Compound circular curves consist of two or more simple circular curves that bend in the same direction with different radii but share a common tangent. Their centers lie on the same side of this common tangent. Compound curves allow for better fitting of transportation centerlines to complex terrain. To calculate the elements of a two-centered compound circular curve, trigonometric relationships are used based on the common deflection angle and tangent lengths of the component curves. The curves are then projected from their points of curvature or point of compound curvature using standard calculation tables.

1. The Twelfth Lecture
1
Surveying Engineering
Compound Circular Curves:
A compound curve consists of two or more simple curves having different radii bending
in the same direction and lying on the same side of the common tangent. Their centers lie on the
same side of the curve.
Use of compound curves permits better fitting of highway and railway center lines to difficult
topographic conditions.
‫ﻧﺼﻔﺎ‬ ‫ﻟﻬﻤﺎ‬ ‫ﺍﻛﺜﺮ‬ ‫ﺃﻭ‬ ‫ﺩﺍﺋﺮﻳﻴﻦ‬ ‫ﻣﻨﺤﻨﻴﻴﻦ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻤﺮﻛﺐ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻳﺘﺄﻟﻒ‬
‫ﻟﻠﻤﻨﺤﻨﻴﻴﻦ‬ ‫ﺍﻻﻧﺤﻨﺎﺀ‬ ‫ﺍﺗﺠﺎﻩ‬ ‫ﻭﻳﻜﻮﻥ‬ ‫ﺍﻛﺜﺮ‬ ‫ﺍﻭ‬ ‫ﻭﻣﺮﻛﺰﻳﻴﻦ‬ ‫ﺍﻛﺜﺮ‬ ‫ﺍﻭ‬ ‫ﻗﻄﺮﻳﻦ‬
‫ﺍﻟﻘﻄﺮﻳﻴﻦ‬ ‫ﻧﺼﻔﻲ‬ ‫ﺍﻥ‬ ‫ﻭﻛﻤﺎ‬ ,‫ﺍﻻﻭﻝ‬ ‫ﺍﻟﺮﺋﻴﺴﻲ‬ ‫ﺍﻟﻤﻤﺎﺱ‬ ‫ﻣﻦ‬ ‫ﻭﺍﺣﺪﺓ‬ ‫ﺟﻬﺔ‬ ‫ﻓﻲ‬
.‫ﺍﻟﻤﺸﺘﺮﻙ‬ ‫ﺍﻟﻤﻤﺎﺱ‬ ‫ﻣﻦ‬ ‫ﻭﺍﺣﺪﺓ‬ ‫ﺟﻬﺔ‬ ‫ﻓﻲ‬ ‫ﻳﻘﻌﺎﻥ‬ ‫ﺍﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫ﻭﻣﺮﻛﺰﻱ‬
‫ﻃﺒﻴﻌﺔ‬ ‫ﺗﺴﺘﺪﻋﻲ‬ ‫ﺍﻟﺘﻲ‬ ‫ﺍﻟﺤﺎﻻﺕ‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺮﻛﺐ‬ ‫ﺍﻟﺪﺍﺋﺮﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻳﺴﺘﺨﺪﻡ‬
‫ﺍﻟﺨﻂ‬ ‫ﻳﺘﺒﻊ‬ ‫ﺍﻥ‬ ‫ﻳﺠﺐ‬ ‫ﺣﻴﺚ‬ ‫ﺍﻟﺠﺒﻠﻴﺔ‬ ‫ﺍﻻﺭﺍﺿﻲ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫ﺍﺳﺘﺨﺪﺍﻣﺔ‬ ‫ﺍﻻﺭﺽ‬
‫ﺍﻟﺘﻘﺎﻃﻌﺎﺕ‬ ‫ﻓﻲ‬ ‫ﻳﺴﺘﺨﺪﻡ‬ ‫ﻭﺍﻳﻀﺎ‬ ,‫ﻭﺍﻟﺠﺒﺎﻝ‬ ‫ﺍﻟﻤﺮﺗﻔﻌﺎﺕ‬ ‫ﺣﺎﻓﺎﺕ‬ ‫ﺍﻟﻤﺮﻛﺰﻱ‬
‫ﻓﻲ‬ ‫ﺍﻟﺘﺪﺭﻳﺠﻲ‬ ‫ﺍﻟﺘﻐﻴﺮ‬ ‫ﻟﻐﺮﺽ‬ ‫ﺍﻟﻤﻌﺰﻭﻟﺔ‬ ‫ﻭﺍﻟﺘﻘﺎﻋﺎﺕ‬ ‫ﺍﻟﻜﺒﻴﺮﺓ‬ ‫ﺍﻟﺴﻄﺤﻴﺔ‬
.‫ﻟﻠﺮﻛﺎﺏ‬ ‫ﻭﺍﻟﺮﺍﺣﺔ‬ ‫ﺍﻟﺴﻼﻣﺔ‬ ‫ﻳﻀﻤﻦ‬ ‫ﺑﺤﻴﺚ‬ ‫ﺍﻻﻧﺤﻨﺎﺀ‬
Two-Centered Compound Circular Curve:-
C.T: Common tangent.
∆: Common deflection angle.
P.C.C: Point of compound curvature.
Ta: Larger tangent length.
.‫ﺍﻟﻤﺸﺘﺮﻙ‬ ‫ﺍﻟﻤﻤﺎﺱ‬
.‫ﺍﻟﻜﻠﻴﺔ‬ ‫ﺍﻻﻧﺤﺮﺍﻑ‬ ‫ﺯﺍﻭﻳﺔ‬
.‫ﺍﻟﻤﺮﻛﺐ‬ ‫ﺍﻟﺘﻘﻮﺱ‬ ‫ﺍﻭ‬ ‫ﺍﻟﺘﺤﺪﺏ‬ ‫ﻣﺤﻄﺔ‬
.(‫ﺍﻟﺨﻠﻔﻲ‬ ‫)ﺍﻟﻤﻤﺎﺱ‬ ‫ﺍﻻﻭﻝ‬ ‫ﺍﻟﺮﺋﻴﺴﻲ‬ ‫ﺍﻟﻤﻤﺎﺱ‬

2. The Twelfth Lecture
2
-:‫ﺍﻻﺗﻲ‬ ‫ﺑﺤﺴﺎﺏ‬ ‫ﻧﻘﻮﻡ‬ ‫(ﻭ‬Tb ,Ta) ‫ﻭﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻻﻭﻝ‬ ‫ﺍﻟﺮﺋﻴﺴﻲ‬ ‫ﺍﻟﻤﻤﺎﺱ‬ ‫ﻃﻮﻝ‬ ‫ﺣﺴﺎﺏ‬ ‫ﻟﻐﺮﺽ‬
.(C.T) ‫ﺍﻟﻤﺸﺘﺮﻙ‬ ‫ﺍﻟﻤﻤﺎﺱ‬ ‫ﻃﻮﻝ‬ 
.(∆) ‫ﺍﻟﻜﻠﻴﺔ‬ ‫ﺍﻻﻧﺤﺮﺍﻑ‬ ‫ﺯﻭﺍﻳﺔ‬ 
...‫ﺗﺤﺴﺐ‬ ‫ﺍﻟﻤﺜﻠﺜﻴﺔ‬ ‫ﺍﻟﺠﻴﻮﺏ‬ ‫ﻗﺎﻧﻮﻥ‬ ‫ﺑﺄﺳﺘﺨﺪﺍﻡ‬
.(P.I₁ →P.I)‫ﺍﻟﻤﺤﻄﺘﻴﻴﻦ‬ ‫ﺑﻴﻦ‬ (a) ‫ﺍﻟﻤﺴﺎﻓﺔ‬ 
.(P.I₂ →P.I)‫ﺍﻟﻤﺤﻄﺘﺘﻴﻦ‬ ‫ﺑﻴﻦ‬ (b) ‫ﺍﻟﻤﺴﺎﻓﺔ‬ 
(P.I) ‫ﻣﺤﻄﺔ‬ ‫ﻣﻦ‬ (P.T ,P.C) ‫ﺍﻟﻤﺤﻄﺘﻴﻦ‬ ‫ﺗﺜﺒﻴﺖ‬ ‫ﻟﻐﺮﺽ‬ (Tb ,Ta) ‫ﺍﻟﻤﻤﺎﺳﺎﻥ‬ ‫ﻳﺴﺘﺨﺪﻡ‬
.‫ﺍﻟﻤﺮﻛﺐ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﺗﺴﻘﻴﻂ‬ ‫ﻋﻨﺪ‬
-:‫ﺍﻻﺗﻲ‬ ‫ﺧﻼﻝ‬ ‫ﻣﻦ‬ ‫ﺍﻟﻤﺮﻛﺐ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﺗﺴﻘﻴﻂ‬ ‫ﻳﺘﻢ‬
.‫ﺑﻪ‬ ‫ﺍﻟﺨﺎﺻﺔ‬ ‫ﺍﻟﺠﺪﺍﻭﻝ‬ ‫ﻭﺑﺎﺳﺘﺨﺪﺍﻡ‬ (P.C) ‫ﻣﺤﻄﺔ‬ ‫ﻣﻦ‬ ‫ﺍﻻﻭﻝ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﺗﺴﻘﻴﻂ‬ -1
.‫ﺑﻪ‬ ‫ﺍﻟﺨﺎﺻﺔ‬ ‫ﺍﻟﺠﺪﺍﻭﻝ‬ ‫ﻭﺑﺎﺳﺘﺨﺪﺍﻡ‬ (P.C.C) ‫ﻣﺤﻄﺔ‬ ‫ﻣﻦ‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﺗﺴﻘﻴﻂ‬ -2
-3
Tb: Smaller tangent length.
R₁, ∆₁, T₁, L₁, C₁ & P.I₁: Elements of the first curve.
R₂, ∆₂, T₂, L₂, C₂ & P.I₂: Elements of the second curve.
Refer to the previous figure, elements of this curve are calculated from the triangle (∆:
P.I-P.I₁-P.I₂) by using sine rule…
C.T = T₁ + T₂
∆ = (∆₁+∆₂)
= =
𝑇₁ + 𝑇₂
sin (180ᵒ ― ∆)
𝑎
sin (∆₂)
𝑏
sin (∆₁)
a =
𝑇₁ + 𝑇₂
sin (180ᵒ ― ∆) × sin (∆₂)
Ta = T₁ + a
b =
𝑇₁ + 𝑇₂
sin (180ᵒ ― ∆) × sin (∆₁)
Tb = T₁ + b
St. P.C = St. P.I – Ta
St. P.C.C = St. P.C + L₁
St. P.T = St. P.C.C + L₂
St. P.I₁ = St. P.C + T₁
St. P.I₂ = St. P.C.C + T₂
.‫ﺍﻻﻭﻝ‬ ‫ﺍﻟﺪﺍﺋﺮﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻋﻨﺎﺻﺮ‬
.‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺪﺍﺋﺮﻱ‬ ‫ﺍﻟﻤﻨﺤﻨﻲ‬ ‫ﻋﻨﺎﺻﺮ‬
.(‫ﺍﻻﻣﺎﻣﻲ‬ ‫)ﺍﻟﻤﻤﺎﺱ‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﺍﻟﺮﺋﻴﺴﻲ‬ ‫ﺍﻟﻤﻤﺎﺱ‬

3. The Twelfth Lecture
3
Example: Compute stations for a compound circular curve, if you know ∆ = 38ᵒ 20′, ∆₂= 41ᵒ₁
30′, R₁=500m, R₂=750m and Station P.I=28+50.
Solution:
T₁ = R₁ = 500 tan = 173.79mtan
∆₁
2 ×
38ᵒ 20′
2
L₁ = = = 334.54m
𝜋 𝑅₁ ∆₁ᵒ
180ᵒ
𝜋 × 500 × 38.333
180
T₂ = R₂ = 750 tan = 284.15mtan
∆₂
2 ×
41ᵒ 30′
2
L₂ = = = 543.23m
𝜋 𝑅₂ ∆₂ᵒ
180ᵒ
𝜋 × 750 × 41.5
180
∆ = ∆₁ + ∆₂ = + =38ᵒ 20′ 31ᵒ 40′ 79ᵒ 50′
a = = = 308.28 m
𝑇₁ + 𝑇₂
sin (180ᵒ ― ∆) × sin (∆₂)
173.79 + 334.54
sin (180ᵒ ― 79ᵒ 50′) × sin (31ᵒ 40′)
b = = = 288.56 m
𝑇₁ + 𝑇₂
sin (180ᵒ ― ∆) × sin (∆₁)
173.79 + 334.54
sin (180ᵒ ― 79ᵒ 50′) × sin (38ᵒ 20′)
Ta = T₁ + a = 173.79 + 308.28 = 482.07 m
Tb = T₁ + b = 288.56 + 284.15 = 572.71 m
St. P.C = St. P.I – Ta = (28+50) – (4+82.07) = (23+67.93)
St. P.I₁ = St. P.C + T₁ = (23+67.93) + (1+73.79) = (25+41.72)
St. P.C.C = St. P.C + L₁ = (23+67.93) + (3+43.52) = (27+02.45)
St. P.I₂ = St. P.C.C + T₂ = (27+02.45) + (2+48.15) = (29+86.60)
St. P.T = St. P.C.C + L₂ = (27+02.45) + (5+43.23) = (32+45.68)