# Golden Spiral

By Murray Bourne, 04 Sep 2011

Fern [image source]

Here's a recent question from reader Pehr in Sweden:

Hi,

First of all, wonderful site. Love it.

I've been studying the polar coordinates section hoping to expand my knowledge on the equiangular spiral. The interactive tools are great, though I'm having a hard time to derivate the exact mathematical solution to why the resulting function for the golden spiral is

r = ae^(b(theta))

For some background on Pehr's question, see:

Polar Coordinates, Curves in Polar Coordinates and Equi-angular Spiral

Spirals are common in nature and have inspired mathematicians for centuries.

Aloe spiral [Image source]

Spiral galaxy NGC 5194

[Image courtesy NASA]

## Logarithmic Spirals

The Golden Spiral that Pehr is asking about is a special case of the logarithmic spiral.

Logarithmic spirals grow such that the angle of a line from the center of the spiral to the tangent to the curve at that point is constant. This is why they are also known as "equi-angular" spirals.

To see what this means, the 3 acute angles marked in the following fern image are approximately 80°

Equi-angular fern

We normally use functions in Polar Coordinates when describing spirals. Otherwise, if we use ordinary rectangular coordinates, the formulas become very complex.

The formula for a logarithmic spiral using polar coordinates is:

*r* = *ae*^{θ cot b}

where

*r* is the distance from the origin (or "pole")

*a* is a constant

θ is the angle (in radians) from the horizontal axis. So the coordinates of a point on the curve in polar coordinates is given by (*r*, θ).

*b* is the angle (in radians - the "equal" angle) that the line from the center of the spiral makes with the tangent to the spiral. In the fern case above, *b* ≈ 1.4 radians (≈ 80°).

As a consequence of the way we defined the logarithmic spiral, the ratio of the distances from the center to each spiral arm of an adjacent pair is constant.

Spiral arms at a constant ratio

The ratio

distance to the first arm: distance to second arm

= 29:69

≈ 0.42

The other ratio

distance to the second arm: distance to third arm

= 69:154

≈ 0.45

We see the ratios are almost the same. (In an actual logarithmic spiral, they are exactly the same. Choosing the start point for the fern is not an exact science!)

## Golden Spiral

The Golden Spiral is a special case of the logarithmic spiral.

We can write the general logarithmic spiral as a function in polar coordinates using *t* as follows:

*r*(*t*) = *ae*^{t cot b}

**Note: **Normally, we use θ for the independent variable, but we often use *t* as we can think of the spiral being traced out over time. Besides, it's easier to type!

The **Golden Spiral** has the special property such that for every 1/4 turn (90° or π/2 in radians), the distance from the center of the spiral increases by the golden ratio φ = 1.6180.

For this to occur, cot *b* must take the value (which comes from solving our function):

Using this value, and taking the simple case where *a* = 1, our function becomes:

*r*(*t*) = *e*^{0.30635t}

We'll use the excellent free graphing tool GeoGebra from here on.

## Setting up the Golden Spiral using GeoGebra

Now if we graph our function on ordinary rectangular coordinate axes in GeoGebra, we get the following exponential curve. Note that *r* increases at an ever-increasing rate (it gets steeper) as *t* increases.

But to see a spiral, we need to graph the curve using **polar coordinates**.

To convert the polar form (which we've got) to rectangular form (which we need for the graph) in Geogebra, we need to set up and graph the following function:

*a*(*t*) = (*r*(*t*) cos(*t*), *r*(*t*) sin(*t*))

Let's substitute a few important values to see what this expression means. Starting at *t* = 0, we get the starting point of the curve:

*a*(0) = (*r*(0) cos(0), *r*(0) sin(0))

= (1×1, 1×0)

= (1, 0)

So it means we start 1 unit from the origin along the positive *x*-axis. You can see the starting point in the following graph of the spiral.

Next, we rotate a quarter turn and find at *t* = π/2,

*a*(π/2) = (*r*(π/2) cos(π/2), *r*(π/2) sin(π/2))

= (1.618×0, 1.618×1)

= (0, 1.618)

Note that we are now 1.618 units from the origin up the *y*-axis. That is, φ = 1.6180 times the distance we started from.

Another rotation of a quarter turn brings us to *t* = π, where:

*a*(π) = (*r*(π) cos(π), *r*(π) sin(π))

= (-2.618×1, -2.618×0)

= (-2.618, 0)

We are now 2.618 units away from the origin along the negative *x-*axis, or φ = 1.6180 times the distance from the origin we were at the last quarter turn.

Note:

φ^{2} = 2.6180

We could work out our next position, along the negative *y*-axis, by just multiplying this last value by φ = 1.6180, giving us:

φ^{3} = 4.23606...

So the spiral will cut the *y*-axis at (0, -4.236).

One more quarter turn will bring us to φ^{4} = 6.85410... units along the positive *y*-axis, that is (6.854, 0).

We can see these values are correct on our spiral graph above.

If we keep going, we'll get a spiral as follows (this is 2 complete revolutions, or 4π = 720°):

As an aside, since in this problem

cot *b* = 0.30635

then

*b* = arccot 0.30635 = 1.274 radians or around 73°

This is the angle our spiral arms make with a line from the center of the spiral. You can see on the graph above each spiral arm makes an angle of 73° with the *x*-axis (and *y*-axis, or any line from the center).

## Approximating the Golden Spiral using arcs of a circle

We can obtain a spiral that looks quite similar to the Golden Spiral by using arcs of circles that increase in size by the Golden Ratio, as follows.

We start with a 1×1 square and draw an arc, center C, through 2 corners such that the sides of the square are tangent to the arc (that is, they touch once only).

Next, we place a square with side length φ = 1.6180 above our first square and construct another circular arc, center E, as before:

Our next square goes on the left and has sides φ^{2} = 2.6180 = 1 + φ.

We continue the pattern (we've gone another complete round) and get a spiral which looks quite a lot like our Golden Spiral from before.

## How close is our approximation?

Wikipedia's article on Golden Spiral has an image which claims the above spiral and the Golden Spiral are very close in shape.

Here's that image:

The caption for the image states:

Approximate and true golden spirals: the green spiral is made from quarter-circles tangent to the interior of each square, while the red spiral is a golden spiral, a special type of logarithmic spiral. Overlapping portions appear yellow. The length of the side of a larger square to the next smaller square is in the golden ratio.

Can we re-create this ?

In the image below, the red curve is first part of the Golden Spiral we constructed above, whereas the green curve is based on the quarter-turn approximation we were just working on.

The point F is the "right most" point on the spiral, which will be my starting point for the quarter-turn arc. The point J is the highest point of this portion of the spiral.

Point A is the intersection of the horizontal and vertical lines passing through F and J respectively and this will be the center of my arc.

Now, the arc GF is not at all close to the related portion of the spiral FJ.

Let's do another step and see if the next part is any better.

As you can see, it's worse (as expected, since we have moved further away from the origin and the spiral arm is getting bigger).

Clearly, this is never going to work.

However, in my earlier Golden Spiral I was using:

*r*(*t*) = *e*^{0.30635t}

The constant *a*, had value 1.

If we want our approximating arcs to be a good fit for the actual Golden Spiral, we need to use a value of (probably not surprisingly)

*a* = φ = 1.618103399..

This gives us the following curves, similar to the graph in Wikipedia.

The red curve is the Golden Spiral,

*r*(*t*) = 1.618013 *e*^{0.30635t}

The green curve is the collection of circular arcs.

The side length of the squares (in pixels) are shown and we can see they are approximately in the ratio 1.618013...

## Golden Spiral in the media

From Wolfram's Mathworld:

In the Season 4 episode "Masterpiece" (2008) of the CBS-TV crime drama "Criminal Minds," the agents of the FBI Behavioral Analysis Unit are confronted by a serial killer who uses the Fibonacci number sequence to determine the number of victims for each of his killing episodes. In this episode, character Dr. Reid also notices that locations of the killings lie on the graph of a golden spiral, and going to the center of the spiral allows Reid to determine the location of the killer's base of operations.

Here's more interesting information from Wolfram's Mathworld:

## For the geeks - design using Golden Spiral

It is believed by many that designs using the Golden Ratio and Golden Spiral are pleasing to the eye.

Even Twitter recently re-designed their main page using the Golden Spiral.

Here is a great article by a guy who has contructed a golden spiral without images. (Mostly for those interested in Web design)

Golden Spiral without images - using CSS and jQuery

As he suggests in the article, an elephant's trunk is also close to the Golden Spiral.

Elephant trunk - almost a Golden Spiral

## Conclusion

The Golden Spiral is an interesting topic - one worth pursuing not only for the pleasant designs involved, but also the interesting math behind them.

I hope that helps to answer your question, Pehr!

See the 17 Comments below.

9 Sep 2011 at 10:44 pm [Comment permalink]

[...] Golden Spiral (squareCircleZ) [...]

3 Jul 2012 at 5:28 pm [Comment permalink]

How would one construct an accurate 3D golden spiral like you find in a pine cone?

4 Jul 2012 at 1:19 pm [Comment permalink]

Hi Dave. Interesting question! This PDF may give you some starting points: The Natural 3D Spiral.

If not, this search may help.

5 Jun 2014 at 7:25 am [Comment permalink]

I want to draw a spiral such as the one in the elephant trunk in GeoGebra, but inside a pinecone. I can't find the way. Any ideas?

5 Jun 2014 at 10:47 am [Comment permalink]

@Mary: GeoGebra allows you to place an image in the background, and then you could use the formula given in the post for your spiral. Is that what you are trying to do?

11 Jan 2016 at 11:25 pm [Comment permalink]

I found the article that the red curve:

, which you claim is golden spiral doesn't fit perfectly the square area split (like on the picture).

See. http://www.emis.de/journals/NNJ/Sharp_v4n1-pt03.html

"If the spiral were to touch the sides of the rectangle, the line from the pole would need to make a tangent angle of 72.9676° with the side of the rectangle."

It says that an additional rotation (~3.75°) is needed to make it perfect.

Would you confirm that is the case?

and if so, would you know the math to calculate the exact value of that angle?

16 Jan 2016 at 8:59 am [Comment permalink]

@Maciek: I haven't come across that issue before. Perhaps the author of that article could provide the missing steps for you?

15 Aug 2017 at 4:55 am [Comment permalink]

[…] If you want to create your Fibo spiral, look here: Golden Spiral […]

19 Sep 2017 at 3:13 am [Comment permalink]

Like to have a symbolic solution for θ

in Equiangluar Spiral expression d = exp ^(θ . cot( αT ) . cos (θ )

to determine the spiral angle for a given d (x-coordinate) on spiral

This is for a 3D design purpose. αT is the Spiral TANGENT angle

Can I paste an image HERE to explain more ? How

Thanks

19 Sep 2017 at 12:55 pm [Comment permalink]

@Kanil: I don't believe there is a "nice" solution for theta in this case. You would need to do it numerically. That is, create an array of (&theta, d) values by substituting θ, and then do reverse look up to find θ for a given d.

27 Dec 2017 at 4:13 pm [Comment permalink]

-- a(0) = (r(0) cos(0), r(0) sin(0))

--

-- = (1×1, 0×0)

--

-- = (1, 0)

I don't get this, how is r(0) = 1, then r(0) = 0?

What is r?

28 Dec 2017 at 8:40 am [Comment permalink]

Hello

Thanks for pointing out the typo which I have now corrected. The middle line should have been this:

= (1×1, 1×0)

It is now!

16 Dec 2018 at 8:33 am [Comment permalink]

Hello, two questions:

In the final diagram of the two overlapping spirals, Why are the side lengths for all the squares such huge numbers? For example, the smallest square looks like it should have a side length of around two but it is marked 67.

Second question: for the final diagram, the squares do not have the same growth factor, i.e. it isn't like in the first example where you obtained all the squares by multiplying the previous one's side lengths by phi. So since the squares are all different sizes, how did you know what size to make them?

Thank you.

16 Dec 2018 at 11:18 am [Comment permalink]

@Jennifer: The side lengths of the squares in that overlapping spiral diagram are in

pixels(since such units are easy to measure in image editing software). Sorry about the confusion - I've updated the post to say this is the case.In the first one I was

startingwith a (unit) square then multiplying subsequently by φ, and constructing a spiral from circular arc based on those square. In that final one, I'm going the other way round -startingwith a spiral based on an exponential function (the red one) and then constructing squares around that spiral, then measuring those squares to see if they do share the common ratio. (I also added the green "based on circular arcs" spiral for comparison.)Hope it helps!

(BTW - Reader's comments only appear once I've approved them, to remove spam.)

17 Dec 2018 at 7:37 am [Comment permalink]

Thank you for your previous response. However, I'm still kind of confused about how you were able to construct the squares in the final graph though. If you didn't have a standard growth factor, then how did you calculate how big each square should be? Especially since the red spiral goes slightly outside some of the squares, so you couldn't have just plotted points on the widest parts of the spiral....

17 Dec 2018 at 2:48 pm [Comment permalink]

@Jennifer: Good on you for thinking this through carefully!

Let's consider the smallest square, marked 67px wide. What it appears I did (it's been 7 years since I wrote this piece, so I'm not exactly sure now) is to draw a vertical line through the right-most part of the spiral, and then a horizontal line through the top most part. The square was completed based on those extremities.

Then for the second largest square, the left side vertical line from the first square is extended to meet the horizontal drawn from the left-most extremity of the spiral. Completing that gives the 109px square. The red and green curves are close to each other - I probably chose the actual point where I'd get a square, so the green quarter-circle would fit each time.

13 Dec 2019 at 8:31 pm [Comment permalink]

[…] Golden Spiral is a shape that makes sure of the balance of a picture. As a result, a picture achieves a more […]