- 1. Highway Geometric design Houssam Ihsan Siyoufi
- 2. Q1) • An urban freeway is being designed with the following characteristics: 11-ft lanes; no lateral clearance; a ramp density of 4.5/mi; 7% trucks and no RVs; PHF = 0.90; rolling terrain. Determine the minimum number of lanes required for the freeway to achieve a LOS of C or better if the demand peak-hour directional demand volume is 3400 veh/h.
- 3. Step 1: Specify input data • DHV = 3400 veh/h • PHF = 0.90 • PT = 0.07 • PR = 0 • Lane width = 11 ft • Lateral clearance = 0 • Rolling terrain • TRD = 4.5 ramps per mile Assume 4 lane highway ( 2 in each direction)
- 4. Step 2: compute the value of free-flow speed (FFS) • We either measure from field test or we use the equation of 𝐹𝐹𝑆=75.4–𝑓𝐿𝑊–𝑓𝐿𝐶–3.22𝑇𝑅𝐷0.84 1. Get the adjustment factor for lane width from table 9.1 (slide number 48) our lane width is 11ft so our 𝑓𝐿𝑊 =1.9 Table 9.1 (Garber & Hoel2013)
- 5. 2. For our 𝑓𝐿𝐶 which is our lateral clearance, go to table 9.2 (slide 49). We assumed 4 lane freeway which means we have 2 lanes in each direction and we are given that there is zero lateral clearance. From the table our 𝑓𝐿𝐶=3.6 • Table 9.2 (Garber & Hoel2013)
- 6. Now we can calculate FFS • For TRD we are given that we have a ramp density of 4.5/mile • 𝐹𝐹𝑆=75.4–𝑓𝐿𝑊–𝑓𝐿𝐶–3.22𝑇𝑅𝐷0.84 • 𝐹𝐹𝑆=75.4–1.9–3.6–3.22∗ 4.50.84 = 58.1 mi/h
- 7. Step 3: select FFS curve • It is based on what value we calculated FFS which was 58.1 mi/h • For FFS = 58.51 mi/h (≥ 57.5 < 62.5) → use the 60 mi/h speed curve; breakpoint = 1600 pc/h/ln. • Use slide number 51
- 8. Step 4: compute the demand flow rate (vp) • FOR 𝑓𝑝 which the adjustment factor for unfamiliar drivers, in the question given we are not told who are the type of drivers, so if it is not given, we go to base condition of freeway and assume all are regular commuters (regular users) for that 𝑓𝑝 =1 • 𝑣𝑝= 𝑉 𝑃𝐻𝐹×𝑓𝑃×𝑁×𝑓𝐻𝑉 = 3400 0.9×1×2×𝑓𝐻𝑉 • The only unknown is 𝑓𝐻𝑉, the 𝑓𝐻𝑉 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) • The terrain that we have is rolling (given) • Since 𝑃𝑇 = 0.07 then 𝐸𝑇=2.5 while 𝑃𝑅=0 then 𝐸𝑅=2
- 9. • the 𝑓𝐻𝑉 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) = the 𝑓𝐻𝑉 = 1 1+0.07 2.5−1 +0(2−1) =0.905 • 𝑣𝑝= 𝑉 𝑃𝐻𝐹×𝑓𝑃×𝑁×𝑓𝐻𝑉 = 𝑣𝑝= 3400 0.9×1×2×0.905 =2087 pc/h/ln • 𝒗𝒑=2087pc/h/ln • Remember in this question we used the case 1 extended grade segment » Not too long or too steep grades » Area is level, rolling, or mountainous » Use Table 9.3
- 10. Step 5: estimate average passenger car speed (S) • We must check first if our 𝑣𝑝 is either greater or less than the breakpoint , if 𝑣𝑝 is greater than breakpoint we use the following equations given in slide number 16, if not we use the FFS as our average car speed. • The breakpoint we got from step 3 is equal to 1600 pc/h/ln • The 𝑣𝑝=2087 pc/h/ln • So 𝑣𝑝> breakpoint – so we compute 𝑆𝐵𝑃 • 𝑆𝐵𝑃−60=60-0.00001816(𝑣𝑝 − 1600)2= 60 – 0.00001816( 2087- 1600)2 =55.69 mi/h
- 11. Step 6: compute the density (D) • D= 𝑣𝑝 𝑆 = 2087 55.69 =37.47 pc/mi/ln • Step 7: estimate LOS For D = 37.47 pc/mi/ln (>35-45) → LOS = E • LOS on Basic Freeway Segments (pg461) • Use 6 lane highway (3/direction)
- 12. Step 2: compute the value of free-flow speed (FFS) • We either measure from field test or we use the equation of 𝐹𝐹𝑆=75.4–𝑓𝐿𝑊–𝑓𝐿𝐶–3.22𝑇𝑅𝐷0.84 1. Get the adjustment factor for lane width from table 9.1 (slide number 48) our lane width is 11ft so our 𝑓𝐿𝑊 =1.9 Table 9.1 (Garber & Hoel2013)
- 13. In getting flc there is a change from 2 lanes to 3 lanes 2. For our 𝑓𝐿𝐶 which is our lateral clearance, go to table 9.2 (slide 49). We assumed 6 lane freeway which means we have 3 lanes in each direction and we are given that there is zero lateral clearance. From the table our 𝑓𝐿𝐶=2.4 Table 9.2 (Garber & Hoel2013)
- 14. Now we can calculate FFS • For TRD we are given that we have a ramp density of 4.5/mile • 𝐹𝐹𝑆=75.4–𝑓𝐿𝑊–𝑓𝐿𝐶–3.22𝑇𝑅𝐷0.84 • 𝐹𝐹𝑆=75.4–1.9–2.4–3.22∗ 4.50.84 = 59.71 mi/h
- 15. Step 3: select FFS curve • It is based on what value we calculated FFS which was 59.71 mi/h • For FFS = 59.71 mi/h (≥ 57.5 < 62.5) → use the 60 mi/h speed curve; breakpoint = 1600 pc/h/ln. • Use slide number 51
- 16. Step 4: compute the demand flow rate (vp) • 𝑣𝑝= 𝑉 𝑃𝐻𝐹×𝑓𝑃×𝑁×𝑓𝐻𝑉 = 3400 0.9×1×𝟑×0.905 =1392 pc/h/ln • The only change is the number of lanes (N)
- 17. Step 5: estimate average passenger car speed (S) • We must check first if our 𝑣𝑝 is either greater or less than the breakpoint , if 𝑣𝑝 is greater than breakpoint we use the following equations given in slide number 16, if not we use the FFS as our average car speed. • The breakpoint we got from step 3 is equal to 1600 pc/h/ln • The 𝑣𝑝=1392 pc/h/ln • So 𝑣𝑝 < breakpoint – so we compute so our S is 60mi/h from step 3 •
- 18. Step 6: compute the density (D) • D= 𝑣𝑝 𝑆 = 1392 60 =23.20 pc/mi/ln • Step 7: estimate LOS For the computed Density = 23.20 pc/mi/ln (>18-26) → LOS = C • LOS on Basic Freeway Segments (pg461) • Therefore, use six-lane highway (3 lanes per direction)
- 19. Q2) • A long section of a four-lane undivided multilane highway is in level terrain. A section of 5 mile is however followed by a 5% grade, 2.0 mi in length. If the demand hourly volume is 1500 veh/h, estimate the level of service on each of the upgrade and downgrade if the roadway has the following characteristics: percentage trucks and buses = 10%; percentage RVs = 4%; PHF = 0.95; base free-flow speed = 60 mi/h; average lane width = 11 ft; lateral clearance = 4 ft at both roadsides; and access point density = 15/mi on each side of the roadway
- 20. Step 1: specify input data • N = 2 lanes/direction • Median type = Undivided • DHV = 1500 veh/h/direction • PT = 0.10 • PR = 0.04 • PHF = 0.95 • BFFS = 60 mi/h • Level terrain • Lane width = 11 ft • Lateral clearance => 4 ft on both road sides • Average access point density = 15 points/mile
- 21. Step 2: compute the value of free-flow speed FFS=BFFS- 𝑓𝑙𝑤 − 𝑓𝐿𝐶 − 𝑓𝐴- 𝑓𝑀 1. Lane width = 11 ft → fLW = 1.9 (Table 9.1)
- 22. 2. Total lateral clearance = 4 (right side shoulder) + 6 (undivided road) = 10 ft → fLC = 0.4 (Table 9.7) (slide 58) TLC= LCR + LCL =6+4=10 » LCR = right-side lateral clearance (shoulder); max = 6 ft » LCL = left-side lateral clearance (median) » LCL = 6 ft(if median > 6ft; including TWLTL) » LCL = 6 ft(undivided roads)
- 23. 3. 𝑓𝑚 =? median = undivided → fM = 1.6 (Table 9.8) (slide number 59)
- 24. 4. Average access point density = 15 points/mile → fA = 3.75 (Table 9.9)
- 25. • FFS=BFFS- 𝑓𝑙𝑤 − 𝑓𝐿𝐶 − 𝑓𝐴- 𝑓𝑀 • 60 - 1.9 - 0.4 – 3.75 - 1.6= 52.35 mi/h • FFS=52.35 mi/h
- 26. Step 3: select FFS curve • For FFS = 52.35 mi/h (≥ 47.5 < 52.5) → use the 50 mi/h speed curve; breakpoint = 1400 pc/h/ln Recommended FFS Curves & Flow-Rate Ranges for Multilane Highways
- 27. From step 4-7 we will calculate the result once for the upgrade and once for the downgrade • Upgrade : Step 4: compute the value of flow rate vp ofp = 1.0 (assume mainly commuter traffic) oPT = 0.1; ET = 3.5 (Table 9.4; G=5% & L = 2 mi) oPR = 0.04; ER = 3.5 (Table 9.5; G=5% & L = 2 mi
- 28. Table 9.4 (Garber & Hoel2013) slide 53 Table 9.5 (Garber & Hoel2013)
- 29. • 𝑣𝑝= 𝑉 𝑃𝐻𝐹×𝑓𝑃×𝑁×𝑓𝐻𝑉 = 1500 0.95×1×2×𝑓𝐻𝑉 • the 𝑓𝐻𝑉 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) = 1 1+0.1 3.5−1 +0.4(3.5−1) =0.74 • 𝑣𝑝= 𝑉 𝑃𝐻𝐹×𝑓𝑃×𝑁×𝑓𝐻𝑉 = 1500 0.95×1×2×0.74 =1067pc/h/ln • Remember we used case 2 • Case 2: Specific grade • » L > 0.25 mi & G > 3% or L > 0.5 mi & G = 2-3% • » For upgrades: use Table 9.4 for ET & Table 9.5 for ER • » For downgrades: use Table 9.6 for ET & ER = ER for level grade
- 30. Step 5: estimate average passenger car speed (S) • Equal to FFS at low flow rates (up to the breakpoint of vp = 1400 pc/h/ln): • Beyond the breakpoint, use Figure 9.7or the relevant equations in slide number 27 • 1067<1400pc/h/ln • 𝑣𝑝 < 1400 pc/h/ln → S = 50 mi/h
- 31. Step 6: compute the density (D) • D= 𝑣𝑝 𝑆 = 1067 50 =21.34 mi/h • Step 7: estimate LOS –Threshold of LOS F changes with FFS: – Begins with Density= 40 pc/mi/ln for 60 mi/h and increases to 45 pc/mi/ln for 45 mi/h –Otherwise, LOS criteria are similar to those on basic freeway segments or use Figure 9.8 or Table 9.10
- 32. • For D = 21.34 pc/mi/ln → Table 9.10 → LOS = C • Table 9.10 (Garber & Hoel2013)
- 33. Downgrade • Step 4: compute the value of flow rate (𝑣𝑝) fp = 1.0 (assume mainly commuter traffic) PT = 0.1; ET = 1.5 (Table 9.6; G=5% & L = 2 mi) PR = 0.04; ER = 1.2 (Table 9.3; level terrain) Note: use level terrain for PR on downgrades (ref. pg. 470, textbook) Our grade is 5% and our L=2 mil which is less than 4, while our proportion of trucks and buses= 10% (Pt=0.1)
- 34. Table 9.6 slide number 55 Et=1.5
- 35. Table 9.3 (Garber & Hoel2013) ER=1.2
- 36. • 𝑣𝑝= 𝑉 𝑃𝐻𝐹×𝑓𝑃×𝑁×𝑓𝐻𝑉 = 1500 0.95×1×2×𝑓𝐻𝑉 • the 𝑓𝐻𝑉 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) = 1 1+0.1 1.5−1 +0.4(1.2−1) =0.945 • 𝑣𝑝= 𝑉 𝑃𝐻𝐹×𝑓𝑃×𝑁×𝑓𝐻𝑉 = 1500 0.95×1×2×0.945 =835pc/h/ln
- 37. Step 5: estimate average passenger car speed (S) • 835>1400 • 𝑣𝑝 < 1400 pc/h/ln → S = 50 mi/h •Step 6: compute the density (D) •D= 𝑣𝑝 𝑆 = 835 50 =16.7pc/mi/ln •
- 38. Step 7: estimate LOS • –Threshold of LOS F changes with FFS: • – Begins with D= 40 pc/mi/ln for 60 mi/h and increases to 45 pc/mi/ln for 45 mi/h • –Otherwise, LOS criteria are similar to those on basic freeway segments or use Figure 9.8or Table 9.10 • Our density is 16.7 pc/mi/ln • For D = 16.7 pc/mi/ln → Table 9.10 → LOS = B
- 39. Table 9.10 (Garber & Hoel2013)
- 40. Tutorial 6
- 41. Q1 • An existing Class II two-lane highway is to be analyzed to determine the level of service in the peak direction given the following information: Peak hourly volume in the analysis direction: 900 veh/h; Peak hourly volume in the opposing direction: 400 veh/h; Trucks: 12% of total volume; Recreational vehicles: 2% of total volume; PHF: 0.95; Lane width: 12 ft; Shoulder width: 10 ft; Access points per mile: 20; Terrain: rolling; Base free flow speed: 60 mi/h; No passing zones: 40% of analysis segment length
- 42. Step 1: specify input data • Class II Highway • No passing zones = 40% • BFFS ( base free flow speed)= 60 mi/h • Access points per mile = 20 • Peak hourly volume in the analysis direction = 900 veh/h • Peak hourly volume in the opposing direction = 400 veh/h • PHF (peak hour factor)= 0.95 • PT = 0.12 (percentage of trucks =12%) • PR = 0.02 (percentage of recreational vehicles= 2%) • Lane width = 12 ft • Shoulder width = 10 ft • Rolling terrain
- 43. Step 2: compute the demand adjustment for PTSF • For a class II two–lane highway, we analyze level of service by PTSF ( percent time spent following) • Percent time spent following another vehicle (PTSF): ❖ It is the average percentage of time that vehicles are traveling behind slower vehicles (time headway between consecutive vehicles is less than 3 s) • We do computation of the demand adjustment for PTSF in; 1. Analysis direction 2. Opposing direction
- 44. Analysis Direction • We are given Peak hourly volume in the analysis direction = 900 veh/h • We are give PHF=0.95 • 𝑣𝑖,𝑃𝑇𝑆𝐹= 𝑉𝑖 𝑃𝐻𝐹×𝑓𝑔,𝑃𝑇𝑆𝐹×𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 900 0.95×𝑓𝑔,𝑃𝑇𝑆𝐹×𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 • The 2 unknowns are 𝑓𝑔,𝑃𝑇𝑆𝐹 & 𝑓𝐻𝑉 ,𝑃𝑇𝑆𝐹 • 𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) • The 𝑃𝑇 = 0.12 and the 𝑃𝑅 =0.02
- 45. • for 𝑷𝑻 = 0.12 we go to table 9.23 in slide number 77, remember our terrain is Rolling , in this table no interpolation 𝑬𝑻=1
- 46. 𝑃𝑅=0.02 from the same table we get 𝐸𝑅 • 𝑃𝑅 =0.02 our 𝐸𝑇 =1 • 𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) = 1 1±0.12 1−1 +0.02(1−1) =1 • For 𝑓𝑔,𝑃𝑇𝑆𝐹 = 1 from table 9.21 (slide number 75) the 𝑣= 𝑉 𝑃𝐻𝐹 = 900 0.95 =947
- 47. • 𝑣𝑖,𝑃𝑇𝑆𝐹= 𝑉𝑖 𝑃𝐻𝐹×𝑓𝑔,𝑃𝑇𝑆𝐹×𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 900 0.95×1×1 =947.37pc/h • For table 9.21 interpolate, • Table 9.23 do not interpolate
- 48. Opposing direction • We are given Peak hourly volume in the opposite direction = 400 veh/h • We are give PHF=0.95 • 𝑣𝑖,𝑃𝑇𝑆𝐹= 𝑉𝑖 𝑃𝐻𝐹×𝑓𝑔,𝑃𝑇𝑆𝐹×𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 400 0.95×𝑓𝑔,𝑃𝑇𝑆𝐹×𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 • The 2 unknowns are 𝑓𝑔,𝑃𝑇𝑆𝐹 & 𝑓𝐻𝑉 ,𝑃𝑇𝑆𝐹 • 𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) • The 𝑃𝑇 = 0.12 and the 𝑃𝑅 =0.02
- 49. • for 𝑷𝑻 = 0.12 we go to table 9.23 in slide number 77, remember our terrain is Rolling , in this table no interpolation, 𝑬𝑻=1.6
- 50. 𝑃𝑅=0.02 from the same table we get 𝐸𝑅 • 𝑃𝑅 =0.02 our 𝐸𝑇 =1 • 𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) = 1 1±0.12 1.6−1 +0.02(1−1) =0.93 • For 𝑓𝑔,𝑃𝑇𝑆𝐹 = 0.91 from table 9.21 the 𝑣= 𝑉 𝑃𝐻𝐹 = 400 0.95 = 421 • Remember for this table we interpolate.
- 51. How to interpolate • Our directional demand flow was calculated to be 421 and it is between 400 and 500. The 400 has a value of 0.9 while the 500 has a value of 0.96. • X1=400, X2=421 which we want its Y2 , X3=500 • Y1=0.9 ,Y2=? , Y3=0.96 • Y2= (𝑋2−𝑋1)×(𝑌3−𝑌1) (𝑋3−𝑋1) + 𝑌1 = 0.9126 𝑤ℎ𝑖𝑐ℎ 𝑤𝑒 𝑠𝑎𝑦 𝑖𝑡 𝑖𝑠 0.91 (interpolation to the nearest 0.01).
- 52. • 𝑣𝑖,𝑃𝑇𝑆𝐹= 𝑉𝑖 𝑃𝐻𝐹×𝑓𝑔,𝑃𝑇𝑆𝐹×𝑓𝐻𝑉,𝑃𝑇𝑆𝐹 = 400 0.95×0.91×0.93 =497.52pc/h • For table 9.21 interpolate, • Table 9.23 do not interpolate
- 53. Step 3: Estimate PTSF • 𝑃𝑇𝑆𝐹𝑑= 𝐵𝑃𝑇𝑆𝐹𝑑 + 𝑓𝑛𝑝,𝑃𝑇𝑆𝐹 × { 𝑣𝑑,𝑃𝑇𝑆𝐹 𝑣𝑑,𝑃𝑇𝑆𝐹±𝑣𝑜,𝑃𝑇𝑆𝐹 }
- 54. • First we get BPTS𝐹𝑑 • so a=-0.0027 (Table 9.26, Use 𝑣𝒐 𝑷𝑻𝑺𝑭= 497.52; interpolate to nearest 0.0001) (slide 80) • And b=0.897 (Table 9.26, Use 𝑣𝒐 𝑷𝑻𝑺𝑭 = 497.52; interpolate to nearest 0.001) same slide 80 • 𝐵𝑃𝑇𝑆𝐹𝑑= 100 × (1 − 𝑒−0.0027(947.37)0.897 )=71.71%
- 55. We do the interpolation between 400 and 600 because our opposing demand flow rate=498 from step 2
- 56. 𝑣𝑇,𝑃𝑇𝑆𝐹= 𝑣𝑑,𝑃𝑇𝑆𝐹+ 𝑣𝑜,𝑃𝑇𝑆𝐹=947.37+497.52=144.89 pc/h • 𝑣𝑑,𝑃𝑇𝑆𝐹 𝑣𝑇,𝑃𝑇𝑆𝐹 = 947.37 144.89 =0.66=66% ≈ 70% •So the split is directional =70/30
- 57. 𝑓𝑛𝑝,𝑃𝑇𝑆𝐹? • From table 9.25 slide number 79, the no passing zone was given 40%, the computed 𝐯𝐓,𝐏𝐓𝐒𝐅=144.89 pc/h • 𝑓𝑛𝑝,𝑃𝑇𝑆𝐹=17.2 (by interpolation)
- 59. 𝑃𝑇𝑆𝐹𝑑= 𝐵𝑃𝑇𝑆𝐹𝑑 + 𝑓𝑛𝑝,𝑃𝑇𝑆𝐹 × { 𝑣𝑑,𝑃𝑇𝑆𝐹 𝑣𝑑,𝑃𝑇𝑆𝐹±𝑣𝑜,𝑃𝑇𝑆𝐹 } • 𝑃𝑇𝑆𝐹𝑑= 71.71+17.2× { 947.37 947.37±497.52 }=82.99% •
- 60. Step 4 Determine Level of Service (LOS) • LOS = D (Table 9.11) slide 65
- 61. Q2) • An existing Class III two-lane highway is to be analyzed to determine the level of service in the peak direction given the following information: Peak hourly volume in the analysis direction: 900 veh/h; Peak hourly volume in the opposing direction: 720 veh/h; Trucks: 10% of total volume; Recreational vehicles: 2% of total volume; PHF: 0.94; Lane width: 12 ft; Shoulder width: 2 ft; Access points per mile: 30; Terrain: level; Measured free flow speed: 45 mi/h; No passing zones: 60% of analysis segment length
- 62. Step 1: specify input data • Class III Highway • Peak hourly volume in the analysis direction = 900 veh/h • Peak hourly volume in the opposing direction = 720 veh/h • PT = 0.10 • PR = 0.02 • PHF: 0.94 • Lane width = 12 ft • Shoulder width = 2 ft • Access points per mile = 30 • Terrain: level Measured FFS = 45 mi/h • No passing zones = 60%
- 63. Step 2: Compute demand adjustments for ATS • Analysis Direction • Opposite Direction
- 64. Main direction • 𝑣𝑑,𝐴𝑇𝑆= 𝑉𝑑 𝑃𝐻𝐹×𝑓𝑔,𝐴𝑇𝑆×𝑓𝐻𝑉,𝐴𝑇𝑆 • The 2 unknowns are 𝑓𝑔,𝐴𝑇𝑆 & 𝑓𝐻𝑉 ,𝐴𝑇𝑆 • 𝑓𝐻𝑉,𝐴𝑇𝑆 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) • 𝑃𝑇 = 0.10 so from table 9.16, where our terrain is level in slide 70 so our 𝐸𝑇= 1 , v= 𝑉 𝑃𝐻𝐹 = 900 0.94 =957 • 𝑃𝑅=0.02 from same table 𝐸𝑅=1
- 65. 𝑓𝐻𝑉,𝐴𝑇𝑆 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) • 𝑓𝐻𝑉,𝐴𝑇𝑆 = 1 1+0.10 1.1−1 +0.02(1−1) =1
- 66. Remember for 𝑓𝐻𝑉,𝐴𝑇𝑆 • For general terrain, upgrade and downgrades where trucks do travel at a crawling speed (lecture notes slide 38). • You can use table 9.16, 9.17 , 9.18 • For downgrades where trucks travel at crawling speed • 𝑓𝐻𝑉,𝐴𝑇𝑆 = 1 1+𝑃𝑇𝐶 × 𝑃𝑇𝐶 𝐸𝑇𝐶−1 +(1−𝑃𝑇𝐶 )×𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1 • 𝑃𝑇𝐶= proportion of trucks at crawl speed, to 𝐸𝑇 use table 9.19
- 67. • For 𝑓𝑔,𝐴𝑇𝑆= 1 from table 9.14 the 𝑣= 𝑉 𝑃𝐻𝐹 = 900 0.94 =957 pc/h • INTERPOLATE to nearest 0.01 • Level or rolling terrain and downgrade (table 9.14) • If Upgrades (G≥ 3% & L≥ 0.6 mi) (Table 9.15)
- 68. Table 9.14
- 70. Opposite direction • 𝑣𝑜,𝐴𝑇𝑆= 𝑉𝑜 𝑃𝐻𝐹×𝑓𝑔,𝐴𝑇𝑆×𝑓𝐻𝑉,𝐴𝑇𝑆 • The 2 unknowns are 𝑓𝑔,𝐴𝑇𝑆 & 𝑓𝐻𝑉 ,𝐴𝑇𝑆 • 𝑓𝐻𝑉,𝐴𝑇𝑆 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) • 𝑃𝑇 = 0.10 so from table 9.16, where our terrain is level in slide 70 so our 𝑬𝑻= 1.1 , v= 𝑉 𝑃𝐻𝐹 = 720 0.94 =766 (interpolate nearest 0.1 • 𝑃𝑅=0.02 from same table 𝐸𝑅=1
- 72. • 𝑓𝐻𝑉,𝐴𝑇𝑆 = 1 1+𝑃𝑇 𝐸𝑇−1 + 𝑃𝑅(𝐸𝑅−1) = 𝑓𝐻𝑉,𝐴𝑇𝑆 = 1 1+0.10 1.1−1 +0(1−1) =0.99 • For 𝑓𝑔,𝐴𝑇𝑆= 1 from table 9.14 the 𝑣= 𝑉 𝑃𝐻𝐹 = 720 0.94 =766 pc/h • INTERPOLATE to nearest 0.01 • 𝑣𝑜,𝐴𝑇𝑆= 𝑉𝑜 𝑃𝐻𝐹×𝑓𝑔,𝐴𝑇𝑆×𝑓𝐻𝑉,𝐴𝑇𝑆 = 720 0.94×1×0.99 =773.69pc/h
- 73. Step 3 : Estimate ATS • 𝐴𝑇𝑆𝑑=FFS – 0.00776(𝑣𝐷,𝐴𝑇𝑆 + 𝑣0,𝐴𝑇𝑆) -𝑓𝑛𝑝,𝐴𝑇𝑆 • 𝑓𝑛𝑝,𝐴𝑇𝑆= adjustment factor for ATS determination for the percentage of no-passing zones in the analysis direction (Table 9.20) • 0.9 (Table 9.20; FFS = 45 mi/h; No passing zones = 60%; 𝑣0,𝐴𝑇𝑆=773.69 pc/h; interpolate to nearest 0.1) • 𝐴𝑇𝑆𝑑=FFS – 0.00776(𝑣𝐷,𝐴𝑇𝑆 + 𝑣0,𝐴𝑇𝑆) -𝑓𝑛𝑝,𝐴𝑇𝑆= • 𝑨𝑻𝑺𝒅=45 – 0.00776(957.57+773.69) – 0.9 = 30.67mi/h
- 74. Table 9.20 has two tables, in this example we do not use this table
- 75. Table 9.20
- 76. Step 4 : Determine LOS • PFF= 𝐴𝑇𝑆 𝐹𝐹𝑆 = 30.67 45 = 0.68 = 68% • LOS= D (TABLE 9.11)
- 77. TABLE 9.11