# How to calculate the rated speed of a DC motor given the no load speed, rated current and rated torque?

I am working on a mobile robotics project and I'm beginning to learn about DC motors. I am looking at a particular geared motor, and the given parameters are the no load speed, the rated torque, rated voltage and rated current. I am confused on how to find the rated speed (the speed the motor will run if given the rated load) using these parameters. The data sheet for the motor is not provided, but the motor itself can be found at this link: https://www.amazon.com/Greartisan-Electric-Reduction-Centric-Diameter/dp/B071GTTSV3?th=1.

I know that torque and speed are inversely proportional at a given voltage, and I have tried to calculate it using that relationship, but I think that my method is flawed. I was looking at another motor's data sheet, and I was not getting the result that was stated on that data sheet.

• It says "200 RPM" in the Color field.... But this, ladies and gentlemen, is why you don't buy components without datasheets from dubious sellers via marketplaces like Amazon and Aliexpress, etc. Unless you're a hobbyist that just wants something cheap and actual specifications are unimportant. Commented Dec 14, 2021 at 18:47
• What to check for when buying an electronic component or module - take note that very few EEs will buy crucial parts or equipment from such sources. The link explains why. Commented Dec 14, 2021 at 19:02
• If serious project, then don't buy garbage. Commented Dec 14, 2021 at 19:04
• And note that at least in the US (and probably most other developed countries) there are electronics surplus houses sell cheap stuff with datasheets. The level of quality is usually better than what Amazon or AliExpress will sell you, but you still have to live with the fact it's made for some application other than yours. Commented Dec 14, 2021 at 20:00
• Problem is the specs don't say whether the rated current current and torque are at the same operating point, or if the 200 rpm is under no load or at rated load (or something else). And can you believe the figures anyway? Best way to resolve this uncertainty is to simply buy the motor and test it. The unit appears to be well made - pity they couldn't provide better specs! Commented Dec 14, 2021 at 20:14

The product description includes:

Specifications:
Rated Voltage: 12V
Speed: 200RPM
Material: Metal
Rated Torque: 2.2Kg.cm
Reduction Ratio: 1:24
Rated Current: 0.5A

The rated input power is 12 V x 0.5 A = 6 watts.

If you assume that 200 RPM is the rated speed, at 200 RPM and 2.2 Kg torque, the output power would be 200 RPM x 0.216 Nm / 9.55 = 4.52 watts. The efficiency would then be 4.52/6 = 75%. That seems high. The 24:1 speed reducer is likely to have quite a bit of friction losses.

If you assume that the efficiency is 50%, you would have 3 watts of mechanical power at rated torque and 3 W x 9.55 / 0.216 Nm = 133 RPM. I would hope for that but be prepared for less.

• Where does the 9.55 come from in the output power calculation? Commented Dec 14, 2021 at 20:13
• 60/(2*Pi). RPM is not an SI measurement, but you try convincing suppliers to use radians/second.
– user16324
Commented Dec 14, 2021 at 20:20
• @user_1818839 to better than 2% - watts = kg.m X RPM :-). It just happens that various mixed unit items almost cancel out correctly. This is an awesomelybl.useful.firmuls to store in your head. Commented Dec 16, 2021 at 2:53