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From: OSTexo
Hello,
Benfarmer,
Connection of the whorls is one option, I'm curious how the points are generated in the seed script? Are they actually points along the spiral or something else? I'm not the best mathematician so you'll probably go way over my head quickly. I found this link and the seed head optimal packing and the subcurves created from the mathematically closest points looks interesting.
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibnat.html#seedsIMA50
Ideally what I'm looking for is a way to intersperse the curves between the seed points, or if the seed points are derived from the curves then I could just double up the amount of points to create the interspersed curves if that makes sense.
Futagoza,
It could be that's what I was remembering, the only challenge being that I need the combination of points and curves, and not knowing how the points are derived, or even if there is any correlation between the curves and points is making it difficult to wrap my head around.
Thank you both for your insight.
From: bemfarmer
Each seed is placed using polar coordinates, consisting of an angle from the x axis, and a radius, or distance from the origin..
The radius is converted to Cartesian coordinates, x and y, using cos and sin.
An angle 360 degrees is the same as an angle of 2*PI radians.
222.4922 degrees plus 137.5078 degrees, equals 360 degrees.
Point 1 is located at 222.4922 degrees = (-1) * 137.5078 degrees. Radius = sqrt(1).
Point 2 is located at (-2) * 137.5078 degrees. Radius = sqrt(2).
Point 3 is located at (-3) * 137.5078 degrees. Radius = sqrt(3).
- Brian
From: bemfarmer
Here is a redone, fast, Fibonacci "Sunflower" script, with option to do the Fermat spiral passing through the
seed points. The spiral works with up to several thousand seeds. With unchecked spiral checkbox, a test with
100,000 seed points worked. The calculation used 2*pi*phi, avoiding dividing by 2 and then multiplying by 2,
so the seed points are very slightly different than the old script. The script uses the fast "Hilbert Curve" style of
code in the .htm file. A possible drawback is hanging up if a huge number of seeds are used.
I was unable to get the Fibonacci seeds calculated in a separate Function.
I was unable to puzzle out calculating forming the curve segments for each seed, with nested loops.
Seeds may start at any integer count, leaving a donut hole, but the spiral always starts at the origin.
- Brian
Attachments:
FibonacciFermat01.zip
From: OSTexo
Hello benfarmer,
Thanks a million for this, I really appreciate it. I don't think I'll be approaching 100K seeds anytime soon.
From: Frenchy Pilou (PILOU)
No possibility to use randomly these points for deform the curve in the Z direction for example ?
From: bemfarmer
Here is version 02, with the curve plotted from seed to seed, with nested For loops.
(See .htm file for some explanation.)
Index (j+1) is equal to the seedID number.
Index i in fraction formula (i / (numpoints-1)) is added to j to yield t.
Nesting the For loops was easy, although I could not figure it out last week :-)
This scheme would permit ramping the value of z up and down between (randomly) selected (sequential) seeds.
The i fraction could be used for the ramp.
(This code has not been done yet.)
- Brian
Now to make it play music...
Attachments:
FibonacciFermat02.zip
From: Frenchy Pilou (PILOU)
Keep the good work!
From: Frenchy Pilou (PILOU)
French Version ;)
http://moiscript.weebly.com/geacuteomeacutetries-par-bemfarmer.html
Here on the left Points of the Fibo_Fermat are used for the _voronoi by Max Smirnov ! :)
@Bemfarmer
Can you Add the "Z" altitude on the right Fermat Curve above ? (because I am not dexterous with this language! :(
//>>>> Insert z value formula here, with z value changing depending upon
// randomly selected j value. z value could ramp up from 0 to the next seed,
// and then ramp down to 0 to the next seed.
// Pass z value to Polar2Cartesian function.
From: bemfarmer
Hi Pilou,
I'll have a look...
- Brian
From: 864kD1998 (BOB1998)
How do you start drawing the Fibonacci flower using only a compass and straight edge?
From: Frenchy Pilou (PILOU)
each square has a quarter circle
(and with colors you can see 2 Fibonacci flowers intricated : a colored one and a black one! :)
Magic geometry!
From: Frenchy Pilou (PILOU)
@Michael
During the "triming" circles / squares i had this message ? (beta v4 17 october 2018 )
(i just drawn circles and squares! ;)
http://moiscript.weebly.com/uploads/3/9/3/8/3938813/alert_message.3dm
From: amur (STEFAN)
Ha, good reminder! So i did a MoI doodle too, based on the Fibonacci Spiral. :-)
Regards
Stefan
From: mkdm
Beatiful! Very elegant :)
From: amur (STEFAN)
Thank you Marco! :-)
Regards
Stefan
From: amur (STEFAN)
BTW, for people who like to calculate the Golden Ratio for an image size, for rendering, layout etc. ...
There is a nice free OS X Widget available. :-)
http://www.thismanslife.co.uk/projects/phiculator/
Regards
Stefan
From: 864kD1998 (BOB1998)
do 360 / phi^3 you get the approximation 85 degrees = 1
then times phi you get this very interesting number 137.5 degrees = phi
2phi + 1 = phi^3 or 1/phi^-3
222.508 would be phi^2 and 275 would be 2 phi Radius = 1
try this experiment: get three equal size sphere magnets (I used Neodymium Iron Boron) and mark the center theta for the first two magnets . Arrange the two magnets on either poles of the third magnet without touching each other (the first two magnets) you get the angle from the center of the two magnets that intersects approximately 85 degrees by 137.5 by 137.5 . The arrangement of the three magnets should look like Y
From: mimitot
222.508 would be phi^2 and 275 might be 2 phi Radius = 1
do this test: get three equal length sphere magnets (I used Neodymium Iron Boron) and mark the center theta for the first magnets . set up the 2 magnets on either poles of the third magnet without touching every different (the first magnets) you get the angle from the center of the two magnets that intersects about 85 stages by 137.five by 137.5 . The association of the three magnets need to appear like Y
From: 864kD1998 (BOB1998)
Ah , thank you for making my explanations more clear. :)
From: 864kD1998 (BOB1998)
hello is there a way to draw a 137.5 by 85 degree angle using only a compass and a straight edge?
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