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):

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

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.

## 113 Comments

It was useful for me

thank you for this calculation

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

exccelent

It’s been very useful, thank you soo much

thank you for this calculation

Very well explained. Thanks

Well done

Good.,

Very useful for me

Wow excellent

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!

I love this thing, thank you very much!

Thank you very much!

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

Very usefull thanks for explanations

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

This is very easy to understand. Thank you.

I like it.thanks alot

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

Thanks

got me out of a hole thanks

great

Thanks

that was great, thanks!

NICE DESCRIPTION

Really great support to us

Merci beaucoup!

Very explanatory

thank you so much. its very useful to me

Incredible breakdown, thanks a lot.

very useful and very well exlained…

thanks a lot.

good explanation

GREAT JOB

EASY TO UNDERSTAND…….

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

thank you so much

It was helpful….Thanks

Very good. Help me in school.

interesting

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

thanks its to important for me….

Superb

Thanks

Explained in easy way to understand.

Thank you very much

WONDERFULL

Excellent explanation!

very useful formulas

that gives clear understanding

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!

great calculation!

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

Tnx a lot

THANK YOU FOR GREAT INFORMATION

Excellent and clear explanation.

superbly explained thanks a lot

Wow excellen

Very helpful. Excellent. Tanx

Wonderful Explanation

Really helpful. Well educative. Thanks so much.

easy to understand

thanks

Thanks for this fantastic article

Classy aritlce dude

super sir

Excellent Explanation.

Wonderful & very useful Explanation. Thanks.

i like this, it’s very useful yet understandable

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

Thanks…

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

Thank u

Thnks bro

Superb explanation…

Awesome…

Thank you very much…

THANKS FOR SIMPLE CALCULATION FORMULAE EASY TO UNDERSTAND

Nice work, showing your class.

Excellent

It is useful..

Excellent sir

Very much usefull

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

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

Really it is helpful.

Very good details. Thanks.

easy understanding.good thank you

Thank you! Exactly, what I was looking for!

Great stuff

Very clearly understood.thanks for the information

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

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! 😀

NICE AND SIMPLIFIED PRESENTATION

Thanks nice explained

good job

Thanks good job

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!!!

realy good

I Want to learn more from you.

Simple and straight forward explanation

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?

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

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

Fantastic analysis

Keep it up

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 .

Thank you for this

is the calculation applicable for both DC and AC motors ?

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

thankyou for explanation

Very useful! Thanks✅

Thank you so much.

Nice one, remembering my secondary school days

Very nice