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I'm having some conceptual difficulties with a derivation presented in the Modern Robotics text by Park and Lynch. Here are the relevant parts: enter image description here enter image description here enter image description here

At the beginning of the book, they insist that ${\it all}$ frames used in the book are stationary, ${\it even}$ body-fixed frames. This has been extremely confusing for me, as we are constantly talking about body frames evolving in time, and there are many animations on the YouTube videos that depict embedded body-fixed frames that are ${\it not}$ stationary.

I assume they are trying to say that all body-fixed frames follow a sequence of stationary frames? Anyway...

My main confusion comes from the $[\omega_b] v_b$ term in their $F = m a$ statement. What is this term? Suppose a planar rigid body is simultaneously translating and rotating with, say, $v_b = [1,0,0]$ and $\omega_b = [0,0,1]$. The center of mass does not accelerate for this motion. Clearly $\dot{v}_b = 0$, but $[\omega_b]v_b = [0,1,0]$, so the equations tell me that a force must be present in the body y-direction to support this motion.

What am I missing?


EDIT:

Here is a possibility that makes sense to me. Suppose that the frame $\{{\bf b}_1, {\bf b}_2, {\bf b}_3\}$ is affixed to the body at its CG. Then, we can use this basis to describe the velocity of the CG of the body: $\bar{\bf v} = \bar{v}_i {\bf b}_i$. Now, the acceleration of the CG using this basis becomes $\bar{\bf a} = \dot{\bar{v}}_i {\bf b}_i + \boldsymbol{\omega} \times \bar{\bf v}$. Now if everything is represented on the $\{{\bf b}_i \}$ basis including the force ${\bf F} = F_i {\bf b}_i$ and angular velocity $\boldsymbol{\omega} = \omega_i {\bf b}_i$, then the matrix form of the three equations you get from ${\bf F} = m \bar{\bf a}$ becomes that obtained by the book.

This derivation is clean, but it assumes a $\textbf{non-inertial}$, $\textbf{non-stationary}$ frame $\{ {\bf b}_i \}$.


EDIT 2: Here are the confusing remarks in question about stationary versus non-stationary frames: enter image description here enter image description here


EDIT 3: Interpretation of (8.22)

Consider a wheel rolling without slip to the right. Mark two neighboring points, $A$ and $B$. enter image description here

These two material points follow their own flows through space, as in the perspective of system dynamics or fluid mechanics. Their paths look something like this: enter image description here

Velocity profile of disk at any time looks something like this: enter image description here

The interpretation of (8.22) is similar to the Eulerian description of fluid mechanics. If we choose the inertial point coincident with material point $A$, then at the next time instance, $B$ will occupy that point in space. Therefore, $\dot{v}_b$ is upwards, and clearly $[\omega_b] v_b$ is downwards. The two terms cancel to tell us at the current time instance that the acceleration of material point $A$ is zero, as expected.

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    $\begingroup$ From the book, All frames in this book are stationary, inertial, frames. When we refer to a body frame {b}, we mean a motionless frame that is instantaneously coincident with a frame that is fixed to a (possibly moving) body. This is important to keep in mind, since you may have had a dynamics course that used non-inertial moving frames attached to rotating bodies. Do not confuse these with the stationary, inertial, body frames of this book. $\endgroup$
    – CroCo
    Dec 23, 2021 at 16:56
  • $\begingroup$ cont., I'm reading this book now and it is confusing when they say all frames are stationary as I quoted above. This remark is not totally clear to me what they infer. $\endgroup$
    – CroCo
    Dec 23, 2021 at 16:58
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    $\begingroup$ My understanding to their remark when they say stationary to the body frame, they mean it is aligned with the inertial frame but translated to a fixed point in the rigid body so that they can track the orientation; technically, there are two frames attached to the body frame. Not sure though. $\endgroup$
    – CroCo
    Dec 23, 2021 at 17:05
  • $\begingroup$ Hi CroCo. Thanks for the reply. I too am not sure about the insistence about body frames being stationary. I am trying to understand their perspective and why they want to stress that; it seems to make things unnecessarily difficult. I imagine there is some point they are trying to get across that is getting lost in translation. I think they might want us to imagine at an instant that the body is passing through a stationary frame because when you get to the part about twists, you need to imagine how the body is moving at that instant, and maybe they don't want us rotating the body frame in $\endgroup$
    – HelpMe
    Dec 23, 2021 at 17:24
  • $\begingroup$ our heads while we are doing that. Check out my edit though, I think I figured out another perspective for how they are deriving their formula for a single rigid body. Feel free to message me if you have any other ideas. $\endgroup$
    – HelpMe
    Dec 23, 2021 at 17:24

1 Answer 1

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The key thing to remember is that none of the equations used in the Modern Robotics textbook use "body-fixed frames". The {b} frame is defined as a "body frame" which is instantaneously coincident (same position & orientation) with a body-fixed frame attached to the body (as stated in the grey box in your second edit).

This means that any body-fixed frame attached to any given body in a system can move as much as you want, but for every instant of time in which you need to formulate an equation, you will be using a fixed, inertial, instantaneous copy of the frame which the authors have (somewhat confusingly) labelled as the "body frame".

The second point that should be made is that the calculations you did in Edit 1 were for a stationary frame, not a rotating frame. Otherwise, you would have to also include one additional term in the velocity calculation and two additional terms in the acceleration calculation to deal with the rotating frame (see https://en.wikipedia.org/wiki/Centrifugal_force).

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  • $\begingroup$ Exactly. The nomenclature of body frame is confusing, as modern robotics draws a distinction between basis vectors (orientation) and points of summations, not to be grouped together. All basis vectors must of the non-rotating kind to get equations of motion, but the points of summation might ride along a body, such as a center of mass. The gray box above Fig 3.2 in the question says so explicitly. $\endgroup$ Jan 22 at 0:28
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    $\begingroup$ I finally figured out the interpretation of (8.22) by looking at the example of a wheel rolling without slip. For example, if you look at the material point at the center of the wheel, it is definitely not accelerating, but the terms of (8.22) are non-zero. The way you must interpret (8.22) is to imagine a fixed location in space with material passing through it, much like the Eulerian description of fluid mechanics. In this perspective, every material point is following a flow, and at the next time instance a new material point will occupy the inertial point you were looking at. $\endgroup$
    – HelpMe
    Jan 29 at 17:42
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    $\begingroup$ So in the context of the wheel moving to the right at constant axle speed, if we look at the center point, then at the next time instance, a new material point will occupy that same point in space with a velocity that is to the right and up. Hence, $\dot{v_b}$ is up, while $[\omega_b] v_b$ is down, and the two terms cancel to tell us that the acceleration of the center of the wheel is zero. $\endgroup$
    – HelpMe
    Jan 29 at 17:45
  • $\begingroup$ Happy to hear that you've got your head wrapped around it now. Once you can take a step back and see the fact that twists and wrenches are vector fields, and not vectors, it makes all of the screw theory math much clearer (I find). $\endgroup$ Jan 31 at 19:01

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