Good science project does not stop with building a motor. It is very important to measure different electrical and mechanical parameters of your motor and calculate unknown values using the following helpful formulas.

We will use the International System of Units (SI). This is modern metric system that is officially accepted in electrical engineering in the USA.

One of the most important laws of physics is the fundamental Ohm’s Law. It states that current through the conductor is directly proportional to applied voltage and is expressed as:

I = V / R

where I – current, measured in amperes (A);
V – applied voltage, measured in volts (V);
R – resistance, measured in ohms (Ω).

This formula could be used in many cases. You may calculate the resistance of your motor by measuring the consumed current and applied voltage. For any given resistance (in the motors it is basically the resistance of the coil) this formula explains that the current can be controlled by applied voltage.

The consumed electrical power of the motor is defined by the following formula:

Pin = I * V

where Pin – input power, measured in watts (W);
I – current, measured in amperes (A);
V – applied voltage, measured in volts (V).

Motors supposed to do some work and two important values define how powerful the motor is. It is motor speed and torque – the turning force of the motor. Output mechanical power of the motor could be calculated by using the following formula:

Pout = τ * ω

where Pout – output power, measured in watts (W);
τ – torque, measured in Newton meters (N•m);
ω – angular speed, measured in radians per second (rad/s).

It is easy to calculate angular speed if you know rotational speed of the motor in rpm:

ω = rpm * 2π / 60

where ω – angular speed, measured in radians per second (rad/s);
rpm – rotational speed in revolutions per minute;
π – mathematical constant pi (3.14).
60 – number of seconds in a minute.

If the motor has 100% efficiency all electrical power is converted to mechanical energy. However such motors do not exist. Even precision made small industrial motors such as one we use as a generator in generator kit have maximum efficiency of 50-60%. Motors built from our kits usually have maximum efficiency of about 15% (see Experiments section on how we estimated this).

Don’t be disappointed with 15% maximum efficiency. All our kits are intended for education and not designed for real applications. This efficiency is not bad at all – it is actually much better than most of other self made designs on Internet can provide. The motors have enough torque and speed to do all kinds of experiments and calculations.

Measuring the torque of the motor is a challenging task. It requires special expensive equipment. Therefore we suggest calculating it.

Efficiency of the motor is calculated as mechanical output power divided by electrical input power:

E = Pout / Pin

therefore

Pout = Pin * E

after substitution we get

τ * ω = I * V * E

τ * rpm * 2π / 60 = I * V * E

and the formula for calculating torque will be

τ = (I * V * E *60) / (rpm * 2π)

Connect the motor to the load. Using the motor from generator kit is the best way to do it. Why do you need to connect the motor to the load? Well, if there is no load – there is no torque.

Measure current, voltage and rpm. Now you can calculate the torque for this load at this speed assuming that you know efficiency of the motor.

Our estimated 15% efficiency represents maximum efficiency of the motor which occurs only at a certain speed. Efficiency may be anywhere between zero and the maximum; in our example below 1000 rpm may not be the optimal speed so the for the sake of calculations you may use 10% efficiency (E = 0.1).

Example: speed is 1000 rpm, voltage is 6 Volts, and current is 220 mA (0.22 A):

τ = (0.22 * 6 * 0.1 * 60) / (1000 * 2 * 3.14) = 0.00126 N•m

As the result is small usually it is expressed in milliNewton meters (mN•m). There is 1000 mN•m in 1 N•m, so the calculated torque is 1.26 mN•m. It could be also converted further to still common gram force centimeters (g-cm) by multiplying the result by 10.2, i.e. the torque is 12.86 g-cm.

In our example input electrical power of the motor is 0.22 A x 6 V = 1.32 W, output mechanical power is 1000 rpm x 2 x 3.14 x 0.00126 N•m /60 = 0.132 W.

Motor torque changes with the speed. At no load you have maximum speed and zero torque. Load adds mechanical resistance. The motor starts to consume more current to overcome this resistance and the speed decreases. If you increase the load at some point motor stops (this is called stall). When it occurs the torque is at maximum and it is called stall torque. While it is hard to measure stall torque without special tools you can find this value by plotting speed-torque graph. You need to take at least two measurements with different loads to find the stall torque.

How accurate is the torque calculation? While voltage, current and speed could be accurately measured, efficiency of the motor may not be correct. It depends on the accuracy of your assembly, sensor position, friction, alignment of the motor and generator axles etc. If you want to get meaningful numbers you might use a second generator kit as explained in Torque and Efficiency Calculation section.

Speed, torque, power and efficiency of the motors are not constant values. Usually the manufacturer provides the following data in a table like this one (sample data from one of the motors used in generator kit):

RF-500TB-12560 data

Also the manufacturers usually provide power curves for the motor at nominal voltage:

RF-500TB-12560 power curves

These curves are generated by plotting motor speed, consumed current, and efficiency as functions of the motor torque. Sometimes there might be also a curve representing mechanical output power.

As you can see from the graph speed and current are linear functions of torque so you might need only two measurements to draw these graphs. Efficiency and power will need more data. Usually for small motors maximum power is at 50% of stall torque (approximately 50% of no load speed). Maximum efficiency may be 10-30% of motor stall torque (70-90% of no load speed).

While it is technically better to follow the same format and create similar curves for your motor it is not absolutely necessary for a good science project. You may take all measurements, calculate unknown values and plot the graphs where for example speed and torque are represented as functions of applied voltage or current etc.

Simple formulas and calculations described here are essential for calculating most common motor parameters. However this is a simplified approach that does not take into consideration many factors. If you want to extend your research further – see Links section and search the Internet. There is tons of information with more complex calculations.

See other pages at Experiments section on how to measure motor parameters.

114 Comments

  1. thank you for this calculation

  2. The best explanation i’ve got so far in around 10 webpages

  3. exccelent

  4. thank you for this calculation

  5. Very well explained. Thanks

  6. Well done

  7. Good.,

  8. Very useful for me

  9. Wow excellent

  10. Hi..
    The information was useful. Can you please send me calculation for what motor to be best suited to lift or move a load of 150 kgs.

    • Unless the questions refer to the motors built from our kits we cannot answer them or provide any calculations. Sorry guys – please do your own research!

      Thanks for understanding and best of luck!

  11. I love this thing, thank you very much!

  12. Thank you very much!

  13. Very useful! …. I searched this thing thousand times… Finally got a simple explanation… Thanks 🙂

  14. Very usefull thanks for explanations

  15. i enjoy reading this topics of motors. Thank you very much

  16. This is very easy to understand. Thank you.

  17. I like it.thanks alot

  18. It’s been a while since I’ve had to use this calculation, and I just couldn’t remember it !!

    Thanks

  19. got me out of a hole thanks

  20. Thanks

  21. that was great, thanks!

  22. Very explanatory

  23. thank you so much. its very useful to me

  24. very useful and very well exlained…

    • thanks a lot.

  25. good explanation

  26. GREAT JOB
    EASY TO UNDERSTAND…….

  27. its most important for engineers n its simple formula for all students

  28. thank you so much

  29. It was helpful….Thanks

  30. Very good. Help me in school.

  31. interesting

  32. Hello
    Are you sure about this formula?
    angular speed if you know rotational speed of the motor in rpm:

    ω = rpm * 2π / 60

    I think that should 2πr or π * Diameter since you are calculating the circumference of the disk
    π is defined as Circumference / Diameter.
    Hence Circumference = π * Diameter or π * 2r

    • If you are measuring linear speed then circumference matters. However angular speed defines how the angle changes over the time so the circumference is irrelevant and the formula is correct.

    • The formula is just unit conversion from rpm to rad/s, but the measure still remains an angular velocity.

    • 1 revolution = 2pi rad
      1 minute = 60 second
      rpm = 1revolution/1minute = 2pi/60

  33. thanks its to important for me….

  34. Thanks

  35. WONDERFULL

  36. Excellent explanation!

  37. very useful formulas

  38. that gives clear understanding

  39. why the efficiency is between 0 to 15 percent .? why not between 0 to 100 percent. how do you say 15% is the maximum efficiency of the motor ?

    • The calculations are for the motors we have on our site. Explore the site!

  40. great calculation!

  41. Wow! what a wonderful and most important calculation, really I appreciate the effort . bandle of thanks

  42. Tnx a lot

  43. THANK YOU FOR GREAT INFORMATION

  44. Excellent and clear explanation.

  45. superbly explained thanks a lot

  46. Wow excellen

  47. Very helpful. Excellent. Tanx

  48. Wonderful Explanation

  49. Really helpful. Well educative. Thanks so much.

  50. easy to understand
    thanks

  51. Thanks for this fantastic article

  52. Classy aritlce dude

  53. Excellent Explanation.

  54. Wonderful & very useful Explanation. Thanks.

  55. i like this, it’s very useful yet understandable

  56. It’s actually a great and helpful piece of information. I am satisfied that you simply shared this helpful info with us.

  57. delicious and interesting explanations. I’ve been lost the best like this!

  58. Superb explanation…
    Awesome…
    Thank you very much…

  59. THANKS FOR SIMPLE CALCULATION FORMULAE EASY TO UNDERSTAND

  60. Nice work, showing your class.

  61. Excellent sir
    Very much usefull

  62. I was trying to work on a quadcopter project and trying to calculate how much force is produced from the motors to overcome gravity and mass of quad to lift it up,thanks alot

  63. Excellent explanation. Your page deserves to be the top on Google search results. Thanks a lot. 😉

  64. Really it is helpful.

  65. Very good details. Thanks.

  66. easy understanding.good thank you

  67. Thank you! Exactly, what I was looking for!

  68. Great stuff

  69. Very clearly understood.thanks for the information

  70. Outstanding job. thank you for your dedication. It was very helpful for me 🙂

  71. I feel lucky to find your website. I would feel much luckier and happier if i found myself a gf of my dreams but hey i got you, so anyways thanks for the knowledge! 😀

  72. Thanks nice explained

  73. Thanks good job

  74. Well explained methodically yet concise and easy to follow. Thank you so much. Was able to generate a specifications sheet to make an order from a motor manufacturer using these. Great work!!!

  75. I Want to learn more from you.

  76. Simple and straight forward explanation

  77. This was very straight forward, but I have a question.
    If you were to know the efficiency, the rpm and the torque, can you find the number of watts the motor will need?

  78. How to calculate electrical losses within the motor due to resistance?

  79. How to check my motor torque , with 1400 rpm , motor current 1.2 amp , shaft size 28 mm , kw rating is 0.2 kw

  80. Fantastic analysis
    Keep it up

  81. Thanks for sharing valuable knowledge. Note : pls check again, I think the formula is reverse, the correct one is rpm = (omega) X 2(pi) X 60 .

  82. Thank you for this

  83. is the calculation applicable for both DC and AC motors ?

    • No, it is for DC motors even though some formulas apply to both.

  84. Very useful! Thanks✅

  85. Thank you so much.

  86. Short and clear explantion


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