= Wrench pitch
{title2=$h=\mathbf F\cdot\boldsymbol\tau/\|\mathbf F\|^2$}
For a <wrench (mechanics)> with nonzero resultant <force> $\mathbf F$ and moment $\boldsymbol\tau_O$ about a stated point $O$, its <wrench pitch> is
$$
\boxed{h=\frac{\mathbf F\cdot\boldsymbol\tau_O}{\|\mathbf F\|^2}.}
$$
It has units of length: <torque> divided by <force>. It measures the signed moment remaining parallel to the resultant, per unit <force>. The <force>/moment representation and its frame transformations are developed in the primary robotics text’s wrench supplement.<Modern Robotics: Mechanics, Planning, and Control by Kevin M. Lynch and Frank C. Park>[§3.4, Wrenches, transcript: force/moment six-vector, reference-frame change and power pairing]{ref}
The <wrench pitch> does not depend on the moment origin. Moving from $O$ to $P$, with displacement $\mathbf r=\overrightarrow{OP}$ expressed in the same axes, gives $\boldsymbol\tau_P=\boldsymbol\tau_O-\mathbf r\times\mathbf F$. Taking the <dot product> with $\mathbf F$ removes the added term because $\mathbf F\cdot(\mathbf r\times\mathbf F)=0$. A common rotation of both vectors also preserves the ratio. Scaling every applied load by the same nonzero scalar leaves $h$ unchanged.
The <central axis of a wrench> makes the meaning visible. Define
$$
\mathbf r_0=\frac{\mathbf F\times\boldsymbol\tau_O}{\|\mathbf F\|^2}.
$$
The vector triple-product identity gives
$$
\mathbf r_0\times\mathbf F=\boldsymbol\tau_O-h\mathbf F,
\qquad \boldsymbol\tau_{\mathbf r_0}=h\mathbf F.
$$
Every point on the line $\mathbf r_0+s\widehat{\mathbf F}$ has that same parallel residual moment. This line also minimizes the moment magnitude over all choices of origin: the perpendicular part can be removed, while the invariant parallel part cannot. The minimum is $|h|\|\mathbf F\|$.
For an original example, let $\mathbf F=(0,0,20)\,\mathrm N$ and $\boldsymbol\tau_O=(4,-2,6)\,\mathrm{N\,m}$. Then $h=0.30\,\mathrm m$ and $\mathbf r_0=(0.10,0.20,0)\,\mathrm m$. Translating the moment origin to that point leaves $(0,0,6)\,\mathrm{N\,m}$: a $20\,\mathrm N$ axial <force> together with a $6\,\mathrm{N\,m}$ axial <force couple>. No relocation of a single <force> can eliminate this remaining couple.
If $h=0$ with $\mathbf F\ne0$, the central-axis moment vanishes and one <force> on that line reproduces the external load system. If $\mathbf F=0$ and the moment is nonzero, the load is a pure <force couple>; the formula for $h$ is undefined. An infinite-pitch description is a limiting screw-theory convention, not a finite value supplied by this quotient. Both zero <force> and zero moment give the zero wrench and no distinguished axis.
A proposed load-sensor check combines a known <force> on an offset line with a known parallel couple. Recompute moments about several origins and compare the resulting $h$ values. Errors in origin translation can corrupt the moment components while the exact invariant remains fixed; near zero resultant <force>, the division by $\|\mathbf F\|^2$ makes pitch sensitive to measurement error. This geometric load parameter is not automatically a physical screw’s thread spacing.
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