Question

to build a robot capable of handling 150lbs.how do I pick a motor considering all of the specs I see are for example rpm=2500 volts=12 gear ratio = 10:14.
I am new to this so all the info I can get will help.Thanks

Answer 1

First of all every motor specified by its speed and torque.

I am assuming that you are asking about the drive motors and your robot is traversing on a horizontal plane. To start with you need to have some basic data on your requirements regarding the robot. These are:

Total Weight of the Robot (in Kg), Velocity required (v), Acceleration required (a), Number of drive motors (N) and the radius of Robot Wheel (r).

Now there are two main relations you need to understand:

v=2∗pi∗r∗RPS-----------(1)

--------------------(2)

In (2) T is the Torque and we can replace F with "m∗a" where m is the total weight of the robot in Kg.

So now according to your requirement some you provided and some I assumed.

Speed of motor = G.R. * speed of Wheel ( assumed 2.8 m/s)

= 10/14 * (2.8) = 2 m/s

Weight of my Robot = 68.0389 Kg, Speed of motor required = 2 m/s, No of drive wheels = 4, Acceleration required = 0.5 m/s^2 , Radius of each wheel = 0.0625m

Now using above data in (1) to solve for RPS (Rotations per second), we get

RPS=(2m/s)/(2∗pi∗0.0625m)

Convert RPS to RPM (Rotation per minute) by Multiplying RPS by 60, we get,

RPM = 305 — required at any instant at for the robot to run at the desired velocity.

Now we need to find the Torque of the motor required. To find this we will use (2). We get,

T=(68.0389∗0.5∗0.0625)/4 = 0.53153 N*m

We also have to compensate for the motor’s gear and wheel slip. To do that, we do,

(T∗100)/Efficiency

Assuming efficiency to be 60% is fair. So after calculating, we get,

T = 0.8859 Nm — required at any instant for our robot to accelerate at the desired rate.

Small DC Geared motors are not rated in Nm as it is a very big unit. To convert it to more reasonable unit of kgf-cm, we do,

T =(100*T)/9.8— Torque in kgf-cm

T = 9.0398 Kgf-cm

The calculated RPM and Torque are the continuous output we need from the motor but DC motors are not specified in this manner. They are specified by their No-load Speeds and their Stall Torque.

It’s a rule of thumb that the stall torque of the motor should be 4 times your calculated torque. And no-load speed of the motor should be higher by 25% or more.

Thus we need a motor of:

No load Speed = 380RPM

Stall Torque = 36.16 Kg-cm

[ Product of speed and torque of motor is constant. If one reduces the other must increase. Also, T is proportional to current drawn by the motor Speed of the motor is proportional to supplied voltage ]

If you noticed that 25% and 4 times seems related you are right!

In a DC geared motor, torque and speed are related by the below graph (Forget the units!)

Therefore, if our Stall Torque is 4 times more than our required torque and our motor is running at the required torque, then, our motor will be operating at 75% of its No-load Speed at that time. Makes sense right!

It doesn’t stop here. If you want, you can also calculate the continuous current that your motor will draw at any point in your application. For that obtain the power that your motor will be delivering at any time by using

Power=Speed∗Torque

From the above equation of equation of power in DC circuits, we can obtain the current draw by the motor in terms of Torque and Speed.

Current=(Torque∗Speed)/Voltage

It has to be noted that I have not considered any inclination for my robot to move which makes life easier.

Also we have ignored the role of Rolling friction which also makes our calculation simplified. But taking torque 4 times the calculated value accounts for every kind of practical opposition you can think of.

I hope you have a clear understanding of the subject now. Following this logical methodology will save you a lot of time and money.