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var w = 800, midx=w/2
var h = 600, midy=h/2, y_invert = h-1 - midy
screenres w,h
dim as v2 p = (100,0)
'' these 3 variables double precision will keep length remarkably stable
var iangle = cdbl(3.14159/5) 'worst-case scenario i could find after minutes of searching
var cosa=cos(iangle), sina=sin(iangle)
while inkey=""
pset (midx + p.x, y_invert - p.y)
for i as long = 1 to 100000
var x2=p.x
p.x=p.x*cosa-p.y*sina
p.y=p.y*cosa+x2*sina
next
sleep 1
Wend