Voltage, Current, Power, Acceleration Relationship by: Steve
I have to argue with some of what was said. Say you have 10 12-volt bateries arranged in series giving you a 120-volt system and your motor is rated for an input range of 100-150 volts, then adding two more batteries in series would allow you to now have more charge on hand, essentially giving you more power and range, but may actually reduce the current delivered to your motor for an identical speed (not necessarily a bad thing).
Lets remember the relationship of voltage, current, and power. Power is really what you want in order to go fast (accelerate). We measure most household appliances in Watts (1 Joule/second), while we measure cars in horsepower (746 Watts). Here's the relationship: Power=Current*Potential (1Watt=1Amp*1Volt). Raising the voltage with the same battery array just lowers current and keeps the power the same. It comes down to finding the best voltage for the motor your using to draw current most effectively.
Here's a scenario. You have two battery arrays, which can both hold the same amount of charge and run a 70HP (52220W) motor.
Amps=Watts/Volts
120 Volts=435 Amps 144 Volts=363 Amps
Also, amps are not drawn in a time period. They represent charge per unit time (Coulumb/sec). The following is the relationship between power and acceleration:
F=ma(of car as whole), E=Fd, P=F*d/t, so:
POWER=mad/t, and since d=.5at^2 (basic kinematics)
P=.5m(a^2)t
As t apporaches 1/infinity and mass is held constant, power is proportional to half of the square of acceleration. More power equals good.
