field guide

Thrust vector from attitude angles

Mapping attitude angles to a body thrust vector in the NED frame - how a desired orientation becomes the direction the rotors push.

updated 2026-06-28 guidance-control kinematics

Given total thrust magnitude $T$ and attitude (roll $\phi$ , pitch $\theta$ , yaw $\psi$ ), how do you find the inertial-frame components of the force vector perpendicular to the quadcopter’s plane?

The setup

The quadcopter’s “plane” is the body-fixed xy-plane — the arms form an X or + in this plane. The vector perpendicular to it is the body-fixed z-axis, pointing up through the centre of mass (by convention).

All four rotors spin to produce thrust along this axis. In body coordinates:

$\mathbf{F}_B = \begin{bmatrix} 0 \\ 0 \\ T \end{bmatrix}$

Total thrust $T$ is purely along body-z. There is no body-x or body-y thrust component — those come from tilting this vector via attitude.

To find the components in the world/inertial frame, we need to rotate $\mathbf{F}_B$ by the quadcopter’s current orientation. That orientation is described by the Euler angles $(\phi, \theta, \psi)$ and the rotation matrix that maps body→world.

The rotation matri·x (derived)

The world frame is NED (north-east-down) or ENU (east-north-up) depending on your convention — the math is the same either way, the axes just swap labels.

We build the rotation matrix from body to world as a sequence of three intrinsic rotations, applied in ZYX order (yaw → pitch → roll). This is the standard aeronautical convention.

Step 1 — Yaw $\psi$ about body z-axis:

$R_z(\psi) = \begin{bmatrix} c_\psi & -s_\psi & 0 \\ s_\psi & c_\psi & 0 \\ 0 & 0 & 1 \end{bmatrix}$

This rotates the body’s x-y plane around the vertical axis — pointing the nose north/south/east/west.

Step 2 — Pitch $\theta$ about the (now-rotated) body y-axis:

$R_y(\theta) = \begin{bmatrix} c_\theta & 0 & s_\theta \\ 0 & 1 & 0 \\ -s_\theta & 0 & c_\theta \end{bmatrix}$

Tilts the nose up or down.

Step 3 — Roll $\phi$ about the (now-rotated) body x-axis:

$R_x(\phi) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & c_\phi & -s_\phi \\ 0 & s_\phi & c_\phi \end{bmatrix}$

Banks left or right.

The full rotation is the product of these three matrices, applied right to left (the first rotation is rightmost):

$R_{BW} = R_z(\psi) \, R_y(\theta) \, R_x(\phi)$

Which expands to:

$R_{BW} = \begin{bmatrix} c_\theta c_\psi & s_\phi s_\theta c_\psi - c_\phi s_\psi & c_\phi s_\theta c_\psi + s_\phi s_\psi \\ c_\theta s_\psi & s_\phi s_\theta s_\psi + c_\phi c_\psi & c_\phi s_\theta s_\psi - s_\phi c_\psi \\ -s_\theta & s_\phi c_\theta & c_\phi c_\theta \end{bmatrix}$

Applying to the thrust vector

$\mathbf{F}_W = R_{BW} \cdot \mathbf{F}_B = R_{BW} \cdot \begin{bmatrix}0\\0\\T\end{bmatrix}$

Since only the third column of $R_{BW}$ multiplies against the non-zero entry of $\mathbf{F}_B$ , we get:

$\boxed{ \begin{aligned} F_x &= T \cdot (c_\phi s_\theta c_\psi + s_\phi s_\psi) \\ F_y &= T \cdot (c_\phi s_\theta s_\psi - s_\phi c_\psi) \\ F_z &= T \cdot (c_\phi c_\theta) \end{aligned}}$

These are equations (15)–(17) from the quadcopter PID tuning paper in the vault.

Why derive the full matrix?

The third column is all you need right now — you only care about one body-axis vector (body-z → world). But the full matrix will reappear when you need to:

The rotation matrix $R_{BW}$ is how you implement the coordinate transform that rotating-coordinate-frames describes for any vector measured from a rotating platform. The $R_{BW} \cdot [0,0,T]^T$ operation is the discrete version of the carrying term $\boldsymbol\omega \times \mathbf{q}$ applied to the body-z thrust vector — same geometry, matrix form.

Go the other way (world → body): invert the rotation (which for a rotation matrix is just the transpose $R_{WB} = R_{BW}^T$ ). You need this when taking a world-frame target direction (e.g. “fly north-east”) and decomposing it into body-relative pitch/roll commands.
Convert angular velocities ( $p, q, r$ ) to Euler angle rates ( $\dot{\phi}, \dot{\theta}, \dot{\psi}$ ) for your attitude estimator — that uses a different matrix but comes from the same rotation sequence. The body-rate cross-coupling — when the gyroscopic term $\boldsymbol\omega \times (I\boldsymbol\omega)$ matters and when it doesn’t — is covered in euler-body-rate-cross-coupling.
Rotate arbitrary vectors, like wind, magnetic field, or GPS velocity into body frame for sensor fusion.

So yes, deriving it once and keeping it in a reference note is worthwhile. You’ll use different columns/rows of it as the quadcopter code grows.

In code

void thrust_body_to_world(float T, float phi, float theta, float psi,
                          float *fx, float *fy, float *fz) {
    float c_phi = cosf(phi), s_phi = sinf(phi);
    float c_th  = cosf(theta), s_th = sinf(theta);
    float c_psi = cosf(psi), s_psi = sinf(psi);

    *fx = T * (c_phi * s_th * c_psi + s_phi * s_psi);
    *fy = T * (c_phi * s_th * s_psi - s_phi * c_psi);
    *fz = T * (c_phi * c_th);
}

Feed $\phi, \theta, \psi$ from your complementary filter (MPU-6050 → quaternion → Euler angles, or direct Madgwick/Mahony output) and $T$ from your control law. The resulting $(f_x, f_y, f_z)$ goes into your position controller or state estimator as the applied force in world coordinates.

Common pitfalls

Euler angle convention matters. ZYX (yaw → pitch → roll) is standard for aerospace. If your code uses XYZ or some other order, the matrix changes. Verify against a known test case (e.g. $\phi=0, \theta=30^\circ, \psi=0$ → thrust should tilt forward, giving positive $F_x$ and reduced $F_z$ ).
Gimbal lock at $\theta = \pm 90^\circ$ . Euler angles become degenerate. If your quadcopter ever pitches straight vertical (unlikely in normal flight but possible in a crash), switch to quaternions in the estimator and extract the third column directly from the rotation matrix without going through Euler angles.
Sign conventions. Make sure your attitude estimator and your controller use the same sign convention for roll/pitch. Inconsistent sign flips will cause the quad to fly in the wrong direction and likely crash.