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 PI_{1} and the outgoing tangent at PI_{2}.
T_{i} is the distance from the BC_{1} to the PI; T_{o} the distance from the PI to the EC_{2}.
There are seven parts to a compound curve: T_{i}, T_{o}, R_{1}, R_{2}, Δ_{1}, Δ_{2}, and Δ. Because the tangents deflect Δ at the PI and the compound curve transitions from the incoming to the outgoing tangent:
Δ = Δ_{1}+ Δ_{2} | Equation D-1 |
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 O_{1}-BC_{1}-PI-EC_{2}-O_{2}, 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 eqaul 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 Δ_{2} using Equation D-1
Determine the distance from the PI_{1} to the PI which is a side of the triangle.
The distance between PI_{1} and PI_{2} is the sum of the curve tangents. Using the Law of Sines and the known T_{1}, we can compute T_{2}.
Using T_{2} and Δ_{2}, R_{2} 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" respectivley.
Determine the radius of each curve,
Sketch:
Becasue 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 of the other lines can be determined.
Updated sketch with bearings:
Compute Latitudes and Departures:
Line | Bearing | Length | Latitude | Departure |
O_{1}-BC_{1} | North | R_{1} | R_{1} | 0.00 |
BC_{1}-PI | East | 463.64 | 0.00 | 463.64 |
PI-EC_{2} | S 5°00'00"E | 405.60 | -405.60 x cos(5°00'00") | +405.60 x sin(5°00'00") |
EC_{2}-O_{2} | S 85°00'00"W | R_{2} | -R_{2} x cos(85°00'00") | -R_{2} x sin(85°00'00") |
O_{2}-O_{1} | S 30°00'00"W | R_{1}-R_{2} | -(R_{1}-R_{2}) x cos(30°00'00") | -(R_{1}-R_{2}) x sin(30°00'00") |
Sum and reduce the Latitudes:
Sum and reduce the Departures:
We have two equatiosn with unkowns R_{1} and R_{2}. Solve them simultaneously. We'll use substitution.
Solve the Latitide summation for R_{1}
Substiute the equation for R_{1} in the Departures summation and solve R_{2}:
Solve R_{1}:
With R_{1} and R_{2} computed, we have two geometric attributes for each curve. Using Chapter C 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 discussion in Chapter C Section 2.
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.
Following are radial chord tables for each curve in Example (2) (computations aren't shown).
Curve 1 | |||
Sta | Defl ang | Chord | |
EC | 54+10.64 | 15º00'00" | 301.58 |
54+00 | 14º28'36" | 291.29 | |
53+00 | 9º33'34" | 193.51 | |
52+00 | 4º38'32" | 94.31 | |
BC | 51+05.59 | 0º00'00" | 0.00 |
Curve 2 | |||
Sta | Defl ang | Chord | |
EC | 58+12.43 | 27º30'00" | 386.54 |
58+00 | 26º38'57" | 375.47 | |
57+00 | 19º48'17" | 283.63 | |
56+00 | 12º57'37" | 187.77 | |
55+00 | 6º06'57" | 89.109 | |
BC | 54+10.64 | 0º00'00" | 0.00 |
Stakeout procedure
Measure T_{i} and T_{o} from the PI along the tangents to set the BC_{1} and EC_{2}.
Stake Curve 1
Setup instrument on BC_{1}.
Sight PI with 0º00'00" on the horizontal circle.
Stake curve points using the curve table.
Stake Curve 2
Setup on PCC
Sight BC_{1} with -(Δ_{1}/2) = -15º00'00"=345°00'00" on the horizontal circle, Figure D-14a.
Inverse the scope to sight away from the BC_{1}, Figure D-14b.
Rotating right to a horizontal angle of 0º00'00" orients intrument tangent to the curves at the PCC, Figure D14c.
Stake curve points using the curve table.
Last point should match the previously set EC_{2}.
a. Sight BC_{1} with -(Δ_{1}/2) reading |
b. Inverse telescope |
c. Rotate to 00°00'00" |
Figure D-14 Orienting at PCC |