I am reading the paper "Motion control of wheeled mobile robots"

at page 5, as we have the equation (34.7), i wonder how it contains $− dc(s)\dot s$ in side. As the triangle law of vector: $\vec{OP}=\vec{OP_s} +\vec{P_sP}$. that is enough. Taking derivative, we should get (34.7) without $− dc(s)\dot s$


1 Answer 1


You made a simple mistake while calculating the derivative.The equation is: $\vec{OP}= OP_s \vec {i_s} + P_sP \vec {j_s}$

The derivative should be $\partial \vec{OP}/ \partial t =\partial({OP_s}\vec {i_s})/\partial t + \partial (P_sP \vec {j_s})/\partial t = \partial({OP_s}\vec {i_s})/\partial t $ but it's given that $d\vec{OP_s} / dt = \partial ({OP_s}\vec {i_s}) /\partial t = \dot s \vec {i_s}$.

Now for the second term:

$\partial (P_sP \vec {j_s})/\partial t = (\partial P_sP/ \partial t) \vec{j_s} + (P_sP) \partial \vec {j_s} / \partial t$

So you have $ (\partial P_sP/ \partial t) \vec{j_s} = \dot d \vec{j_s}$

And $ P_sP \partial \vec {j_s} / \partial t = d (\partial \vec {j_s} / \partial s) * \partial s/ \partial t = d \partial \vec {j_s} / \partial s * \dot s$. to calculate the derivative of the normal vector at that point you can check out this book eq. (3.4) and (3.6), added to this you curve is planar and the vectors $\vec {i_s}, \vec {j_s}$ are in the same plane so you end up to eq. (4.1) which is the eq. you where searching for.

  • $\begingroup$ 1/$\overrightarrow{OP}$ is a vector while $\overrightarrow{OP_{s}}\overrightarrow{i_{s}}$ is scalar (scalar product of 2 vectors results a scalar quantity). how do you form your original equation?not triangle law? 2/ derivative of $\overrightarrow{OPs}\overrightarrow{i_{s}}$ (a scalar quantity) now is a vector (a sum of 2 vectors a vector) and i honestly dont know how you calculate derivative (it does not fit product rule of derivative) i am confused here and could not go further $\endgroup$
    – abcd
    Commented Oct 1, 2019 at 2:32
  • $\begingroup$ @abcd I didn't double check the equations that I wrote, and they had some dummy mistakes. Checkout the updated version, and also how to actually calculate the partial derivative. $\endgroup$
    – nionios
    Commented Oct 1, 2019 at 7:50
  • $\begingroup$ Thank nionios, i mistakenly treated the vectors as constant and resulted partial derivative as zero $\endgroup$
    – abcd
    Commented Oct 9, 2019 at 13:31

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