Skip to content

Inertia

This section introduces the concept of inertia and how it affects the motion of a drone. A multirotor resists changes in both linear and angular motion, and this resistance appears through two quantities: mass, which governs translational dynamics, and the moment of inertia, which governs rotational dynamics. Understanding both is essential for predicting how the drone accelerates and how it responds to control inputs.


Mass

Mass represents how much the drone resists changes in velocity along any translational axis. It depends only on the total amount of matter and is the same in every direction. In other words, it does not matter whether the motion is upward, forward, or sideways, the vehicle has a single mass value.

Mass

Exercise 1

We can estimate the total mass of the drone by summing the mass of its components:

Determine the total mass by adding the component masses. The values can be obtained from the technical specifications on Bitcraze’s website (links provided above).

Solution

The total mass is given by:

\[ \begin{align} m &= m_{cf} + m_{fd} \\ m &= 37 + 1.6 \\ m &= 38.6 \, g \end{align} \]

In SI units, the approximate value is:

\[ m \approx 0.04 \, ~\text{kg} \]

Moment of Inertia

The moment of inertia represents how much the drone resists changes in angular velocity about a particular axis. Unlike mass, it depends not only on the amount of matter but also on how that matter is distributed relative to the axis of rotation.

Since the drone can rotate around three axes, it has three corresponding moments of inertia.

Moment of Inertia

Exercise 2

We can estimate the drone’s moments of inertia using a simplified model1:

Determine the total moment of inertia of the drone about each axis. The dimensions and masses can be obtained from the technical specifications on Bitcraze’s website (links provided above).

Solution

The battery has mass \(9.10 ~\text{g}\) and dimensions \(3.3 \times 2.0 \times 0.8 ~\text{cm}\):

\[ \left\{ \begin{align} m_b &= 9.10 ~\text{g}\\ a &= 3.3 ~\text{cm}\\ b &= 2.0 ~\text{cm}\\ c &= 0.8 ~\text{cm} \end{align} \right. \]

Using the standard formulas for the moment of inertia of a rectangular block, we can calculate the battery moment of inertia around each axis:

Moment of Inertia Battery

\[ \begin{align} I_{b_{xx}} &= \frac{m_b}{12} (b^2 + c^2) & I_{b_{yy}} &= \frac{m_b}{12} (a^2 + c^2) & I_{b_{zz}} &= \frac{m_b}{12} (a^2 + b^2) \\ I_{b_{xx}} &= \frac{9.1}{12} (2.0^2 + 0.8^2) & I_{b_{yy}} &= \frac{9.1}{12} (3.3^2 + 0.8^2) & I_{b_{zz}} &= \frac{9.1}{12} (3.3^2 + 2.0^2) \\ I_{b_{xx}} &= 3.52 ~\text{g·cm}^2 & I_{b_{yy}} &= 8.74 ~\text{g·cm}^2 & I_{b_{zz}} &= 12.29 ~\text{g·cm}^2 \end{align} \]

The motors (\(2.30 ~\text{g}\)) with propellers (\(1.34 ~\text{g}\)) and mounts (\(0.33 ~\text{g}\)) has a total mass of \(3.97 ~\text{g}\) each and are located \(10 ~\text{cm}\) apart diagonally:

\[ \left\{ \begin{align} m_m &= 3.97 ~\text{g}\\ l_x &= 5 \frac{\sqrt{2}}{2} ~\text{cm} \\ l_y &= 5 \frac{\sqrt{2}}{2} ~\text{cm} \\ l_z &= 5 ~\text{cm} \end{align} \right. \]

Considering them as point masses, we can calculate its moment of inertia around each axis:

Moment of Inertia Motors

\[ \begin{align} I_{m_{xx}} &= m_m l_x^2 & I_{m_{yy}} &= m_m l_y^2 & I_{m_{zz}} &= m_m l_z^2 \\ I_{m_{xx}} &= 3.97 \left( 5\frac{\sqrt{2}}{2} \right)^2 & I_{m_{yy}} &= 3.97 \left( 5\frac{\sqrt{2}}{2} \right)^2 & I_{m_{zz}} &= 3.97 \cdot 5^2 \\ I_{m_{xx}} &= 49.62 ~\text{g·cm}^2 & I_{m_{yy}} &= 49.62 ~\text{g·cm}^2 & I_{m_{zz}} &= 99.25 ~\text{g·cm}^2 \end{align} \]

herefore, the total moments of inertia are the sum of the moments of inertia of the battery and the moments of inertia of the four motors:

\[ \begin{align} I_{xx} &= I_{b_{xx}} + 4 I_{m_{xx}} & I_{yy} &= I_{b_{yy}} + 4 I_{m_{yy}} & I_{zz} &= I_{b_{zz}} + 4 I_{m_{zz}} \\ I_{xx} &= 3.52 + 4 \cdot 49.62 & I_{yy} &= 8.74 + 4 \cdot 49.62 & I_{zz} &= 12.29 + 4 \cdot 99.25 \\ I_{xx} &= 202.02 ~\text{g·cm}^2 & I_{yy} &= 207.24 ~\text{g·cm}^2 & I_{zz} &= 408.29 ~\text{g·cm}^2 \end{align} \]

In SI units, the approximate values ​​are:

\[ \left\{ \begin{align} I_{xx} &\approx 2 \times 10^{-5} \, ~\text{kg·m}^2 \\ I_{yy} &\approx 2 \times 10^{-5} \, ~\text{kg·m}^2 \\ I_{zz} &\approx 4 \times 10^{-5} \, ~\text{kg·m}^2 \end{align} \right. \]


  1. We are neglecting the PCB structure, as its mass is small compared to the battery and motors, and it lies close to the center of mass.