http://caltek.net/dan/connectivity/phibiz/ekgtree/ekgtree6.gif
I can use basic gnuplot syntax and I've plotted a few basic curves but
I can't work out how to plot one of the two curves shown above. Each
decays by golden section - each smaller wave compares to the previous
by golden section relation in amplitude and wavelength.
Can anyone here do this ?
I am wanting to accurately plot this for import into a softsynth.
Here it is:
#############################################
maxn = 10
phi = 0.5*(-1+sqrt(5))
s(n) = (1-phi**(n+1))/(1-phi)
c(x,n) = (x >= (n==0 ? 0 : pi*s(n-1))) && (x < pi*s(n)) \
? phi**n*sin((x-pi*s(n-1))/phi**n ) \
: 1/0
set samples 1000
unset key
set xr [0:pi*s(maxn)]
unset tics
unset border
plot for [n=0:maxn] c(x,n) w l lt -1, for [n=0:maxn] -c(x,n) w l lt -1
############################################
I'm sure there is a more elegant recursive solution, but I can't be
assed to find it right now.
An interesting optical illusion: the width of the 3rd, 4th, etc. lobes
combined is exactly equal to that of the first, but it seems much
longer.
Péter Juhász
If you're interested then this is the page the graphic is from ...
http://caltek.net/dan/connectivity/phibiz/ekgtree/ekgtree.html ... the
context of Caduceus and biological systems is covered in great depth
here - http://goldenmean.info - http://www.fractalfield.com
On Oct 22, 5:16 pm, Péter Juhász <peter.juhas...@gmail.com> wrote:
Modify the function definition like this:
c(x,n) = (x >= (n==0 ? 0 : pi*s(n-1))) && (x < pi*s(n)) \
? (-1)**n*phi**n*sin((x-pi*s(n-1))/phi**n ) \
: 1/0
Note that there is no "one of two sine waves" here: there are
individual lobes of several sine curves, each scaled and translated
appropriately so that they appear to form a single curve.
If all you need is a table of values instead of a graphical plot, use
"set table".
Please also note that I view this problem as a mildly interesting
mathematical distraction only - I distance myself from the content of
the linked sites.
Péter Juhász
http://portal.djbarney.org/node/71#Here_it_is:_Actually_Plotting_the_Caduceus_Curve