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The most frequent uses of a personal computer have been for word processing and accounting. Another use for computer's is simulation in which a system under study is modeled as a series of equations. An electric vehicle can be such a system.
For the purpose of this simulation, we will simulate an elecric vehicle as a vehicle with four wheels, and electric motor, controller and batteries. Each of these components will be modeled as a series of equations that best describe the characteristics of each of the parts. In addition, english units for measurement will be used. i.e. distance is in miles, speed is in miles per hour, acceleration is in miles per hour per second, weight is in pounds, area is in square feet, torque is in pound-feet, power is in horsepower or watts, and time is in seconds or hours.
The vehicle itself has two main characteristics. It has mass and it occupies space. Becuse of its mass it has weight. Effort is needed to accelerate it, effort is needed to overcome frictional losses due to its weight, and effort is needed to raise this weight (as when going uphill). Because it occupies space, effort is needed to push air out of the way when it moves.
The force needed to push air out of the way and then move it back again is proportional to the density of the air, the frontal area of the vehicle (perpendicular to the direction of travel), the vehicl's drag coefficient, and the velocity of air past the vehicle squared. The velocity of air past the vehicle is the sum of the vehicle's speed and the speed of any head wind. (A tail wind is subtracted.) The area is reduced by reducing a vehicles height and/or width. Its drag coefficient is reduced by streamlining. (Removing things that stick out and create air turbulence.) At normal temperature and pressure, the density of air is about .00327 slugs per cubic foot. Therefore:
Force of Air = .00327 • area • dragcoef • speed • speed.
[Force of Air in pounds at sea level, area in square feet, and speed in mph]
Ballpark range of dragcoef:
0.11 = Aptera
0.19 = EV1
0.2 = very streamlined vehicle
0.3 = streamelined vehicle
0.5 = typical vehicle
0.7 = pickup, SUV or van
The force needed to overcome tire/wheel bearing resistance is dependent on vehicle weight and speed. It consists of two parts: 1. a part that is independent of speed and, 2. and a part that is proportional to speed. Both parts are proportional to vehicle weight. Most of the loss is due to changing the shape of the tire as it rotates against the ground. Additional vehicle weight causes it to change more. Additional tire pressure causes it to chnge less. The type of rubber used to make the tire also affects its rolling resistance. Bearing losses are much less then the tire losses. They depend on the size of the wheel bearing, the type of grease used, and the vehicle weight and speed. It is assumed that the breaks are in good shape and not rubbing on the disk or drum.
Force rolling = weight • (X + Y • speed)
For older tires, (X + Y • 50) was about 0.02 [at 50 mph]. Today, the new radials are about half that [0.01], and special tires have been made that have half the rolling resistance of regular radials [0.005]. I do not have the numbers at slow speeds so I will assume rolling resistance is constant with speed.
If the vehicle goes uphill, force is needed to raise the vehicle. If the hill has a grade, then:
Force hill = weight • grade
Freeways generally do not exceed a 6% grade. City streets seldom exceed 12%, but there are exceptions!
The difference between the motor force and the above losses is the force availiable to accelerate the vehicle. This force is proportional to the vehicle's mass and to its acceleration. Note that near the surface of the Earth, a dropped object in free fall will gain speed at the rate of 22 mph/sec.
Force acceleration = weight • acceleration / 22
The vehicle's speed is increased by acceleration. i.e. if the acceleration is 4 mph/sec, then each second the speed would be increased by 4 mph.
Speed = Speed + acceleration • time
The distance the vehicle travels is increased by speed. i.e. if the speed is 36 mph, then each second the distance would be increased by 1/100 of a mile. (3600 seconds = 1 hour)
Distance = Distance + speed • time
An electric motor is used to overcome the losses caused by force of air (wind resistance), force of tires and bearings (rolling resistance) and hill force. Any force left over accelerates the vehicle. If the electric motor force is insufficient, the behicle decelerates. A useful way to measure vehicle losses is to set the motor force to zero (disengage the motor) and let the vehicle coast. Total losses equals the weight of the vehicle times the rate of deceleration.
Deceleration (MPH/sec) = (MPH1 - MPH2)/(Time2 - Time1)
Total loss = weight • Deceleration / 22
The RPM of the electric motor is proportional to vehicle speed, the gear ratios of transmission and differential, and inversely proportional to tire size. Revolutions per mile can be obtained for most tires. Revolutions per mile equls tire RPM at 60 miles per hour. Total gear ratio is the product of differential and transmission gear ratios.
Tire RPM = (tire revolutions/mile) • Speed / 60
Motor RPM = tire RPM • (total gear ratio)
The Motor Force on the vehicle is proportional to motor torque, total gear ratio, and inversely proportional to tire size. [A tire with an effective radius of 1 foot makes about 840 revolutions per mile.] It is motor force that pushes the vehicle.
Motor Force = Motor Torque • (Total Gear Ratio) • (tire revolutions/mile) / 840
For an Advanced DC Motor model number 203-06-4001, the following approximation exists throughout its normal operating range:
Motor Torque = (Amps - 85) / 5
Motor Volts = (RPM / 286) * SQRT(Motor Torque + 10)
(Note: that SQRT means square root.)
The Motor Volts must be less than the available battery voltage. In other words, during acceleration, the vehicle's controller msintains a constant current to the motor until available battery voltage [minus losses] is applied to the motor. Then the motor torque (and current) decrease with increasing RPM:
Motor Torque = SQRT(286 * Motor Volts / RPM) - 10
Battery voltage is dependent on battery charge, temperature, and rate of discharge. If state of battery charge and temperature is neglected by assuming a full charge and optimum temperature, the drop in battery voltage is basically due to internal and external resistance of the battery whch is about 0.05 ohms for a typical 120 volt high discharge battery. For twenty large 6 volt deep cycle batteries in series:
Battery voltage = 126 - .05 • Amps
For twenty 8 volt batteries in series [same weight as 6 volt batteries]:
Battery voltage = 168 - .09 • Amps
The controller consumes several watts of power for its internal processing, and drops about a volt through its switching transistor or free wheeling diode, and drops additional voltage across its internal resistance.
Simulation of an electric vehicle uses all of the above equations. Simulation is easiest if the calculations each second are made for the next second. At any time parameters may be changed. i.e. motor current, total gear ratio, etc. Vehicle parameters such as weight should not be changed during simulation (unless something falls out!).
Useful Numbers: (all close, some exact)
1 kw-hr = 3.6 MJ = 3413 BTU = 1.34 HP-hr
1 HP=550 ft-lb/sec=375 lb-mph=torque • RPM / 5252=746 watts
1 g = 22 mph/sec = 32.16 ft/sec/sec = 9.80 m/sec2