1. A reverse circular curve consists of two circular arcs bending in opposite directions that share a common tangent. They are commonly used in railway sidings and roads meant for low speeds. 2. To calculate a reverse curve, the first curve is dropped from its point of commencement (P.C.) using tables and methods. Then the second curve is dropped from the point of reverse curvature (P.R.C.). 3. An example problem calculates the main stations of a horizontal reverse circular curve given the elements of the two curves and their radii, central angles, and stations of points of curvature, reverse curvature, and tangency.

1. The Thirteenth Lecture
1
.‫ﺍﻟﺒﺴﻴﻄﻴﻦ‬ ‫ﺍﻟﺪﺍﺋﺮﻳﻴﻦ‬ ‫ﺍﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫ﻋﻨﺎﺻﺮ‬ ‫ﺣﺴﺎﺏ‬ ‫ﻓﻲ‬ ‫ﺍﻟﺴﺎﺑﻘﺔ‬ ‫ﺍﻟﻘﻮﺍﻧﻴﻦ‬ ‫ﻧﻔﺲ‬ ‫ﻳﺴﺘﺨﺪﻡ‬
Surveying Engineering
Reverse Circular Curves:
A reverse curve is made up of two arcs having equal or different radius bending in the
opposite direction with a common tangent at their junction. Reverse curves are used when the
straights are parallel or intersect at a very small angle.
They are commonly used in railway sidings and sometimes on railway tracks and roads meant
for low speeds. They should be avoided as far as possible on main lines and highways where
speeds are necessarily high.
Calculation of Reverse Circular Curve:-
Elements calculate two simple circular curves using the same previous methods…
T = R tan
∆
2
L =
𝜋 𝑅 ∆ᵒ
180ᵒ
C = 2 R sin
∆
2
‫ﻣﻨﺤﻨﻴﻴﻦ‬ ‫ﻣﻦ‬ (‫)ﺍﻟﻤﻘﻠﻮﺏ‬ ‫ﺍﻟﻤﻌﻜﻮﺱ‬ ‫ﺍﻟﺪﺍﺋﺮﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻳﺘﺄﻟﻒ‬
‫ﻭﻳﻜﻮﻥ‬ ‫ﺑﻪ‬ ‫ﺧﺎﺹ‬ ‫ﻭﻣﺮﻛﺰ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫ﻣﻨﻬﻤﺎ‬ ‫ﻟﻜﻞ‬ ‫ﺑﺴﻴﻄﻴﻦ‬ ‫ﺩﺍﺋﺮﻳﻴﻦ‬
‫ﻧﺼﻒ‬ ‫ﺍﻥ‬ ‫ﻛﻤﺎ‬ ,‫ﺍﻻﺧﺮ‬ ‫ﺍﻧﺤﻨﺎﺀ‬ ‫ﺍﺗﺠﺎﺓ‬ ‫ﺑﻌﻜﺲ‬ ‫ﺍﺣﺪﻫﻤﺎ‬ ‫ﺍﻧﺤﻨﺎﺀ‬ ‫ﺍﺗﺠﺎﻩ‬
‫ﺍﻟﻴﻤﻴﻦ‬ ‫)ﺍﻟﻰ‬ ‫ﺍﻟﻤﺸﺘﺮﻙ‬ ‫ﺍﻟﻤﻤﺎﺱ‬ ‫ﻣﻦ‬ ‫ﺟﻬﺔ‬ ‫ﻓﻲ‬ ‫ﺍﺣﺪﻫﻤﺎ‬ ‫ﻣﺮﻛﺰ‬ ‫ﻗﻄﺮ‬
‫ﺍﻟﻤﻤﺎﺱ‬ ‫ﻣﻦ‬ ‫ﺍﻻﺧﺮﻯ‬ ‫ﺍﻟﺠﻬﺔ‬ ‫ﻓﻲ‬ ‫ﺍﻻﺧﺮ‬ ‫ﻭﻣﺮﻛﺰ‬ ‫ﻗﻄﺮ‬ ‫ﻭﻧﺼﻒ‬ (‫ﻣﺜﻼ‬
.(‫ﻣﺜﻼ‬ ‫ﺍﻟﻴﺴﺎﺭ‬ ‫)ﺍﻟﻰ‬ ‫ﺍﻟﻤﺸﺘﺮﻙ‬
‫ﺍﻟﺤﺪﻳﺪﻳﺔ‬ ‫ﺍﻟﺴﻜﻚ‬ ‫ﻓﻲ‬ ‫ﻋﺎﺩﺓ‬ ‫ﺍﻟﻤﻌﻜﻮﺱ‬ ‫ﺍﻟﺪﺍﺋﺮﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻳﺴﺘﺨﺪﻡ‬
‫ﺍﻟﻄﺮﻕ‬ ‫ﻓﻲ‬ ‫ﻭﺧﺎﺻﺘﺎ‬ ‫ﺍﻟﻤﻨﺨﻔﻀﺔ‬ ‫ﻟﻠﺴﺮﻋﺎﺕ‬ ‫ﺍﻟﻤﺨﺼﺼﺔ‬ ‫ﻭﺍﻟﻄﺮﻕ‬
‫ﺍﻟﺮﺋﻴﺴﻴﺔ‬ ‫ﺍﻟﺨﻄﻮﻁ‬ ‫ﻋﻠﻰ‬ ‫ﺍﻹﻣﻜﺎﻥ‬ ‫ﻗﺪﺭ‬ ‫ﺗﺠﻨﺒﻬﺎ‬ ‫ﻳﺠﺐ‬ .‫ﺍﻟﻮﻋﺮﺓ‬ ‫ﺍﻟﺠﺒﻠﻴﺔ‬
‫ﺑﺎﻟﻀﺮﻭﺭﺓ‬ ‫ﻋﺎﻟﻴﺔ‬ ‫ﺍﻟﺴﺮﻋﺎﺕ‬ ‫ﺗﻜﻮﻥ‬ ‫ﺣﻴﺚ‬ ‫ﺍﻟﺴﺮﻳﻌﺔ‬ ‫ﻭﺍﻟﻄﺮﻕ‬

2. The Thirteenth Lecture
2
‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﺗﺴﻘﻴﻂ‬ ‫ﺛﻢ‬ (P.C) ‫ﻣﻦ‬ ‫ﺍﺑﺘﺪﺍﺀﺍ‬ ‫ﺍﻻﻭﻝ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﺑﺘﺴﻘﻴﻂ‬ ‫ﺍﻟﻤﻌﻜﻮﺱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﺗﺴﻘﻴﻂ‬ ‫ﻳﺘﻢ‬
‫ﻣﻊ‬ ‫ﻣﻨﻬﺎ‬ ‫ﻟﻜﻞ‬ ‫ﻟﻠﺘﺴﻘﻴﻂ‬ ‫ﺍﻟﻤﺨﺘﺎﺭﺓ‬ ‫ﻭﺍﻟﻄﺮﻳﻘﺔ‬ ‫ﺍﻟﺠﺪﺍﻭﻝ‬ ‫ﻭﺑﺎﺳﺘﺨﺪﺍﻡ‬ (P.R.C) ‫ﻣﻦ‬ ‫ﺍﺑﺘﺪﺍﺀﺍ‬ ‫ﺍﻟﺜﺎﻧﻲ‬
‫ﺍﻟﻨﻘﺎﻁ‬ ‫ﻣﻮﺍﻗﻊ‬ ‫ﻟﺘﻌﻴﻴﻦ‬ ‫ﺳﺘﺜﺒﺖ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺍﻟﻤﻤﺎﺳﺔ‬ ‫ﺍﻟﺰﺍﻭﻳﺔ‬ ‫ﻭﻣﺮﺍﻋﺎﺓ‬ ‫ﻣﻨﻬﻤﺎ‬ ‫ﻟﻜﻞ‬ ‫ﺍﻟﺘﺴﻘﻴﻂ‬ ‫ﺍﺗﺠﺎﻩ‬ ‫ﻣﻼﺣﻈﺔ‬
.‫ﺍﻟﻤﻨﺤﻨﻲ‬
P.R.C.: Point of Reverse Curvature
R₁, ∆₁, T₁, L₁, C₁ & P.I₁: Elements of the first curve.
R₂, ∆₂, T₂, L₂, C₂ & P.I₂: Elements of the second curve.
St. P.C = St. P.I₁ – T₁
St. P.R.C = St. P.C + L₁
St. P.T = St. P.R.C + L₂
St. P.I₂ = St. P.R.C + T₂
Example: - Compute the main stations of reverse circular horizontal curve, if (C₁=244m, ∆₁=42ᵒ
36′, St. P.R.C=51+42.50, T₂=108m & R₂=325m).
Solution: -
C₁ = 2 R₁ → R₁ = = = 335.86msin
∆₁
2
C₁
2 sin
∆₁
2
244
2 sin 21ᵒ 18′
T₁ = R₁ = 355.86 = 130.95mtan
∆₁
2 × tan 21ᵒ 18′
L₁ = = = 249.95m
𝜋 𝑅₁ ∆₁ᵒ
180ᵒ
3.14156 × 335.86 × 21ᵒ 18′
180ᵒ
T₂ = R₂ → = = = → =tan
∆₂
2 tan
∆₂
2
T₂
R₂
108
325 18ᵒ 23′ ∆₂ 36ᵒ 46′
L₂ = = = 208.552m
𝜋 𝑅₂ ∆₂ᵒ
180ᵒ
3.14156 × 325 × 36ᵒ 46′
180ᵒ
C₂ = 2 R₂ = 2 325 = 204.99msin
∆₂
2 × × sin 18ᵒ 23′
St. P.C = St. P.I₁ – T₁ = (51+42.50) – (2+49.72) = (48+92.78)
St. P.R.C = St. P.C + L₁ = (51+42.50) + (2+08.55) = (53+51.05)
St. P.T = St. P.R.C + L₂ = (48+92.78) + (1+30.95) = (50+23.73)
St. P.I₂ = St. P.R.C + T₂ = (51+42.50) + (1+08) = (52+50.50)
.‫ﺍﻻﻭﻝ‬ ‫ﺍﻟﺪﺍﺋﺮﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻋﻨﺎﺻﺮ‬
.‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺪﺍﺋﺮﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻋﻨﺎﺻﺮ‬
.‫ﺍﻟﻤﻌﻜﻮﺱ‬ ‫ﺍﻟﺘﻘﻮﺱ‬ ‫ﺍﻭ‬ ‫ﺍﻟﺘﺤﺪﺏ‬ ‫ﻧﻘﻄﺔ‬