cube: Specimen conductance matrices

In RobinHankin/ResistorArray: Electrical Properties of Resistor Networks

Specimen conductance matrices

Description

Various conductance matrices for simple resistor configurations including a skeleton cube

Usage

cube(x=1)
octahedron(x=1)
tetrahedron(x=1)
dodecahedron(x=1)
icosahedron(x=1)

Arguments

x	Resistance of each edge. See details section

Details

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 resistance 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].

Author(s)

Robin K. S. Hankin

References

F. J. van Steenwijk “Equivalent resistors of polyhedral resistive structures”, American Journal of Physics, 66(1), January 1988.

Examples

 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) 

RobinHankin/ResistorArray documentation built on Jan. 17, 2024, 5:05 p.m.