2. Compound curve

a. Nomenclature

Starting with two tangent lines intersecting at a PI, a compound curve is fit between them, Figure D-9.

Figure D-9 Compound curve

Figure D-10 shows major parts of the compound curve.

Figure D-10 Compound curve parts

The PCC (Point of Curve to Curve) is the EC of the first curve and BC of the second. A tangent line at the PCC intersects the incoming tangent at the PI₁ and the outgoing tangent at PI₂.

T_i is the distance from the BC₁ to the PI; T_o the distance from the PI to the EC₂.

There are seven parts to a compound curve: T_i, T_o, R₁, R₂, Δ₁, Δ₂, and Δ. Because the tangents deflect Δ at the PI and the compound curve transitions from the incoming to the outgoing tangent:

That means there are six independent parts. For a unique solution four parts must be fixed, including at least one angle and at least two distances.

Because they share a point (PCC) and tangent line, once all six parts are determined, the compound curve can be computed as two simple curves, Figure D-11. Individual curve attributes can be computed using the single curve equations in Chapter C.

Figure D-11 Two simple curves

Depending on which parts are initially fixed, there are different ways to compute the remaining parts.

One way is solving the vertex triangle, Figure D-12, created by the three PIs.

Figure D-12 Vertex triangle

Another method is to use a closed loop traverse through points O₁-BC₁-PI-EC₂-O₂, Figure D-13.

Figure D-13 Traverse method

The direction of one line is assumed to be in a cardinal direction and the others written in terms of 90° angles and Δs. Distances are Rs and Ts. Latitudes and departures are computed, summed, are set equal to zero to force closure. This results in two equations in two unknowns which can then be simultaneously solved.

b. Examples

(1) Example 1

A PI is located at station 38+00.00 with a left deflection of 72°00'00"L. The compound curve begins at sta 33+50.00. The first curve has a 700.00 ft radius and 30°00'00" central angle.

Determine the radius and central angle of the second curve and the length of both curves.

Sketch:

We'll try a vertex triangle solution. Isolate the triangle and label the tangents:

Since we have the radius and central angle of the first curve, we can compute its tangent:

Compute Δ₂ using Equation D-1

Determine the distance from the PI₁ to the PI which is a side of the triangle.

The distance between PI₁ and PI₂ is the sum of the curve tangents. Using the Law of Sines and the known T₁, we can compute T₂.

Using T₂ and Δ₂, R₂ can be determined.

Finally, compute each curve's length.

(2) Example 2 

A PI is located at 55+69.23 with a deflection angle between tangents of 85°00'00"R. The compound curve must begin 463.64 ft before the PI and end 405.60 ft after. Central angles of the two curves are 30°00'00" and 55°00'00", respectively.

Determine the radius of each curve,

Sketch:

Because we only have angles of the vertex triangle with no distances (and insufficient given data to compute any), the curve system can't be solved that way. Instead, we will use the traverse method.

To start, the curve system is rotated to make the initial radial line run North. Then using right angles and Δs, bearings (or azimuths) of the other lines can be determined.

Updated sketch with bearings:

Compute Latitudes and Departures:

 Sum and reduce the Latitudes:

Sum and reduce the Departures:

We have two equations with unknowns R₁ and R₂. Solve them simultaneously. We'll use substitution.

Solve the Latitude summation for R₁

Substitute the equation for R₁ in the Departures summation and solve R₂:

Solve R₁:

With R₁ and R₂ computed, we have two geometric attributes for each curve. Using Chapter C. Horizontal Curves equations, their remaining attributes can be determined. 

c. Stationing

As with a single curve, there are two paths to the EC from the BC. One goes up and down the original tangents, the other along the curves. Whether a station equation is used at the EC depends on how the entire alignment is stationed. Refer to the Station Equation discussion in Chapter C. Horizontal Curves.

Equations to compute the curves' endpoints are:

		Equation D-2
		Equation D-3
		Equation D-4
		Equation D-5

 The stationing for Example (2)  is:

e. Radial stakeout

Each curve's table is computed as described in Chapter C. Horizontal Curves.

Following are radial chord tables for each curve in Example (2) (computations aren't shown).

Stakeout procedure

Measure T_i and T_o from the PI along the tangents to set the BC₁ and EC₂.

Stake Curve 1

Setup instrument on BC₁.

Sight PI with 0º00'00" on the horizontal circle.

Stake curve points using the curve table.

Stake Curve 2

Setup on PCC

Sight BC₁ with -(Δ₁/2) = -15º00'00"=345°00'00" on the horizontal circle, Figure D-14a.

Inverse the scope to sight away from the BC₁, Figure D-14b.

Rotating right to a horizontal angle of 0º00'00" orients instrument tangent to the curves at the PCC, Figure D14c.

Stake curve points using the curve table.

Last point should match the previously set EC₂.

a. Sight BC1 with  -(Δ1/2) reading

b. Inverse telescope

c. Rotate to 00°00'00"

Figure D-14 Orienting at PCC