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The center of pressure locates the resultant of a distributed fluid pressure load relative to its moment. On a planar surface carrying parallel pressure forces, the first-moment ratio in center of pressure applies. On an airfoil, pressure variation and its resultant can change with the flow state, so the point is not a permanently fixed geometric centre. NASA's primary account explicitly distinguishes a moving center of pressure from retaining a force and pitching moment about an aerodynamic center.^[ref]

For an original hydrostatic plate example, take a vertical rectangle of width $w$, extending from the free surface to depth $H$ in a constant-density fluid. If the same atmospheric pressure acts on its other face, the net gauge pressure at depth $y$ is $p(y)=\rho gy$. This hydrostatic relation follows from vertical force balance.^[ref] The resultant and its depth areFor $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $w=1\mathrm{m}$ and $H=2\mathrm{m}$, the net force is $19.62\mathrm{kN}$ at depth $1.333\mathrm{m}$. The area centroid is only one metre below the surface; the increasing pressure shifts the load's centroid downward.

A tank demonstration can infer the resultant line from the force and hinge moment on a small plate while changing the water depth. The plate's own gravitational moment and the pressure on both faces must be included. This tests a specified pressure-load model, not a general claim that the point always lies below every geometric centre.

For aerodynamic net loads, differences of pressures on opposite faces can have both signs. The convex-hull property proved for nonnegative contact pressure then need not apply, and a small resultant with a finite moment makes its location ill-conditioned. A curved surface or a general wrench may also require a residual couple rather than a single force that removes every component of torque. The resultant convention and moment reference are therefore part of the definition.

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 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 $\mathbf{F}\ne 0$ and ${\mathbf{\tau }}_O$ be the force and moment about origin $O$. DefineThen the central axis, expressed relative to $O$, isHere $h$ 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.^[ref]

For an original derivation, write the moment at a point displaced by $\mathbf{r}$ as ${\mathbf{\tau }}_{\mathbf{r}}={\mathbf{\tau }}_O-\mathbf{r}\times \mathbf{F}$. The triple-product identity gives ${\mathbf{r}}_0\times \mathbf{F}={\mathbf{\tau }}_O-h\mathbf{F}$, henceThe two terms are perpendicular. ThereforeThis proves both the minimum-moment property and the uniqueness of the minimizing line: the minimum magnitude is $|h|\Vert \mathbf{F}\Vert$, 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 ${\mathbf{\tau }}_O=(2,3,-4)\mathrm{N}\mathrm{m}$. 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 $(2,0,0)\mathrm{N}\mathrm{m}$; at a point $0.10\mathrm{m}$ perpendicularly from it, its magnitude is $\sqrt{2^2+1^2}\approx 2.236\mathrm{N}\mathrm{m}$. At $O$, it is $\sqrt{29}\mathrm{N}\mathrm{m}$.

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.

A ground-plane transformation of a measured contact wrench combines a change of coordinate axes with a change of the point about which moments are reported. A six-component wrench is incomplete without both pieces of metadata. Drake's primary SpatialForce reference explicitly tracks the application point and expressed-in frame separately from the stored force and torque values.^[ref]

Let $R$ rotate sensor-frame components into ground-frame components, and let $p$ be the ground-frame vector from ground origin $O$ to sensor origin $S$. For measured force $f_S$ and moment $M_S$ about $S$, the equivalent result about $O$ isThe plus sign uses the stated vector from the new origin to the old one. A shift expressed with the opposite vector uses the opposite sign. The offset and force must be expressed in the same axes before taking the cross product.

For a planar normal-pressure reconstruction with $f_{G,z}>0$, computeThe special parallel-axis height correction in center of pressure is recovered with $p=(0,0,h)$ and $R=I$.

For an original rotated-sensor example, use a 90-degree rotation about $z$, sensor readings $f_S=(10,0,100)\mathrm{N}$ and $M_S=(1,2,3)\mathrm{N}\mathrm{m}$, and $p=(0.05,-0.02,0.10)\mathrm{m}$. Rotation gives $f_G=(0,10,100)\mathrm{N}$ and $R{M}_S=(-2,1,3)\mathrm{N}\mathrm{m}$. The translated moment is $M_O=(-5,-4,3.5)\mathrm{N}\mathrm{m}$, yielding $(x_P,y_P)=(0.040,-0.050)\mathrm{m}$. The yaw moment remaining at that point is $3.1\mathrm{N}\mathrm{m}$, so translating to the centroid does not erase the entire wrench.

A bench check applies known loads at known ground locations and repeats after changing sensor orientation. Geometry, tare loads, units and axis signs must all be included. A correct rotation with the wrong moment origin can still give an incorrect pressure location; a small normal load also amplifies reconstruction uncertainty.

The pitching moment coefficient expresses an aerodynamic pitching torque relative to a chosen flow-load and geometry scale:Here $M$ is the moment about a stated reference point and axis, $q_{\infty }$ the freestream dynamic pressure, $S$ a reference area, and $\bar{c}$ a reference chord length, commonly the mean aerodynamic chord for a finite wing. The denominator has units $\mathrm{Pa}{\mathrm{m}}^2\mathrm{m}=\mathrm{N}\mathrm{m}$, so $C_m$ is dimensionless. The reference point, axis, sign, area and length are all necessary metadata.

An original wind-tunnel example uses $\rho =1.2kg/{\mathrm{m}}^3$, $U_{\infty }=30m/s$, $S=0.80{\mathrm{m}}^2$ and $\bar{c}=0.25\mathrm{m}$. Then $q_{\infty }=540\mathrm{Pa}$ and $q_{\infty }S\bar{c}=108\mathrm{N}\mathrm{m}$. A measured nose-down moment $M=-5.4\mathrm{N}\mathrm{m}$ gives $C_m=-0.050$ when nose-up is positive. At twice the speed, the same coefficient would imply $M=-21.6\mathrm{N}\mathrm{m}$. That scaling is conditional on equal coefficients; a changed Reynolds number, Mach number or flow state can change the coefficient itself.

The reference-point transformation of a pitching moment coefficient follows the wrench translation law. If ${\mathbf{r}}_{AB}$ points from old reference $A$ to new reference $B$ in common axes,Project onto the chosen pitch axis before normalizing. In the chord-line model used in aerodynamic center, take $x$ positive aft, nose-up moment positive, and resultant $N$ perpendicular to the chord. ThenFor $C_m(x_A)=-0.050$, $C_N=0.60$ and an aft reference shift of $0.10\bar{c}$, the new coefficient is $+0.010$. The loads did not change; their reported moment did. Replacing $C_N$ by a lift coefficient requires the appropriate axis transformation or an explicitly justified small-angle approximation.

The aerodynamic center is a reference location about which moment is approximately independent of angle of attack in the stated regime. The center of pressure instead places a resultant so that the relevant moment is zero. NASA distinguishes these locations and reports the near-quarter-chord aerodynamic center as a low-speed approximation for many airfoils.^[ref] An approximately constant nonzero $C_m$ at the aerodynamic center is therefore entirely possible.

For sectional data, moment per unit span $m^{\prime }$ is often normalized by $q_{\infty }{c}^2$. That is consistent with using unit span and area $c$ times that span, but should not be interchanged silently with a finite-wing coefficient using total area and mean aerodynamic chord. A comparison between datasets requires matching these conventions.

A proposed apparatus test mounts a model on a force-and-moment balance, measures freestream speed and density, subtracts support tare, and steps angle of attack while retaining a fixed moment reference. Translating the measured wrench to another reference tests the moment-shift relation without changing the flow. Repeat measurements should track balance uncertainty and the sensitivity of $q_{\infty }$ to speed: a fractional speed error contributes approximately twice that fractional error to dynamic pressure when density is fixed.

The principle of transmissibility states that moving a force along its line of action preserves its external mechanical effect on an ideal rigid body. The force vector is unchanged, and its torque about every reference point is unchanged. The statement concerns external rigid-body motion or equilibrium; it does not assert unchanged internal deformation.

For an original proof, a nonzero force $\mathbf{F}$ is applied at $\mathbf{r}$. Move it to $\mathbf{r}+s\hat{\mathbf{F}}$, where $\hat{\mathbf{F}}=\mathbf{F}/\Vert \mathbf{F}\Vert$ and $s$ is a signed distance. The new moment isThe last cross product vanishes because its vectors are parallel. This is the same moment geometry used in the primary torque definition.^[ref]

For a numerical example, let $\mathbf{F}=(0,10,0)\mathrm{N}$ act at $(0.30,0.20,0)\mathrm{m}$. Its moment is $(0,0,3)\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80,0)$ preserves that moment because the displacement follows the force direction. Moving it instead to $(0.50,0.20,0)$ gives $(0,0,5)\mathrm{N}\mathrm{m}$. The second move needs an additional compensating couple to preserve the original wrench.

A one-dimensional elastic counterexample shows the rigid-body limitation. Consider a uniform bar along $0\le x\le L$, fixed at the left end, with a rightward axial force $F$ applied to a cross-section at $x=a$. In a small-strain axial model, internal force is $F$ on $0<x<a$ and zero on $a<x<L$. Its right-end displacement iswhere $E$ is Young's modulus and $A$ the section area. Applying the same force at $a=L$ instead gives $FL/(EA)$, although both applied forces share one line of action and give the same external resultant and moment.

With stipulated $F=100\mathrm{N}$, $L=1\mathrm{m}$, $E=200\mathrm{GPa}$ and $A=1\mathrm{m}{\mathrm{m}}^2$, the end displacement is $0.25\mathrm{mm}$ for $a=L/2$ and $0.50\mathrm{mm}$ for $a=L$. The linear stress–strain relation underlying that calculation is described in the primary elasticity section.^[ref]

A proposed comparison therefore measures support reaction and body deformation separately when changing load position along a common line. Equal reactions and moments can coexist with different local stress and displacement. The external free-body diagram may be equivalent while the internal loading model differs.

A reference-point transformation of a pitching moment coefficient changes the location about which the aerodynamic torque is reported while leaving the physical loads unchanged. The pitching moment coefficient therefore requires a reference point as well as a sign, axis and normalization.

Let ${\mathbf{r}}_{AB}$ point from the old reference $A$ to the new reference $B$, with force $\mathbf{F}$ and moment vectors expressed in common axes. The wrench translation law isIt follows immediately by replacing each old force arm ${\mathbf{r}}_i$ with ${\mathbf{r}}_i-{\mathbf{r}}_{AB}$ before summing their cross products. The primary Drake spatial-force documentation gives the same old-to-new offset convention and unchanged force component.^[ref]

For a planar example, choose $x$ positive aft, $z$ positive upward and the right-handed pitch axis so positive moment is nose-up. Write $\mathbf{F}=(X,0,N)$ and ${\mathbf{r}}_{AB}=(\Delta x,0,\Delta z)$. Dividing the pitch component by $q_{\infty }S\bar{c}$ givesHere $q_{\infty }$ is dynamic pressure, $S$ the reference area and $\bar{c}$ the reference chord. The equation assumes those normalizations remain the same.

In a constructed case, $C_{m,A}=-0.04$, $C_N=0.70$ and an aft shift $\Delta x=0.15\bar{c}$ at the same height give $C_{m,B}=0.065$. Adding an upward shift $0.02\bar{c}$ with $C_X=0.05$ changes it to 0.064. The aerodynamic loading has not changed between these reports.

A chordwise zero-moment point, when $C_N\ne 0$, satisfiesThis is a center of pressure construction, not the definition of an aerodynamic center. If the normal force vanishes while the moment does not, no finite chordwise shift removes that moment. NASA distinguishes the force-plus-moment representation from placing a resultant at the center of pressure.^[ref] Replacing $C_N$ by a lift coefficient additionally requires transforming between chord-normal and freestream-normal axes; changing a reference point alone performs no such rotation.

A residual contact torque at the center of pressure is the moment left after translating a planar contact's resultant force to its normal-pressure centroid. Let the resultant wrench at an origin in $z=0$ be force $F=(F_x,F_y,N)$ and moment $M=(M_x,M_y,M_z)$, with $N>0$. Put $r_P=(x_P,y_P,0)$, where $x_P=-M_y/N$ and $y_P=M_x/N$. Translation givesThe two horizontal moment components vanish, but the yaw component need not. The zero-moment-point construction for flat robot contact removes moments perpendicular to the surface normal; it does not assert that every component is zero.^[ref]

A useful invariant makes the limitation exact. Since $F\cdot (r_P\times F)=0$,If this scalar is nonzero, no change of the moment origin can eliminate the entire moment vector. Replacing the whole contact by only one force at the center of pressure would discard that information.

For an original two-contact example, put equal normal loads of $100\mathrm{N}$ at $(+0.10,0)$ and $(-0.10,0)$ metres. Add tangential forces $(0,+15,0)\mathrm{N}$ at the positive position and $(0,-15,0)\mathrm{N}$ at the negative position. The total force is $(0,0,200)\mathrm{N}$, the pressure centroid is the origin, and the opposing tangential forces give a pure yaw couple of $3\mathrm{N}\mathrm{m}$. A friction coefficient of at least $0.15$ permits the stipulated tangential magnitudes at these loads.

A normal-load-only reconstruction would identify the correct centroid while missing the couple. A sensor measuring all force and moment components can distinguish the two situations. The origin, surface orientation and axis signs must be retained when comparing measurements. Near zero normal load, the centroid reconstruction itself becomes singular, independently of whether a tangential moment remains.

A wrench represents a system of loads on a rigid body by its resultant force and resultant torque about a stated reference point. In the moment-first convention, writeThe ${\mathbf{C}}_j$ are applied force couples. The six entries package two different physical units: newtons for force and newton-metres for torque. The primary robotics text and its author video use this representation to transform measured forces and moments between reference frames.^[ref]

An original example has a force $(0,10,0)\mathrm{N}$ applied at $(0.30,0,0)\mathrm{m}$, plus a free couple $(0,0,2)\mathrm{N}\mathrm{m}$. Its resultant torque about the origin is $(0,0,5)\mathrm{N}\mathrm{m}$. Shift the reference point to $P=(0.10,0,0)\mathrm{m}$ without rotating the axes. ThenThe physical loading has not changed, but its torque coordinates depend on the reference point. A rotation of axes additionally rotates the vector components.

The dual velocity object is a rigid-body twist, ordered as $(\mathbf{\omega },{\mathbf{v}}_O)$. Their pairing is mechanical power:Since ${\mathbf{v}}_P={\mathbf{v}}_O+\mathbf{\omega }\times {\mathbf{r}}_{OP}$, substituting the shifted torque cancels the extra scalar triple product and preserves power. This explains why changing the force-sensor frame must transform torque and force together.

A nonzero resultant force does not always replace the whole wrench by one force alone. Define wrench pitchThen ${\mathbf{r}}_0\times \mathbf{F}={\mathbf{\tau }}_O-h\mathbf{F}$. A force along the central line through ${\mathbf{r}}_0$ leaves the parallel residual couple $h\mathbf{F}$. For $\mathbf{F}=(0,0,10)\mathrm{N}$ and ${\mathbf{\tau }}_O=(2,0,3)\mathrm{N}\mathrm{m}$, ${\mathbf{r}}_0=(0,0.20,0)\mathrm{m}$ and $h=0.30\mathrm{m}$. The remaining axial couple is $(0,0,3)\mathrm{N}\mathrm{m}$. If the resultant force is zero, a nonzero pure couple must instead be retained directly.

A proposed check uses a calibrated force-torque sensor and known lever arms to compare the reported six components at two reference points. Agreement tests the transformation and loading model; it does not establish that two externally equivalent loads produce identical internal stress in a deformable part.

For a wrench with nonzero resultant force $\mathbf{F}$ and moment ${\mathbf{\tau }}_O$ about a stated point $O$, its wrench pitch isIt 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.^[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 ${\mathbf{\tau }}_P={\mathbf{\tau }}_O-\mathbf{r}\times \mathbf{F}$. 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. DefineThe vector triple-product identity givesEvery point on the line ${\mathbf{r}}_0+s\hat{\mathbf{F}}$ 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 $|h|\Vert \mathbf{F}\Vert$.

For an original example, let $\mathbf{F}=(0,0,20)\mathrm{N}$ and ${\mathbf{\tau }}_O=(4,-2,6)\mathrm{N}\mathrm{m}$. 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 $(0,0,6)\mathrm{N}\mathrm{m}$: a $20\mathrm{N}$ axial force together with a $6\mathrm{N}\mathrm{m}$ axial force couple. No relocation of a single force can eliminate this remaining couple.

If $h=0$ with $\mathbf{F}\ne 0$, 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 $\Vert \mathbf{F}{\Vert }^2$ makes pitch sensitive to measurement error. This geometric load parameter is not automatically a physical screw’s thread spacing.