# Sizing motors

This article is dedicated to motor sizing. Althrough it is dedicated to robotics and electrical motors, it can easily be extended to any application or other kind of motors.

## Problem specifications

Let’s consider a wheeled robot that climb up an inclinated plane:

Figure a.

The wheeled robot illustrated on figure a. have the following specifications:

• weight $m$[$Kg$],
• wheel diameter $D$[$m$],
• maximum velocity of the robot $v$[$m.s^{-1}$],
• maximum acceleration of the robot $a$[$m.s^{-2}$],
• angle of the greater positive slope to climb up $\alpha$[$rad$],
• gear ratio between the motor and the wheel $R$.
• efficiency of the gear box $\eta$.

Properties to determine :

• torque on wheel shaft $\tau_{wheel}$[$N.m$],
• torque on motor shaft $\tau_{motor}$[$N.m$],
• angular velocity of wheel shaft $\omega_{wheel}$[$rad.s^{-1}$ and $rpm$],
• angular velocity of motor shaft $\omega_{motor}$[$rad.s^{-1}$ and $rpm$],
• power of the motor $P_{motor}$[$W$].

## Angular velocities

Let’s first calculate the angular velocity of the wheel:

$\omega_{wheel} [rad.s^{-1}]=\frac{2v}{D}$

The angular velocity of the motor shaft is given by:

$\omega_{motor} [rad.s^{-1}]=R.\omega_{wheel} =R.\frac{2v}{D}$

The angular velocities converted in rotations per minutes are given by:

$\begin{array}{r c l} \omega_{wheel} [rpm] &=& \frac{60.v}{D.pi} \\ \omega_{motor} [rpm] &=& R. \frac{60.v}{D.pi}\end{array}$

## Torque on wheels shaft

Torque is more tricky to calculate. The fundamental principle of dynamics gives us:

$\sum \vec{F_i} = m.\vec {a}$

Forces acting on the robot are gravity and actuator. The force inducted by the gravity is $\vec {F_{Gravity}}=m.\vec{g}$. When projected on the direction of motion (see figure a.), the force inducted by the gravity is given by:

$\| \vec{F_g} \|=m.g.sin(\alpha)$

The fundamental principle of dynamics can be rewritten as:

$m.a = \|\vec{F_m}\| + \|\vec{F_g}\| = \|\vec{F_m}\| - m.g.sin(\alpha)$

$\|\vec{F_m}\|$ can be deducted :

$\|\vec{F_m}\|=m.a + m.g.sin(\alpha)=m.(a + g.sin(\alpha))$

Thus, the torque on wheel’s shaft is given by:

$\tau_{wheel} = \|\vec{F_m}\|.\frac{D}{2}= \frac{m.D.(a + g.sin(\alpha))}{2}$

## Torque on motor shaft

Considering the gear box specifications gives us:

$\tau_{motor} = \frac {\tau_{wheel}}{R.\eta} = \frac{m.D.(a + g.sin(\alpha))}{2.R.\eta}$

## Power

The power produced by the motor is given by the product of the angular speed by the torque on the motor’s shaft. We can thus calculate the motor’s minimal power:

$P_{motor}=\omega_{wheel}.\tau_{motor}=\frac{m.v.(a + g.sin(\alpha))}{\eta}$