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Let $R_t$ be the horizontal rotation matrix through angle $t$. Define the screw motionIts linear part is an orthogonal matrix with determinant one, and its remaining part is a translation. Thus it is a proper Euclidean motion of Euclidean three-space, with ${\Phi }_s{\Phi }_t={\Phi }_{s+t}$.

In horizontal-vector notation the given parametric surface satisfies $\mathbf{r}(u,v)=(R_v(1+\cos u,\sin u),v)$. ThereforeThe defining circle in each horizontal plane is entirely parametrized by $u$, so $\mathbf{r}$ covers $X$. The displayed equality and ${\Phi }_{-t}={\Phi }_t^{-1}$ prove ${\Phi }_t(X)=X$. This one-parameter subgroup preserves the dot products and cross products defining $I$, $II$ and the chosen unit normal. Consequently their coordinate coefficients, and the Gaussian curvature, depend only on $u$.

Use the unit normal from part (c), and orient ${\gamma }_a$ by increasing $v$. Write $T={\gamma }_a^{\prime }/|{\gamma }_a^{\prime }|$ for its unit tangent vector. The signed geodesic curvature convention used below iswhere $s$ is arc length. The second equality follows by differentiating $T$: the term parallel to ${\gamma }_a^{\prime }$ disappears in the scalar triple product. It makes the correction for a parameter which is not arc-length parametrization explicit.

Put $C=\cos a$, $S=\sin a$, $D=1+S^2$, and $G=3+2C$. The speed is the constant $\sqrt{G}$. At $v=0$,soTaking its dot product with $N=(C,S,-S)/\sqrt{D}$ yields $-S(3+2C)/\sqrt{D}=-SG/\sqrt{D}$. ThereforeThe screw motion of part (b) makes this independent of $v$. Thus ${\gamma }_a$ is a geodesic precisely when $a$ is an integer multiple of $\pi$. Reversing the direction of traversal or the unit normal changes the sign of signed geodesic curvature; its magnitude is the absolute value of the displayed expression. The signed convention is needed in part (e).

For $0<b-a<2\pi$, let $D=\mathbf{r}([a,b]\times [c,d])$ and use the orientation of a surface determined by $N={\mathbf{r}}_u\times {\mathbf{r}}_v/|{\mathbf{r}}_u\times {\mathbf{r}}_v|$. Its positive boundary orientation follows the bottom side in increasing $u$, the right side in increasing $v$, the top side in decreasing $u$, and the left side in decreasing $v$. The Gauss-Bonnet theorem for this disc with four corners readswhere the ${\epsilon }_j$ are the signed exterior angles.

The bottom and top sides have cancelling geodesic curvature integrals: a screw motion takes one to the other with opposite boundary orientation. Explicitly, along increasing $u$ at fixed $v$, the speed is one and the signed geodesic curvature is $-\sin u/\sqrt{1+{\sin }^{2}u}$ on both sides. The right and left sides contribute, respectively, the integral along ${\gamma }_b$ and minus the integral along ${\gamma }_a$.

Although the coordinate sides are generally not perpendicular, the corner contributions still sum to $2\pi$. Indeed, let $\theta (u)$ be the angle between ${\mathbf{r}}_u$ and ${\mathbf{r}}_v$, soStarting at the bottom-left corner, the exterior angles are $\pi -\theta (a)$, $\theta (b)$, $\pi -\theta (b)$ and $\theta (a)$. Their sum is $2\pi$. Cancelling this from the Gauss-Bonnet theorem leaveswhich is the identity in part (e). For a rectangle covering more than one turn, apply the same argument on the parameter rectangle with its induced Riemannian metric, or partition it into narrower rectangles and cancel the artificial internal boundary contributions. This gives the version counting multiplicity, as required by the correction in part (e).

The screw pitch of a rigid-body twist is its signed axial translation per unit rotation angle about the instantaneous motion axis. For angular velocity $\mathbf{\omega }\ne 0$ and linear twist component $\mathbf{v}$ in the same reference frame,The units are length per radian. This mechanics convention differs from using the word pitch for an axial distance per complete revolution.^[ref]

To derive the formula, write the instantaneous velocity field as $\mathbf{u}(\mathbf{x})=\mathbf{v}+\mathbf{\omega }\times \mathbf{x}$. At a point $\mathbf{q}$ on the instantaneous axis, the velocity is parallel to $\mathbf{\omega }$, so $\mathbf{u}(\mathbf{q})=h\mathbf{\omega }$. Taking its scalar product with $\mathbf{\omega }$ removes the cross-product term and gives the displayed result. A suitable closest axis point iswhich verifies $\mathbf{u}(\mathbf{q})=h\mathbf{\omega }$ directly.

Although $\mathbf{v}$ depends on the chosen origin, $h$ does not. Translating that origin by $\mathbf{a}$ replaces $\mathbf{v}$ by $\mathbf{v}+\mathbf{\omega }\times \mathbf{a}$; the added term is perpendicular to $\mathbf{\omega }$. A common rotation of the coordinate axes preserves the scalar product and the norm as well. Multiplying the entire rigid-body twist by a nonzero scalar changes the motion rate but preserves $h$.

For an original mechanism with constant axial advance $2\mathrm{mm}$ per full turn,At five revolutions per second, $\dot{\theta }=10\pi rad/s$ and the axial speed is $h\dot{\theta }=10mm/s$. The advance per revolution is $2\pi h$, which corresponds to screw lead. In a multiple-start screw thread, the spacing between neighbouring threads need not equal that lead, so a thread's engineering pitch must not silently replace the twist parameter.

When $h=0$ and $\mathbf{\omega }\ne 0$, the instantaneous motion is pure rotation about an axis, even if points away from that axis have nonzero linear velocity. When $\mathbf{\omega }=0$ and $\mathbf{v}\ne 0$, the motion is pure rigid-body translation and the displayed quotient is undefined; an infinite-pitch convention can describe that limiting case. A finite nonzero constant pitch produces a screw motion, with off-axis point paths following helices.

A rigid-body twist represents an instantaneous rigid body velocity field by a pair of three-dimensional vectors, conventionally ordered $(\mathbf{\omega },\mathbf{v})$. In a stated reference frame, $\mathbf{\omega }$ is angular velocity and $\mathbf{v}$ is the velocity field evaluated at that frame's origin. Every point at position $\mathbf{x}$ has instantaneous velocityThe linear component therefore need not be the center of mass velocity. Both components must use the same coordinate frame; their physical units also differ.^[ref]

The formula follows by differentiating the point position $\mathbf{x}=R(t)\mathbf{X}+\mathbf{p}(t)$, where $R$ is a rotation matrix and $\mathbf{X}$ is fixed in the body. The skew-symmetric matrix $\dot{R}{R}^T$ acts as the cross product with $\mathbf{\omega }$. HenceAt the body's chosen origin $\mathbf{x}=\mathbf{p}$, this correctly returns $\dot{\mathbf{p}}$. Moving the reference origin by a fixed vector $\mathbf{a}$, without changing its axes, gives ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{\omega }\times \mathbf{a}$, with the same angular velocity. The numbers change while the physical velocity field is unchanged.

When $\mathbf{\omega }\ne 0$, define the screw pitch and a point on the instantaneous axis byThe vector triple-product identity gives $\mathbf{u}(\mathbf{q})=h\mathbf{\omega }$. Points on $\mathbf{q}+\lambda \mathbf{\omega }$ share that axial velocity; off-axis points additionally rotate about it. A constant rigid-body twist therefore generates a screw motion. Zero angular velocity instead gives pure rigid-body translation, for which these divisions are not defined.

For an original numerical example take $\mathbf{\omega }=(0,0,2)rad/s$ and $\mathbf{v}=(1,0,0.4)m/s$. At $(0,1,0)\mathrm{m}$ the velocity is $(-1,0,0.4)m/s$. The axis passes through $\mathbf{q}=(0,0.5,0)\mathrm{m}$, with axial speed $0.4m/s$ and screw pitch $h=0.2m/rad$. If the twist stays constant for a revolution, the axial advance is $2\pi h\simeq 1.257\mathrm{m}$.

A twist describes instantaneous motion; it is not generally a six-coordinate finite pose or its ordinary derivative. Its reference-frame convention matters when using a robot's measured joint rates to predict endpoint velocity.