pch=15 symbol of size=10 mm

Denis White wrote:

How do I determine cex in

  points (x, y, pch=15, cex=?)

to get a square whose side length is, say, 10 mm?

thanks

The easy way would be:

points(6,3,pch=15,cex=10/(par("cin")[1]*25.4))

That is, divide your desired width by the current width in inches multiplied by
25.4,

but this won't _really_ give you a 10 mm square.  The hardcopy output you get
depends upon the size of the image you generate (e.g. the "bounding box" in
Postscript).  However, you could fudge it if you always use the same size
output...

Jim

Thanks for response from Jim Lemon.  I tried 10/25.4/par("csi") and
several other ideas before submitting to r-help in frustration.
Below is a function that generates regular polygons, filled or
borders, of n sides (n>8 => circle), with "diameter" prescribed
in mm, for use alone or with apply.

############################################################

ngon <- function (xydc, n=4, type=1)
  # draw or fill regular polygon
  # xydc a four element vector with
  #   x and y of center, d diameter in mm, and c color
  # n number of sides of polygon, n>8 => circle
  #   if n odd, vertex at (0,y), else midpoint of side
  # type=1 => interior filled, type=2 => edge
  # type=3 => both
{
  u <- par("usr")
  p <- par("pin")
  d <- as.numeric(xydc[3])
  inch <- d/25.4
  rad <- inch*((u[2]-u[1])/p[1])/2
  ys <- inch*((u[4]-u[3])/p[2])/2/rad
  if (n > 8) n <- d*4 + 1
  th <- pi*2/n
  costh <- cos (th)
  sinth <- sin (th)
  x <- y <- rep (0,n+1)
  if (n %% 2) {
    x[1] <- 0
    y[1] <- rad
    }
  else {
    x[1] <- -rad*sin(th/2)
    y[1] <-  rad*cos(th/2)
    }
  for (i in 2:(n+1)) {
    xl <- x[i-1]
    yl <- y[i-1]
    x[i] <- xl*costh - yl*sinth
    y[i] <- xl*sinth + yl*costh
    }
  x <- x + as.numeric(xydc[1])
  y <- y*ys + as.numeric(xydc[2])
  if (type %% 2) polygon (x,y,col=xydc[4],border=0)
  if (type %/% 2) lines (x,y,col=xydc[4])
  invisible()
}
 
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