Explanation L S = 1 + 00.00 f t D = 2 ∘ 30 ’ , I = 10 ∘ 30 ’ and P I s t a t i o n 36 + 44.5 f t Spiral angle Δ s = L s D 2 Δ s = 1 × 2 ∘ 30 ' 2 = 1 ∘ 15 ' Where, LS is the arc distance from the TS to the SC or CS to ST. D is the degree of curvature. Approximate formulas for computing X and Y: X = L S ( 100 − 0.003046 Δ s 2 ) X = 1 × ( 100 − 0.003046 × 1 ∘ 15 ' 2 ) = 99.992385 f t Y = L S ( 0.58178 Δ s − 0.003046 Δ s 3 ) Y = 1 × ( 0.58178 × 1 ∘ 15 ' − 0.003046 × 1 ∘ 15 ' 3 ) = 0.72127 f t Radius R = 5729.58 2 ∘ 30 ' = 2291.832 Length of curve L = R I L = 2291.832 × 10 ∘ 30 ' × π 180 ∘ = 420.00 f t Spiral throw o = Y − R ( 1 − cos Δ s ) o = 0.72127 f t − 2291.832 × ( 1 − cos 1 ∘ 30 ' ) = 0.175876 Distance H H = X − R sin Δ s + ( R + o ) tan I 2 H = 99.992385 f t − 2291.832 sin 1 ∘ 15 ' + ( 2291.832 + 0.175876 ) tan 10 ∘ 30 ' 2 H = 260.60227 f t P I s t a t i o n = 36 + 44

Classification of curve notes to stake out the alignment from the TS to ST at full stations in the form of a table using different methods.

ELEMENTARY SURVEYING (LOOSELEAF)

15th Edition

ISBN: 9780134604701

Author: GHILANI

Publisher: PEARSON

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Chapter 24, Problem 24.31P

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Classification of curve notes to stake out the alignment from the TS to ST at full stations in the form of a table using different methods.

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