Because of all the variables, load, rpms, the best way to get a good answer is by measuring the current and voltage applied to a 6v starter. Then measure current and voltage when starter is connected to 12v. However that wouldn't be 100% accurate either, because we couldn't account for the condition, CCA, of two different batteries.
I will still go out on a limb and say for the most part, a starter motor's max current is when you first hit it with voltage and it's just starts to turn. That max current is determined by all the resistance in the circuit, which mostly the internal resistance of motor.
Then if you double the voltage, the max starting current will double as well. Simplify ohm's law: and what some call watt's law, which watt didn't write.
E = I x R : P = I x E
1 v = 1 ohm x 1 amp: 1 watt = 1 a x 1 v
2 v = 1 ohm x 2 amps: 4 watts = 2a x 2v
In theory power is 4x. I'm sure if we could conduct an experiment, the power may be less than 4x because of other variables, load, CCA, RPM's.
On the other hand, applying 12v to a 6v starter, the starter spins faster. The faster you crank an engine requires more power from starter.
If someone has an ammeter, volt meter, a good 6v and 12v battery, post your results.
I'm sticking with power staying around 4x, give or take a little.
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Today's Featured Article - Talk of the Town: How to Remove a Broken Bolt - by Staff. Another neat discussion from the Tractor Talk Discussion Forum. The discussion started out with the following post: "I have an aluminum steering gear housing with a bolt broken off in it. The bolt is about a 3/8" x 1 1/2" bolt. I've already drilled the center of the bolt out with about 7/64" drill bit the entire length of the bolt. Only one end of the bolt is visible. I tried to use an easy out but it wasn't budging and I didn't want t
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