6. Example 2.14:

A right hand compound curve, with radius of the first arc as 200 m and the radius of the second arc as 250 m, has a total deflection angle is 800. If the chainage of the point of intersection is 1504.80 m and the deflection angle of the first arc is 500, determine the chainages of the starting point, the point of compound curve and the point of tangency.

This chunk introduces the basic parameters of a compound curve. A compound curve consists of two circular arcs that are joined at a point of compound curvature (PCC). Here, we have two arcs with different radii and a specific total deflection angle, which helps determine the position of the curve relative to the tangent lines.

Think of riding a bike where the path curves to the right first before smoothing out further down. Each curve in the path can be likened to the arcs in our example.

Let, BA and BC be the back tangent and forward tangent. Δ = α + ꞵ = 800 Deflection angle of the second arc ꞵ = Δ - α = 800 - 500 = 300 MP = MT = t = R tan α/2 = 200 tan 50/2 = 93.26 m NP = NT = t = R tan ꞵ/2 = 200 tan 30/2 = 67.00 m MN = MP + NPN = 93.26 + 67.00 = 160.26 m.

This chunk provides the calculations necessary to find the tangent lengths (MP and NP) leading to the points where the curve begins and ends. The tangent lengths are important as they establish the straight paths from which the curves begin. The angles (α and ꞵ) are key to calculating the necessary tangent lengths using trigonometric functions.

Imagine you're drawing a curved road on a map. Before you start the curve, you need to draw straight lines that show how the road turns and bends.

In triangle BMN, ∠BMN + ∠BNM + ∠MBN = 1800 ∠MBN = 1800 – ∠BMN – ∠BNM = 1800 – 500 – 300 = 1000 Applying sine rule in triangle BMN. (BN / sin α) = MN / sin MBN BN = MN sin α / sin MBN = 160.26 sin 50 / sin 100 = 124.66 m.

Here, we apply the sine rule to relate the sides and angles in triangle BMN. This rule allows us to find unknown side lengths based on the known angles, which is crucial in geometry and applications such as navigation and surveying.

This is similar to using a map to find how far you would travel on a diagonal path between two streets, knowing the angles between the streets.

Tangent length BT = (T1) = BM + MT = 81.23 + 93.26 = 174.49 m Tangent length BT = (T2) = BN + NT = 124.66 + 67.00 = 191.66 m Length of the first curve = R α (π / 180) = 200 * 50 (π / 180) = 174.53 m Length of the Second curve = R ꞵ (π / 180) = 200 * 30 (π / 180) = 130.90 m Chainage at point of curve (T1) = Chainage at the point of intersection – tangent length (BT1) = 1504.80 – 174.53 = 1330.27 m. Chainage at point of PCC = chainage at the point of curve + length of first curve = 1330.27 + 174.53 = 1504.8 m Chainage at point of tangency (T2) = chainage at the point of PCC + length of second curve = 1504.8 + 130.90 = 1635.7 m.

This chunk covers the final steps of calculating the chainages at various points: the initial point of the curve (T1), the point of compound curvature (PCC), and the final point of tangency (T2). These calculations are crucial for accurately laying out the path in real-world applications, such as roads or railways.

Think about planning a new road route. You need to know exactly where to mark the starting point, where the road curves into a bend, and where it returns to a straight path after the curve.

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