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CircularCurve ArcImplementation Calculation Example RoadDesign # Calculation example – Road design– Circular arc implementation 2

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How to implement on the ground the main components (start, middle, end points) of a circular curve with known radius R, tangent to two main alignments E1 and E2, where node K (intersecting point of the two main tangents) is not accessible to occupy.

**Known** **values**: Curve radius **R**, Tangents **E1** and **E2**.

**Solution**: Positions of main components **A**, **M**, **T** on the ground.

**Step 1**: Define two auxiliary points **B** and **C** reciprocally visible along tangents **E1** and **E2** respectively, and measure length **(BC)**.

**Step 2**: Measure angles **δ _{1}** and

**Step 3**: Applying sine law on triangle **KBC**:

Calculate **(KB)** and **(KC)**:

**Step 4**: Calculate lengths **(KA)** and **(KT)** along tangents by entering known values for angle **α** and radius **R** on the following functions:

**Step 5**: Calculate lengths **(BA)** and **(CT)**, occupy points **B** and **C** and define points **A** and **T** respectively:

**Step 6**: To define arc’s middle point **M**, apply sine law on triangle **BKD** and calculate length **(BD)** along **BC** line:

**Step 7**: Define and occupy point **D** on the ground by measuring length **(BD)** along direction **BC**, define angle **ω** and direction of line **KM** and finally define middle point **M** by measuring length **(DM)** along direction of line **KM**.

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Given the rate and direction of full dip of a plan...