Mechanics: Kinematics

$m = 6$ for spatial mechanisms, $m = 3$ for planar ones.

$N$ counts links including ground, $J$ the joints, $f_i$ the dof of joint $i$.

A single holonomic constraint between joints removes one dof.

Seen in Practice exam 3872.

A planar robot reaches a 2-D set of positions, no matter how many joints it has.

Extra joints add redundancy, not workspace dimension.

Seen in Practice exam 3872.

The right-handed cyclic order $x \to y \to z \to x$.

Reversing the order flips the sign.

To complete a frame from two orthonormal vectors $a$ and $b$, take $c = a \times b$; the result is automatically right-handed.

Always verify a cross product with $a^{T}c = 0$, $b^{T}c = 0$ and $\lVert c \rVert = 1$. Ten seconds, catches every sign slip.

Seen in Practice exam 3872.

Column $j$ is the target frame's $j$-th axis, written in the reference frame's coordinates.

This is not algebra: read the arrows off the drawing, one column at a time.

Easiest method is relative. Ask whether each axis of $W_2$ is parallel or anti-parallel to the matching axis of $W_1$, rather than judging each against an imagined baseline.

A superscript names the frame the coordinates are expressed in, so $r_p^{W_1}$ is the vector to $p$ written in $W_1$. Never write a bare vector without its frame.

Sanity check: a point further from a frame's origin must have a larger magnitude in that frame. Catches swapped answers instantly.

Write the three axes as columns in any reference you trust, then take the determinant. Works regardless of how the figure is drawn.

Curl rule if you prefer geometry: point the right hand's fingers along $+x$, curl them toward $+y$; the thumb gives $+z$ for a right-handed frame.

To repair a left-handed frame, flip exactly ONE axis. Flipping two is just a rotation and changes nothing.

Counting flips between two frames: an even number means the same handedness, an odd number means opposite.

Every question in 2.1 is this one equation solved for a different slot.

$p^{*}$ is the point in WORLD coordinates, $p$ is the same physical point in BODY coordinates, $R$ is the body orientation in world, $r_C$ is the body origin in world.

Point in world: $p^{*} = Rp + r_C$. Point in body: $p = R^{T}(p^{*} - r_C)$. Body origin or centre of mass: $r_C = p^{*} - Rp$. Pose: state both $\{r_C, R\}$.

Inverses go on the LEFT, because that is the side $R$ sits on. Matrix multiplication does not commute, so you cannot divide across as in scalar algebra.

$R$ always multiplies a BODY-frame vector. If $R$ sits next to something, that something is in body coordinates.

Label every vector with its frame before starting. Most errors here are a mislabel, not bad algebra.

Seen in Practice exam 3872.

Rotate $\alpha$ about $z$, then $\beta$ about the NEW $x$, then $\gamma$ about the NEW $z$.

Three parameters instead of nine, at the cost of singularities.

Singular when $\beta = k\pi$: the first and third axes line up and only $\alpha + \gamma$ is recoverable.

The $y$-axis never appears, so a rotation about $y$ must be reached indirectly by first rotating a usable axis onto it.

A unit rotation axis $\omega$ paired with the rotation angle $\phi$: axis-angle.

Invert by negating the axis, $-\omega\phi$, or equivalently by negating the angle.

Example: $+90°$ about $y$ is $[0,1,0]^{T}\tfrac{\pi}{2}$; $-90°$ about $y$ is $[0,-1,0]^{T}\tfrac{\pi}{2}$.

Rotating by $-\alpha$ undoes a rotation by $\alpha$, so it is the inverse.

Because $R$ is orthonormal the inverse is just the transpose. Never invert a rotation matrix by hand.

Sanity check on any transform: rotations preserve length, so a step that changes a distance is wrong.

These two conditions define $SO(3)$. Both are needed.

Columns must be mutually orthogonal AND of unit length: $a^{T}b = 0$, $a^{T}a = b^{T}b = 1$.

$\det R = -1$ is a reflection, not a rotation: it flips handedness.

Orthonormal columns give $R^{-1} = R^{T}$, so inverting a rotation is free.

$R$ has 9 entries and 6 constraints, leaving 3 independent parameters.

Reads a unit quaternion $Q = [\eta, \nu^{T}]^{T}$ backwards: $\eta$ is the scalar part, $\nu$ the vector part.

The factor 2 is the whole point. The quaternion stores the HALF angle, so $\arccos\eta$ gives $\tfrac{\phi}{2}$ and you must double it. Forgetting this is the most common error in both directions.

Normalising $\nu$ recovers the unit axis; the length of $\nu$ is $\sin\tfrac{\phi}{2}$, which carries no axis information.

Every rotation has exactly two unit quaternions, $Q$ and $-Q$. Negating all four entries flips both the axis and the angle, which lands on the same rotation. Convention keeps $\eta$ positive.

Worked example: $Q = [\tfrac{\sqrt2}{2}, -\tfrac{\sqrt2}{2}, 0, 0]^{T}$ gives $\phi = \tfrac{\pi}{2}$ and $\omega = [-1,0,0]^{T}$, a $90°$ rotation about $-x$.

Roll is $\gamma$ about $x$, pitch is $\beta$ about $y$, yaw is $\alpha$ about $z$.

Two valid readings of the same product. RIGHT to LEFT: successive rotations about the FIXED initial axes. LEFT to RIGHT: successive rotations about the ROTATED axes.

Singular at $\beta = \pm\tfrac{\pi}{2}$, where roll and yaw act about the same physical axis. This is gimbal lock.

The rotation axis is untouched by its own rotation, so row 1 and column 1 are the identity.

The minus sign sits in the upper of the two off-diagonal slots, same as $R_z$.

Row 2 and column 2 are the identity: the $y$-axis is the rotation axis.

$R_y$ is the odd one out. Its minus sign is bottom-left, while $R_x$ and $R_z$ carry theirs top-right, a consequence of the cyclic order $x \to y \to z$.

At $\alpha = \pm\tfrac{\pi}{2}$ the cosines vanish, leaving only the two off-diagonal entries and the axis row.

Row 3 and column 3 are the identity: the $z$-axis is the rotation axis.

This is also the planar rotation, $SO(2)$, padded out to three dimensions.

A rotation by angle $\phi$ about unit axis $\omega$, as a four-vector of unit length.

Remember the HALF angle. It is the single most missed detail here.

Inverse: $Q = [\eta, \nu^{T}]^{T}$ gives $Q^{-1} = [\eta, -\nu^{T}]^{T}$. Negate the vector part only, which flips the axis direction.

For $\phi = \pi$ the scalar part vanishes and $Q = [0, \omega^{T}]^{T}$.

$Q$ and $-Q$ describe the same rotation, so every orientation has two unit quaternions.

Free of singularities, which is why game engines store rotations this way.

Seen in Practice exam 3872.

Packs rotation and translation into one $4\times4$ matrix that composes by multiplication.

Rotation block top-left, translation vector top-right, bottom row always $[0,0,0,1]$.

16 entries, 12 of them variable (parameters), 6 degrees of freedom once the 6 orthonormality conditions are applied.

Vectors are right-multiplied, so a chain of transforms is read RIGHT to LEFT for the vector and left to right for the frames.

Not the plain transpose. The rotation block transposes, but the translation becomes $-R^{T} r$.

Same structure as $p = R^{T}(p^{*} - r_C)$ from 2.1, written as one matrix.

ALWAYS check the rotation block first. That single check picks the branch, and picking the wrong one is the main way these questions go wrong.

Translation branch (prismatic joints, $R = I$): $\omega = 0_3$, so $v$ carries the unit-length condition. $\phi$ is the length of the displacement and $v$ is its normalised direction.

Rotation branch (revolute joints): $\bar\omega$ is the normalised rotation axis, $\phi$ is the LENGTH of the un-normalised angular velocity vector, and $v$ comes from the cross product with any point $p$ on the rotation axis.

Why $v$ is not simply $p$: the screw axis is a direction in six dimensions, not a position. A rotation about an axis that misses the origin induces a linear velocity at the origin, and $-(\bar\omega \times p)$ is exactly that.

Any point on the axis works for $p$; coordinates along the axis direction cancel in the cross product.

The equation $v = -(\omega \times r_C) + \dot{r}_C$ is the SAME formula with a term kept for a body that also translates. With $\dot{r}_C = 0_3$ the two are identical; $p$ and $r_C$ both just mean a point given in the world frame.

Either branch: stack $\xi = [\omega^{T}, v^{T}]^{T}$, six entries, angular part on top. The answer is always the pair $\xi\phi$.

Worked example (rotation branch): $\omega = [0,0,\pi]^{T}$ with $p = [1,0,0]^{T}$ gives $\bar\omega = [0,0,1]^{T}$, $\phi = \pi$, $v = -([0,0,1]^{T} \times [1,0,0]^{T}) = [0,-1,0]^{T}$, so $\xi\phi = [0,0,1,0,-1,0]^{T}\pi$.

$\xi$ is a screw axis; $\phi$ is the distance travelled around and along it.

Either $\omega$ has unit length (general motion), or for a PURE TRANSLATION $\omega = 0_3$ and $v$ has unit length.

Pure translation example: displacement of 1 along $x$ is $\xi\phi = [0,0,0,1,0,0]^{T}\cdot 1$.

Interpretation: apply the velocities in $\xi$ for $\phi$ seconds and you get the rigid transformation.

$\xi$ is the DIRECTION (the screw axis) and $\phi$ is the AMOUNT (how far you travel around and along it). Same split as $\omega\phi$ in 2.2: a unit direction times a scalar.

Look at the rotation block FIRST. If $R = I$ there is no rotation, so $\omega = 0_3$ and the motion is a pure translation.

Pure translation: $\phi = \lVert r \rVert$ and $v = r / \lVert r \rVert$, where $r$ is the translation vector.

Example: $T$ with $R = I$ and $r = [1,0,1]^{T}$ gives $\phi = \sqrt{2}$, $v = [\tfrac{\sqrt2}{2}, 0, \tfrac{\sqrt2}{2}]^{T}$, so $\xi\phi = [0,0,0,\tfrac{\sqrt2}{2},0,\tfrac{\sqrt2}{2}]^{T}\sqrt{2}$.

Check: multiplying $\xi$ by $\phi$ must return the original translation in the lower three entries.

Two cases, and the exam asks for both. General motion: the angular part is the unit one. Pure translation: $\omega$ is zero and cannot be normalised, so the job passes to $v$.

Something must be unit length so that $\phi$ carries the magnitude unambiguously. Otherwise the same motion could be written with infinitely many $(\xi, \phi)$ pairs.

Geometric reading: $\xi$ is the bolt, $\phi$ is how far you drive it in.

Velocity reading: apply the velocities packed into $\xi$ for $\phi$ seconds and you get the rigid transformation.

Four parameters per joint: twist $\alpha$, length $a$, offset $d$, angle $\theta$.

With $\alpha = 0$, a pure $\theta$ rotation sits in the top-left $R_z$ block and $d$ shifts along $z$.

Seen in Practice exam 3872.

Maps joint coordinates $q$ to the end-effector pose $x$.

Always single-valued: one configuration gives one pose.

Seen in Practice exam 3872.

One screw axis $\xi_i$ per joint, no chain of link frames needed.

$M$ is the home configuration of the end-effector, at $\phi = 0$.

Seen in Practice exam 3872.

Turns a desired end-effector pose into joint coordinates.

For an $n$-joint arm the solution lives in $\mathbb{R}^{n}$; it may have many or no solutions.

Seen in Practice exam 3872.

Joint velocities needed to deliver a wanted end-effector velocity.

Blows up near singularities, where $J^{-1}$ does not exist.

Seen in Practice exam 3872.

A square Jacobian loses rank; the arm cannot move in some direction.

Rows or columns become linearly dependent (e.g. one is a multiple of another).

Seen in Practice exam 3872.

Forward differential kinematics: joint velocities map to end-effector velocity.

Each column is the partial derivative of the pose with respect to one joint.

Seen in Practice exam 3872.

The transposed Jacobian maps an end-effector force to joint torques at rest.

It is the duality partner of $\dot{X} = J\dot{q}$.

Seen in Practice exam 3872.

The rate of change of acceleration; high jerk stresses actuators and wastes energy.

It affects smoothness, not the maximum velocity.

Seen in Practice exam 3872.

Two position boundary conditions need a degree-one polynomial, so two coefficients.

Plug in $q(0)$ and $q(T)$ to solve for $a_0$ and $a_1$.

Seen in Practice exam 3872.

The linear part of the screw for a body rotating about an axis through $r_C$.

It is the velocity of the body point momentarily at the world origin.

Seen in Practice exam 3872.