1. Exercise 1: Basic Design Criteria 1) Route 17 is a minor arterial and it has a design ADT of 4000. Assume rolling terrain. Determine the following information: Design Speed Drawing # for typical section Degree of curve or radius Min. SSD Min. R/W required Foreslope and min. ditch depth Max. percent grade Range of K value 50 mph D-62 7 1/2 400 ft 150 ft 6:1, 2ft 5% 90-110

2. Exercise 1: Basic Design Criteria 2) You are realigning a portion of I-44, assuming rolling terrain, determine the following: Design Speed Drawing # for typical section Degree of curve or radius Min. SSD Min. R/W required Foreslope and min. ditch depth Max. percent grade Range of K value What type of R/W access do you think I-44 requires? Fully Controlled 70 mph D-61 3 625 ft 250 ft dual 6:1, 4 ft 4% 150-220

3. Exercise 1: Basic Design Criteria 3) You are designing a 4-lane principal arterial, find the following design criteria. This is an urban setting. Design Speed Median Width Degree of curve or radius Min. SSD C&G Protected left turn lanes? Lane width, parking lane Sidewalk Max. % grade Range of K value Width of shoulder Superelevation rate 40 mph 14 10 275 ft None Required Yes 12 ft, none 5 5% 60-70 10 or 12 0.04 or 4%

4. Exercise 2: Horizontal Alignment, Simple and Spiral Curves Given Information to fix substandard curve: Road is principal arterial with ADT=1500 Back bearing is N 85^ 49’ 13” E Ahead bearing is S 56^ 39’ 24” E  = 37^ 31’ 23” (RT) A)Design curve, find L & T B)If the PC station starts at 15+63.34, what are the PI & PT stations for the curve you designed? C)When does MoDOT require spiral curves? PI = 1563.34 + 389.00 = 1952.34 PI sta = 19+52.34 PT = 1563.34 + 750 = 2313.34 PT sta = 23+13.34 1.When degree of curve > 2 2.V >50 mph 3.ADT > 400 L = 750 ft & T = 389.00 ft

5. Exercise 2: Horizontal Alignment, Simple and Spiral Curves D)Design a spiral curve based on MoDOT Design Criteria 1)What’s the transition length, Ls? 2)If the curve is 617.88 ft long, what’s the overall length of the SCS? 3)If the station at the SC point is 16+50.42, what are the stations at TS and ST points? Ls=192 617.88 +192+192=1001.88 ft TS= 1650.42 – 192 = 1458.42, so TS station = 14+58.42 CS = 1650.42 + 617.88 = 2268.30, so CS station = 22+68.30 ST = 2268.30 + 192 = 2460.30, so ST station = 24+60.30

6. Exercise 3: Horizontal Alignment, Simple and Spiral Curves Given information to design a temporary bypass: Minor arterial with ADT = 1500 20 ft requirement between the EOP of main line and EOP of temporary bypass d = 500 ft in length on each end of bypass A)Find , R, and L for this reverse curve B)If the first PC station on the bypass starts at 20+00, what are the stations for the PRC and PT of the first reverse curve you designed? D = 10^ 3’ 29” (10.058^), R = 1431.47 ft, L = 251.29 ft PRC = 2000.00 + 251.29 = 2251.29, so PRC station = 22+51.29 PT = 2251.29 + 251.29 = 2502.58, so PT station = 25+02.58

7. Exercise 4: Horizontal Alignment, Superelevation Fix the substandard superelevation transitions on the following project without re-aligning the curve: Principal arterial with ADT = 22940, NC(%) = 2, rural flat terrain Existing Curve Data PC station = 388+72.21D (degree of curve) = 3 PT station = 394+42.77 L = 570.56 ft PI station = 391+60.46 T = 288.25 ft 2.Find the SE rate and runoff length, specify the standard plan used Standard Drawing 203.21 pg 3/5 Runoff length, L = 356 ft

8. Exercise 4: Horizontal Alignment, Superelevation Fix the substandard superelevation transitions on the following project without re-aligning the curve: Principal arterial with ADT = 22940, rural setting, NC(%) = 2 Existing Curve Data PC station = 388+72.21D (degree of curve) = 3 PT station = 394+42.77 L = 570.56 ft PI station = 391+60.46 T = 288.25 ft 3.Calculate the superelevation transition points. Before PC StationSE Transition Section After PT Station 385+44.75A-A397+70.23 386+34.88B-B396+80.10 387+25.01C-C395+89.97 389+90.88D-D393+24.10

9. Exercise 5: Horizontal Alignment, Horizontal Sight Distance Check for horizontal sight distance: 2-lane minor arterial with ADT = 1890, rural setting on a rolling terrain, assume lane width = 12’ Existing Curve Data PC station = 0+00  = 29^ 30’ 6” (RT)D = 4^46’29” PT station = 6+17.88L = 570.56 ft PI station = 3+15.65T = 288.25 ft 1.Find the SSD for a sight of obstruction of 25 ft 2.Determine the distance of obstruction M based on Figure 4-04.2 & 4-04.3 SSD = 489.49 ft Round to 490 ft according to Fig. 4-01.1 M = 23.55 ft

10. Exercise 6: Vertical Alignment Design a crest curve that will meet MoDOT standards based on the given information: Road is principal arterial with ADT=19,800 on a rolling terrain Divided highway Plus grade = 1.03% Minus grade = 2.99% PI station = 683+50.00 with elevation = 676.99 ft 1)Based on the length of curve designed, calculate SSD, PSD and K value if appropriate 2)What is the station at the VPC and VPT? 3)Calculate the high point of the curve VPC station = 677 + 45VPC elevation = 670.76 ft VPT station = 689 + 55VPT elevation = 658.90 ft Xm = 310.02 ft HP station = 680+55.02; HP elevation = 675.55 ft For “my” L 1210 ft, K = 301 and SSD = 650 ft

11. Exercise 7: At-Grade Intersections (Handout) Given: Route 123 Principal Arterial Design ADT = 18000, Rolling Terrain Design Speed = 60 mph T = 12% 4-Lane divided Width = 12 ft Shoulder width = 8 ft Route A Minor Arterial Design ADT = 2500 Rolling Terrain Design Speed = 50 mph T = 6% 2-lane undivided Lane width = 12 ft Shoulder width = 8 ft Product Volume Std. Dwg. to be used Driveway Type Min. Radius Min. Surface Width “d” along main road “d” along minor road SSD along main road SSD along minor road Find: P.V.=18000x2500 = 45,000,000 203.65 Type V 90 ft 24 ft 260 ft 220 ft 1050-1300 ft 400-475 ft

12. Comprehensive Lab: Horizontal, Vertical Alignment and Sight Distance Given: Rural Collector, 2-12’ lanes (undivided) Rolling Terrain Design ADT = 770, Construction ADT = 530 Design Speed = 40 mph DHV = 12% T = 7% NC = -2% Project Description: Bridge replacement and re-alignment. Project starts at station 226+30, and ends at station 244+69.35. Total length of project = 1839.35 ft. Beginning tie-in bearing is S 48°40’57” E and ending tie-in bearing is S 41°01’12” E 1 st Curve2 nd Curve Incoming Grade-0.3%+5.07% Outgoing Grade+5.07%+1.06% VPI station & elevation236+30, 721.14240+22.74, 741.04

13. Comprehensive Lab: Horizontal, Vertical Alignment and Sight Distance 1. Find L, R, and T for each curve Curve 1Curve 2 R = 2721.67 ft D = 2.105° T = 301.22 ft  = 12.6.31° RT  = 4.9684° LT R = 6919.22 ft D = 0.8281º T = 300.19 ft 2. Find PC, PI, and PT stationing for each curve Curve 1 Data PC Station227+79.84 PI Station230+81.06 PT Station233+79.84 Curve 2 Data 237+99.00 240+99.19 243+99.00

14. Comprehensive Lab: Horizontal, Vertical Alignment and Sight Distance 3. Find superelevation transition points Curve 1 Curve 2: According to Standard Plan 203.20 page 25 no superelevation is required.

15. Comprehensive Lab: Horizontal, Vertical Alignment and Sight Distance 4. Find VPI and VPT stationing and elevations for e/vertical curve Vertical Curve 1Vertical Curve 2 A = 5.37% VPC sta = 234+60.00 VPC elevation = 721.65 ft L = 340 ft for K = 63.3 VPT sta = 238+00.00 VPT elevation = 729.76 ft A = 4.01% L = 320 ft SSD = 325.71 and V = 40 mph VPC sta = 238+62.74 VPC elevation = 732.93 ft VPT sta = 241+82.74 VPT elevation = 742.74 ft

16. Comprehensive Lab: Horizontal, Vertical Alignment and Sight Distance 5. Calculate the low and/or high point for each vertical curve Vertical Curve 1Vertical Curve 2 X m = 18.99 ft from VPC E 18.99 = 721.62 ft VPT station is low point = 241+82.74 Elevation = 742.74 ft