CE 515 Railroad Engineering
Hump Yard Design
Source: Lecture Notes – CE 353
by Reg Souleyrette
“Transportation exists to conquer space and time -”
Classification (Hump) Yard
Photo: www.bilderberg.org/railways
Factors to Consider
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Size of yard (number of tracks/length)
Resistance
Acceleration on grade
Maximum impact speed
Safety
Classification (Hump) Yard
Source: Dr. Souleyrette’s Lecture Notes
General Guidelines
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Hump grades: 4% (100-200 ft.)
Transition: 1.5%
Switching: 1.2%
Classification track: 0.1-0.5%
Spacing: 14-18 feet on centers
Turnouts: #7-10
Retarders
Photo: www.wikipedia.org
Photo: www.sigrail.com
Source: railpictures.net
Too Much?
Engineering It All
• Energy balance equation:
KE1 + Y1 = KE2 +Y2 – (MKX + SW + CR + WR + ER)
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KE: Kinetic Energy (v2/(2g))
Y: Elevation head, (ft)
X: Horizontal distance, (ft)
MK: Static rolling resistance, (lb/ton) (typ. 2-18)
SW: Losses due to passing through switch, (ft) (typ. 0.02-0.06)
CR: Curve losses, (ft) (typ. 0.025 ft/º of angle)
WR: Wind loss (air resistance), (ft) (next slide)
ER: Energy extracted by retarders, (ft) (next slide)
Energy Losses
• Air resistance (Davis equation):
KAV2 * X
Wn
• Retarders:
– Variable, up to 0.11 ft. of head/ft. of retarder
– Typical minimum length of 20 ft.
– Double if retarders on both rails
Vertical/Horizontal Curves
• Vertical Curves
– Minimum length (ft.): L = A * C
• A: Algebraic difference in grades, %
• C: Constant dependent on curve type
– C = 15 for hump crest
– C = 40 for other crests
– C = 60 for sag curves
• Horizontal Curves
– Maximum of 12.5º
Car Velocity
• Consider headway to allow throwing of
switches
• Vs = Lc + H * Vh
Lc
– Vs: Velocity at switch
– Vh: Velocity at hump (release)
– Lc: Length of car (avg. 60 ft.)
– H: Headway (typ. 60 ft.)
• Coupling velocity of 6 ft/s (4 mph)
Examples
• Grade leading to hump = +1.0%
• Grade after hump = -3.5%
– Min. Length L = A * C
– L = (1.0 – (-3.5)) * 15
– L = (4.5) * 15 = 67.5 feet
Examples
• Grade after hump = -3.5%
• Grade leading to switches = -1.5%
– Min. Length L = A * C
– L = (-3.5 – (-1.5)) * 60
– L = (-2) * 60 = -120 ft.  120 ft.
Sample Calculations
Point Length (ft) Gradient (%) Elev. Chg. (ft) Sw (ft) Cr (ft) Mk (ft) Net (ft)
A
130
-3.7
4.81
0
0
1.17
3.64
B
120
-1.3
1.56 0.03 0.16
1.08
0.29
C
KE
v (ft/s)
0.76
7.00
4.40
16.83
4.69
17.38
• KEA = v2/(2g) = (7)2/(2*32.2) = 0.76 ft
• Elev. Chg. = -X * (%)/100 = -130 * (-3.7)/100 = 4.81 ft
• MK loss = X * MK/2000 = 130 * 18/2000 = 1.17 ft
• Net = EC – SW – CR – MK = 4.81 – 0 – 0 – 1.17 = 3.64 ft
• KEB = KEA + Net = 0.76 + 3.64 = 4.40 ft
Other Calculations
• CR = 0.025 * º of central angle
• WR: As discussed previously
Side Note: Targets
Source: wikipedia.org
Questions?