= Central axis of a wrench
{title2=$\mathbf r=\mathbf F\times\boldsymbol\tau_O/\|\mathbf F\|^2+s\widehat{\mathbf F}$}
The <central axis of a wrench> is the line of reference points about which the resultant <torque> is parallel to the nonzero resultant <force>. It converts a general <wrench (mechanics)> into a <force> along one line plus a parallel <force couple>. A nonzero parallel couple cannot be removed merely by moving the <force>’s application point.
Let $\mathbf F\ne0$ and $\boldsymbol\tau_O$ be the <force> and moment about origin $O$. Define
$$
h=\frac{\mathbf F\cdot\boldsymbol\tau_O}{\|\mathbf F\|^2},\qquad
\mathbf r_0=\frac{\mathbf F\times\boldsymbol\tau_O}{\|\mathbf F\|^2}.
$$
Then the central axis, expressed relative to $O$, is
$$
\boxed{\mathcal L=\{\mathbf r_0+s\widehat{\mathbf F}:s\in\mathbb R\}.}
$$
Here $h$ is the <wrench pitch>. Since $\mathbf r_0\cdot\mathbf F=0$, $\mathbf r_0$ is the point on the axis closest to $O$. The primary robotics wrench representation supplies the <force>/moment translation law used in this construction.<Modern Robotics: Mechanics, Planning, and Control by Kevin M. Lynch and Frank C. Park>[§3.4, Wrenches, transcript: moment as position cross force and frame transformation]{ref}
For an original derivation, write the moment at a point displaced by $\mathbf r$ as $\boldsymbol\tau_{\mathbf r}=\boldsymbol\tau_O-\mathbf r\times\mathbf F$. The triple-product identity gives $\mathbf r_0\times\mathbf F=\boldsymbol\tau_O-h\mathbf F$, hence
$$
\boldsymbol\tau_{\mathbf r}=h\mathbf F-(\mathbf r-\mathbf r_0)\times\mathbf F.
$$
The two terms are perpendicular. Therefore
$$
\boxed{\|\boldsymbol\tau_{\mathbf r}\|^2=h^2\|\mathbf F\|^2+\|\mathbf F\|^2\operatorname{dist}(\mathbf r,\mathcal L)^2.}
$$
This proves both the minimum-moment property and the uniqueness of the minimizing line: the minimum magnitude is $|h|\|\mathbf F\|$, attained exactly on $\mathcal L$. It also quantifies the moment penalty for a perpendicular displacement from that line.
For a numerical example, take $\mathbf F=(10,0,0)\,\mathrm N$ and $\boldsymbol\tau_O=(2,3,-4)\,\mathrm{N\,m}$. Then $h=0.20\,\mathrm m$, $\mathbf r_0=(0,0.40,0.30)\,\mathrm m$, and the axis is parallel to the $x$ direction. The closest-axis distance from the original origin is $0.50\,\mathrm m$. On the axis the moment is $(2,0,0)\,\mathrm{N\,m}$; at a point $0.10\,\mathrm m$ perpendicularly from it, its magnitude is $\sqrt{2^2+1^2}\approx2.236\,\mathrm{N\,m}$. At $O$, it is $\sqrt{29}\,\mathrm{N\,m}$.
If $h=0$, a single <force> on the central axis represents the wrench. If $\mathbf F=0$ and the moment is nonzero, every origin has the same pure couple and this construction has no unique finite axis. With both zero, no direction is selected at all. Near zero <force>, the denominator also makes a reconstructed axis sensitive to measurement uncertainty; an enormous calculated offset need not be a well-resolved physical location.
A proposed six-component load-sensor test applies a known <force> on an offset line and a known parallel couple, then computes the axis from measurements about several reference points. Correctly translated axes should describe the same geometric line even though their coordinate vectors differ. Subtract tare loads and retain axis conventions before comparing the resulting moments; minimum moment does not mean zero moment unless the wrench pitch is zero.
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