3 →
R3 denote a rigid motion of
R3, and let
v denote an arbitrary vector of
R3. Now let
v0 denote the vector f(
0). We may now define the rigid motion rot:
R3 →
R3 via rot(
v) := f(
v) −
v0 for all
v in
R3. This guarantees that rot(
0) =
0. Since rot(
0) =
0 (and being a motion of
R3, rot preserves orientation), rot must be a
rotation.
If rot is equal to the identity
I, then in this very simple case, the screw motion is merely translation by the original vector
v0 and does not involve a rotation. From here on we assume rot is not the identity
I. We know that each rotation of
R3 must have an axis (a bi-infinite straight line) that is pointwise fixed by the rotation. Call that line
L. Therefore the original rigid motion f satisfies rot(
v) = f(
v) −
v0 for all
v in
R3, and hence :f(
v) = rot(
v) +
v0 for all
v in
R3. This means that f is the result of applying a rotation rot, followed by a translation by
v0. Finally, we may uniquely resolve the vector
v0 into a sum of two vectors, one parallel to the line
L, and the other perpendicular to
L, as follows: :
v0 =
vL +
v⊥. We may now add the translation vector
v0 by first adding
v⊥, and next adding
vL, as follows: :f(
v) = (rot(
v) +
v⊥) +
vL Now, the first portion (rot(
v) +
v⊥) of f(
v) may be thought of a taking place purely within any given plane
P perpendicular to
L, where
v⊥ is a vector in the plane
P. In fact, define :f1(
v) := rot(
v) +
v⊥, where both the rotation rot and the vector
v⊥ are thought of as lying in the plane
P. Then, there is a unique point
x in the plane
P such that f1(
x) = rot(
x) +
v⊥ = x. Or in other words, a unique point
x in
P that is fixed by the mapping f1. Since we want rot(
x) +
v⊥ =
x, or in other words (
I - rot)
x =
v⊥ (where
I is the identity on
P), we may easily solve this uniquely for
x by writing :
x = (
I - rot)−1(
v⊥). This makes sense as long as (
I − rot) is invertible, or in other words as long as rot is not equal to the identity
I, and we excluded that case above. Since
f1:
P →
P is a rigid motion of a plane with a fixed point (
f1(
x) =
x), it must be a rotation of
P. Hence the effect on
R3 of mapping any
v to rot(
v) +
v⊥ is just a rotation (by the same angle as rot) about the line
L1 that is parallel to
L, and passes through the point
x. Recall that he original rigid motion
f on
R3 is given by f(
v) = (rot(
v) +
v⊥). This implies that the original rigid motion
f is the same as first rotating
R3 about the line
L1, and following this by translation by the vector
vL (which is in the same direction as
L1). Conclusion: every rigid motion of
R3 is the result of a rotation of
R3 followed by a translation along the axis of the rotation — which is the definition of a screw motion. -->
Geometric argument Let be an orientation-preserving rigid motion of
R3. The set of these transformations is a subgroup of
Euclidean motions known as the special Euclidean group SE(3). These rigid motions are defined by transformations of
x in
R3 given by : D(\mathbf{x})=A(\mathbf{x}) + \mathbf{d} consisting of a three-dimensional rotation
A followed by a translation by the vector
d. A three-dimensional
rotation A has a unique axis that defines a line
L. Let the unit vector along this line be
S so that the translation vector
d can be resolved into a sum of two vectors, one parallel and one perpendicular to the axis
L, that is, :\mathbf{d}=\mathbf{d}_L + \mathbf{d}_{\perp},\quad \mathbf{d}_L =(\mathbf{d}\cdot\mathbf{S})\mathbf{S}, \quad \mathbf{d}_{\perp}=\mathbf{d}- \mathbf{d}_L. In this case, the rigid motion takes the form : D(\mathbf{x})=(A(\mathbf{x}) + \mathbf{d}_{\perp}) + \mathbf{d}_L. Now, the orientation preserving rigid motion
D* =
A(
x) +
d⊥ transforms all the points of
R3 so that they remain in planes perpendicular to
L. For a rigid motion of this type there is a unique point
c in the plane
P perpendicular to
L through
0, such that : D^*(\mathbf{C})=A(\mathbf{C})+\mathbf{d}_{\perp}=\mathbf{C}. The point
C can be calculated as : \mathbf{C}=[I-A]^{-1}\mathbf{d}_{\perp}, because
d⊥ does not have a component in the direction of the axis of
A. A rigid motion
D* with a fixed point must be a rotation of around the axis
Lc through the point
c. Therefore, the rigid motion : D(\mathbf{x})=D^*(\mathbf{x}) + \mathbf{d}_L, consists of a rotation about the line
Lc followed by a translation by the vector
dL in the direction of the line
Lc. Conclusion: every rigid motion of
R3 is the result of a rotation of
R3 about a line
Lc followed by a translation in the direction of the line. The combination of a rotation about a line and translation along the line is called a screw motion.
Computing a point on the screw axis A point
C on the screw axis satisfies the equation: : D^*(\mathbf{C})=A(\mathbf{C})+\mathbf{d}_{\perp}=\mathbf{C}. Solve this equation for
C using
Cayley's formula for a rotation matrix : [A]=[I-B]^{-1}[I+B], where [B] is the skew-symmetric matrix constructed from
Rodrigues' vector : \mathbf{b}=\tan\frac{\phi}{2}\mathbf{S}, such that :[B]\mathbf{y}=\mathbf{b}\times\mathbf{y}. Use this form of the rotation
A to obtain :\mathbf{C} =[I-B]^{-1}[I+B]\mathbf{C} + \mathbf{d}_{\perp},\quad [I-B]\mathbf{C} =[I+B]\mathbf{C} + [I-B]\mathbf{d}_{\perp}, which becomes : -2[B]\mathbf{C} =[I-B]\mathbf{d}_{\perp}. This equation can be solved for
C on the screw axis
P(t) to obtain, : \mathbf{C} = \frac{\mathbf{b}\times\mathbf{d} - \mathbf{b}\times(\mathbf{b}\times\mathbf{d})}{2\mathbf{b}\cdot\mathbf{b}}. The screw axis of this spatial displacement has the
Plücker coordinates . ==Dual quaternion==