Chapter 2 Solutions
Exercise 2.1.
The first point placed has n degrees of freedom, the next one has one constraint so n −1 degrees of freedom,
the next has two constraints, etc. So n + (n − 1) + (n − 2) + . . . 1 = n(n + 1)/2. (Get this by summing the
outermost pair in the sequence, n + 1 = n + 1, then the pair (n − 1) + 2 = n + 1, etc., and observe that
there are n/2 such pairs.) n of these freedoms are the linear freedoms of placing the first point; the other
n(n −1)/2 are rotational freedoms. After choosing the first point, the next point is on the sphere S
n−1
, the
next is on S
n−2
, etc., so the topology of the space is R
n
× S
n−1
× S
n−2
× . . . × S
1
.
Exercise 2.2.
(a) The shoulder is a spherical joint (four dof), the elbow has one dof, the wrist has two dof, and between
the elbow and the wrist there is one more dof (rotation of the forearm about the axis of the forearm).
Therefore the arm has seven dof.
(b) Placing the palm at a fixed position and orientation in space puts six constraints on the arm (the six
dof of a rigid body). Keeping the center of the shoulder joint stationary, there is only one dof left: the
arc of a circle on which the tip of the elbow can lie. This is one dof, so the arm must have started with
seven dof before six constraints were placed on it.
Exercise 2.3.
Treat the shoulder as a spherical joint (three dof) between the torso and the upper arm bone (humerus),
and assume the carpal bones just beyond the wrist joint form a rigid body. Then the closed-chain linkage
of the forearm between the humerus and the carpal bones, which includes only the radius and the ulna as
links, must have four dof, since our solution in the previous exercise tells us that the arm has seven dof.
We know that each of the radius and the ulna must have at least one joint at the proximal (closer to the torso)
and distal (closer to the hand) ends of forearm, so there are at least four joints between the humerus and
carpal bones. There could be as many as six: three at the elbow (humeroradial, humeroulnar, and proximal
radioulnar) and three at the wrist (radiocarpal, ulnocarpal, and distal radioulnar). Without knowing more
about the anatomy of the arm, we cannot say for sure.
If we assume the maximum number of joints, six, in the forearm closed chain, then the arm has J = 7 joints
(the three-dof S joint at the shoulder and the six forearm joints mentioned above) and N = 5 links (the
torso “ground,” the humerus, the ulna, the radius, and the carpal bones). By Gr¨ubler’s formula,
7 = 6(N − 1 − J) +
7
X
i=1
f
i
= −18 + freedoms of the six forearm joints.
Therefore the six forearm joints must have a total of 25 freedoms. These joints, averaging more than four
freedoms each, are not standard joints we have studied. They are stabilized by a complex of ligaments joining
the bones.
If we assume the minimum number of joints, four, in the forearm closed chain, then the arm has J = 5 joints
and N = 5 links. By Gr¨ubler’s formula,
7 = 6(N − 1 − J) +
5
X
i=1
f
i
= −6 + 3 + freedoms of the four forearm joints.
Therefore there must be a total of 10 freedoms at the four forearm joints. These could potentially be joints
we have studied, such as two universal joints at the elbow (four dof) and two spherical joints at the wrist
(six dof).
The problem is to show correct general reasoning, not to demonstrate a detailed understanding of arm
anatomy!
Exercise 2.4.
Once the hands firmly grip the steering wheel, each arm has n−6 dof if the wheel is stationary. The mobility
5
