0.1 Derived copy of michell trusses  (Page 3/5)

and

because $b$ is parallel to ${\omega }_b\frac{b}{\left|b\right|}$ (and likewise for $c$ ) and $\mathbf{f}$ is balanced. According to Lemma 1, there is a truss $R$ with equilibrating $\mathbf{e}.$ Let $S$ be the truss consisting of the collection of beams $(0,b)$ and $(0,c)$ with weights ${\omega }_b$ and ${\omega }_c$ respectively. We claim $T=R+S$ equilibrates $\mathbf{f}.$ Let $\phi$ be a continuous vector field. Then

which then implies [link] .

The following lemma shows when the volume of the parallelpiped spanned by three vectors can be made to be nonzero by sheering theparallelpiped in a fixed direction by an appropriate amount. An example in two dimensions is drawn below.

Lemma 3 Let $a,b,c$ be distinct, nonzero points in Euclidean space and $v$ a nonzero vector. Then there exists $t\in \mathbb{R}$ such that $v\in \mathrm{Span}\{a+tv,b+tv,c+tv\}.$

If $\nu \in \mathrm{Span}\{a,b,c\}$ , then just take $t=0$ . If any of the vectors, say $a$ and $b$ , are linearly dependent, then for some constant, $n$ , $b=na$ . For any nonzero $t$ , $a+t\nu$ is linearly independent from $b+t\nu$ because $b+t\nu =an+t\nu$ is not equal to $n(a+t\nu )=an+nt\nu )$ since $nt\nu \ne t\nu$ because $\nu$ and $t$ are nonzero. Applying this argument again to $b$ and $c$ if necessary, we can get three vectors, $a+t\nu ,b+t\nu ,$ and $c+t\nu$ , which are guaranteed to be linearly independent. Because the span of any three linearly independent vectors is all of ${\mathbb{R}}^3$ , every vector is contained in $\mathrm{Span}\{a,b,c\}$ , so it must be that $\nu \in \mathrm{Span}\{a,b,c\}$ , as desired.

Combining the above lemmas, we have

Theorem 1 A necessary and sufficient condition that a point force field $\mathbf{f}=f_1\delta \left(a_1\right)+\cdots +f_{\nu }\delta \left(a_{\nu }\right)$ be balanced is that it is equilibrated by a truss $T.$

(Necessity) Suppose $\mathbf{f}$ is equilibrated by the truss $T.$ Let $\nu \in {\mathbb{R}}^3$ and $\phi \left(x\right)=\nu$ for all $x\in {\mathbb{R}}^3.$ Then [link] implies

because $\phi$ is a constant vector field. Since $\nu$ was arbitrary, it can be chosen to be

so that

and conclude that

which implies [link] . For $w\in {\mathbb{R}}^3,$ define the skew symmetric matrix

It is easy to check that $\widehat{w}\nu =w\times \nu .$ Let $\nu \in {\mathbb{R}}^3$ and $\phi \left(x\right)=\widehat{x}\nu .$ Then

because the cross product $w_i(a_i-b_i)\times \nu$ is perpendicular to the vector $a_i-b_i$ , which is parallel to $\frac{a_i-b_i}{|a_i-b_i|}$ so their dot product is zero. By [link] , this implies

Since $\nu$ is arbitrary, we infer [link] . Thus $\mathbf{f}$ is balanced.

(Sufficiency) We proceed by induction on $\nu .$ If $\nu =2,$ then $\mathbf{f}$ is equilibrated by the truss $T$ guaranteed in Lemma 1. If $\nu =3$ and $a_1,a_2$ and $a_3$ do not all lie on a line, then $\mathbf{f}$ is equilibrated by the truss $T$ guaranteed in Lemma 2. If $\nu =3$ and $a_1,a_2$ and $a_3$ lie on a line, consider the equivalent point force field $\mathbf{g}=f_1\delta \left(a_1\right)+f_2\delta \left(a_2\right)+f_3\delta \left(a_3\right)+0\delta \left(a_4\right)$ where $a_4$ is a arbitrary point chosen off the line containing $a_1,a_2$ and $a_3.$ The result for $\nu =4$ proves that $\mathbf{g}$ is equilibrated by a truss $T.$ However, since $(\mathbf{g},\phi )=(\mathbf{f},\phi )$ for all continuous vector fields $\phi ,$ this implies that $T$ also equilibrates $\mathbf{f}.$ Assume that sufficiency holds for $\nu \ge 3.$ If $f_{\nu +1}=0,$ then we are done. Otherwise, according to Lemma 3, there is $\rho \in \mathbb{R}$ so that