Absolutely! Let's break down each part of your notes step by step: --- ### 1. **Diagram and Notations** - **TP1, TP2**: Tangent Points 1 & 2 (where the curve begins and ends) - **IP**: Intersection Point (where two tangents meet, before the curve) - **E**: Midpoint of the circular curve #### **Deflection Angle (ϕ)** - The angle between the two tangents at the intersection point (IP). - Here: \( \phi = 105^\circ - 45^\circ = 60^\circ \) --- ### 2. **Curve Design for 80 km/h Speed** #### **Given:** - Maximum superelevation, \( e_{\text{max}} = 8\% = 0.08 \) - Maximum desirable side friction, \( f = 0.16 \) #### **Formula for Minimum Radius \((R)\) of Curve:** \[ \frac{V^2}{127R} = e + f \] Where: - \( V \) = speed (km/h) - \( R \) = radius (m) - \( e \) = superelevation - \( f \) = side friction factor #### **Substitute the values:** \[ \frac{(80)^2}{127R} = 0.08 + 0.16 \] \[ \frac{6400}{127R} = 0.24 \] \[ 127R \times 0.24 = 6400 \] \[ R = \frac{6400}{127 \times 0.24} = \frac{6400}{30.48} \approx 209.97 \approx 210 \text{ m} \] --- ### 3. **Length of Curve** - Formula: \( \text{Length of curve} = R \times \phi \) (but \(\phi\) must be in radians) - Convert degrees to radians: \( \phi \) radians = \( \phi^\circ \times \frac{\pi}{180} \) So, \[ \text{Length} = R \times \frac{\pi}{180} \times \phi \] \[ = 210 \times \frac{\pi}{180} \times 60 \] \[ = 210 \times \frac{3.14}{180} \times 60 \] \[ = 210 \times 1.047 \] \[ = 219.91 \approx 220 \text{ m} \] --- ### 4. **Length of Arc (T\(_1\)E or T\(_2\)E)** Since E is the midpoint, \[ \text{Arc T}_1\text{E} = \frac{1}{2} (\text{T}_1\text{E}_1) = \frac{1}{2} \times 220 = 110 \text{ m} \] --- ## **Summary Table** | Parameter | Value/Formula | Result | |----------------------|-------------------------------------|-------------| | Radius of Curve (R) | \( V^2 / [127(e+f)] \) | 210 m | | Length of Curve | \( R \times \phi \) (in radians) | 220 m | | Length of half-arc | \( \frac{1}{2} \times \text{curve} \) | 110 m | --- ### **Let me know if you need more clarification on any step or formula!**

Certainly! Let's walk through the detailed explanation step by step, ensuring clarity at each stage. --- ### **Step 1: Understanding the Basic Data** - **Tangents**: - First tangent bearing: \(45^\circ\) - Second tangent bearing: \(105^\circ\) - **Deflection angle (\(\phi\))**: \[ \phi = 105^\circ - 45^\circ = 60^\circ \] This is the angle between the two tangents at the intersection point (IP). --- ### **Step 2: Determining the Minimum Radius of the Curve** **Given:** - Speed \(V = 80\, \text{km/h}\) - Maximum superelevation \(e_{\max} = 8\% = 0.08\) - Maximum side friction factor \(f = 0.16\) **Using the standard formula for minimum radius \(R\):** \[ \frac{V^2}{127 R} = e + f \] Rearranged: \[ R = \frac{V^2}{127 (e + f)} \] **Calculate:** \[ R = \frac{80^2}{127 \times (0.08 + 0.16)} = \frac{6400}{127 \times 0.24} = \frac{6400}{30.48} \approx 209.97\, \text{m} \] **Rounded:** \[ \boxed{ R \approx 210\, \text{m} } \] --- ### **Step 3: Length of the Curve** **Transform the deflection angle \(\phi\) into radians:** \[ \phi = 60^\circ = \frac{\pi}{3} \text{ radians} \] **Calculate arc length \(L\):** \[ L = R \times \phi = 210 \times \frac{\pi}{3} = 210 \times 1.0472 \approx 219.91\, \text{m} \] **Approximate to:** \[ \boxed{ L \approx 220\, \text{m} } \] --- ### **Step 4: Mid-Arc (Half of the Curve)** Half-arc length: \[ L/2 \approx 110\, \text{m} \] This is the point where trees or obstructions are often considered. --- ### **Step 5: Mid-ordinate Calculation (Obstruction Clearance)** The **mid-ordinate** is the perpendicular distance from the chord (straight line connecting the start and end points of the curve) to the arc at its midpoint. **Formula for mid-ordinate \(m\):** \[ m = R \times \left(1 - \cos \frac{\phi}{2}\right) \] **Compute:** \[ \frac{\phi}{2} = 30^\circ \] \[ m = 210 \times \left(1 - \cos 30^\circ\right) \] \[ \cos 30^\circ \approx 0.8660 \] \[ m = 210 \times (1 - 0.8660) = 210 \times 0.1340 \approx 28.14\, \text{m} \] **Interpretation:** - Any obstacle (like trees) placed at the midpoint must be set back at least **28.14 meters** from the centerline to avoid collision, considering the curvature. --- ### **Step 6: Position of Trees and Required Setback** Assuming: - Road width = 7 m (carriageway) + shoulders, etc. - The trees are located **1.25 m** from the edge of the roadway. **Distance from centerline to trees:** \[ \frac{\text{Road width}}{2} + 1.25\, \text{m} \] Suppose total half-road width is about 5 m (assuming a typical lane and shoulder), then: \[ \text{Distance from centerline} \approx 5 + 1.25 = 6.25\, \text{m} \] **Required clearance at mid-curve:** - Obstructions must be **at least 28.14 m** from the centerline. **Current position:** - Trees are only **6.25 m** from the centerline. **Therefore, the trees need to be cut back:** \[ 28.14\, \text{m} - 6.25\, \text{m} \approx 21.89\, \text{m} \] **Rounded:** \[ \boxed{ \text{Cutback distance} \approx 22\, \text{m} } \] --- ### **Summary of Key Results:** | Parameter | Value | Remarks | |------------------------------|----------------------------|------------------------------------------------------| | Minimum radius \(R\) | **210 m** | Based on speed and safety factors | | Length of curve \(L\) | **220 m** | Arc length for the curve | | Mid-ordinate (clearance) | **28.14 m** | Distance from centerline to obstruction at midpoint | | Tree setback required | **~22 m** | To ensure clearance and safety | --- ### **Optional: Visualization Help** Imagine the curve as a segment of a circle with radius 210 m. The mid-ordinate is the maximum distance from the chord (the straight line between start and end points) to the curve at its midpoint. Any obstruction at the midpoint must be placed beyond this distance from the centerline to avoid collision, hence the need for the 22 m setback. --- **Would you like me to prepare a simple sketch illustrating the mid-ordinate and the clearance?**