CE 515 Railroad Engineering

Hump Yard DesignSource: 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

• 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

• 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

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Engineering It All

• Energy balance equation:KE1 + Y1 = KE2 +Y2 – (MKX + SW + CR + WR + ER)

• 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 * XWn

• 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

• 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

Point Length (ft) Gradient (%) Elev. Chg. (ft) Sw (ft) Cr (ft) Mk (ft) Net (ft) KE v (ft/s) A 0.76 7.00 130 -3.7 4.81 0 0 1.17 3.64 B 4.40 16.83 120 -1.3 1.56 0.03 0.16 1.08 0.29 C 4.69 17.38

Other Calculations

• CR = 0.025 * º of central angle

• WR: As discussed previously

Side Note: Targets

Source: wikipedia.org

Questions?