A quick ohms law lesson:
Where E=volts dc (Vdc), I=current in Amps (A), and R=resistance in ohms
E=I*R
I=E/R
R=E/I
Since power is a function of work we can say that Power (W) in watts is: E*I since in simplified terms both E and I represent work and R is the resistance to work.
Therefore:
In a rheostat calculation:
Fan current draw= 0.1A (substitute your fan current but 0.1 is common)
Voltage is 12Vdc
From this we can assume that a 0.1A fan rates about 120 ohms in actual resistance: 12/0.1=120
Since fan motor resistance is relatively constant we can assume the following:
12Vdc/R1(fan resistance)+R2 (rheostat resistance)=I
Thence: I*E=P
So: A fan rated at 0.1A with a rheostat set to 0 ohms with a 12Vdc supply will run as stated. If the rheostat is set to 25 ohms then we have: 25(R1)+120(R2)=(145t) 12(Vdc)/145t)=.08.
.08*12=.96W. By ohms law we see a voltage drop of 2.0Vdc across the rheostat and the remaining 10.0 across the fan.
If the rheostat is increased to 50 ohms:
50+120=170ohms 12/170=.07A .07*12=.84W Voltage drop across the rheostat: 3.5 across the fan: the remaining 8.5Vdc
If the rheostat is increased to 100 ohms:
100+120=220ohms 12/220=.05A .05*12=.6W Voltage drop across the rheostat = 5Vdc with the remaining 7Vdc dropping across the fan motor.
Of course your fan motor may draw more or less current, but it is simple to substitute you actual values into the above equations.
The point is, your fan should run fine with the Jameco 5watt 100 ohm rheostat mentioned above by Empireking01.
<EDIT> I didn't check the math, I was doing it on the fly so if I missed something, just hollar.