# Fibonacci At Any Scale

 Blog entry by Grumpy posted 08-06-2011 05:02 AM 6817 reads 8 times favorited 11 comments

I thought it was about time I made myself a ‘Fibonacci Gauge’ and like most of us would I looked up on how everyone else did it. There were some great efforts made by fellow Jocks on the subject;
David made a good crack at it
http://lumberjocks.com/David/blog/1639
so did FJPetruso
http://lumberjocks.com/projects/37696
and my good buddy Karson
http://lumberjocks.com/projects/24528

Karson has his Excel spreadsheet for calculating gauges at any size. This is what got me thinking.
There must be a simple way to design a gauge of any size and as sure as pigs don’t fly there is.
This is my version of Fibonacci’s famous work (not to forget to give Pythagoras a big mention, after all he started it)
The ‘Golden Ratio’ as we all know is 1.61803399, the three dimensional one that is most pleasing to the eye.
To be used as a rule of thumb and not the be all and end all of proportions in design.

I have used FJPetruso’s labelling (A to H) on his project, this should help if anyone wants to compare blogs. I have also worked out the relationships between the points and they are shown on the diagram

The length of all the pieces and the distance between the pivots are all relative to Fibonacci’s workings as you can see from the above analysis.
Now to the meat of it all
First you need to decide on the width your calipers need to be at. Once you do that everything else will fall into place. Example
I will do this in millimeters to keep it simple.
Lets say you want a caliper around 200mm wide, of course it will measure greater or lesser widths but it’s a general starting point.
So the distance FH=200mm
Now calculate GH where GH=FH/2.61803399, so GH=200/2.61803399=76.4mm
Now calculate FG where FG=1.61803399 x GH, so FG=1.61803399×76.4=123.6mm
So GH+FG should = FH which it does.
From this we can calculate the distance between the pivots & points of the outside arms of the calipers, AF & AH where they are each equal to twice the length of FG plus the length of GH or more simply FH+FG
So AF&AH=FH+FG, =200+123.6=323.6mm
So AF&AH=323.6mm, the distance between the pivots and points on the outside arms.
Now to the inside arms
The distance between the pivots on the short arm CE is equal to the length of FG which we already know to be 123.6mm
The distance between the pivots on the long arm at BE is also 123.6mm.
All that remains is the distance BG between the pivot & the end point
So BG=FH which we know already to be 200mm in this example.
So now I can decide on the size of the calipers I want to make to fit the project.

In summary then
Distance FH you decide
Distance GH=FH/2.61803399 (where 2.61803399 = 1+ the Fibonacci golden ratio)
Distance outside arms = (2xFG)+GH
Distance of Short Arm CE=FG
Distance between longer inside arm pivots BE =FG
Distance longer inside arm BG=FH (the length you decided in the first place)

There is a quick and easy way to do it fairly closely in you head, surely not you say.LOL
In the above example take 200 as one dimension of your project
If you multiply by 0.6 and 0.4 you get very close to the Golden Ratio
EG: Length 200m, Height 200×0.6=120mm, Depth 200×0.4=80mm
Using the exact ‘Golden Ratio’ I would have come up with Length 200m, Height 200×0.6=123.6mm, Depth 200×0.4=76.4mm
There is not much in it at all and it is easy to remember. Like Kilometers to miles just multiply by 0.6 & you have a rough answer
Now all I have to do is go and make my Fibonacci Gauge.
If anyone finds a problem with the logic please let me know ASAP

-- Grumpy - "Always look on the bright side of life"- Monty Python