.5\ \text{m}\\ \text{RL of EVC} &= 1300 + \frac{(0.5)}{100}\times 100 = 1300.5\ \text{m} \end{aligned}

Step 3: RL of curve points

Using the tangent-offset form of the parabola (xx measured from BVC, grades in %):

yx=RLBVC+g1x100+(g2−g1)x2200Ly_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}

with g1=−3.5%g_1 = -3.5\%, g2=0.5%g_2 = 0.5\% and L=200L = 200 m, the offset is y=(4)x2200×200y = \frac{(4)x^2}{200\times 200}. The pegs are taken at multiples of 30 m.

Chainage (m)x from BVC (m)Tangent RL (m)Offset y (m)Curve RL (m)
250001303.5000.0001303.500
2520201302.800+0.0401302.840
2550501301.750+0.2501302.000
2580801300.700+0.6401301.340
26101101299.650+1.2101300.860
26401401298.600+1.9601300.560
26701701297.550+2.8901300.440
27002001296.500+4.0001300.500

Check: the last curve RL (1300.5 m) equals the RL of EVC (1300.5 m), and at the IP chainage the curve lies 1 m above the IP, at RL 1301 m.

Step 4: Lowest point

At the lowest point the grade is zero, so x=−g1Lg2−g1=3.5×2004=175x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 200}{4} = 175 m from BVC.

  • Chainage = 2500 + 175 = 2675 m
  • RL = 1303.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 1300.438 m

Answer: L=200L = 200 m, BVC at 2500 m (RL 1303.5 m), EVC at 2700 m (RL 1300.5 m), lowest point at chainage 2675 m (RL 1300.438 m); curve levels as tabulated.

✓ Answer checked by Claude Opus 5.5 (high effort)

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