Various conductance matrices for simple resistor configurations including a skeleton cube
cube(x=1)
octahedron(x=1)
tetrahedron(x=1)
dodecahedron(x=1)
icosahedron(x=1)Resistance of each edge. See details section
Function cube() returns an eight-by-eight conductance matrix
for a skeleton cube of 12 resistors. Each row/column corresponds to
one of the 8 vertices that are the electrical nodes of the compound
resistor.
In one orientation, node 1 has position 000, node 2 position 001, node 3 position 101, node 4 position 100, node 5 position 010, node 6 position 011, node 7 position 111, and node 8 position 110.
In cube(), x is a vector of twelve elements (a scalar
argument is interpreted as the resistance of each resistor)
representing the twelve resistances of a skeleton cube. In the
orientation described below, the elements of x correspond to
\(R_{12}\), \(R_{14}\), \(R_{15}\),
\(R_{23}\), \(R_{26}\), \(R_{34}\),
\(R_{37}\), \(R_{48}\), \(R_{56}\),
\(R_{58}\), \(R_{67}\), \(R_{78}\) (here
\(R_{ij}\) is the resistancd between node \(i\) and
\(j\)). This series is obtainable by reading the rows given by
platonic("cube"). The pattern is general: edges are ordered
first by the row number \(i\), then column number \(j\).
In octahedron(), x is a vector of twelve elements (again
scalar argument is interpreted as the resistance of each resistor)
representing the twelve resistances of a skeleton octahedron. If node 1
is “top” and node 6 is “bottom”, the elements of x
correspond to
\(R_{12}\), \(R_{13}\), \(R_{14}\),
\(R_{15}\), \(R_{23}\), \(R_{25}\),
\(R_{26}\), \(R_{34}\), \(R_{36}\),
\(R_{45}\), \(R_{46}\), \(R_{56}\).
This may be read off from the rows of platonic("octahedron").
To do a Wheatstone bridge, use tetrahedron() with one of the
resistances Inf. As a worked example, let us determine the
resistance of a Wheatstone bridge with four resistances one ohm and
one of two ohms; the two-ohm resistor is one of the ones touching the
earthed node.
To do this, first draw a tetrahedron with four nodes. Then say we
want the resistance between node 1 and node 3; thus edge 1-3 is the
infinite one. platonic("tetrahedron") gives us the order of
the edges: 12, 13, 14, 23, 24, 34. Thus the conductance matrix is
given by jj <- tetrahedron(c(2,Inf,1,1,1,1)) and the resistance
is given by resistance(jj,1,3) [compare the analytical answer
of 117/99 ohms].
F. J. van Steenwijk “Equivalent resistors of polyhedral resistive structures”, American Journal of Physics, 66(1), January 1988.
# NOT RUN {
resistance(cube(),1,7) #known to be 5/6 ohm
resistance(cube(),1,2) #known to be 7/12 ohm
resistance(octahedron(),1,6) #known to be 1/2 ohm
resistance(octahedron(),1,5) #known to be 5/12 ohm
resistance(dodecahedron(),1,5)
# }
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