Here is the math and theory.
The fan speed will drop roughly in proportion to voltage (I think). So will the current draw of the fan. A lot of fans are marked with the current and/or watts on the label. If it's watts, divide by the voltage, 12, to get current. Power = volts X amps. Or amps = Power / volts.
Resistance = volts / amps. That is, 12/ amps gives you the resistance of the fan motor.
If you want 1/4 speed, you need to drop 3/4 of the voltage (12) with the resistor and leave 1/4 (df 12)for the fan. That means the series resistor will need to be 3 times whatever the resistance of the fan motor calculates to.
Let F be the fractional speed you need. For instance, if you want 2000 RPM from a fan that spins at 3000 RPM at 12V, then F is 2000/3000 =0.66. The voltage across the fan will need to be F X 12. The voltage accoss the resistor will be whats left, (1 - F) X 12. That means the ratio of the dropping resistor to the fans resistance will need to be (1-F)/ F
Let R be the resiistance of the fan motor.
Then ( (1-F)/ F ) X R is the resistor you need.
Let G = (1-F) . (1-F ) X 12 = G X 12 is the voltage accoss the resistor.
When the fan is large and draws a lot of current, you may need to watch out that the resistor can handle it without burning up.
The wattage the resistor will disappate is
power = voltage squared / resistance =
(G X 12 X G X 12 )/ ( G/F X R) =
(G X F X 144)/R = ( (1-F) X F X 144 )/ R
The maximum power dissapated occurs when F = 1/2.
A rule of thumb is that you will need a resistor rated at twice the calculated wattage to avoid it getting so hot as to burn up anything it might touch. All the wattage rating for a resistor means is that the resistor will not go up in flames in a reasonably cool environment. IAC, you need a resistor that can handle 1/2 the wattage of the fan, worst case. If the resistor is larger than, or smalller than, the fans resistance, the resistor will disappate less power than at 1/2 voltage
I thought this was going to be more straightforward! I hope I didn't screw up at 3 AM.