Here we are concerned with the motion of a vehicle, i.e., its acceleration, deceleration, and the amount of weight it can carry. Also we are concerned with the selection and sizing of propulsion systems for a vehicle.
To look at propulsion and resistance, use Newton's laws of motion and
Concerned with net force on a vehicle or the difference between what
is available from the propulsion system and the resistance encountered.
There is a circular problem. Since force and resistance are functions of velocity, velocity is a function of acceleration, acceleration is a function of force. So you need to know V to compute F, F to compute a, a to compute V.
Resistance to Motion
- Elements of resistance to motion
1) Friction, not a function of velocity--fixed in quantity, between wheel and surface, bearings, etc., friction depends upon (weight, form, and type of surface)
2) Losses varying with speed, i.e., sway, flange friction, bumps, etc.
3) Air resistance depends upon cross section area, shape, length, etc. of vehicle,varies with the square of speed.
Note curve can drop when vehicle reaches a more drag-free regime, i.e., break sound barrier, hydrofoil.
4) Grade resistance, from going up or down grade can be positive or
negative; doesn't vary with speed.
Grade resistance is equal to the component of vehicle weight which is
parallel to the grade line.
Grade resistance =
W = weight of vehicle
G = percent grade
5) Curve Resistance: From friction of rail flanges against rails as a train goes around a cure, etc.. Railroad cure resistance is 0.8 lb. per ton of car per degree of curvature. I.e., a 100 ton car on a 2 degree curve has a resistance of 160 lb..
or R = K1 + K2 * V + K3*V2 + W * G/100
Resistance equations for various types of vehicles.
Rflat ground = (1.3wn+29n) + bwnV + CAV2
Rt = (1.3wn+29n) + bwnV + CAV2 + 20wn*G
Rt = total resistance in lb. including grade resistance
V = speed - mph.
w = weight per axle (tons)
n = number of axles;
Note, wv = w*n = gross vehicle weight
b = coefficient of moving friction: .03 locomotives; .045 freight cars
C = drag coefficient of air .0017 streamlined locomotives, .0025 other locomotives, .0005 for trailing freight cars and .00034 for trailing passenger cars.
A = cross sectional area of vehicle, 120 sq. ft. for locomotives, 90 sq. ft. for freight cars and 120 sq. ft. for passenger cars.
G = % grade (upgrade +, downgrade -)
. . for Davis formula
K1 = 1.3wn + 29n
K2 = bwn
K3 = CA
Modified Davis Equation (FE booklet)
Road vehicle resistance
for Trucks on a gradient
R = 7.6T + .09TV + .002AV2 + 10*T*G
T = weight of vehicle in 1000 lb.
V = velocity = mph.
A = cross sectional area - ft.2
R = resistance in lb.
G = percent grade
for Autos on a gradient
R = 10T + .1TV + .0026CAV2 + 10*T*G
C = air resistance parameter
Auto C = .4 to .5, new autos C = .35
Convertible C = .6 to .65
Bus C = .6 to .7
Comparison of equations
These equations are all similar in form, but show different effects
from the various components of motion.
|Truck||Auto||Bus||Four Axle Locomotive|
(T is weight in kips in all of the above.)
Need propulsive force to overcome resistance, accelerate vehicle. We are interested in the force needed to accelerate a given mass of vehicle at a given rate. Again, simple physics used:
F = ma
The rate at which work is done is power.
P = dw from work equation dw = F(x)dx/dt
P = F(x)dx/dt; units in the English system are ft. - lb/sec.
when F is less than R, the vehicle decelerates.
Force applied over a distance, x, is work:
Fuel consumption is directly related to the amount of work done.
Fuel = consumption rate * work done
The fuel consumption rate could be given in gallons of fuel per ft.-lb. For example, 1 gallon of gasoline contains 120,000 BTU and 1 BTU is equivalent to 778 ft.-lb. of work. Therefore 1 gallon of gasoline has the energy equivalent of 9.33 * 107 ft.-lb. of work.
1) A 200 ton, 4 axle locomotive has the following characteristics:
b = .03, C = .0025, A = 120 ft.2
What is the resistance at 12 mph? 50 mph? 70 mph? 100 mph?
Weight/axle = 50 tons
at 12 mph:
R = 1.3 * 200 + 29 * 4 + .03 * V * 200 + .0025 * 120 * V2
R = 376 + 6V + .3V2
R = 376 + 72 + 43.2 = 491.2 lb.
at 50 mph:
R = 260 + 116 + .03 * 200 * 50 + .0025 * 120 * 50 * 50
R = 376 + 300 + 750 = 1426 lb.
at 70 mph:
R = 376 + 6 * 70 + .3 * 702
R = 376 + 420 + 1470 = 2266 lb.
at 100 mph:
R = 376 + 600 + 3000 = 3976 lb.
2) What is the maximum traction force of a 2500-HP locomotive with an efficiency of .83 at 12 mph? 50 mph? 70 mph? 100 mph?
at 12 mph:
1 HP = 550 ft.-lb./sec = 550 * 3600/5280 = 375 lb.-mi./hr.
F = 375 * P(HP) * E/V(mph) = 375 * 2,500 * .83/12
F = 64,800 lb.
at 50 mph: F = 15,600 lb.
at 70 mph: F = 11,100 lb.
at 100 mph: F = 7,800 lb.
3) At what speed does the resistance of the locomotive equal its maximum propulsive force? (i.e., what is the maximum speed assuming proper gearing and suitable track?)
F = R
778,100/V = 376 + 6V + .3V2
Solving the above (by trial and error),
Vmax = 128 mph.
4) At what gradient will the locomotive coast downhill at a constant 50 mph?
R50 mph = Rgrade
1426 = 20 * wn * g
g = 1426/(20 * 200) = 0.356
g = -0.36%
5) The locomotive moving alone travels 10 miles at a constant 50 mph, on level track. If the energy conversion rate is 30%, what is the fuel consumption rate?
At a constant speed a force equal to the resistance must be applied: (zero acceleration)
F = R = 1426 lb. (from Davis equation)
Work = 1426 * 5280 * 10 ft. lb.
Fuel used = 1426 * 5280 * 10/(.3 * 9.33 * 107) = 2.69 gallons
Rate = .269 gal./mile = 3.72 mpg
6) Very small automobile
Engine power = 12 hp.
Weight = 1200 lb.
Cross sectional area = 20 ft.2
Standard suspension, aerodynamic characteristics
Wind resistance parameter: C = 0.4
Engine efficiency = 0.3
Flat road, no curves
Question: How fast is the maximum steady state speed and what is the fuel economy at that speed?
R = Ra + Rr
R = 0.01T + 0.0001TV + 0.0026CAV2
= 0.01(1200) + 0.0001(1200)V + 0.0026 * 20 * .4 * V2
R = 12.0 + 0.12V + 0.0208V2
At steady state speed
F = R Propulsive force = R
so substituting F for M in the horsepower relationship
R*V = 375HP = 4500
Substituting for R
12V + 0.12V2 + 0.0208V3 = 4500
Which can be solved for V (trial and error)
V ~ 55.1 mph
R = 81.7 lb.
Is this car fuel efficient?
Let's use these relationships.
1 btu = 778 ft.-lb.
1 gallon of gas = 120,000 btu
Fuel conversion efficiency = 0.3
Drive the car for one mile:
7) What is the maximum acceleration from a speed of 30 mph of a 250 hp truck operating up a 5% grade? The truck weighs 20,000 lbs., is 85% efficient and has a cross-sectional area of 100 sq. ft.. (15)
Torque: twisting moment applied to a wheel
Torque = F * b, units are ft.-lb.
Horsepower = 0.00019*t*N
t = torque of engine
N = rpm of engine
For a fixed gearing, torque is directly proportional to tractive effort.
Tractive effort = tGE/r
t = torque at output shaft of engine
G = total gear ratio between output shaft and axle
E = drive line efficiency
r = radius of wheel under loaded conditions
concern with propulsion system which has two elements: prime mover and transmission.
For prime mover we can look at horsepower and torque curves
The above curve is typical of a diesel engine; for a gasoline engine, maximum torque is reached at a higher RPM.
This difference is why diesel engines are good for weight hauling, since they can apply high torque at low speeds, as used in locomotives. Gasoline engines are good for rapid acceleration, as used in automobiles.
Electric motors, Horsepower v. Engine Speed
Curves can have a number of different shapes depending upon requirements; i.e., need high starting torque--you can get practically any shape you want depending on the design of the engine and transmission.
Advantages of straight electric locomotive: you can temporarily overload the locomotive when starting out and start up a heavy train.
Same idea used in diesel-electric locomotive (and also heavy off-road trucks).
Steam locomotive did not have nearly the same starting torque as a diesel. Diesel can pull more than a steam locomotive with the same horsepower. This and lower maintenance and operating costs led to the replacement of steam locomotives by diesel-electrics.
Transmissions. Want to get torque and power of engine transmitted to the wheels.
Purpose of transmissions
transfer power from engine to drive wheels while producing a reduction
in engine speed and providing flexible control.
Vehicle movement is concerned is the variation of net propulsive force with speed (unfortunately, this is not a constant).
Also, net force = propulsive force - resistance.
Maximum propulsive force limited by: characteristics of the propulsion system, i.e., for above at low speeds motors and generators heat up. To maintain high torque and prevent stalling.
Also adhesion of wheels to ground limits maximum pull.
F = µ N
µ = coefficient of friction between wheel and rail, roughly .30
N = weight on one wheel
Minimum tractive effort is determined by maximum safe operating speed of motor, i.e., red line, above that engine tears apart from centrifugal force.
Look at propulsive force of vehicle as it varies with speed and also resistance.
Note when F = R the vehicle is at maximum speed (i.e. net force = 0)
and there is no force available to accelerate the vehicle.
Use curves such as this to describe motion of vehicle.
Note: It is analytically impossible to predict speed, distance, acceleration of most vehicles because:
You can, however, solve for maximum speed, maximum grade, maximum weight, etc. This has to be analyzed by simulation.
Use small increments of time, Dt. Using t, beginning--find v, then F, then solve for new a, find v, . . ., etc.
Go to speed limit.
Two types of limits on speed.
- Artificial (i.e. speed limit)
- Curvature (i.e. maximum safe speed around a curve)
V curve = 429.53 * (F+e)/degree curvature
Deceleration of vehicles usually occurs at constant rate.
Vehicle motion then simulated for a short period of time t.
K1, K2, K3
Side friction factor