# Confused on DC motor rating formula

I am testing small DC Motors for toy cars. In determining the watt rating, I was informed of a formula used for them. I simply cannot find any reason for the introduction of 100 000 in the equation.

$$P = \dfrac{\dfrac{RPM_{max}}{2} \cdot \dfrac{Torque_{max}}{2}}{100000}~~[W]$$

The motors are simple FC 130/FK 180 types. Designed for 12VDC power.

Here is an example of a test someone has done:

FC130 - rated 21,500 RPM/12V, and 170 gcm torque

$$P = \dfrac{\dfrac{21500~RPM}{2} \cdot \dfrac{170~gcm}{2}}{100000} =\\ = \dfrac{10750~RPM \cdot 85~gcm}{100000} =\\ = \dfrac {913750}{100000} = 9.1~W$$ or: $$P = \dfrac{9.1~W}{746} = 0.012~HP$$

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

Anyone know where the formula comes from and why the added 100000 is used?

• DCR (resistance) of motor determines max current can be as high as 10x rated current for rated HP for small motors. Less efficient motors will be 5~8x depending..... This DCR also determines max start torque. Your calculations ought to prove this. It may be designers choice to raise DCR to reduce brush currents on startup/ If you convert RPM to revs/s, gm to kg for KGS units and other factors it may lead to the approx 100k, but I have not seen this before. Normal current is not used for HP rather rated current at some T('C) rise and rated mechanical load at like 80% of no load RPM – Tony Stewart EE75 Apr 7 '17 at 6:46
• I strongly suspect the 100,000 is a unit conversion factor for whatever torque scale was used wherever you found that formula... – Trevor_G Apr 7 '17 at 12:08

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.

I strongly suspect that 100,000 is a crudely rounded and combined constant plus unit conversion factor to convert the torque into Nm.

A rounded combination of

• 100 for cm to m,
• 101.971621298 for gram-force to N
• 9.5488 from equation P = Torque (N.m) x Speed (RPM) / 9.5488

Rounded to 100 * 100 * 10 to get you to 100,000.

Actual divisor should be ~ 97370.66175, but 100,000 is probably close enough in the scheme of things since the formula is a wing and a prayer anyway.