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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 • • • • • 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 Source: railpictures.net Too Much? 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 * 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?