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pch=15 symbol of size=10 mm

Denis White wrote:

            
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

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