Unit 3: Multi Rotor Dynamics - Subjective Questions

A multirotor has six degrees of freedom (6-DOF):

• Translational motion: position along the inertial-frame axes , , and .
• Rotational motion: roll , pitch , and yaw about the body-frame axes.

Full rigid-body dynamics are required because translational and rotational motions are coupled. For example, pitching the vehicle changes the direction of its thrust and therefore produces horizontal acceleration. A complete model includes:

• Translational position and velocity.
• Vehicle orientation and angular velocity.
• Gravity, rotor thrust, aerodynamic drag, and external forces.
• Control torque, aerodynamic torque, and gyroscopic effects.

Thus, a 6-DOF model describes both the position and attitude of the multirotor under applied forces and torques.

• The inertial frame is fixed relative to the Earth. Position, velocity, gravity, and the flight path are commonly expressed in this frame.
• The body frame is attached to the multirotor. Its axes rotate and translate with the vehicle. Rotor thrust, body angular velocity, and control torques are naturally expressed in this frame.

A rotation matrix transforms a body-frame vector into the inertial frame:

The inverse transformation is:

Correct frame transformation is essential because total rotor thrust usually acts along the body vertical axis, whereas gravity acts along the inertial vertical axis.

Using Newton's second law in the inertial frame,

Let be the inertial position, the vehicle mass, the body-to-inertial rotation matrix, and the total thrust magnitude. If thrust acts along the body unit vector , the inertial thrust is . Including gravity, drag, and disturbance forces gives

Therefore,

This equation shows that attitude controls the direction of thrust, while rotor speed controls its magnitude. Sign changes may be required when a different vertical-axis convention is used.

The rotational motion in the body frame is governed by Euler's equation:

where:

• is the inertia matrix.
• is the body angular velocity.
• is the applied torque.
• represents nonlinear gyroscopic coupling.

Solving for angular acceleration gives

Roll and pitch torques are mainly generated by differential rotor thrust, while yaw torque is produced by differences in rotor reaction torque.

A spinning propeller accelerates air and generates an axial thrust. A commonly used static model is

where is rotor thrust, is the thrust coefficient, and is rotor angular speed.

The aerodynamic resistance acting on the propeller produces a reaction torque on the airframe:

where is the moment coefficient. The sign of depends on the rotor's direction of rotation.

• Increasing increases both thrust and reaction torque.
• Counter-rotating rotor pairs cancel yaw torque when their speeds are balanced.
• Differential changes in rotor speeds generate roll, pitch, and yaw control moments.

• Roll torque: Increase thrust on one lateral side and decrease thrust on the opposite side. The thrust difference acting through arm length creates a torque about the body -axis.
• Pitch torque: Increase thrust at one longitudinal end and decrease it at the opposite end, producing torque about the body -axis.
• Yaw torque: Create an imbalance between the reaction torques of clockwise and counterclockwise rotors, producing torque about the body -axis.

For two opposing rotors, a typical roll or pitch moment is

The yaw moment may be written as

where or represents the rotor's spin direction. Collective equal-speed changes primarily alter total thrust without producing roll or pitch.

For a simplified plus-configuration quadrotor, let each rotor produce , and let be the arm length. One possible allocation model is

The first row gives collective thrust. The next two rows give roll and pitch moments from thrust differences. The final row gives yaw moment from reaction torques. Rotor numbering and signs vary with geometry and spin convention, but the allocation principle remains the same. Inverting or pseudo-inverting this matrix gives the rotor commands needed for a desired force and moment vector, subject to rotor-speed limits.

For steady hover in still air:

• Linear velocity and acceleration are zero.
• Angular velocity and angular acceleration are zero.
• The vehicle is normally level.
• Total upward thrust balances weight.

The force equilibrium is

The moment equilibrium is

For a symmetric quadrotor with identical rotors,

All rotor reaction torques must also cancel. If thrust is smaller than , the vehicle descends; if it is greater than , the vehicle accelerates upward. Disturbances require corrective thrust and moment commands even when the desired state is hover.

During level hover, the four equal rotor thrusts balance the vehicle weight:

Using ,

Therefore, the required angular speed of each rotor is

This expression assumes:

• All four rotors and propellers are identical.
• The vehicle is level.
• Rotor interaction and aerodynamic disturbances are neglected.
• The thrust coefficient is constant.

The result shows that hover speed increases with vehicle mass and decreases as the rotor thrust coefficient increases.

A small roll or pitch disturbance tilts the total thrust vector. The thrust can then be resolved into vertical and horizontal components:

where is the tilt angle.

If the total thrust remains equal to , then . Consequently, the vehicle begins to lose altitude while also accelerating horizontally. To preserve altitude, thrust must be increased to

An attitude controller must generate a restoring moment to remove the unwanted tilt. This demonstrates that hover is sensitive to attitude disturbances and normally requires closed-loop stabilization.

To enter forward flight, the controller commands a pitch angle in the desired direction. This rotates the total thrust vector so that it has:

• A vertical component that supports the vehicle's weight.
• A horizontal component that accelerates the vehicle forward.

For pitch angle ,

To maintain constant altitude,

During acceleration, the forward component exceeds aerodynamic drag. As speed increases, drag increases. Steady forward flight is reached when

and the net horizontal acceleration becomes zero. The vehicle must generally increase total thrust during the transition to prevent altitude loss.

Let the multirotor tilt by angle from the vertical. Resolving thrust gives

for constant altitude, and

for horizontal motion. From vertical equilibrium,

Substituting into the horizontal equation,

Therefore,

This ideal result shows that horizontal acceleration grows with tilt angle. In practice, maximum thrust, aerodynamic drag, attitude limits, and control margins restrict the achievable acceleration.

Level hover:

• The thrust vector is vertical.
• Total thrust balances weight: .
• Horizontal speed is zero in still air.
• Net force and net torque are zero.

Steady forward flight:

• The vehicle is tilted, so thrust has vertical and horizontal components.
• Vertical balance requires .
• Horizontal balance requires .
• Total thrust is greater than weight because only part of it acts vertically.

Thus, hover requires thrust mainly for weight support, whereas steady forward flight requires additional thrust to overcome aerodynamic drag while still supporting the vehicle's weight.

At tilt angle , the vertical component of thrust is . To maintain altitude at maximum thrust,

Hence,

and

A real solution requires . The formula represents an ideal limit. Practical controllers use a smaller angle because thrust reserve is needed for disturbances, altitude correction, actuator uncertainty, and maneuvering.

At the thrust limit, the vertical component must support weight:

The horizontal force is

Using ,

Therefore, neglecting drag, the maximum horizontal acceleration is

This result illustrates the thrust trade-off: part of the available thrust must always support weight, and only the remaining horizontal component can accelerate the vehicle.

The motors and propellers can produce only a finite maximum thrust. This limit affects flight in several ways:

• Climb: Upward acceleration is limited by .
• Tilted flight: Tilting reduces the vertical thrust component, so more total thrust is required to maintain altitude.
• Horizontal acceleration: Only the horizontal component of the remaining thrust can accelerate the vehicle.
• Disturbance rejection: If motors are already near saturation, there may be insufficient reserve to oppose wind or recover altitude.
• Attitude control: Rotor saturation can prevent the required differential thrust and therefore reduce available control torque.

A safe design maintains a thrust margin above hover rather than operating continuously near .

Aerodynamic drag is the force that opposes the motion of the multirotor relative to the surrounding air. A common model is

where:

• is air density.
• is the drag coefficient.
• is the reference area.
• is velocity relative to the air.
• The negative sign indicates that drag opposes motion.

In one-dimensional motion, the drag magnitude is

Because drag increases approximately with the square of speed, progressively more horizontal thrust is required as the vehicle flies faster.

At maximum steady speed, horizontal acceleration is zero, so available horizontal thrust equals aerodynamic drag:

Using the quadratic drag model,

Solving for maximum velocity gives

If altitude must be maintained and total thrust is limited to , then

Therefore,

This is an ideal estimate and neglects changes in propeller performance, vehicle attitude, and drag coefficient.

Linear drag model:

• Drag is proportional to velocity.
• It is mathematically simple.
• It is often useful for low-speed operation and controller design.

Quadratic drag model:

• Drag magnitude is proportional to the square of speed.
• It more accurately represents moderate- and high-speed aerodynamic resistance.
• It introduces stronger nonlinearity into the dynamic model.

The appropriate model depends on speed range, vehicle geometry, airflow conditions, and the accuracy required.

For steady, level forward flight, vertical and horizontal accelerations are zero. At the maximum-thrust condition, resolve into components:

With quadratic drag,

The maximum feasible tilt while maintaining altitude is

The corresponding horizontal thrust is

Equating this to drag gives

Thus,

These equations connect mass, thrust capability, tilt angle, aerodynamic drag, and maximum velocity.