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ENGINEERING SURVEYS CONTENTS = Circular Curves: General Background = Circular Curve Geometry = Vertical Curves: General Background = Geometric Properties of the Parabola = Computation of the High or Low Point on a Vertical Curve = Procedure for computing a Vertical Curve = Circular Curve Geometry A = Tangent Deflection Angle Nie = Curve Deflection Angle FIGURE 13-2 Geometry of the circle Formulas: T = Rtan > (13-1) OA C => 2R sin “> (13-2) M = R(1 - sos 4) (13-3) _ 1 4 . E = R( ap 1) = R( sec an 1) (alternate) (13-4) 27rRA = <= ss 713-5) 360 D _ 100 | iy 5,729.58° 360° 27R a R (15-6) ip A A pee Ss Se _ = =. 13-7 rn 7m = sz 100 (2 Jr ( ) Procedure For Example13-2: ha AL S| R= 400m Pl at 0 + 241.782 A A T= Rtan> L = 20h (55 = 400 tan 6°25'30" = 2m x 400 x — = 45.044 m = 89.710m EC 0+ 286.448 Pl atO + 241.782 =-T 45.044 BC = 0 + 196.738 +L 89.710 EC = 0 + 286.448 FIGURE 13-5 Sketch for Example 13-2 ROUTE SURVEYS AND HIGHWAY CURVES c=27R FIGURE 13-3 Relationship between the degree of curve (D) and the circle 10 Example13-1: Given the following information : A = 16°38 R = 1,000ft Pl at 6 + 26.57 Calculate the station of the BC and EC; also calculate lengths C M, and & FIGURE 13-4 Sketch for Example 13-1. Note: To aid in comprehension, the magnitude of the angle has been exaggerated in this section = Geometric Properties of the Parabola Crest Curve FIGURE 13-17 Geometric properties of the parabola 14 Procedure For Example13-3: 5 A=1T2T35° Pl at 14 + 87.33 D=6 R = 5,729.58/D = 954.93 ft T = Rtan A/2 = 954.93 tan 5.679861° = 94.98 ft L = 100A/D = 100 X 11.359722/6 = 189.33 ft or L = 27RA/360 = 27 X 954.93 X 11.359722/360 = 189.33 ft Plat 14 + 87.33 —F 94.98 BC = 13 + 92.35 +L 1 89.33 EC = 15 + 81.68 FIGURE 13-6 Sketch forExample 13-3000 eee = Vertical Curves: General Backgrounc PVI PVI Forward Stations FIGURE 13-15 Vertical curve terminology (profile view shown) (b) FIGURE 13-16 Types of vertical curves. (a) Sag curve. (b) Crest curve = Procedure for computing a Vertical Curve ° FIGURE 13-18 Tangent at curve low point Compute the algebraic difference in grades: A = & — R1- . Compute the chainage of the BVC and EVC. If the chainage of the PVI is known, } L is simply subtracted and added to the PVI chainage. . Compute the distance from the BVC to the high or low point (if applicable) using Equation 13-15 and determine the station of the high or low point. Compute the tangent gradeline elevation of the BVC and the EVC. Low Point PVI Tangent Through Low Point Slope = 2 ax + g, = 0 5. 6. Compute the tangent gradeline elevation for each required station. Compute the midpoint of chord elevation: Elevation of BVC + elevation of EVC 2 . Compute the tangent offset (d) at the PVI (i.e., distance VM in Figure 13-17): difference in elevation of PVI wi and mid-point of chord 2 . Compute the tangent offset for each individ- ual station (see line ax in Figure 13-19): d(x) (L/2)?’ an 2 where x is the distance from the BVC or EVC (whichever is closer) to the required station. Tangent offset = (13-16) . Compute the elevation on the curve at each required station by combining the tangent offsets with the appropriate tangent gradeline elevations (add for sag curves and subtract for crest curves). 1. A = 0.018 — (—0.032) = 0.05 7. Tangent offset at PVI (a 2. PVI — 3L = BVC; BVC at (30 + 30) — 150 difference in elevation of PVI WY = 28 + 80.00 d= and mid-point of chord 2 Pvil + 34 = EVC;EVC at (30 + 30) + 150 469.67 — 465.92 = = 1.875 ft = 31 + 80.00 2 Evc — BVC = L;(31 + 80) — (28 + 80) 8. Tangent offsets are computed by multiply- = 300;Check ing the distance ratio squared (x/{L/2)*, by the maximum tangent offset (¢). See 3. Elevation of PVI = 465.92 ft Table 13-4. 150 ft at 3.2 percent = 4.80 (see Figure 9. The computed tangent offsets are added 13-19) (in this exarnple) to the tangent elevation in Elevation BVC = 470.72ft order to determine the curve elevation (see Elevation PV! = 465.92 ft Table 13-4). 150 ft at 1.8 percent = 2.70 ElevationEVC = 468.62ft 4. Location of low point is calculated using Equation 13-15: J x x= 2982 200 0 seo.90% Grom the BYE 5. Tangent gradeline computations are en- tered in Table 13-4. For example: Elevation at 29 + 00 = 470.72 — (0.032 x 20) = 470.72 — 0.64 = 470.08ft 6. Midchord elevation: FIGURE 13-19 Sketch for e 13-6 470.72 (BVC) “ 468.62 (EVC) _ “aart Examp!l 
