Appendix B
Railway Planning Example
This appendix contains an example of railway planning. Use this plan for the operation of any rail system.
SITUATION
B-1. Plan for the operation of a rail system to move supplies in a theater of operations. The target date for the initiation of service is on 1 December. Route all rail tonnages originating in the port to the railhead over the main line of the system as shown in Figure B-1.
Figure B-1. Hypothetical Rail System for Planning
Note 1. All tonnages are expressed and computed in STONs.
Note 2. All computations resulting in a fraction are raised to the next higher whole number.
PLANNING DATA
B-2. Planning depends on the size and type of rails, condition of crossties, rail and ballast, washout and rockslide potential, number of single and double main lines, and the availability of sidings or passing tracks.
TRACK
B-3. If there is a usable double track, trains may operate in both directions without delays in schedules. However, the unit often takes the usable parts of a damaged double track to make one single main line with good passing tracks. Computations in this appendix are based on a single track.
| Number | Single track (unless otherwise stated) | |
| Gauge | Standard (56.5 inches) | |
| Condition | All divisions: Good to fair | |
| Percent of Grade | All divisions: 1.5 percent or less | |
| Ruling Curve | All divisions: 5 degrees | |
Weather |
All divisions:
Summer: +60oF to +95oF |
|
| Passing Tracks | First divisions - 15 Second division - 9 Third division - 11 Fourth division - 14 |
MOTIVE POWER
B-4. Motive power consists of all self-propelling equipment found on a railroad. The most common motive power refers to locomotives.
Road Engines
B-5. US Army 0-6-6-0, 120 tons, diesel-electric locomotive.
Switch Engines
B-6. US Army 0-4-4-0, 60 tons, diesel-electric locomotive.
ROLLING STOCK
B-7. Rolling stock refers to a collection of a large group of railway cars.
| Boxcars | 40-ton rated capacity | |
| Gondolas | 40-ton rated capacity | |
| Flatcars | 50-ton rated capacity |
FIRST COMPUTATION
B-8. Determine the train density for each of the four railway divisions.
| TD = | (NPT +1) 2 |
X | 24 X S LD |
| S = | 10 mph (refer to Table 10-2) | ||
| STEP 1. | |||||||||
| First Division: | TD = | (15 + 1) 2 |
X | 24 X 10 130 |
= | 16 2 |
X | 240 130 |
|
| = | 3,840 260 |
= 14 + of 15 trains | |||||||
| STEP 2. | |||||||||
| Second Division: | TD = | (9 + 1) 2 |
X | 24 X 10 100 |
= | 10 2 |
X | 240 100 |
|
| = | 2,400 200 |
= 12 trains | |||||||
| STEP 3. | |||||||||
| Third Division: | TD = | (11 + 1) 2 |
X | 24 X 10 110 |
= | 12 2 |
X | 240 110 |
|
| = | 2,880 220 |
= 13 + 14 trains | |||||||
| STEP 4. | |||||||||
| Forth Division: | TD = | (14 + 1) 2 |
X | 24 X 10 120 |
= | 15 2 |
X | 240 120 |
|
| = | 3,600 240 |
= 15 trains | |||||||
SECOND COMPUTATION
B-9. Determine the end delivery tonnage of this rail line during winter months using single-engine operation. You must use the following formulas:
- EDT = NDT of most restrictive division
- NDT = NTL X TD
- NTL = GTL X .50
- GTL = DBP X WF
RR + GR + CR - DBP = CTE -- (Total weight of engine in STONs x 20 pound per STON)
- CTE = STE
2 - CTE = Weight on drives (lb)
25% adhesion factor
STEP 1. Compute the starting tractive effort.
| STE = | Weight on drivers (lb) 4 |
|
| = | 240,000 4 |
= 60,000 pounds |
STEP 2. Compute the continuous tractive effort.
| STE = | STE 2 |
|
| = | 60,000 2 |
= 3,000 pounds |
STEP 3. Compute the drawbar pull of the road engine.
| DBP = | CTE-- (Total weight of engine in STONs x 20 pounds per STON) |
| = | 30,000 -- (120 X 20) |
| = | 30,000 -- 2,400 = 27,600 pounds |
STEP 4. Compute the gross trailing load.
| GTL = | DBP X WF RR + GR + CR |
Where:
DBP = 27,600 pounds (preceding calculations)
WF = 80 percent (see Table 10-1)
RR = 6 (see paragraph 10-4)
GR = 1.5 percent X 20 = 30
CR = 5 degrees X 0.8 = 4
| GTL = | 27,600 pounds X .*80 6 + 30 + 4 |
||
| = | 22,080 40 |
= | 552 STONs |
STEP 5. Compute the net trainload.
| NTL = | GTL X .50 |
| = | 552 X .50 = 276 STONs |
STEP 6. Compute the EDT of the system by determining the NDT of the most restrictive division.
NTL X TD = NDT
| First Division | 276 X 15 = 4,140 STONs |
| Second Division | 276 X 12 = 3,312 STONs |
| Third Division | 276 X 14 = 3,864 STONs |
| Fourth Division | 276 X 15 = 4,140 STONs |
| EDT = NDT of second division (most restrictive) | |
| EDT = 3,312 STONs | |
THIRD COMPUTATION
B-10. Determine the rolling stock requirements for this rail system when operating at maximum capacity during winter months using single-engine operation. Each type of freight car will move the following percentages of the end delivery tonnage:
| Boxcars | 50 percent of EDT | |
| Gondolas | 25 percent of EDT | |
| Flatcars | 25 percent of EDT |
STEP 1. Compute the portion of the EDT to be moved in each type of railcar:
| Boxcars: | EDT X 50 percent = 3,312 X .50 = 1,656 STONs | |
| Gondolas: | EDT X 25 percent = 3,312 X .25 = 828 STONs | |
| Flatcars: | EDT X 25 percent = 3,312 X .25 = 828 STONs |
STEP 2. Compute rolling stock requirements for one day's dispatch. You must apply the following formulas:
| Total cars required = | EDT (by type car) X TAT X 1.1 Average payload for type of car |
| 1 DD = | EDT (by type car) Average payload for type car |
| Note: Average payload in tons per type car = | Rated capacity 2 |
Therefore, 1 day's dispatch for all types of cars is computed as follows:
| Boxcars: | 1 DD = | 1,656 20 |
= 82 + or 83 cars |
| Gondolas: | 1 DD = | 828 20 |
= 41 + or 42 cars |
| Flatcars: | 1 DD = | 828 25 |
= 33 + or 34 cars |
| Total cars in 1 DD = 159 cars | |||
Rolling stock requirements are based on a TAT of 11 days (see paragraph 10-47). Therefore, total rolling stock requirements are computed as follows: 1 DD X TAT = cars required X 1.1 (reserve factor) = total cars required (also see Figure B-2).
1 DD X TAT = cars required X 1.1 (reserve factor) = total cars required
| Boxcars: | 83 X 11 = 913 X 1.1 = | 1,004+ or | 1,005 cars |
| Gondolas: | 42 X 11 = 462 X 1.1 = | 508+ or | 509 cars |
| Flatcars: | 34 X 11 = 374 X 1.1 = | 411+ or | 412 cars |
| Total rolling stock requirements: 1,926 cars | |||
Figure B-2. Determination of Turnaround Time in Days
FOURTH COMPUTATION
B-11. Determine the road and switch engine requirements for the operation of the system at maximum capacity during winter months using single engine operation.
STEP 1. Compute for road engines required.
| Number of road engines = TD X | (RT + TT) 24 |
X 2 X 1.2 |
COMPUTE FOR FACTORS
B-12. Compute the running time for each division.
| RT | |||
| TD | (Length of div ÷ avg speed) | ||
| First division: | 15 | 130 ÷ 10 = 13 | |
| Second division: | 12 | 100 ÷ 10 = 10 | |
| Third division: | 14 | 110 ÷ 10 = 11 | |
| Forth division: | 15 | 120 ÷ 10 = 12 |
Note: Average value for terminal time of diesel-electric motive power is 3. Average value for steam is 8.
COMPUTE REQUIREMENTS
B-13. The following computations shows the number of road engines (per division) required for operation over a given railway division.
| First division: | 15 X | (13 + 3) 24 |
X 2 X 1.2 = 36 X | 16 24 |
= | 576 24 |
| Second division: | 12 X | (10 + 3) 24 |
X 2 X 1.2 = 28.8 X | 13 24 |
= | 274.4 24 |
| Third division: | 15 X | (13 + 3) 24 |
X 2 X 1.2 = 36 X | 14 24 |
= | 504 24 |
| Forth division: | 15 X | (12 + 3) 24 |
X 2 X 1.2 = 36 X | 15 24 |
= | 540 24 |
Total road engines required = 24 + 16 + 21 + 23 = 84 road engines
STEP 2. Compute for switch engines.
| Cars dispached and recieved per day |
Cars passing per day |
Computation factor (refer to par 10-4C(2)) |
Switch engines required |
||||
| Port | |||||||
| Terminal | 159 X 2 | ÷ | 67 | = | 4+ or 5 | ||
| 2nd div: | |||||||
| Terminal | 159 X 2 | ÷ | 100 | = | 3+ or 4 | ||
| 3rd div: | |||||||
| Terminal | 159 X 2 | ÷ | 100 | = | 3+ or 4 | ||
| 4th div: | |||||||
| Terminal | 159 X 2 | ÷ | 100 | = | 3+ or 4 | ||
| Railhead | 159 X 2 | ÷ | 67 | = | 4+ or 5 | ||
| Subtotal | = | 22 | |||||
| + 20 percent reserve (4+ or 5) | = | 5 | |||||
| Total | = | 27 | |||||
FIFTH COMPUTATION
B-14. Determine the number of switch and road crews required to support this rail system.
STEP 1. Compute for road crews required.
| Road crews = TD X 2 X | (RT + 3) 12 |
X 1.25 |
COMPUTE FOR FACTORS
B-15. Compute the running time for each division.
| RT | |||
| TD | (Length of div ÷ avg speed) | ||
| First division: | 15 | 130 ÷ 10 = 13 | |
| Second division: | 12 | 100 ÷ 10 = 10 | |
| Third division: | 14 | 110 ÷ 10 = 11 | |
| Forth division: | 15 | 120 ÷ 10 = 12 |
COMPUTE FOR ROAD CREW REQUIREMENTS
B-16. The following computations shows the number of road crews (per division) required for operation over a given railway division.
| First division: Road crews = | 15 X 2 X | (13 + 3) 12 |
X 1.25 | = 37.5 X | 16 12 |
= | 600 12 |
= 50 crews |
| Second division: Road crews = | 12 X 2 X | (10 + 3) 12 |
X 1.25 | = 30 X | 13 12 |
= | 390 12 |
= 32+ of 33 crews |
| Third division: Road crews = | 14 X 2 X | (13 + 3) 12 |
X 1.25 | = 35 X | 14 12 |
= | 490 12 |
= 40+ or 41 crews |
| Forth division: Road crews = | 15 X 2 X | (12 + 3) 12 |
X 1.25 | = 37.5 X | 15 12 |
= | 562.5 12 |
= 46+ or 47 crews |
Total road crews required = 50 + 33 + 41 + 47 = 171 road crews
STEP 2. Compute for switch engine crews required (do not include reserve switch engines).
| Switch crews = SE X 2 X 1.25 | ||
| Port area: | Switch crews = 5 X 2 X 1.25 = | 12+ or 13 |
| Second division terminal: | Switch crews = 4 X 2 X 1.25 = | 10 |
| Third division terminal: | Switch crews = 4 X 2 X 1.25 = | 10 |
| Fourth division terminal: | Switch crews = 4 X 2 X 1.25 = | 10 |
| Railhead: | Switch crews = 5 X 2 X 1.25 = | 12+ or 13 |
| Total switch crews required = | 56 | |
STEP 3. Determine total number of switch and road crews required.
| Road crews = | 171 |
| Switch crews = | 56 |
| Total switch and road crews = | 227 |
SIXTH COMPUTATION
B-17. Determine the monthly engine fuel, lubricants, and repair parts requirements for the operation of this system.
STEP 1. Compute fuel requirements for road engines.
| TD | X | Two-way travel |
X | LD | = | Train miles per day |
|
| First division: | 15 | X | 2 | X | 130 | = | 3,900 |
| Second division: | 12 | X | 2 | X | 100 | = | 2,400 |
| Third division | 14 | X | 2 | X | 110 | = | 3,080 |
| Fourth division | 15 | X | 2 | X | 120 | = | 3,600 |
| Total train miles per day = | 12,980 | ||||||
12,980 train miles per day X 2.5 gal per train mile = 32,450 per day
32,450 gallons per day X 30 days = 973,500 gallons per month
| 973,500 + 48,675 1,022,175 |
gallons per month 5-percent reserve total gallons per month |
STEP 2. Compute fuel requirements for switch engines (do not include reserve switch engines).
| Number of switch engines |
X | Hours per day operation |
X | Rate of fuel consumption (gallons per hour per locomotive) |
= | Daily require- ment (gallons) |
| 22 | X | 20 | X | 8 | = | 3,520 |
3,520 gallons per day X 30 days = 105,600 gallons per month
| 105,600 + 5,280 110,880 |
gallons per month 5-percent reserve total gallons per month |
STEP 3. Compute total fuel requirements.
| 1,022,175 + 110,880 1,133,055 |
road engine requirements per month in gallons switch engine requirements per month in gallons total requirements per month in gallons |
STEP 4. Compute monthly lubricant requirements in STONs.
| TD | X | Two-way travel |
X | Lubricants (STONs per month per train per day |
= | Lubricants (STONs per month) |
|
| First division: | 15 | X | 2 | X | .5 | = | 15 |
| Second division: | 12 | X | 2 | X | .5 | = | 12 |
| Third division | 14 | X | 2 | X | .5 | = | 14 |
| Fourth division | 15 | X | 2 | X | .5 | = | 15 |
| Total lubricants per month = | 56 STONs | ||||||
STEP 5. Compute monthly repair parts requirements in STONs.
| TD | X | Two-way travel |
X | Repair parts (STONs per month per train per day |
= | Repair parts (STONs per month) |
|
| First division: | 15 | X | 2 | X | 1.5 | = | 45 |
| Second division: | 12 | X | 2 | X | 1.5 | = | 36 |
| Third division | 14 | X | 2 | X | 1.5 | = | 42 |
| Fourth division | 15 | X | 2 | X | 1.5 | = | 45 |
| Total spare parts per month = | 168 STONs | ||||||
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