Screw theory

The pitch of a pure screw relates rotation about an axis to translation along that axis.

A spatial displacement of a rigid body can be defined by a rotation about a line and a translation along the same line, called a screw displacement. This is known as Chasles' theorem. The six parameters that define a screw displacement are the four independent components of the Plücker vector that defines the screw axis, together with the rotation angle about and linear slide along this line, and form a pair of vectors called a screw. For comparison, the six parameters that define a spatial displacement can also be given by three Euler angles that define the rotation and the three components of the translation vector.

Let a screw be an ordered pair

${\displaystyle {\mathsf {S}}=(\mathbf {S} ,\mathbf {V} ),}$

where S and V are three-dimensional real vectors. The sum and difference of these ordered pairs are computed componentwise. Screws are often called dual vectors.

Now, introduce the ordered pair of real numbers â = (a, b) called a dual scalar. Let the addition and subtraction of these numbers be componentwise, and define multiplication as

${\displaystyle {\hat {a}}{\hat {c}}=(a,b)(c,d)=(ac,ad+bc).\!}$

The multiplication of a screw S = (S, V) by the dual scalar â = (a, b) is computed componentwise to be,

${\displaystyle {\hat {a}}{\mathsf {S}}=(a,b)(\mathbf {S} ,\mathbf {V} )=(a\mathbf {S} ,a\mathbf {V} +b\mathbf {S} ).\!}$

Finally, introduce the dot and cross products of screws by the formulas:

${\displaystyle {\mathsf {S}}\cdot {\mathsf {T}}=(\mathbf {S} ,\mathbf {V} )\cdot (\mathbf {T} ,\mathbf {W} )=(\mathbf {S} \cdot \mathbf {T} ,\,\,\mathbf {S} \cdot \mathbf {W} +\mathbf {V} \cdot \mathbf {T} ),}$

which is a dual scalar, and

${\displaystyle {\mathsf {S}}\times {\mathsf {T}}=(\mathbf {S} ,\mathbf {V} )\times (\mathbf {T} ,\mathbf {W} )=(\mathbf {S} \times \mathbf {T} ,\,\,\mathbf {S} \times \mathbf {W} +\mathbf {V} \times \mathbf {T} ).}$

which is a screw. The dot and cross products of screws satisfy the identities of vector algebra, and allow computations that directly parallel computations in the algebra of vectors.

Let the dual scalar ẑ = (φ, d) define a dual angle, then the infinite series definitions of sine and cosine yield the relations

${\displaystyle \sin {\hat {z}}=(\sin \varphi ,d\cos \varphi ),\,\,\,\cos {\hat {z}}=(\cos \varphi ,-d\sin \varphi ),\!}$

which are also dual scalars. In general, the function of a dual variable is defined to be f(ẑ) = (f(φ), df′(φ)), where f′(φ) is the derivative of f(φ).

These definitions allow the following results:

• Unit screws are Plücker coordinates of a line and satisfy the relation

${\displaystyle |{\mathsf {S}}|={\sqrt {{\mathsf {S}}\cdot {\mathsf {S}}}}=1;}$

• Let ẑ = (φ, d) be the dual angle, where φ is the angle between the axes of S and T around their common normal, and d is the distance between these axes along the common normal, then

${\displaystyle {\mathsf {S}}\cdot {\mathsf {T}}=|{\mathsf {S}}||{\mathsf {T}}|\cos {\hat {z}};}$

• Let N be the unit screw that defines the common normal to the axes of S and T, and ẑ = (φ, d) is the dual angle between these axes, then

${\displaystyle {\mathsf {S}}\times {\mathsf {T}}=|{\mathsf {S}}||{\mathsf {T}}|\sin {\hat {z}}{\mathsf {N}}.}$

A common example of a screw is the wrench associated with a force acting on a rigid body. Let P be the point of application of the force F and let P be the vector locating this point in a fixed frame. The wrench W = (F, P×F) is a screw. The resultant force and moment obtained from all the forces F_i, i = 1,...,n, acting on a rigid body is simply the sum of the individual wrenches W_i, that is

${\displaystyle {\mathsf {R}}=\sum _{i=1}^{n}{\mathsf {W}}_{i}=\sum _{i=1}^{n}(\mathbf {F} _{i},\mathbf {P} _{i}\times \mathbf {F} _{i}).}$

Notice that the case of two equal but opposite forces F and −F acting at points A and B respectively, yields the resultant

${\displaystyle {\mathsf {R}}=(\mathbf {F} -\mathbf {F} ,\mathbf {A} \times \mathbf {F} -\mathbf {B} \times \mathbf {F} )=(0,(\mathbf {A} -\mathbf {B} )\times \mathbf {F} ).}$

This shows that screws of the form

${\displaystyle {\mathsf {M}}=(0,\mathbf {M} ),}$

can be interpreted as pure moments.

In order to define the twist of a rigid body, we must consider its movement defined by the parameterized set of spatial displacements, D(t)=([A(t)],d(t)), where [A] is a rotation matrix and d is a translation vector. This causes a point p that is fixed in moving body coordinates to trace a curve P(t) in the fixed frame given by,

${\displaystyle \mathbf {P} (t)=[A(t)]\mathbf {p} +\mathbf {d} (t).}$

The velocity of P is

${\displaystyle \mathbf {V} _{P}(t)=\left[{\frac {dA(t)}{dt}}\right]\mathbf {p} +\mathbf {v} (t),}$

where v is velocity of the origin of the moving frame, that is dd/dt. Now substitute p =  [A^T](P − d) into this equation to obtain,

${\displaystyle \mathbf {V} _{P}(t)=[\Omega ]\mathbf {P} +\mathbf {v} -[\Omega ]\mathbf {d} \quad {\text{or}}\quad \mathbf {V} _{P}(t)=\mathbf {\omega } \times \mathbf {P} +\mathbf {v} +\mathbf {d} \times \mathbf {\omega } ,}$

where [Ω] = [dA/dt][A^T] is the angular velocity matrix and ω is the angular velocity vector.

The screw

${\displaystyle {\mathsf {T}}=({\vec {\omega }},\mathbf {v} +\mathbf {d} \times {\vec {\omega }}),\!}$

is the twist of the moving body. The vector V = v + d × ω is the velocity of the point in the body that corresponds with the origin of the fixed frame.

There are two important special cases: (i) when d is constant, that is v = 0, then the twist is a pure rotation about a line, then the twist is

${\displaystyle {\mathsf {L}}=(\omega ,\mathbf {d} \times \omega ),}$

and (ii) when [Ω] = 0, that is the body does not rotate but only slides in the direction v, then the twist is a pure slide given by

${\displaystyle {\mathsf {T}}=(0,\mathbf {v} ).}$

The coordinate transformations for screws are easily understood by beginning with the coordinate transformations of the Plücker vector of line, which in turn are obtained from the transformations of the coordinate of points on the line.

Let the displacement of a body be defined by D = ([A], d), where [A] is the rotation matrix and d is the translation vector. Consider the line in the body defined by the two points p and q, which has the Plücker coordinates,

${\displaystyle {\mathsf {q}}=(\mathbf {q} -\mathbf {p} ,\mathbf {p} \times \mathbf {q} ),}$

then in the fixed frame we have the transformed point coordinates P = [A]p + d and Q = [A]q + d, which yield.

${\displaystyle {\mathsf {Q}}=(\mathbf {Q} -\mathbf {P} ,\mathbf {P} \times \mathbf {Q} )=([A](\mathbf {q} -\mathbf {p} ),[A](\mathbf {p} \times \mathbf {q} )+\mathbf {d} \times [A](\mathbf {q} -\mathbf {p} ))}$

Thus, a spatial displacement defines a transformation for Plücker coordinates of lines given by

${\displaystyle {\begin{Bmatrix}\mathbf {Q} -\mathbf {P} \\\mathbf {P} \times \mathbf {Q} \end{Bmatrix}}={\begin{bmatrix}A&0\\DA&A\end{bmatrix}}{\begin{Bmatrix}\mathbf {q} -\mathbf {p} \\\mathbf {p} \times \mathbf {q} \end{Bmatrix}}.}$

The matrix [D] is the skew-symmetric matrix that performs the cross product operation, that is [D]y = d × y.

The 6×6 matrix obtained from the spatial displacement D = ([A], d) can be assembled into the dual matrix

${\displaystyle [{\hat {A}}]=([A],[DA]),}$

which operates on a screw s = (s.v) to obtain,

${\displaystyle {\mathsf {S}}=[{\hat {A}}]{\mathsf {s}},\quad (\mathbf {S} ,\mathbf {V} )=([A],[DA])(\mathbf {s} ,\mathbf {v} )=([A]\mathbf {s} ,[A]\mathbf {v} +[DA]\mathbf {s} ).}$

The dual matrix [Â] = ([A], [DA]) has determinant 1 and is called a dual orthogonal matrix.

Consider the movement of a rigid body defined by the parameterized 4x4 homogeneous transform,

${\displaystyle {\textbf {P}}(t)=[T(t)]{\textbf {p}}={\begin{Bmatrix}{\textbf {P}}\\1\end{Bmatrix}}={\begin{bmatrix}A(t)&{\textbf {d}}(t)\\0&1\end{bmatrix}}{\begin{Bmatrix}{\textbf {p}}\\1\end{Bmatrix}}.}$

This notation does not distinguish between P = (X, Y, Z, 1), and P = (X, Y, Z), which is hopefully clear in context.

The velocity of this movement is defined by computing the velocity of the trajectories of the points in the body,

${\displaystyle {\textbf {V}}_{P}=[{\dot {T}}(t)]{\textbf {p}}={\begin{Bmatrix}{\textbf {V}}_{P}\\0\end{Bmatrix}}={\begin{bmatrix}{\dot {A}}(t)&{\dot {\textbf {d}}}(t)\\0&0\end{bmatrix}}{\begin{Bmatrix}{\textbf {p}}\\1\end{Bmatrix}}.}$

The dot denotes the derivative with respect to time, and because p is constant its derivative is zero.

Substitute the inverse transform for p into the velocity equation to obtain the velocity of P by operating on its trajectory P(t), that is

${\displaystyle {\textbf {V}}_{P}=[{\dot {T}}(t)][T(t)]^{-1}{\textbf {P}}(t)=[S]{\textbf {P}},}$

where

${\displaystyle [S]={\begin{bmatrix}\Omega &-\Omega {\textbf {d}}+{\dot {\textbf {d}}}\\0&0\end{bmatrix}}={\begin{bmatrix}\Omega &\mathbf {d} \times \omega +\mathbf {v} \\0&0\end{bmatrix}}.}$

Recall that [Ω] is the angular velocity matrix. The matrix [S] is an element of the Lie algebra se(3) of the Lie group SE(3) of homogeneous transforms. The components of [S] are the components of the twist screw, and for this reason [S] is also often called a twist.

From the definition of the matrix [S], we can formulate the ordinary differential equation,

${\displaystyle [{\dot {T}}(t)]=[S][T(t)],}$

and ask for the movement [T(t)] that has a constant twist matrix [S]. The solution is the matrix exponential

${\displaystyle [T(t)]=e^{[S]t}.}$

This formulation can be generalized such that given an initial configuration g(0) in SE(n), and a twist ξ in se(n), the homogeneous transformation to a new location and orientation can be computed with the formula,

${\displaystyle g(\theta )=\exp(\xi \theta )g(0),}$

where θ represents the parameters of the transformation.

In transformation geometry, the elemental concept of transformation is the reflection (mathematics). In planar transformations a translation is obtained by reflection in parallel lines, and rotation is obtained by reflection in a pair of intersecting lines. To produce a screw transformation from similar concepts one must use planes in space: the parallel planes must be perpendicular to the screw axis, which is the line of intersection of the intersecting planes that generate the rotation of the screw. Thus four reflections in planes effect a screw transformation. The tradition of inversive geometry borrows some of the ideas of projective geometry and provides a language of transformation that does not depend on analytic geometry.

The combination of a translation with a rotation effected by a screw displacement can be illustrated with the exponential mapping. This idea in transformation geometry was advanced by Sophus Lie more than a century ago. Even earlier, William Rowan Hamilton displayed the versor form of unit quaternions as exp(a r)= cos a + r sin a. The idea is also in Euler's formula parametrizing the unit circle in the complex plane.

Since ε² = 0 for dual numbers, exp(aε) = 1 + aε, all other terms of the exponential series vanishing.

Let F = {1 + εr : r ∈ H}, ε² = 0. Note that F is stable under the rotation q → p ⁻¹ qp and under the translation (1 + εr)(1 + εs) = 1 + ε (r + s) for any vector quaternions r and s. F is a 3-flat in the eight-dimensional space of dual quaternions. This 3-flat F represents space, and the homography constructed, restricted to F, is a screw displacement of space.

Let a be half the angle of the desired turn about axis r, and br half the displacement on the screw axis. Then form z = exp((a + bε)r ) and z* = exp((a − bε)r). Now the homography is

${\displaystyle [q\$ :\ 1]{\begin{pmatrix}z&0\\0&z^{*}\end{pmatrix}}=[qz\ :\ z^{*}]\thicksim [(z^{*})^{-1}qz\ :\ 1].}

The inverse for z* is

${\displaystyle {\frac {1}{\exp(ar-b\varepsilon r)}}=(e^{ar}e^{-br\varepsilon })^{-1}=e^{br\varepsilon }e^{-ar},}$

so, the homography sends q to

${\displaystyle (e^{b\varepsilon }e^{-ar})q(e^{ar}e^{b\varepsilon r})=e^{b\varepsilon r}(e^{-ar}qe^{ar})e^{b\varepsilon r}=e^{2b\varepsilon r}(e^{-ar}qe^{ar}).}$

Now for any quaternion vector p, p* = −p, let q = 1 + pε ∈ F  where the required rotation and translation are effected.

William Kingdon Clifford initiated the use of dual quaternions for kinematics, followed by Aleksandr Kotelnikov, Eduard Study (Geometrie der Dynamen), and Wilhelm Blaschke. However, the point of view of Sophus Lie has recurred.[12] In 1940, Julian Coolidge described the use of dual quaternions for screw displacements on page 261 of A History of Geometrical Methods. He notes the 1885 contribution of Arthur Buchheim.[13] Coolidge based his description simply on the tools Hamilton had used for real quaternions.

Evidently the group of units of the ring of dual quaternions is a Lie group. A subgroup has Lie algebra generated by the parameters a r and b s, where a, b ∈ R, and r, s ∈ H. These six parameters generate a subgroup of the units, the unit sphere. Of course it includes F and the 3-sphere of versors.

Consider the set of forces F₁, F₂ ... F_n act on the points X₁, X₂ ... X_n in a rigid body. The trajectories of X_i, i = 1,...,n are defined by the movement of the rigid body with rotation [A(t)] and the translation d(t) of a reference point in the body, given by

${\displaystyle \mathbf {X} _{i}(t)=[A(t)]\mathbf {x} _{i}+\mathbf {d} (t)\quad i=1,\ldots ,n,}$

where x_i are coordinates in the moving body.

The velocity of each point X_i is

${\displaystyle \mathbf {V} _{i}={\vec {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+\mathbf {v} ,}$

where ω is the angular velocity vector and v is the derivative of d(t).

The work by the forces over the displacement δr_i=v_iδt of each point is given by

${\displaystyle \delta W=\mathbf {F} _{1}\cdot \mathbf {V} _{1}\delta t+\mathbf {F} _{2}\cdot \mathbf {V} _{2}\delta t+\cdots +\mathbf {F} _{n}\cdot \mathbf {V} _{n}\delta t.}$

Define the velocities of each point in terms of the twist of the moving body to obtain

${\displaystyle \delta W=\sum _{i=1}^{n}\mathbf {F} _{i}\cdot ({\vec {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+\mathbf {v} )\delta t.}$

Expand this equation and collect coefficients of ω and v to obtain

${\displaystyle {\begin{aligned}\delta W&=\left(\sum _{i=1}^{n}\mathbf {F} _{i}\right)\cdot \mathbf {d} \times {\vec {\omega }}\delta t+\left(\sum _{i=1}^{n}\mathbf {F} _{i}\right)\cdot \mathbf {v} \delta t+\left(\sum _{i=1}^{n}\mathbf {X} _{i}\times \mathbf {F} _{i}\right)\cdot {\vec {\omega }}\delta t\\[4pt]&=\left(\sum _{i=1}^{n}\mathbf {F} _{i}\right)\cdot (\mathbf {v} +\mathbf {d} \times {\vec {\omega }})\delta t+\left(\sum _{i=1}^{n}\mathbf {X} _{i}\times \mathbf {F} _{i}\right)\cdot {\vec {\omega }}\delta t.\end{aligned}}}$

Introduce the twist of the moving body and the wrench acting on it given by

${\displaystyle {\mathsf {T}}=({\vec {\omega }},\mathbf {d} \times {\vec {\omega }}+\mathbf {v} )=(\mathbf {T} ,\mathbf {T} ^{\circ }),\quad {\mathsf {W}}=\left(\sum _{i=1}^{n}\mathbf {F} _{i},\sum _{i=1}^{n}\mathbf {X} _{i}\times \mathbf {F} _{i}\right)=(\mathbf {W} ,\mathbf {W} ^{\circ }),}$

then work takes the form

${\displaystyle \delta W=(\mathbf {W} \cdot \mathbf {T} ^{\circ }+\mathbf {W} ^{\circ }\cdot \mathbf {T} )\delta t.}$

The 6×6 matrix [Π] is used to simplify the calculation of work using screws, so that

${\displaystyle \delta W=(\mathbf {W} \cdot \mathbf {T} ^{\circ }+\mathbf {W} ^{\circ }\cdot \mathbf {T} )\delta t={\mathsf {W}}[\Pi ]{\mathsf {T}}\delta t,}$

where

${\displaystyle [\Pi ]={\begin{bmatrix}0&I\\I&0\end{bmatrix}},}$

and [I] is the 3×3 identity matrix.

• Screw axis
• Newton–Euler equations uses screws to describe rigid body motions and loading.
• Twist (mathematics)
• Twist (rational trigonometry)

• Joe Rooney William Kingdon Clifford, Department of Design and Innovation, the Open University, London.
• Ravi Banavar notes on Robotics, Geometry and Control