# Missing centrifugal acceleration

I am trying to get correct equations for acceleration of a point in reference frame A, given position, velocity and acceleration in rotating reference frame B.

Let $\mathbf{x}_A(t)$, $\mathbf{v}_A(t)$, $\mathbf{a}_A(t)$ denote the position, velocity and acceleration of the point in reference frame A respectively. Let $\mathbf{x}_B(t)$, $\mathbf{v}_B(t)$, $\mathbf{a}_B(t)$ denote the position, velocity and acceleration of the point in rotating reference frame B respectively. If rotation of reference frame A into reference frame B is presented with rotation matrix $\mathbf{R}(t)$: \begin{equation} \mathbf{x}_A(t) = \mathbf{R}(t)\mathbf{x}_B(t) \end{equation} If derivative is taken from previous equation, velocity of the point in reference frame A is : \begin{align} \mathbf{v}_A(t)=\mathbf{\dot{x}}_A(t) &= \mathbf{\dot{R}}(t)\mathbf{x}_B(t) + \mathbf{R}(t)\mathbf{\dot{x}}_B(t)\\ &=\mathbf{\dot{R}}(t)\mathbf{x}_B(t) + \mathbf{R}(t)\mathbf{v}_B(t) \end{align} If derivative is taken from previous equation, acceleration of the point in reference frame A is : \begin{align} \mathbf{a}_B(t)=\mathbf{\ddot{x}}_B(t) &= \mathbf{\ddot{R}}(t)\mathbf{x}_B(t) + \mathbf{\dot{R}}(t)\mathbf{\dot{x}}_B(t) + \mathbf{\dot{R}}(t)\mathbf{v}_B(t) + \mathbf{R}(t)\mathbf{\dot{v}}_B(t)\\ &= \mathbf{\ddot{R}}(t)\mathbf{x}_B(t) + 2\mathbf{\dot{R}}(t)\mathbf{v}_B(t) + \mathbf{R}(t)\mathbf{a}_B(t)\\ \end{align} In most textbooks, there is cross product presentation of equations for position (not sure about correctness of below equations): \begin{equation} \mathbf{x}_A(t) = \mathbf{\phi}(t)\times\mathbf{x}_B(t) \end{equation} velocity: \begin{equation} \mathbf{v}_A(t) = \mathbf{\phi}(t)\times\mathbf{v}_B(t) + \mathbf{\Omega}(t)\times\mathbf{x}_B(t) \end{equation} acceleration: \begin{equation} \mathbf{a}_A(t) = \mathbf{\phi}(t)\times\mathbf{a}_B(t) + 2\mathbf{\Omega}(t)\times\mathbf{v}_B(t) + \mathbf{\dot{\Omega}}(t)\times\mathbf{x}_B(t) \end{equation} Previous equations I derived using derivative property of cross product $\frac{d}{dt}(\mathbf{a}\times\mathbf{b}) = \left(\frac{d}{dt}\mathbf{a}\right)\times\mathbf{b} + \mathbf{a}\times\left(\frac{d}{dt}\mathbf{b}\right)$

From both of these sets of equations, I'm missing centrifugal acceleration $\mathbf{\Omega}(t)\times\mathbf{\Omega}(t)\times\mathbf{x}_B(t)$. I know I may have lack of some fundamental knowledge and I hope you could help me with that.

• Have you had a gander at the Wiki article? – BMS Nov 23 '13 at 15:55
• @BMS I get correct formula just by substitution $\dot{\mathbf{\Omega}} = \left[\dot{\mathbf{\Omega}}\right] + \mathbf{\Omega}\times \mathbf{\Omega}$, but as I understand it this is not correct... This formula applies only to vectors in rotating reference frame, and $\mathbf{\Omega}$ is not one of them.. or is it..? – Slaven Glumac Nov 23 '13 at 19:16

It's not missing, it's in the $\ddot{R}(t)$ matrix.

It doesn't show up on its own when you do the calculation with matrices instead just vectors however.

## Vector equations

The first equation should be written

$$\mathbf{x}_A(t) = \mathbf{\phi}(t) \times \mathbf{x}_B(t)$$

(just reverse where you have $A$ and $B$) since given the position in frame $B$ you have to find where it has rotated to in frame $A$.

For the velocity & acceleration relationships, what we usually find is actually $v_A(t)$ and $a_A(t)$ and in the end re-arrange the terms to get $v_B(t)$ and $a_B(t)$, substituting the above relationship as necessary.

We define a new operator that gives us the velocity in $A$ given the position in $B$; let's call it $\hat{v}$. As described in the wikipedia link in one of the above user's comments, what the operator should be is this:

$$\hat{v} \mathbf{x} = \frac{d \mathbf{x}}{d t} + \mathbf{\Omega} \times \mathbf{x}$$

with the acceleration operator as

$$\hat{a} \mathbf{x} = \hat{v}^2 \mathbf{x} = \hat{v} (\hat{v} \mathbf{x})$$.

Conceptually, you just replace $d/dt$ with $d/dt + \mathbf{\Omega} \times$ and apply this to $x_B (t)$ to get $v_A(t)$ and $a_A(t)$.

• Could you tell me where did I go wrong with my vector calculation then? Thank you for the above answer. – Slaven Glumac Nov 23 '13 at 19:30
• answer updated. – ghollisjr Nov 23 '13 at 23:47
• I had a change of mind in notation. That is why I made that mistake about A and B coordinate system. P.S. This is not a homework (I don't know why someone tagged it as such), I'm trying to learn this myself in order to understand robotics simulator better. Thank you for these pointers. – Slaven Glumac Nov 24 '13 at 9:34