Velocity Kinematics

Notation convention warning of this lecture: if a vector has no superscript, it means this vector is expressed in the world frame by default; otherwise, the frame the coordinates of this vector is with respect to will be specified on its superscript.

Derivative of Rotation Matrix

Consider a time-varying rotation matrix \(\boldsymbol{R}=\boldsymbol{R}(t)\). Based on the property of rotation matrix, we always have

\[\boldsymbol{R}(t) \boldsymbol{R}^{T}(t)=\boldsymbol{I}\]

Differentiating the above equation with respect to time gives

\[\dot{\boldsymbol{R}}(t) \boldsymbol{R}^{T}(t)+\boldsymbol{R}(t) \dot{\boldsymbol{R}}^{T}(t)=\boldsymbol{0}\]

Let’s define a new matrix

(4)\[\boldsymbol{S}=\dot{\boldsymbol{R}} \boldsymbol{R}^{T}\]

Then,

\[ \boldsymbol{S}+\boldsymbol{S}^{T}=\boldsymbol{0} \]

Any matrix that satisfies the above equation is called skew-symmetric matrix (a square matrix whose transpose equals its negative).

From (4), we have

(5)\[\dot{\boldsymbol{R}}=\boldsymbol{S} \boldsymbol{R}\]

Now, let’s find out what is \(\boldsymbol{S}\) and its physics interpretation.

Suppose \({\boldsymbol{R}}\) represents a rotation of a moving body frame \(O'-x'y'z'\) with respect to a fixed reference frame \(O-xyz\). Let’s consider a point \(P\), rigidly fixed to the body frame (thus moving as body is moving), with its body frame coordinate \(\boldsymbol{p}^{\prime}\). Then, its coordinate in the reference frame \(\boldsymbol{p}\) is

(6)\[\boldsymbol{p}(t)=\boldsymbol{R}(t) \boldsymbol{p}^{\prime}\]

which is time varying.

Taking the derivative to both sides of (6) yields

(7)\[\dot{\boldsymbol{p}}=\dot{\boldsymbol{R}} \boldsymbol{p}^{\prime}=\boldsymbol{S} \boldsymbol{R} \boldsymbol{p}^{\prime}\]

where we have used (4).

Recall in your mechanics courses, given the angular velocity \(\boldsymbol{\omega}\) of a moving body with respect to a reference frame, any point \(P\) fixed on this body has a velocity in the reference frame as

(8)\[ \dot{\boldsymbol{p}}=\boldsymbol{\omega} \times \boldsymbol{p}=\boldsymbol{\omega} \times \boldsymbol{R} \boldsymbol{p}^{\prime} \]

where \(\boldsymbol{\omega}=\left[\begin{array}{lll}\omega_{x} & \omega_{y} & \omega_{z}\end{array}\right]^{T}\) is the angular velocity of the moving body, expressed in the reference frame.

If we compare (7) and (8), we can find

\[\begin{split}\boldsymbol{S}=[\boldsymbol{\omega} \times]=\boldsymbol{S}(\boldsymbol{\omega})=\left[\begin{array}{ccc} 0 & -\omega_{z} & \omega_{y} \\ \omega_{z} & 0 & -\omega_{x} \\ -\omega_{y} & \omega_{x} & 0 \end{array}\right]\end{split}\]

Thus, this skew-symmetric matrix \(\boldsymbol{S}\) can be obtained directly from the body’s angular velocity \(\boldsymbol{\omega}\). We typically write it as \(\boldsymbol{S}(\boldsymbol{\omega})\).

In sum, we can conclude the derivative of a rotation matrix is

\[\dot{\boldsymbol{R}}=\boldsymbol{S}(\boldsymbol{\omega}) \boldsymbol{R}\]

Note that the above angular velocity \(\boldsymbol{\omega}\) is expressed in the reference frame!

One property of \(\boldsymbol{S}(\boldsymbol{\omega})\): For any rotation matrix \(\boldsymbol{R}\), one has

(9)\[\boldsymbol{R} \boldsymbol{S}(\boldsymbol{\omega}) \boldsymbol{R}^{T}=\boldsymbol{S}(\boldsymbol{R} \boldsymbol{\omega})\]

Derivative of Rotation + Translation

Consider a transformation mapping the coordinate mapping of a point \(P\) (NOT necessarily fixed on a moving body) from the body frame \(O_1-x_1y_1z_1\) to reference frame \(O_0-x_0y_0z_0\)

\[\boldsymbol{p}^{0}=\boldsymbol{o}_{1}^{0}(t)+\boldsymbol{R}_{1}^{0}(t) \boldsymbol{p}^{1}\]

Differentiating the above equation with respect to time yields

(10)\[\begin{split} \begin{aligned} \dot{\boldsymbol{p}}^{0}=\dot{\boldsymbol{o}}_{1}^{0}+\boldsymbol{R}_{1}^{0} \dot{\boldsymbol{p}}^{1}+\dot{\boldsymbol{R}}_{1}^{0} \boldsymbol{p}^{1}&=\dot{\boldsymbol{o}}_{1}^{0}+\boldsymbol{R}_{1}^{0} \dot{{\boldsymbol{p}}}^{1}+\boldsymbol{S}\left(\boldsymbol{\omega}_{1}^{0}\right) \boldsymbol{R}_{1}^{0} \boldsymbol{p}^{1}\\ &=\dot{\boldsymbol{o}}_{1}^{0}+\boldsymbol{R}_{1}^{0} \dot{{\boldsymbol{p}}}^{1}+ \boldsymbol{\omega}_{1}^{0} \times \boldsymbol{p}^{0} \end{aligned} \end{split}\]

In (10), when \(P\) is moving on the body, it will have a local velocity \(\dot{{\boldsymbol{p}}}^{1}\) in the in the body frame. Note that the second line is derived due to \(\boldsymbol{R}_{1}^{0} \boldsymbol{p}^{1}=\boldsymbol{p}^{0}\) and \(\boldsymbol{S}\left(\boldsymbol{\omega}_{1}^{0}\right)=[\omega_{1}^{0} \times] \).

Notice that, if \(\boldsymbol{p}^{1}\) is fixed in the moving body frame , (10) becomes

\[\dot{\boldsymbol{p}}^{0}=\dot{\boldsymbol{o}}_{1}^{0}+\boldsymbol{\omega}_{1}^{0} \times \boldsymbol{p}^{0}\]

Apply to links of a robot arm

Consider the Link \(i\) of a robot arm in Fig. 53 (the frames follow the DH convention). Let \(\boldsymbol{r}_{i-1, i}^{i-1}\) denote the position of the origin of Frame \(i\) from the origin of Frame \(i-1\), expressed in Frame \(i-1\).

Before we proceed, a quick question: how to obtain \(\boldsymbol{p}_{i-1}\) and \(\boldsymbol{R}_{i-1}\)? and how to obtain \(\boldsymbol{r}_{i-1, i}^{i-1}\)?

Fig. 53 Link \(i\) of a robot arm

Linear (Translational) Velocity

The position of Frame \(i\) is

(11)\[\boldsymbol{p}_{i}=\boldsymbol{p}_{i-1}+\boldsymbol{R}_{i-1} \boldsymbol{r}_{i-1, i}^{i-1}\]

Similar to (10), taking the derivative of both sides in the above equation with respect to time \(t\), we have

(12)\[\begin{split} \begin{aligned} \dot{\boldsymbol{p}}_{i}&=\dot{\boldsymbol{p}}_{i-1}+\boldsymbol{R}_{i-1} \dot{\boldsymbol{r}}_{i-1, i}^{i-1}+\boldsymbol{\omega}_{i-1} \times \boldsymbol{R}_{i-1} \boldsymbol{r}_{i-1, i}^{i-1}\\&=\dot{\boldsymbol{p}}_{i-1}+\boldsymbol{v}_{i-1, i}+\boldsymbol{\omega}_{i-1} \times \boldsymbol{r}_{i-1, i} \end{aligned} \end{split}\]

Here, we define \(\boldsymbol{v}_{i-1, i}=\boldsymbol{R}_{i-1}\dot{\boldsymbol{r}}_{i-1, i}^{i-1}\), which is local translational velocity of Frame \(i\) in Frame \(i-1\), but expressed in the reference frame.

If joint \(i\) is revolute joint,

(13)\[\boldsymbol{v}_{i-1, i}=\boldsymbol{\omega}_{i-1, i}\times\boldsymbol{r}_{i-1, i}\]

where \(\boldsymbol{\omega}_{i-1, i}\) is relative angular velocity of Frame \(i\) in Frame \(i-1\). More specifically,

(14)\[\boldsymbol{\omega}_{i-1, i}=\dot{\vartheta}\boldsymbol{z}_{i-1}\]

What is \(\boldsymbol{z}_{i-1}\)? This is question for you!

Combine (12), (13) and (14), we have: If Joint \(i\) is revolute

\[\begin{aligned} \dot{\boldsymbol{p}}_{i} =\dot{\boldsymbol{p}}_{i-1}+(\dot{\vartheta}_{i} \boldsymbol{z}_{i-1}+\boldsymbol{\omega}_{i-1}) \times \boldsymbol{r}_{i-1, i} \end{aligned}\]

If joint \(i\) is prismatic joint,

(15)\[\boldsymbol{v}_{i-1, i}=\dot{d_i}\boldsymbol{z}_{i-1}\]

Combine (12), (13) and (15), we have: If Joint \(i\) is prismatic

\[\begin{aligned} \dot{\boldsymbol{p}}_{i} =\dot{\boldsymbol{p}}_{i-1}+\dot{d}_{i} \boldsymbol{z}_{i-1} +\boldsymbol{\omega}_{i-1}\times \boldsymbol{r}_{i-1, i} \end{aligned}\]

Angular Velocity

Next, let’s consider the rotation of Frame \(i\)

\[\boldsymbol{R}_{i}=\boldsymbol{R}_{i-1} \boldsymbol{R}_{i}^{i-1}\]

Its time derivative is

(16)\[\boldsymbol{S}\left(\boldsymbol{\omega}_{i}\right) \boldsymbol{R}_{i}=\boldsymbol{S}\left(\boldsymbol{\omega}_{i-1}\right) \boldsymbol{R}_{i}+\boldsymbol{R}_{i-1} \boldsymbol{S}\left(\boldsymbol{\omega}_{i-1, i}^{i-1}\right) \boldsymbol{R}_{i}^{i-1}\]

where \(\boldsymbol{\omega}_{i-1, i}^{i-1}\) denotes the angular velocity of Frame \(i\) with respect to Frame \(i-1\) expressed in Frame \(i-1\). The second term on the right-hand side of (16) can be rewritten as

\[\boldsymbol{R}_{i-1} \boldsymbol{S}\left(\boldsymbol{\omega}_{i-1, i}^{i-1}\right) \boldsymbol{R}_{i}^{i-1}=\boldsymbol{R}_{i-1} \boldsymbol{S}\left(\boldsymbol{\omega}_{i-1, i}^{i-1}\right) \boldsymbol{R}_{i-1}^{T} \boldsymbol{R}_{i-1} \boldsymbol{R}_{i}^{i-1}=\boldsymbol{S}\left(\boldsymbol{R}_{i-1} \boldsymbol{\omega}_{i-1, i}^{i-1}\right) \boldsymbol{R}_{i}\]

by recalling the property of the skew-symmetric matrix in (9). Then,

(17)\[\boldsymbol{S}\left(\boldsymbol{\omega}_{i}\right) \boldsymbol{R}_{i}=\boldsymbol{S}\left(\boldsymbol{\omega}_{i-1}\right) \boldsymbol{R}_{i}+\boldsymbol{S}\left(\boldsymbol{R}_{i-1} \boldsymbol{\omega}_{i-1, i}^{i-1}\right) \boldsymbol{R}_{i}\]

If we cancel out all \( \boldsymbol{R}_{i}\) on both sides of (17), we can get

(18)\[\boldsymbol{\omega}_{i}=\boldsymbol{\omega}_{i-1}+\boldsymbol{R}_{i-1} \boldsymbol{\omega}_{i-1, i}^{i-1}=\boldsymbol{\omega}_{i-1}+\boldsymbol{\omega}_{i-1, i}\]

Thus, the angular velocity \(\boldsymbol{\omega}_{i}\) of Link \(i\) depends only on the angular velocity \(\boldsymbol{\omega}_{i-1}\) of Link \(i-1\) and their relative angular velocity \(\boldsymbol{\omega}_{i-1, i}\).

If joint \(i\) is revolute joint,

\[ \boldsymbol{\omega}_{i} =\boldsymbol{\omega}_{i-1}+\dot{\vartheta}_{i} \boldsymbol{z}_{i-1} \]

If joint \(i\) is prismatic.

\[ \boldsymbol{\omega}_{i} =\boldsymbol{\omega}_{i-1} \]

Summary

Considering different joint types for Joint \(i\) at a robot arm, according to (12) and (18), we can obtain

Important

If Joint \(i\) is prismatic

\[\begin{split}\begin{aligned} \boldsymbol{\omega}_{i} & =\boldsymbol{\omega}_{i-1} \\ \dot{\boldsymbol{p}}_{i} & =\dot{\boldsymbol{p}}_{i-1}+\dot{d}_{i} \boldsymbol{z}_{i-1}+\boldsymbol{\omega}_{i-1} \times \boldsymbol{r}_{i-1, i} \end{aligned}\end{split}\]

If Joint \(i\) is revolute

\[\begin{split}\begin{aligned} \boldsymbol{\omega}_{i} & =\boldsymbol{\omega}_{i-1}+\dot{\vartheta}_{i} \boldsymbol{z}_{i-1} \\ \dot{\boldsymbol{p}}_{i} & =\dot{\boldsymbol{p}}_{i-1}+\boldsymbol{\omega}_{i} \times \boldsymbol{r}_{i-1, i} \end{aligned}\end{split}\]