Reverse Curves - Mypdh.engineer (2024)

A reverse curve is composed of two or more simple curves turning in opposite directions. Their points of intersection lie on opposite ends of a common tangent, and the PT of the first curve is coincident with the PC of the second. This point is called the point of reverse curvature (PRC).

Reverse curves are useful when laying out such thingsas pipelines, flumes, and levees. The surveyor may also use them on low-speed roads and railroads. They cannot be used on high-speed roads or railroads since they cannot be properly superelevated at the PRC. They are sometimes used on canals, but only with extreme caution, since they make the canal difficult to navigate and contribute to erosion.

Reverse Curve Data
The computation of reverse curves presents three basic problems. The first is where the reverse curve is to be laid out between two successive PIs. (See figure 12.) In this case, the surveyor performs the computations in exactly the same manner as a compound curve between successive PIs. The second is where the curve is to be laid out so it connects two parallel tangents (figure 13). The third problem is where the reverse curve is to be laid out so that it connects diverging tangents (figure 14).

Connecting Parallel Tangents
Figure 13 illustrates a reverse curve connecting two parallel tangents. The PC and PT are located as follows.

1. Measure p, the perpendicular distance between tangents.
2. Locate the PRC and measure m₁ and m₂. (If conditions permit, the PRC can be at the midpoint between the two tangents. This will reduce computation, since both arcs will be identical.)
3. Determine R₁.
4. Compute I₁.
   Cos I₁ = (R₁ – m₁) / R₁
5. Compute L1 from
   L₁ = R₁ Sin I₁
   R₂ ,I₂ ,and L₂ are determined in the same way as R₁, I₁, and L₁. If the PRC is to be the midpoint, the values for arc 2 will be the same as for arc 1.
6. Stake each of the arcs the same as a simple curve. If necessary, the surveyor can easily determine other curve components. For example, the surveyor needs a reverse curve to connect two parallel tangents. No obstructions exist so it can be made up of two equal arcs. The degree of curve for both must be 5°. The surveyor measures the distance p and finds it to be 225.00 feet.
   m₁ = m₂ and L₁= L₂
   R₁= R₂ and I₁= I₂
   R₁ = 50 ft / (Sin ¹/₂ D) = 50 ft / 0.043619 = 1,146.29 ft
   Cos I₁= (R₁ – m₁) / R₁ = 1,033.79 / 1,146.29 = 0.901857
   L₁ = R₁ Sin I₁ = 1,146.29 x 0.432086 = 495.30 ft
7. The PC and PT are located by measuring off L₁ and L₂.

Connecting Diverging Tangents
The connection of two diverging tangents by a reverse curve is illustrated in figure 14. Due to possible obstruction or topographic consideration, one simple curve could not be used between the tangents. The PT has been moved back beyond the PI. However, the I angle still exists as in a simple curve. The controlling dimensions in this curve are the distance Ts to locate the PT and the values of R₁ and R₂, which are computed from the specified degree of curve for each arc.

1. Measure I at the PI.
2. Measure Ts to locate the PT as the point where the curve is to join the forward tangent. In some cases, the PT position will be specified, but Ts must still be measured for the computations.
3. Perform the following calculations:
   Determine R₁ and R₂. If practical, have R₁ equal R₂.
   Angle s = 180-(90+I)=90-I
   m = T_s (Tan I)
   L = T_s / Cos I
   angle e = I₁ (by similar triangles)
   angle f = I₁ (by similar triangles)
   therefore, I₂ = I + I₁
   n = (R₂ – m) Sin e
   p = (R₂ – m) Cos e
   Determine g by establishing the value of I₁.
   Cos I₁ = (R₁ + p) / (R1 + R₂)
   Knowing Cos I₁, determine Sin I₁.
   g = (R₁ + R₂) Sin I₁
   T_L = g + n + L
4. Measure T_L from the PI to locate the PC.
5. Stake arc 1 to PRC from PC.
6. Set instrument at the PT and verify the PRC (invert the telescope, sight on PI, plunge, and turn angle I₂/2).
7. Stake arc 2 to the PRC from PT.
   For example, in figure 14, a reverse curve is to connect two diverging tangents with both arcs having a 5-degree curve. The surveyor locates the PI and measures the I angle as 41 degrees. The PT location is specified and the Ts is measured as 550 feet.
   R₁ = R₂ = 50 ft / Sin ¹/₂ D = 50 / 0.043619 = 1,146.29
   Angle s = 90° – I = 49°
   m = T_s Tan I = 550 x 0.869287 = 478.11 ft
   L = T_s / Cos I = 550 / 0.754710 = 728.76 ft
   n = (R₂ – m) Sin I = (1,146.29 – 478.11) 0.656059 = 438.37 ft
   p = (R₂ – m) Cos I = (1,146.29 – 478.11) 0.754710 = 504.28 ft
   Cos I₁ = (R₁ + p) / (R₁ + R₂) = (1,146.29 + 504.28) / (1,146.29 + 1,146.29) = 0.719962
   I₁ = 43° 57’
   g = (R₁ + R₂) sin I₁ = (2,292.58) 0.694030 = 1,591.12 ft
   T_L = g + n + L = 1,591.12 + 438.37 + 728.76 = 2,758.25 ft
   The PC is located by measuring T_L. The curve is staked using 5-degree curve computations.