8. Example 2.16:

Two straights BA and AC are intersected by a line EF so that angles BEF and EFC are 1400 and 1450, respectively. The radius of the first curve is 600 m and the second curve is 400 m. If the chainages of the intersection point A is 3415 m, compute the chainages of the tangent points and the point of compound curvature.

In this problem, we are working with two straight lines or tangents (AB and AC) that intersect at point A. There is a line (EF) that connects the two tangents, and it creates two angles at point A. The angles BEF and EFC represent how much the curve will bend between straight paths, which is critical when designing roads or railways to ensure safety and efficiency. The goal is to find out where the curves (which have specified radii) will end relative to the point of intersection.

Imagine you're at a road intersection where two straight roads meet. You want to add a curved road that connects these two straight roads smoothly. The angles at the intersection and how sharp the curves are (the radii of the curves) will affect how the new road connects to the existing roads.

In the given problem, the angles are calculated as follows:

• Angle AEF = α = 1800 - Angle BEF = 1800 - 1400 = 400
• Angle AFE = β = 1800 - Angle EFC = 1800 - 1450 = 350

To solve the problem, we first need to convert the given angles into more usable forms. Using basic geometry, the angles created after subtracting the known angles from 180 degrees gives us α and β, the angles needed to describe how the curves will bend. This is crucial since these angles will dictate how the curve will be laid out on the ground.

Think of a pizza cut into slices. The angle of each slice determines how much pizza you get. In the same way, by adjusting the angles of the new curve at the intersection, we can control how the road will flow from one straight section to another.

The tangent lengths are calculated as follows:

• ET = EN = t = R tan α / 2 = 600 tan 40/2 = 218.40 m
• FT = FN = t = R tan β / 2 = 400 tan 35/2 = 126.12 m
• EF = EN + FN = 218.40 + 126.12 = 344.52 m

Each tangent length represents the straight segment of the curved road before it begins to turn. The formulas used include radius (R) and the angles we calculated earlier to find how long those line segments will be. Adding up the lengths of these tangent segments gives us the total length of the 'straight' before turning begins.

Imagine driving on a straight road and then approaching a curve. The distance you travel straight before you start to turn represents the tangent length. It’s important to know this distance for designing safe roads and for ensuring drivers can smoothly transition into the curve.

In the triangle AEF, applying the sine rule allows us to find the distances AE and AF, which are needed for further calculations to establish the tangent points.

The sine rule involves using the sides of triangles and their corresponding angles to find unknown lengths. By applying this rule in our scenario, we derive distances AE and AF, which help us figure out how far each tangent point is from the intersection at A, essential for mapping out the road layout.

If you are trying to measure a boat's distance to dock while navigating, it’s like using the sine rule to determine how far away the dock is at various angles — this helps you plot the safest and most efficient route to reach it.

The lengths of the curves are computed as follows:

• Length of the first curve = π R α / 180 = 600 * 40 π / 180 = 418.90 m
• Length of the Second curve = π R β / 180 = 400 * 35 π / 180 = 244.35 m

The length of each curve is calculated using the formula involving the radius and the angles. This gives a precise measurement of how long the curves will span, which is necessary for construction and road safety assessments.

Think of it like unrolling a ribbon for each curve. The radius represents how much curvature there is, while the angle tells you how ‘tight’ or ‘loose’ the ribbon is wound. Knowing how long each piece of ribbon needs to be is key to ensuring they fit nicely without overlapping or leaving gaps.

The final computations for chainage are as follows:

• Chainage at point of curve (T1) = Chainage at the point of intersection - tangent length (T1) = 3415 – 423 = 2992 m.
• Chainage at point PCC (N) = Chainage at the point of curve + Length of first curve = 2992 + 418.90 = 3410.9 m.
• Chainage at point of tangency (T2) = Chainage at the point of PCC + Length of second curve = 3410.9 + 244.35 = 3655.25 m.

Finally, we calculate the exact points along the road where each curve starts and ends, which are crucial for construction. The chainages tell us the position along a roadway, while considering the lengths of tangents and curves ensures an accurate representation of the road layout.

Think of these chainage points as milestones on a race track. Each milestone shows you where to make your turns and how far you've come, ensuring that participants know exactly where they are on the track when racing.

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