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I want to verify a theory stating that a golden mean spiral plotted on a polar graph will have intersection points at specific values.

golden mean spiral plotted on polar graph

There should be 8 concentric circles, each at the same distance as the previous one. The radial lines should be in 10-degree increments, and the spiral should start from 0 on the first circle in the middle and end on circle #8 at 360.

The golden mean spiral should progress independently of the grid lines. If all these requirements are met, NSolve should reveal where the intersection points are (the theory claims - 0deg at circle #1 - 120deg at circle #2 - 190deg at circle #3 - 240deg at circle #4 - 280deg at circle #5 - 360deg at circle #8).

If the initial stretch of the spiral proves the theory wrong, trying to stretch the spiral between any 2 intersection points might do it.

enter image description here

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That's not a golden spiral. This might be PolarPlot[GoldenRatio^(2 n/\[Pi]), {n, 0, 2 \[Pi]}]. Can you quote or link to the theorem? –  wxffles Oct 14 '13 at 23:17
    
Once properly normalized, the 120, 240, and 360 crossings are correct as stated. This is because the exponents for 2,4,8 are in ratio 1,2,3. The other crossings that you give are only approximately on target. For 3rd one it comes to around 190.2, and 5th is at 278.63 degrees. 6th is 310.2. –  Daniel Lichtblau Oct 15 '13 at 20:54
    
Is it possible to show this on the graphic? –  Bo C. Oct 15 '13 at 21:02
    
Please do not edit it just to add "Unsolved" or "Solved". We don't do status updates here. If an answer has solved your problem, you can accept it by clicking the tick mark/check mark. –  rm -rf Oct 16 '13 at 6:33
    
The question remains unsolved; some of the answers are incomplete, others do not reflect the problem correctly, but still they get voted up as useful. A wrong answer is not useful, thus the reason for marking it "Unsolved". –  Bo C. Oct 16 '13 at 8:51

3 Answers 3

Here's my attempt at the graph.

radial = Range[10, 360, 10] /. {
 120 :> {120, Black}, 190 :> {190, Black}, 240 :> {240, Black}, 
 280 :> {280, Black}, 360 :> {360, Black}
 } /. k_Integer :> k Degree;
circular = GoldenRatio^(2 {0, 120, 190, 240, 280, 320, 360} Degree/π);

PolarPlot[GoldenRatio^(2 n/π), {n, 0, 2 π}, 
  PolarAxes -> {True, False}, PolarTicks -> Range[0, 350, 10] Degree, 
  PolarGridLines -> {radial, circular}, PlotRange -> All]

golden

I'm still not convinced about the theorem though.

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The theorem is from a book I read some time ago, I'll search to see if it's published somewhere online. The golden mean spiral should progress from its starting point independently of the grid lines (maybe a second rule would be to finish at 360 on the 8th circle, not 7th). We are checking to see at which angles it intersects the circular grid lines (with NSolve?); these angles are afterwards emphasized as radial black lines. –  Bo C. Oct 15 '13 at 6:06
    
Right now the code doesn't reflect the problem correctly. The circles are built on top of the spiral and the distance between them is not equal. –  Bo C. Oct 15 '13 at 14:54
    
That's why I'd like to know exactly what the theorem is. As far as I can tell, the curve does not go through equally spaced points at those angles (aside from my ability to count to eight). –  wxffles Oct 15 '13 at 19:55
    
This is exactly the theorem, you got it right: allegedly, the curve goes through equally spaced circles at those angles. This may be untrue, but I can't prove it; I just began learning Mathematica. And it is a wild ride! –  Bo C. Oct 15 '13 at 20:49

Here you can control the axes quite nicely:

tickmarks = Range[0, 350, 10];
circles = Range[0, 200, 18];
ListPolarPlot[Table[GoldenRatio*n, {n, 100}], Joined -> True, 
      PolarAxes -> {True, False}, 
      PolarTicks -> {tickmarks Degree, Automatic},
      PolarGridLines -> {tickmarks Degree, circles}, PlotRange -> All]

enter image description here

Control the locations of the ticks using the variable tickmarks and the location of the circles using the variable circles.

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There should be 8 radial grid lines; circles = Range[0, 200, 20] approximates it. The spiral should start from the 0 edge of first circle (radial grid). I'm not sure if this is really a golden mean logarithmic spiral. We need to check whether this spiral intersects the circular grid lines at nodes where the circular and radial lines also intersect. I guess that in order to do this, we'll have to also define that the spiral should end at 360 on the outermost circle (radial grid #8). –  Bo C. Oct 15 '13 at 6:20
    
I wasn't trying to prove or reprove the theorem, but to give control over the axes... btw -- the increments in the Range don't need to be integers. –  bill s Oct 15 '13 at 15:10
    
Using the GoldenRatio^(2 n/π), {n, 0, 2 π} from the other answer complicates even more what I have until now. Also, I'm racking my brains trying to move the spiral starting point on the first circle with no luck... –  Bo C. Oct 15 '13 at 15:22

The problem is that we don't know the definition of this "Golden Mean spiral"

I think here $$ r(\theta) = 8^\frac{\theta}{2\pi} $$

r[t_] := 8^(t/(2 \[Pi]));
tickmarks = Range[0, 350, 10];
circles = Range[8];
angles = {120, 190, 240, 280} Degree;

PolarPlot[r[t], {t, 0, 2 \[Pi]}, PolarAxes -> Automatic, 
 PolarTicks -> {tickmarks Degree, Automatic}, 
 PolarGridLines -> {Join[tickmarks Degree, {#, Red} & /@ angles], circles}, 
 Epilog -> {PointSize[Medium], Point[{r[#] Cos[#], r[#] Sin[#]} & /@ angles]}]

enter image description here

Points at $120^{\circ}$ and $240^{\circ}$ have exact matching

r[120. Degree]
r[240. Degree]

2

4

However, angles $190^{\circ}$ and $280^{\circ}$ isn't exact

r[190. Degree]
r[280. Degree]

2.99661

5.03968

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I'm not sure what the formula for representing it is, but the Golden Mean curve is a logarithmic spiral having as exponent the GoldenRatio or "\[Phi]": (1 + Sqrt[5])/2 mathworld.wolfram.com/GoldenSpiral.html –  Bo C. Oct 15 '13 at 20:56
    
@Bogdan I know, but in your book definition may be different. I solve equation for $120^\circ$-point and get $8$ as exponent. –  ybeltukov Oct 15 '13 at 21:00
    
The code doesn't correspond with the drawing, the spiral is missing. Assuming we have the right Golden spiral, can the intersection points be represented on the graphic? –  Bo C. Oct 15 '13 at 23:21
    
@Bogdan Sorry, there was a typo: not f[t] but r[t]. Of course, see update. –  ybeltukov Oct 16 '13 at 9:46
    
That makes sense, thank you! This looks just like the original drawing, but is that actually a Golden Mean spiral? It is logarithmic indeed, but it doesn't seem to get wider (or further from its origin) by a factor of φ for every quarter turn it makes. –  Bo C. Oct 16 '13 at 16:33

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