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A directional tipping margin is the distance an interior projection of the center of mass can move along one specified direction before reaching a boundary of its support polygon. It differs from the nearest-edge static tipping margin, which protects against movement in every horizontal direction. For a bounded polygon $P=\{x:a_i\cdot x\le b_i\}$, interior point $c$, and unit direction $u$, the ray $c+su$ stays inside exactly while all edge inequalities hold. HenceEdges with nonpositive denominator do not limit positive motion along that ray. The underlying gravity-only feasibility criterion is the nonnegative-reaction convex-hull condition in support polygon.^[ref]

For an original triangular footprint, take $x\ge 0$, $y\ge 0$, $x+y\le 1$ in metres and $c=(0.20,0.30)\mathrm{m}$. Moving right gives $s_{\max }=0.50\mathrm{m}$; moving downward gives $0.30\mathrm{m}$; moving along $u=(1,1)/\sqrt{2}$ gives $0.50/\sqrt{2}\approx 0.3536\mathrm{m}$. The all-direction margin is only $0.20\mathrm{m}$, set by the left edge. A displacement allowance restricted to the right can therefore be larger than the isotropic allowance without contradiction.

The same geometry can enter a planar load calculation. A horizontal force $Fu$ applied at height $h$ shifts the effective support-resultant point by $(Fh/mg)u$ under the corresponding static torque balance. The directional threshold is $F_{\mathrm{tip}}=mg{s}_{\max }/h$, subject also to friction and the other contact assumptions. With $m=10\mathrm{kg}$ and $h=0.50\mathrm{m}$, the rightward example gives $98.1\mathrm{N}$, whereas the nearest-edge worst-direction value is $39.24\mathrm{N}$.

A measured pull direction and contact layout are therefore part of the specification. Changes in contact set, applied couples, or slipping require the fuller analysis in tipping of a rigid body; this geometric directional reserve alone is not a dynamical stability certificate.

A static tipping margin is the distance of an interior center of mass projection $c$ from the boundary of a horizontal support polygon $P$. It quantifies the horizontal displacement reserve in the gravity-only rigid-contact model. For a convex polygon written as $P=\{x:a_i\cdot x\le b_i\}$, with nonzero edge normals $a_i$, the margin isThis is the radius of the largest Euclidean ball about $c$ contained in $P$: each normalized edge inequality bounds how far the projection can move toward that edge. The admissible contact-reaction argument is proved in support polygon and in the primary equilibrium model.^[ref]

For an original rectangular footprint, let $P=[-0.30,0.30]\times [-0.20,0.20]$ in metres and $c=(0.10,0.05)\mathrm{m}$. Its four edge distances are $0.20$, $0.40$, $0.15$, and $0.25\mathrm{m}$, so $d=0.15\mathrm{m}$. Any horizontal projection perturbation of norm at most $0.05\mathrm{m}$ leaves a margin of at least $0.10\mathrm{m}$. This follows because each normalized edge gap changes by at most the perturbation norm.

Unsigned distance alone must not classify an exterior point as stable. At $c=(0.40,0.05)\mathrm{m}$, distance to the polygon boundary is still positive, namely $0.10\mathrm{m}$, but gravity-only balance is impossible. The minimum signed edge gap is then negative. For an exterior point, that minimum is a useful feasibility indicator; its magnitude is not generally the Euclidean distance to the polygon, especially beyond a corner.

A directional tipping margin can exceed the nearest-edge margin when the allowed displacement direction is restricted. The scalar $d$ protects against every horizontal direction, so it is the appropriate geometric reserve only for an isotropic displacement bound. It omits center of mass height, static friction, applied moments, and contact compliance. Under a specified horizontal load, tipping of a rigid body instead compares the load's overturning torque with the restoring gravitational torque.

A support polygon is the convex hull of a body's supporting contact positions, projected onto a horizontal plane. For a rigid body resting on that plane in a uniform gravitational field, with gravity as the only applied load besides support reactions, it describes which horizontal center of mass positions can admit static equilibrium. Widely separated feet can support a center of mass above empty space between them: the footprint's convex hull, rather than the material under every point, matters.^[ref]

The geometry follows directly from force and torque balance. Let the contact positions be $(x_i,y_i,0)$ and their upward normal forces be $N_i\ge 0$. The inequality expresses unilateral contact: an ordinary unbonded support can push the body upward but cannot pull it downward. For total mass $m>0$, gravitational acceleration $g>0$ and horizontal center of mass projection $c=(x_c,y_c)$, balance givesDivide by $mg$ and write ${\lambda }_i=N_i/(mg)$. ThenThus $c$ must be a convex combination of the contacts, which is exactly membership in their convex hull. Conversely, any such convex combination supplies nonnegative reactions $N_i=mg{\lambda }_i$ satisfying this model's balance equations. The center of pressure of these vertical contact forces coincides with $c$. This proves existence of allowable reactions; it does not determine every reaction uniquely when there are more contacts than independent balance equations.

For an original three-foot example, take contact positions $(0,0)$, $(1,0)$ and $(0,1)$ in metres. The support polygon is the triangle described by $x\ge 0$, $y\ge 0$, $x+y\le 1$. A $10\mathrm{kg}$ body with $c=(0.30,0.20)\mathrm{m}$ requires the convex combination coefficientsWith $g=9.81m/{\mathrm{s}}^2$, the corresponding normal forces are $49.05$, $29.43$ and $19.62\mathrm{N}$. Moving the projection to $(0.70,0.50)\mathrm{m}$ would require ${\lambda }_1=-0.20$, a downward reaction at the first foot. No set of nonnegative normal forces can hold that configuration at rest under the stated loads.

A simple test uses a rigid plate on three small supports, a movable known mass, and three load cells measuring the upward reactions. Include the plate's own mass and center of mass when calculating the combined projection. As that projection approaches the edge $x+y=1$, the first support's reading approaches zero. Beyond the edge, maintaining the same contact configuration would require the forbidden downward reaction; the observed motion or changed contact set marks the failure of the original static equilibrium. This is a proposed apparatus test of the balance equations, not reported experimental data.

The distance from an interior projection to the polygon's nearest edge is a static tipping margin; here it is $\min (0.30,0.20,0.50/\sqrt{2})=0.20\mathrm{m}$. A boundary projection has zero margin: some supports unload, and an arbitrarily small outward shift can make balance impossible. A strictly interior projection gives a positive geometric margin against small horizontal shifts under the same model. It is not a general certificate of dynamical stability for a flexible or articulated system.

This explains a constraint in a quadrupedal robot's gait. Four feet at the corners of a square give a square support polygon. Lifting the corner $(1,1)$ leaves the triangle $x\ge 0$, $y\ge 0$, $x+y\le 1$. A center of mass projection at $(0.60,0.60)$ was inside the square but is outside that triangle, so the body must shift or use dynamics to lift that foot without losing the assumed balance. The contact set must be updated when computing the support polygon.

The assumptions matter when applying the test. Nonhorizontal contacts require the full balance equations and admissible contact forces, including Coulomb friction; a projected hull alone need not decide feasibility on irregular terrain.^[ref] Additional horizontal loads also change the balance problem. For example, a horizontal force $F$ applied at center of mass height $h$ produces an overturning torque $Fh$ about the relevant edge. If the inward horizontal distance to that edge is $d$, incipient tipping of a rigid body satisfies $Fh=mgd$. Whether the body slides first depends on static friction. The gravity-only support polygon contains no information about $h$ or that additional load.

Tipping of a rigid body begins when a supported configuration can no longer balance its applied forces and torques with admissible contact forces, and rotation about a support edge becomes possible. For gravity alone on a horizontal surface, the center of mass projection must lie in the support polygon. A boundary projection marks incipient unloading of contacts; it is not a general guarantee about subsequent dynamics. The unilateral-contact and force/torque conditions are explicit in the primary robotics treatment.^[ref]

A planar horizontal pull gives a useful distinction between tipping and sliding. Let the gravitational weight be $mg$, the inward distance of the center of mass projection from the candidate tipping edge be $d>0$, and the pull's application height be $h>0$. Taking torque about that edge gives the limiting balanceThe contact forces elsewhere on the base then unload. With coefficient of static friction $\mu$, the horizontal sliding limit in this same model is $F_{\mathrm{slide}}=\mu mg$. Tipping can occur first only if $d/h<\mu$; the geometric support polygon by itself does not decide this competition.

For an original example, choose $m=20\mathrm{kg}$, $d=0.15\mathrm{m}$, $h=0.60\mathrm{m}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The tipping threshold is $49.05\mathrm{N}$. With $\mu =0.40$, sliding would require $78.48\mathrm{N}$, so tipping is the first predicted loss of this planar static configuration. With $\mu =0.15$, sliding starts at $29.43\mathrm{N}$ instead. Equal thresholds require a coupled contact analysis rather than a strict choice between the two labels.

A bench comparison can use a rigid box, a horizontal force gauge at a known height, and a slow force increase while observing the first loss of full-base contact or translation. Include the box's actual combined center of mass, not just its visible geometric centre. The static tipping margin measures the gravity-only geometric reserve; a pull, anchoring, changing contact set, or nonhorizontal ground changes the governing balance problem. The full friction constraints must then be retained.^[ref]