Question

A circular curve having a radius of curvature of \(1000\) m is set out by connecting two straights with a deflection angle of \(60^{\circ}\). The apex distance, in \(m\), is . (Rounded off to three decimal places)

Hint:

Use the geometry of the curve centre, the point of intersection, and the tangent point to relate the radius and half the deflection angle to the apex distance.

Answer 1

Correct Answer: 154.701

Step 1: Understand the geometry of the curve.
When two straight roads meet at a point called the point of intersection (PI), a circular curve is inserted between them to give a smooth turn. The angle between the two straights, measured as the change in direction, is the deflection angle \(\Delta\).
The apex distance (also called the external distance) is the length measured from the point of intersection straight along the bisector of \(\Delta\) up to the curve. It tells a surveyor how far back from the PI the curve actually bulges.

Step 2: Write the formula for apex distance.
Consider the centre of the curve O, radius R, and the point of intersection I. The line OI bisects the deflection angle, so the angle between OI and either tangent line is \(\Delta/2\).
In the right triangle formed by O, the tangent point, and I, the distance OI equals \(R/\cos(\Delta/2)\), because R is the side adjacent to the angle \(\Delta/2\) and OI is the hypotenuse.
The apex distance E is the part of OI that lies beyond the curve, so
\[ E = OI - R = R\left(\sec\frac{\Delta}{2} - 1\right) \]

Step 3: Substitute the given values.
Here \(R = 1000\) m and \(\Delta = 60^{\circ}\), so \(\Delta/2 = 30^{\circ}\).
\[ \cos 30^{\circ} = 0.866025 \]
\[ \sec 30^{\circ} = \frac{1}{0.866025} = 1.154701 \]
\[ E = 1000 \times (1.154701 - 1) = 1000 \times 0.154701 \]

Final Answer:
\[ \boxed{E = 154.701 \text{ m}} \]