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Full Version: Fibonacci Spiral?

Show messages:  1-15  16-29

From: Frenchy Pilou (PILOU)
23 Oct 2018   [#16] In reply to [#15]
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)
23 Oct 2018   [#17]
@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)
23 Oct 2018   [#18]
Ha, good reminder! So i did a MoI doodle too, based on the Fibonacci Spiral. :-)

fibonacci

Regards
Stefan
From: mkdm
24 Oct 2018   [#19] In reply to [#18]
Beatiful! Very elegant :)
From: amur (STEFAN)
24 Oct 2018   [#20]
Thank you Marco! :-)

Regards
Stefan
From: amur (STEFAN)
24 Oct 2018   [#21]
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)
25 Oct 2018   [#22] In reply to [#7]
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
29 Oct 2018   [#23]
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)
29 Oct 2018   [#24] In reply to [#23]
Ah , thank you for making my explanations more clear. :)
From: 864kD1998 (BOB1998)
4 Dec 2018   [#25]
hello is there a way to draw a 137.5 by 85 degree angle using only a compass and a straight edge?
From: Michael Gibson
4 Dec 2018   [#26] In reply to [#25]
Hi Bob - I'm sorry I don't understand your question, is that something you're trying to draw in MoI ?

- Michael
From: 864kD1998 (BOB1998)
4 Dec 2018   [#27] In reply to [#26]
Yes, I'm trying to construct a Golden angle of 137.5 by 137.5 by 85 degrees approximate in a circle using a compass and a straight edge.Which can be done on a piece of paper.
From: bemfarmer
4 Dec 2018   [#28] In reply to [#27]
I recall from math class that the square root of 5 can be so constructed...
- Brian
From: bemfarmer
4 Dec 2018   [#29] In reply to [#28]
According to this link, the answer is YES :-)
https://www.johndcook.com/blog/2017/05/01/golden-angle/

But the method is not given...

- Brian

After reading google links on "Constructible Angles," I'll conclude that construction of 137.5 degrees is NOT possible with compass and straigtedge...

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