Handbook of Civil Engineering Calculations

Tangent-Offset Method

0.0001 Jc^2 , ft

Station x, ft (m) (m) 0.046*, ft (m) y, ft (m)

52 + 60 O (O) O (O) O (O) O (O)

53 + 00 40(12.19) 0.16(0.05) 1.84(0.56) -1.68 (-0.51)

53 + 50 90(27.43) 0.81(0.25) 4.14(1.26) -3.33 (-1.01)

54 + 00 140(42.67) 1.96(0.60) 6.44(1.96) -4.48 (-1.37)

54 + 50 190 (57.91) 3.61 (1.10) 8.74 (2.66) -5.13 (-1.56)

55 + 00 240(73.15) 5.76(1.76) 11.04(3.36) -5.28 (-1.61)

55 + 50 290 (88.39) 8.41 (2.56) 13.34 (4.07) -4.93 (-1.50)

55 + 80 320(97.54) 10.24(3.12) 14.72(4.49) -4.48 (-1.37)

6. Locate the sag S
   Since the grade is zero at this point, Eq. 42 yields G 5 = rxs + Ga = O; therefore xs = -GJr =
   -(-0.046/0.0002) = 230 ft (70.1 m); station of sag = (52 + 60) + (2 + 30) = 54 + 90; Gav =
   V 2 G 0 = -0.023; ys = -0.023(230) =•- 5.29 ft (1.61 m); elevation of sag = 303.66 - 5.29 =
   298.37 ft (90.943 m).


7. Verify the location of the sag
   Do this by ascertaining that the offsets of the PC and PT from the tangent through S 9
   which is horizontal, satisfy the tangent-offset principle. From the preceding results, tan-
   gent offset of PC = 5.29 ft (1.612 m); tangent offset of PT = 5.29 - 4.48 = 0.81 ft (0.247
   m); distance to PC = 230 ft (70.1 m); distance to PT = 320 - 230 = 90 ft (27.4 m);
   5.29/0.81 = 6.53; 2302 /90^2 = 6.53. Therefore, the results are verified.

LOCATION OFA SINGLE STATION ON A

PARABOLICARC

The PC of a vertical parabolic curve is at station 22 + OO of elevation 165.30, and the
grade at the PC is + 3.2 percent. The elevation of the station 24 + OO is 168.90 ft (51.481
m). What is the elevation of station 25 + 50?

Calculation Procedure:

1. Compute the offset of P 1 from the tangent through the PC
   Refer to Fig. 21. The tangent-offset principle offers the simplest method of solution. Thus
   Jc 1 = 200 ft (61.0 m); yi = 168.90 - 165.30 = 3.60 ft (1.097 m); Q 1 T 1 = 200(0.032) = 6.40
   ft (1.951 m); P 1 T 1 = 6.40 - 3.60 = 2.80 ft (0.853 m).


2. Compute the offset of P 2 from the tangent through the PC;
   find the elevation of P 2
   Thus Jc 2 = 350 ft (106.7 m); P 2 T 2 I(PiT 1 ) = Jcf/jc^2 ; P 2 T 2 = 2.80(350/20O)^2 = 8.575 ft (2.6137
   m); Q 2 T 2 = 350(0.032) = 11.2 ft (3.41 m); ^ 2 P 2 = 11.2 - 8.575 = 2.625 ft (0.8001 m); ele-
   vation OfP 2 = 165.30 + 2.625 = 167.925 ft (51.184 m).