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\layout Chapter

Equations\SpecialChar \ldots{}

\layout Standard

This document used to contain figures and other stuf, now it's only for
 testing editing of equations.
\layout Section

Planar geometry
\layout Standard

Note that the asymmetry has been introduced via the parameter 
\begin_inset Formula $\gamma$
\end_inset 

 according to the following transformation
\begin_inset Formula \[
\left\{ \begin{array}{c}
r=\frac{r_{1}+r_{2}}{2}\\
\gamma=\frac{r_{1}-r_{2}}{r_{2}+r_{1}}\end{array}\right\} \,\leftrightarrow\,\left\{ \begin{array}{c}
r_{1}=\left(1+\gamma\right)r\\
r_{2}=\left(1-\gamma\right)r\end{array}\right\} \]

\end_inset 

where 
\begin_inset Formula $r_{1}$
\end_inset 

 and 
\begin_inset Formula $r_{2}$
\end_inset 

 are the distances from 
\begin_inset Formula $A$
\end_inset 

 to the left and right feet respectively.
 In case of possible future use, I provide some relationsships as follows:
\begin_inset Formula \begin{eqnarray*}
r_{1}+r_{2} & = & 2r\\
r_{1}-r_{2} & = & 2\gamma\\
r_{1}*r_{2} & = & \left(1-\gamma^{2}\right)r^{2}\\
\frac{r_{1}}{r_{2}} & = & \frac{1+\gamma}{1-\gamma}\\
r_{1}^{2}+r_{2}^{2} & = & 2\left(1+\gamma^{2}\right)r^{2}\\
r_{1}^{2}-r_{2}^{2} & = & 4\gamma r^{2}\end{eqnarray*}

\end_inset 


\layout Section

Planar equilibria
\layout Paragraph*

More general definition of ground force
\layout Standard

Assume that the ground force on foot 
\begin_inset Formula $P_{i}$
\end_inset 

 is 
\begin_inset Formula \begin{eqnarray*}
\overline{f}^{G_{l}}\left(\overline{r}^{NP_{l}};\,\overline{r}^{NC_{l}^{0}},\overline{\overline{K}}^{l}\right)=\,\,\,\,\,\,\,\,\,\,\\
=\begin{cases}
\overline{f}^{Gnd}\left(\overline{r}^{NP_{l}};\,\overline{r}^{NC_{l}^{0}},\overline{\overline{K}}^{l}\right) & \textrm{if}\,\textrm{ }\left(\overline{r}^{NC_{l}^{0}}-\overline{r}^{NP_{l}}\right)\cdot\overline{n}_{3}>0\\
\overline{0} & \textrm{otherwise}\end{cases}\end{eqnarray*}

\end_inset 

where 
\begin_inset Formula \[
\overline{f}^{Gnd}\left(\overline{r};\,\overline{r}^{0},\overline{\overline{K}}\right)=\overline{\overline{K}}\cdot\overline{r}\]

\end_inset 

, 
\begin_inset Formula $\overline{\overline{K}}^{l}=k^{l}\overline{n}_{3}\overline{n}_{3}$
\end_inset 

.
 Note that only the static case is considered here, so any velocity dependant
 arguments have been removed.
 Furthermore, the ground force is uni-lateral, i\SpecialChar \@.
e\SpecialChar \@.
 here it is always directed
 out of the surface.
\layout Standard

Let us be convenient and consider the centre of mass as foot 
\begin_inset Formula $0$
\end_inset 

, i\SpecialChar \@.
e\SpecialChar \@.
 let 
\begin_inset Formula $P_{0}=B$
\end_inset 

 and denote the corresponding force 
\begin_inset Formula $\overline{f}^{G_{0}}\left(\overline{r}\right)=-mg\overline{n}_{3}$
\end_inset 

.
 Then the force equilibrium constraint can be written as follows
\begin_inset Formula \begin{equation}
\sum_{l=0}^{L}\overline{f}^{G_{l}}=\overline{0}\label{eq: force equlibrium}\end{equation}

\end_inset 

and the moment equilibrium constraint around an aribtrary point 
\begin_inset Formula $A$
\end_inset 

 can be written as follows
\begin_inset Formula \begin{equation}
\sum_{l=0}^{L}\overline{r}^{AP_{l}}\times\overline{f}^{G_{l}}=\overline{0}.\label{eq: moment equilibrium}\end{equation}

\end_inset 


\layout Standard

Depending on the number of feet in contact with ground, these constraints
 look different.
 For instance, the case without any feet att all in contact with the ground,
 results in
\begin_inset Formula \[
\sum_{l=0}^{L}\overline{f}^{G_{l}}=\overline{f}^{G_{0}}=mg\overline{n}_{3}\neq\overline{0}\]

\end_inset 

and hence no equlibria are possible.
\layout Standard

Assume that only one foot, say foot 
\begin_inset Formula $l$
\end_inset 

, supports the robot, i\SpecialChar \@.
e\SpecialChar \@.

\begin_inset Formula \[
\overline{f}^{Gnd_{i}}=\begin{cases}
\overline{f}^{Gnd_{l}} & \textrm{if}\, i=l\\
\overline{0} & otherwise\end{cases},\]

\end_inset 

 then (
\begin_inset LatexCommand \ref{eq: force equlibrium}

\end_inset 

) and (
\begin_inset LatexCommand \ref{eq: moment equilibrium}

\end_inset 

) reduce to
\begin_inset Formula \begin{equation}
\overline{f}^{Gnd_{l}}=-\overline{f}^{G_{0}}\label{eq: force equlibrium, w only foot l}\end{equation}

\end_inset 

and
\begin_inset Formula \begin{equation}
\overline{r}^{AP_{l}}\times\overline{f}^{G_{l}}=-\overline{r}^{AP_{0}}\times\overline{f}^{G_{0}}.\label{eq:moment equlibrium, w only foot l}\end{equation}

\end_inset 

respectively.
 By letting the arbitrary point 
\begin_inset Formula $A$
\end_inset 

 in (
\begin_inset LatexCommand \ref{eq:moment equlibrium, w only foot l}

\end_inset 

) coincide with 
\begin_inset Formula $P_{0}$
\end_inset 

, i\SpecialChar \@.
e\SpecialChar \@.
 
\begin_inset Formula $B,$
\end_inset 

 we see that 
\begin_inset Formula $\overline{f}^{G_{l}}=\overline{f}^{G_{0}}\shortparallel\overline{r}^{BP_{l}}$
\end_inset 

, i\SpecialChar \@.
e\SpecialChar \@.
 the foot 
\begin_inset Formula $l$
\end_inset 

 must be directly below the centre of mass.
 The distance will of course be given by the solution to (
\begin_inset LatexCommand \ref{eq: force equlibrium, w only foot l}

\end_inset 

).
 This kind of equilibrium, with only a single point in contact with the
 ground will from now on be refered to as an 
\emph on 
extreme equilibrium
\emph default 
.
 Since the centre of mass is directly over the only supporting foot, it
 is easy to see from figure\SpecialChar ~

\begin_inset LatexCommand \ref{Fig: Extreme equilibrium, existance}

\end_inset 

 that in the planar configuration, the equlibrium angle 
\begin_inset Formula $\alpha^{*}$
\end_inset 

must be one of 
\begin_inset Formula $-\alpha_{1}$
\end_inset 

 and 
\begin_inset Formula $\alpha_{2}$
\end_inset 

.
 In figure\SpecialChar ~

\begin_inset LatexCommand \ref{Fig: Extreme equilibrium, existance}

\end_inset 

, the equlibrium is 
\begin_inset Formula $\alpha^{*}=\alpha_{2}$
\end_inset 

.
\layout Standard

The reasoning above assumed that there was only one foot supporting the
 robot, and thus showed that any equlibrium must be directly above that
 foot.
 This reasoning can be reversed.
 Assume that there exists an equlibrium where the centre of mass is directly
 above foot\SpecialChar ~
two, one of two feet.
 Then, by letting 
\begin_inset Formula $A=P_{2}$
\end_inset 

 in (
\begin_inset LatexCommand \ref{eq: moment equilibrium}

\end_inset 

) we get
\begin_inset Formula \[
\underbrace{\overline{r}^{P_{2}P_{0}}\times\overline{f}^{G_{0}}}_{=\overline{0}}+\overline{r}^{P_{2}P_{1}}\times\overline{f}^{G_{1}}+\underbrace{\overline{r}^{P_{2}P_{2}}\times\overline{f}^{G_{l}}}_{=\overline{0}}=\overline{r}^{P_{2}P_{1}}\times\overline{f}^{G_{1}}=\overline{0.}\]

\end_inset 

When can this be satisfied? A general solution is given by 
\begin_inset Formula $\overline{f}^{G_{1}}=f^{1}\overline{r}^{P_{2}P_{1}}$
\end_inset 

 where 
\begin_inset Formula $f^{1}\in\Re$
\end_inset 

.
 The most typical solution is of course 
\begin_inset Formula $f^{1}=0$
\end_inset 

, i\SpecialChar \@.
e\SpecialChar \@.
 
\begin_inset Formula $\overline{f}^{G_{1}}=\overline{0}$
\end_inset 

, which corresponds to only one supporting foot.
 However, if 
\begin_inset Formula $\overline{r}^{P_{2}P_{1}}=\overline{0}$
\end_inset 

 the equation is also satisfied, corresponding to two feet in the same place.
 A third alternative (illustrated in figure\SpecialChar ~
) when both 
\begin_inset Formula $f^{1}\neq0$
\end_inset 

 and 
\begin_inset Formula $\overline{r}^{P_{2}P_{1}}\neq\overline{0}$
\end_inset 

 can sometimes also exist.
 First note that since ground forces by assumption must be directed upwards,
 it follows that 
\begin_inset Formula $f^{1}\overline{r}^{P_{2}P_{1}}$
\end_inset 

 also must be directed upwards.
 Then, if 
\begin_inset Formula $f^{1}>0$
\end_inset 

 it follows that 
\begin_inset Formula $P_{1}$
\end_inset 

 must be located somewhere 
\emph on 
below
\emph default 
 
\begin_inset Formula $P_{2}$
\end_inset 

, and similarly if 
\begin_inset Formula $f^{1}>0$
\end_inset 

, that 
\begin_inset Formula $P_{1}$
\end_inset 

 must be located somewhere 
\emph on 
above
\emph default 
 
\begin_inset Formula $P_{2}$
\end_inset 

.
 Inserting this into the force equlibrium (
\begin_inset LatexCommand \ref{eq: force equlibrium}

\end_inset 

) and scalar multiplying with a vertical and horizontal unit vector produces
 the following equation:
\begin_inset Formula \[
^{N}f^{Gnd}\left(\overline{r}^{NC_{2}};\,\overline{r}^{NC_{2}^{0}},\overline{\overline{K}}^{2}\right)=\left[\begin{array}{c}
0\\
0\\
mg\end{array}\right]-f^{1}\left(n\cdot b^{T}\right)\{^{B}r^{P_{2}P_{1}}\}\]

\end_inset 

Sigh\SpecialChar \ldots{}
this is more complicated than I thought...
 Anyway, at least when only vertical forces are allowed, the third alternative
 is impossible and the second disregarded since the foot feet can then just
 be considered as one foot.
 
\layout Standard

In a configuration with more than two feet, we will get the same result
 for a foot on the edge of a convex support area .
 Note however that there can exist an equilibrium where the centre of mass
 is above one of the supporting feet.
 Consider for instance a symmetric and planar case with three eqally spaced
 supporting feet.
 If the foot stiffnesses are equal, and the ground horizontal, then it should
 be obivous that the centre of mass directly above middle foot is an equlibrium.
 The definition of an extreme equlibrium is therefore refined:
\layout Standard

The 
\emph on 
projected\SpecialChar ~
support\SpecialChar ~
region
\emph default 
 is a unique region in any normal plane (w\SpecialChar \@.
r\SpecialChar \@.
t\SpecialChar \@.
 gravity), defined as the convex
 hull of the projection of all supporting feet into that plane.
 
\layout Standard

The 
\emph on 
extreme\SpecialChar ~
equlibrium
\emph default 
 is any equlibrium, where the projection of the centre of mass along the
 direction of gravity belongs to the boundary of the projected support region.
 A 
\emph on 
singular extreme equlibrium
\emph default 
 is an extreme equlibrium where the centre of mass is directly above one
 of the supporting feet.
\layout Standard

From the results above, a singular extreme equlibrium can only be supported
 by one foot.
\the_end