Confused on DC motor rating formula
That formula makes a lot of assumptions, and I suspect the 100,000 constant value is just a 'fudge factor' to get a somewhat accurate answer. It is based on the fact that mechanical power = torque * rpm, and most DC motors have a performance curve that looks like this:-



Maximum power output occurs at 50% of both rpm and torque (which may explain the redundant /2 terms in your formula). This applies to any normal permanent magnet DC motor, large or small - good or bad. 

At maximum theoretical output the motor is slightly less than 50% efficient. A high quality motor will generally have high peak efficiency, so it may not be designed to be run at such low efficiency (and high dissipation) because it doesn't need to. Without knowing the design operating point you can't accurately calculate the motor's input power rating.     

However small 'toy' motors are usually made from poorer quality materials, so their maximum efficiency is usually not much better than 50%. Therefore they usually are designed to run at close to maximum theoretical output, and dissipate ~50% of the input power continuously (or at least for the short time the motor is expected to be used for). 

Since rpm * torque tells you the power output, and at 50% efficiency the motor will be drawing twice as much power as it produces, all you need is the right 'fudge factor' to get the input power rating. However whether the particular constant used in your formula is appropriate is another matter. I suggest only using it as a rough guide, and do your own tests to see how much power your motor can handle in your application.


  These motors normally draw only 0.15 Amps. I cannot see them pulling 9
  watts.


9W is 0.75A at 12V. A motor that 'normally' only draws 0.15 Amps may draw 5 times that at maximum power. For example the Mabuchi FK180SH-14180 draws 0.07A no-load and 1.7A stalled, so it should draw ~0.85A at maximum output.