I understand that most car starter motors' power is rated between 0.5kw to 1.5kw. Doesn't that mean that they are supposed to draw 40-120 amps? (500 w / 12 volts etc.. )? Yet when they start, they draw hundreads of amps for the split second they run. Why does that happen? Are the motors "overclocked" during that period?
It takes a LOT of power to get the rotating assembly - crank, pistons (or rotors), etc. - moving. For reference, try turning your engine over with a breaker bar on the crank. It's not super-easy (though some of that is due compression).
All the parts in the rotating assembly - crankshaft, connecting rods, pistons, valves, camshafts, timing chain - add up to a very, very heavy piece of metal that must be moved by a fairly small electric motor (starter) to start the car. Not only that, they have to get moving pretty quickly for the combustion cycle to take over. That takes a lot of power.
You can work backwards from your numbers using Ohm's law (V=I*R) and the definition of power (P=I^2 * R). The significant factor here is resistance, which is huge in this context.
So, the short answer: Metal parts are heavy, and take a lot of energy to move. This is one reason that things like lightweight alloys and composites are so important in high-efficiency designs: by reducing the weight of the moving parts, we reduce the energy required to move them. All of that surplus goes to the output, making your car / bike / jet pack / spaceship faster.
See the comments for better information.
All electric motors consume more current at startup compared to steady state. Check out the label on your fridge for example (or look at this one): the max current on the label is 2-3 times higher than the value you'd obtain from power to voltage ratio.
The reason behind this lies in the properties of electric motors. Approximately, such motors have torque proportional to the current and speed proportional to voltage. When the motor starts, you need much more torque to get it running than you'd need in steady state to keep it running. Hence you need more current.
By the way, a lot of cars have even more powerful starters (e.g. a Landcruiser has a 2.5 kW one). That's over 200 A in steady state. Multiply that by 2 or 3 to get the startup current, and you'll get around 500 A that the battery must be able to provide.
A characteristic of electric motors is that they produce the highest torque when stationary, coupled to this is a very high initial current 400 to 600A for cars and commercial starter motors can exceed 1000A.
Once they start to rotate the current demand goes down - remember that the pinion / flywheel ratio is 10 to 1 or more so when the engine is being turned at 500rpm then the starter is doing 5000...
Consider the following model of an electric DC motor
- Va = 12Volts in a car
- Ra = the ohms resistance of windings, cables,battery etc.
- La = inductance (consider it to be zero in first approximation)
- Ia = current through motor
- Vc = Voltage electromagneticall induced into motor (proportional to rotation speed wa)
The rated power of a motor is conventionally defined as the availble output power (≈Vc*ia) at some combination of speed and torque. Under normal continous operation the input power (=Va*ia) will be a bit higher than the output power.
But startup is not "normal continuous operation".
As a first approximation we can treat the inductance as zero. The current drawn by a DC motor then depends on three things, the supply voltage Va, The resistance of the windings Ra, and the "back EMF" Vc which in turn depends on the motors rotation speed. Power delivered into the back EMF (= Vc * ia) is mostly delivered to the load while power delivered into the winding resistance ( =iaiaRa) is wasted as heat in the windings.
Due to intertia in both the motor and the load the initial rotation speed is zero, hence initially the current in the motor is limited only by the winding resistance, the motor draws far more current than normal and all of the power entering the motor is wasted as heat.
Gradually as the load and motor come up to speed Vc increases, thus V_Ra decreases, thus Ia (= (Va-Vc)/Ra) decreases as well and the motor transitions to normal continuous operation. If the engineers did their job right then the motor should reach a safe operating speed before it overheats.
In the case of a car hopefully the engine then starts and the starter motor is disconnected.
A typical starter motor is an induction motor which can produce high torque on starting. It has a stator coil and a rotor coil. The stator coil is composed of many turns of copper wire that are fixed to the inside of the motor housing. The rotor coil is composed of many turns of copper wire that are fixed to the rotor shaft. When the starter is turned on, the 12 Volt (V) car battery sends current to the starter motor. At this instant the resistance (R) of the motor is just the resistance of the copper wire that makes up the stator and rotor coils and is therefore low (less than 0.05 Ohms). The initial starting current (I) is therefor high (greater than 240 Amps; From Ohms Law I=V/R =12/0.05). this is the peak starting current and will only last a fraction of a second. As the starter motor rotor begins to turn, the electric fields of the stator and rotor coils will interact to produce a "back EMF" which is an internal voltage which opposes the input voltage from the battery. The starter motor's motion will be opposed by the mechanical force required to turn the engine over until it starts. Starter motors are matched to the engines that they are required to turn, so that they would only be required to turn the engine for a few seconds. The current requires by the starter motor during these few seconds will drop to about half of the peak current mentioned above.
During the starting, the starter motor draws so much power that the voltage collapses somewhat (caused by the internal resistance of the battery). By this, the nominal power P = U I of the starter corresponds to a current I that is higher than what you compute with U = 12V (e.g., if the voltage is drained down to 6V, the current is twice as high to have the same power). Also, note that the power corresponding to the loss of voltage and the same current produce heat in the battery...