Parabolic Curve Elevations Computed Directly- 123docz.net

13.14 ComPutation of tHe HiGH oR tHe low Point

13.15.1 Parabolic Curve Elevations Computed Directly

In addition to the tangent offset method shown earlier, vertical curve elevations can also be computed directly from the general equation.

y = ax2 + bx + c, where a = 1g2 - g12/2L L = horizontal length of vertical curve b = g1

c = elevation at BVC

x = horizontal distance from BVC

y = elevation on the curve at distance x from the BVC This technique is illustrated in Table 13.2 using the data from Example 13.6.

13.16 SPiRal CuRveS: GeneRal BaCkGRound

A spiral is a curve with a uniformly changing radius. Spirals are used in highway and rail- road alignment to overcome the abrupt change in direction that occurs when the alignment changes from a tangent to a circular curve, and vice versa. The length of the spiral curve is also used for the transition from normally crowned pavement to fully superel- evated (banked) pavement.

Figure 13.22 illustrates how the spiral curve is inserted between tangent and circular curve alignment. You can see that, at the beginning of the spiral (T.S. = tangent to spiral), the radius of the spiral is the radius of the tangent line (infinitely large), and that the radius of the spiral curve decreases at a uniform rate until, at the point where the circular curve begins (S.C. = spiral to curve), the radius of the spiral equals the radius of the circular curve. In the previous section, we noted that the parabola, which is used in vertical alignment, had the important property of a uniform rate of change of slope. Here, we find that the spiral, used in horizontal alignment, has a uniform rate of change of radius (curvature).

This property permits the driver to leave a tangent section of highway at relatively high rates of speed without experiencing problems with safety or comfort.

Figure 13.23 illustrates how the circular curve is moved inward (toward the center of the curve), leaving room for the insertion of a spiral at either end of the shortened circular

Table 13.2 Parabolic curve elevations computed from the equation Y = ax2 + bx + c Station Distance from bVC ax2 bx c y (elevation on the Curve)

BVC 28 + 80 0 470.72

29 + 00 20 0.03 -0.64 470.72 470.11

30 + 00 120 1.20 -3.84 470.72 468.08

PVI 30 + 30 150 1.88 -4.80 470.72 467.80

Low 30 + 72 192 3.07 -6.14 470.72 467.65

31 + 00 220 4.03 -7.04 470.72 467.71

curve. The amount that the circular curve is shifted in from the main tangent line is known as P. This shift results in the curve center (O) being at the distance 1R + P2 from the main tangent lines.

The spirals illustrated in this text reflect the common practice of using equal spirals to join the ends of a circular or compound curve to the main tangents. For more complex

Figure 13.22 Spiral curves.

Figure 13.23 Shifting the circular curve to make room for the insertion of spirals. (a) Circular curve joining two tangents.

spiral applications, such as unequal spirals and spirals joining circular arcs, refer to a text on route surveying. This text shows excerpts from spiral tables (see Tables 13.3–13.5).

The agency responsible for design and construction of transportation facilities in each U.S. state and Canadian province prepares and publishes similar tables for use by their personnel. A wide variety of spirals, both geometric and empirical, have been used to develop spiral tables. Geometric spirals include the cubic parabola and the clothoid curve, and empirical spirals include the American Railway Engineering Association (AREA) 10-chord spiral, used by many railroads. Generally, the use of tables is giving way to computer programs for spiral solutions. All spirals give essentially the same appearance when staked out in the field.

13.17 SPiRal CuRve ComPutationS

Usually, data for a spiral computation are obtained as follows (refer to Figure 13.24):

1. ∆ is determined in the field.

2. R or D (degree of curve) is given by design considerations (usually defined by design speed, but sometimes by property constraints).

3. Stationing (chainage) of PI is determined in the field.

4. Ls is chosen with respect to design speed and the number of traffic lanes.

Figure 13.23 (Continued ) (b) Circular curve shifted inward (toward curve center) to make room for the insertion of spiral curves at either end of the circular curve.

Table 13.3 Spiral Tables for Ls=150 feet D∆sRpR+pqLTST∆sXcYcD/10 Ls 7°30′5°37′30″763.94371.2268765.170574.9759100.030550.04595.62500°149.864.910.00500 8°00′6°00′00″716.19721.3085717.505774.9726100.057550.05236.00000°149.845.230.00533 30′6°22′30″674.06801.3902675.458274.9691100.064950.05906.37500°149.815.560.00567 9°00′6°45′00″636.61981.4719638.091774.9653100.072850.06626.75000°149.795.880.00600 30′7°07′30″603.11351.5536604.667174.9614100.081150.07387.12500°149.776.210.00633 10°00′7°30′00″572.95781.6352574.593074.9572100.089950.08177.50000°149.746.540.00667 30′7°52′30″545.67411.7169547.391074.9528100.099150.09017.87500°149.726.860.00700 11°00′8°15′00″520.87071.7985522.669274.9482100.108850.09898.25000°149.697.190.00733 30′8°37′30″498.22421.8802500.104474.9434100.119050.10828.62500°149.667.510.00767 12°00′9°00′00″477.46481.9618479.426674.9384100.129550.11789.00000°149.637.840.00800 13°00′9°45′00″440.73682.1249442.861774.9277100.152150.13839.75000°149.578.490.00867 14°00′10°30′00″409.25562.2880411.543674.9161100.176550.160510.50000°149.509.140.00933 15°00′11°15′00″381.97192.4510384.422974.9037100.202750.184311.25000°149.429.790.01000 16°00′12°00′00″358.09862.6139360.712574.8905100.230750.209812.00000°149.3410.440.01067 17°00′12°45′00″337.03402.7767339.810774.8764100.260650.237012.75000°149.2611.090.01133 18°00′13°30′00″318.30992.9394321.249374.8614100.292450.265913.50000°149.1711.730.01200 19°00′14°15′00″301.55673.1020304.658774.8456100.325950.296414.25000°149.0712.380.01267 20°00′15°00′00″286.47893.2645289.743474.8290100.361450.328715.00000°148.9813.030.01333 21°00′15°45′00″272.83703.4269276.263974.8115100.398750.362715.75000°148.8713.670.01400 22°00′16°30′00″260.43543.5891264.024574.7932100.437950.398316.50000°148.7614.310.01467 23°00′17°15′00″249.11213.7512252.863374.7740100.479050.435717.25000°148.6514.960.01533 24°00′18°00′00″238.73243.9132242.645674.7539100.521950.474818.00000°148.5315.600.01600 25°00′18°45′00″229.18314.0750233.258174.7331100.566850.515718.75000°148.4016.260.01667 26°00′19°30′00′220.36844.2367224.605174.7114100.613550.558219.50000°148.2716.880.01733 Source: Spiral Tables (foot units), courtesy Ministry of Transportation, Ontario.

Table 13.4 Spiral curve lengths and superelevation rates: superelevation (e) maximum of 0.06, typical for northern climates* V=30V=40V=50V=60V=70V=80 De

L, ft e

L, ft 2 lane4 lane2 lane4 lane2 lane4 lane2 lane4 lane2 lane4 lane2 lane4 lane 0°15′NC 0 0NC 0 0NC 0 0NC 0 0NC 0 0RC250250 0°30′NC 0 0NC 0 0NC 0 0RC200200RC200200.023250250 0°45′NC 0 0NC 0 0RC150150.021200200.026200200.033250250 1°00′NC 0 0RC150150.020150150.027200200.033200200.041250250 1°30′RC100100.020150150.028150150.036200200.044200200.053250300 2°00′RC100100.026150150.035150150.044200200.052200250.059250300 2°30′.020100100.031150150.040150150.050200200.057200300.060250300 3°00′.023100100.035150150.044150200.054200250.060200300Dmax =2°30′ 3°30′.026100100.038150150.048150200.057200250Dmax =3°00′ 4°00′.029100100.041150150.051150200.059200250 5°00′.034100100.046150150.056150200.060200250 6°00′.038100100.050150200.059150250Dmax =4°30′ 7°00′.041100150.054150200.060150250 8°00′.043100150.056150200Dmax =7°00′ 9°00′.046100150.058150200 10°00′.048100150.059150200 11°00′.050100150.060150200 12°00′.052100150Dmax =11°00′ 13°00′.053100150 14°00′.055100150 16°00′.058100200 18°00′.059150200 20°00′.060150200 21°00′.060150200 Dmax =21°00′ Source: Ministry of Transportation, Ontario. Legend: V, design speed, mph; e, rate of superelevation, feet per foot of pavement width; L, length of superelevation runoff or spiral curve; NC, normal crown section; RC, remove adverse crown, superelevate at normal crown slope; D, degree of circular curve; *Above the heavy line, spirals are not required, but superelevation is to be run off in distances shown.

Table 13.5 Spiral curve lengths (feet) and superelevation rates: superelevation (e) maximum of 0.100, typical for southern climates* V=30V=40V=50V=60V=70 lllll DRe2 lane4 lanee2 lane4 lanee2 lane4 lanee2 lane4 lanee2 lane4 lane 0°15′22918′NC00NC00NC00NC00RC200200 0°30′11459′NC00NC00RC150150RC175175RC200200 0°45′7639′NC00RC125125RC1501500.0181751750.020200200 1°00′5730′NC00RC1251250.0181501500.0221751750.028200200 1°30′3820′RC1001000.0201251250.0271501500.0341751750.042200200 2°00′2865′RC1001000.0271251250.0361501500.0461751900.055200250 2°30′2292′0.0201001000.0331251250.0451501600.0591752400.069210310 3°00′1910′0.0241001000.0381251250.0541501900.0701902800.083250370 3°30′1637′0.0271001000.0451251400.0631502300.0812203300.096290430 4°00′1432′0.0301001000.0501251600.0701702500.0902403600.100300450 5°00′1146′0.0381001000.0601301900.0832003000.099270400Dmax =3°9′ 6°00′955′0.0441001200.0681402100.0932203300.100270400 7°00′819′0.0501001400.0761602400.097230350Dmax =5°5′ 8°00′716′0.0551001500.0841802600.100240360 9°00′637′0.0611101600.0891902800.100240360 10°00′573′0.0651201800.093200290Dmax =8.3° 11°00′521′0.0701301900.096200300 12°00′477′0.0741302000.098210310 13°00′441′0.0781402100.099210310 14°00′409′0.0821502200.100210320 16°00′358′0.087160240Dmax =13.4° 18°00′318′0.093170250 20°00′286′0.096170260 22°00′260′0.099180270 24.8°231′0.100180270 Dmax =24.8° *NC, normal crown section; RC, remove adverse crown, superelevate at normal crown slope. Spirals desirable but not such essential above the heavy line. Lengths rounded in multiples of 25 or 50 ft permit simpler calculations. The higher maximum e value (0.100) in this table permits a sharper maximum curvature.

All other spiral parameters can be determined by computation and/or by use of spiral tables.

Tangent to spiral 1see Figure 11.242: Ts = 1R + P2tan ∆

2 + q (13.22) Spiral tangent deflection: ∆s = LsD

200 (13.23) In circular curves, ∆ = LD/100 [see Equation (13.7)]. Because the spiral has a uniformly changing D, the spiral angle 1∆s2 = the length of the spiral (Ls) in stations times the average degree of curve (D/2). The total length of the curve system is comprised of the circular curve and the two spiral curves, that is,

Total length: L = Lc + 2Ls (13.24)

Figure 13.24 Summary of spiral geometry and spiral symbols.

See Figure 13.24, where L is the total length of the curve system.

Total deflection: ∆ = ∆c + 2∆s (13.25) See Figure 13.24.

Spiral deflection: us = ∆s

3 1approximate2 (13.26)

us is the total spiral deflection angle; compare to circular curves, where the deflection angle is ∆/2. The approximate formula in Equation (13.26) gives realistic results for the vast majority of spiral problems. For example, when ∆s is as large as 21° (which is seldom the case), the correction to ∆s is approximately +30″.

Lc = 2πR∆c

360 1foot or meter units2 (13.27)

Lc = 100∆c

D 1foot units2 (13.28)

∆s = 90 π * Ls

R (13.29)

f = a l

Ls b2us (13.30) Equation (13.30) comes from the spiral definition, where f is the deflection angle for any distance l, and f and us are in the same units. Practically,

f′ = l21us * 602 Ls2

where f′ is the deflection angle in minutes for any distance measured from the T.S.

or the S.T.

Other values, such as x, y, P, q, ST, and LT, are found routinely in spiral tables issued by state and provincial highway agencies and can also be found in route surveying texts.

For the past few years, solutions to these problems have been achieved almost exclusively by the use of computers and appropriate coordinate geometry computer software.

13.18 SPiRal layout PRoCeduRe SummaRy

Refer to Figure 13.24. The following list summarizes the spiral layout procedure using traditional techniques:

1. Select Ls (foot units) in conjunction with the design speed, number of traffic lanes, and sharpness of the circular curve (radius or D).

2. From the spiral tables, determine P, q, x, y, and so on.

3. Compute the spiral tangent (Ts) using Equation (13.22) and the circular tangent (Tc) using Equation (13.1).

4. Compute the spiral angle 1∆s2. Use Equation (13.33) for foot units (see Table 13.3).

5. Prepare a list of relevant layout stations. This list should include all horizontal alignment key points (e.g., T.S., S.C., C.S., and S.T.), as well as all vertical alignment key points, such as grade points, BVC, low point, and EVC.

6. Calculate the deflection angles. See Equation (13.30).

7. From the established PI, measure out the Ts distance to locate the T.S. and S.T.

8. (a) From the T.S., turn off the spiral deflection 1us = 1/3 ∆s approximately2, measure out the LC, and thus locate the S.C.

(b) From the T.S., measure out the LT distance along the main tangent and locate the spiral PI (SPI); the spiral angle 1∆s2 can now be turned, and the ST distance can be measured out to locate the S.C.

9. From the S.C., measure out the circular tangent (Tc) along the line SPI1–SC to estab- lish the CPI.

10. Steps 8 and 9 are repeated, starting at the S.T. instead of at the T.S.

11. The key points are verified by checking all angles and redundant distances. Some sur- veyors prefer to locate the CPI by intersecting the two tangent lines (i.e., lines through SPI1 and S.C. and through SPI2 and C.S.). The locations can be verified by checking the angle ∆c and by checking the two tangents (Tc).

12. Only after all key control points have been verified can the deflection angle stakeout commence. The lower-chainage spiral is run in from the T.S., whereas the higher- chainage spiral is run in from the S.T. The circular curve can be run in from the S.C., although it is common practice to run half the circular curve from the S.C. and the other half from the C.S. so that any acceptable errors that accumulate can be iso- lated in the middle of the circular arc where they will be less troublesome, relatively speaking.

When employing layout procedures using polar layouts by total station, or position- ing layouts using GPS techniques, the surveyor must first compute the position coordinates of all layout points and then upload these coordinates into the instrument’s controller.

example 13.7 Illustrative Spiral Problem (Foot Units) You are given the following data:

∆ = 25°45′ RT V = 40mph D = 9°

PI at 36 + 17.42

Two@lane highway, 24 ft wide

Compute the stations for all key curve points (in foot units) and for all relevant spiral and circular curve deflections.

Solution

1. From Table 13.5,

Ls = 150ft

e = 0.058 1superelevation rate2 2. From Table 13.4,

∆s = 6.75000°

R = 636.6198ft P = 1.4719 ft R + P = 638.0917ft

q= 74.9653 ft LT = 100.0728ft ST = 50.0662 ft

X= 149.79ft Y= 5.88 ft You can also use Equation (13.23) to determine

∆s = LsD 200

= 150

200 * 9 = 6.75000°

3. From Equation (13.22), we have Ts = 1R + p2 tan ∆

2 + q

= 638.0917 tan 12°52.5′ + 74.9653= 220.82 ft 4. ∆c = ∆ - 2∆s

= 25°45′ - 216°45′2 = 12°15′

5. From Equation (13.7), we have Lc= 100∆c

= 1100 * 12.252

9 = 136.11 ft 6. Key station computation:

PI at 36 + 17.42 Ts 2 20.82 TS = 33+ 96.60 +Ls 1 50.00 SC = 35+ 46.60 +Lc 1 36.11 CS = 36+ 82.71 +Ls 1 50.00 ST = 38+ 32.71

7. us = ∆s

= 6.753000°

3 = 2.25° = 2°15′00″

8. Circular curve deflections. See Table 13.6

13.19 aPPRoximate Solution foR SPiRal PRoBlemS

It is possible to lay out spirals by using the approximate relationships illustrated in Figure 13.25. Because Ls≈LC, the following can be assumed:

Ls = sin us Y = Ls sin us (13.33)

Table 13.6 Curve System Deflection Angles

Station

Distance* from T.S. (or S.T.)

l (ft) l 2

U s° * 60 L s2

l2(U s * 60) L s2

Deflection angle

(Minutes) Deflection

T.S. 33 + 96.60 0 0 0.006 0 0°00′00″

34 + 00 3.4 11.6 0.006 0.070 0°00′04″

34 + 50 53.4 2,851.4 0.006 17.108 0°17′06″

35 + 00 103.4 10,691.6 0.006 64.149 1°04′09″

S.C. 35 + 45.60 150 22,500 0.006 135 us = 2°15′00″

Circular Curve Data Deflection angle (Cumulative) S.C. 35 + 46.60 ∆ c= 12° 15′,∆ c

2 = 6° 07′30″ 0°00.00′ 0°00′00″

35 + 50 Deflection for 3.40′ + 9.18′ 0°09.18′ 0°09′11″

36 + 00 Deflection for 50′ + 135′ 2°24.18′ 2°24′11″

36 + 50 Deflection for 32.71′ + 88.32′ 4°39.18′ 4°39′11″

C.S. 36 + 82.71 6°07.50′ 6°07′30″

C.S. 36 + 82.71 150 22,500 0.006 135 us = 2°15″00″

37 + 00 132.71 17,611.9 0.006 105.672 1°45′40″

37 + 50 82.71 6,840.9 0.006 41.046 0°41′03″

38 + 00 32.71 1,069.9 0.006 6.420 0°06′25″

S.T. 38 + 32.71 0 0 0.006 0 0°00′00″

*Note that l is measured from the S.T.

X2 = Ls2 - Y2

X = 2Ls2 - Y2 (13.34) q = 1

2 X (13.35) p = 1

4 Y (13.36) Using the sine law, we obtain the following:

LT = sin 2

3 ∆s * sin Ls

sin ∆s (13.37)

Using the sine law yields the following:

ST = sin 1

3 ∆s * Ls

sin ∆s (13.38)

Figure 13.25 Sketch for approximate formulas.

For comparison, the values from Example 13.7 are compared with the values obtained by the approximate methods:

Precise Methods Approximate Methods Parameter Example 13.7 (ft) Example 13.7 (ft)

Y 5.88 5.89

X 149.79 149.88

Q 74.97 74.94

P 1.47 1.47

LT 100.07 100.13

ST 50.07 50.10

You can see from the data summary that the precise and approximate values for Y, X, q, P, LT, and ST are quite similar. The largest discrepancy shows up in the X value, which is not required for spiral layout. The larger the ∆s value, the larger is the discrepancy between the precise and approximate values. For the normal range of spirals in use, the approximate method is adequate for the layout of an asphalt-surfaced ditched highway. For curbed highways or elevated highways, precise methods should be employed.

13.20 SuPeRelevation: GeneRal BaCkGRound

If a vehicle travels too fast on a horizontal curve, the vehicle may either skid off the road or overturn. The factors that cause this phenomenon are based on the radius of curvature and the velocity of the vehicle: The sharper the curve and the higher the velocity, the larger will be the centrifugal force requirement. Two factors can be called on to help stabilize the radius and velocity factors: (1) side friction, which is always present to some degree between the vehicle tires and the pavement, and (2) superelevation (e), which is a banking of the pavement toward the center of the curve.

The side friction factor ( f ) has been found to vary linearly with velocity. Design values for f range from 0.16 at 30 mph (50 km/hr) to 0.11 at 80 mph (130 km/hr). Superelevation must satisfy normal driving practices and climatic conditions. In practice, values for superelevation range from 0.125 (i.e., 0.125 ft/ft or 12.5 percent cross-slope) in warmer southern U.S. states to 0.06 in the northern U.S. states and Canadian provinces. Typical values for superelevation can be found in Tables 13.4 and 13.5.

13.21 SuPeRelevation deSiGn

Figure 13.26 illustrates how the length of spiral (Ls) is used to change the pavement cross-slope from normal crown to full superelevation. Figure 13.26(b) illustrates that the pavement can be revolved about the centerline, which is the usual case, or the pavement can be revolved about the inside or outside edges, a technique that is often encoun- tered on divided four-lane highways where a narrow median restricts drainage profile manipulation.

See your own state or provincial highway agency’s construction manuals for more on this topic.

Figure 13.26 Methods of attaining superelevation for spiraled curves. (From Geometric Design Standards for Ontario Highways, Ministry of Transportation, Ontario)

Review QueStionS

13.1. Why are curves used in roadway horizontal and vertical alignments?

13.2. Curves can be established by occupying cL or offset stations and then turning off appropriate deflection angles, or curves can be established by occupying a central control station and then turning off angles and measuring out distances—as determined through coordinates analyses.

What are the advantages and disadvantages of each technique?

13.3. Why do chord and arc lengths for the same curve interval sometimes appear to be equal in value?

13.4. What characteristic do parabolic curves and spiral curves have in common?

13.5. Describe techniques that can be used to check the accuracy of the layout of a set of inter- change curves using polar techniques (i.e., angle/distance layout from a central control station).

13.6. How can the centre line of a road be recovered from offset stakes?

PRoBlemS

13.1. Given PI at 7 + 33.33, ∆ = 29°42′, and R = 700 ft, compute the tangent (T ) and the length of arc (L).

13.2. Given PI at 9 + 71.83, ∆ = 7°10′, and D= 8°, compute the tangent (T ) and the length of arc (L).

13.3. From the data in Problem 13.1, compute the stationing of the BC and EC.

13.4. From the data in Problem 13.2, compute the stationing of the BC and EC.

13.5. A straight-line route survey, which had PIs at 3 + 81.27 1∆ = 12°30′2 and at 5 + 42.30 1∆ = 10°56′2, later had 600-ft-radius circular curves inserted at each PI. Compute the BC and EC stationing (chainage) for each curve.

13.6. Given PI at 3 + 333.013, ∆ = 12°47′, and R = 300 m, compute the deflections for even 20-m stations.

13.7. Given PI at 4 + 212.352, ∆ = 24°24′20″, and R = 500 m, compute E (external), M (midor- dinate), and the stations of the BC and EC.

13.8. Given PI at 15 + 78.28, ∆ = 36°10′30″ RT, and R = 1,150 ft, compute the deflections for even 100-ft stations.

13.9. The radius of a simple curve is 453 m. The tangent deflection angle is 43° 47′. The first sub- chord is 6.987 m, and a standard chord is 20 m. Compute the: (1) length of the main chord, (2) respective deflection angles of the first sub-chord and a standard chord.

13.10. The radius of a simple curve is 123 m. The tangent deflection angle is 33° 30′. The first sub- chord is 4.987 m, and a standard chord is 25 m. Compute the: (1) length of the main chord, (2) respective deflection angles of the first sub-chord and a standard chord.

13.11. The radius of a simple curve is 321 m. The tangent deflection angle is 13° 15′. The first sub- chord is 0.987 m, and a standard chord is 25 m. Compute the length of the last chord.

13.12. Two highway tangents must be joined with a circular curve of radius 1,000 ft (see Figure 13.27 The PI is inaccessible because its location falls in a river. Point A is established near the river on the back tangent, and point B is established near the river on the forward tangent. Distance AB is measured to be 615.27 ft. Angle α = 51°31′20″, and angle β = 32°02′45″. Perform the calculations required to locate the BC and the EC in the field.

Figure 13.27 Sketch for Problem 13.12.

Figure 13.28 Sketch for Problem 13.13.

13.13. Two street curb lines intersect with ∆ = 71°36′ (see Figure 13.28). A curb radius must be se- lected so that an existing catch basin (CB) will abut the future curb. The curb side of the catch basin cL is located from point V: V to CB = 8.713 m and angle E@V@CB = 21°41′. Compute the radius that will permit the curb to abut the existing catch basin.

13.14. Given the following compound curve data: R1 = 200 ft, R2= 300 ft, ∆1= 44°26′, and

∆2= 45°18′, compute T1 and T2 (see Figure 13.15).

13.15. Given the following vertical curve data: PVI at 7 + 25.712, L = 100 m, g1 = -3.2%, g2 = +1.8%, and elevation of PVI = 210.440, compute the elevations of the curve low point and even 20-m stations.

13.16. Given the following vertical curve data: PVI at 19+ 00, L = 500 ft, g1 = +2.5%, g2= -1%, and elevation at PVI = 723.86 ft, compute the elevations of the curve summit and even full stations (i.e., 100-ft even stations).

13.17. Given the following spiral curve data: D= 8°, V = 40 mph, ∆ = 16°44′, and PI at 11 + 66.18, determine the value of each key spiral and circular curve component (Ls, R, P, q, LT, ST, X, Y, ∆s, ∆c, Tc, and Lc) and determine the stationing (chainage) of the T.S., S.C., C.S., and S.T.

13.18. Given the same data as in Problem 13.17, use the approximate equations [Equations (13.33)–

(13.38)] to compute X, Y, q, P, LT, and ST. Enter these values and the equivalent values determined in Problem 13.17 in a table to compare the results.

13.20. Use the data from Problem 13.17 to compute the deflections for the curve system (spirals and circular curves) at even 50-ft stations.

HigHway ConstruCtion surveys