cube {ResistorArray} | R Documentation |
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 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].
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)