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Proportional Dividers - Golden Section and Generalized #2: Proportional Divider (4 arms) - General Equation

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Blog entry by Mark Whitsitt posted 11-22-2010 08:31 AM 10532 reads 1 time favorited 6 comments Add to Favorites Watch
« Part 1: Phi (Golden Ratio) Rule (Ruler, Scale, etc.) Part 2 of Proportional Dividers - Golden Section and Generalized series Part 3: Revised General Equation »

Ok, while I’m messing with the Golden Ratio and mathematics, I thought I’d also try to figure out how to make a simple cross beam proportional divider. It’s been a while since High School Geometry, so I had to go back and reprove the relationships between proportional isosceles triangles to myself. I also couldn’t find this specific formula for making a divider anywhere on the net in a reasonable amount of time, so I just re-derived it.

(I believe this is all correct, but don’t use this for any mission-critical or potentially injurious tasks until you’ve tried it out! grin)

What we’re talking about is essentially two arms of the same length, joined somewhere between their ends so that they can pivot around that point, and the distance between the arms at one end is some proportion of the distance between the arms at the other end. E.g. given a ratio (proportion) of 1:2, if I make the arms 2 inches apart at one end, the arms are 1 inch apart at the other end. The result is a divider having 4 arms in the shape of an “X”

(For those who are interested, this divider creates two similar isosceles triangles, and since the legs of the triangles are fixed, the law of cosines says that the bases of the triangles must be proportional however far apart they are set.)

This formula is generalized so that you can make a divider of any length and any proportion, including a cross-beam Golden Ratio divider. Just decide how long you want the overall divider to be, decide what proportion you want, and plug those numbers into the formula to determine how far from one end you need to put the pivot on the arms of the divider.

Enjoy!

Mark

BTW, if you use ratios, greater than 1, you are actually calculating the distance Y, not X…

-- -- "there are many good reasons to use old hand tools, but moral superiority is NOT one of them..."



6 comments so far

View TopamaxSurvivor's profile

TopamaxSurvivor

17654 posts in 3136 days


#1 posted 11-22-2010 08:45 AM

Therefore, you must be using .618 for R?

-- Bob in WW ~ "some old things are lovely, warm still with life ... of the forgotten men who made them." - D.H. Lawrence

View BertFlores58's profile

BertFlores58

1684 posts in 2382 days


#2 posted 11-22-2010 09:21 AM

Actually, for the golden ratio… the angle is 31.7 degrees… If you set a protractor on this angle and extend the radius… THEN any x and y (ordinate axis) values at any point intersecting on the extended radius will be the sides of a golden triangle. Say 13 and 8 is the nearest whole value that will give the .618 ratio. The easiest solution, I used for this is using excel spreadsheets.. of course in the computer. However in a practical way.. 13 and 8 plotting (even without a protractor) will do.

-- Bert

View Mark Whitsitt's profile

Mark Whitsitt

86 posts in 2440 days


#3 posted 11-22-2010 04:43 PM

TS,

Yes, if you’re making a Golden Ratio divider, R = 0.618 = Phi^-1

Mark

-- -- "there are many good reasons to use old hand tools, but moral superiority is NOT one of them..."

View Mark Whitsitt's profile

Mark Whitsitt

86 posts in 2440 days


#4 posted 11-22-2010 06:12 PM

Bert, you are correct!

To be clear, the angle you’re talking about is the angle “a” formed by the x-axis (abscissa) and a line with the formula:

y=(R*x)+0.

The angle “a” can be found using arctan(y/x) or tan^-1(y/x).

In this case (for a Golden Rectangle),
y/x = R = 0.618,
and angle “a” = 31.7 degrees.

As the diagram shows, you can draw perpendiculars from any point on the line to each of the axes to determine the dimensions of a Golden Rectangle. (In fact, the rectangle in the diagram is a Golden Rectangle)

Your method of using a protractor and extending the radii is really useful! To extend your idea and generalize the protractor approach, simply calculate arctan( R ), where R is any ratio you want to obtain, and set your protractor to the resulting angle!

Thanks for pointing this out!

Mark!

(P.S. yes, I know that Phi, the Golden Ratio, is an irrational number, and all equal signs ”=” should be construed as “approximately equals”, but when was the last time you laid out a dimension to the closest 100th? 8^) )

-- -- "there are many good reasons to use old hand tools, but moral superiority is NOT one of them..."

View BertFlores58's profile

BertFlores58

1684 posts in 2382 days


#5 posted 11-23-2010 08:08 AM

Additionally, if you have a polar to rectangular calculator (mostly all scientific calculator has) then it would be easy to have values of x,y being the rectangular axis by converting in the calculator with given R, 31.7 degrees as the polar coordinate.

Thanks too Mark for the analysis….

-- Bert

View TopamaxSurvivor's profile

TopamaxSurvivor

17654 posts in 3136 days


#6 posted 11-24-2010 03:05 AM

These dividers look like they would be infinitely easier to get perfect than Fibonacci Calipers ;-))

-- Bob in WW ~ "some old things are lovely, warm still with life ... of the forgotten men who made them." - D.H. Lawrence

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