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:

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% ( ft.) Transition: 1.5% Switching: 1.2% Classification track: % Spacing: feet on centers Turnouts: #7-10

Retarders Photo:

Too Much? Source: railpictures.net

Engineering It All Energy balance equation: KE 1 + Y 1 = KE 2 +Y 2 – (M K X + S W + C R + W R + E R ) KE: Kinetic Energy (v 2 /(2g)) Y: Elevation head, (ft) X: Horizontal distance, (ft) M K : Static rolling resistance, (lb/ton) (typ. 2-18) S W : Losses due to passing through switch, (ft) (typ ) C R : Curve losses, (ft) (typ ft/º of angle) W R : Wind loss (air resistance), (ft) (next slide) E R : Energy extracted by retarders, (ft) (next slide)

Energy Losses Air resistance (Davis equation): KAV 2 * 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 V s = L c + H * V h L c – V s : Velocity at switch – V h : Velocity at hump (release) – L c : 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 KE A = v 2 /(2g) = (7) 2 /(2*32.2) = 0.76 ft Elev. Chg. = -X * (%)/100 = -130 * (-3.7)/100 = 4.81 ft M K loss = X * M K /2000 = 130 * 18/2000 = 1.17 ft Net = EC – S W – C R – M K = 4.81 – 0 – 0 – 1.17 = 3.64 ft KE B = KE A + Net = = 4.40 ft PointLength (ft)Gradient (%)Elev. Chg. (ft)Sw (ft)Cr (ft)Mk (ft)Net (ft) KE v (ft/s) A B C

Other Calculations C R = * º of central angle W R : As discussed previously

Side Note: Targets Source: wikipedia.org

Questions?