I have seen many references that state that the Golden Spiral can be approximated via creating squares whose sides increase with the Fibonacci sequence. Specifically (and as shown in the figure), you can stack a series of squares whose sides increase with the Fibonacci numbers. Drawing an arc that is tangent to the sides of the squares and connecting them approximates a Golden Spiral.
I thought I would put this to the test and see how close they are (surprisingly, I did not find many references that superimposed the two types of curves).
What I found what that the Fibonacci spiral does a terrible job of approximating a Golden Spiral. In the figure (link below), the blue curve is the "true" Golden Spiral given by $r = \varphi^{2 \theta / \pi}$. I start with $\theta = 0$ along the positive $x$-axis and the initial radius of the spiral is 1. The purple dotted squares are the Fibonacci squares and the purple spiral is generated by drawing arcs within them.
Perhaps this (poor) approximation is to be expected -- hoping that someone could please verify?
I find it hard to believe that this Fibonacci method could be a good approximation. The Golden spiral changes by a factor of $\varphi$ with each turn of $\pi / 2$ degrees. However, the radius of curvature in each Fibonacci square remains the same for the turn of $\pi / 2$ degrees.
Here is the code from Mathematica used to generate the drawing:
% Define the Golden Spiral.
r2 = GoldenRatio^((2 / Pi) theta )
% The code lines below generate the Graphics elements of the squares and the arcs within the squares:
l1 = Line[{{0, 0}, {1, 0}, {1, 1}, {0, 1}}];
c1 = Circle[{0, 0}, 1, {0, Pi/2}];
l11 = Line[{{0, 0}, {-1, 0}, {-1, 1}, {0, 1}}];
c11 = Circle[{0, 0}, 1, {Pi/2, Pi}];
l2 = Line[{{-1, 0}, {1, 0}, {1, -2}, {-1, -2}, {-1, 0}}];
c2 = Circle[{1, 0}, 2, {Pi, 3 Pi/2}];
l3 = Line[{{1, 1}, {4, 1}, {4, -2}, {1, -2}}];
c3 = Circle[{1, 1}, 3, {3 Pi/2, 2 Pi}];
l5 = Line[{{4, 1}, {4, 6}, {-1, 6}, {-1, 1}}];
c5 = Circle[{-1, 1}, 5, {0, Pi/2}];
l8 = Line[{{-1, 6}, {-9, 6}, {-9, -2}, {-1, -2}}];
c8 = Circle[{-1, -2}, 8, {Pi/2, Pi}];
l13 = Line[{{-9, -2}, {-9, -15}, {4, -15}, {4, -2}}];
c13 = Circle[{4, -2}, 13, {Pi, 3 Pi /2}];
l21 = Line[{{4, -15}, {25, -15}, {25, 6}, {4, 6}}];
c21 = Circle[{4, 6}, 21, {3 Pi/2, 2 Pi}];
% Plot the Fibonacci Spiral and superimpose the Fibonacci spiral graphical elements.
polarplot34 = PolarPlot[r2, {theta, 0, 4 Pi},
BaseStyle -> {FontSize -> 14, FontFamily -> "Latin Modern Roman"},
Epilog -> {
{Dashed, Purple, l1}, {Purple, c1},
{Dashed, Purple, l11}, {Purple, c11},
{Dashed, Purple, l2}, {Purple, c2},
{Dashed, Purple, l3}, {Purple, c3},
{Dashed, Purple, l5}, {Purple, c5},
{Dashed, Purple, l8}, {Purple, c8},
{Dashed, Purple, l13}, {Purple, c13},
{Dashed, Purple, l21}, {Purple, c21}
}
]
2019-09-07 - UPDATE: I realized that if we start the Golden Spiral and the Fibonacci spiral with radius 1, then, of course, the approximation will be terrible. The ratio of subsequent Fibonacci numbers, $F(n+1) / F(n)$, only closely start to approximate $\varphi$ at values of $n > 5$. Therefore, if you start with approximating a Golden Spiral with arcs in smaller Fibonacci squares, the two spirals will be way off in the beginning and that difference will only be amplified as $\theta$ increases.
Another way to think about it is to normalize the Golden Spiral so that it starts at a corner of a much larger Fibonacci square. After all, the Golden Spiral does not require any specific ``starting point"; it is simply a figure that increases its radius by $\varphi$ every time $\theta$ increases by $\pi / 2$.
I recreated this using Mathematica as shown in Figure below. Here I have drawn the spiral created with Fibonacci squares in purple, as before. But now, the Golden Spiral (in blue) starts at the position (25,6) corresponding to the corner of the Fibonacci square of side length 34. As you can see by comparing to the first figure, this approximation is much better.
