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90 Degrees of Separation:
The Golden Rectangle & Logarithmic Spiral
Part II

©Lorena Loo

This is a consequence of the three successive vertices of the rectangular spiral all lying on the vertices of a right angle triangle and the fact that the two successive diagonals of the rectangular spiral intersect at the pole at 90º. But the rectangular spiral itself is nothing more than a succession of such right angle triangles. This means that any 3 successive vertices of the spiral will also have a distance that grows by the same constant multiplicative growth factor G of the spiral itself as we spiral outwards. Or decreases by a factor of G as we spiral inwards. The distances of successive vertices of the rectangular spiral therefore decrease their distance from the pole successively by the same growth factor G of the spiral itself as it spirals inwards, increases by that as it spirals outwards.

Property 3 has now been demonstrated.
In the next diagram, beginning with the outermost arm, we subdivide the rectangle so that each of the first three arms of the spiral form the diagonals of a square. None of those three squares overlap. Had we continued this procedure beyond the third arm of the spiral, the new squares would overlap existing ones. There are also two thin blue diagonals of two different blue rectangles, one with vertex the top left corner of the blue rectangle enclosing the spiral, the other which has as one of its vertices the top right corner of that rectangle.
© Copyright Lorena Loo
We can now formulate the distance r of any vertex of a rectangular spiral of growth factor G from its pole in terms of G and the rotational angle    .

     



Later on in this article, we will look at the general equation for a logarithmic spiral.

A Spiral With a View

In the diagram below, the rectangular spiral from Part I has been rotated through a 45º counterclockwise about its pole. It is the same spiral, just a different persepctive. In addition, we have enclosed the spiral with a rectangle that just fits around it.
We can now determine the equations of the two blue diagonals. They are:


     



For these two blue diagonals to be orthogonal (intersect at 90º), the product of their slopes must equal -1.


i.e.                                or   



Solving for G we get,                   but as G must be greater than 1, there

is only one solution to the equation which satisfies orthogonality and that is

                   which is precisely the value of     . In other words, the growth

factor of our spiral must be     . When this happens, the rectangle just enclosing this spiral becomes the golden rectangle and our diagram above then looks like this:
An Equation From Here to There

We can use property 3 to tabulate the distances of the vertices of our rectangular spiral from its pole as a function of rotational angle about the pole. We know that the successive vertices are separated by 90º or pi/2 radians. We also know that successive vertices grow by a multiplicative factor of G, the growth factor, as we move outwards along the spiral. That allows us to compile the following table:

In a rectangle, the angle at each of the corners or vertices is 90º. Any two successive arms of this spiral then can be viewed as half a rectangle. So you can think of a rectangular spiral as consisting of whirling half rectangles as depicted below where the thick lines represent the arms of the spiral.
Let's add the Cartesian co-ordinate system to our diagram, locating the origin (0,0) at the point depicted in the diagram below. Remember, G is the growth factor of our spiral. There is more information in the diagram than is necessary but it has been added for those interested.
From the diagram, it looks as though the two sets of diagonals, that of the rectangular spiral and that of the golden rectangles, intersect at the same point, namely the pole. But just because it looks that way does not mean this is actually the case. We can however prove this is exactly the case, that what appears to be so is indeed so.

Here are the equations for the two diagonals (thin blue lines) of the golden rectangles.

    



Solving for x and y, the intersection point of the two diagonals of golden rectangles is

    



(Author Note: Add      x (1) and     x (2) to solve for y, then substitute the

value of y to solve for x in (1).)