Golden rectangle.

• Jan 19th 2010, 03:44 PM
artofstoo
Golden rectangle.
Hello all,
I'm wondering if someone here could kindly give me a hand to solve a small problem.
I'm trying to draft a golden spiral but I keep finding myself snagged mid-way through the drawing.

I'll explain my procedure first, leading into the issue i'm having.

I start by creating the largest square of the sequence using inches, I pick any size to fit within the page i'm working on... so anywhere from 4"x4" - 9x9.
Once i've done that I divide the length in half and draw my diagonal line from the mid-point of the length to one of the corners on the adjacent side. Then while using a compass pivoting on the mid-point, I create my arc starting at the corner traveling down to level with the length of the square.

Now, to my understanding i've created the golden rectangle.

So, I then proceed on to creating the second, smaller square by using the measurement of the length I had just created (the golden mean of the first square)
Now i'm left with two squares, one big and one small and one small golden rectangle. Using the same method as in creating the second square, I create the third square within the smaller rectangle, which then leaves me with an even smaller rectangle.
Repeating this pattern once or twice more I hit a point where the rectangle i'm left with is too narrow to continue on ( it's proportions are outside of being a golden rectangle ) ...For example. The next square in the sequence I have to create may be 3"x3".. the rectangle would be 3"x8". leaving me with a 5 unit remainder and if I was to treat the 3"x3" square as the 1,1 of the fibonacci sequence then I would create another 3"x3" square however, that creates a 6"x3" rectangle, still leaving me with a 3"x2" square...

*the measurements I used are just for a rough example.

-So, what the heck am i'm doing wrong? lol. What i'm hoping to get is a 9 square sequence for this spiral.
I would also like to mention that i've gone over my measurements many times now and all of my boxes are completely square.... my only guess is that my arc may be too wide but I don't understand how that would be if I was accurate with the measurement of the first square....

Any help will be greatly appreciated :) !!!
• Jan 19th 2010, 08:19 PM
Soroban
Hello, artofstoo!

Quote:

I'm trying to draft a golden spiral but I keep finding myself snagged
mid-way through the drawing.

I'll explain my procedure first, leading into the issue i'm having.

I start by creating the largest square of the sequence using inches.
I pick any size to fit within the page i'm working on, anywhere from 4"x4" - 9x9.

Once i've done that I divide the length in half and draw my diagonal line from the midpoint
of the side to one of the corners on the adjacent side.

Then while using a compass pivoting on the midpoint, I create my arc
starting at the corner traveling down to level with the length of the square.

Now, to my understanding i've created the golden rectangle.

So, I then proceed on to creating the second, smaller square by using
the measurement of the length I had just created (the golden mean of the first square)

Now i'm left with two squares, one big and one small and one small golden rectangle.
Using the same method as in creating the second square, I create the third square
within the smaller rectangle, which then leaves me with an even smaller rectangle.

Repeating this pattern once or twice more I hit a point where the rectangle i'm left with
is too narrow to continue on. (It's proportions are outside of being a Golden Rectangle).

Code:

      D    1    C       * - - - - - *       |          /|  *       |        / |    *     1 |        /  |    *       |      /  |       |      /    |      *       * - - * - - * - -  * - -       A    M    B    P

We have square $ABCD\!:\;\;AB = BC = CD = DA = 1$
$M$ is the midpoint of $AB\!:\;\;MB = \tfrac{1}{2}$

In right triangle $CBM\!:\!:\;MC^{\:\!2} \:=\:1^2 + \left(\tfrac{1}{2}\right)^2 \:=\:\frac{5}{4} \quad\Rightarrow\quad MC = \tfrac{\sqrt{5}}{2}$

With center $M$ and radius $MC$, draw arc $CP$, cutting $AB$ extended at $P.$

Therefore: . $MP \:=\:\tfrac{1}{2} + \tfrac{\sqrt{5}}{2} \;=\;\frac{1+\sqrt{5}}{2}$ . . . the Golden Mean.

There should be no integers after the initial square.

$\text{We have a }\text{1-by-}\phi\text{ rectangle }APQD\;\text{ . . . a Golden Rectangle.}$

Code:

      : - - - - - φ - - - - - :       D            C        Q     - *-------------*---------*     : |            |        |     : |            |        |     1 |            |        |     : |          R *---------* S     : |            |        |     - *-------------*---------*       A            B        P
Cut off square $ABCD$
and the remaining rectangle $CQPB$ is a Golden Rectangle.

Cut off square $CQSR$
and the remaining rectangle $RSPB$ is a Golden Rectangle.

In theory, we can repeat this process forever.
. . The ratio of the sides will always be $1\!:\!\phi.$

• Jan 20th 2010, 12:20 AM
artofstoo
Thanks for the response Soroban, Fancy ASCII illustration too btw haha.

The problem i am having though is when I do repeat the process.

Lets say I cut off R,S,B,P to get:

Code:

      : - - - - - φ - - - - - :       D                  C            Q     - *-------------*---------*     : |                  |            |     : |                  |            |     1 |                  |  t        |     : |                R *---------* S     : |                  |    |        |     - *-------------*---------*       A                  B    u        P
Now I have R,T,B,U .. The problem i'm getting usually happens while squaring off this section on either this step or on the following step ( R,T,"V,W"? ) ...somehow I keep finding myself left with more space then what I should have.

I assume, obviously, that my measurements have to be off in order for this to happen. What I don't get is how they could be off in the first place unless the "negative space" ,i'll call it, is the summation of some earlier lines being short a millimeter or two.
I'll try to illustrate what i'm being left with -

Code:

                          ^                                        ^                           ^                                        ^                         C*                                        *Q                           |                                          |                           R            T                          |                           *---------*-------------------*S                           |              |                          |                         X*----------*Y                        |                           |              |                          |                           |              |                          |                         V*----------*W                        |                           |              |                          |                           |              |                          |     <<*------------ *---------*-------------------*       A                  B            U                            P
Alright, So this illustration shows the rectangle zoomed in to the R,T,B,U sector.
now, lets say I had just created the cut from T-U and I shifted my attention to R,T,B,U...the next line I need to create is V,W, so I take the length of B,U and bring it over to B,R to get B,V and I create my square V,W,B,U.... Now, notice R,T,V,W... it's an usual shape.
-The cut for V,W should actually be where I placed X,Y... giving me the horizontal rectangle R,T,X,Y which actually looks a lot closer to the dimensions of a golden rectangle even in this illustration.
However, I keep getting that V,W line...which I should call "checkmate" lol because I can't follow it up with another cut. When I do I end up creating an adjacent square to V,W,B,U and of the same dimension, leaving me with the "negative space" I was explaining earlier, similar to the shape of R,T,X,Y...
...it's definitely not an infinite spiral i'm drawing lol. If anything B,U,V,W and X,Y,V,W should both = 1 leaving me with nothing correct?

So whats potentially causing this problem?
my lines are accurate to 1/16 of an inch.
• Jan 20th 2010, 12:40 AM
artofstoo
yes code reformatting!.
-i'll have to edit that later.