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

 Blog entry by Mark Whitsitt posted 11-22-2010 08:31 AM 11475 reads 1 time favorited 6 comments
 « 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

 TopamaxSurvivor18241 posts in 3638 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 BertFlores581694 posts in 2884 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 Mark Whitsitt86 posts in 2941 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..." Mark Whitsitt86 posts in 2941 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..." BertFlores581694 posts in 2884 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 TopamaxSurvivor18241 posts in 3638 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

## Have your say...

 You must be signed in to post the comments.

DISCLAIMER: Any posts on LJ are posted by individuals acting in their own right and do not necessarily reflect the views of LJ. LJ will not be held liable for the actions of any user.

 Bathroom renovation London 04-13-2018 01:29 AM Bathroom Vanity 02-01-2018 09:49 PM folding chair hangers 01-29-2018 01:02 AM Building a built in bookshelf with arch top 01-19-2018 05:48 PM Building a Japanese Timber Frame Inspired front door 12-15-2017 10:54 PM Building a Timber Frame Staircase with through tenons held with wedges 12-07-2017 09:08 PM bathroom vanity re-do 06-07-2017 01:30 AM Bedroom redo 05-10-2017 02:11 AM Dresser 05-10-2017 02:04 AM Bungalow Porch 01-20-2017 11:04 AM
The text passage is a blog entry discussing the creation of a proportional divider, a tool used for transferring measurements in a fixed ratio. It includes a description of the tool, the mathematical principles behind it, and a discussion in the comments section about specific ratios and methods for creating the divider.
no_question

the passage does not contain a direct mathematical question that requires an answer. it is more of an informative piece explaining the construction and principles of a proportional divider. the comments section does contain some questions, but they are not mathematical in nature and are more about clarifying the content of the blog post.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Proportional Dividers - Golden Section and Generalized #2: Proportional Divider (4 arms) - General Equation

 Blog entry by Mark Whitsitt posted 11-22-2010 08:31 AM 11475 reads 1 time favorited 6 comments
 « 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

 TopamaxSurvivor18241 posts in 3638 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 BertFlores581694 posts in 2884 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 Mark Whitsitt86 posts in 2941 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..." Mark Whitsitt86 posts in 2941 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..." BertFlores581694 posts in 2884 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 TopamaxSurvivor18241 posts in 3638 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

## Have your say...

 You must be signed in to post the comments.

DISCLAIMER: Any posts on LJ are posted by individuals acting in their own right and do not necessarily reflect the views of LJ. LJ will not be held liable for the actions of any user.

 Bathroom renovation London 04-13-2018 01:29 AM Bathroom Vanity 02-01-2018 09:49 PM folding chair hangers 01-29-2018 01:02 AM Building a built in bookshelf with arch top 01-19-2018 05:48 PM Building a Japanese Timber Frame Inspired front door 12-15-2017 10:54 PM Building a Timber Frame Staircase with through tenons held with wedges 12-07-2017 09:08 PM bathroom vanity re-do 06-07-2017 01:30 AM Bedroom redo 05-10-2017 02:11 AM Dresser 05-10-2017 02:04 AM Bungalow Porch 01-20-2017 11:04 AM
user
The text passage is a blog entry discussing the creation of a proportional divider, a tool used for transferring measurements in a fixed ratio. It includes a description of the tool, the mathematical principles behind it, and a discussion in the comments section about specific ratios and methods for creating the divider.

TYPE: NO_QUESTION

The passage does not contain a direct mathematical question that requires an answer. It is more of an informative piece explaining the construction and principles of a proportional divider. The comments section does contain some questions, but they are not mathematical in nature and are more about clarifying the content of the blog post.