Example 2: Try This On Your Own......... The common tangent of a compound curve is 630m. If the central angles are 35° for the first curve and 50° for the second, find the degree of the second curve and its external distance given that the first curve is a 3º curve. Where: PCC = point of common curvature L = L₁ + L₂ = length of curve T₁+T₂ = common tangent distance 1 = 1₂ + 1₂ = central angle/angle of intersection common longent 28, sample figures LC PCC R₂

Structural Analysis
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Chapter2: Loads On Structures
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topic: horizontal curves (with figure please  need a detailed explanation per step  thankyou so much!)

Example 2: Try This On Your Own...
The common tangent of a compound curve is 630m. If the central angles are 35° for the first curve and 50° for
the second, find the degree of the second curve and its external distance given that the first curve is a 3º curve.
Where:
PCC = point of common curvature
L = L₁ + L₂ = length of curve
T₁+T₂ = common tangent distance
I= I₁ + 1₂ = central angle/angle of intersection
common langent
20₁
sample figures
LC
PCC
LC₂
R₂
PT
Transcribed Image Text:Example 2: Try This On Your Own... The common tangent of a compound curve is 630m. If the central angles are 35° for the first curve and 50° for the second, find the degree of the second curve and its external distance given that the first curve is a 3º curve. Where: PCC = point of common curvature L = L₁ + L₂ = length of curve T₁+T₂ = common tangent distance I= I₁ + 1₂ = central angle/angle of intersection common langent 20₁ sample figures LC PCC LC₂ R₂ PT
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