4.1 Michell trusses  (Page 3/4)

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 $v\in \mathrm{Span}\{a+tv,b+tv,c+tv\}.$

The scalar triple product

is constant in $t$ if $v\in \mathrm{Span}\{a,b,c\}$ and is nonzero for all but one $t$ otherwise. In the former case, setting $t=0$ while in the latter choosing $t$ so that $D\left(t\right)\ne 0$ and consequently $\mathrm{Span}\{a+tv,b+tv,c+tv\}={\mathbb{R}}^3$ then implies the lemma.

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 balanced by the truss $T.$ Let $v\in {\mathbb{R}}^3$ and $\phi \left(x\right)=v$ for all $x\in {\mathbb{R}}^3.$ Then ( ) implies

Since $v$ was arbitrary, we infer ( ). For $w\in {\mathbb{R}}^3,$ define the skew symmetric matrix

One then checks that $\widehat{w}v=w\times v.$ Let $v\in {\mathbb{R}}^3$ and $\phi \left(x\right)=\widehat{x}v.$ Then

By ( ), this implies

Since $v$ is arbitrary, we infer ( ). 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 . 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 . 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 , there is $\rho \in \mathbb{R}$ so that

Let $b=a_{\nu +1}+\rho f_{\nu +1}$ and let

Define the point force field $\mathbf{g}=g_1\delta \left(b_1\right)+\cdots +g_{\nu }\delta \left(b_{\nu }\right)$ by

We claim that $\mathbf{g}$ is balanced; by ( )

and by ( )

By induction, there exists a truss $R$ equilibrating $\mathbf{g}.$ Let $S$ be the truss consisting of the collection of beams $a_1-b,a_2-b,a_3-b$ and $b-a_{\nu +1}$ with weights ${\Omega }_1,{\Omega }_2,{\Omega }_3$ and $|f_{\nu }|$ resp. Arguing as in lemma , we find that $T=R+S$ equilibrates $\mathbf{f}.$

Some economical trusses

After proving the question of existence we turn to the question of economy. We say that a truss $T$ is economical if it satisfies

whenever $\delta S=\delta T.$ That is to say, the cost of $T$ is less than or equal to that of any truss which equilibrates the same force system as $T.$ We begin by describing some global statements about economical trusses that can proven by choosing special test vector fields $\phi$ in ( ). Then we make local perturbations on trusses with corners to find a necessary conditionfor economy.

Some global properties

A surprising and easily proven fact is

Lemma 4 A truss is economical if the weights of the beams are all of the same sign. Such a truss lies in the closed convex hull of the support of the point forcesit equilibrates.