03 Train Resistance and Tractive Power (Railway Engineering Lectures هندسة السكك الحديدية & Dr. Walied A. Elsaigh)

1. CE 435 RAILWAY ENGINEERING
Part II. Resistance and Tractive Effort
Text
Practical Guide to Railway Engineering, AREMA
Page 52 to 59

2. • What are the two primary aspects of transportation energy
efficiency?
• Resistance
How much work is required to move something.
• Energy efficiency
How efficiently energy is converted into useful work.

3. Comparison of truck vs. rail energy efficiency
What distance can each mode transport a given amount of freight for
a given amount of energy?
i.e. how far can we transport one ton
of freight with one gallon of diesel fuel?
Rail is over 3 times
more efficient than truck

4. • 1 Train = 1 ton of freight carried 435 miles on 1 gallon of diesel fuel
• 1 Train = 280 truckloads
• 1 Train = 3 to 4 times more fuel efficient
than trucks

5. Trains VS. Trucks
Trains require less energy
to move than trucks
because of smaller resistive
forces:
•Lower friction factor (0.18)
–Smaller contact area
–Steel wheels on steel rails
–Rubber tires on pavement
•Less wind resistance per
pound of load
•Size of contact area?

6. Energy Utilization
The two primary aspects of transportation energy
efficiency:
• Resistance – How much work is required to move something
• Energy Efficiency – How well energy is converted into useful work

7. Train Resistance
For our purposes, Resistance is the combination of forces that
work against a train’s movement.
Early measurement of railcar resistance simply involved piling
on weight, w, and determining how much was needed to make
the car move.
If the weight, w, is great enough to overcome the static friction,
the car will start to move.

8. What is resistance and how is it measured?
Resistance is typically measured
in “pounds per ton” (in U.S.)

9. Train Resistance
 Two Elements:
Rolling Resistance (On Level Tangent Track)
 Friction related
 wheel, track quality, track modulus
 Aerodynamics of equipment
Grade & Curve Resistance
 Track Profile & Alignment related
 Change in potential energy head

10. Sources of rail vehicle resistance
A varies with weight ("journal" or "bearing" resistance)
B varies directly with velocity ("flange" resistance)
C varies with the square of velocity (air resistance)
The general expression for train resistance is thus:
R = AW + BV + CV2
where: R equals total resistance
W = weight
V = velocity
A = resistances that vary with axle
load (includes bearing friction,
rolling resistance and track
resistance)
B = resistances that vary directly
with speed (primarily flange friction
and effects of sway and oscillation)
C = resistances
that vary as the
square of speed
(affected by
aerodynamics of
the train)
Cross-section of the
vehicle, streamlining
of the front & rear,
and surface
smoothness all affect
air resistance

11. SPEED (mph)
RESISTANCE(lbs./ton)Speed and Resistance by Transport Mode
Rail uniquely combines
High Speed and Low Resistance
Airplanes
Trucks
Boat

12. Rolling Resistance
•Rolling Resistance
• The resistance for train movement on straight and level track can be
determined by W. J. Davis formula:
R=A+BV+CV²
Where
R= Train rolling resistance
A= Rolling resistance component independent of train speed
B= Train resistance dependant on speed
C= Drag coefficient based on the shape of the front of the train and other
features affecting air turbulance
V= Train speed
• The Train resistance on level tangent track is the sum of locomotive and
train resistances.

13. Speed and resistance for conventional freight train
At low speeds, journal resistance dominates, but as speed increases air resistance
is increasingly the most important term
C
B
A

14. Resistance versus speed for a 10,000 ton train
 Train resistance is calculated by multiplying the resistance per ton at each
speed, by the total tonnage of the train.

15. Measurement of train resistance
 Substantial research early in the 20th century led to the development
of a general formula for train resistance.
 Developed by W.J. Davis, it is still sometimes referred to as the “Davis”
equation.
 Ru = 1.3 + 29/w + bV + CAV2
/wn
where:
Ru = Unit resistance in lbs. per ton
w = weight per axle (Weight, tons/ Number of axle)
n = Number of axles
V= velocity mi/h
b = an experimental friction coefficient for flanges, shock, etc.
A = cross-sectional area of vehicle
C = drag coefficient based on the shape of the front of the train and
other features affecting air turbulence etc.
 The Davis Equation has been substantially updated to reflect modern
developments, but its basic form is the same.

16. “Davis” equation (Total Resistance)
R =1.3 W + 29N + bWV + CAV2
R = Resistance in lbs
W = Weight, tons
N = Number of axles
V= velocity mi/h
b = an experimental friction coefficient.
A = cross-sectional area of vehicle
C = drag coefficient based on the shape of the front of the train
and other features affecting air turbulence etc.
 See AREMA Manual for Railway Engineering Chapter 16 -
2.1.3 for latest version of formula.

17. Recommended values for “Davis” equation
CbA sq.ftEquipment Type
0.0024
0.0024
0.0024
0.03
0.03
0.03
105
110
120
Locomotives -50 tons
70 tons
100 tons and over
0.00050.04585-90Freight cars
0.000340.03120Passenger car

18. Grade Resistance
1% Grade = Approximately 20 lbf / ton

19. Grade Resistance is the force required to overcome
gradient and is equal to 20 lb. per ton per percent grade.
This force is derived from the resolution of force vectors
and is independent of train speed. An up grade produces
a resistive force while a down grade produces an
accelerating (negative resistive) force.
Grade Resistance

20. Grade Resistance
 F = (W X CB) / AB = 20
lbs/ton for each percent of
grade
 where:
 W = weight of car
 CB = distance between C and
B
 AB = distance between A and
B
 example:
 Grade Resistance = 20 * 0.5 =
10 lbs/ton = 10 lbs/ton *
10,000 tons = 100,000 lbs

21. Curve Resistance
10 = 0.8 lbf / ton

22. Curve Resistance
Curve Resistance is an estimate of the added resistance a
locomotive or car must overcome when operating through
a horizontal curve. The exact details of the mechanics
contributing to curve resistance are not easy to define.
It is generally accepted in the railway industry that curve
resistance is approximately the same as a 0.04% up grade
per degree of curvature (which equals 0.8 lb. per ton per
degree of curvature) for standard gauge tracks.

23. Curve Resistance
 Unit curve resistance values determined by test and experiment
 Conservative values suitable for a wide range of conditions: 0.8 - 1.0
lbs/ton/degree of curve
 Equivalent grade resistance - divide curve resistance by unit grade
resistance
 For our example:
 Equivalent grade resistance = 0.8/20 = 0.04 of the resistance offered
by a 1% grade
 Curve resistance = (4 * 0.04) * 20 = 3.2 lbs/ton ; 3.2 lbs/ton * 10,000
tons = 32,000 lbs

24. Tractive Effort or Tractive Force, is the amount of force
at the wheels available for moving a train.
Tractive effort is the amount of force in foot-pounds that
the motive power must produce to move a train.
Tractive Effort
Tractive effort is simply the force available, at the roadway
surface, to perform work.

25. POWER!
Q: How do railroads overcome such large resistances?

26. Physics of power, force and speed
 Work = force x distance W = F x D
 Power is the rate at which work is done
Power = work/time P = W/T = FD/T
P = FV
 The relationship between
Force and Distance/Time
(i.e. speed) is thus inverse
Force
Speed
The higher the speed,
the less force available
Curves like this are used
to describe the performance
capabilities of locomotives

27. P = F V
P= (1/375) F V
P = Power, HP
F = Tractive effort , lb
V= velocity m/h

28. Speed/tractive effort curve for a
modern locomotive
Tractive effort is measured in pounds of force available as a function of speed
At low speed, tractive
effort is limited by
adhesion, not power

29. Train resistance and tractive effort are both measurements of force (typically in pounds
in North America) so we can simply overlay the curves
Tractive effort & train resistance curves
Q: What is the maximum speed
possible for this train with this
locomotive?
A: About 58 mph. This is referred
to as the “balancing speed”.
Tractive force = resistance
35,000 lbs.

30. Force available for acceleration declines with speed until the balancing speed is reached,
where it is zero. Consequently, the rate of acceleration declines with speed.
The difference between tractive effort and resistance
is the force available for acceleration
Q: How much force is available for
acceleration at 15 mph?
A: 135,000 - 15,000 = 120,000 lbs.
Q: How much force is available for
acceleration at 35 mph?
A: 59,000 - 21,000 = 38,000 lbs.

31. Example 1
A train composed of
2-diesl locomotives, each weighing 136 tons, having 4-axles.
20-empty freight cars, each weighing 30 tons, with 4 –axles.
At a speed of 50 mi/h on a level straight route.
calculate train resistance.

32. Example 2
[No. of Locomotives]
•How many 5,000 horsepower locomotives at 85%
efficiency will it take to provide the pull of the 77,440
lb. if the resistance of each engine is 1,200 lb.?
375 X HP x e
V
TE=

33. 375 X HP x e
V
TE=
375 X (5,000 X N) x 0.85
7077,440 + 1,200 X N=
N = 3.59 Take 4 Locomotives

35. Thank You