Vertical Alignment in Railway Track Geometric Design by Dr. Walied A. Elsaigh

              
Dr. Walied A. Elsaigh
Dr. Walied A. Elsaigh
            
            
welsaigh@ksu.edu.sa
welsaigh@ksu.edu.sa
      Asst. Prof. of Civil Engineering
      Asst. Prof. of Civil Engineering
 
               CE 435
               CE 435
Railway Engineering
Railway Engineering
 
Geometric Design
Geometric Design
of Railway Track
of Railway Track
 
Railway vertical curves – old formula:
 
L = D / R
D = algebraic difference of grade (ft. per 100-ft. station)
R = rate of change per 100-ft. station
0.05 ft. per station for crest on main track
0.10 ft. per station for sag on main track
Secondary line may be twice those for main line
 
Vertical Alignment
Vertical Alignment
Curve Length
Curve Length
 
  Vertical Alignment
  Vertical Alignment
Curve Length
Curve Length
 
Old railway formula developed in 1880’s for “hook and pin”
couplers in those days
Present day couplers can accommodate shorter vertical
curves
New formula 
developed in recent years:
 
L = 2.15 V
2
 D / A
V = train speed in mph
D = algebraic difference of grade in decimal
A = vertical acceleration in ft./sec
2
0.1 ft./ sec
2
 
for freight, 0.6 ft./ sec
2
 
for passenger or transit
 
Types of Crest and Sag Vertical Curves
 
Properties of Typical Vertical Curve
 
EXAMPLE-1
 
A plus 3.0 percent grade intersects a minus 2.0
percent grade at station 3 + 20 and at an
elevation of 320.40 ft. Given that a 180-ft length
of curve is utilized,
(1)
Determine the 
station
 and 
elevation
 of the PVC
and PVT.
(2)
Calculate elevations at every even 25-ft station
(3)
Compute the station and elevation of the high
point of the curve
 
Solution
 
EXAMPLE  (continued)
 
E
PVC
 = E
PVI
 - (G
1
/100)(L/2) = 320.40 - 0.03(90) = 317.70 ft
E
PVT
 = E
PVI
 - (G
2
/100)(L/2) = 320.40 - 0.02(90) = 318.60 ft
 
Location of high point:
 
High point Sta = PVC Sta + X
m
 = 230 + 108 = 338 → Sta 3+
38
Elevation of high point:
 
 
 
 
 
 
 
Calculations for point elevations at even
25-ft stations along the vertical curve
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Explore the vertical alignment of railway tracks through Dr. Walied A. Elsaigh's insights on curve length formulas, types of crest and sag vertical curves, properties of typical vertical curves, and a detailed example with solutions for calculating elevations at various stations along the curve. Learn about traditional and modern formulas for designing vertical curves in railway engineering.

  • Railway Engineering
  • Vertical Alignment
  • Dr. Walied A. Elsaigh
  • Geometric Design
  • Civil Engineering

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  1. Vertical Alignment Picture Geometric Design of Railway Track Dr. Walied A. Elsaigh welsaigh@ksu.edu.sa Asst. Prof. of Civil Engineering CE 435 Railway Engineering

  2. Vertical Alignment Curve Length Railway vertical curves old formula: L = D / R D = algebraic difference of grade (ft. per 100-ft. station) R = rate of change per 100-ft. station 0.05 ft. per station for crest on main track 0.10 ft. per station for sag on main track Secondary line may be twice those for main line

  3. Vertical Alignment Curve Length Old railway formula developed in 1880 s for hook and pin couplers in those days Present day couplers can accommodate shorter vertical curves New formula developed in recent years: L = 2.15 V2 D / A V = train speed in mph D = algebraic difference of grade in decimal A = vertical acceleration in ft./sec2 0.1 ft./ sec2for freight, 0.6 ft./ sec2for passenger or transit

  4. Types of Crest and Sag Vertical Curves

  5. Properties of Typical Vertical Curve

  6. EXAMPLE-1 A plus 3.0 percent grade intersects a minus 2.0 percent grade at station 3 + 20 and at an elevation of 320.40 ft. Given that a 180-ft length of curve is utilized, (1)Determine the station and elevation of the PVC and PVT. (2)Calculate elevations at every even 25-ft station (3)Compute the station and elevation of the high point of the curve

  7. Solution

  8. EXAMPLE (continued) EPVC = EPVI - (G1/100)(L/2) = 320.40 - 0.03(90) = 317.70 ft EPVT = EPVI - (G2/100)(L/2) = 320.40 - 0.02(90) = 318.60 ft Location of high point: High point Sta = PVC Sta + Xm= 230 + 108 = 338 Sta 3+ 38 Elevation of high point:

  9. Calculations for point elevations at even 25-ft stations along the vertical curve Final Elevation x Elevation on Station y on Curve (feet) Initial Tangent (Elev on tan - y) 318.24 318.77 319.12 319.30 319.30 319.13 318.79 318.60 2 + 50 2 + 75 3 + 00 3 + 25 3 + 50 3 + 75 4 + 00 4 + 10 20 45 70 95 120 145 170 180 318.30 319.05 319.80 320.55 321.30 322.05 322.80 323.10 -0.06 -0.28 -0.68 -1.25 -2.00 -2.92 -4.01 -4.50

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