Kinematics

Kinematics is the geometry of motion, without forces. Forward kinematics maps joint angles to the foot position. Inverse kinematics maps a foot position back to joint angles. The Jacobian links joint velocities to foot velocities. This page derives all three for the Strido leg, including its knee linkage.

In Strido

Leg model: packages/strido-core/src/strido/leg.py (foot_position, inverse_kinematics, calf_from_crank, crank_from_calf, transmission_angles), mirrored in C++ in firmware/strido_esp32s3/lib/strido_core/.
Dimensions: hardware/kinematics.yaml (legs_linkage, legs_base), measured from the CAD. Limits: hardware/joint_limits.yaml.
Practical guide: Kinematics and gaits shows how the gait uses these functions.

The leg as a kinematic chain

Each leg has three revolute joints (leg.py docstring; the conventions match the Unitree Go2):

Joint	Axis	Zero	Positive
hip $q_{1}$	$+ x$	leg vertical	foot towards $+ y$
thigh $q_{2}$	$+ y$	femur straight down	knee backwards
calf $q_{3}$	$+ y$ , relative to femur	calf in line with femur	negative = knee bent

From hardware/kinematics.yaml (front-left leg, rounded): the hip is at $h = (0.1197, 0.039, 0) m$ in base_link, the leg plane sits $d = 38.3 mm$ outboard of the hip ( $40.2 mm$ on the rear legs), the thigh is $l_{1} = 130.0 mm$ and the calf is $l_{2} = 124.1 mm$ to the foot centre.

The thigh and calf move in a vertical leg plane. The hip joint tilts that plane about $x$ . So the problem splits into two easier ones: a planar two-link arm, and one rotation.

Forward kinematics

In the leg plane, let $u$ point forward and $v$ point up, both measured from the thigh pivot. Write $s_{2} = \sin q_{2}$ and $s_{23} = \sin (q_{2} + q_{3})$ , and similarly for cosines. The knee and the foot are at:

k = [\begin{matrix} - l_{1} s_{2} \\ - l_{1} c_{2} \end{matrix}], [\begin{matrix} u \\ v \end{matrix}] = [\begin{matrix} - l_{1} s_{2} - l_{2} s_{23} \\ - l_{1} c_{2} - l_{2} c_{23} \end{matrix}]

The angles add because $q_{3}$ is measured relative to the femur, so the calf's absolute angle is $q_{2} + q_{3}$ . The signs are negative because zero means "pointing down" and positive $q_{2}$ swings the foot backwards (towards $- u$ ).

Next, the hip rotation. The foot in the hip frame, before abduction, is $(u, d, v)$ . Rotating about $x$ by $q_{1}$ and adding the hip position gives:

p = h + R_{x} (q_{1}) [\begin{matrix} u \\ d \\ v \end{matrix}] = h + [\begin{matrix} u \\ d \cos q_{1} - v \sin q_{1} \\ d \sin q_{1} + v \cos q_{1} \end{matrix}]

This is foot_position line for line. For the frame conventions, see frames and rotations.

Inverse kinematics

We are given a foot target $p$ and need $(q_{1}, q_{2}, q_{3})$ . Let $r = p - h = (r_{x}, r_{y}, r_{z})$ .

Hip abduction

A rotation about $x$ leaves the $x$ component alone and preserves length in the $y$ – $z$ plane:

r_{y}^{2} + r_{z}^{2} = d^{2} + v^{2}

The foot is below the hip, so we take the negative root:

v = - \sqrt{r_{y}^{2} + r_{z}^{2} - d^{2}}

If the term under the root is not positive, the target lies inside the cylinder of radius $d$ around the hip axis and cannot be reached. The direction of $(r_{y}, r_{z})$ is the direction of $(d, v)$ rotated by $q_{1}$ , so:

q_{1} = atan 2 (r_{z}, r_{y}) - atan 2 (v, d)

The planar two-link problem: law of cosines

Now set $u = r_{x}$ and solve the two-link arm. The thigh, the calf and the line from pivot to foot form a triangle with sides $l_{1}$ , $l_{2}$ and $ρ = \sqrt{u^{2} + v^{2}}$ . The law of cosines at the knee gives the knee angle. With our convention (the knee interior angle is $π + q_{3}$ ):

ρ^{2} = l_{1}^{2} + l_{2}^{2} + 2 l_{1} l_{2} \cos q_{3} ⟹ \cos q_{3} = \frac{u^{2} + v^{2} - l_{1}^{2} - l_{2}^{2}}{2 l_{1} l_{2}}

There are two solutions, $\pm \arccos$ , for knee-forward and knee-backward. Strido uses $q_{3} = - \arccos (\cdot)$ , a bent knee pointing backwards. If $| \cos q_{3} | > 1$ , the target is out of reach.

The thigh angle is the direction to the foot minus the angle the calf adds inside the triangle:

q_{2} = atan 2 (- u, - v) - atan 2 (l_{2} \sin q_{3}, l_{1} + l_{2} \cos q_{3})

The minus signs in the first term come from the "zero = straight down, positive = backwards" convention. Use atan2, never atan, so that the quadrant is correct.

The knee four-bar

The knee servo does not drive $q_{3}$ directly. A parallelogram (arm $24.0 mm$ , short link $28.3 mm$ , crank $24.0 mm$ ) copies the servo angle onto a crank $ψ$ at the thigh pivot. A four-bar (crank $26.9 mm$ , long link $131 mm$ , tibia crank $24.0 mm$ , and the femur $130.0 mm$ as ground link) then turns the tibia. Because the four-bar's ground link is the femur itself:

q_{3} = f (q_{2}, ψ)

Moving the thigh alone also moves the knee. crank_from_calf inverts this geometrically. The far end of the crank lies on a circle of radius $26.9 mm$ around the pivot, and also on a circle of radius $131 mm$ (the long link) around the tibia pin. Intersecting the two circles gives two candidates, and the assembly branch picks one. circle_intersection returns None when the circles do not meet. That means the linkage cannot be assembled in that pose.

Rendering diagram…

The Jacobian

The Jacobian maps joint rates to foot velocity, $\dot{p} = J (q) \dot{q}$ . Differentiating the planar forward kinematics:

J_{plane} = \frac{\partial (u, v)}{\partial (q_{2}, q_{3})} = [\begin{matrix} - l_{1} c_{2} - l_{2} c_{23} & - l_{2} c_{23} \\ l_{1} s_{2} + l_{2} s_{23} & l_{2} s_{23} \end{matrix}], det J_{plane} = - l_{1} l_{2} \sin q_{3}

The full $3 \times 3$ Jacobian adds the hip column, $\partial p / \partial q_{1} = \hat{x} \times (p - h)$ , and rotates the planar columns by $R_{x} (q_{1})$ . The Jacobian also governs forces through $τ = J^{⊤} F$ (see statics and dynamics).

For the drive, the chain rule adds one more factor. The knee servo's rate is $\dot{ψ}$ , and

{\dot{q}}_{3} = \frac{\partial f}{\partial q_{2}} {\dot{q}}_{2} + \frac{\partial f}{\partial ψ} \dot{ψ}

Workspace and singularities

Ignoring joint limits, the planar foot reaches an annulus:

| l_{1} - l_{2} | \leq ρ \leq l_{1} + l_{2} \Rightarrow 5.9 mm \leq ρ \leq 254.1 mm

The joint limits and self-collisions cut this down. hardware/joint_limits.yaml limits $q_{3}$ to $[- 150^{\circ}, - 45^{\circ}]$ and the front $q_{2}$ to $[- 45^{\circ}, 125^{\circ}]$ , found by sweeping the posed CAD for collisions on a $5^{\circ}$ grid.

A singularity is a pose where $det J = 0$ . There the foot cannot move in some direction, however fast the joints turn, and inverse kinematics becomes ill-conditioned. For the planar leg this happens at $\sin q_{3} = 0$ : the leg fully straight ( $q_{3} = 0$ , on the outer workspace boundary) or fully folded. The calf limit keeps $| \sin q_{3} | \geq \sin 30^{\circ} = 0.5$ (at the $- 150^{\circ}$ end; at least $0.71$ at the $- 45^{\circ}$ end), so the leg never gets close. This is one reason the stand pose keeps the knee well bent.

The four-bar has its own singularities. When the long link lines up with the crank or with the tibia crank (a transmission angle near $0$ or $180^{\circ}$ ), $\partial f / \partial ψ$ blows up or vanishes. transmission_angles measures both angles, and joint_limits.yaml requires at least $25^{\circ}$ (transmission_min_deg).

Worked example: the stand pose

The stand pose puts each foot straight below its hip, $0.17 m$ down (stand_pose(height=0.17) in strido/gait.py). For the front-left leg, $r = (0, 0.0383, - 0.170) m$ .

Hip. $v = - \sqrt{{0.0383}^{2} + {0.170}^{2} - {0.0383}^{2}} = - 0.170$ , so $q_{1} = atan 2 (- 0.170, 0.0383) - atan 2 (- 0.170, 0.0383) = 0$ .

Knee. With $u = 0$ :

\cos q_{3} = \frac{{0.170}^{2} - {0.1300}^{2} - {0.1241}^{2}}{2 \cdot 0.1300 \cdot 0.1241} = \frac{- 0.00339}{0.03226} = - 0.105 \Rightarrow q_{3} = - {96.0}^{\circ}

Thigh. $l_{2} \sin q_{3} = - 0.1234$ and $l_{1} + l_{2} \cos q_{3} = 0.1170$ , so

q_{2} = atan 2 (0, 0.170) - atan 2 (- 0.1234, 0.1170) = 0 + {46.5}^{\circ} = {46.5}^{\circ}

These match stand_pose(), which returns $(0^{\circ}, {46.53}^{\circ}, - {96.03}^{\circ})$ for every leg. The calf's absolute angle is $q_{2} + q_{3} = - {49.5}^{\circ}$ : the knee sits behind and above the foot, and the calf leans forward to reach it.

Jacobian. At this pose,

J_{plane} = [\begin{matrix} - 0.170 & - 0.0806 \\ 0 & - 0.0943 \end{matrix}] m / rad, det = 0.0160 m^{2}

A thigh rate of $1 rad / s$ moves the foot $0.17 m / s$ backwards and not at all vertically. To lift the foot you mostly need the knee.

Linkage. leg.py gives a crank angle of $ψ = {38.9}^{\circ}$ and transmission angles of $84^{\circ}$ and $86^{\circ}$ , close to the ideal $90^{\circ}$ . The sensitivity is $\partial q_{3} / \partial ψ = - 1.12$ : the knee turns ${1.12}^{\circ}$ for each degree of crank.

Check yourself

Why can the hip angle be solved before, and separately from, the thigh and knee?
What does $| \cos q_{3} | > 1$ mean physically?
At the stand pose, which joint mainly moves the foot forward, and which mainly moves it vertically?
Why do the transmission angles matter even when $det J_{plane}$ is far from zero?

Answers

The hip axis is the $x$ axis, and $q_{2}$ , $q_{3}$ never change the foot's distance from it. So $r_{y}^{2} + r_{z}^{2}$ depends only on the planar depth $v$ and the fixed offset $d$ .
The target is farther than $l_{1} + l_{2}$ or closer than $| l_{1} - l_{2} |$ , so no triangle exists.
The thigh moves it forward and back ( $J_{11} = - 0.170$ , $J_{21} = 0$ ). Only the knee column has a vertical component.
They describe the drive linkage, not the leg. Near $0^{\circ}$ or $180^{\circ}$ the knee servo loses leverage and the crank–knee map becomes ill-conditioned, whatever the leg Jacobian says.

Kinematics ​

The leg as a kinematic chain ​

Forward kinematics ​

Inverse kinematics ​

Hip abduction ​

The planar two-link problem: law of cosines ​

The knee four-bar ​

The Jacobian ​

Workspace and singularities ​

Worked example: the stand pose ​

Check yourself ​

Further reading ​