Contact mechanics

Under construction.

Rigid body statics: manipulator equilibrium

\dot q = \ddot q = 0

(1)
J^T F = -Q

Duality of force and velocity

(2)
J^T F = -Q
(3)
J \dot q = \dot x

Equivalent forces

Notes in this section are based on chapter 5 of Mason's Mechanics of Robotic Manipulation.

Imagine we apply a force f at a particle in a rigid body at the location x. The force may cause translational acceleration of the body, but may also cause rotational acceleration due to the interaction of internal forces in the body. Choose some reference point O in the body (any one you like). Define the moment of force or torque about this point O to be

(4)
n_o = (x - O) \times f.

Notice that torque is a vector; the magnitude is proportional to the distance of the line of force through x to the point O, and proportional to the magnitude of the force. The direction of the torque vector is perpendicular to the line from O to x and perpendicular to the line of force; the direction may be found using the right-hand rule.

There may be many external forces acting on particles in the system. The net force on the system is the sum of forces, and the net torque around some point to be the sum of the torques around O. It turns out that the motion of the body depends only on the net force and the net torque. (You can choose any reference point you like; there is likely to be a different net torque, but the rotation is also described about the new reference point. The resulting motion of the body is the same either way.)

Line of action

A force acting on a particle is a vector; we just need three numbers to tell us everything we want to know. The effect of a force acting on a rigid body depends on where the force is applied, since the location and angle of the force are factors in computing the torque.

We define a line of action to be the directed line through the particle at which a force is applied, parallel to the force vector.

Notice that all forces of the same magnitude acting along the same line of force generate the same torque about a given point; we say that any two such forces are equivalent.

Resultant of two forces on intersecting lines of action

If two forces are applied to the same rigid body, along two different lines of action, what is the net effect on the body? If the lines of action intersect, just slide the two forces to the intersection point; this will not change the effect of the forces. Now we have two forces acting on a single point; add the force vectors to find a single resultant force. This resultant force is equivalent to the two original forces.

Couples

A couple is a system of forces whose total force is zero. Given a couple, it is easy to construct an equivalent system with just two forces. We have a few theorems:

Theorem 5.1 from Mason: For any reference point O, any system of forces is equivalent to a single force through Q, plus a couple.

Theorem 5.2 from Mason: Every system of forces is equivalent to a system of just two forces.

Wrenches

Poinsot's theorem: Every system of forces is equivalent to a single force, plus a couple with moment parallel to the force.

Mason points out that Poinsot's theorem is equivalent to Chasles's theorem, and gives a formal definition of a wrench, which gives the force and couple in terms of a screw axis. It turns out that the components of a wrench w are very familiar; it's just a force vector together with a moment around some origin:

(5)
w = (f, n_O)

Any system of forces applied to a rigid body has an equivalent wrench. Assume that several fingers are grasping an object. Each finger applies a wrench to the object about some origin. We can sum the wrenches to find a resultant wrench on the object.

Example: supporting a triangle with two frictionless fingers

More about polyhedral convex cones (PCCs): edge representation

We've already seen polyhedral convex cones in the context of form closure for analyzing grasps. Each finger added a constraint on the possible velocities and angular velocities of the body (the possible twists of the body); each constraint was a half-space with boundary thorough the origin of twist space. The intersection of these half spaces was a polyhedral convex cone.

The polyhedral convex cone in twist space was given by the normals to the faces of the cone. One could also describe a PCC by giving the edges to the cone, and this will turn out to be useful in our analysis of how forces are applied at contacts. In this section, we review matrix representations of polyhedral convex cones. Our discussion is based on Goldman and Tucker's 1956 paper, and a 2002 IJRR paper by Balkcom and Trinkle.

Assume matrix F is given. The polar of F is defined as the set of solutions to the matrix inequality F g \le 0, where the inequality applies element-by-element:

(6)
\mathrm{polar}(F) = \{ g: F g \le 0\}.

Any g \in \mathrm{polar}(F) makes a non-positive dot product with each row of F. Each row of F may be interpreted as the normal to the plane bounding a half-space. This plane contains the origin and is included in the half-space described by the corresponding inequality. Thus solutions lie in the intersection of the half-spaces, and \mathrm{polar}(F) is therefore a polyhedral convex cone with apex at the origin. We say that the inequality F g \le 0 is a face normal representation of the cone. This is the representation of a polyhedral convex cone that we have previously seen.

Now assume matrix G is given and define the positive linear span of G:

(7)
\pos(G) = \{ g: g = G z \mathrm{~for~some~} z \ge 0 \}.

Any vector g \in \pos(G) is in the positive linear span of the columns of G, and we say that the equation g = G z together with the inequality constraint z \ge 0 is a span representation of a polyhedral cone. The columns of G are referred to as generators.

Contact wrenches

Assume we contact an object with a frictionless finger. The finger can push on the object, but not pull on it. There is a line of action for the force applied by the finger; the magnitude of the force must be positive. In wrench space, there is a ray of possible wrenches that can be applied by the finger.

Now assume there are multiple fingers. Each finger applies a wrench from a ray in wrench space; the direction of the ray depends on the contact normal and the location and orientation of this normal with respect to the origin. Assume a unit force is applied by each finger; let w_1, w_2, w_3, \ldots w_n be the associated wrenches.

The net wrench w is

(8)
w = w_1 + w_2 + \ldots + w_n

Now assume each contact can apply an arbitrary positive force c_i \ge 0; for example, to resist an externally applied wrench due to gravity.

The net wrench w is

(9)
w = c_1 w_1 + c _2 w_2 + \ldots + c_ n w_n

Notice that w sits in a polyhedral convex cone in wrench space. If the force applied by gravity can be counterbalanced by w, then the object is in equilibrium. Also, we can write the equation in matrix format:

(10)
w = W c

where c is a positive vector. In fact, notice that W, which is computed using cross products, is in fact J^T.

Friction

Contact interactions; contact modes

The direction and magnitude of the frictional force depends on the velocity at the contact point and the force at the contact point. Therefore, if we want to understand what forces are acting on the body, we have to clearly specify the current velocity of the body. We label each of the interactions of interest:

Interaction Normal velocity Tangential velocity Normal force Tangential force
Sliding left zero negative positive positive
sliding right zero positive positive negative
Rolling
Separation
Approach

Example of computations in wrench space

(example from section 4 of Balkcom and Trinkle)

page_revision: 20, last_edited: 1236005746|%e %b %Y, %H:%M %Z (%O ago)
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