Friction Cone and Antipodal Grasps

A parallel-jaw gripper holds a coffee mug with two flat pads and nothing else — no glue, no hooks, no suction. The only thing stopping the mug from sliding out between the fingers is friction. The friction cone turns the question “will it slip?” into a piece of geometry: does a force vector point inside a cone?

Put two of those cones on opposite sides of an object and you get the antipodal grasp, the condition behind almost every two-finger grasp planner: squeeze along a line that lies inside both cones, and the object stays put. Squeeze along a line that doesn’t, and no amount of grip force will save you.

---

1. Coulomb Friction at One Contact

When a fingertip presses on an object, the force it applies at the contact point splits into two parts:

• The normal component \(f_n\), along the contact normal \(\mathbf{n}\), the unit vector pointing into the object’s surface. This is the push.
• The tangential component \(\mathbf{f}_t\), in the tangent plane of the surface. This is the part trying to drag the surface sideways.

Coulomb’s law of dry friction says the contact sticks (does not slide) as long as the tangential force does not exceed \(\mu\) times the normal force:

\[\|\mathbf{f}_t\| \le \mu\, f_n, \qquad f_n \ge 0\]

Two things are hidden in this inequality.

• \(\mu\) is the coefficient of static friction, a property of the pair of materials in contact. A rubber pad on cardboard is grippy; a plastic pad on wet glass is not.
• \(f_n \ge 0\): a finger can only push, never pull. Contacts are unilateral.

2. The Friction Cone

Now read the same inequality geometrically. Write the contact force as \(\mathbf{f} = f_n\mathbf{n} + \mathbf{f}_t\), and call \(\varphi\) its angle away from the normal. Then \(\tan\varphi = \|\mathbf{f}_t\| / f_n\), and Coulomb’s law becomes simply

\[\tan\varphi \le \mu \quad\Longleftrightarrow\quad \varphi \le \alpha = \arctan\mu\]

So the set of every force the contact can transmit without slipping is a cone: apex at the contact point, axis along the normal, half-angle \(\alpha = \arctan\mu\).

\[\mathcal{FC} = \left\{\, f_n\mathbf{n} + \mathbf{f}_t \;:\; \|\mathbf{f}_t\| \le \mu\, f_n \,\right\}\]

The crucial property is that only the direction of the force matters, not its magnitude. Pushing twice as hard doubles \(f_n\) and \(\|\mathbf{f}_t\|\) together, so the angle \(\varphi\) never changes. A force inside the cone sticks however hard you push; a force outside it slips however hard you push.

Here is how quickly the cone closes as friction drops:

In 2D the cone is just a wedge with two edges. In 3D it is a circular cone, which is a nonlinear constraint, so grasp-analysis code often replaces it with an inscribed pyramid of \(m\) edges:

\[\mathbf{f} = \sum_{j=1}^{m} \lambda_j\,\big(\mathbf{n} + \mu\,\mathbf{t}_j\big), \qquad \lambda_j \ge 0\]

where the \(\mathbf{t}_j\) are \(m\) unit vectors evenly spaced around the tangent plane. Every allowable force is now a non-negative combination of the edges, which turns grasp checks into linear programs. Because the pyramid sits inside the true cone, the approximation is conservative.

3. Two Fingers: The Antipodal Condition

A parallel-jaw gripper touches the object at two points, \(\mathbf{p}_1\) and \(\mathbf{p}_2\), and squeezes. With no other load on the object, the two contact forces must balance. Two forces balance only if they are equal, opposite, and collinear, so both must act along the line through the contacts:

\[\mathbf{f}_1 = s\,\hat{\mathbf{d}}, \qquad \mathbf{f}_2 = -s\,\hat{\mathbf{d}}, \qquad \hat{\mathbf{d}} = \frac{\mathbf{p}_2 - \mathbf{p}_1}{\|\mathbf{p}_2 - \mathbf{p}_1\|}, \qquad s > 0\]

Each force must sit inside its own friction cone. Writing \(\mathbf{n}_1, \mathbf{n}_2\) for the inward normals at the two contacts:

\[\theta_1 = \arccos\big(\mathbf{n}_1\cdot\hat{\mathbf{d}}\big) \le \alpha, \qquad \theta_2 = \arccos\big(\mathbf{n}_2\cdot(-\hat{\mathbf{d}})\big) \le \alpha\]

In words: the line connecting the two contacts must lie inside both friction cones. That is the antipodal condition. Nguyen (1988) showed that for two fingers it is also sufficient for force closure: if the line lies strictly inside both cones, the grasp can resist any external force or torque on the object, given enough grip force.

The name comes from the globe: antipodal points are diametrically opposite each other. On a sphere every such pair has both normals exactly along the connecting line, \(\theta_1 = \theta_2 = 0\) — the squeeze itself needs no friction at all.

There is one caveat in 3D. Two point contacts, however grippy, cannot resist a torque about the line joining them. Hold a pencil by its two ends between thumb and finger and it spins freely. Real fingers have soft pads that touch over a small patch, and that patch resists the twist with torsional friction — the soft-finger contact model. With soft fingers, the same antipodal condition gives force closure in 3D.

4. Force Closure vs. Form Closure

It helps to contrast this with form closure, where geometry alone locks the object in place, like a peg in a snug hole. Without friction you need at least 4 contacts to immobilize a planar object and 7 for a 3D one. Force closure trades contacts for friction: two soft, frictional fingers do the job, which is why a two-finger gripper works at all.

The price is that force closure is conditional. It holds only while every contact force stays inside its cone, so it depends on \(\mu\), on the contact normals, and (as we’ll see next) on how hard you squeeze. Form closure depends on none of these.

5. A Worked Example: Squeezing a Can Off-Center

Look down from above at a can of radius \(R = 30\) mm. The gripper closes horizontally, but the perception system got the can’s center slightly wrong, so the closing line passes \(e = 10\) mm off the center. Put the center at the origin; the closing line is \(y = e\).

Step 1: Contacts and Normals

The jaws touch the circle where the line \(y = e\) meets it:

\[\mathbf{p}_{1,2} = \Big(\mp\sqrt{R^2 - e^2},\; e\Big) = (\mp 28.28,\; 10)\ \text{mm}\]

On a circle, the inward normal points straight at the center:

\[\mathbf{n}_1 = \frac{-\mathbf{p}_1}{R} = \frac{(28.28,\; -10)}{30} = (0.943,\; -0.333)\]

The closing direction is \(\hat{\mathbf{d}} = (1, 0)\).

Step 2: The Angle Between the Squeeze and the Normal

\[\cos\theta_1 = \mathbf{n}_1 \cdot \hat{\mathbf{d}} = 0.943 \quad\Longrightarrow\quad \theta_1 = 19.47^\circ\]

By symmetry \(\theta_2 = \theta_1\). (In general, for a circle, \(\sin\theta = e/R\).)

Step 3: Compare with the Cone

• Grippy pads, \(\mu = 0.5\): \(\alpha = 26.57^\circ\). Since \(19.47^\circ < 26.57^\circ\), the closing line is inside both cones. The grasp is antipodal, with \(7.1^\circ\) to spare.
• Slippery can, \(\mu = 0.2\): \(\alpha = 11.31^\circ\). Since \(19.47^\circ > 11.31^\circ\), the line is outside both cones. The squeeze itself pushes the can sideways out of the jaws.

Equivalently, a squeeze at angle \(\theta\) needs \(\mu \ge \tan\theta\):

\[\mu_{\text{required}} = \tan\theta = \frac{e}{\sqrt{R^2 - e^2}} = \frac{10}{28.28} = 0.354\]

Turning it around, the largest offset the gripper can tolerate is \(e_{\max} = R\sin\alpha = R\mu/\sqrt{1+\mu^2}\): 13.4 mm for \(\mu = 0.5\) but only 5.9 mm for \(\mu = 0.2\). The slippery can demands placement 2.3× more accurate.

Step 4: Now Lift It

So far the only force was the squeeze. Now pick the can up: mass \(m = 0.5\) kg, weight \(mg = 4.905\) N, shared by the two pads, so each contact must also supply \(2.45\) N of vertical friction. At each pad the normal force is \(N\), and the tangential force now has two perpendicular parts:

• In-plane: \(N\tan\theta\), the sideways part of the tilted squeeze.
• Vertical: \(mg/2\), holding the weight.

Coulomb’s law says their combined magnitude must stay under \(\mu N\):

\[\sqrt{(N\tan\theta)^2 + \left(\tfrac{mg}{2}\right)^2} \;\le\; \mu N \quad\Longrightarrow\quad N \;\ge\; \frac{mg/2}{\sqrt{\mu^2 - \tan^2\theta}}\]

Plug in \(\mu = 0.5\):

• Centered (\(\theta = 0\)): \(N \ge 2.4525 / 0.5 = 4.91\) N.
• Off by 10 mm (\(\tan^2\theta = 0.125\)): \(\mu^2 - \tan^2\theta = 0.25 - 0.125 = 0.125\), so \(N \ge 2.4525 / 0.354 = 6.94\) N — 41% more grip for the same can. The jaw itself must squeeze with \(s = N/\cos\theta = 7.36\) N.

Push the offset further and watch what happens:

The off-center squeeze eats the friction budget that would otherwise hold the weight. Near the edge of the cone the required grip force diverges: a grasp that is technically antipodal can still need more force than the gripper has, or more than the object can take.

6. What This Means for a Grasp Planner

A planner never sees the true normals. It estimates them from a depth map or a point cloud, then checks the antipodal condition with those estimates. If the estimated normal is off by \(\delta\), the true angle can be off by up to \(\delta\) too, so the test that actually matters is

\[\theta_{\text{true}} \;\le\; \theta_{\text{est}} + \delta \;\le\; \alpha\]

The useful quantity is therefore the margin \(\alpha - \theta_{\text{est}}\). A grasp accepted with \(20^\circ\) of margin survives a \(15^\circ\) normal error; one accepted right at the edge of the cone survives nothing. Good planners rank candidates by margin, or plan with a deliberately narrower cone (\(\mu_{\text{plan}} < \mu_{\text{true}}\)). Grasp-quality metrics such as the Ferrari–Canny \(\epsilon\) metric — the radius of the largest ball of disturbance forces and torques the grasp can resist with a unit budget of grip force — generalize the same idea to many contacts.

Two more unknowns hide in that inequality. The planner has to guess \(\mu\): a camera that cannot tell a rubber ball from a glass ornament has to assume the narrower cone. And surface normals are derivatives of depth, so a few millimeters of depth noise across a small contact patch can turn into tens of degrees of normal error, which eats the whole margin.

Real grippers bend the rules a little. An object resting on a table is free to move as the jaws close: it often rotates or slides until its contacts become antipodal on their own. That is why real grippers frequently tolerate misalignments larger than \(\arctan\mu\) would suggest. But the forgiveness comes from the object moving, and the price is that you no longer know exactly where it sits in the hand.

Intuitively: The Wet Bar of Soap

Squeeze a dry bar of soap between thumb and finger and it stays put even if your fingers are a little off. The friction is high, the cones are wide, and almost any squeeze line fits inside them.

Now wet it. \(\mu\) drops to maybe 0.1, and each cone shrinks to a needle of \(\pm 6^\circ\). The slightest misalignment — a gently curved face, a finger a millimeter off — puts the squeeze line outside the cone, and the part of your own grip force that runs along the surface shoots the soap out of your hand.

Squeezing harder makes it worse, not better. You scale the normal and tangential parts together, so the angle never changes and the sideways push only grows. That is the friction cone in one sentence: for the squeeze itself, direction is everything and strength is nothing. A grasp whose line of action points outside the cone can only be fixed by better geometry or more friction, which is exactly why a grasping robot cares so much about seeing the surface normal accurately.